Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Proving that for $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge \frac{1}{2}$
For $\{ a,b\} \subset \Bbb{R^{+}}$; $a+b=1 \implies a^2+b^2 \ge
\frac{1}{2}$
I'm trying to prove this in the following way, but I'm not sure if it's correct. Could anyone please check it and see if it's okay?
$a+b=1 \implies (a... | Your proof is correct, but could be made a little briefer, with a gain in clarity.
Here's a proof using the same ideas as yours, but I think a little easier to read . . .
\begin{align*}
&a+b=1\\[4pt]
\implies\;&(a+b)^2 = 1\\[4pt]
\implies\;&(a+b)^2 + (a-b)^2 \ge 1\\[4pt]
\implies\;&2a^2 + 2b^2 \ge 1\\[4pt]
\implies\;... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
Convergence of the series $\sum_{n = - \infty}^{\infty} (\sqrt{(a+n)^2+b^2} - |n| )$ Given $a, b \in \mathbb{R}$ with $b > 0$, is the series
\begin{equation}
\sum_{n = - \infty}^{\infty} (\sqrt{(a+n)^2+b^2} - |n| )
\end{equation}
convergent or divergent?
If we drop out the $n=0$ term and fold the remaining sum, the qu... | $$
\begin{align}
\lim_{n \to \infty }\sqrt{a^2+2an+n^2+b^2}-n=\lim_{n \to \infty }\frac{\left ( a^2+2an+b^2 \right )}{\sqrt{a^2+2an+n^2+b^2}+n}=a
\end{align}
$$
evidently divergent for
$$
\begin{align}
a\neq 0
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding the Paramertric Coordinates of the Maximum and Minimum points of a Certain Curve I was studying for some quizzes when a wild question appears. It goes like this:
Find the paramertric coordinates of the maximum and minimum points of
the following curves. The curve is $x = 3 \cos \theta$ ;$\space$ $ y = 4 \si... | The max/min values occur at the values of $\theta$ for which
$$ \frac{dy}{dx}=\frac{dy/d\theta}{dx/d\theta}=0 $$
which occurs when
$$ \frac{dy}{d\theta}=4\cos\theta=0 $$
So the solution is
$$\theta=\pm\frac{\pi}{2}$$
(or, more generally)
$$\theta=2\pi n\pm\frac{\pi}{2}$$
But $\theta=\pm\dfrac{\pi}{2}$ will be sufficien... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2406825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$x^2 + 3x + \frac{3}x + \frac{1}{x^2} = 26$ and $x$ can be written as $a + \sqrt{b}$ where $a$ and $b$ are positive integers, then find $a + b$. If $x$ satisfies $x^2 + 3x + \frac{3}x + \frac{1}{x^2} = 26$ and $x$ can be written as $a + \sqrt{b}$ where $a$ and $b$ are positive integers, then find $a + b$.
This equation... | We can factor this equation as follows:
$$x^4+3x^3-26x^2+3x+1 = (x^2-4x+1) (x^2+7x+1).$$
One can check that $2\pm \sqrt3$ are solutions to the $(x^2-4x+1)$,
alse we can check that $\dfrac{-7\pm 3\sqrt5}{2}$ are solutions to the $(x^2+7x+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2407700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
What are some mathematically interesting computations involving matrices? I am helping designing a course module that teaches basic python programming to applied math undergraduates. As a result, I'm looking for examples of mathematically interesting computations involving matrices.
Preferably these examples would be ... | If $(a,b,c)$ is a Pythagorean triple (i.e. positive integers such that $a^2+b^2=c^2$), then
$$\underset{:=A}{\underbrace{\begin{pmatrix}
1 & -2 & 2\\
2 & -1 & 2\\
2 & -2 & 3
\end{pmatrix}}}\begin{pmatrix}
a\\
b\\
c
\end{pmatrix}$$
is also a Pythagorean triple. In addition, if the initial triple is primitive (i.e. $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 26,
"answer_id": 4
} |
Line integral $\int_K \frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy$ Solve $\int_K \frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy$ =?
K: $x^2+y^2=1$ is oriented positively.
My attempt:
$\int_K \frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy = |x=cost, y=sint,x'=- sint,y'=cost | $ = *
I put for $dx = -sint,dy=cost$
$= \int_0^{2\p... | Another approach is
$$\frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy=-\dfrac{x\,dy-y\,dx}{x^2+y^2}=-d\left(\arctan\dfrac{y}{x}\right)$$
so
$$\int_K \frac{y}{x^2+y^2} dx -\frac{x}{x^2+y^2} dy=-\int_K d\left(\arctan\dfrac{y}{x}\right)=-\int_0^{2\pi}dt=-2\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
use **similarity transformation** to find the matrix $ \ B \ $ with respect to the basis $ \ \{(1,2) , \ (3,5) \} \ \ of \ \ \mathbb{R}^2 $ . Consider the linear operator $ \ T(x,y)=(5x,x-y) \ \ on \ \ \mathbb{R}^2 $ . Find the matrix $ \ A \ $ of $ \ T \ $ relative to the standard basis on $ \ \mathbb{R}^2 \ $ . The... | You made a "silly" minor mistake $$P=\left(
\begin{array}{ll}
1 & 2 \\
3 & 5 \\
\end{array}
\right)$$
If you use that then $$ P^{-1}AP=\left(
\begin{array}{cc}
-5 & 2 \\
3 & -1 \\
\end{array}
\right).\left(
\begin{array}{cc}
5 & 0 \\
1 & -1 \\
\end{array}
\right).\left(
\begin{array}{cc}
1 & 2 \\
3 & 5 \\
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2408852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\int\frac{x^b}{1+x^a}~dx$ for $a,b\in\Bbb N$ In this previous answer, MV showed that for $n\in\Bbb N$,
$$\int\frac1{1+x^n}~dx=C-\frac1n\sum_{k=1}^n\left(\frac12 x_{kr}\log(x^2-2x_{kr}x+1)-x_{ki}\arctan\left(\frac{x-x_{kr}}{x_{ki}}\right)\right)$$
where
$$x_{kr}=\cos \left(\frac{(2k-1)\pi}{n}\right)$$
$$x_{k... | Let $\gamma=\exp(2\pi i/a)$.
The polynomial $Q(x)=x^a+1$ has $a$ simple roots $z_k=\gamma^{k+1/2}$,
$0\leq k<a$. Since $z_k^a=-1$, we have $Q'(z_k)=a z_k^{a-1}=-a z_k^{-1}$, so
$$ \frac{1}{Q(x)} =
\sum_{k=0}^{a-1} \frac{1}{Q'(z_k)} \frac{1}{x-z_k} =
-\frac{1}{a} \sum_{k=0}^{a-1} \frac{z_k}{x-z_k} $$
For the n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2409312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 0
} |
Write $(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$ as a sum of (three) squares of quadratic forms The quartic form
$$(x^2 + y^2 + z^2)^2 - 3 ( x^3 y + y^3 z + z^3 x)$$
is non-negative for all real $x$, $y$, $z$, as one can check (with some effort). A theorem of Hilbert implies that there exist quadratic forms $... | $$(\,\sum\limits_{cyc}\,x^{\,2}\,)^{\,2}- 3(\,\sum\limits_{cyc}\,x^{\,3}y\,)= \frac{1}{6}\left ( \sum\limits_{cyc}\,(\,-\,3\,yz+ 3\,xy+ y^{\,2}+ z^{\,2}- 2\,x^{\,2}\,)^{\,2} \right )$$
$$(\,\sum\limits_{cyc}\,x^{\,2}\,)^{\,2}- 3(\,\sum\limits_{cyc}\,x^{\,3}y\,)= \frac{1}{2}\left ( \sum\limits_{cyc}\,(\,-\,yz- zx+ 2\,xy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2410994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$ Evaluate
$$I=\int \frac{\cos 2x \: dx}{3 \sin x+4 \cos x}$$
My Try:
$$I=\int \frac{(\cos x-\sin x)(\cos x+\sin x)dx}{3 \sin x+4 \cos x}$$
we have $$\cos x-\sin x=\frac{1}{25}(3 \sin x+4 \cos x)+\frac{7}{25}(3 \cos x-4 \sin x)$$ and
$$\cos x+\sin x=\frac{7}{25}... | Note that $$3\sin x+4\cos x=5\Big( \dfrac{3}{5}\sin x+\dfrac{4}{5}\cos x \Big)$$ and $$\cos2x=1-2\sin^2x$$
also $\arcsin(4/5)=\arccos(3/5)=0.92$
combining all this, your integral becomes $\frac{1}{5}\int \frac{1 -2\sin^2x\: dx}{\sin(x+0.92)}$.
From here, writing $x=(x+0.92)-0.92$ in numerator and basic integration form... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2415567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Ordered pairs of positive integers $2^{2x} - y^2 = 60$ How many ordered pairs of positive integers are there that satisfy $2^{2x} - y^2 = 60$?
This can be rewritten as $(2^x)^2-y^2 = 60$ and then $(2^x+y)(2^x-y) = 60$. Then since $2^x$ is always positive and so $2^x+y$ and $2^x-y$ are both positive and the first is al... | No you are not right.
You have neglected the case $\{ 2^x-y, 2^x+y \} = \{ 6, 10 \}$
; which gives you $(x,y)=(3,2)$.
Notice that:
*
*$(2^x+y)$ and $(2^x-y)$ have the same pairity;
*product of $(2^x+y)$ and $(2^x-y)$ is even; so at least one of them is even;
*by the above two remarks it follows that
both o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2416692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 1
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Find the sum of this series? Show that the series $\displaystyle \sum_{n=0}^{\infty}\frac{x}{(1+x^2)^n} $ converges $\forall x\in \mathbb{R}$ and find the sum $\forall x\in \mathbb{R}$
So, I've shown that the series converges $\forall x\in \mathbb{R}$ and, through testing, suspect that the sum is equal to $\frac{x}{1+x... | Alternatively: If $x=0$, then $S=0$. If $x\ne 0$, then:
$$S=\sum_{n=0}^{\infty} \frac{x}{(1+x^2)^{n}}\Rightarrow$$
$$(1+x^2)S=\sum_{n=0}^{\infty} \frac{x}{(1+x^2)^{n-1}}=x(1+x^2)+\sum_{n=1}^{\infty} \frac{x}{(1+x^2)^{n-1}}=x^2(1+x^2)+\sum_{n=0}^{\infty} \frac{x}{(1+x^2)^n}=x(1+x^2)+S \Rightarrow$$
$$S=\frac{x(1+x^2)}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2417825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Are there any known methods to compute series $\sum_{k=0}^{\infty}2^k \big(\sum_{n=0}^{\infty}\frac{(-1)^n}{(n2^{k+1}+(2k+1))^3}\big)$? I would like to ask if there are any known methods to compute series like this one ?
$$\sum_{k=0}^{\infty}2^k \bigg(\sum_{n=0}^{\infty}\frac{(-1)^n}{(n2^{k+1}+(2k+1))^3}\bigg)$$
And t... | We have
$\displaystyle \sum\limits_{k=0}^\infty 2^k \sum\limits_{n=0}^\infty \frac{(-1)^n}{(n2^{k+1}+2k+1)^3} =\sum\limits_{k=0}^\infty \frac{2^k}{(2k+1)^3} - \sum\limits_{k=0}^\infty \left(\frac{1}{2^{2k+3}}\sum\limits_{n=1}^\infty \frac{(-1)^{n-1}}{(n+\frac{2k+1}{2^{k+1}})^3}\right)$
and
$\displaystyle\sum\limits_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
The ellipse $x^2+2y^2=2.$ Question: The ellipse $x^2+2y^2=2$ is given by $x=a\cos{t}, \quad y=b\sin{t}, \quad t\in[0,2\pi],$ for what values of $a$ and $b$?
a) $(a,b)=(2,1)$
b) $(a,b)=(1,2)$
c) $(a,b)=(1,\sqrt{2})$
d) None of the above.
Attempt: Substitute the trig-values for $x$ & $y$ in the elliptic equation:
$$x^2+2... | In order for $x^2+2y^2=2$ with $x=a\cos(t)$, and $y=b\sin(t)$ we have
$$(a\cos(t))^2+2(b\sin(t))^2=2\iff a^2\cos^2(t)+2b^2\sin^2(t)=2$$
Now we want $a^2=2b^2=2$ since then we can factor out the coefficients and use $\cos^2(t)+\sin^2(t)=1$. Hence, $a=\sqrt 2$ and $b=1$.
Also, just the first error I see in your solution ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2421949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Prove by induction that $ \ \forall n \ge 2, \ 2^{n+1} > n^{2} + 3$ Question:
Prove by induction that $$ \ \forall n \ge 2, \ 2^{n+1} > n^{2} + 3$$
My attempt:
Base case is trivial.
Suppose that $ n \ge 2$ and $\ 2^{n+1} > n^{2} + 3$
WTS $ \ 2^{n+2} > (n+1)^{2} + 3$
$ 2^{n+2} = 2.2^{n+1}> 2(n^2 + 3) = 2n^2 + 6 =... | $$n^2-2n+2=(n^2-2n+1)+1=\dots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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The License Plate Problem The 7-digit license plate of a car happens to be a palindrome. Its first digit on the far left is twice the digit next to it. The 3rd digit from the right is odd and 5 more than 2nd digit from the left. The middle digit is the average of the first three digits on the far left. What is this lic... | The license plate is of the form $abcdcba$, where each $a,b,c,d$ is a digit.
From the information in the question, we know
$$
a=2b \quad\text{and}\quad c=5+b \quad\text{and}\quad c\ \text{is odd}
\quad\text{and}\quad d=(a+b+c)/3.
$$
Since $c$ is odd and $c=5+b$, we know $b$ is even. Thus $b$ is one of $0,2,4,6,8$. It c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2422952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Definite integral $\int_0^1\sqrt{x^2+1}\, dx$. Use trig substitution? I have an integral:
$$\int_0^1\sqrt{x^2+1}\, dx$$
but I have gotten stuck. Here's the work I have done already:
I'm not sure where to go from here. Using a trig identity doesn't seem like it would work. Integration by parts doesn't work (at least... | Hint:)
Since $\cosh^2u=\sinh^2u+1$ so let $x=\sinh u$.
It's better to do this
$$\int\sec^3t dt=\int\dfrac{\cos t}{(1-\sin^2t)^2}dt$$
Edit:
\begin{align}
I
&= \int\sec^3t dt \\
&= \int\dfrac{1}{\cos^3t} dt \\
&= \int\dfrac{\cos t}{\cos^4t} dt \\
&= \int\dfrac{\cos t}{(1-\sin^2t)^2}dt
\end{align}
with substitution $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
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Generator of an ideal in the polynomial ring $ F[x] $ for a field $ F $
Let $ F $ be a field and consider the polynomial ring $ F[x] $.
I know that $ F[x] $ is a principal ideal domain and so the ideal $\langle x^{n - 2} - x^{n - 1}, x^{n - 3} - x^{n - 1}, \dots, x - x^{n - 1}, 1 - x^{n - 1} \rangle $ is generated ... | We know that every element in the ideal $\langle 1-x\rangle$ is a multiple of $(1-x)$, so:
$$f(x) = (1-x) \sum_{i = 0}^m a_{i} x^{i}$$
Generate $1-x$ simply by subtracting the second-to-last element in the basis from the last:
$$(1-x^{n-1}) - (x-x^{n-1}) = 1-x$$
We're mostly done:
$$f(x) = [(1-x^{n-1}) - (x-x^{n-1})] \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving $\frac12\cdot\frac34\cdot\dots\cdot\frac{2n-1}{2n}\leq\frac1{\sqrt{3n+1}} ,\;\forall n\in\mathbb{N}$ using induction Base case. Let $n=1$, then $\frac12\leq\frac1{\sqrt{3+1}}$.
Induction step. Let's assume the inequality is true for some $k\in\mathbb{N}$.
We need to show that it's true for $k+1$, i.e. $\frac12\... | Taking squares it is equivalent to $(2k+1)^2(3k+4)\leq (3k+1)(2k+2)^2$, exapanding everything you get $12k^3+12k^2+3k+16k^2+16k+4\leq 12k^3+24k^2+12k+4k^2+8k+4$ which is equivalent to $19k\leq 20k$ which is trivially true since $k\geq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2423691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Help finding the Marginal PDF of Y given a Density Function of Two Variables Problem:
The join pdf of $(X,Y)$ is given by
$
f(x,y) = \begin{cases}
\Big( \frac{5}{32} \Big) x^2(4 - y) & \text{for } x < y < 2x, 0 < x < 2 \\
0 & \text{otherwise} \\
\end{cases}
$
where $k = \frac{5}{32}$. Find the marginal pdf of $Y$.
Answ... | It seems like you have a good idea of what to do but just made a tiny mistake, I'll clear that up and leave you the work of finishing up.
The problem is that the set of possible values of $y$ depends on the value of $x$. You are not taking this into account. To do so you can write the joint pdf in terms of characteris... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2425919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How to Calculate the derivative of $\frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}}$ I'm sorry to bother you with this easy problem. But I'm working alone and totally confused.
It is the 1378th Problem from "Problems in Mathematical Analysis" written by Demidovich. The standard answer is
$$\frac{60(1+x)^{99}(1+6x)}{(1-2x)^{... | You can just take the derivative of the logarithm:
$$y = \frac{(1+x)^{100}}{(1-2x)^{40}(1+2x)^{60}} \implies \ln y = 100\ln(1+x)-40\ln(1-2x)-60\ln(1+2x)$$
$$\frac{y'}{y} = (\ln y)' = \frac{100}{1+x} + \frac{80}{1-2x} - \frac{120}{1+2x}$$
\begin{align}y' &= y\left(\frac{100}{1+x} + \frac{80}{1-2x} - \frac{120}{1+2x} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2426394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Find $\lim_\limits{x\to 0} \frac{\sin x}{e^x -1 -\sin x}$ I have to find $$\lim_\limits{x\to 0} \dfrac{\sin x}{e^x -1 -\sin x}.$$
By using L'Hopital, $$\lim_\limits{x\to 0} \frac{\sin x}{e^x -1 -\sin x}= \lim_\limits{x\to 0} \frac{\cos x}{e^x -\cos x}.$$
Wolfram now states that $\nexists$ because $\lim_\limits{x\to 0^+... | Hint:
\begin{align}
& \lim_{x\to 0} \frac{\sin x}{e^x -1 -\sin x} \\[6pt]
= {} & \lim_{x\to 0} \frac{\sin x}{1+x+\frac{x^2}{2}+o(x^3) -1 -(x-\frac{x^3}{6}+o(x^5))} \\[6pt]
= {} & \lim_{x\to 0} \frac{\sin x}{\frac{x^2}{2}+o(x^3) } \\[6pt]
= {} & \lim_{x\to 0} \frac{x}{\frac{x^2}{2}+o(x^3) } \\[6pt]
= {} & \lim_{x\to 0} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2429273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 4
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Evaluating this line integral Evaluate the following line integral
$\int 2x+y \,dx +xy\,dy$ from $(-1,2)$ to $(2,5)$ where $y=x^2 +1$
When I integrated it w.r.t $x$ I found the integral
$$\int 2x^4 +3x^2 +2x+1 \, dx=\frac{141}{5}$$
from $x=-1$ to $x=2$
But when I tried it w.r.t $y$ I got a confusion with evaluating th... | Your evaluation
$$\int_C (2x+y)dx +xydy=\int_{x=-1}^2 ((2x+x^2+1) +x(x^2+1)(2x))dx=\frac{141}{5}$$
is correct. Using $y$ as a parameter is more difficult, but you should obtain the same result.
\begin{align*}
\int_C (2x+y)dx +xydy&=\int_{y=2}^1 ((2f(y)+y)f'(y) +f(y)y)dy
+\int_{y=1}^5 ((2g(y)+y)g'(y) +g(y)y)dy\\
&=\int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2430613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$0.x3+0.3x=\frac{7}{9} \implies x=?$ An old book problem I was trying to solve some of problems in an old very old book , which not signed
$$0.3\times x ,\\0.\overline{3}\times x \\ 0.\overline{3x} $$ to solve below equation . I tried like below .
My question : How many way we can read this question ?
Am I right ?
$$... | $$\quad{0.\overline{x3}+0.\overline{3x}=\frac{7}{9}\\0.\overline{x}\times 3+0.\overline{3x}=\frac{7}{9}\\
0.\overline{x3}+0.\overline{3}\times x=\frac{7}{9}
\\0.\overline{x}\times 3+0.\overline{3}\times x=\frac{7}{9}\\or \\0.3\times x+0.x\times 3=\frac{7}{9}\\0.(3x)+0.(x3)=\frac{7}{9}}$$ is there an other possible read... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2431516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solve the equation $ z^2 + \left\vert z \right\vert = 0 $, where $z$ is a complex number. I've tried solving this, but I'm stuck at one point.
Here's what I did:
Let $ z = x + yi $, where $x, y \in \mathbf R$
Then , $ (x + yi)^2 + \sqrt{x^2 + y^2} = 0 $
Or, $x^2 + {(yi)}^2 + 2xyi + \sqrt{x^2 + y^2} = 0 $
Or, $ x^2 - y^... | In $x^2 = -\sqrt{x^2}$, note that $x$ is real, so you have a non-negative on the left, and a non-positive on the right. Therefore they must both be $0$.
Alternative solution:
$$
z^2=-|z|\\
|z|^2=|z|\\
|z|=0\quad \text{or}\quad|z|=1
$$
then for $|z|=1$, solve $z^2+1=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
Show that $(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2$
In the list of questions proposed in the "Meeting for Training for the Brazilian Olympiad", 2013. No answer provided. Could solve some problems in that list but got stuck in this one. My developments are going into v... | Let $c = -(a+b)$. Consider following cubic polynomial having $a, b, c$ as roots:
$$t^3-At^2 + Bt - C \stackrel{def}{=} (t-a)(t-b)(t-c)
\quad\text{ where }\quad
\begin{cases}
A &= a + b + c = 0\\
B &= ab+bc+ca = -(a^2+b^2+ab)\\
C &= abc = -ab(a+b)
\end{cases}$$
When $t$ is one of $a, b, c$, we have $t^3 = C - Bt$. This... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2432992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Let $a, b, c$ be positive real numbers such that $a^3 + b^3 = c^3$. Prove that $a^2 + b^2 - c^2 \gt 6(c - a)(c - b)$ .
Let $a$, $b$ and $c$ be positive real numbers such that $a^3 + b^3 = c^3$. Prove that $$a^2 + b^2 - c^2 \gt 6(c - a)(c - b).$$
This problem is from the Indian MO 2009, however, I do not know how to ... | We need to prove that
$$a^2+b^2-\frac{a^3+b^3}{c}>6(c-a)(c-b)$$ or
$$\frac{a^2(c-a)}{c}+\frac{b^2(c-b)}{c}>6(c-a)(c-b)$$ or
$$\frac{(c^3-b^3)(c-a)}{ac}+\frac{(c^3-a^3)(c-b)}{bc}>6(c-a)(c-b)$$ or
$$\frac{c^2+bc+b^2}{ac}+\frac{c^2+ac+c^2}{bc}>6.$$
Now, by AM-GM
$$\frac{c^2+bc+b^2}{ac}+\frac{c^2+ac+c^2}{bc}\geq2\sqrt{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2433718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $ax+by=7$, .
If
$$ax+by=7$$
$$ax^2+by^2=49$$
$$ax^3+by^3=133$$
$$ax^4+by^4=406$$
then find the value of
$$2014(x+y-xy) - 100(a+b)$$
My attempt:
$$ax^2+by^2=49$$
$$ax^2+by^2=(ax+by)^2$$
$$ax^2+by^2=a^2x^2+2abxy+b^2y^2$$
$$ax^2-a^2x^2+by^2-b^2y^2=2abxy$$
$$ax^2(1-a)+by^2(1-b)=2abxy$$
| Let $\frac{ax}{ax+by}=\alpha$ and $\frac{by}{ax+by}=\beta$.
Thus, $$\alpha+\beta=1,$$
$$\alpha x+\beta y=7,$$
$$\alpha x^2+\beta y^2=19$$ and
$$\alpha x^3+\beta y^3=58.$$
Hence, $$(x+y)(\alpha x+\beta y)=7(x+y)$$ or
$$19+xy(\alpha+\beta)=7(x+y)$$ or
$$19+xy=7(x+y).$$
In another hand
$$(x+y)(\alpha x^2+\beta y^2)=19(x+y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2440302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How can I solve this equation for real numbers? How can I solve this equation for real numbers?
$$(x+2)^4+x^4=82.$$
I tried $(x+2)^4+x^4-82=
2x^4 + 8 x^3 + 24 x^2 + 32 x - 66=0$
It is very difficult to continue.
| Let $x+2=a$ and $x=b$.
Thus, $a-b=2$ and we have
$$a^4+b^4=82$$ or
$$(a^2+b^2)^2-2a^2b^2=82$$ or
$$((a-b)^2+2ab)^2-2a^2b^2=82$$ or
$$(4+2ab)^2-2a^2b^2=82$$ or $$a^2b^2+8ab-33=0,$$ which gives
$ab=-11$, $x(x+2)+11=0,$ which is impossible or
$ab=-3$, $x(x+2)=3,$ which gives the answer:
$$\{1,-3\}.$$
We can end also your ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2441965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Solve the radical equation $ x\sqrt{x^2+5} + (2x+1)\sqrt{4x^2+4x+6}=0.$
Solve the following equation:
$$ x\sqrt{x^2+5} + (2x+1)\sqrt{4x^2+4x+6}=0.$$
I wanted to solve this equation. First I tried to change the equations under the roots to the complete square to simplify them out, but it just became more complicated... | Assuming you want real roots, you can use this trick . . .
Let $f\colon \mathbb{R} \to \mathbb{R}$ be given by $f(t) = t\sqrt{t^2+5}$.
Then $f$ is an odd function.
Also, $f$ is strictly increasing, hence $f$ is one-to-one.
Then, letting $u=2x+1$,
\begin{align*}
&x\sqrt{x^2+5}+(2x+1)\sqrt{4x^2 + 4x + 6}=0
\qquad\qqu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2443864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Closed form of $\sum\limits_{k=0}^{n} {2n+1 \choose k}$ I am stuck with this problem: $\sum_\limits{k=0}^{n} {2n+1 \choose k}$.
I used Wolframalpha for the answer and is $4^n$. So i searched for a similar way to expressed it which is:
$$4^n=(1+1)^{2n}=\sum_{k=0}^{2n}{2n \choose k}$$
Is there a way to prove that $\sum_... | Since $$\sum_{k = 0}^n \binom{2n + 1}{k} = \frac{1}{2}\left(\sum_{k = 0}^n \binom{2n + 1}{k} + \sum_{k = 0}^n \binom{2n + 1}{2n+1-k}\right) = \frac{1}{2}\sum_{k = 0}^{2n + 1} \binom{2n + 1}{k}$$
Then we have:
$$\sum_{k = 0}^n \binom{2n + 1}{k} = \frac{1}{2}\sum_{k = 0}^{2n + 1} \binom{2n + 1}{k} = \frac{1}{2} \cdot 2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2444518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Range of the function $f(x) = \frac{2x}{x^{2}+1}$ Plotting the function $f(x) = \frac{2x}{x^{2}+1}$ we can see the range is $[-1,1]$.
Now I was told to do $f(x)=y$ and isolate $x$ doing this I have, using the quadratic formula $x=\frac{1\pm \sqrt{1-y^2}}{y}$ then $$range f(x)= domain \frac{1\pm \sqrt{1-y^2}}{y}$$.
But ... | $y = \frac {2x}{x^2+1}\\
yx^2 - 2x + y = 0$
And then you use the quadratic formula to find values of $y,$ such that $x$ can be a real root of the polynomial.
All good.
What happens at $y = 0$?
$yx^2 - 2x + y = 0$ ceases to be a quadratic and the quadratic formula is not applicable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2448328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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For what primenumbers, $p$, is $\sqrt{5p +49}$ an integer? For what primenumbers, $p$, is $\sqrt{5p +49}$ an integer?
I managed to figure out:
$5p +49 = n^2$
$5p = (n+7)(n-7)$
But can't think of anything more than that, anyone able to solve this without bruteforcing?
For primenumbers under $10^7$, the solutions are $3$... | You're close.
You have $5p=(n+7)(n-7)$ for some integer $n$. Without loss of generality we may assume that $n$ is nonnegative, if $n$ were negative we could render $(n+7)(n-7)=(|n|-7)(|n|+7)$ where $|n|$ is nonnegative. Since $n+7$ is thereby positive we need $n-7$ positive too, so the nonnegative $n$ is at least $8$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2449644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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I need some help with this bernoulli's equation that is given in my assignment Solve:
$$y′−\frac{6y}{x}=\frac{y^5}{x^{13}}$$
I have tried again and again but my answer is still wrong. Can someone show me the steps to solving this question?
| For Bernoulli equation of the form
$$y^\prime+p(x)y=q(x)y^\beta$$
we use substituation $u=y^{1-\beta}$. So with $\beta=5$ we have $u=y^{1-5}=\dfrac{1}{y^4}$ then $u'=-\dfrac{4y'}{y^5}$ then
\begin{eqnarray*}
-\frac{y^5u^\prime}{4}-6\frac{y}{x} &=& \frac{1}{x^{13}}y^5\\
-\frac{u^\prime}{4}-6\frac{1}{x}\frac{1}{y^4} &=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2452304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Sum of series $\frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \frac{x^4}{1-x^8} +\ldots$
The problem is to find the sum of series $$\frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \frac{x^4}{1-x^8} +\ldots$$
to infinite terms, if $|x|<1$
I tried taking $\frac{x}{1-x^2}$ outside but could not solve it.
How to proceed further?
| Notice $$\frac{y}{1-y^2} = \frac{(1+y)-1}{1-y^2} = \frac{1}{1-y} - \frac{1}{1-y^2}$$
The sum at hand is a telescoping one. The partial sums has the form:
$$\sum_{n=1}^p
\frac{x^{2^{n-1}}}{1-x^{2^n}} =
\sum_{n=1}^p
\left[\frac{1}{1-x^{2^{n-1}}} - \frac{1}{1-x^{2^n}}\right]
= \frac{1}{1-x} - \frac{1}{1-x^{2^p}}$$
When $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2455780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Solve the recurrence $T(n) = 2$ if $n=0$
$T(n) = 9T(n-1)-56n +63$ if $n>=1$
Repeated substitution
$k=1$
$T(n) = 9T(n-1)-56n+63 $
$k =2$
$T(n) = 81T(n-2) -560n + 1134$
$k =3$
$T(n) = 729T(n-3) -5096n + 15309$
I cant find the pattern for the n term and the integer
For now i just have
T(n) = $9^k(n-k)$
| Let's apply generating functions. It's not always the easiest method, but it solidly works. From
$$T_n=9T_{n-1}-56n+63$$
the generating function is
$$f(x)=\sum\limits_{n=0}T_nx^n \tag{1}$$
$$f(x)=T_0+\sum\limits_{n=1}T_nx^n=T_0+\sum\limits_{n=1}\left(9T_{n-1}-56n+63\right)x^n=\\
T_0+9\sum\limits_{n=1}T_{n-1}x^n-56\sum\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2457918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Finding an integral formula for intertwined recursive sequences I saw on Wikipedia that the arithmetic-geometric mean of two numbers is given by
$$\operatorname{agm}(x,y)=\frac{\pi}{2}\Bigg[\int_0^{\pi/2}\frac{d\theta}{\sqrt{x^2\cos^2\theta+y^2\sin^2\theta}}\Bigg]^{-1}$$
and I understand the proof of this little gem...... | For any $a,b>0$, let $HG(a,b)=HG\left(\frac{2ab}{a+b},\sqrt{ab}\right)$ and for any $x>1$ let $f(x)=HG(1,x)$.
We have:
$$f(x) = HG(1,x) = HG\left(\frac{2x}{1+x},\sqrt{x}\right) = \frac{2x}{1+x}\,HG\left(1,\frac{1+x}{2\sqrt{x}}\right)\\= \frac{2x}{1+x}\,f\left(\frac{1+x}{2\sqrt{x}}\right)\tag{A} $$
and we may notice tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2459643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find n'th derivative of polynomial and smallest possible degree Let $p(x)$ be a polynomial of degree strictly less than 100 and such that it does not have $x^3−x$ as a factor.If
$\frac{d^{100}}{d x^{100}} \frac{p(x)}{x^3-x} =\frac{f(x)}{g(x)}$
for some polynomials $f(x)$ and $g(x)$ then find the smallest possible degre... | Using the division algorithm we have
$$
\frac{p(x)}{x^3 − x}= q(x) +\frac{r(x)}{x^3 − x}$$
As the degree of $q(x)$ is strictly less than that of $p(x)$ its $100-$th derivative is certainly zero. As $x^3 − x$ is not a factor of $p(x)$ one may assume (without loss of generality) that $x^2 − 1$ divides $r(x)$. In that cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2460756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Simple Probability Distribution Question Alice flips a fair coin n times and Bob flips another fair coin n + 1 times, resulting in independent X $\sim$ Bin(n, 1/2) and Y $\sim$ Bin(n + 1, 1/2).
Can someone please explain to me, in simple english, why this statement is true?
$$P(X<Y)=P(n−X<n+1−Y)$$
The random variable... | Binomial coefficients have a symmetry that can be written
$$\binom{n}{k} = \binom{n}{n-k}$$
$Y$ is a random variable following a binomial distribution with parameters $n+1$ and $1/2$. When the second parameter is $1/2$, the symmetry in binomial coefficients extends to a symmetry in the probability mass function:
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Area enclosed by quarter circle arcs in a square
The problem I have here asks to find the area $A$ enclosed by the two quarter circle arcs in terms of $a$. The first circle has radius $a/2$, and the second has radius $a$. This is what I did.
Firstly I determined that $$A=\frac{a^2}{8}(2\theta_1-\sin(2\theta_1))+\frac... |
Use the cosine rule
\begin{eqnarray*}
\cos ( A ) =\frac{b^2+c^2-a^2}{2bc}.
\end{eqnarray*}
\begin{eqnarray*}
\cos ( \theta_1 ) =\frac{(\sqrt{2}a)^2+(\frac{a}{2})^2-a^2}{2(\sqrt{2}a)(\frac{a}{2})}=\frac{5}{4 \sqrt{2}}
\end{eqnarray*}
\begin{eqnarray*}
\cos ( \theta_2 ) =\frac{(\sqrt{2}a)^2+a^2-(\frac{a}{2})^2}{2(\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $ab+bc+ca+abc=4$, then $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c$ Let $a,b,c$ be positive reals such that $ab+bc+ca+abc=4$.
Then prove
$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c $
So high guys im a high schooler trying to solve this inequality. I did a few things such as try to set a> b>c and then try to ge... | Both inequalities we can prove by the Contradiction method.
Indeed, for the right inequality let $a+b+c<3$, $a=kx$, $b=ky$ and $c=kz$,
where $k>0$ and $x+y+z=3$.
Thus, $k(x+y+z)<3$, which gives $k<1$.
Hence, $4=ab+ac+bc+abc=k^2(xy+xz+yz)+k^3xyz<xy+xz+yz+xyz$,
which is contradiction because we'll prove now that
$$4\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2461934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Napier analogy and algebra in triangle.
If in a $\triangle{ABC}$, we define $x=\tan\frac{B-C}{2}\tan\frac{A}{2}$, $y=\tan\frac{C-A}{2}\tan\frac{B}{2}$ and $z=\tan\frac{A-B}{2}\tan\frac{C}{2}$, then show that $x+y+z=-xyz$.
My attempts:
By Napier analogy,
$x=\frac{b-c}{b+c},\ y=\frac{c-a}{c+a},\ z=\frac{a-b}{a+b}$
Then... | $$\sum_{cyc}\frac{a-b}{a+b}=\sum_{cyc}\frac{(a-b)(c^2+ab+ac+bc)}{\prod\limits_{cyc}(a+b)}=$$
$$=\frac{\sum\limits_{cyc}c^2(a-b)}{\prod\limits_{cyc}(a+b)}=\frac{(a-b)(a-c)(b-c)}{\prod\limits_{cyc}(a+b)}=-\prod_{cyc}\frac{a-b}{a+b}$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Limits of the strings What is the limit of this:
$a_n=\dfrac{(1!+2!+\dots+n!)}{(2 \cdot n)!}$, where n tends to infinity?
I would also like an intuitive explanation in addition to the logical one. Thanks in advance!
P.s.: Do you know where can I find the math API on this site?
| By induction prove that,
$$a_n\le \frac{1}{n}$$
*
*for $n=1$
$$a_1 = \frac{1}{2}\le \frac{1}{1}$$
Assume that
$$a_n=\dfrac{(1!+2!+\dots+n!)}{(2 \cdot n)!}\le \frac{1}{n} $$
$$a_{n+1}=\dfrac{(1!+2!+\dots+n!)+(n+1)!}{2(n+1)(2n+1)(2 \cdot n)!}\\=\frac{1}{2(n+1)(2n+1)}\dfrac{(1!+2!+\dots+n!)}{(2 \cdot n)!}+\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Calculate the triple integral $\iiint_D \sqrt{x^2+y^2+z^2}\, dV$.
Calculate the triple integral of
$$\iiint_D \sqrt{x^2+y^2+z^2}\, dV$$ where $D$ is bounded by
(1) $x^2+y^2+z^2=2ay$ and (2) $y=\sqrt{x^2+z^2}$.
So far I was thinking that if I made $y$ turn to $z$ and $z$ turn to $y$ I could use cilyndrical coordinates... | Yes! Made $y$ turn to $z$ and $z$ turn to $y$, then $z=\sqrt{x^2+y^2}$ turns one time about $z-$axis. This gives us - with spherical coordinates - $0\leq\varphi\leq2\pi$, also $z=a$ shows $0\leq\theta\leq\dfrac{\pi}{4}$ and with $x^2+y^2+z^2=2az$ we have $\rho\leq2a\cos\theta$, hence
$$\iiint_D \sqrt{x^2+y^2+z^2} dV = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2464937",
"timestamp": "2023-03-29T00:00:00",
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minimum value of a two variable problem Suppose that $x$ and $y$ are in $(−2, 2)$ and $xy = −1$. The minimum value of
$\frac4{4-x^2}+\frac9{9-y^2}$ is ?
So I have tried to manipulate this equation but what I got is a complex equation with square roots in it...
| Let $x^2=\frac{2}{3}a$ and $y^2=\frac{3}{2}b$.
Hence, $ab=1$ and by C-S and AM-GM we obtain:
$$\frac{4}{4-x^2}+\frac{9}{9-y^2}=\frac{4}{4-\frac{2}{3}a}+\frac{9}{9-\frac{3}{2}b}=$$
$$=6\left(\frac{1}{6-a}+\frac{1}{6-b}\right)\geq\frac{6(1+1)^2}{6-a+6-b}=$$
$$=\frac{24}{12-a-b}\geq\frac{24}{12-2}=\frac{12}{5}.$$
The equa... | {
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Summation of $\dfrac{1}{1+1^2+1^4}+ \dfrac{2}{1+2^2+2^4}+\dfrac{3}{1+3^2+3^4}...$
Find the sum of $\dfrac{1}{1+1^2+1^4}+ \dfrac{2}{1+2^2+2^4}+\dfrac{3}{1+3^2+3^4}...$ till n terms.
The $i^{th}$ term is given by $\dfrac{i}{1+i^2+i^4}$. I did a similar problem previously but there I was able to decompose the fraction ... | You can get the partial fraction decomposition by writing $x^4+x^2+1$ as the product of two quadratics $(Ax^2+Bx+C)(Dx^2+Ex+F)$, and doing some casework to get the variables. Since the coefficient of $x^4$ is $1$, and $1$'s only factor is $1$, then both $A$ and $D$ equal $1$. We then have:
$$(x^2+Bx+C)(x^2+Ex+F)$$
The ... | {
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Solve $a^2+b^2+c^2+d^2+2ab+2bc+2cd-2ca-2ad-2db=N^2$ I am looking for a characterization of the solutions of
$$a^2+b^2+c^2+d^2+2ab+2bc+2cd-2ca-2ad-2db=N^2$$ in positive integers.
| In case you typed the thing correctly, it is integrally equivalent to $x^2 + 4yz.$ Setting this to zero is easy, setting it to a square not too bad.
Alright, this is exactly the comment by Joffan, with $x = a+b-c-d$ and $y=b$ and $z=c.$
$$
\left(
\begin{array}{rrrr}
1 & 0 & 0 & 0 \\
1 & 1 & 0 & 0 \\
-1 & 0 & 1 & 0 \... | {
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prove that $\lim_{(x,y)\to (0,0)}\frac{x^2y-xy^2}{x^2+y^1}=0$ I need prove that
$$\lim_{(x,y)\to (0,0)}\frac{x^2y-xy^2}{x^2+y^2}=0$$
Can I use it?
if $\sqrt{x^2+y^2}< \delta$ then $|\frac{x^2y-xy^2}{x^2+y^2}|<\epsilon$
$$\left |\frac{x^2y-xy^2}{x^2+y^2} \right |=\frac{x^2 \left |y \right | -\left |x \right |y^2}{x^2+y... | The first inequality you're using has a flaw, as it is : $ |xy|\leqslant x^2+y^2$ and after that you can continue on.
In my answer,I assume that you have made a typo and in the denominator you have $y^2$, since I see you using it in the rest of your attempt.
Showing you some approaches here :
Re-writing the Limit usin... | {
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Closed form $\int_0^\infty\left(\frac{\tanh x}{x^2}-\frac{1}{xe^{2x}}\right)dx=12\log A-\frac{4}{3}\log 2$ Evaluate
$$\int_0^\infty\left(\frac{\tanh x}{x^2}-\frac{1}{xe^{2x}}\right)dx$$
I haven't been able to find references to the indefinite integral of the $\tanh$ term except for some similar forms that had solutions... | Let $$I(a,b) = \int_{0}^{\infty} \left(\frac{\tanh (x)}{x} -e^{-bx} \right)\frac{e^{-ax}}{x} \, dx, $$ where $a, b >0$.
I will show that $$I(a,b) = 8 \psi^{(-2)} \left(\frac{a+2}{4} \right)- 8 \psi^{(-2)}\left(\frac{a}{4} \right)-a \log(a)+a + 2 a \log (2)+ \log(a+b)- 2 \log(4 \pi),$$ where $\psi^{(-2)}(x)$ is the poly... | {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
minimize $x^4 - 6x^2 y^2 + y^4$ given $x^2 + y^2 \leq 1$ I have a constrained optimization problem. Can we maximize / minimize this function on the unit sphere?
$$ f(x,y,z) = x^4 - 6 x^2 y^2 + y^4 \quad\text{given that}\quad x^2 + y^2 + z^2 = 1$$
One idea could be to use the Cauchy-Schwartz inequality. Since I forg... | Another approach would be to use cylindrical coordinates. Then your expression is $r^4\cdot g(\theta)$ with $0\leq r\leq1$, so understanding $g$'s extrema, together with extremal values of $0$ and $1$ for $r^4$ would tell you the extreme values of $f$. Specifically,
$$\begin{align}
g(\theta)&=\cos^4(\theta)-6\cos^2(\t... | {
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Inductive proof of sum of falling factorials with Stirling number coefficients Concrete Mathematics (page 262 2nd ed.) demonstrates $x^n = \sum_k \left\{ \begin{matrix} n \\ k \\ \end{matrix} \right\} x^{\underline{k}}$ using a proof by induction:
$$
\begin{align}
x \sum_k \left\{ \begin{matrix} n-1 \\ k \\ \end{matr... | It is important to make the summation ranges explicit ( to be sure (to be sure))
\begin{eqnarray*}
x \sum_{k=1}^{n-1} \left\{ \begin{matrix} n-1 \\ k \\ \end{matrix} \right\} x^{\underline{k}} &=&
\sum_{k=1}^{n-1} \left\{ \begin{matrix} n-1 \\ k \\ \end{matrix} \right\} x^{\underline{k+1}} +
\sum_{k=1}^{n-1} \left\... | {
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"source": "stackexchange",
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If $a$, $b$ and $n$ are positive integers such that $\frac{1}{a} + \frac{1}{b} = \frac{1}{n}$ then what is the total number of pair of $(a,b)$?
If $a$, $b$ and $n$ are positive integers such that $\frac{1}{a} + \frac{1}{b} = \frac{1}{n}$ then what is the total number of pair of $(a,b)$? What if $a$, $b$ and $n$ are no... | The answer is infinitely many pairs.
Let $a = b$. Then,
$$\frac{1}{a} + \frac{1}{b} = \frac{1}{a} + \frac{1}{a} = \frac{2}{a}$$
From which we get $n = \frac{a}{2}$. This gives $a = b = 2n$ as sufficient to fulfill the equation. Since the last equation is itself fulfilled by infinitely many different values of $a$, it f... | {
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Implicit differentiation question with a ladder and a house. A $13$ ft ladder is leaning against a house when its base starts to slide away from the house, at the instant when the base is $12$ ft from the house, the base is moving at a rate of $5$ ft/sec.
Question:
Let $ \theta $ be the angle between the ladder and the... | With $\theta = \arctan \frac{y}{x}$ we get
$$d \theta =d \arctan \frac{y}{x} = \frac{ d (\frac{y}{x})}{1+ (\frac{y}{x})^2}= \frac{ x \cdot d y - y \cdot d x}{x^2 + y^2}$$
Now use $x^2 + y^2 = l^2= $ const. and get
$$ x dx + y d y = 0$$
Substituting from this $d y = -\frac{x}{y} dx$ into the first equality we get
$$ d... | {
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If $a,b,c$ are rational roots for $x^3+a x^2+b x+c=0$, find the value of a, b, c
Given: $a,b,c$ are rational roots for the equation $x^3+a x^2+b x+c=0$
Find: all possible values of $a,b,c$.
Question used within the preparation for an entrance exam.
My attempt: from Vieta's formulas, we know that (1) $a+b+c=-a$, (2) $... | The way the question is phrased, it doesn't say that $a$, $b$, $c$ have to be all the roots of the equation (or in case some of them are equal, as in $(0,0,0)$, that they will list the roots with the correct multiplicities). In other words, Viet doesn't have to apply as $abc = -c$, it could be $aac = -c$ (if, for examp... | {
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"source": "stackexchange",
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Compute $\lim_{n\rightarrow\infty}\left(\frac{n+1}{n}\right)^{n^2}\cdot\frac{1}{e^n}.$ NOTE 1: L'Hospitals and Taylor expansions are not allowed.
NOTE 2: I really appreciate if someone would correct my attempt, however any other easier method only involving single variable calculus (excluding the concepts in NOTE 1) ar... | Let
$$a=\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} \quad ;\quad b=\lim_{n \to \infty} \left(1-\frac 1n\right)^{n^2}{e^n}$$
We've;
\begin{align}
c= \frac ab
=\frac{\displaystyle\lim_{n \to \infty} \left(1+\frac 1n\right)^{n^2} \frac{1}{e^n} }{\displaystyle\lim_{n \to \infty} \left(1-\frac 1n\right)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 4
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Find $\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}$ I am just trying to calculate
$$\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}.$$
To do this I apply formula for sum of fourth powers of $n$ number. My result: $$\lim_{n\to\infty}\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n... | If
$a_n > 0,
s_n = \sum_{k=1}^n a_n,
r_n = s_{n+1}/s_n$,
then,
if $a_{n+1}/a_n \to 1$,
$r_n \to 1$.
Proof:
$r_n-1
=a_{n+1}/s_n
$.
Since
$a_{n+1}/a_n \to 1$,
then,
for any fixed $m$,
$a_{n+k}/a_n \to 1$
for $1 \le k \le m$.
Therefore,
for any fixed $m$,
$\dfrac{a_{n+1}}{s_n}
\lt \dfrac{a_{n+1}}{\sum_{k=0}^{m-1} a_{n-k}... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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Integrate $f(x)=\frac{1+\sqrt{x+1}}{1-\sqrt{x+1}}$, for $x>-1$ Substituting $t^2=x+1$ I get $2t \ dt=dx.$ So therefore $$\int\frac{2t(1+t)}{1-t}dt=-2\int\frac{t^2+t}{t-1}dt.$$
Dividing the polynomials I get $$-2\int t+2+\frac{2}{t-1}dt=-2\int t \ dt-4\int1 \ dt-4\int\frac{1}{t-1}dt.$$
First integral equals $-t^2+C_1,$ ... | Note that
$$
\frac{1+\sqrt{x+1}}{1-\sqrt{x+1}}
=
\frac{1+\sqrt{x+1}}{1-\sqrt{x+1}}\cdot \frac{1+\sqrt{x+1}}{1+\sqrt{x+1}}
=
\frac{1+2\sqrt{x+1}+(x+1)}{1-(x+1)}
\\
=
-\frac{2+2\sqrt{x+1}+ x}{x}
=
-\frac{2}{x}
-\frac{2\sqrt{x+1}}{x}
-1
$$
| {
"language": "en",
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How to evaluate the sum $\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$ How to evaluate the sum: $$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right)$$
Can anyone help me,I really appreciate it.
| It is evident that
$$S=\sum_{n=1}^{\infty}\frac{1}{(2n+1)(2n+2)}\left(1+\frac{1}{2}+...+\frac{1}{n}\right) = \frac{\pi^2}{12} - \ln^{2}2$$
and can be evaluated by following the pattern:
Consider the series
$$S(x) = \sum_{n=1}^{\infty}\frac{H_{n} \, x^{2n+2}}{(2n+1)(2n+2)}$$
which upon differentiation leads to $S(0) = 0... | {
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"question_score": "14",
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Evaluate $\lim_{n\rightarrow \infty} \Gamma(n+\frac{1}{2})/ \left( \sqrt{2n\pi} \Gamma(n) \right)$ using Stirling's formula. I am working on the limit
$$
\displaystyle\lim_{n\rightarrow \infty} \frac{\Gamma(n+\frac{1}{2})}{ \sqrt{2n\pi}\, \Gamma(n)}\,.
$$
I am thinking I may be able to use Stirling's formula, but they... | By approaching this in a more general way, first we establish the following lemma.
Lemma: Let $p,q\in\mathbb{R}^{+}$. Then,
$$
\lim_{p\rightarrow \infty} \frac{\Gamma\left(\frac{p+q}{2}\right)}{\Gamma\left(\frac{p}{2}\right)p^{q/2}}
= \frac{1}{2^{q/2}}\,.
$$
Proof: It is easy to verify using standard calculus ... | {
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"source": "stackexchange",
"question_score": "3",
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Prove the following expression Prove that if $a,b,c$ are not equal:
$$\dfrac {(x-a)(x-b)}{(c-a)(c-b)}.c+ \dfrac {(x-b)(x-c)}{(a-b)(a-c)}.a+\dfrac {(x-a)(x-c)}{(b-a)(b-c)}.b=x$$
My Attempt:
Proving $ \dfrac {(x-a)(x-b)}{(c-a)(c-b)}.c+ \dfrac {(x-b)(x-c)}{(a-b)(a-c)}.a+\dfrac {(x-a)(x-c)}{(b-a)(b-c)}.b=x$ is equivalent t... | Let $f(x) = \dfrac {(x-a)(x-b)}{(c-a)(c-b)}.c+ \dfrac {(x-b)(x-c)}{(a-b)(a-c)}.a+\dfrac {(x-a)(x-c)}{(b-a)(b-c)}.b-x = 0$
Note that $f(x)$ seems to be a quadratic equation in $x$ and can have no more than two roots. But $f(a) = f(b) = f(c) = 0$ (as Lord Shark stated in comment), meaning that $a,b,c$ are three roots. ... | {
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Show that the polynomial $x^8 -x^7+x^2-x+15$ has no real root I am not getting on how to approach this problem. Clearly, this polynomial can have atleast 2 real roots. And using Descartes's rule of signs, it can have a maximum of 4 positive real roots. But after that, how should I proceed ?
Any help would be highly app... | We have that $x^8-x^7$ and $x^2-x$ are bigger then zero for $x<0$ because $x^8>0$ and $-x^7>0$ similar logic for $x^2$ and $-x$ so $p(x)>15$ for $x<0$.
For $0<x<1$ we have that $x^8-x^7+x^2-x+15>-x^7-x+15>-1-1+15>13$ so $p(x)>13$
For $x\geq1$ we have that $x^8\geq x^7$ and $x^2\geq x$ so $x^8-x^7+x^2-x+15\geq 15$ so $p... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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We play with $3$ dice. Which sum of pips on the dice is more probable, $10$ or $13$? (Question isn't as long as it looks.)
We play with $3$ dice. Which sum of pips on the dice is more
probable, $10$ or $13$?
Without doing any calculation, I would say that it is more probable to have $10$ as sum of pips because we o... | By the stars-and-bars formula, if $n,r$ are positive integers, the equation
$$x_1 + \cdots + x_n = r$$
has exactly ${\large{\binom{r-1}{n-1}}}$ solutions in positive integers $x_1,...,x_n$.
Let the values of the dice rolls be represented as an ordered triple $(d_1,d_2,d_3)$.
First count the number of triples $(d_1,d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximizing a polynomial expression involving the sidelengths of an inscribed pentagon
Given that $ABCDE$ is a convex pentagon and is inscribed in a circle of radius $1$ unit with $AE$ as diameter. If $AB=a,BC=b ,CD=c$ and $DE=d$, then maximum possible integral value of $a^2+b^2+c^2+d^2+abc+bcd$ is ?
What I tried
$AE ... | We will use the fact: If a bunch of circular chords can be packed to "fit" a half-circle, so do any rearrangement of them.
Let $\mathcal{E} = a^2+b^2+c^2+d^2 + bc(a+d)$ be the expression at hand.
Choose points $P, Q, R$ on the half-circle such that
$AP = b$, $PQ = c$, $QR = a$, $RE = d$.
Let $u = AQ$, $v = QE$ and $\a... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to efficiently compute $\,25^{37}\pmod{55}$? Most efficient way to do this question?
$25^{37}\cong (mod 55)$
Attempt:
$25^2 \cong 20 (mod55)$
Then I don't know the best way to continue... I know how to brute force it but I'm just curious what the best way is?
| First, $55 = 5 \cdot 11$ and $(5,11) = 1$, so we will use the Chinese remainder theorem to assemble the result after working modulo $5$ and $11$, separately.
Modulo $5$, $25 \cong 0$, so $25^{37} \cong 0 \pmod{5}$.
Modulo $11$, $25 \cong 3$ and (using Fermat's little theorem) $37 \cong 7 \pmod{10}$, so we compute
$$ 2... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Induction proof: $2^n + 3^n ≡ 5^n (mod 6)$ I'm trying to prove that $2^n + 3^n ≡ 5^n\ (mod\ 6)$ using induction.
$n=1$:
$2+3≡5\ (mod\ 6)$
$n=k$:
$2^k + 3^k ≡ 5^k\ (mod\ 6)$
$n=k+1$:
$2^{k+1} + 3^{k+1} ≡ 5^{k+1}\ (mod\ 6)$
$2*2^k + 3*3^k ≡ 5*5^k\ (mod\ 6)$
$6\ |\ 5*5^k - 2*2^k - 3*3^k$
$6\ |\ (2+3)*5^k - 2*2^k - 3*3^k$
... | Without induction:
$$5^n=(2+3)^n=2^n+3^n+6k.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2497601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
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Find the equation of the circle circumscribing the triangle formed by the three points $(a,0,0)$, $(0,b,0)$ and $(0,0,c)$. Assuming the equation of the circle circumscribing the triangle formed by the three given points is given by the sphere through the three points and the plane through the three points.
Plane is g... | First, let's work out the plane of the circle.
The three triangle vertices are
$$\vec{a} = \left [ \begin{array}{c} a \\ 0 \\ 0 \end{array} \right ], \qquad \vec{b} = \left [ \begin{array}{c} 0 \\ b \\ 0 \end{array} \right ], \qquad \vec{c} = \left [ \begin{array}{c} 0 \\ 0 \\ c \end{array} \right ] \tag{1}\label{na1}$... | {
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"url": "https://math.stackexchange.com/questions/2501105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to see the order relation for $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}$ and $\frac{n^4}{n^7}$ as $n$ very large? I have two sequences $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}$ and $\frac{n^4}{n^7}$. How do I check whether these two sequence have a fixed order relation when $n$ getting larger? Say,... | Let $n$ be positive and large. Then
$$
\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2} > \frac{n^4}{n^7}
\\
\Longleftrightarrow
\\
(n^4+99999n^3-99n^2+3)n^7 > n^4(n^7+9999n^5-9999n^2)
\\
\Longleftrightarrow
\\
n^{11}+99999n^{10}-99n^{9}+3n^7 > n^{11}+9999n^{9}-9999n^{6}
\\
\Longleftrightarrow
\\
99999n^{10}-99n^{9}+3... | {
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"url": "https://math.stackexchange.com/questions/2501548",
"timestamp": "2023-03-29T00:00:00",
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} |
Change of basis of linear transformation $ℝ^2 \to ℝ^3$ Linear transformation $T:ℝ^2\to ℝ^3$ in bases
$\left\{ \begin{bmatrix} 1 \\ 1 \end{bmatrix}, \begin{bmatrix} 1 \\ 3 \end{bmatrix}\right\}$ and $\left\{ \begin{bmatrix} 2 \\ 1 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ 0 \\ 1 \end{bmatrix},\begin{bmatrix} 1 \\ -1 \\ 1 ... | let $x_1, x_2$ be the canonical basis in \mathbb R^2.
and
$y_1, y_2, y_3$ be the canonical basis in \mathbb R^3.
$T(x_1 + x_2) = T(x_1) + T(x_2) = $$(2\cdot 2 + 1\cdot 1 + 8\cdot 1) y_1 + (2\cdot 1 + 1\cdot 0 + 8\cdot -1) y_2 + (2+1+8) y_3\\
13 y_1 - 6y_2 + 11y_3$
$T(x_1) + 3T(x_2) = 12y_1 + 4y_2 + 7y_3\\
2T(x_2) = -1y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2501801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Inequality estimate involving series Given $b_{j}>0$ and $b_{j}=b_{j+1}$ for $j=1,\ldots,n$, with $n\ge 2$ an integer and for some $\Psi>1$, does the following estimate or inequality hold?
\begin{equation}
\sum_{j=1}^n j(j-1)b_{j}|\Psi|^{j-2}~~\le~~\frac{ \left(\sum_{j=1}^n jb_j|\Psi|^{j-1}\right)^2}{\left(1+\sum_{j=1}... | Substituting $x=|\Psi|$, consider a polynomial $p(x)=1+\sum_{j=1}^n b_j x^j$. Then the first inequality your asking is $$p’’(x)p(x)\le p’(x)^2$$ and the second is $$x^4(x-1)\left(\frac{p’’(x)}{p(x)}-\frac{p’(x)^2}{p(x)^2}\right)\le C\ln p(x)+D.$$
Moreover, since all $b_j=b$ are equal and $x\ne 1$, we have $p(x)=1-b +b\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$P(z)=(z^n-1)$ has roots $1, a_1,a_2,...,a_{n-1} $. Prove that $(1-a_1)(1-a_2)...(1-a_{n-1})=n$ $P(z)=(z^n-1)$ has roots $1, a_1,a_2,...,a_{n-1} $. Prove that $(1-a_1)(1-a_2)...(1-a_{n-1})=n$
No idea how to prove this. Would really appreciate any help, hints...
Thanks in advance :)
| Since
$P(z)= z^n - 1, \tag 1$
we have
$P(z) = (z - 1)(z^{n - 1} + z^{n - 2} + \ldots + z + 1) = (z - 1)Q(z), \tag 2$
where
$Q(z) = z^{n - 1} + z^{n - 2} + \ldots + z + 1 = \sum_0^{n - 1} z^i \tag 3$
it then follows that if $1 \ne b \in \Bbb C$ satisfies
$P(b) = (b - 1)(b^{n - 1} + b^{n - 2} + \ldots + b + 1) = (b - 1)Q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2510286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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How to solve $a^3 + 39 ab^2 - 18 = 0$, $3a^2 b + 13 b^3 - 5 = 0$ In this answer, the user @123 has claimed that by solving the system
$$\begin{cases}a^3 + 39 ab^2 - 18 = 0 \\ 3a^2 b + 13 b^3 - 5 = 0\end{cases}$$
we give $ a = \dfrac 32$ and $ b = \dfrac12$. Could anyone explain for me that how one can solve such a syst... | Setting $a=bt$ then we get
$$b^3+t^3+39b^3t-18=0$$
$$3b^3t^2+13b^3-5=0$$ eliminating $b^3$ then we get
$$5t^3-54t^2+195t-234=0$$
one solution is $t=3$ then you will get $a=3b$
plugging this in the given equation we get $$a=\frac{3}{2},b=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2511064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Computing the determinant of a large matrix? How would I go about computing the determinant of large matrices, such as $6 \times 6$. I believe that I need to use multilinear maps, but I am not sure how I can go about computing the determinant in a nice and efficient way.
Can anyone show me how I can determine the dete... | You can use Laplace's expansion of the determinant.
More precisely, begin expanding by the last row, then some row manipulation and expanding by convenient rows/columns ends up in a $2\times2$ determinant:
\begin{align}
\begin{vmatrix}
0&0&1&1&1&1 \\ 1&0&0&0&0&1 \\ 1&0&1&1&1&1 \\
0&1&1&1&0&1 \\ 0&1&0&1&0&0 \\ 0&0&\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2511554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 1
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How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$? How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$?
I've spent quite some time on this problem. So far, I've simplified the right-hand side to $\f... | We have
\begin{eqnarray*}
1^4+\cdots + n^4 &=& \frac{1}{5} n^5+\frac{1}{2}n^4+\frac{1}{3}n^3+\frac{1}{30}n) \\
&=& \frac{1}{30}(6 n^5+15n^4+10n^3-n). \\
\end{eqnarray*}
So
\begin{eqnarray*}
1^4+\cdots + n^4 +(n+1)^4 &=& \frac{1}{30}(6 n^5+15n^4+10n^3-n +30n^4+120n^3+180n^2+120n+30). \\
\end{eqnarray*}
Now check this r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2515600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Prove that, if $a,b,c$ are positive real numbers $\frac {a} {2a+b+c}+ \frac {b} {2b+a+c}+\frac {c} {2c+a+b}<\frac{19}{25}$
Prove that, if $a,b,c$ are positive real numbers:
$$\frac {a} {2a+b+c}+ \frac {b} {2b+a+c}+\frac {c} {2c+a+b}<\frac{19}{25}$$
Can the proof be written without the need for high mathematics?
| The inequality is homogeneous, i.e. $(a,b,c)$ satisfies the inequality if and only if $(ta,tb,tc)$ $\forall t>0$ satisfies the inequality. So we can assume WLOG $a+b+c=1$.
$$\sum_{\text{cyc}}\frac{a}{2a+b+c}=\sum_{\text{cyc}}\frac{a}{a+1}=$$
$$=\sum_{\text{cyc}}\left(1-\frac{1}{a+1}\right)=3-\sum_{\text{cyc}}\frac{1}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2522184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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$\left\lfloor \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}\right\rfloor =?$ $$\left\lfloor \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}\right\rfloor =?$$
I can't find $$? \leq \frac{1}{\sqrt{1}}+\frac{1}{... | Since $$\frac1{\sqrt{n+1}}<2(\sqrt{n+1}-\sqrt{n})=\frac2{\sqrt{n+1}+\sqrt{n}}<\frac1{\sqrt{n}},$$
we have $$2(\sqrt{n+1}-\sqrt{2})<\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}}<2(\sqrt{n}-\sqrt{1}),$$ i.e.
$$2\sqrt{n+1}-2\sqrt{2}+1<\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2522888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
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Residue theorem for a rational function
Evaluate
$$\int_{-\infty}^{\infty}\frac{x^4}{1+x^8}\mathrm{d} x.$$
My concern: One of the consequences of the residue theorem states that, given a polynomial of the form $P/Q$ such that the degree of $Q$ exceeds $P$ by at least two, the integral can then be expressed as $$\... | You can use the following reasoning.
$$\frac{x^4}{x^8+1}=\frac{x^4}{(x^4+\sqrt2x^2+1)(x^4-\sqrt2x^2+1)}=$$
$$=\frac{1}{2\sqrt2}\left(\frac{x^2}{x^4-\sqrt2x^2+1}-\frac{x^2}{x^4+\sqrt2x^2+1}\right)=$$
$$=\frac{1}{2\sqrt2}\left(\tfrac{x^2}{\left(x^2+\sqrt{2-\sqrt2}x+1\right)\left(x^2-\sqrt{2-\sqrt2}x+1\right)}-\tfrac{x^2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2524879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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For each $a \in \mathbb{R}$ evaluate $ \lim\limits_{n \to \infty}\left(\begin{smallmatrix}1&\frac{a}{n}\\\frac{-a}{n}&1\end{smallmatrix}\right)^n$
If $a \in \mathbb{R}$, evaluate $$ \lim_{n \to \infty}\left(\begin{matrix} 1&\frac{a}{n}\\\frac{-a}{n}&1\end{matrix}\right)^{n}$$
My attempt: Let $$A = \left(\begin{matrix... | Your answer is correct as well. but there is a short way to reach it. Merely by putting ourselves in the complex plan where we identify
$$ 1 \equiv \left(\begin{matrix} 1& 0\\0&1 \end{matrix}\right)~~~\text{and}~~~~ i \equiv \left(\begin{matrix} 0& 1\\-1&0 \end{matrix}\right).$$
Thus, $$\begin{align}\displaystyl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 2,
"answer_id": 0
} |
Simplify $\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$
Which one is equivalent to: $$\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$$
1)$\dfrac{\cos(x)}{1+\sin(x)}$
2)$\dfrac{\cos(x)}{1-\sin(x)}$
3)$\dfrac{1-\sin(x)}{\cos(x)}$
4)$\dfrac{1+\sin(x)}{\cos(x)}$
Of course we can find the correct choice by testing eac... |
First Method: Using the identities
$$\cos a+ \cos b =2\cos\left(\frac{a+b}2\right)\cos\left(\frac{a-b}2\right)\tag{I}$$
and
$$\cos a + \sin a =\sqrt2\cos\left(a-\frac{\pi}4\right)\tag{II}$$
we have, \begin{align}\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}&= \frac{1+\sqrt2\cos\left(x+\frac{\pi}4\right)}{1+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Bounding $\|f(a)-f(b)\|$ for $a,b$ in the unit ball Let $f:\Bbb{R}^3 \to \Bbb{R}$ be given by
$$f(x,y,z) = x^2 +\cos(xyz) -z^2$$
I want to show that for any $a,b \in B(0,1)$ (the unit open ball around the origin) $$\|f(a)-f(b)\|\le 2\sqrt{5}$$
In a addition I wonder if this is the best bound I could get?
My attempt: ... | Your approach is fine.
By the mean value theorem exists $ a \in B(0,1) $ such that
$$ \|f(b)-f(a)\| \le \|\nabla f(a)\|\|b-a\| < 2 \|\nabla f(a)\| $$
Having in mind that $ x^2 + y^2 + z^2 < 1 $ we can bound $ \|\nabla f(a)\| $ as follows
$$ \begin{align}
\|\nabla f(a)\| &= \sqrt{\left(x^2 \left(y^2+z^2\right)+y^2 z^2\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526872",
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"source": "stackexchange",
"question_score": "1",
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Derivating with respect the parameter calcule $\int_0^{\frac{\pi}{2}}\frac{\arctan(a\tan x)}{\tan x}dx$ Derivating with respect the parameter calcule $$\int_0^{\frac{\pi}{2}}\frac{\arctan(a\tan x)}{\tan x}dx$$
| To complement the post by @MarkViola, let us show
\begin{align}
\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2x}= \frac{\pi}{2(a+1)}
\end{align}
assuming $a>0$.
Observe
\begin{align}
\int^{\pi/2}_0 \frac{dx}{1+a^2\tan^2x} = \int^{\pi/2}_0 \frac{\sec^2 x- \tan^2x}{1+a^2\tan^2x}\ dx
\end{align}
which means
\begin{align}
\int^{\pi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $6^{273} + 8^{273}\pmod{49}$
The number $6^{273} + 8^{273}$ divided by $49$ has a remainder, what is its value?
I used the totient function to compute for modulo 49.
$6^{42}$ and $8^{42}$ are $-1$ and $1$ mod $49$ respectively, $273/49$ is equal to $5$ with a remainder of $21$.
We would then look for the remain... | Observe that
\begin{align*}
6^{273}+8^{273} & \equiv (7-1)^{273}+(7+1)^{273} \pmod{7^2}\\
& \equiv \sum_{r=0}^{136}\binom{273}{2r}7^{273-2r}\pmod{7^2}\\
& \equiv \binom{273}{272}7\pmod{7^2}\\
& \equiv 7(273) \pmod{7^2}\\
& \equiv 0 \pmod{7^2}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
$\sum_{k=\frac{n}{2}}^{n}\binom{k}{n-k}=F_n$?
Is it true:
$$\sum_{k=\frac{n}{2}}^{n}\binom{k}{n-k}=F_n? $$ where $F_n$ is a
Fibonacci number.
Wolfram alpha says that the answer must be $F_{n+1}$.
I've seen many posts with this question but none of them was helpful to me. And this time the sum is different. Can anyb... | It depends. If you assume that the Fibonacci numbers start with $F_1=1$ and $F_2=1$, then Wolfram alpha is correct, i.e.
$$
\sum_{k=\frac{n}{2}}^n \binom{k}{n-k}=F_{n+1}.
$$
If you assume that the Fibonacci numbers start with $F_1=0$ and $F_2=1$, then we have
$$
\sum_{k=\frac{n}{2}}^n \binom{k}{n-k}=F_{n}.
$$
Actually,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find this maximum of the $\frac{\sqrt{3}}{4}x^2+\frac{\sqrt{(9-x^2)(x^2-1)}}{4}$ Let $x\in \mathbb{R}$, find the function maximum of the value
$$f(x)=\dfrac{\sqrt{3}}{4}x^2+\dfrac{\sqrt{(9-x^2)(x^2-1)}}{4}$$
my attemp
$$x^2=5+4\sin{t},t\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$$
then
$$f=\dfrac{5\sqrt{3}}{4}+2\sin{\... | There's always the classic, but perhaps computationally icky derivative approach.$$f(x)=\dfrac{\sqrt{3}}{4}x^2+\dfrac{\sqrt{(9-x^2)(x^2-1)}}{4}$$
$$f'(x)= \frac{\sqrt3}2 x + \frac{20 x - 4 x^3}{
8 \sqrt{(9 - x^2) (-1 + x^2)}} = \frac 12 x \, \Big(\sqrt{3} + \frac{5 - x^2}{\sqrt{- x^4 - 10 x^2 - 9 }}\Big)$$
Let $f'(x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2539062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Does there exist a closed form for the sinc function series $\sum_{n=1}^\infty \frac{\sin\sqrt{n^2+1}}{\sqrt{n^2+1}}$? Here I want to get the closed form solution of the following summation
$$
\sum_{n=1}^\infty \frac{\sin\sqrt{n^2+1}}{\sqrt{n^2+1}} \qquad(1)
$$
Or the more general form ($x$ be an arbitrary real number,... | To go beyond the limitation $-2\pi<x<2\pi$ for the sine series, we can use the representation G&R (6.677.6) (or Ederlyi TI p.57 1.13.47)
\begin{equation}
\frac{\sin x\sqrt{n^2+a^2}}{\sqrt{n^2+a^2}}=\int_0^x J_0\left( n\sqrt{x^2-t^2} \right)\cos at\,dt\tag{1}
\end{equation}
valid for $x>0$. (For $x<0$, we will use the ... | {
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"url": "https://math.stackexchange.com/questions/2539532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Solve $h_n = 3h_{n-1} -4n$ using generating functions Solve $h_n = 3h_{n-1} -4n$, where $h_0 = 2$ using generating functions.
I am struggling to figure out how to solve this using generating functions. I know the answer should be $h_n = -3^n +2n + 3$.
Here is the method that both my textbook and professor used.
Let $g... | The best approach to get the generating function of a recurrence
is to include directly in it the initial conditions.
In your case, we have
$$
\left\{ \matrix{
h_{\,n} = 0\quad \left| {\;n < 0} \right. \hfill \cr
h_{\,n} = 3h_{\,n - 1} - 4n + 2\left[ {n = 0} \right] \hfill \cr} \right.
$$
where $[P]$ denotes t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2543730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find value of $x^2+y^2$ using given expression
Consider two real numbers $x , y$ such that $$\left(x^2+1\right)\left(y^2+1\right)+9=6\left(x+y\right)$$ Hence find the value of $x^2+y^2$.
At first I tried to factorise the condition but the $(xy)^2$ created much problems. I also tried to create a complete square to get... | Hint. You can rewrite this as $$(y^2 + 1) x^2 - 6x + (y^2 - 6y + 10) = 0.$$
Taken as a quadratic function of $x$ this has discriminant $$36 - 4(y^2 + 1)(y^2 - 6y+10) = -4 (y^2 - 3y + 1)^2;$$ in particular, the only way to get a real solution is with $y^2 - 3y + 1 = 0$, i.e. $y = \frac{3 \pm \sqrt{5}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2546895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to factorize polynomials to the 5th degree? I have the polynomial:
$$2x^5-x^4+10x^3-5x^2+8x-4$$
and I know that the final result is:
$$(2x-1)(x^4+5x^2+4) = (2x-1)(x^2+1)(x^2+4)$$
But how would you do it step by step? I've seen a couple of videos and blogs about it, but they mostly use examples, where their is a com... | Laboring under the assumption that such a polynomial is not irreducible and can be factored (which is clear in this case since you already have a factorization), we would generally start by looking for rational roots.
The rational root theorem asserts that if $z = \frac{a}{b}$ is a rational root of the polynomial (in l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2549116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Find the sum of the double series $\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)} $ First, the original problem follows,
$$\sum_{k=1}^\infty \frac{H_{k+1}}{k(k+1)}$$
where $$H_{k}=\sum_{j=1}^k \frac{1}{j}$$ is the $k$-th partial sum of harmonic series.
Using the following identity,
$$H_{k+1} = \sum_{j... | We show here how to solve the original problem.
Partial fraction decomposition gives
$$\frac{1}{k (k+1)}=\frac{1}{k}-\frac{1}{k+1}\tag{1}$$
Hence the sum can be written as
$$s = \sum _{k=1}^{\infty } \left(\frac{1}{k}-\frac{1}{k+1}\right) H_{k+1}$$
$$=\sum _{k=1}^{\infty } \left(\frac{H_{k+1}}{k}-\frac{H_{k+1}}{k+1}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2551905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Transformation matrices in $\mathbb{R}^2 \to \mathbb{R}^2$ Let $v_1=(1,1)$ and $v_2=(-1,1)$ vectors in $\mathbb{R}^2$. They are clearly linearly independent since each is not an scalar multiple of the other. The following information about a linear transformation $f: \mathbb{R}^2 \to \mathbb{R}^2$ is given: $$f(v_1)=10... | Recall that $v_1$ and $v_2$ are linearly independent eigenvectors of $f$ since:
$$f(v_1)=10\cdot v_1$$
$$f(v_2)=4\cdot v_2$$
The vectors $v_1$ and $v_2$ form a basis of $\mathbb{R}^2$ consisting of eigenvectors of $f$. So $_vF_v$ is similar to a diagonal matrix whose elements in the diagonal are the eigenvalues $\lamb... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove for any integer $a$ show that $a$ and $a^{4n+1}$ have the same last digit
For any integer $a$, show that $a$ and $a^{4n+1}$ have the same last digit
I know that if $a^{4n+1} \equiv a\pmod{10}$ then $10|a^{4n+1}-a$, so $2|a^{4n+1}-a$ and $5|a^{4n+1}-a$, but I'm not sure where to go from here.
| Here's a more elementary argument. Note that mod $10$ we have
$$a^{4n+1}-a = a(a^{4n}-1) = a(a^{2n}-1)(a^{2n}+1) \equiv a(a^{2n}-1)(a^{2n}-9) =a(a^n-1)(a^n+1)(a^n-3)(a^n+3) \pmod{10}.$$
If $\gcd(a,10)=1$ or $10$, then $a^n \equiv 0, \pm1 \mbox{ or }\pm3 \pmod{10}$, making the above product congruent to $0$ mod $10$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2557373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Checking for convergence of $\sum_{n=0}^\infty\frac{(-1)^nn}{n^2+1}$ I have the following series $\sum_{n=0}^\infty\frac{(-1)^nn}{n^2+1}$ which I am checking for convergence.
My working:
$\lim_{n\to\infty}|a_n|=0$ where $a_n=\frac{(-1)^nn}{n^2+1}$.
Doesn't that mean $\sum_{n=0}^\infty\frac{(-1)^nn}{n^2+1}$ is absolute... | The sum of each
even-odd pair is
$\begin{array}\\
\dfrac{2n}{(2n)^2+1}-\dfrac{2n+1}{(2n+1)^2+1}
&=\dfrac{2n((2n+1)^2+1)-(2n+1)(4n^2+1)}{((2n)^2+1)((2n+1)^2+1)}\\
&=\dfrac{4n^2+2n-1}{(4n^2+1)(4n^2+4n+2)}\\
&\lt \dfrac1{8n^2}\\
\end{array}
$
and the sum of these converges.
Since
$\dfrac{2n}{(2n)^2+1}
\to 0$,
the sum of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
What is the coefficient of $x^{9}$ in the expansion of $(1+x)^{12}(1+x)^{4}$? Answer in the textbook: $\sum_{k=0}^{4} \binom{12}{9-k}\binom{4}{k}$
Can I just multiply the two terms together to get $(1+x)^{16}$ then apply the binomial coefficient theorem?
I don't know how they got the sum at the end, I was only taught h... | The factorization $(1+x)^{12}(1+x)^4$ is not accidental here. Let's expand by using Binomial Theorem:
$$\sum_{k=0}^{12}\binom{12}{k}x^k\cdot\sum_{m=0}^{4}\binom{4}{m}x^m.$$
What about $x^9$ in this expansion? We get this power when $k=9,m=0$, $k=8,m=1$, $k=7,m=2$, $k=6,m=3$ and finally $k=5,m=4$. Then our coefficient i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2558914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Is the proof I am using, sufficient/ correct for the system of equation? I want to ask MSE to confirm the correctness of the alternate solution and its mistake.
I know possible solution: https://math.stackexchange.com/a/2557094/456510
If $x,y,z\in {\mathbb R}$, Solve the system equation:
$$
\left\lbrace\begin{array}{... | Good job. Just too verbose.
You are correct into saying you can assume $x$, $y$ and $z$ all positive (there will be a corresponding solution with their negatives). The case when two are positive and one negative cannot appear, nor can the case of two negative and one positive, because the positivity of the left-hand si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2559814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What can we learn as we reduce the fraction at the end of the quadratic formula process? In the final throes of the quadratic formula, you reduce a fraction. Consider the following two examples.
*
*$y = 6x^2 + 11x + 3$; the quadratic formula reveals the roots $x = -4/12$ or $x = -18/12$. The parts of the first fract... | Suppose that we have some quadratic $q(x) = Ax^2 + Bx + C$, where $A, B, C$ are integers with a common factor $k$. Since they have a common factor $k$, there exist integers $a, b, c$ such that $A = ak$, $B = bk$, $C = ck$. Applying the quadratic formula to $q(x)$ gives
$$ \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} = \frac{-bk ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Number of possible matrices with the following property? I have a $8\times 8$ matrix, where every element of the matrix is either $0$, $1$ or $2$, so I define $A$ as a $8\times 8$ ternary matrix.
The matrix needs to have these properties:
*
*In the rows $1,2$ and $3$ the number $1$ must appear exactly once and no nu... | Hint:
The top-left $3\times3$ minor is the identity or a permutation of the identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Taylor series for $\sqrt{1+x^2}$ I want to expand
$$f(x)=\sqrt{1+x^2}$$ in powers of $x-2$
I started by getting the maclaurin series
$$\sqrt{1+x}=1+\frac{1}{2}x+\frac{1}{2} \left( \frac{-1}{2} \right) \frac{x^2}{2!} + \frac{1}{2} \left( \frac{-1}{2}\right) \left(\frac{-3}{2}\right)\frac{x^3}{3!}$$
$$\sqrt{1+x^2}=1+\f... | The Generalized binomial formula for $a\in \Bbb R\setminus \{\Bbb N\}$ says
$$(1+x)^a= \sum_{k=1}^{\infty}{a\choose k} x^k$$
where: $$ \color{blue}{{a\choose k} =\frac{a(a-1)\cdots(a-k+1)}{k!}}$$
see here https://en.wikipedia.org/wiki/Binomial_series
just take $a=1/2$ and replace $x$ by $x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2563704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Proof of Wilsons theorem How to understand the grouping of integers from $2$ to $p-2$ in the proof af Wilsons theorem?
I don't understand how you can group the integers $2$ to $p−2$ into $(p-3)/2$ and next how you become the $2*3*...=1 (mod p)$
| I think the best way to illustrate what's happening is to apply the proof to a specific value of $p$. Take, for example, $p = 11$.
We can solve $2x \equiv 1 \pmod {11}$ with $x = 6$. We can solve $3x \equiv 1 \pmod {11}$ with $x = 4$. We can solve $5x = 1 \pmod {11}$ with $x \equiv 9$. We can solve $7x \equiv 1 \pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2565184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Quadratic equation estimating changes in b and c If $x^2 + bx + c=0$ is an equation the roots are
$$
\begin{aligned}
x&=\frac{-b + \sqrt{b^2 - 4c}}{2}\\
x&=\frac{-b - \sqrt{b^2 - 4c}}{2}
\end {aligned}$$
What is the effect of small changes in $b$ or $c$?
For example if I have $\sqrt{a}$ then to estimate a small change ... | Let us consider $$x_1=\frac{-b + \sqrt{b^2 - 4c}}{2} \qquad \text{and} \qquad x_2=\frac{-b - \sqrt{b^2 - 4c}}{2}$$ and use Taylor expansions around, say, $b_0$ or $c=c_0$.
For a change in $b$, what we would obtain is
$$x_1=\frac{1}{2} \left(\sqrt{b_0^2-4 c}-b_0\right)+(b-b_0) \left(\frac{b_0}{2 \sqrt{b_0^2-4
c}}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2566028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Solving the Differential equation: $y'=\frac{2}{x}y+x^3$ We have the differential equation $$y'=\frac{2}{x}y+x^3$$ and we know $x \in (0, \infty)$.
My attempt with variation of constants
\begin{align}
\phi(x) &= \exp \left(\int \frac{2}{x} dx \right) \\
&= \exp(2\ln|x|) \\
&= x^2c
\end{align}
and
\begin{align}
\psi(x)... | If $y'-p(x)y=0$ is a homogeneous equation then its general solution is given by $y=Ce^{\int p(x)dx}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2569625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Show that the sequence $a_n = \frac{(n+1)^2 -n^2}{n}$ converges and give its limit. can somebody tell me if I did this somewhat correctly? First I estimated the limit of the sequence by calculating the first few results of the sequence and it looks like it is converging towards $2$. Using the Cauchy criterion we then h... | Note that $$a_n = \dfrac{(n+1)^2-n^2}{n} = \dfrac{(n^2+2n+1)-n^2}{n} = 2 + \dfrac{1}{n}\,.$$ Therefore, $$\lim a_n = \lim {}\biggl( 2 + \dfrac{1}{n} \biggr) = \lim 2 + \lim \dfrac{1}{n} = 2 + 0 = 2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2571003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"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.