Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Evaluation of $\int_{0}^{2}\frac{(2x-2)dx}{2x-x^2}$ Evaluate $$I=\int_{0}^{2}\frac{(2x-2)dx}{2x-x^2}$$
I used two different methods to solve this.
Method $1.$ Using the property that if $f(2a-x)=-f(x)$ Then $$\int_{0}^{2a}f(x)dx=0$$
Now $$f(x)=\frac{(2x-2)}{2x-x^2}$$ So
$$f(2-x)=\frac{2(2-x)-2}{2(2-x)-(2-x)^2}=\frac{... | The second gives the correct answer, but both solutions are, strictly speaking, incorrect: Since the integrand has singularities at both ends of the interval of integral, we define the value of the integral to be
$$\int_0^2 \frac{(2 x - 2)}{(2 x - x^2)} dx = \lim_{a \searrow 0} \int_a^c \frac{(2 x - 2)}{(2 x - x^2)} dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrate $\frac{\sin^3 \frac{x}{2}}{\cos \frac{x}{2} \sqrt{\cos x+\cos^2 x+\cos^3 x}}$ Evaluate $$\int \frac{\sin^3 \frac{x}{2}}{\cos \frac{x}{2} \sqrt{\cos x+\cos^2 x+\cos^3 x}}dx$$ I saw terms like $1+t+t^2$ in the denominator , so I thought of $t^3-1$ and then converting back into half angle but it doesn't help me.... | I might be wrong but if you combine these two identities :
$$ \sin^2(x) = 1 - \cos^2(x) $$
$$ \cos(x) = \cos(\frac{x}{2} + \frac{x}{2}) = \cos^2(\frac{x}{2}) - \sin^2(\frac{x}{2}) = 2\cos^2(\frac{x}{2}) - 1 $$
You can expand the expressions until only the sine function in the numerator remains and everything else is po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Resolving $\sec{x}\left(\sin^3x + \sin x \cos^2x\right)=\tan{x}$ Going steadily through my book, I found this exercise to resolve
$$ \sec{x}\left(\sin^3x + \sin x \cos^2x\right)=\tan{x}$$
Here's how I resolve it ($LHS$) and again bear with me as I truly reverting to a feeling of vulnerability, like a child actually
A... | So the book does about the same thing as you but in a different order (your solution is hence correct too).
By the trigonometric one we get $\sin^2 x= (1-\cos^2 x) $. Thus
$$\frac{\sin^3 x}{\cos x}=\frac{\sin x (\sin^2 x)}{\cos x}= \frac{\sin x(1-\cos^2 x)}{\cos x} = \frac{\sin x-\sin x\cos^2 x}{\cos x}= \frac{\sin x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7}$
Evaluation of $$\sin \frac{\pi}{7}\cdot \sin \frac{2\pi}{7}\cdot \sin \frac{3\pi}{7} = $$
$\bf{My\; Try::}$ I have solved Using Direct formula::
$$\sin \frac{\pi}{n}\cdot \sin \frac{2\pi}{n}\cdot......\sin \frac{(n-1)\pi}{n} = \frac{... | This is a Gauss sum in disguise, but it can be computed through elementary geometry, without resorting to algebraic number theory. Let us consider a regular heptagon $P_1 P_2 P_3 P_4 P_5 P_6 P_7$ with center $O$ and take $A=P_1, B=P_5, C=P_6$. The angles of $ABC$ are $\frac{\pi}{7},\frac{2\pi}{7},\frac{4\pi}{7}$. By Eu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Points $P_i$ on an ellipse such that angle $P_iOP_{i+1}=\frac{\pi}{n}$ Consider an ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ with $O$ as the origin. $n$ points denoted as $P_1,P_2,\cdots$ are taken on the ellipse such that angle $P_iOP_{i+1}=\frac{\pi}{n}$ where $i\in(1,n-1)$. Find the value of: $$\sum_{i=1}^n\frac{1}{OP... | Better to set up a system of two equations:
$$
3\cos\theta_2=r_2\cos(\pi/n)
\quad\hbox{and}\quad
2\sin\theta_2=r_2\sin(\pi/n),
$$
where $r_2=OP_2$.
From that you get, after eliminating $\theta_2$:
$$
{1\over r_2^2}={\cos^2(\pi/n)\over9}+{\sin^2(\pi/n)\over4},
$$
and in general:
$$
{1\over r_{k+1}^2}={\cos^2(k\pi/n)\ov... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Diagonalisation proof
Suppose the nth pass through a manufacturing process is modelled by the linear equations $x_n=A^nx_0$, where $x_0$ is the initial state of the system and
$$A=\frac{1}{5} \begin{bmatrix} 3 & 2 \\ 2 & 3 \end{bmatrix}$$
Show that
$$A^n= \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \fra... | Diagonalize the matrix $A$:
$$
A=\begin{bmatrix}
\frac{3}{5}&\frac{2}{5}\\
\frac{2}{5}&\frac{3}{5}
\end{bmatrix}=
\begin{bmatrix}
-1&1\\
1&1
\end{bmatrix}
\begin{bmatrix}
\frac{1}{5}&0\\
0&1
\end{bmatrix}
\begin{bmatrix}
-\frac{1}{2}&\frac{1}{2}\\
\frac{1}{2}&\frac{1}{2}
\end{bmatrix}=PDP^{-1}
$$
So we have:
$$
A^n=(PD... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a$ and $b$ are roots of $x^4+x^3-1=0$, $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$. I have to prove that:
If $a$ and $b$ are two roots of $x^4+x^3-1=0$, then $ab$ is a root of $x^6+x^4+x^3-x^2-1=0$.
I tried this :
$a$ and $b$ are root of $x^4+x^3-1=0$ means :
$\begin{cases}
a^4+a^3-1=0\\
b^4+b^3-1=0
\end{cases}$
whic... |
Lemma: If $\alpha,\beta,\gamma$ are the roots of an irreducible polynomial over $\mathbb{Q}$ with degree $3$, $\alpha\beta$ is a root of:
$$ q(x)=(x-\alpha\beta)(x-\alpha\gamma)(x-\beta\gamma)\in\mathbb{Q}[x]. $$
Proof: The coefficients of $q(x)$ are symmetric polynomials in $\alpha,\beta,\gamma$, hence $q(x)\in\m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Check equivalence of quadratic forms over finite fields How to check whether the two quadratic forms \begin{equation} x_1^2 + x_2^2 \quad \text{(I)}\end{equation} and
\begin{equation} 2x_1x_2 \quad \text{(II)} \end{equation}
are equivalent on each of the spaces $\mathbb{F}_3^2\, \text{and}\,\mathbb{F}_5^2$?
I know t... | The characteristic of the field makes a big difference in the computations. I would start by comparing the discriminates of the two matrices.
Over both fields, the discriminants are $\pm 1$. The question is: do $\pm1$ differ by a square factor in both fields or not?
Over the field of five elements, $1\cdot 3^2\equiv -1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Another of $\frac{1^2}{1^2}+\frac{1-2^2+3^2-4^2}{1+2^2+3^2+4^2}+\cdots=\frac{\pi}{4}$ type expressable in cube? Gregory and Leibniz formula
(1)
$$-\sum_{m=1}^{\infty}\frac{(-1)^m}{2m-1}=\frac{\pi}{4}$$
We found another series equivalent to (1)
This is expressed in term of square numbers
$$-\sum_{m}^{\infty}\frac{\sum_{... | In term of cube numbers:
$$\frac{1^3}{1^3}+\frac{1^3-2^3}{1^3+2^3}+\frac{1^3-2^3+3^3}{1^3+2^3+3^3}+\frac{1^3-2^3+3^3-4^3}{1^3+2^3+3^3+4^3}+\cdots=3-\frac{\pi^2}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1789914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is the smallest number $n$ , such that $n\uparrow^4 n>3\uparrow^5 3$ holds?
What is the smallest number $n$, such that $$n\uparrow^4 n>3\uparrow^5 3$$ holds ?
$\uparrow$ stands for Knut's up-arrow-notation and is defined as follows
$a\uparrow b=a^b$
$$a\uparrow \uparrow b=a\uparrow a\uparrow ...\uparrow a\uparro... | In general, for $k \ge 2$ the smallest $n$ such that $n \uparrow^k n > 3 \uparrow^{k+1} 3$ is $n = (3 \uparrow^k 3) - 2$.
Set $S = 3 \uparrow^k 3$. We will prove:
Lemma: For all $0 \le i \le k$ and $j \ge 2$, $(S-2) \uparrow^i j > 3 \uparrow^i (j+2)$.
Proof by induction on $i$ and then on $j$.
Base case: $i=0$. Using t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1793555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the value of $\frac{a^2}{a^4+a^2+1}$ if $\frac{a}{a^2+a+1}=\frac{1}{6}$ Is there an easy to solve the problem? The way I did it is to find the value of $a$ from the second expression and then use it to find the value of the first expression. I believe there must be an simple and elegant approach to tackle the prob... | The second way is also a bit weird, you can divide by $a$ and $a^2$ and notice that $(a + \frac{1}{a})^2 = a^2 + \frac{1}{a^2} + 2$.
So you have $$\frac{1}{t+1}=\frac{1}{6}$$ and $$\frac{1}{t^2-1}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
} |
Geometrical Description of $ \arg\left(\frac{z+1+i}{z-1-i} \right) = \pm \frac{\pi}{2} $
The question is in an Argand Diagram, $P$ is a point represented by the complex number. Give a geometrical description of the locus of $P$ as $z$ satisfies the equation:
$$ \arg\left(\frac{z+1+i}{z-1-i} \right) = \pm \frac{\pi}{2... | Let's do it by another way and see if the results are the same (which would be good).
What you have inside the $\arg(\cdot)$, $\frac{z+z_0}{z-z_0}$ with $z_0=1+i$ is a particular case of a Möbius transformation
let's break it into pieces.
You can rewrite $\frac{z+z_0}{z-z_0}$ as
$$\frac{z+z_0}{z-z_0}=\frac{z-z_0}{z-z_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1799004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Definite integration of an algebraic expression
Evaluate $$\int_{0}^{1}\frac{1-x}{1+x}\frac{dx}{\sqrt{x+x^2+x^3}}$$
I think none of the properties of definite integral will be useful here so I think I will have to integrate. But I am unable to do so. Some hints on how to integrate. Thanks.
| Let $u = \sqrt{x+1+x^{-1}}$, we have $2 u du = (1 - x^{-2})dx$ and
$$
\int_0^1 \frac{1-x}{1+x}\frac{dx}{\sqrt{x+x^2+x^3}}
= \int_\infty^\sqrt{3} \frac{1-x}{1+x}\frac{2u du}{(x-x^{-1})u}
= 2 \int_\sqrt{3}^\infty \frac{du}{x+2+x^{-1}}\\
= 2 \int_\sqrt{3}^\infty \frac{du}{u^2+1}
= 2 \left[\tan^{-1}u\right]_{\sqrt{3}}^\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $a =\cos (\frac{2 \pi}{7})+i \sin (\frac{2 \pi}{7}) $ then construct a quadratic equation. If $a =\cos (\frac{2 \pi}{7})+i \sin (\frac{2 \pi}{7}) $, then find a quadratic equation whose roots are $\alpha = a + a^2 + a^4$ and $ \beta = a^3 + a^5 + a^6$.
Using the fact that sum of $7th$ roots of unity is $0$, I obta... | There is an obvious typo in the question $\beta$ should be $a^3+a^5+a^6$. Since $a$ is a complex 7th root of unity we have $1+a+a^2+a^3+a^4+a^5+a^6=0$, so $\alpha+\beta=-1$.
Multiplying out and using $a^7=1$ we have $\alpha\beta=3+a+a^2+a^3+a^4+a^5+a^6=2$. So $\alpha,\beta$ are the roots of the quadratic $x^2-x+2=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many ways can 2m objects be paired and assigned to m boxes? In how many ways can $2m$ objects be paired and assigned to $m$ boxes?
In that post the questioner claims that there are $\frac{(2m)!}{2^m}$ to pair $2m$ and assign them to $m$ boxes.
My attempt :-
First we need to find number of ways to pair $2m$ objects... | Line up the $m$ boxes in some order from left to right. We choose two of the $2m$ objects to place in the first box, choose two of the remaining $2m - 2$ boxes in the second box, choose two of the remaining $2m - 4$ boxes in the third box, and so forth until we place the final two objects in the $m$th box. We can do ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
$a$ and $b$ be positive real numbers such that $a + b = 1$. Prove that $a^a \cdot b^b+a^b\cdot b^a≤1$
Let $a$ and $b$ be positive real numbers such that $a + b = 1$. Prove that
$$a^a\cdot b^b+a^b \cdot b^a≤1$$
| By the Jensen's inequality,
$$
\log[a^ab^b]=a\log a+b\log b\leq\log[a^2+b^2]\implies a^ab^b\leq a^2+b^2.
$$
Similarly,
$$
\log[a^bb^a]=a\log b+b\log a\leq \log[ab+ba]\implies a^bb^a\leq 2ab.
$$
Summing then gives:
$$
a^ab^b+a^bb^a\leq a^2+b^2+2ab=(a+b)^2=1.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to verify $(1+\frac{1}{n})^2(1-\frac{1}{n^2})^{n-1}\geq \exp(\frac{1}{n})$ How to verify this inequality? Assuming that $n\in \mathbb{N}^+$, we have:
$$\left(1+\frac{1}{n}\right)^2\left(1-\frac{1}{n^2}\right)^{n-1}\geq \exp\left(\frac{1}{n}\right).$$
| If we take the logarithm of
$$\left(1+\frac{1}{n}\right)^2 \left(1-\frac{1}{n^2}\right)^{n-1} $$
we get:
$$ \log\left(\frac{1+\frac{1}{n}}{1-\frac{1}{n}}\right)+n \log\left(1-\frac{1}{n^2}\right)= 2\,\text{arctanh}\left(\frac{1}{n}\right)+n\log\left(1-\frac{1}{n^2}\right) $$
and since in a neighbourhood of the origin:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Can someone help me with this question of finding x as exponent? The equation is:
$$6^{x+1} - 6^x = 3^{x+4} - 3^x$$
I need to find x. I forgot how to use logarithm laws. Help would be appreciated. Thanks.
| Divide both sides by $3^x$, which is positive for all $x\in\mathbb R$:
$$6^{x+1} - 6^x = 3^{x+4} - 3^x$$
$$\iff 3^{x}\cdot 3\cdot 2^{x+1}-3^x\cdot 2^x=81 \cdot 3^x - 3^x$$
$$\iff 3\cdot 2^{x+1}-2^x=81-1$$
$$\iff 6\cdot 2^x-2^x=80\iff 5\cdot 2^x =80$$
$$\iff 2^x =16\iff x=\log_2(16)=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Let $n \in \mathbb{N}$. Proving that $13$ divides $(4^{2n+1} + 3^{n+2})$ Let $n \in \mathbb{N}$. Prove that $13 \mid (4^{2n+1} + 3^{n+2} ). $
Attempt: I wanted to show that $(4^{2n+1} + 3^{n+2} ) \mod 13 = 0. $ For the first term, I have $4^{2n+1} \mod 13 = (4^{2n} \cdot 4) \mod 13 = \bigg( ( 4^{2n} \mod 13) \cdot ( 4... | We prove that $13$ divides $4^{2n+1}+3^{n+2}$ by induction on $n$. If $n=1$, then $4^3+3^3=64+27=91$ is a multiple of 13. So the assertion holds if $n=1$.
Now assume that the assertion holds for $n=k$ (this assumption is called the inductive hypothesis). Thus, $13$ divides $4^{2k+1}+3^{k+2}$. We prove that the ass... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How to find the range of the given function?
Find the range of $$f(x)=\dfrac{x^2+14x+9}{x^2+2x+3}$$ where $x\in \mathbb R$
I thought of finding derivative but this will get too complicated so i am completely blank.
Thanks in advance!
| Its not that hard to differentiate.
$\frac {d}{dx} \dfrac {x^2 + 14x + 9}{x^2 + 2x + 3} = 0\\
\dfrac {(x^2 + 2x + 3)(2x + 14) - (2x + 2)(x^2 + 14x + 9)}{(x^2 + 2x + 3)^2} = 0$
Since the denominator is clearly not $0.$ We can multiply through by the denominator.
$2x^3 + 4x^2 + 6x + 14x^2 + 28x + 42 - 2x^3 - 2x^2 - 28x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Evaluating $\int_{0}^\infty \frac{\log x \, dx}{\sqrt x(x^2+a^2)^2}$ using contour integration I need your help with this integral:
$$\int_{0}^\infty \frac{\log x \, dx}{\sqrt x(x^2+a^2)^2}$$
where $a>0$. I have tried some complex integration methods, but none seems adequate for this particular one.
Is there a specific... | We integrate $$f(z) = \frac{\log z}{\sqrt{z}(z^2+a^2)^2}$$ along a
keyhole contour with the branch cut of the logarithm on the positive
real axis and its argument between $0$ and $2\pi.$ There is a small
circle of radius $\epsilon$ enclosing the origin and the large circle
has radius $R.$
Along the straight s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Prove $\forall n \geq 10, 2^n > n^3$ Prove $\forall n \geq 10, 2^n > n^3$
base case: $n = 10$
$2^{10} = 1024$
$10^3 = 1000$
$1024 > 1024$.
So $P(k)$ holds for $k = n$. We seek to show $P(k+1)$ holds:
We know $2^k > k^3$.
$2^k+3k^2+3k+1>k^3+3k^2+3k+1 = (k+1)^3$
$\iff 2^k+3k^2+3k+1>(k+1)^3$
Let us compare $k^3$ and $3k^... | Your proof is valid. You segue is clunky but acceptable. It's a matter of personal style but I'd do the following.
Induction step: Assume for some $k$ that $2^k > k^3$.
If we could show that $k^3 > 3k^2 + 3k + 1$ we'd have:
$2^{k+1} = 2^k + 2^k > k^3 + k^3 > k^3 + 3k^2 + 3k + 1 = (k+1)^3$
and we'd be done.
So it suf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
$f_n(x) = \cos \left( \sqrt{x^2 + \frac{1}{n} }\right)$ is uniformly convergent I need to prove that this sequence of functions:
$$f_n(x) = \cos \left( \sqrt{x^2 + \frac{1}{n} }\right)$$
converges uniformly on $[0,1]$.
This is a question regarding this one. There's an answer that suggests these hints:
$$\cos\alpha-\co... | Here's a different idea: The mean value theorem gives $\cos b - \cos a = -\sin c\cdot(b-a),$ which implies $|\cos b - \cos a| \le |b-a|.$ Thus
$$|\cos \sqrt {x^2+1/n} - \cos x| \le |\sqrt {x^2+1/n} - x|.$$
So we need only show $\sqrt {x^2+1/n} \to x$ uniformly on $[0,1].$ For this it's good to know that if $a,b \ge 0,$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Range of $xyz\;,$ If $x+y+z=4$ and $x^2+y^2+z^2=6$
If $x,y,z\in \mathbb{R}$ and $x+y+z=4$ and $x^2+y^2+z^2=6\;,$ Then range of $xyz$
$\bf{My\; Try::}$Using $$(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$$
So we get $$16=6+2(xy+yz+zx)\Rightarrow xy+yz+zx = -5$$ and given $x+y+z=4$
Now let $xyz=c\;,$ Now leyt $t=x,y,z$ be the roo... | By CS inequality we get
$$|x^3+y^3+z^3|\le\sqrt{x^2+y^2+z^2}\sqrt{x^4+y^4+z^4}\tag{1}$$
Where
\begin{align*}
x^4+y^4+z^4&=(x^2+y^2+z^2)^2-2[(xy+yz+xz)^2-2xy^2z-2xyz^2-2x^2yz]\\
&=36-2[5^2-2xyz(4)]\\
x^4+y^4+z^4&=16xyz-14\tag{2}
\end{align*}
Now, from the identity
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)$$
follo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Fourier Series of $\cos(ax)$ I would like to ask some help on this problem..
01) Expand the following function in fourier series.
$f(x)=cosax,−π<x<π$ where '$a$' is not an integer. Hence, Show that
$\frac{1}{sint} = \frac{1}{t} + \sum_{n=0}^{\infty}(-1)^{n}\left \{ \frac{1}{t-n\pi} + \frac{1}{t+ n\pi}\right \}$
Where '... | Write
$$f(x) = a_0 + 2 \sum_{k=1}^{\infty} a_k \cos{k x} $$
because $f$ is even. Then
$$\begin{align}a_k &= \frac1{2 \pi} \int_{-\pi}^{\pi} dx \, \cos{a x} \, \cos{k x} = \frac1{4 \pi} \int_{-\pi}^{\pi} dx \, \left [\cos{(k-a)x} + \cos{(k+a) x} \right ]\end{align}$$
or
$$\begin{align}a_k &= \frac1{2 \pi} \left [\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1818734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do I prove $\int^{\pi}_{0} \frac{\cos nx}{1+\cos\alpha \cos x} \mathrm{d}x = \frac{\pi}{\sin \alpha} (\tan \alpha - \sec \alpha)^n $? The result I wish to show is that for $n \in \mathbb{Z}$, $$\int^{\pi}_{0} \frac{\cos nx}{1+\cos\alpha \cos x} \mathrm{d}x = \frac{\pi}{\sin \alpha} (\tan \alpha - \sec \alpha)^n $$... | One approach is to use contour integration.
Assuming that $0 <\alpha < \pi $ and $n \in \mathbb{Z}_{\geq 0}$,
$$ \begin{align}\int_{0}^{\pi} \frac{\cos nx}{1+ \cos \alpha \cos x} \, dx &= \frac{1}{2} \, \text{Re} \int_{-\pi}^{\pi} \frac{e^{inx}}{1+\cos \alpha \cos x} \, dx \\ &= \frac{1}{2} \, \text{Re} \int_{|z|=1} \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1820446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
$\lim_{x \to 0} \frac{\sin(a + b)x + \sin(a - b)x + sin(2ax)}{\cos^2 bx - cos^2 ax}$ Question : -
$$\lim_{x \to 0} \frac{\sin(a + b)x + \sin(a - b)x + \sin(2ax)}{\cos^2 bx - \cos^2 ax}$$
My attempt :-
$$\lim_{x \to 0} \frac{2* \sin ax * \cos bx + 2*\sin ax * \cos (ax)}{(\cos bx - \cos ax)*(\cos bx +\cos ax)}$$
$$\... | I'll assume that $a^2-b^2\ne0$ and $a\ne0$.
Notice that $\cos^2(bx)-\cos^2(ax)=(\cos(bx)+\cos(ax))(\cos(bx)-\cos(ax))$ and that the first factor has limit $2$, so it can be set away momentarily. Then
$$
\cos(bx)-\cos(ax)=1-\frac{b^2x^2}{2}+o(x^2)-1+\frac{a^2x^2}{2}+o(x^2)
=\frac{a^2-b^2}{2}x^2+o(x^2)
$$
and so the deno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
General integral of a linear system of ODEs I want to find the solution of
\begin{cases}
x'= -5x-y+e^t \\
y'= 2x-3y
\end{cases}
$$A = \begin{bmatrix} -5 & -1 \\ 2 & -3 \end{bmatrix}$$
I calculated the exponential matrix $e^{tA}$ and found the solution of the homogeneous system
$$\begin{bmatrix} x\\ y\end{bmatrix} = e... | Laplace-transforming the system of ODEs,
$$\begin{bmatrix}s+5 & 1\\ -2 & s+3\end{bmatrix} \begin{bmatrix} X (s) \\ Y (s)\end{bmatrix} = \begin{bmatrix}\frac{1}{s-1}\\ 0\end{bmatrix}$$
Hence,
$$\begin{array}{rl} \begin{bmatrix} X (s) \\ Y (s)\end{bmatrix} &= \begin{bmatrix}s+5 & 1\\ -2 & s+3\end{bmatrix}^{-1} \begin{bma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1823008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How would you find the roots of $x^3-3x-1 = 0$ I'm not too sure how to tackle this problem. Supposedly, the roots of the equation are $2\cos\left(\frac {\pi}{9}\right),-2\cos\left(\frac {2\pi}{9}\right)$ and $-2\cos\left(\frac {4\pi}{9}\right)$
How do I start? The Cosines seem especially scary...
| Let $x=t+t^{-1}$ in $y=x^3-3x-1$. So we have $$\left(t+\frac {1}{t}\right)^{3}-3\left(t+\frac {1}{t}\right)-1=0$$
Expanding that and simplifying, we get $\frac {t^{6}-t^{3}+1}{t^{3}}$ and multiplying both the numerator and the denominator by $(t^3+1)$ we get $\frac {t^{9}+1}{t^{3}(t^{3}+1)}$.
We let $t=\cos(\alpha)+i\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $ Prove:
$$ 1+2q+3q^2+...+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2} $$
Hypothesis:
$$ F(x) = 1+2q+3q^2+...+xq^{x-1} = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} $$
Proof:
$$ P1 | F(x) = \frac{1-(x+1)q^x+xq^{x+1}}{(1-q)^2} + (x+1)q^x = \frac{1-(x... | You can also prove by induction. Assume true for $n$, and show for $n+1$.
$$1+2q+3q^2+\cdots+nq^{n-1} = \frac{1-(n+1)q^n+nq^{n+1}}{(1-q)^2}.$$
Then,
$$1+2q+3q^2+\cdots+nq^{n-1}+(n+1)q^n = \frac{1-(n+1)q^n+nq^{n+1}+(n+1)q^n(1-q)^2}{(1-q)^2}= \frac{1-(n+2)q^{n+1}+(n+1)q^{n+2}}{(1-q)^2}.$$
Thus the result holds for $n+1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
In $\triangle ABC$ with $A = \frac{\pi}{4}$, what is the range of $\tan B\tan C$?
In a $\triangle ABC\;,$ If $\displaystyle A=\frac{\pi}{4}\;,$ and $\tan B\cdot \tan C = p\;,$ Then range of $p$
$\bf{My\; Try::}$ For a $\triangle ABC\;, A+B+C=\pi.$ So we get $\displaystyle A+B=\frac{3\pi}{4}$
So $$\tan(A+B)=-1\Rightar... | http://www.askiitians.com/forums/Algebra/let-a-b-c-be-three-angles-such-that-a-4-and_104701.htm
$A + B + C = \pi$
Also $A = \pi/4\Rightarrow B + C = 3 \pi/4\Rightarrow 0 < B, C < 3\pi/4$
Now $\tan B \tan C = P\Rightarrow \dfrac{\sin B \sin C}{\cos B \cos C} = p/1$
Applying componendo and dividendo, we get
$$\dfrac{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826678",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Proving that $3 + 3 \times 5 + 3 \times 5^2 + \cdots+ 3 \times 5^n = [3(5^{n+1} - 1)] / 4$ whenever $n \geq 0$
Use induction to show that $$3 + 3 \times 5 + 3 \times 5^2 + \cdots+ 3 \times 5^n= \frac{3(5^{n+1} - 1)}{4} $$whenever $n$ is a non-negative integer.
I know I need a base-case where $n = 0$:
$$3 \times 5^0 =... | Let $S(n)$ be the statement: $3+3\times{5}+3\times{5^{2}}+\cdots+3\times{5^{n}}=\dfrac{3(5^{n+1}-1)}{4}$; $n\geq{0}$
Basis step: $S(0)$:
LHS: $3\times{5^{0}}=3\times{1}$
$\hspace{23 mm}=3$
RHS: $\dfrac{3(5^{(0)+1}-1)}{4}=\dfrac{3(5^{1}-1)}{4}$
$\hspace{37.5 mm}=\dfrac{3\times{4}}{4}$
$\hspace{37.5 mm}=3$
$\hspace{75 mm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
prove $\lim_{n\to\infty}{\frac{a^n}{n!}}=0$ I have the proof but i don't understand one part. The proof (for $a>1$) goes as follows:
there exists $k \in N$ such that$a<k, \frac{a}{k}<1$ and since $\frac{a}{k+i}<\frac{a}{k}, i \in N$
$$\frac{a^n}{n!}=\frac{a}{1} \cdot \frac{a}{2} \cdot...\cdot\frac{a}{k}\cdot\frac{a}{k+... | We are given :$$\frac { a^{ n } }{ n! } =\overset { k }{ \overbrace { \frac { a }{ 1 } \cdot \frac { a }{ 2 } \cdot ...\cdot \frac { a }{ k } } } \cdot \overset { n-k }{ \overbrace { \frac { a }{ k+1 } \cdot ...\cdot \frac { a }{ n } } } $$
Now you know:$$\frac { a }{ k+1 } <\frac { a }{ k } \\ \frac { a }{ k+2 } <... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to find number of solutions of an equation? Given $n$, how to count the number of solutions to the equation $$x + 2y + 2z = n$$ where $x, y, z, n$ are non-negative integers?
| We need the coefficient of $x^n$ in $(1+x+x^2+\cdots)(1+x^2+x^4+\cdots)^2$.
This can be written as
\begin{align*}
\frac{1}{(1-x)(1-x^2)^2} &= \frac{1}{(1-x)^3(1+x)^2} \\
&= \frac{1/4}{(1-x)^3}+\frac{1/4}{(1-x)^2}+\frac{3/16}{1-x}+\frac{1/8}{(1+x)^2}+\frac{3/16}{1+x}
\end{align*}
The required coefficient is
$$ \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let
$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$
Prove that $I=J={\pi \over 2\sqrt3}$
Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\... | Noting that $$
J \stackrel{x\mapsto\frac{1}{x}}{=} \int_0^{\infty} \frac{x^6}{x^8+x^4+1} d x
$$
On average,
$$
\begin{aligned}
2 J & =\int_0^{\infty} \frac{x^6+1}{x^8+x^4+1} d x \\
& =\int_0^{\infty} \frac{\left(x^2+1\right)\left(x^4-x^2+1\right)}{\left(x^4+x^2+1\right)\left(x^4-x^2+1\right)} d x \\
& =\int_0^{\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 9,
"answer_id": 8
} |
Finding Angle of Irregular Quadrilateral I have a linkage with 4 vertices and 3 bars. One point is at origin and one point is at point X,Y. The lengths of the 3 bars are known. In order to keep the solutions less than infinite, I assume the two angles are equal.
Given this info how can I solve for A?
| Let $O$ be the origin, $A,B$ be the vertices with angel $A$, and $C$ be the point $(X,Y)$. We know $OC=\sqrt{X^2+Y^2}$. Let $OA$ intersect $CB$ at $D$. Then, $DAB$ is isosceles with base angles equal to $\pi-A$. Thus, $DA=DB=\dfrac a{-2\cos A}$. Also, using Cosine's Law on $DOC$, we have
$$X^2+Y^2=OC^2=(a-\frac a{2\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$5^{th}$ degree polynomial expression
$p(x)$ is a $5$ degree polynomial such that
$p(1)=1,p(2)=1,p(3)=2,p(4)=3,p(5)=5,p(6)=8,$ then $p(7)$
$\bf{My\; Try::}$ Here We can not write the given polynomial as $p(x)=x$
and $p(x)=ax^5+bx^4+cx^3+dx^2+ex+f$ for a very complex system of equation,
plz hel me how can i solve that... | HINT:
Let $$\dfrac{p(x)}{(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)}=\sum_{i=1}^6\dfrac{A_i}{x-i}$$
Multiply both sides by $(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)$ and put $x=1,2,3,4,5,6$ one by one in the resultant identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 4
} |
How to solve asymptotic expansion: $\sqrt{1-2x+x^2+o(x^3)}$ Determinate the best asymptotic expansion for $x \to 0$ for:
$$\sqrt{1-2x+x^2+o(x^3)}$$
How should I procede?
In other exercise I never had the $o(x^3)$ in the equation but was the maximum order to consider.
| Hint. One may recall the Taylor series expansion, as $u \to 0$,
$$
\sqrt{1+u}=1+\frac{u}2-\frac{u^2}8+\frac{u^3}{16}+o(u^3).
$$ Setting, as $x \to 0$,
$$
u=-2x+x^2+o(x^3)
$$ it gives
$$
1+\frac{-2x+x^2+o(x^3)}2-\frac{(-2x+x^2+o(x^3))^2}8+\frac{(-2x)^3}{16}+o(x^3)
$$ or
$$
1+\frac{-2x+x^2}2-\frac{4x^2-4x^3}8+\frac{-8x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
An erroneous application of the Counting Theorem to a regular hexagon? I'm trying to count the unique orbits of a regular hexagon such that each vertex is either Black or White and each edge is either Red, Gree, or Blue. The group I've chosen to act on the hexagon is the dihedral group $D_7$, $$\{e,r,r^2,r^3,r^4,r^5,r^... | I now redo the calculation (correctly, I hope), using the notation of the original question. The symmetry group of the hexagon is the dihedral group $D_6$,
$$ D_6 = \langle r,s \mid r^6 = s^2 = 1, r^s = r^{-1} \rangle, $$
where $r$ is a rotation by $\frac{2\pi}{6}$, and $s$ a reflection in an axis connecting two oppo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Algebraic Manipulation Given that
$ a^2 - b^2 = 60 $
and
$a - b = 6 $
$a + b = 10$
Find value of $a\cdot b$
I tried
$(a-b)^2 = 6^2 \longrightarrow a^2 - 2ab + b^2 = 36$
| Hint. One may observe that
$$
2a=(a-b)+(a+b)=6+10=16
$$
$$
2b=(a+b)-(a-b)=10-6=4.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the value of $ [1/ 3] + [2/ 3] + [4/3] + [8/3] +\cdots+ [2^{100} / 3]$ Assume that [x] is the floor function. I am not able to find any patterns in the numbers obtained. Any suggestions?
$$[1/ 3] + [2/ 3] + [4/3] + [8/3] +\cdots+ [2^{100} / 3]$$
| I think I have an answer. Look at $\frac{1}{3}$ in binary format and you will notice that it is $0.\overline{01}$.
This tells us something, namely that $[2^{2i}/3]+[2^{2i+1}/3] = 4^i-1$. This immediately tells me that your sum reduces to:
$$\sum \limits_{i=0}^{50} [2^{2i}/3]+\sum \limits_{i=0}^{49} [2^{2i+1}/3]=\sum \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
} |
Let $f = 2x^4 + 2(a - 1)x^3 + (a^2 + 3)x^2 + bx + c.$ ,Find out $a, b, c ∈ R$ and its roots knowing that all roots are real.
Let $f = 2x^4 + 2(a - 1)x^3 + (a^2 + 3)x^2 + bx + c.$
Find out $a, b, c ∈ R$ and its roots knowing that all roots are real.
The first thing that came into my mind was to use vieta's formula... | Hint: Since $f(x)$ has $4$ roots, the equation $f''(x)=0$ has at least one solution. Note that
$$
f''(x)=24x^2+12(a-1)x+2(a^2+3)
$$
The discriminate is
$$
\Delta=12^2(a-1)^2-4\cdot 24\cdot 2(a^2+3)=-48(a+3)^2
$$
In order to have a root we must have that $a=-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
I want to show that $\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$ I want to show that
$$\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$$
Expand $(x^4-x+\pi)^2=x^4-2x^3+2x^2-2x\pi+\pi{x^2}+\pi^2$
Let see (substitution of $y=x^2$)
$$\int_{-\infty}... | I think that this is a time for complex analysis....
$\int \frac {(x^2 - x + \pi)^2}{(x^2 x + 1)^2} dx$
$\int \frac {(x^2 - x + \pi)^2}{(x+a)^2(x-\bar a)^2(x- b)^2(x-\bar b)^2} dx$
where $a,b,\bar a, \bar b$ are the complex roots of $(x^4-x^2 + 1)$
The Cauchy integral theorem says:
$\oint \frac {f(a)}{(z-a)^2} dz = 2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1836306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 4
} |
Solve in integers the equation $\left\lfloor\frac{x^2-y^3}{x+y^2} \right\rfloor=1+x-y$
Solve in integers the equation
$$\left\lfloor\frac{x^2-y^3}{x+y^2} \right\rfloor=1+x-y$$
My attempt:
I used http://www.wolframalpha.com/:
$x=-2; y=\{3,4,5,6,7,..\}$ or $x=-1, y=\{-10,-9,...\}$.
1) Let $\lfloor\frac{x^2-y^3}{x+y^... | We have
\begin{align}
\lfloor \frac{x^2 - y^3}{x + y^2} \rfloor &= \lfloor \frac{x^2 + xy^2 - xy - y^3}{x + y^2} - \frac{xy^2 - xy}{x + y^2}\rfloor \\
&= \lfloor \frac{(x-y)(x + y^2)}{x + y^2} - \frac{xy^2 - xy}{x + y^2}\rfloor \\
&= \lfloor x - y - \frac{xy^2 - xy}{x + y^2}\rfloor \\
&= x - y + \lfloor - \frac{xy^2 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove by induction $3+3 \cdot 5+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$ My question is:
Prove by induction that $$3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$$ whenever $n$ is a nonnegative integer.
I'm stuck at the basis step.
If I started with $1$. I get the right hand side is $1... | Let $S(n)$ be the statement: $3+3\cdot{5}+\cdots+3\cdot{5^{n}}=\dfrac{3\hspace{1 mm}(5^{n+1}-1)}{4}$; $n\geq{0}$
Basis step: $S(0)$:
LHS: $3\cdot{5^{(0)}}=3$
RHS: $\dfrac{3\hspace{1 mm}(5^{(0)+1}-1)}{4}=\dfrac{3\hspace{1 mm}(5^{1}-1)}{4}$
$\hspace{35.5 mm}=\dfrac{3(4)}{4}$
$\hspace{35.5 mm}=3$
$\hspace{52.5 mm}$LHS $=$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1838161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Intersection of a nested interval of $A_n=\left[3-{\frac{1}{\sqrt{n}},3+\frac{1}{3^n}}\right]$
$A_n=\left[3-{\frac{1}{\sqrt{n}},3+\frac{1}{3^n}}\right]$
What is $\bigcap_{n=1}^{\infty}A_n$
Since every set becomes a subset of the next set, is it correct to say that the intersection of all sets $n$ is just:
$\left[2,... | $$3+\frac{1}{3^n}-(3+\frac{1}{3^{n-1}}) = \frac{-2}{3^{n}} < 0$$ for all $ n \ge 2$ (1)
$$3-{\frac{1}{\sqrt{n}}}-(3-{\frac{1}{\sqrt{n-1}}}) = \frac{\sqrt{n}-\sqrt{n-1}}{\sqrt{n-1}\sqrt{n}} > 0 $$ for all $ n \ge 2$(2)
$$\lim_{x\to \infty}3+\frac{1}{3^n} = 3$$
$$\lim_{x\to \infty}3-{\frac{1}{\sqrt{n}}} = 3$$
Therefore B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$a\equiv b\pmod{n}\iff a/x\equiv b/x\pmod{n/\gcd(x,n)}$ for integers $a,b,x~(x\neq 0)$ and $n\in\Bbb Z^+$? I'm trying to prove/disprove the following:
If $a,b,x$ be three integers (where $x\neq 0$) such that $x\mid a,b$ and $n$ be a positive integer, then the following congruence holds:
$$a\equiv b\pmod{n}\iff a/x\equ... | Assume that $a \equiv b \pmod n$. Then there exist an integer $k$ such that $a-b = kn$. Now as $x$ divides the RHS we will have $\frac{x}{gcd(x,n)} \mid k$. So now we have:
$$\frac{a}{x} - \frac{b}{x} = \frac{k}{\frac{x}{gcd(x,n)}} \cdot \frac{n}{gcd(x,n)}$$
So as $\frac{k}{\frac{x}{gcd(x,n)}} \in \mathbb{Z}$ we have ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to solve $x<\frac{1}{x+2}$ Need some help with:
$$x<\frac{1}{x+2}$$
This is what I have done:
$$Domain: x\neq-2$$
$$x(x+2)<1$$
$$x^2+2x-1<0$$
$$x_{1,2} = \frac{-2\pm\sqrt{4+4}}{2} = \frac{2 \pm \sqrt{8}}{2} = \frac{-2\pm2\sqrt{2}}{2}$$
What about now?
| Hint 1: You already did a mistake: When you multiply by $(x+2)$ the inequality changes if $x+2<0$.
So first split the problem in two cases: $x+2 >0$ and $x+2<0$.
Hint 2 Factor $x^2+2x-1=(x-x_1)(x-x_2)$ and disscuss the sign of each linear term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Some doubts with the sign of a derivative Good evening to everyone. The derivative is defined in the following order:
$$ \frac{d}{dx} f(x)=\frac{{-x^2-x+11}}{\left(x+3\right)^2}e^{2-x}\:$$ for $ x < -3 $ $$\:\frac{d}{dx}f(x)=\frac{{x^2+x-11}}{\left(x+3\right)^2}e^{2-x} $$ for $ -3< x< 2 $ and $$ \frac{d}{dx}f(x) = \fra... | We have a root for the first derivative at around $x\approx-3.854$ the other is negligible as the derivative is not defined with the particular function at that point.
With $x<-\frac{1+3\sqrt{5}}{2}, f'(x)<0 $. Thus, $f(x)$ is decreasing for $x\in(-\infty,-\frac{1+3\sqrt{5}}{2})$ and increasing for $x\in( -\frac{1+3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is something wrong with this solution for $\sin 2x = \sin x$? I have this question. What are the solutions for $$
\sin 2x = \sin x; \\ 0 \le x < 2 \pi $$
My method:
$$ \sin 2x - \sin x = 0 $$
I apply the formula $$ \sin a - \sin b = 2\sin \left(\frac{a-b}{2} \right) \cos\left(\frac{a+b}{2} \right)$$
So:
$$ 2\sin\le... | If we need to solve $$\sin \frac{x}{2} \cos \frac{3x}{2} = 0$$ for $0 \le x < 2\pi$, then the inequality condition is equivalent to $$0 \le x/2 < \pi,$$ and $$0 \le 3x/2 \le 3\pi.$$ Thus $\sin x/2 = 0$ admits only $x = 0$ in first interval; and in the second interval, we have solutions $$\frac{3x}{2} \in \left\{ \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1845034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
How to perform the following integration $\int \frac{\cos 5x+5\cos 3x+10\cos x}{\cos 6x+ 6\cos 4x+ 15\cos 2x +10}dx$ $$\int \frac{\cos 5x+5\cos 3x+10\cos x}{\cos 6x+ 6\cos 4x+ 15\cos 2x +10}dx$$
How to simplify the expression given? I tried using formulas for $\cos 2x$ and $\cos 3x$. I also tried the using $\cos x + \c... | Using $$\cos 6x+6\cos 4x+15\cos 2x+10 = (\cos 6x+\cos 4x)+5(\cos 4x+\cos 2x)+10(1+\cos 2x)$$
So $$ = 2\cos 5x\cos x+5\cdot 2\cos 3x\cos x+20\cos^2 x = 2\cos x(\cos 5x+5\cos 3x+10\cos x)$$
So $$\int \frac{\cos 5x+5\cos 3x+10\cos x}{\cos 6x+ 6\cos 4x+ 15\cos 2x +10}dx=\frac{1}{2}\int \frac{1}{\cos x}dx$$
So $$ = \frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Limit of a sequence defined as a definite integral Given two sequences $$a_n=\int_{0}^{1}(1-x^2)^ndx$$ and $$b_n=\int_{0}^{1}(1-x^3)^ndx$$ where $n$ is a natural number, then find the value of $$\lim_{n \to \infty}(10\sqrt[n]{a_n}+5\sqrt[n]b_n)$$
I have no starts. Looks good though. Some hints please. Thanks.
| In the same spirit as previous answers, using the gamma function, we should get $$a_n=\frac{\sqrt{\pi }}2\frac{ \Gamma (n+1)}{ \Gamma \left(n+\frac{3}{2}\right)}$$ $$b_n=\Gamma \left(\frac{4}{3}\right)\frac{ \Gamma (n+1)}{\Gamma
\left(n+\frac{4}{3}\right)}$$ Taking logarithms and using Stirling approximation $$\log(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Proving that $1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 } $ by induction
Prove that
$$1\cdot 2+2\cdot 3+\cdots+n\left( n+1 \right) =\frac { n\left( n+1 \right) \left( n+2 \right) }{ 3 }. $$
I can get to $1/3(k+1)(k+2) + (k+1)(k+2)$ but then finishing off an... | $$\sum\limits_{i=1}^{n}{i(i+1)=}\sum\limits_{i=1}^{n}{{{i}^{2}}+\sum\limits_{i=1}^{n}{i=}}\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}=\frac{n(n+1)(n+2)}{3}$$
for $n=1$we have $1(2)=\frac{1(1+1)(1+2)}{3}$
let
$$1(2)+2(3)+3(4)+...+k(k+1)=\frac 13 k(k+1)(k+2)$$
we show
$$1(2)+2(3)+3(4)+...+(k+1)(k+2)=\frac 13 (k+1)(k+2)(k+3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
$a$ and $b$ are factors of $6^6$ and $a$ is a factor of $b$ How many pairs of ($a$,$b$) of positive integers are there such that $a$ and $b$ are factors of $6^6$ and $a$ is a factor of $b$?
What I tried
I know $6^6$ an be broken down into $(2)^6 (3)^6$
If $a$ is a factor of $b$,and if $a=1$,there will be $18$ groups.
S... | It suffices to consider powers of $2,3$ separately.
If $v_2(a)=r$ then $r≤v_2(b)≤6$. Of course, $v_2(a)\in \{0,1,2,3,4,5,6\}$. If $v_2(a)=0$ there are $7$ possibilities for $v_2(b)$. If $v_2(a)=1$ there are $6$ possibilities for $v_2(b)$, and so on. Thus, considering only powers of $2$, we get $$7+6+5+4+3+2+1=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The number of irrational roots of $(x^2 -3x +1)(x^2 +3x +2)(x^2 -9x + 20) = -30$ is___?
The number of irrational roots of $(x^2 -3x +1)(x^2 +3x +2)(x^2 -9x + 20) = -30$ is___?
My Approach :
Multiplying the above equation and then apply Descartis rule is very lengthy method. Also I tried it by taking factors of $3... | I'm not sure if this is a good method for your question, but we have
$$(x^2-3x+1)(x+1)(x+2)(x-4)(x-5)=-30$$
rearranging
$$(x^2-3x+1)(x+1)(x-4)(x+2)(x-5)=-30$$
$$(t+1)(t-4)(t-10)=-30$$
where $t=x^2-3x$.
Now, one sees that $t=5$ works. So,
$$(t+1)(t-4)(t-10)+30$$
is divisible by $t-5$ to have
$$(t+1)(t-4)(t-10)+30=(t-5)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Constant such that $\max\left(\frac{3}{3-2c},\frac{3a}{3-2d},\frac{3b}{3-2e}\right)\geq k\cdot\frac{2+3a+4b}{9-c-2d-3e}$ What is the greatest constant $k>0$ such that
$$\max\left(\frac{3}{3-2c},\frac{3a}{3-2d},\frac{3b}{3-2e}\right)\geq k\cdot\frac{2+3a+4b}{9-c-2d-3e}$$
for any $0\leq b\leq a\leq 1$ and $0\leq c\leq d\... | Yes, $\color{red}{k=\frac 17}$ works.
First,
$$\frac{3}{3-2c}\cdot\frac{9-c-2d-3e}{2+3a+4b}\ge \frac{3}{3-2\cdot 0}\cdot\frac{9-1-2\cdot 1-3\cdot 1}{2+3\cdot 1+4\cdot 1}=\frac 13$$
Second, when
$$\frac{3a}{3-2d}\ge \frac{3}{3-2c}\implies a\ge\frac{3-2d}{3-2c}\ge \frac{3-2\cdot 1}{3-2\cdot 0}=\frac 13$$
we have
$$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Doubt for substitution in two variables limit I have an exercise on my book like that
$\lim_{(x,y)\to (0,1)}f(x,y)=\frac {(y^2-1)(9x^2+2)\log\left(1+x^5\right)}{x^4+(y-1)^6}$
then it says without other explanations:
for $(x,y)\to(0,1)$ you have
$\left\vert f(x,y)\right\vert \le 5\left\vert\frac{(y-1)x^5}{x^4}\right\ve... | When estimating the absolute value of a fraction, you needn't limit yourself to algebraic equalities, but may also make the (absolute value of) the numerator larger and the denominator smaller. If $x \neq 0$, for example,
\begin{align*}
\left\lvert\frac{(y^{2} - 1) (9x^{2}+ 2) \log(1 + x^{5})}{x^{4} + (y - 1)^{6}}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How many integers $\leq N$ are divisible by $2,3$ but not divisible by their powers? How many integers in the range $\leq N$ are divisible by both $2$ and $3$ but are not divisible by whole powers $>1$ of $2$ and $3$ i.e. not divisible by $2^2,3^2, 2^3,3^3, \ldots ?$
I hope by using the inclusion–exclusion principle ... | We want to count those numbers divisible by $6=2\cdot3$, but not those divisible by $12=2^2\cdot3$ or $18=2\cdot3^2$. However, if we count both those numbers divisible by $12$ and those divisible by $18$, we've counted those divisible by $36$ twice. Therefore, the count should be
$$
\left\lfloor\frac N6\right\rfloor-\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proof of the square root inequality $2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}$ I stumbled on the following inequality: For all $n\geq 1,$
$$2\sqrt{n+1}-2\sqrt{n}<\frac{1}{\sqrt{n}}<2\sqrt{n}-2\sqrt{n-1}.$$
However I cannot find the proof of this anywhere.
Any ideas how to proceed?
Edit: I posted a... | Let $f(x)=2\sqrt{x}$. Using mean value theorem we get
$$\frac{f(n+1)-f(n)}{(n+1)-n} = f'(c)$$
for some $c \in (n, n+1)$. Equivalently
$$2\sqrt{n+1} - 2\sqrt{n} = \frac{1}{\sqrt c}.$$
Since $c>n$, $$\frac{1}{\sqrt c} < \frac{1}{\sqrt{n}},$$
therefore
$$2\sqrt{n+1}-2\sqrt{n} < \frac 1{\sqrt n}.$$
Right inequality can be ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Show that $x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0$
Show that for any real number $x$:
$$x^2 \sin{x} + x \cos{x} + x^2 + \frac{1}{2} > 0.$$
$\bf{My\; Try::}$ Using $a\sin x+b\cos x\geq -\sqrt{a^2+b^2}$
So $$x^2\sin x+x\cos x\geq -\sqrt{x^4+x^2}=-x\sqrt{1+x^2}$$
and $$4x^4+4x^2+1>4x^4+4x^2\Rightarrow (2x^2+... | Let us check for $$x^2(1+\sin x)+x\cos x-y=0$$
As $x$ is real, the discriminant $$\cos^2x+4(1+\sin x)y\ge0$$
Now $1+\sin x\ge0$
Check what if $1+\sin x=0?$
Else $4y\ge\sin x-1\ge-1-(-1)$
Observe that the equality cannot occur as $1+\sin x\ne0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Matrix decomposition into square positive integer matrices This is an attempt at an analogy with prime numbers. Let's consider only square matrices with positive integer entries. Which of them are 'prime' and how to decompose such a matrix in general?
To illustrate, there is a product of two general $2 \times 2$ matric... | I now understand that an algorithm for the general case is unlikely. Searching the web, I only found several articles, concerned with positive integer matrix decomposition into binary matrices.
I tried to consider a particular case, which seems to be the most simple:
$$C=\left[ \begin{matrix} 2 & c_b \\ c_a & c \en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1864244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int\sin^{7}x\cos^4{x}\,dx$
$$\int \sin^{7}x\cos^4{x}\,dx$$
\begin{align*}
\int \sin^{7}x\cos^4{x}\,dx&= \int(\sin^{2}x)^3 \cos^4{x}\sin x \,dx\\
&=\int(1-\cos^{2}x)^{3}\cos^4{x}\sin x\,dx,\quad u=\cos x, du=-\sin x\,dx\\
&=-\int(1-u^{2})^3u^4{x}\,du\\
&=-\int (1-3u^2+3u^4-u^6)u^4\,du\\
&=u^4-3u^6+3u^8-u^{1... | The derivative is $$\sin x(-\cos^4x-3\cos^6x+3\cos^8x-\cos^{10}x)$$
$$=-\sin x\cos^4x(-1-3\cos^2x+3\cos^4x+\cos^6x)$$
Now,
$$-1-3\cos^2x+3\cos^4x+\cos^6x=-(1-\cos^2x)^3=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Notice that the sum of the powers of $2$ from $1$, which is $2^0$, to $2^{n-1}$ is equal to ${2^n}{-1}$. Please explain in quotations!
"Notice that the sum of the powers of $2$ from $1$, which is $2^0$, to $2^{n-1}$ is equal to $2^n-1$." In a very simple case, for $n = 3, 1 + 2 + 4 = 7 = 8 - 1$.
| $1 + 2+ 4 + 8 + ......... + 2^{n-2} + 2^{n-1}=$
$1 + 1 + 2+ 4 + 8 + ......... + 2^{n-2} + 2^{n-1} - 1=$
$2 + 2+ 4 + 8 + ......... + 2^{n-2} + 2^{n-1} - 1=$
$4+ 4 + 8 + 16 + 32 + ......... + 2^{n-2} + 2^{n-1} - 1=$
$8 + 8 + 16 + 32 + ......... + 2^{n-2} + 2^{n-1} - 1=$
$16 + 16 + 32 + ......... + 2^{n-2} + 2^{n-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
find $f\in L^2([0,\pi])$ such that its $L^2$ distance to $\sin(x)$ and $\cos(x)$ are both bounded by specific constants I want to find all $f\in L^2([0,\pi])$ such that
$$
\begin{align}
\int_0^\pi\lvert f(x)-\sin(x)\rvert^2\,dx &\le \frac{4\pi}{9}\\
\int_0^\pi\lvert f(x)-\cos(x)\rvert^2\,dx &\le \frac{\pi}{9}\\
\end{al... | If $f(x)$ is real-valued, take $\dfrac{1}{3}$ times the first inequality plus $\dfrac{2}{3}$ times the second inequality to get:
$$\int_{0}^{\pi}\dfrac{1}{3}(f(x)-\sin x)^2+\dfrac{2}{3}(f(x)-\cos x)^2\,dx \le \dfrac{1}{3} \cdot \dfrac{4\pi}{9} + \dfrac{2}{3}\cdot\dfrac{\pi}{9}$$
$$\int_{0}^{\pi}\left[f(x)^2-2f(x)\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Show $\ln\left(\frac{n+(-1)^{n}\sqrt{n}+a}{n+(-1)^{n}\sqrt{n}+b} \right)=\frac{a-b}{n}+\mathcal{O}\left(\frac{1}{n^2} \right) $
I would like to prove the following:
$$\ln\left(\dfrac{n+(-1)^{n}\sqrt{n}+a}{n+(-1)^{n}\sqrt{n}+b} \right)=\dfrac{a-b}{n}+\mathcal{O}\left(\dfrac{1}{n^2} \right) $$
My attempt
i tried thi... | Your approach until
$$
\ln\left(\left( 1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{a}{n}\right)\left(1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{b}{n}\right)^{-1} \right) \\$$
is correct.
After that, we use $\ln (1+x) = x - \frac {x^2}2 + \frac{x^3}3+O(x^4)$ with $|x|< 1 $.
We have
$$
\ln \left( 1+\dfrac{(-1)^{n}}{\sqrt{n}}+\dfrac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Knowing that $a,b,c \in ℝ^*_+$ prove that $\frac{a+b}{a+b-c},\frac{b+c}{b+c-a},\frac{c+a}{c+a-b} $ don't belong simultaneously to the interval $(1,2)$ I have to solve the following problem but I don't know how to :
Knowing that $a,b,c \in ℝ^*_+$ prove that $\frac{a+b}{a+b-c},\frac{b+c}{b+c-a},\frac{c+a}{c+a-b} $ don't ... | Assume that they all belong to $(1,2)$ then
$$ \frac{a+b-c}{a+b},\quad\frac{a-b+c}{a+c},\quad \frac{-a+b+c}{b+c} $$
all belong to $\left(\frac{1}{2},1\right)$, hence
$$ \frac{c}{a+b},\quad \frac{b}{a+c},\quad \frac{a}{b+c} $$
all belong to $\left(0,\frac{1}{2}\right)$ and
$$ 2c<(a+b),\quad 2b<(a+c),\quad 2a<(b+c). $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the determinant of this $n \times n$ matrix in a clever way? Is there any clever and short way to find out the determinant of the following matrix?
\begin{bmatrix}
b_1 & b_2 & b_3 & \cdots & b_{n-1} & 0 \\
a_1 & 0 & 0 & \cdots & 0 & b_1 \\
0 & a_2 & 0 & \cdots & 0 & b_2\\
\vd... | Think of the given matrix as a block matrix
$$\left[\begin{array}{ccccc|c}
b_1 & b_2 & b_3 & \cdots & b_{n-1} & 0 \\ \hline
a_1 & 0 & 0 & \cdots & 0 & b_1 \\
0 & a_2 & 0 & \cdots & 0 & b_2\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & a_{n-1} & b_{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proving an expression is an integer
Prove that $$A^2 \cdot \dfrac{\sum_{m=1}^{n}(a^{3m}-b^{3m})}{\sum_{m=1}^n(a^m-b^m)}-3A^2$$ is an integer if $a=\frac{k+\sqrt{k^2-4}}{2}, b=\frac{k-\sqrt{k^2-4}}{2}$, where $k>2$ is a positive integer, and $A = \dfrac{1}{\sqrt{k^2-4}}$ for all positive integers $n$.
I thought about... | A possible way to the solution.
First, we have the following:
If $$u_n=\frac{a^3}{a^3-1}(a^{3n}-1)-\frac{b^3}{b^3-1}(b^{3n}-1)$$
$$v_n=\frac{a}{a-1}(a^{n}-1)-\frac{b}{b-1}(b^{n}-1)$$
then if $S_n$ is your expression:
$$S_n=A^2\frac{u_n}{v_n}-3A^2$$
In the following, I use at several places that $ab=1$ (and also $a^3b^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Is this lot drawing fair? Sorry for a stupid question, but it is bugging me a lot.
Let's say there are $30$ classmates in my class and one of us has to clean the classroom. No one wants to do that. So we decided to draw a lot - thirty pieces of paper in a hat, one of which is with "X" on it. The one who draws "X" has ... | The second person's chance is still $1/30$. Indeed, Person 2 draws an X exactly when Person 1 does NOT draw an X, and so:
\begin{align*}
\Pr[P_2 = X]
&= \Pr[P_1 \neq X] \cdot \Pr[P_2 = X \mid P_1 \neq X] \\
&= \frac{29}{30} \cdot \frac{1}{29} \\
&= \frac{1}{30}
\end{align*}
Likewise, Person 3 draws an X exactly when th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 5
} |
Show that $AB-BA\ne C$ for every real $3\times 3$ matrices $A$ and $B$ and some specific $3\times3$ matrix $C$
Prove that there cannot exist real $3 \times 3$ matrices $A$ and $B$ such that
$$AB-BA= \begin{bmatrix}
1 & 0 & 1 \\
0 & 1 & 0 \\
0 & 1 & 1\end{bmatrix}$$
I have no idea to solve it as both $A$ and $B$ are ... | You can employ the trace for this: For any two $3\times 3$-matrices $A$, $B$ we have ${\rm Tr}(AB) = {\rm Tr}(BA)$. Therefore, ${\rm Tr}(AB-BA) = 0$ for all $A$ and $B$. Since ${\rm Tr}(\begin{pmatrix}1 & 0& 1\\0 & 1 & 0\\0 & 1 & 1\end{pmatrix}) = 3\neq 0$, there exist no $A$, $B$ such that $AB-BA = \begin{pmatrix}1 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1876659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding roots of a mathematical equation If $x^3 + 3^x = 0$ what will be the value of $x$?
what will be the easiest way to find it mathematically?
| $$x^3 + 3^x = 0 \iff x^3 = -3^x$$
If $x$ is real, $$\implies x = -3^{\frac{x}{3}} \implies3 ^{-\frac{x}{3}} x = -1 \implies xe^{-\frac{x}{3} \ln3} = -1 \implies -\frac{x}{3} \ln(3) e^{-\frac{x}{3} \ln3} = \frac{\ln 3}{3}$$
So $$-\frac{x}{3} \ln 3 = W(\frac{\ln 3}{3})$$
$$x = - \frac{3W(\frac{\ln 3}{3})}{\ln 3 }$$
If x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to solve the limit $\lim _{n \to \infty }\left[n^2\left(\left(1+\frac{1}{n\left(n+2\right)}\right)^n-\frac{n+1}{n}\right)\right]$? Hi I got an examination at the school which was so arduous that I'm stumped.
This problem is the toughest for me :
$$\lim _{n \to \infty }\left[n^2\left(\left(1+\frac{1}{n\left(n+2\righ... | From the binomial theorem,
$$ \left(1+\frac{1}{n(n+2)}\right)^n = \sum\limits_{k=0}^{n}{{n\choose k}\frac{1}{n^k(n+2)^k}} = 1 + n\frac{1}{n(n+2)} + \frac{n(n-1)}{2}\frac{1}{n^2(n+2)^2} + \sum\limits_{k=3}^{n}{{n\choose k}\frac{1}{n^k(n+2)^k}}. $$
Notice that ${n\choose k}\le n^k$, and hence ${n\choose k}\frac{1}{n^k(n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1877995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove inequality $\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}\ge3$
Let $a,b,c>0$ and $a+b+c=3$. Prove that
$$\frac{a+1}{b^2+1}+\frac{b+1}{c^2+1}+\frac{c+1}{a^2+1}\ge3$$
My work so far:
I use AM-GM:
$$\frac{a+1}{b^2+1}=\frac{a}{b^2+1}+\frac{1}{b^2+1}=$$
$$=a-\frac{ab^2}{b^2+1}+1-\frac{b^2}{b^2+1}\ge a+1-\fra... | You are almost there :).
Inequality $$(a+b+c)^2\geq 3(ab+bc+ac)$$ holds, since it is equivalent with $$(a-b)^2+(b-c)^2+(c-a)^2\geq 0.$$ Therefore, $$\frac92-\frac{ab+bc+ca}{2}\geq \frac92-\frac{(a+b+c)^2}{6}=3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Help with the limit of an integral I am trying to evaluate this limit
$$\lim \limits_{x \to \infty} \frac {1}{\ln x} \int_{0}^{x^2} \frac{t^5-t^2+8}{2t^6+t^2+4} dt=? $$
Any help will be appreciated.
| Let $f(x)=\int_{0}^{x^2} \frac{t^5-t^2+8}{2t^6+t^2+4} dt$. It is easy to conclude $\lim_{x\to\infty} f(x)=\infty$. Now use L'Hopitals rule:
$$\lim \limits_{x \to \infty} \frac {1}{\ln x} \int_{0}^{x^2} \frac{t^5-t^2+8}{2t^6+t^2+4} dt=\lim \limits_{x \to \infty}\frac{2x\cdot \frac{x^{10}-x^4+8}{2x^{12}+x^4+4}}{\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Basic Joint probability distribution calculation example? (X, Y) has the Joint probability distribution:
\begin{array}{|c|c|c|}
\hline
X / Y & 1 & 2 \\ \hline
1& \frac{1}{4}& \frac{1}{3} \\ \hline
2& \frac{1}{6}& a \\ \hline
\end{array}
So $P(X=1\mid Y=1)=\frac{3}{5}$ and $P(X=2\mid Y=2)=\frac{3}{7}$.
My challeng... | $$P(X=1\mid Y=1)=\frac{P(X=1\wedge Y=1)}{P(Y=1)}=\frac{\frac14}{\frac14+\frac16}=\frac35$$
$$a=1-\left(\frac14+\frac13+\frac16\right)=\frac14$$
$$P(X=2\mid Y=2)=\frac{P(X=2\wedge Y=2)}{P(Y=2)}=\frac{\frac13}{\frac13+a}=\frac{\frac13}{\frac13+\frac14}=\frac37$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $f(x+1)+f(x-1)=\sqrt 3 f(x), \forall x$ then $f$ is periodic.
If $f: \mathbb{R} \rightarrow \mathbb{R}$ such that
$f(x+1)+f(x-1)=\sqrt 3 f(x), \forall x$ then $f$ is periodic.
I tried to replace $x$ by $x+1, x-1$ in the equality,to get something like $f(x + k)=f(x)$ but without success.
Any help is appreciated.... | $$f(x+1) +f(x-1)=\sqrt3 f(x)$$
Let $x=x+1$ and $x=x-1$
$$f(x+2) + f(x)=\sqrt3 f(x+1)$$
$$f(x) + f(x - 2)=\sqrt3 f(x-1)$$
Adding the above we get
$$f(x+2) + f(x - 2)+ 2f(x)=\sqrt3(f(x+1) +f(x-1))$$
$$=3f(x)$$
$$f(x+2) + f(x - 2)=f(x)$$
Letting $x=x+2$ in above
$$f(x+4) + f(x)=f(x+2)=f(x)-f(x-2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
irrationality proof of $\sqrt{n}+\sqrt{n+1}$ for any $n>0$ irrationality proof of $\sqrt{n}+\sqrt{n+1}$ for any $n>0$
My attempt:
$\sqrt{n}+\sqrt{n+1}=\frac{p}{q}$
$2n+1+2\sqrt{n(n+1)}=\frac{p^2}{q^2}$
Now we have to show $2n+1+2\sqrt{n(n+1)}$ cannot be a perfect squere but I don't know how.
| $2n + 1 + 2\sqrt{n(n+1)}$ is irrational if and only if $\sqrt{n(n+1)} = \sqrt{n^2 + n}$ is irrational. Then observe that
$$n^2 < n^2 + n < n^2 + 2n + 1 = (n+1)^2,$$
which implies that indeed $n(n+1)$ is not a perfect square. But the square roots of positive integers are either positive integers or irrational, by the ra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1886124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Find $a^{2013} + b^{2013} + c^{2013}$ Problem Statement
Let $f(x) = x^3 + ax^2 + bx + c$ and $g(x) = x^3 + bx^2 + cx + a$ where $a,b,c$ are integers with $c\not=0$
Suppose that the following conditions hold:
*
*$f(1)=0$
*the roots of $g(x)=0$ are the squares of the roots of $f(x)=0$
$$\text{Find the value of} \: ... | If the roots of $f$ are $1,u,v$, then the roots of $g$ are $1, u^2,v^2$.
Thus we obtain by Vieta
$$ \begin{align}-1-u-v&=a=-u^2v^2\\
u+v+uv&=b=-1-u^2-v^2\\
-uv&=c=u^2+v^2+u^2v^2\end{align}$$
In particular,
$$ \tag1a+b+c = (-1-u-v)+(u+v+uv)-(-uv)=-1$$
and
$$\tag2 a^2=1+u^2+v^2+2u+2v+2uv=-b+2b=b$$
and $$ \tag3c^2=u^2v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the sum of the first 21 terms of an Arithmetic Progression(A.P)?
Question. If the sum of first $12$ terms of an A.P. is equal to to the sum of the first $18$ terms of the same A.P., find the sum of the first $21$ terms of the same A.P.
$a=$ first term
$d=$ difference
I know now, $2(2a + 11d) = 3(2a + 17d)... | An arithmetic progression such that the sum of the first $12$ terms equals the sum of the first $18$ terms:
\begin{multline}
(-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \\
(-17) + (-15) + (-13) + (-11) + (-9) + (-7) = -216,\\
\end{multline}
\begin{multline}
(-29) + (-27) + (-25) + (-23) + (-21) + (-19) +{} \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1890813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\frac{1}{2}\leq\int_0^2 \frac{dx}{2+x^2}\leq\frac{5}{6}$? How to prove the following? $$\frac{1}{2}\leq\int_0^2 \frac{dx}{2+x^2}\leq\frac{5}{6}$$
Ok I know we can directly integrate the function.The answer is given by Wolfram Alpha
But is there any shorter/elegant method to obtain the left and right boun... | If $0\le x\le 1$ then $\dfrac 1 3 \le \dfrac 1 {2+1^2} \le \dfrac 1 {2+x^2} \le \dfrac 1 {2+0^2} = \dfrac 1 2$.
If $1\le x \le 2$ then $\dfrac 1 6 \le\dfrac 1 {2+x^2} \le \dfrac 1 3 $.
Therefore
$$
\frac 1 3 + \frac 1 6 \le \int_0^1 \frac{dx}{2+x^2} + \int_1^2 \frac{dx}{2+x^2} \le \frac 1 2 + \frac 1 3.
$$
As for Wolfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $112$ divides the integral part of $4^m-(2+\sqrt{2})^m$
Let $m$ be a positive odd integer not divisible by $3$. Prove that $$112 \mid \left [ 4^m-(2+\sqrt{2})^m\right].$$
I wasn't sure how to simplify the expression since we have the integer part $[x]$ and radicals. Using the Binomial Theorem we have $$(2+... | Since $2-\sqrt{2}\in\left(0,\frac{2}{3}\right)$ and $(2+\sqrt{2})^m+(2-\sqrt{2})^m$ is an integer number, it is enough to study the sequences given by $a_m=(2+\sqrt{2})^m+(2-\sqrt{2})^m$ and $b_m=4^m$. We have:
$$ a_{m+2} = 4a_{m+1}-2a_{m}$$
hence the situation $\!\!\pmod{7}$ and $\!\!\pmod{16}$ is the following:
$$ \b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Number of bit strings of length 8 that do not contain "$100$"? I am thinking the total number of possible strings is $2^8$ and the number of strings with $100$ at the beginning would be $2^8 - 2^3 = 2^5$. Now "$100$" can shift across the string $5$ times going to the right. Is the answer then $2^8 - 2^5 \times 5$?
| A nice technique is the so-called Goulden-Jackson Cluster Method which is a convenient method to derive a generating function for problems of this kind.
We consider words of length $n\geq 0$ built from an alphabet $$\mathcal{V}=\{0,1\}$$ and the set $\mathcal{B}=\{100\}$ of bad words which are not allowed to be part o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Sum of even digits Let $E(n)$ denote the sum of the even digits of $n$.For example,$E(1243)=2+4=6$.
What is the value of $E(1)+E(2)+E(3)+....+E(100)$?
I got the answer after literally adding like
$2+4+6+8+2...+6+8=400$ (for all $100$ numbers)
Is there an easy tricky method for solving this?
| The sum of all the even digits in the numbers $1$ to $100$ is required and both inclusive. Observe that $100$ only two null even digits which doesn't add any significant value. So we remove them from our addition list. We make a list with the following numbers:-
$01, 02, 03, ..., 10, 11, ..., 98, 99$
There are $2\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How substitution is used in second order ODE? I am trying to determine how the equation was simplified which was shown (not completely) in textbook on ordinary differential equations.
The following equation,
\begin{equation}
\frac{d^2 Y}{d x^2} + Y(aY^{2} + b)^{-2}\frac{dY}{dx}+Y=0
\label{1}
\end{equation}
can be simp... | Let $w=\dfrac{dY}{dx}$ ,
Then $\dfrac{d^2Y}{dx^2}=\dfrac{dw}{dx}=\dfrac{dw}{dY}\dfrac{dY}{dx}=w\dfrac{dw}{dY}$
$\therefore w\dfrac{dw}{dY}+Y(aY^2+b)^{-2}w+Y=0$
$w\dfrac{dw}{dY}=-\dfrac{Yw}{(aY^2+b)^2}-Y$
Let $z=-\dfrac{Y^2}{2}$ ,
Then $\dfrac{dw}{dY}=\dfrac{dw}{dz}\dfrac{dz}{dY}=-Y\dfrac{dw}{dz}$
$\therefore-Yw\dfrac{d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1893011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $z$ be a complex number $\ne 0$. What is the absolute value of $z\sqrt{z}$? $\color{red}{\mathbf{EDIT}}$ The question was misinterpreted - it was actually: 'what is the absolute value of $z/\bar{z}$?'; I'am grateful for the answers given on the original problem though and will keep this up as is in case someone els... | The simpler way is to use the polar representation. For $z=\rho e^{i\theta}$ and using only one value of the square root we have:
$$
z\sqrt{z}=\rho e^{i\theta}\sqrt{\rho} e^{i\theta/2}=\rho\sqrt{\rho}e^{i3\theta/2}
$$
so $|z\sqrt{z}|=\rho\sqrt{\rho}=|z|\sqrt{|z|}$ and it is $=1$ only if $|z|=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1894307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Examining convergence of $\sum_{n=1}^{\infty}\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}$ with mean value theorem $$\sum_{n=1}^{\infty}\frac{1}{n}\sqrt{e^{\frac{1}{n}}-e^\frac{1}{n+1}}$$
I would like to examine covergence of this series using mean value theorem.
I would like to check my solution and optionally ... | Doing basically the same as Olivier Oloa in his answer with Taylor series of higher order $$e^{\frac{1}{n }}=1+\frac{1}{n}+\frac{1}{2 n^2}+\frac{1}{6 n^3}+\frac{1}{24
n^4}+O\left(\frac{1}{n^5}\right)$$ $$e^{\frac{1}{n+1 }}=1+\frac{1}{n}-\frac{1}{2 n^2}+\frac{1}{6 n^3}+\frac{1}{24
n^4}+O\left(\frac{1}{n^5}\right)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Two lines through a point, tangent to a curve We are looking for two lines through $(2,8)$ tangent to $y=x^3$. Let's denote the intersection point as $(a, a^3)$ and use the slope equation together with the derivative to get $\frac{a^3-8}{a-2}=3a^2$. This yields a cubic equation. Of course, one of the lines is tangent t... | There's another way, not sure it's quicker in the present case.
Consider a line with slope $t$ passing through the point $(2,8)$. This line will be tangent to the curve if the abscissae equation for intersection points has a root of multiplicity $>1$.
Now the equation of such a line is $\;y=t(x-2)+8$, so the abscissae... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1895466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Compute $\sup\limits_{x\in (0,2)}\left|\operatorname{cosh}\left(x^2-x-1\right)\right|$
$$\rm{ Calculate: }\quad \displaystyle \sup_{x\in (0,2)}\left|\cosh\left(x^2-x-1\right)\right|$$
My proof:
$$f(x)=\cosh\left(x^2-x-1\right)$$
$$f'(x)=\left(2x-1 \right)\sinh\left(x^2-x-1\right) $$
$$f'(x)=0 \iff x=\dfrac{1}{2} \rm{... | Alternate proof: Let $p(x) = x^2 - x - 1.$ Verify that $p([0,2]) = [-5/4,1].$ Now $\cosh x$ is an even function. Therefore
$$\sup_{x\in [0,2]} \cosh (p(x))= \sup_{y\in [-5/4,1]} \cosh y = \sup_{y\in [0,5/4]} \cosh y.$$
On the last interval $\cosh y$ is increasing, therefore the desired answer is $\cosh (5/4).$ (Note th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1896316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the second smallest integer such that its square's last two digits are $ 44 $ Given that the last two digits of $ 12^2 = 144 $ are $ 44, $ find the next integer that have this property.
My approach is two solve the equation $ n^2 \equiv 44 \pmod{100}, $ but I do not know how to proceed to solve that equation.
I t... | We can use your idea: Let $n=10x+y$. If the last digit of $n^2$ is $4$ then the last digit of $n$ is either $2$ or $8$, so $y=2$ ot $y=8$.
First let's assume $y=2$. We then have $40x+4\equiv 44\mod100$, or equivalently $40x+4=44+100t$ for some $t$. For $t=0$ we get $x=1$, so $n=12$, and this is not what we want. For $t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1898485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Show $\lim\left ( 1+ \frac{1}{n} \right )^n = e$ if $e$ is defined by $\int_1^e \frac{1}{x} dx = 1$ I have managed to construct the following bound for $e$, which is defined as the unique positive number such that $\int_1^e \frac{dx}x = 1$.
$$\left ( 1+\frac{1}{n} \right )^n \leq e \leq \left (\frac{n}{n-1} \right )^n$... | Suppose
$$
1=\int_1^x\frac{\mathrm{d}u}u\tag{1}
$$
Using the change of variables $u\mapsto u^n$, we get
$$
1=n\int_1^{x^{1/n}}\frac{\mathrm{d}u}u\tag{2}
$$
Dividing by $n$ and estimating the integral using the width of the interval and the extremes of $u$:
$$
1-x^{-1/n}\le\frac1n\le x^{1/n}-1\tag{3}
$$
or equivalently
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 6,
"answer_id": 3
} |
Evaluate the integral $\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$ This integral looks a lot like an elliptic integral, but with cubes instead of squares:
$$I(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$$
Let's consider $a,b>0$ for now.
$$I(a,a)=\int_0^\infty \frac{dx}{x^3+a^3}=\frac{2 \pi}{3 \sqrt{... | Using the advice from @tired in the comments, we can write:
$$I_1(p)=\frac{2 \pi}{3 \sqrt{3}} \sum_{k=0}^\infty \frac{1}{k!^2} \left(\frac{1}{2}\right)_k \left(\frac{2}{3}\right)_k (1-p)^k=$$
$$=\frac{2 \pi}{3 \sqrt{3}} \sum_{k=0}^\infty \frac{1}{(1)_k} \left(\frac{1}{2}\right)_k \left(\frac{2}{3}\right)_k \frac{(1-p)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Help proving the product of any four consecutive integers is one less than a perfect square Apparently this is a true statement, but I cannot figure out how to prove this. I have tried setting
$$(m)(m + 1)(m + 2)(m + 3) = (m + 4)^2 - 1 $$
but to no avail. Could someone point me in the right direction?
| You can make the algebra simpler by starting from a better place than $m(m+1)(m+2)(m+3)$. Think symmetry!
$$\begin{align}
(m - \tfrac 3 2)(m - \tfrac 1 2)(m + \tfrac 1 2)(m + \tfrac 3 2)
&= (m^2 - \tfrac 9 4)(m^2 - \tfrac 1 4) \\
&= m^4 - \tfrac 5 2 m^2 + \tfrac{9}{16} \\
&= (m^2 - \tfrac 5 4)^2 - 1
\end{align}$$
Final... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 5
} |
Multi-index sum property Exercise 1.2.3.29 in Donald Knuth's The Art of Computer Programming (3e) states the following property of a multi-indexed sum:
$$
\sum_{i=0}^n \sum_{j=0}^i \sum_{k=0}^j a_ia_ja_k = \frac{1}{3}S_3 + \frac{1}{2}S_1S_2 + \frac{1}{6}S_1^3,
$$
where $S_r = \sum_{i=0}^n a_i^r$.
I tried to prove it an... | You can use induction to prove this. Check that the base case $n=0$ holds.
For the inductive step, use a double subscript:
$$S_{n,r}:=\sum_{i=0}^n a_i^r$$
to make the dependence on $n$ explicit. It is clear that $$S_{n+1,r} = S_{n,r} + a_{n+1}^r\quad\text{for}\quad r=1,2,3.\tag1$$ Write $M_n$ as an abbreviation for $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1900441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Product of roots of $ax^2 + (a+3)x + a-3 = 0$ when these are positive integers There is only one real value of $'a'$ for which the quadratic equation $$ax^2 + (a+3)x + a-3 = 0$$ has two positive integral solutions.The product of these two solutions is :
Since the solutions are positive, therefore the product of roots ... | Thanks to @rtmd, I fixed the solution. This only holds when $a=-\frac{3}{7}$. Here is the proof.
$\frac{a-3}{a}, \frac{a+3}{a} \in \mathbb{Z}$ implies $\frac{6}{a} \in \mathbb{Z}$. We can write $a=\frac{6}{n}$ for some integer $n$.
\begin{align*}
\frac{a-3}{a}= 1-\frac{n}{2} \in \mathbb{Z}\\
\frac{a+3}{a}= 1+\frac{n}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $a$, $b$, and $c$ are sides of a triangle, then $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2$.
Let $a,b,c$ be the lengths of the sides of a triangle. Prove that
$$\sum_{\text{cyc}}\frac{a}{b+c}=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}<2\,.$$
Attempt. By clearing the denominators, the required inequality is equiv... | Since $a,b,c$ are the sides of a triangle, there exist $p,q,r\in\mathbb R^+$ such that $a=p+q$, $b=q+r$, $c=r+p$ (Ravi substitution).
$$\sum_{\text{cyc}}\frac{a}{b+c}<2\iff \sum_{\text{cyc}}\frac{p+q}{p+q+2r}<2$$
Now multiply both sides by $(2p+q+r)(p+2q+r)(p+q+2r)$, expand, rearrange.
See WolframAlpha (link) if you wa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1903775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Sequence converging to definite integral Let's define
\begin{equation*}
I_0 := \log\frac{6}{5}
\end{equation*}
and for $k = 1, 2, \ldots, n$
\begin{equation*}
I_k := \frac{1}{k} - 5 I_{k-1}.
\end{equation*}
How the value $I_n$ is linked with the value of $$\int_0^1\frac{x^n}{x+5} \mathrm{d}x \ ?$$
| $I_n+5I_{n-1}=\int_{0}^{1}\frac{x^n}{x+5}dx+5\int_{0}^{1}\frac{x^{n-1}}{x+5}dx=\int_{0}^{1}(\frac{x^n}{x+5}+5\frac{x^{n-1}}{x+5})dx$
$I_n+5I_{n-1}=\int_{0}^{1}x^{n-1}dx=\frac{x^n}{n}|_{0}^{1}=\frac{1}{n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1904317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Trouble understanding proof of the inequality - $(\frac{1}{a}+1)(\frac{1}{b}+1)(\frac{1}{c}+1) \ge 64 $, for $a,b,c > 0$ and $a+b+c = 1$ I was looking into this problem in a book discussing inequalities, However I found the proof quite hard to understand.The problem is as follows:
Let $a,b,c$ be positive numbers with ... | For Step $(3)$,
Using AM-GM inequality
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 3 \sqrt[3]{\frac{1}{a} \cdot \frac{1}{b} \cdot \frac{1}{c}}=\sqrt[3]{\frac{1}{abc}}\tag{3.1}$$
$$\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} \geq 3 \sqrt[3]{\frac{1}{ab} \cdot \frac{1}{bc} \cdot \frac{1}{ca}}=\sqrt[3]{(abc)^2}\tag{3.2}$$
For S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
trouble understanding proof for the inequality: $\frac{a}{(a+1)(b+1)}+\frac{b}{(c+1)(b+1)}+\frac{c}{(a+1)(c+1)} \ge \frac{3}{4}$ I have read this problem in a book discussing inequalities, and I have a few problems in understanding the proof.
The problem is:
Let $a,b,c$ be positive numbers that satisfy $abc=1$, prove t... | The last inequality follows from $a+1\geq 2\sqrt{a}$ (and analogous inequalities for $b$ and $c$), which is true because $(\sqrt{a}-1)^2\geq 0$.
For the step before this, you can rearrange $$1-\frac{2}{(a+1)(b+1)(c+1)} \ge \frac{3}{4}$$ to get $$\frac{1}{4}\geq\frac{2}{(a+1)(b+1)(c+1)}$$ and $$(a+1)(b+1)(c+1)\geq 8.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1905351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Rationalizing denominator with cube roots Rationalize the denominator of $$\frac{6}{\sqrt[3]{4}+\sqrt[3]{16}+\sqrt[3]{64}}$$ and simplify.
I already have an answer. I just want to compare answers with others. Maybe someone has different solutions? Also, I really disagree with the answer found at the back of the questio... | Write $x=\sqrt[3]4$, then the expression is
$$\frac6{x+x^2+x^3}=\frac1x×\frac6{1+x+x^2}$$
Multiply by $x-1$:
$$=\frac1x×\frac{6(x-1)}{(x-1)(x^2+x+1)}$$
$$=\frac1x×\frac{6(x-1)}{x^3-1}$$
$$=\frac{x-1}x×\frac{6}{3}$$
$$=2\left(1-\frac1x\right)$$
$$=2\left(1-\frac{x^2}{x^3}\right)$$
$$=2\left(1-\frac{\sqrt[3]{16}}4\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1907342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove by induction that ${x}^{2k}-{y}^{2k}=(x+y)t$ for all natural numbers $k$ and $t$. Although it is an elementary algebraic result proved easily using algebra but I wanted to prove it inductively.
In my endeavor to do this, I've reached to the step which not only requires the preceding term but also the penultimate ... | Let $C(k)$ be the claim that there exists a polynomial $P_k(x,y)$ in $x$, $y$ such that $$x^{2k} - y^{2k} = (x+y)P_k(x,y).$$ Then a proof by induction might be as follows. For $k = 0$, we have $$x^0 - y^0 = 0 = (x+y) P_0(x,y)$$ the trivial choice $P_0(x,y) = 0$. For $k = 1$, we have $$x^2 - y^2 = (x+y)(x-y),$$ thus ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1908097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.