Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove by induction that $\sum _{r=1}^n \cos((2r-1)\theta) = \frac{\sin(2n\theta)}{2\sin\theta}$ is true $\forall \ n \in \mathbb{Z^+}$
Prove by induction that $$\sum _{r=1}^n \cos((2r-1)\theta) = \frac{\sin(2n\theta)}{2\sin\theta}$$ is true $\forall \ n \in \mathbb{Z^+}$
So my attempt as this is as follows, I've sta... | Starting from where you left off:
$$\frac{\sin (2k\theta)}{2\sin \theta} + \cos ((2k+1)\theta)$$
Let $x=2k\theta$. Then the expression is
$$\frac{\sin x}{2\sin\theta} + \cos (x+\theta)$$
Angle sum identity gives
$$\frac{\sin x}{2\sin\theta} + \cos x\cos\theta - \sin x\sin\theta$$
$$=\frac{\sin x + 2\cos x\cos\theta\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252316",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Given $f(x)=ax^3-ax^2+bx+4$ Find the Value of $a+b$ Let $f(x)=ax^3-ax^2+bx+4$. If $f(x)$ divided by $x^2+1$ then the remainder is $0$. If $f(x)$ divided by $x-4$ then the remainder is $51$. What is the value of $a+b$?
From the problem I know that $f(4)=51$.
Using long division, I found that remainder of $\frac{ax^3-ax^... | This question is very interesting......
Only by this statement, the question can be solved
$f(x)= ax^3-ax^2+bx+4$ is a multiple of $x^2+1$;
Because the roots of the equation $x^2+1$ would be the roots of $f(x)$.
Therefore if $j,k$ are roots of the equation then $j+k=0$ and $j.k =1$.
Let the roots of $f(x)$ be $j,k,l$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Maclaurin series for $f(x) = \frac{1}{1+x+x^2}$ I need to get the Maclaurin series and its radius of the convergence for $f(x) = \dfrac{1}{1+x+x^2}$. I tried to do it manually, with getting the derivatives, but I gave up after some time, because I thought there had to be a better way to solve this. Could anyone help? ... | Alternatively:
$$\begin{align}\frac{1}{1+x+x^2}&=1-(x+x^2)+(x+x^2)^2-(x+x^2)^3+(x+x^2)^4-(x+x^2)^5+\cdots\\
&=1-x+x^2-x^3+x^4-x^5+x^6-x^7+\cdots\\
&\qquad -x^2+2x^3-3x^4+4x^5-5x^6+\cdots\\
&\qquad \ \ \ \ \ x^4-3x^5+6x^6-10x^7+\cdots\\
&\qquad -x^6+4x^7-10x^8+\cdots\\
&\qquad \ \ \ \ \ x^8-5x^9+\cdots\\
&\qquad -x^{10}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3252733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Why does Brent/Salamin algorithm double the digits of $\pi$ with each iteration? The Brent-Salamin-Formula uses the arithmetic-geometric mean to calculate $\pi$.
There are many sophisticated proofs proving very sharp error bounds, for example in Salamin's 1976 paper or in this 2018 paper by Richard Brent, equation (20)... | An easy proof of the quadratic convergence can be found on p. 13-14 of this preprint.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Solving a limit of integral with L'Hopital I have this limit:
$\displaystyle \lim\limits_{x\to \infty}\dfrac{1}{x}\int_{0}^x \dfrac{1}{2+\cos(\mathrm t)}\, \mathrm{dt}$
I said that
Edit:
$-1\leq \cos\mathrm{t}\leq 1 \Rightarrow \dfrac{1}{2+\cos\mathrm{t}} > 0 \text{ and because that expression has no limit as x -> inf... | Note that
\begin{align}
\int^{2\pi n}_0 \frac{dt}{2+\cos t} = n \int^{2\pi}_0 \frac{dt}{2+\cos t}
\end{align}
which means
\begin{align}
\frac{1}{2\pi n}\int^{2\pi n}_0 \frac{dt}{2+\cos t} = \frac{1}{2\pi}\int^{2\pi }_0 \frac{dt}{2+\cos t}.
\end{align}
Hence
\begin{align}
\lim_{n\rightarrow \infty} \frac{1}{2\pi n}\int^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Trig identities analogous to $\tan(\pi/5)+4\sin(\pi/5)=\sqrt{5+2\sqrt{5}}$ The following trig identities have shown up in various questions on MSE:
$$-\tan\frac{\pi}{5}+4\sin\frac{2\pi}{5}=\tan\frac{\pi}{5}+4\sin\frac{\pi}{5}=\sqrt{5+2\sqrt{5}}$$
$$-\tan\frac{2\pi}{7}+4\sin\frac{3\pi}{7}=-\tan\frac{\pi}{7}+4\sin\frac{2... | There is rule to satisfy either to find the identities or to find any particular value of a trigonometric function with that prime in the denominator and a multiple of $ \pi$ in the numerator.
*
*Considering the case of $5$, it is a Fermat prime and hence it can be constructed using straight edges and compass. Becau... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3257939",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
} |
Find the positive integers pair $(x, y)$ such that $xy^2 + y + 1 \mid x^2y + x + y$.
Find the positive integers pair $(x, y)$ such that $xy^2 + y + 1 \mid x^2y + x + y$.
We have that $xy^2 + y + 1 \mid x^2y + x + y \implies xy^2 + y + 1 \mid y(x^2y + x + y) - x(xy^2 + y + 1)$
$\iff xy^2 + y + 1 \mid y^2 - x \implies ... | We know that $(xy^2+y+1) \mid (x^2y+x+y)$. Let $k \in \mathbb{N}$ s.t. $$(x^2y+x+y)=k(xy^2+y+1)$$
We can rewrite this expression as a quadratic in $x$:
$$yx^2+(1-ky^2)x+(-ky-k+y)=0$$
Now, we can see that the discriminant is:
$$\Delta = (ky^2-1)^2+4y(ky+k-y)=(ky^2+1)^2+4y(k-y)$$
We can see that the discriminant must be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3258799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Integrate without expansion? I want to evaluate
$$ \int_0^1 ( 1 - x^2)^{10} dx $$
One way I can do this is by expanding out $(1 - x^2)^{10}$ term by term, but is there a better way to do this?
| When you encounter this type of problem, the first thing to do is to come up with a reduction formula. Let us find the reduction formula for $\int(k^2 - x^2)^n dx $.
Using integration by parts:
$u = (k-x^2)^n $ and $dv = dx$ then $du = n(k^2 - x^2)^{n-1}(-2x)dx $ and $v =x$
Thus,
$$\int(k^2 - x^2)^n dx = x(k^2 - x^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3260371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
How to solve $7^x -5x^3 \equiv 0 \quad \pmod{11}?$ I have to study the solvability of the equation $$ 7^x -5x^3 \equiv 0 \quad \pmod{33} $$ and determine its integer solutions $ x $ with $ 0 \le x \le 110 $.
I started dividing this equation into two equations $$\cases {7^x -5x^3 \equiv 0 \quad \pmod{3} \\ 7^x -5x^3 \eq... | Instead of using the primitive root $2$, note that $7$ is a primitive root mod 11, because $7^2\equiv 5$ and $7^5\equiv -1$. So using 7-indices is a sensible move.
We list the values:
$$
\begin{array}{c|c|c}
\hline
x\pmod{11} & 5x^3\equiv 7^x\pmod{11} & x\pmod{10}
\\\hline
1 & 5&2\\
2 & 7&1\\
3 & 3&4\\
4 & 1&0\\
5 & 9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3261011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving $\arctan (-\sqrt3) = \frac{2\pi}{3}$ using basic methods Background
A recent exam I'm looking at, has the following trigonometric equation:
$$\sin(\pi x) + \sqrt3 \cos(\pi x) = 0, \quad x \in [0, 2]$$
After a simple re-write, we get $$\tan(\pi x) = -\sqrt 3$$
Note, on this exam, the student doesn't have any t... | The fact that $\cos \frac{\pi}{3} = \frac{1}{2}$ is not something that requires memorization, but like many identities in mathematics, it is convenient and efficient to memorize because the proof is more sophisticated than the result.
In an equilateral triangle $\triangle ABC$, draw the altitude $\overline{AD}$ from $A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3262813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
$x^2 + 4x - \lambda +2 \equiv 0 \pmod 7$ I solved this problem but I don't know if my solution is right:
Find the values of $\lambda, 0 \le \lambda \le 6$, such that the
congruence $x^2 + 4x - \lambda +2 \equiv 0 \pmod 7$ has a solution.
Find all the solutions for the minimum value of $\lambda$ for which the
co... | No! There are four values of $\lambda$ such that $x^2+4x-\lambda+2\equiv 0\pmod 7$ has a solution $x\in\mathbb{Z}$.
Note that $x^2+4x-\lambda+2=(x+2)^2-(\lambda+2)$, so you want $\lambda+2$ to be a square, i.e., $\lambda+2\equiv 0,1,2,4\pmod{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3264996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing $\int_{-π/2}^{π/2} \frac{28\cos^2(θ)+10\cos(θ)\sin(θ)-28\sin^2(θ)}{2\cos^4(θ)+3\cos^2(θ)\sin^2(θ)+m\sin^4(θ)}\ dθ$ I am studying the integral
$$I=\int_{-\pi/2}^{\pi/2}
\frac{28\cos^2(\theta)+10\cos(\theta)\sin(\theta)-28\sin^2(\theta)}{2\cos^4(\theta)+3\cos^2(\theta)\sin^2(\theta)+m\sin^4(\theta)}d\theta,$$
... | note that since the function part of the function is odd i.e:
$$f(x)=\frac{28\cos^2x+10\cos x\sin x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}$$
$$f(-x)=\frac{28\cos^2x-10\cos x\sin x-28\sin^2x}{2\cos^4x+3\cos^2x\sin^2x+m\sin^4x}$$
you could notice that the integral can be simplified to:
$$\int_{-\pi/2}^{\pi/2}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3268999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Second order ODE with piecewise term I have the differential equation $\frac{d^2 n}{d z^2} + 2 \frac{dn}{dz} + f(n) = 0$, where $f(n) = \begin{cases} n, \quad 0 \leq n \leq \frac{1}{2} \\ 1 - n, \quad \frac{1}{2} \leq n \leq 1 \end{cases}$.
I am confused about how to handle the $f(n)$ term. I believe it should be solv... | Hint:
Let $u=\dfrac{dn}{dz}$ ,
Then $\dfrac{d^2n}{dz^2}=\dfrac{du}{dz}=\dfrac{du}{dn}\dfrac{dn}{dz}=u\dfrac{du}{dn}$
$\therefore\begin{cases}u\dfrac{du}{dn}+2u+n=0,\quad0\leq n\leq\dfrac{1}{2}\\u\dfrac{du}{dn}+2u+1-n=0,\quad\dfrac{1}{2}\leq n\leq1\end{cases}$
$\begin{cases}\dfrac{du}{dn}=-2-\dfrac{n}{u},\quad0\leq n\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3269266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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To show that $o(o(x^3)-\frac{1}{2}x^2)-o(x^3)=o(x^2) \: as \: x \rightarrow 0.$ Apostol Calculus Example 1 pg.288 I encounter this problem when I was trying to show that
$\sec x=1+\frac{1}{2}x^2+o(x^2) \: as \: x \rightarrow 0.$
We know that $\cos x=1-\frac{1}{2}x^2+o(x^3).$
So $\sec x=\frac{1}{1-\frac{1}{2}x^2+o(x^3... | You are trying to understand in Example 1 that
$$
\frac{1}{\cos x}=\frac{1}{1-\frac12x^2+o(x^3)}=1+\frac12 x^2+o(x^2)
\quad\textrm{as }x\to 0.
$$
Apostol says that this is "from part (e) of Theorem 7.8", i.e.,
As $x\to a$, if $g(x)\to 0$, then
$$
\frac{1}{1+g(x)}={1-g(x)+o(g(x))}.
$$
So if you let $g(x)=-\frac12x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3271730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Determinant with rows $a_1$ to $a_n$ with $-x$ on the diagonal $\left|\begin{matrix}
-x&a_2&\cdots&a_{n}\\
a_1&-x&\cdots&a_{n}\\
a_1&a_2&\cdots&a_{n}\\
\vdots&\vdots&\ddots&\vdots\\
a_1&a_2&\cdots&-x
\end{matrix}\right|$
so i tried to find and expression $D_{n}=kD_{n-1}+tD_{n-2}$
by doing this $\left|\begin{matrix}
-x&... | Subtract the last row from each preceding row to obtain
$$D_n(a_1, \ldots, a_n;x) = \left|\begin{matrix}
-x&a_2&\cdots&a_{n}\\
a_1&-x&\cdots&a_{n}\\
\vdots&\vdots&\ddots&\vdots\\
a_1&a_2&\cdots&-x
\end{matrix}\right| = \left|\begin{matrix}
-x-a_1&0&\cdots&0&a_{n}+x\\
0&-x-a_2&\cdots&0&a_{n}+x\\
\vdots&\vdots&\ddots&\vd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$
$(3x+... | This line
$$3x+7=1^2-2(1)(-\sqrt{x+2})+x+2$$
should be
$$3x+7=1^2-2(1)(\sqrt{x+2})+x+2$$
You put in one too many minuses.
This is my solution.
\begin{align}
\sqrt{3x+7} - \sqrt{x+2} &= 1 \\
(\sqrt{3x+7} + \sqrt{x+2})(\sqrt{3x+7} - \sqrt{x+2})
&= (\sqrt{3x+7} + \sqrt{x+2}) \\
(3x+7) - (x+2) &= \sqrt{3x+7} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3273876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 9,
"answer_id": 8
} |
Show that $\int_{0}^{\pi/2} \frac{\sin x~dx}{1+\sin 2x}=\frac{\coth^{-1} \sqrt{2}}{\sqrt{2}}$ The integral $$\int_{0}^{\pi/2} \frac{\sin x~dx}{1+\sin 2x}=\frac{\coth^{-1} \sqrt{2}}{\sqrt{2}}?$$ can be checked at Mathematica. The question is how to do it by hand?
| $$I=\int_{0}^ {\pi/2}\frac{\sin x~dx}{1+\sin 2x} = \frac{1}{2} \int_{0}^{\pi/2} \frac{(\sin x+\cos x)-(\cos x-\sin x)}{(\sin x+ \cos x)^2} dx=\frac{1}{2}\left (\int_{0}^{\pi/2} \frac{dx}{\sin x+\cos x}-\int_{1}^{1} \frac{dt}{t^2} \right).$$
$$\Rightarrow I=\frac{1}{2\sqrt{2}} \int_{0}^{\pi/2} \mbox{cosec} (x+\pi/4) dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Prove $\lim_{n\to \infty} \sum_{k=0}^{n} 2^{\frac{-kn}{k+n}}=2$ I'm asked to prove
$$\displaystyle\lim_{n\to \infty} \sum_{k=0}^{n} 2^{\frac{-kn}{k+n}}=2$$
Define $$b_{k,n} = \begin{cases}
0,& n \le k\\
2^{\frac{-kn}{k+n}},& n \ge k\\
\end{cases}$$
Our limit is equivalent to finding $\displaystyle\lim_{n\to \infty} \s... | The series can be dominated by $$ b_{k,n} \le 2^{-\frac{k}{2}}$$
For $n < k$ this is obivously true, as we have $b_{k,n} = 0$. For $n\ge k$ we have
$$ n \ge k$$
$$ \frac1n \le \frac 1k$$
$$ \frac1k + \frac1n \le \frac2k$$
$$ \frac{kn}{k+n} = \frac{1}{\frac1k + \frac1n} \ge \frac{k}{2}$$
$$ b_{k,n} = 2^{-\frac{kn}{k+n}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3275963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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What is the least common factor in equation $\frac{5}{x+4}=4+\frac{3}{x-2}$ I am attempting to solve for x $\frac{5}{x+4}=4+\frac{3}{x-2}$
I know that I need to find the least common denominator. In this case, since I cannot see a clear relationship among them all I think it's just the product of all 3 denominators:
$\... | You have a mistake in this step:
$$5x - 10 = (x+4)(x-2)(4) + 3x+12$$
$$5x-10 = (x^2+2x-8)(4) + 3x+12$$
$$5x-10 = 4x^2+\color{red}{8x}-\color{red}{32}+3x+12$$
$$5x-10 = 4x^2+11x-20$$
$$4x^2-6x-10 = 0$$
Alternatively, you can save a step by grouping under the $x+4$ term:
$$(x-2)(5)=(x+4)(x-2)(4)+(x+4)(3)$$
$$5x-10 = (x+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3277024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Closed form for the integral $\int_{-1}^{1} \frac{\ln[(1+x^2)~ -x\sqrt{k+x^2}]}{\sqrt{1-x^2}}\mathrm dx$
Somewhere, the integral
$$\int_{-1}^{1} \frac{\ln[(1+x^2)~ -x\sqrt{k+x^2}]}{\sqrt{1-x^2}}\mathrm dx$$
has been found numerically to be real and finite, if $k \in (-\infty,3].$
Can one find a closed form for this i... | $$I=\int_{-1}^{1} \frac{\ln[(1+x^2) -x\sqrt{k+x^2}]}{\sqrt{1-x^2}}dx$$
Split in two parts and let $x\to-x$ for the $(-1,0)$ interval to see that:
$$ I=\int_{0}^{1} \frac{\ln[(1+x^2) +x\sqrt{k+x^2}]}{\sqrt{1-x^2}}dx+\int_{0}^{1} \frac{\ln[(1+x^2) -x\sqrt{k+x^2}]}{\sqrt{1-x^2}}dx$$
$$=\int_0^1 \frac{\ln(1+(2-k)x^2)}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3279647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to find the value of $\cos\left (\frac{\pi }{28} \right )-\cos\left (\frac{3\pi }{28} \right )+\sin\left ( \frac{5\pi }{28} \right )$? Once , I do a problem $ \sum_{n=1}^{14} \cos\left ( \frac{n^{2}\pi }{14} \right )$
when angle modular by $28$ we've got
$2\left ( \cos\left (\frac{\pi }{14} \right ) - \cos\lef... | Your sum can be written in terms of complex exponentials as
$$ 1+\exp(\pi i/14) + \exp(3 \pi i/14) + \exp(4 \pi i/14) + \exp(8 \pi i/14) + \exp(9 \pi i/14) + \exp(12 \pi i/14) + \exp(16 \pi i/14) + \exp(19 \pi i/14) + \exp(20 \pi i/14) + \exp(24 \pi i/14) + \exp(25 \pi i/14) + \exp(27 \pi i/14)
$$
If $w = e^{i\pi/14}$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Moment of inertia of hemisphere I want to find the moment of inertia of the hemisphere shown in the picture about the $O$ axis.
I am getting different results depending on the approach.
*
*The moment of inertia of the hemisphere about an axis parallel to $O$ passing through its Center of Mass (CoM) is
$
I_{CoM} = \f... | *
*is correct, $\dfrac{83}{320}+\dfrac{5^2}{8^2}=\dfrac{13}{20}.$
*is incorrect, the parallel axis theorem works with an axis through the gravity center;
*is incorrect, there is no account for the $z$ coordinate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Maclaurin series of $\tan(x+x^2)$ to order 3 I know that the maclaurin series of $\tan(x)$ is $\tan(x)=x+\frac{1}{3}x^3+\frac{2}{15}x^5+...$, then shouldn't be $\tan(x+x^2)=(x+x^2)+\frac{1}{3}(x+x^2)^3+\frac{2}{15}(x+x^2)^5+...$?
Mathematica actually gives me $\tan(x+x^2)=x+x^2+\frac{x^3}{3}+o(x^4)$ to order 3.
| Expanding the monomial terms gives Mathematica's result – and you can stop at $(x+x^2)^3$:
$$(x+x^2)+\frac13(x+x^2)^3=x+x^2+\frac13(x^3\color{lightgrey}{+3x^4+3x^5+x^6})=x+x^2+\frac{x^3}3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3280936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show $\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx=\frac{2}{3}G$ Discovered the integral below
$$I=\int_0^{\pi/3} \text{tanh}^{-1}(\sin x)\, dx= \frac23G$$
which looks clean, yet challenging. Have not seen it before. Post it here in case anyone is interested.
Edit:
Here is a solution. Let $J(a)=\int_0^{\frac\pi{3}}\tan... | Use that
$$\operatorname{arctanh}(\sin(x))=2\sum_{n=1}^{\infty} (-1)^{n-1}\frac{\sin((2n-1)x)}{2n-1}, \ 0 <x<2\pi$$and after integration, employ the result in $(3.238)$, page $215$, from the book (Almost) Impossible Integrals, Sums, and Series$\displaystyle \left(\sum_{n=1}^{\infty} (-1)^{n-1} \frac{\sin(\pi/6(4n+1))}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the orthogonal complement The question is, considering
*
*The vector space $\mathcal{P}_3(\mathbb{R})$ of the polynomials with real coefficients of degree $\leq$ 3
*The inner product defined by $\left<p,q\right>=\int_{-1}^{1}pq$
how to find a basis for the orthogonal complement of the space spanned by $... | Let $\textsf{W}=\operatorname{span}(\{x-1,x^2+3\})$ be the subspace of $\textsf{P}_3(\mathbb R)$. Then $p(x)\in \textsf{W}^\perp$ if and only if
$$\langle p(x),x-1\rangle=\int_{-1}^1 p(x)(x-1)dx=0$$
$$\langle p(x),x^2+3\rangle=\int_{-1}^1 p(x)(x^2+3)dx=0$$
at the same time, since $\{x-1,x^2+3\}$ is a basis for $\texts... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3281380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Factoring $-3x^2 - 2xy + 8y^2 + 11x - 8y - 6$ This is quite easy to do if we do the middle term splitting method, but here's another method given in the book:
Only considering terms with powers of $x$ and constants:
$-3x^2 + 11x - 6 = (3x - 2)(-x + 3)$
Only considering terms with powers of y and constants:
$8y^2 - 8y... | So you want: $$-3x^2 - 2xy + 8y^2 + 11x - 8y - 6= (ax+by+c)(dx+ey+f)$$
If we set $y=0$ we get $$\boxed{-3x^2+11x-6 = (ax+c)(dx+f)}$$
and since $$-3x^2+11x-6 = (-3x+2)(x-3)$$ so $a=-3$, $c=2$, $d=1$ and $f=-3$ (or $a=3$, $c=-2$, $d=-1$ and $f=3$.)
Seting $x= 0$ we get $$\boxed{ 8y^2-8y-6 = (by+c)(ey+f)}$$ so ...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factorization in a proof of induction I have to prove the following:
$1+3^3+ ... + (2n+1)^3=(n+1)^2(2n^2+4n+1)$ by induction.
My try:
Base case, $n=1$:
$1+3^3=(2)^2(2\cdot1^2 + 4\cdot1 + 1)$, which is true.
By inductive hypothesis, assume $n=k$:
$1 + 3^3 + ... + (2k+1)^3=(k+1)^2 (2k^2 + 4k + 1)$
For $n=k+1$
$1 + 3^3 + ... | You can get the factor $(k+2)^2$ by the following way.
$$(k+1)^2(2k^2+1)+(2k+3)^2=2k^4+16k^3+47k^2+60k+28=$$
$$=2k^4+8k^3+8k^2+8k^3+32k^2+32k+7k^2+28k+28=$$
$$=(k+2)^2(2k^2+8k+7)=(k+2)^2(2(k+1)^2+4(k+1)+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3289534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Convergence bound for $p\left(n\right) = \text{arctan}\left(\frac{2-\sqrt{4-\left(\frac{2}{n}+\frac{1}{n^2}\right)}}{(2+\frac{1}{n})}\right)$ For my engineering-related work, I ended up with the following expression
$p\left(n\right) = \text{arctan}\left(\frac{2-\sqrt{4-\left(\frac{2}{n}+\frac{1}{n^2}\right)}}{(2+\frac{... | $$ p(n) = \arctan(\frac{2 - \sqrt{4 - \frac{1}{n}\cdot (2+\frac{1}{n})}}{(2+\frac{1}{n})} ) = \arctan(\frac{(2 - \sqrt{4 - \frac{1}{n}\cdot (2+\frac{1}{n})})(2+\sqrt{4 - \frac{1}{n}\cdot (2+\frac{1}{n})})}{(2+\frac{1}{n})(2+\sqrt{4 - \frac{1}{n}\cdot (2+\frac{1}{n})})}) $$
After using $(a-b)(a+b) = a^2 - b^2 $, we have... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Generating prime numbers of the form $\lfloor \sqrt{3} \cdot n \rfloor $ How to prove the following claims ?
Let $b_n=b_{n-1}+\operatorname{lcm}(\lfloor \sqrt{3} \cdot n \rfloor , b_{n-1})$ with $b_1=3$ and $n>1$ . Let $a_n=b_{n+1}/b_n-1$ .
*
*Every term of this sequence $a_i$ is either prime or $1$ .
*Every odd pr... | We prove that the second and the third claims are true.
The second claim is true.
If $a=d\alpha$, $b=d\beta$ and $(a,b)=d$, we have $\mathrm{lcm}(a,b) / b = \alpha = a / d.$
We may rewrite the sequence $a_n$ using above.
$$
a_1=1,$$
$$
a_n=\frac{\lfloor(n+1)\sqrt 3\rfloor}{\left(\lfloor(n+1)\sqrt 3\rfloor, (a_{n-1}+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3290665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solving the diophantine equation $x^3+y^3 = z^6+3$ I've the following problem:
Show that the congruence $x^3+y^3 \equiv z^6+3\pmod{7}$ has no solutions. Hence find all integer solutions if any to $x^3+y^3-z^6-3 = 0.$
We can rearrange the first equation to $z^6 \equiv (x^3+y^3-3) \mod{7}$. But $z^6\equiv 1\mod{7}$ so... | Render any cube $\in\{-1,0,1\}\bmod 9$. Then $z$ can't be a multiple of $3$ because that makes $ x^6+3\equiv 3\bmod 9$. Try $z$ not a multiple of $3$, then $z^6=(z^2)^3$ where $z^2\equiv 1\bmod9$ forcing $z^6\equiv 1\bmod9$. Then $x^3+y^3 =x^6+3\equiv 4\bmod9$. Well, today is not the day Serena Williams ties that t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3292195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Value of $\prod_{n>1} \frac{1}{1-\frac{1}{n^s}}$ or $-\sum_{n=2}^{\infty} \log(1-\frac{1}{n^s} )$ I know
\begin{align}
\prod_{p~is~ prime} \frac{1}{1-\frac{1}{p^2}} = \zeta(2) = \frac{\pi^2}{6}
\end{align}
which has a convergent number.
actually I can even generalized this to
\begin{align}
\prod_{p ~is~ prime} \frac... | For $2k\ge 2$ $$\prod_{n=2}^\infty (1-\frac{1}{n^{2k}})=\prod_{m=1}^{2k} \prod_{n=2}^\infty (1-\frac{e^{2i \pi m/(2k)}}{n}) = \prod_{m=1}^{k}\prod_{n =-\infty, |n|\ge 2}^\infty (1-\frac{e^{2i \pi m/(2k)}}{n}) = \prod_{m=1}^{k} f(e^{2i \pi m/(2k)})$$
where $f(x) =\frac{\sin(\pi x)}{\pi x(1-x^2)} $ and in those products... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many primes are there of the form $x^6+y^6$, where $x,y\in\mathbb{Z}$? I'm trying to find such primes that are of the form $x^6+y^6$; $x,y\in\mathbb{Z}$.
If one of $x$ and $y$ is $0$, say $y=0$ then $x^6+y^6=x^6$ which is not a prime. So assume that $(x,y)\ne(0,0).$
I factored $x^6+y^6$ as follows:
$x^6+y^6=(x^2)^3... | So you know that $x^6 + y^6 = (x^2 + y^2)(x^4 - x^2y^2 + y^4)$, and both are not factorizable over $\mathbb Q$.
If $a, b, c \in \mathbb Z$ and $a = bc$, then if $a$ is prime, either $b = 1$ or $c = 1$.
You already know that $x^2 + y^2 = 1$ if and only if $(x, y) \in \{(0, \pm 1), (\pm 1, 0)\}$. Let us briefly look at t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3294818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find basis for the image and the kernel of a linear map How do I calculate a basis for $\ker (\varphi)$ and $\operatorname{im}(\varphi)$?
Where, given
$$ A = \begin{pmatrix}
1 & -1\\
-1 & 1 \\
\end{pmatrix}$$
we define
$$ \begin{matrix}
\varphi: \mathbb{R}^{2 \times 2} \to \mathbb{R}^{2 \times 2} \\
X \mapsto XA+A^t ... | EDIT: This answer assumes that $\varphi(X) = XA - (XA)^T$, so it only illustrates the method.
Your matrix $M_\varphi$ should be
$$M_B(\varphi) = \begin{pmatrix}
0 & 0 & 0 & 0\\
-1 & 1 & -1 & 1 \\
1 & -1 & 1 & -1\\
0 & 0 & 0 & 0\\
\end{pmatrix}$$
which implies the basis for $\ker\varphi$ is $b_1+b_2, b_1-b_3, b_1+b_4$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int_0^{\frac{\pi}{4}}\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}\,dx$ Find:
$$\int_0^{\frac{\pi}{4}}\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}\,dx$$
I don't know how I starte & evaluate this integral
Wolfram alpha give $=1,28553$
My problem whene I use $t=\cos x$ I get $\arccos x$
Same problem w... | $$\mathcal{J}=\int_0^{\frac{\pi}{4}}\frac{1+\cos x-x^{2}}{(1+x\sin x)\sqrt {1-x^{2}}}\,dx=\int_0^{\frac{\pi}{4}} \color{red}{\frac{1+x\sin{x}}{\cos{x}\sqrt{1-x^2}}} \cdot \color{blue}{\frac{\cos{x} \left(1+\cos x-x^{2}\right)}{(1+x\sin x)^2}\,dx}$$
If $\displaystyle t=\frac{\sin{x}+x}{1+x\sin{x}} \text{ then }\color{bl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sin(x) - \sin(y) = -\frac{1}{3}$, $\cos(x) - \cos(y) = \frac{1}{2}$, what is $\sin(x+y)$? If $\sin(x) - \sin(y) = -\frac{1}{3}$ and $\cos(x) - \cos(y) = \frac{1}{2}$, then what is $\sin(x+y)$?
Attempt:
$$ \sin(x+y) = \sin(x) \cos(y) + \cos(x) \sin(y) $$
If we multiply the two "substraction identities" we get
$$ (\si... | We have
$$\cos \frac{x+y}{2} \sin \frac{x-y}{2}=-\frac{1}{6}$$
$$\sin \frac{x+y}{2} \sin \frac{x-y}{2}=-\frac{1}{4}$$
Thus
$\tan \frac{x+y}{2}=\frac{3}{2}$.
And hence
$$\sin(x+y)=\frac{2\tan\frac{x+y}{2}}{1+\tan^2\frac{x+y}{2}}=\frac{12}{13}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3296374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the constant $k$ from the determinant
Given: $$\begin{vmatrix}(b+c)^2 &a^2&a^2\\b^2 &(c+a)^2&b^2 \\c^2&c^2& (a+b)^2\end{vmatrix}=k(abc)(a+b+c)^3$$ Find $k$.
If I directly open the determinant it will go to long I can't apply most of the row or column operation as they keep making it more complex.
| It's not long:
$$\prod_{cyc}(a+b)^2+2a^2b^2c^2-\sum_{cyc}a^2b^2(a+b)^2=$$
$$\sum_{cyc}\left(a^4b^2+a^4c^2+2a^3b^3+2a^4bc+2a^3b^2c+2a^3c^2b+2a^2b^2c^2+4a^3b^2c+4a^3c^2b+\tfrac{4}{3}a^2b^2c^2\right)+$$
$$+\sum_{cyc}\left(\frac{2}{3}a^2b^2c^2-a^4b^2-a^4c^2-2a^3b^3\right)=$$
$$=\sum_{cyc}(2a^4bc+6a^3b^2c+6a^3c^2b+4a^2b^2c^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3297816",
"timestamp": "2023-03-29T00:00:00",
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How to solve $\int \frac{2x^2-4x+3}{(x-1)^2}\, dx$? My book says the answer to $\int \frac{2x^2-4x+3}{(x-1)^2} \, dx$ is $2x-\frac{1}{x-1}+C$ but symbolab says it is $2x-\frac{2}{x-1}+\frac{4}{x-1}-\frac{3}{x-1}-2+C$. Who is correct and how would I get to the answer? I tried to use u-substitution but that doesn't work ... | Note that $ \int \frac{2x^2-4x+3}{(x-1)^2}\, dx = \int \frac{2\left(x-1\right)^2+1}{\left(x-1\right)^2}\, dx = \int 2+\frac{1}{\left(x-1\right)^2}\, dx = 2x - \frac{1}{\left(x-1\right)} + C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3301198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to show that $\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
$\sqrt{2+\sqrt{3}} = \dfrac{\sqrt{6}+\sqrt{2}}{2}$
How to change $\sqrt{2+\sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$
| If you want to change $\sqrt{2+ \sqrt{3}}$ into $\dfrac{\sqrt{6}+\sqrt{2}}{2}$, then go through the following calculation steps from bottom to top. If you want to change $\dfrac{\sqrt{6}+\sqrt{2}}{2}$ into $\sqrt{2+ \sqrt{3}}$, then go through the following calculation steps from top to bottom. Either way we get equali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3303949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
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Proof by epsilon-delta of $\lim \limits_{x \to 1} 3x^2+1=4$ I have some doubts proving the following:
$$\lim \limits_{x \to 1} 3x^2+1=4$$
My try
$(\forall \varepsilon > 0)(\exists \space \delta > 0): (0<|x-1|< \delta \implies |3x^2+1-4| < \varepsilon)$
$\implies 0<|x-1|< \delta \iff -\delta \lt x -1 \lt \delta \iff -\d... | Given $\varepsilon>0$, your formula $\delta:= \min\{1, \frac{\varepsilon}{9}\}$ works! Let's verify it.
If $|x-1|<\delta$ with $\delta>0$, then
$$|3x^2+1-4|=|3(x-1)(x-1+2)|\leq 3|x-1|(|x-1|+2)<3\delta(\delta+2).$$
Now, by the above definition, $\delta\leq 1$ AND $\delta\leq \frac{\varepsilon}{9}$ which implies
$$3\del... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Finding the adjoint of a differentiation map The integral inner product is defined as
$$\langle p(x),q(x) \rangle = \int_{-1}^1 p(t)q(t)dt$$
on both $\textsf{P}_2(\mathbb{R})$ and $\textsf{P}_1(\mathbb{R})$.
Find the adjoint of the differentiation map
$$\begin{align}
\textsf{T} : \textsf{P}_2(\mathbb{R}) & \to \tex... | Let $p\in P_2(\Bbb R)$ and $q\in P_1(\Bbb R)$ so that
\begin{align*}
p(t) &= a_{2} t^{2} + a_{1} t + a_{0} & q(t) &= b_{1} t + b_{0}
\end{align*}
Then
$$
\langle Tp, q\rangle
= \int_{-1}^1 p^\prime(t)\cdot q(t)\,dt
= 2 \, a_{1} b_{0} + \frac{4}{3} \, a_{2} b_{1}
$$
We may also write $(T^\ast q)(t)=c_{2} t^{2} + c_{1} t... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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what is the value of $9^{ \frac{a}{b} } + 16^{ \frac{b}{a} }$ if $3^{a} = 4^{b}$ I have tried to solve this expression but I'm stuck:
$$9^{ \frac{a}{b} } + 16^{ \frac{b}{a} } = 3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} }$$
and since $3^{a} = 4^{b}$:
$$3^{ \frac{2a}{b} } + 4^{ \frac{2b}{a} } = 4^{ \frac{2b}{b} } + 3^{ \frac... | Here a short way:
*
*$3^a = 4^b \Leftrightarrow 3 = 4^{\frac{b}{a}}$ and $4= 3^{\frac{a}{b}}$
*$\Rightarrow 9^{\frac{a}{b}} = \left(3^{\frac{a}{b}}\right)^2 = 4^2$ and
*$\Rightarrow 16^{\frac{b}{a}} = \left(4^{\frac{b}{a}}\right)^2 = 3^2$
Now, just sum up.
| {
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"url": "https://math.stackexchange.com/questions/3309188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Interesting series involving factorials and logarithms Not sure if the following is easy to prove or not, I obtained the result using probabilistic arguments exclusively.
Define $x_1 = 2; x_2 = 4/3; x_3 = (8\cdot 6) / (7\cdot 5); x_4 = (16\cdot 14 \cdot 12 \cdot 10) / (15\cdot 13 \cdot 11 \cdot 9)$, and so on. Prove t... | Note that, for $k>1, x_k=\dfrac{y_k/y_{k-1}}{z_k/z_{k-1}}$ where $y_k=2^k(2^k-2)\cdot\cdot\cdot2=2^{2^{k-1}}(2^{k-1})!$
and $z_k=(2^k-1)(2^k-3)\cdot\cdot\cdot1=(2^k)!/y_k$.
Thus $x_k=\dfrac{y_k/y_{k-1}}{\dfrac{(2^k)!}{(2^{k-1})!}/\dfrac{y_{k-1}}{y_k}}=
\dfrac{y_k^2}{y_{k-1}^2}\cdot\dfrac{(2^{k-1})!}{(2^k)!}$.
Substitu... | {
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"timestamp": "2023-03-29T00:00:00",
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If $ \sin a +\sin c =2 \sin b $, show that $ \tan\frac{a+b}{2}+\tan\frac{b+c}{2} = 2 \tan\frac{c+a}{2}$ Found the question in the textbook.
I tried many methods of manipulating the identity to be proved but I did not even see a clue of using the given condition(of different forms like
$$
\sin\frac{a+c}{2} \cos\frac{a-c... | $$\tan\dfrac{c+a}2-\tan\dfrac{a+b}2$$
$$=\dfrac{\sin\dfrac{c+a-(a+b)}2}{\cos\dfrac{c+a}2\cos\dfrac{a+b}2}$$
$$=\dfrac{2\sin\dfrac{c-b}2\cos\dfrac{b+c}2}{2\cos\dfrac{c+a}2\cos\dfrac{a+b}2\cos\dfrac{b+c}2}$$
$$=\dfrac{\sin c-\sin b}{2\cos\dfrac{c+a}2\cos\dfrac{a+b}2\cos\dfrac{b+c}2}$$
Similarly,
$$\tan\dfrac{b+c}2-\tan\... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove $ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^4$
Prove that for all $n \in \mathbb{N}$ the inequality $$ \left(\sum \limits_{k=1}^n (2k-1)\frac{k+1}{k}\right) \left( \sum \limits_{k=1}^n (2k-1)\frac{k}{k+1}\right) \le \frac{9}{8}n^... | We can use also the Cassel's inequality:
Let $a$, $b$ and $w$ be sequences of $n$ positive numbers such that $1<m\leq\frac{a_k}{b_k}\leq M$ for any $k.$ Prove that:
$$\sum_{k=1}^nw_ka_k^2\sum_{k=1}^nw_kb_k^2\leq\frac{(M+m)^2}{4Mm}\left(\sum_{k=1}^nw_ka_kb_k\right)^2.$$
This inequality was here:
G.S. WATSON, Seria... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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The natural numbers $a, b, c$, formed by the same $n$ digits $x$, $n$ digits $y$, and $2n$ digits $z$ satisfy $a^2 + b = c$
Given that the natural numbers $a, b, c$ are formed by the same $n$
digits $x$, $n$ digits $y$, and $2n$ digits $z$ respectively. For any
$n \geq 2$, find the digits $x, y, z$ such that $a^2 ... | If we let $1_n= \underbrace{1111....1}_{n}$
$a^2 + b = c$
$\frac {a^2}{1_n} + \frac {b}{1_n} = \frac {c}{1_n}$
$x^2*1_n + y = z*(10^n + 1)$
Let's think about what that means:
If $x^2 = 10j + k$ we have
$10j + k + y \equiv z$. And we carry $j$ or $j+1$
Then we have $j +k (+1)\equiv 0$. Which means $j+k(+1) = 10$.
Thi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3312967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Coordinates of centres of central triangles with respect to the reference triangle In Kimberling's Encyclopedia of Triangle Centers, a lot of centres are described as the centres of certain central triangles of the reference triangle, whether as a main or alternate definition. For example, $X_{164}$ is defined as the i... | This answer is an answer for the very specific question:
How to obtain or verify $X(164)$ knowing it is the $X(1)$-point of the excentral triangle?
Explicitly:
How to obtain the barycentric coordinates of $X(164)$, knowing it is the incenter of the circle through the points with (inhomogeneous) barycentric coordinate... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Chord $x+y=1$ Subtends $45^{\circ}$ at the centre If the Chord $x+y=1$ Subtends $45^{\circ}$ at the centre of the circle $x^2+y^2=a^2$, Find $a$
Method $1$:
We have radius=$a$
If $O$ is the centre and chord be $AB$ we have
$$\angle AOB =45$$
If $OM$ is drawn perpendicular to the chord we have:
$$\angle AOM=22.5$$
In $\... | When $a=1$, the angle subtended is $90º$. This angle will continually decrease until the circle is tangent to the chord, giving a subtended angle of $0º$. Similarly, increasing $a$ will cause the subtended angle to be larger than $90º$, eventually approaching $180º$.
Therefore there is only one solution for $a$. If you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3316133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How do I solve this exponential-surdic equation? I have been trying to solve a problem today, but have had very little to no progress. The problem is:
Find the real values of $x$ satisfying $$(26 + 15\sqrt{3})^x + 6(2 + \sqrt{3})^x + (2 - \sqrt{3})^x - 5(7 + 4\sqrt{3})^x = 5$$
Clear explanations would surely be appreci... | Notice, that $26+15\sqrt3=(2+\sqrt3)^3$ and $7+4\sqrt3=(2+\sqrt3)^2$. Now the equation reads $$(2 + \sqrt{3})^{3x} + 6(2 + \sqrt{3})^x + (2 - \sqrt{3})^x - 5(2 + \sqrt{3})^{2x} = 5$$
Let's multiply both parts of it by a nonzero $(2 + \sqrt{3})^x$. We receive $$(2 + \sqrt{3})^{4x} + 6(2 + \sqrt{3})^{2x} + ((2 - \sqrt{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Derive formula for the area of triangle provided that you know 2 angles and side between them. I am trying to derive the formula for finding the area of the triangle that uses 2 angles and and 1 side between them.
Formula provided by the book:
If we know two angles (call them $A$ and $B$) and the side between
them c... | It is simply use are of triangle is ∆=(1/2)cbSinA=(1/2)abSinB
Therefore ∆^2 = (1/4) acb^2 SinA.SinB
Also ∆=(1/2)acSinC
Dividing both we get Area ∆ =( c^2 SinA.SinB)/2 Sin C
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3317531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this geometric series question? A geometric series has first term 4 and common ratio r, where 0< r<1. The first, second, and fourth terms of this geometric series form three successive terms of an arithmetic series. Find the sum to infinity of the geometric series.
$S = \frac{t_1}{1-r}$ is the sum to infin... | The question says the first term is $4$, so $t_1 = 4$. Thus, since $0 \lt r \lt 1$, you have $r = \frac{-1 + \sqrt{5}}{2}$ giving
$$S = \frac{t_1}{1-r} = \frac{4}{1 - \frac{-1 + \sqrt{5}}{2}} = \frac{8}{2 - (-1 + \sqrt{5})} = \frac{8}{3 - \sqrt{5}} = 2\left(3 + \sqrt{5}\right) \tag{1}\label{eq1}$$
Thanks to J. W. Tanne... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3318689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find equations of the two lines through the origin that make an angle $\tan^{-1}(1/2)$ with $3y=2x$
If two straight lines pass through the origin and makes an angle $\tan^{-1}(1/2)$ with $3y=2x$, then find its equations.
Let $m$ be the gradient of the line then, $$\frac{1}{2}=\frac{m-2/3}{1-2m/3}$$ I don't know wheth... | $tan^{-1}(\frac{1}{2}) = \theta$
$tan(\theta) = \frac{1}{2}$
Line $3y = 2x$ have the slope of $\frac{2}{3}$, suppose this line is called a and the two lines in question is b and c (with $m_{b} > m_{a}, m_{c} < m_{a})$.
So $m_{a} = \frac{2}{3}$ $=>$ $tan(A) = \frac{2}{3}$, $m_{b} = tan(B)$, and $m_{c} = tan{C}$.
Hint :
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $ \frac{ \sqrt[3]{16} - 1}{ \sqrt[3]{27} + \sqrt[3]{4} + \sqrt[3]{2}} $ Simplify
$$ \frac{ \sqrt[3]{16} - 1}{ \sqrt[3]{27} + \sqrt[3]{4} + \sqrt[3]{2}} $$
Attempt:
$$ \frac{ \sqrt[3]{16} - 1}{3 + \sqrt[3]{4} + \sqrt[3]{2}} = \frac{ \sqrt[3]{16} - 1}{ (3 + \sqrt[3]{4}) + \sqrt[3]{2}} \times \frac{ (3 + \sqrt[... | Let $x=\sqrt[3]{2}$ then we have $\frac{x^4-1}{x^2+x+3}$
Multiply by $x-1$
$\frac{x^4-1}{x^2+x+3}\cdot\frac{x-1}{x-1}=\frac{(x^4-1)(x-1)}{x^3+2x-3}=\frac{(x^4-1)(x-1)}{2+2x-3}=\frac{(x^4-1)(x-1)}{2x-1}$
Multiply by $4x^2+2x+1$
$\frac{(x^4-1)(x-1)(4x^2+2x+1)}{(2x-1)(4x^2+2x+1)}=\frac{(x^4-1)(x-1)(4x^2+2x+1)}{8x^3-1}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Convert standard form to factor form Im so confused now
I have no idea how to factor a standard form
i have example shown below
Standard form
$$x^4+6x^3-x^2-6x$$
Factor form
$$x(x+6)(x+1)(x-1)$$
Can you explain a little bit
I have no idea
Thank you
| $$x^4+6x^3-x^2-6x=x^4-x^2+6x^3-6x=x^2(x^2-1)+6x(x^2-1)=$$ $$(x^2-1)x(x+6)=(x-1)(x+1) x(x+6)=x(x-1)(x+1)(x+6).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3321930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Solution verification on homework problem. Separable first order ODE IVP. The answer is supposedly $y^2 = 1 + \sqrt{x^2 - 16}$ I don't know where I went wrong cause I know for a fact that my substitution of $x = 4 \sec(\theta)$ is correct. I know for a fact that after substitution the integral becomes $\int \sec^2(\the... | In integrating the RHS of in$$2y \frac{dy}{dx} = \frac{x}{\sqrt{x^2 - 16}} \space \space \space y(5)=2$$
it is advised to let $$u=x^2-16$$ and you have $$du=2xdx$$
Then you do not have to worry about trig substitution at all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3325621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Calculate how many ways you can give $7$ children $7$ identical candies My try:
Calculate how many ways you can give $7$ children $7$ identical candies if each child got at most 2 candies.
$$x_1+x_2+x_3+x_4+x_5+x_6+x_7=7 \text{ for } x_i \in \left\{ 0,1,2\right\}$$
$$[t^7](1+t+t^2)^7[t^7](\frac{1-t^3}{1-t})^7=[t... | The solution will be confficent of $\,t^7 $ in the
expension of $${(1\, - \ t^3)}^7{(1 - t)}^{-7}$$
I.e coefficient of $t^7$ in $( \,1 - \, {7 \choose 1} t^3\,+{7 \choose 2} t^6)(1\,-t)^{-7}$ $$
= \,{13 \choose 7} \, - 7\,{10 \choose 4}\,+21{7 \choose 1} $$
= 393
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3326485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Angles between vectors of center of two incircles I have two two incircle between rectangle and two
quadrilateral circlein. It's possible to determine exact value of $\phi,$ angles between vectors of center of two circles.
|
Let $|AB|=|CD|=a$,$|BC|=|AD|=b$,
$|O_1E|=r_1$,
$|O_2F|=r_2$,
$\angle O_1AO_2=\phi$,
$\angle O_1AF=\alpha$
$\angle O_2AF=\beta$.
Then
\begin{align}
\tan\alpha&=\frac{b-r_1}{a/2}
\tag{1}\label{1}
,\\
\sin\alpha&=\frac{b-r_1}{b+r_1}
,\\
\tan\alpha&=\frac{\sin\alpha}{\sqrt{1-\sin^2\alpha}}
=\frac{b-r_1}{b+r_1}
\left/
\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Providing divisibility condition given fraction identity If $x,y,z$ are positive integers satisfying
$$\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$$
prove that $20{\,\mid\,}xy$.
My work:
Expanding, we find
$$(xz)^2+(yz)^2=(xy)^2$$
I know the Pythagorean triple formula and I tried applying that, but I couldn't find ... | Suppose $x,y,z$ are positive integers such that
$$\frac{1}{x^2}+\frac{1}{y^2}=\frac{1}{z^2}$$
Equivalently, $x,y,z$ are positive integers such that
$$\qquad\qquad\; x^2y^2=z^2(x^2+y^2)\qquad(\textbf{eq})$$
Aqua has already shown that $5{\,\mid\,}(xy)$.
To show that $4{\,\mid\,}(xy)$, we can argue as follows . . .
If ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3327870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Limit of a function in which square roots are involved $$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}$$
Could someone please help me solve this problem.
I tried multiplying by a unity factor but I end up stuck.
| You can simplify it:
$$\lim_{x \to 0}\frac{x+2-\sqrt{2x+4}}{3x-1+\sqrt{x+1}}=\lim_{x \to 0}\frac{x+2-\sqrt{2}\sqrt{x+2}}{3(x+1)+\sqrt{x+1}-4}=\\
\lim_{x \to 0}\frac{\sqrt{x+2}(\sqrt{x+2}-\sqrt{2})}{3(\sqrt{x+1}-1)(\sqrt{x+1}+\frac43)}=\\
\lim_{x \to 0}\frac{\sqrt{x+2}-\sqrt{2}}{\sqrt{x+1}-1}\cdot \lim_\limits{x\to 0}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Use properties of determinant and show
Let $n$ be a positive integer and
\begin{align} M =
\begin{pmatrix}
n! & (n+1)! & (n+2)! \\
(n+1)! &(n+2)! & (n+3)! \\
(n+2)! & (n+3)! & (n+4)! \\
\end{pmatrix}
\end{align}
Use properties of determinant to show that \begin{align}\left(\frac{|M|}{(n!)^3}- 4\ri... | After you take the common factor $n!$ out of all columns, take $(n+1)$ and $(n+1)(n+2)$ out of 2nd and 3rd columns, respectively:
$$\frac{|M|}{(n!)^3}=\left|\begin{array}{ccc}
1 & \color{red}{n+1} & \color{blue}{(n+1)(n+2)}\\
n+1 & (\color{red}{n+1})(n+2) & \small{\color{blue}{(n+1)(n+2)}(n+3)}\\
(n+1)(n+2) & \small{(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3328655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the minimum $n$ such that $x^2+7=\sum_{k=1}^n f_k(x)^2$ where $f_k(x)\in \mathbb{Q}[x]$ Recently, I have found this problem:
Find the minimum $n \in N$ such that $x^2+7=f_1(x)^2+f_2(x)^2+\cdots+f_n(x)^2$ where $f_1(x),+f_2(x),+\cdots+f_n(x)$ are polynomials with rational coefficients.
I have tried to solve this... | The answer of @Mindlack contains a full solution to the problem. This thread just collects some partial results I got while studying the problem, using only the idea from @Mindlack that $7$ is not the sum of $3$ or less squares of rationals.
Partial results
Notice we can only use polynomials with degree at most $1$, si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Finding all real solutions of $x-8\sqrt{x}+7=0$ Finding all real solutions of $x-8\sqrt{x}+7=0$.
Man, I tried subtituting $x=y^2$ but IDK things got complicated. What is the best way to figure this out? Thanks!
| If
$x - 8\sqrt x + 7 = 0, \tag 1$
then
$x + 7 = 8\sqrt x; \tag 2$
then
$x^2 + 14 x + 49 = (x + 7)^2 = (8\sqrt x)^2 = 64x; \tag 3$
thus
$x^2 - 50x + 49 = 0, \tag 4$
which factors as
$(x - 1)(x - 49) =
x^2 - 50x + 49 = 0; \tag 5$
thus
$x = 1 \; \text{or} \; x = 49; \tag 6$
it is now a simple matter to check that $1$, $4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3330413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 2
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If $x>4$, what is the minimum value of $\frac{x^4}{(x-4)^2}$. If $x>4$, what is the minimum value of $\frac {x^4}{(x-4)^2}$ ?
I have tried using AM-GM Inequality here by letting $y=x-4$ and ended up getting $224$ but that does not seem to be the correct answer. I find out by trial and error that the minimum value is wh... | Let $p=x-4>0$. By AM-GM inequality,
$$
\frac{x^4}{(x-4)^2}
=\frac{(p+4)^4}{p^2}
\ge\frac{\left(2\sqrt{4p\,}\right)^4}{p^2}
=256
$$
and equality holds when $p=4$ or $x=8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3332584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$. If $x$ and $y$ are real numbers such that $4x^2 + y^2 = 4x - 2y + 7$, find the maximum value of $5x+6y$.
I did a little bit of manipulation and got $4(x+1)(x-2) + (y+1)^2=0$. I then got $x=2$ and $y=-1$ which means ... | Let $5x+6y=k$.
Thus, $y=\frac{k-5x}{6}$ and the following equation has real roots:
$$4x^2+\left(\frac{k-5x}{6}\right)^2=7+4x-2\cdot\frac{k-5x}{6}$$ or
$$165x^2-2(5k+102)x+k^2+12k-252=0,$$ which gives
$$(5k+102)^2-169(k^2+12k-252)\geq0$$ or
$$(k-16)(k+23)\leq0$$ or
$$-23\leq k\leq16.$$
For $k=16$ the equality occurs for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
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Solve the equation: $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ A question asks
Solve the equation: $y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$ for positive integers $x$ and $y$.
I tried factoring the LHS by adding $1$ to both sides so we get $(y+1)^3$ in the LHS. But I couldn't get any factorisation for the RHS, neither... | Given:
$$y^3 + 3y^2 + 3y = x^3 + 5x^2 - 19x + 20$$
Let $K(x,y)$ be:
$$K(x,y)=y^3 + 3y^2 + 3y - (x^3 + 5x^2 - 19x + 20)$$
We could write this as:
$$K(x,y)=f(y)-f(x)+g(x)$$
If a root $r$ exists for $g$, then, we would have:
$$K(r,r)=f(r)-f(r)+g(r) $$
We can find g(x) such that:
$$g(x)=K(x,y)-f(y)+f(x) \tag1$$
By inspect... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Calculate $\int\sqrt{(x^2+1)}dx$ Calculate $I$ =$\int\sqrt{(x^2+1)}dx$
I have tried calculating it using integration by parts:
$$f'(x) = 1, f(x) = x$$
$$g(x) = \sqrt{x^2+1}, g'(x) = \frac{x}{\sqrt{x^2+1}}$$
$$\int\sqrt{x^2+1}dx = x\sqrt{x^2+1} - \int\frac{x}{\sqrt{x^2+1}}$$
Then I make the substitution:
$$x^2+1 = u$$
$... | Your IBP should have given $$\int\sqrt{x^2+1}dx=x\sqrt{x^2+1}-\int\frac{x^2dx}{\sqrt{x^2+1}}=x\sqrt{x^2+1}-\int\sqrt{x^2+1}dx+\int\frac{dx}{\sqrt{x^2+1}}.$$Rearranging,$$\int\sqrt{x^2+1}dx=\frac{x\sqrt{x^2+1}+\int\frac{dx}{\sqrt{x^2+1}}}{2}.$$For the last integral, use $x=\tan t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3333556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Finding the sum of the first term and common difference when given sum of first 5 and sum of first 10 the sum of the first 5 terms of an arithmetic series is 110 and the sum of the first 10 terms is 320. How do i go about finding the first term and common difference.
Sn = n/2 [2a+d(n−1)] is the equation for working ou... | $\begin{array}{c| cccccccccccccc}
\text{index} & 1 & 2 & 3 & 4 & 5\\
\text{term} & a & a+d & a+2d & a+3d & a+4d \\
\text{sum} & a & 2a+d & 3a+3d & 4a+6d & \color{red}{5a+10d=110} \\
\text{formula} &&&&& 5\dfrac{(a)+(a+4d)}{2}
\end{array}$
$\begin{array}{c| cccccccccccccc}
\text{index} & 6 & 7 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3334283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Find the sum of the number of all continuous runs of all possible sequences with $2019$ ones and $2019$ zeros I recently had a test that is quite difficult because it is for selecting people to participate important mathematics competition. It has 6 questions like IMO, and the last question is pretty hard. Although I h... | Number of Arrangements with $\boldsymbol{k}$ Runs
Using Stars and Bars, The number of ways to get a sum of $n$ with $k$ positive numbers is $\binom{n-1}{k-1}$.
The number of arrangements with $k$ runs is twice the number of ways (one starting with $0$ and one starting with $1$) to get a sum of $n$ with $\left\lfloor\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3335716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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What is the 94th term of this sequence? $1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,\ldots$
What is the 94th term of the following sequence?
$$1,1,2,2,2,2,3,3,3,3,3,3,4,4,4,4,4,4,4,4,\ldots$$
*
*8
*9
*10
*11
My Attempt: I found that the answer is 3rd option i.e. 94th term is 10. As every number is wr... | Look at the last numbers of the repeating numbers. Notice the arithmetic sequence: $2,4,6,...,2+2(n-1)$, whose sum of $n$ terms is: $S_n=(n+1)n$. So, the general formula is: $a_{S_n}=n$. For example:
$$a_{S_\color{red}1}=a_2=\color{red}1\\
a_{S_\color{red}2}=a_6=\color{red}2\\
\vdots\\
a_{S_n}=a_{94}=?$$
We make up the... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Solve $\sin x + \cos x = \sqrt{1+k}$ for $\sin 2x$, $\sin x-\cos x$, and $\tan x$ in terms of $k$ Given that $\sin x + \cos x = \sqrt{1+k}$, $-1 \le k \le 1$
*
*Find the value of $\sin 2x$ in terms of k
*Given that $x \in (45^{\circ}, 90^{\circ})$ deduce that $\sin x - \cos x = \sqrt{1-k}$
*Hence, show that $\tan... | Try this for the second one,
$$ (\sin x- \cos x)^2= (\sin x+\cos x)^2 -2\sin 2x$$
For the third one, solve the equations below for $\sin x$ and $\cos x$,
$$ \sin x + \cos x = \sqrt{1+k}$$
$$ \sin x- \cos x = \sqrt{1-k}$$
and then plug them into
$$\tan x= \frac{\sin x}{\cos x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3340670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
A matrix of order 8 over $\mathbb{F}_3$ What is an example of an invertible matrix of size 2x2 with coefficients in $\mathbb{F}_3$ that has exact order 8?
I have found by computation that the condition that the 8th power of a matrix
$\begin{bmatrix}a & b\\c & d\end{bmatrix}$ is the identity is
$$
b c (a + d)^2 (a^2 +... | If $A$ satisfies $A^8 = I$, then all of the eigenvalues $\lambda$ of $A$ satisfy $\lambda^8 = 1$. Rearranging and factoring gives
$$0 = \lambda^8 - 1 = (\lambda^4 + 1)(\lambda^4 - 1) .$$ Since we want $A^4 \neq I$ (otherwise the order of $A$ divides $4$), at least one eigenvalue of $A$ must be a solution of $\lambda^4 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3343777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 3
} |
Difficult diophantine equation Solve for integers:
$4n^4+7n^2+3n+6=m^3.$
Hi this is a problem from an Bulgarian olympiad for which I have no idea how to solve.
I figured out using wolfram alpha that $16\cdot m^3-47$ must be a square number.
I would appreciate any solutions. Thank you in advance!
| The idea is going modulo $9$.
Indeed, we will prove that $4n^4 + 7n^2+3n+6$ leaves only remainders $2,5,6$ modulo $9$. None of these are cubes modulo $9$(only $0,1,8$ are), completing the proof that no such integers $n,m$ exist.
For this, we note that if $n \equiv 0 \pmod{3}$ then $4n^4 + 7n^2+3n+6 \equiv 6\pmod{9}$.
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve the following system of equations: $(x-1)^3+2016(x-1)=-1$ and $(y-1)^3+2016(y-1)=1$ This problem came up in my homework for an online class and try as I might, I can't find a solution.
$(x-1)^3+2016(x-1)=-1$
$(y-1)^3+2016(y-1)=1$
Find $x+y$
So far, I've tried letting $(x-1)=a$ and $(y-1)=b$ and then adding the t... | From your point you have that or $a+b=0$ or $a^2-ab+b^2+2016=0$. But $a^2-ab+b^2+2016=(a-\frac b2)^2 +b^2-\frac{b^2}4+2016>0\forall a,b\in\Bbb R$. So you have $a+b=0\implies x+y=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3347540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For each positive integer $k$, there is a positive integer $m\neq 3k$ such that $\varphi(3k)=\varphi(m)$ (with $\varphi(n)$ being Euler's totient)
Prove that for every positive integer $k$, $\exists m\neq 3k, m\in\mathbb{Z}^+$ such that $\varphi(3k)=\varphi(m)$. Here $\varphi(n)$ is Euler's totient function.
I am int... | Partial answer. Euler's totient function is multiplicative, i.e. if $\gcd(m,n)=1$ then $\varphi(m\cdot n)=\varphi(m)\cdot \varphi(n)$.
Let's consider $\gcd(6, k)=1$, then $\gcd(2^2,k)=1$ and $\gcd(3,k)=1$. And:
$$\varphi(3\cdot k)=
\varphi(3)\cdot \varphi(k)=
2\cdot \varphi(k)$$
$$\varphi(2^2\cdot k)=
\varphi(2^2)\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3349893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$. Find the remainder when you divide $x^{81} + x^{49} + x^{25} + x^{9} + x$ by $x^{3}-x$.
Attempt:
Let $f(x) = x^{81} + x^{49} + x^{25} + x^{9} + x$, then the remainder of $f(x)$ divided by $(x-1)$ is $f(1)$, divided by $(x+1)$ is $f... | Hint
$$x^{81} + x^{49} + x^{25} + x^{9} + x=(x^{3}-x)Q(x)+R(x)$$where $$R(x)=ax^2+bx+c$$and $$R(0)=0\\R(1)=5\\R(-1)=-5$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of a sequence depending on a parameter The problem given by our teacher during the class goes as follows: Determine the limit of a sequence $a_n$ depending on the parameter $p$ if $a_n=\sqrt{4n^2+3n+5}-pn-1$.
My attempt of solution:
$$\lim{(\sqrt{4n^2+3n+5}-pn-1)}=\lim{n\left(\sqrt{4+\frac{3}{n}+\frac{5}{n^2}}-p-... | Let's see: $$\lim{(\sqrt{4n^2+3n+5}-pn-1)}=\lim{\frac{(\sqrt{4n^2+3n+5}-pn-1)(\sqrt{4n^2+3n+5}+pn+1)}{\sqrt{4n^2+3n+5}+pn+1}}=\lim{\frac{4n^2+3n+5-(pn+1)^2}{\sqrt{4n^2+3n+5}+pn+1}}=\lim{\frac{n^2(4-p^2)+n(3-2p)+4}{\sqrt{4n^2+3n+5}+pn+1}}=\lim{\frac{n(4-p^2)+3-2p+\frac{4}{n}}{\sqrt{4+\frac{3}{n}+\frac{5}{n^2}}+p+\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3354263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $(x+1)(y+1)(z+1)=144$ in primes "Solve $(x+1)(y+1)(z+1)=144$ in primes".
So far, I have concluded that the solutions are $(x,y,z)=(2,3,11)$ or $(2,5,7)$ and their permutations. I worked like this:
*
*$x \equiv 0\mod 2\Rightarrow x+1=3, 144=2^4*3^2 \Leftrightarrow (y+1)(z+1)=48=2^4*3$
*
*$y \equiv 0\mod 2\Ri... | Assume x <= y <= z. The factors can be 3, 4, 6, 8, 12, 18, 24 and 72 (primes where p+1 divides 144). The smallest factor cannot be 6 or larger since 6^3 >= 216, so we get 3x48 or 4x36. 48 = 4x12 or 6x8, 36 = 6x6. So the solutions (x, y, z) are (2, 3, 11), (2, 5, 7) and (3, 5, 5).
And of course all the permutations.
Y... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Homotopy in $U_{2}$ Let $\mathbb{S}^{1}$ be the unit circle and $U_{2}\subset M_{2\times 2}(\mathbb{C})$ be the unitary group. Let $f,g:\mathbb{S}^{1}\rightarrow U_{2}$ be maps defined by
\begin{align}
f(x)=
\begin{pmatrix}
x & 0\\ 0 & 1
\end{pmatrix}
\quad &\text{and} \quad
g(x)=
\begin{pmatrix}
1 & 0\\ 0 & x
\end{pma... | Your homotopy will not work. Instead, follow the hint: $U_2$ is path-connected and the identity matrix can be connected to the matrix
$$
\begin{pmatrix}
0 & 1\\ 1 & 0
\end{pmatrix}
$$
by a certain path $M_t, t\in [0,1]$, in $U_2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find the projection matrix that projects every vector of $\mathbb{R}^3$ on to the plane of 2 vectors The two vectors that are mentioned above are $$\begin{pmatrix}1 \\1 \\0\end{pmatrix}$$ $$\begin{pmatrix}0 \\ 1 \\1 \end{pmatrix}$$
I am not really sure what i am doing but i tried finding the projection matrix by puttin... | You must have made a mistake, as can be seen by checking $P\begin{pmatrix}1\\1\\0\end{pmatrix}$.
Using your approach, $ A =\begin{pmatrix}1 & 0 \\1 &1 \\0&1\end{pmatrix},$ $A^T=\begin{pmatrix}1 &1 &0 \\0 &1 &1 \end{pmatrix},$
$ A^TA=\begin{pmatrix}2&1\\1&2\end{pmatrix}$, $(A^TA)^{-1}=\dfrac13\begin{pmatrix}2 & -1 \\-... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What did I get wrong when solving $\int\frac{\sqrt{x^2-1}}{x^4}dx$? I'm not sure that this is the problem, but I think I may not know how to find the $\theta$ value when solving an integral problem with trigonometric substitution.
I got $\frac{\sin^3(\sec^{-1}(x))}{3}+C$ for the answer, but the answer should be, $\fra... | You've got the right answer, you've just missed that $$\sin(\sec^{-1}(x))=\sin(\cos^{-1}(1/x))=\sqrt{1-1/x^2}=\frac{1}{x}\sqrt{x^2-1}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Expansion $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$
Expand $(x+{\frac{1}{x}})^4 (x-{\frac{1}{x}})^2$
Right now, I am able to expand this expression by simplifying it to:
$\frac{(x^2+1)^4 (x^2-1)^2}{x^6}$
I used the formula $(a+b)^2$ and $(a-b)^2$ a bunch of times to arrive at the answer. But, is there any simpler/sma... | The following procedure, I believe, is somewhat simpler
$$\begin{equation}\begin{aligned}
\left(x+{\frac{1}{x}}\right)^4 \left(x-{\frac{1}{x}}\right)^2 & = \left(x+{\frac{1}{x}}\right)^2\left(x+{\frac{1}{x}}\right)^2\left(x-{\frac{1}{x}}\right)^2 \\
& = \left(x+{\frac{1}{x}}\right)^2\left(\left(x+{\frac{1}{x}}\right)\l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate this line integral?(complex plane)
There are counter-clockwise two curves, $C_1$ and $C_2$ on complex plane.
$C_1$ ; ${1 \over 3} \leq \vert z \vert \leq 2$ (on the 1st Quadrant.)
$C_2$ ; ${1 \over 3} \leq \vert z \vert \leq 2$ (on the 4th Quadrant)
Calculate $\int_{C_1} {z^3+1 \over z^4 + 4z -1} dz$ ... | The denominator has one root close to $\frac14$ outside the region and a triple of roots close to the roots of $z^3+4=0$, where the conjugate pair $\frac{1\pm i\sqrt3}2\sqrt[3]4$ has one root inside $C_1$ and $C_2$, respectively.
Let $\zeta$ be the root inside $C_1$. Then the integral of $\frac{P(z)}{Q(z)}$ over the ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Help with unsolved indefinite integral ! $\int \!r\ln \left( r \right) \sqrt {ar+{b}^{2}+{r}^{2}}\,{\rm d}r$ I really need to solve this indefinite integral:
$\int \!r\ln \left( r \right) \sqrt {ar+{b}^{2}+{r}^{2}}\,{\rm d}r$
It seems much more complicated than it looks. I have found a integral table with
a integral ... | The integral can be constructed from the following generic cases
$$
\int x \log(x)\sqrt{a^2+x^2} dx=
\frac{1}{3} \left(-\frac{1}{3} \sqrt{a^2+x^2} \left(4 a^2+x^2\right)-a^3 \log(x)+\left(a^2+x^2\right)^{3/2} \log(x)+a^3 \log\left[a
\left(a+\sqrt{a^2+x^2}\right)\right]\right)
$$
$$
\int x \log(x+b)\sqrt{a^2+x^2}dx
=\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is $\operatorname{Var}(X^2) \ge\operatorname{Var}(X − \mu)^2 + \mu^2)$? I wonder why $\operatorname{Var}(X^2) \ge \operatorname{Var}((X − \mu)^2 + \mu^2)$.
IN ADDITION
What factor is the variance of $(X − \mu)^2 + \mu^2$ smaller than the variance of $X^2$?
What I've done so far:
$$\operatorname{Var}((X − \mu)^2 + \... | The result is only conditionally true (I assume you mean $\mu = E(X)$). It is valid assuming that either $E(X) = 0$, $X\ge 0$ or $X\le 0$. If no condition is given, it does not always hold, and we prove this by a simple counterexample below.
$\operatorname{Var}X = E(X^2) - E(X)^2$. Sub in
\begin{align} \mathbf \Delta(X... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3367500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Definite Integral $\int{x^2+1 \over x^4+1}$ Evaluate
$$\int_0^{\infty}{x^2+1 \over x^4+1}$$
I tried using Integration by parts ,
$$\frac{{x^3 \over 3 }+x}{x^4+1}+\int\frac{{x^3 \over 3 }+x}{(x^4+1)^2}.4x^3.dx$$
First term is zero
But it got me no where.
Any hints.
| Another approach: Separate out the integral into
$$
I = \int_0^\infty \frac{x^2}{x^4 + 1}\:dx + \int_0^\infty \frac{1}{x^4 + 1}\:dx
$$
Both take the form of known integral
$$
\int_0^\infty \frac{x^k}{\left(x^n + a\right)^m}\:dx = a^{\frac{k +1}{n} -m}\frac{\Gamma\left(m - \frac{k + 1}{n}\right)\Gamma\left(\frac{k + 1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Indefinite integral of rational expression involving cubic polynomials I was given the following exercise:
$$\int \frac{x^3}{(x^2+1)^3}dx$$
As a tip, my professor suggested using the following substitution: $t=x^2+1$.
Notice that if $t=x^2+1$, then $x^2=t-1$ and therefore $x=(t-1)^\frac{1}{2}$. Then
$$x^3=x^2\cdot... | You forgot that, since $x=(t-1)^{\frac{1}{2}}$, we have also
$$dx=D((t-1)^{\frac{1}{2}})\, dt=\frac{1}{2}(t-1)^{-\frac{1}{2}}\, dt.$$
Therefore, we find
$$\int \frac{x^3}{(x^2+1)^3}dx=\int \frac{(t-1)^\frac{3}{2}}{t^3}\cdot
\color{blue}{\frac{1}{2}(t-1)^{-\frac{1}{2}}}\, dt=
\frac{1}{2}\int \frac{t-1}{t^3}\, dt=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3371713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all real matrices such that $X^{3}-4X^{2}+5X=\begin{pmatrix} 10 & 20 \\ 5 & 10 \end{pmatrix}$ The following question come from the 1998 Romanian Mathematical Competition:
Find all matrices in $M_2(\mathbb R)$ such that $$X^{3}-4X^{2}+5X=\begin{pmatrix} 10 & 20 \\ 5 & 10 \end{pmatrix}$$
Can you guy please help me... | dan_fulea mentioned in another answer that the contestants are not expected to know diagonalisation or Jordan form. So, I will give a more elementary solution below. Let
$$
A=uv^T=\pmatrix{2\\ 1}\pmatrix{1&2}.
$$
The equation in question is equivalent to
$$
X^3-4X^2+5X=5A.\tag{1}
$$
One can easily verify that $A^2=4A$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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If $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, maximise $x+y$. The question is: if $x$ is the remainder when a multiple of $4$ is divided by $6$, and $y$ is the remainder when a multiple of $2$ is divided by $3$, what is the greatest ... | Take any integer $m$ and divide it by $6$. The possible remainders are $0,1,2,3,4,5$. Therefore, if we take $4m$ and divide it by $6$ then the possible remainders are $0,4,2,0,4,2$. The maximum remainder when a multiple of $4$ is divided by $6$ is $4$.
Similarly, if you divide an integer $n$ by $3$ the possible remain... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Proof that $3^{10^n}\equiv 1\pmod{10^n},\, n\ge 2$ This should be rather straightforward, but the goal is to prove that
$$3^{10^n}\equiv 1\pmod{10^n},\, n\ge 2.$$
A possibility is to use
$$\begin{align*}3^{10^{n+1}}-1&=\left(3^{10^n}-1\right)\left(1+\sum_{k=1}^9 3^{10^n k}\right)\\&=\left(3^{10^n}-1\right)\left(3^{9\cd... | Hint:
To get to your goal $3^{10^n}\equiv 1\pmod{10^n},\, n\ge 2,$
can you prove using binomial expansion with $9=10-1$ that $9^{10^{n-1}}\equiv1\mod 10^n $?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3373359",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Prove $\frac{1}{n} \sum^{n}_{k = 1} \left(1 + \frac{1}{k} \right) \geq (n + 1)^{\frac{1}{n}}$ Prove for every $n \in \mathbb{N}$ with $n \geq 1$ that
$$\frac{1}{n} \sum^{n}_{k = 1} \left(1 + \frac{1}{k} \right) \geq (n + 1)^{\frac{1}{n}}$$
What I've done so far:
Since $n \in \mathbb{N}_{\geq 1}$, it follows that $1 + ... | $$\frac{1}{n} \sum^{n}_{k = 1} (1 + \frac{1}{k})\geq [(1 + \frac{1}{1})(1 + \frac{1}{2}) \cdot \cdot \cdot (1 + \frac{1}{n})]^{\frac{1}{n}}$$ $$= [\frac{2 \cdot 3 \cdot \cdot \cdot (n+1)}{n!}]^{\frac{1}{n}}=(n+1)^{\frac{1}{n}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3376162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Inequality with a rational function Let $0< a_1\leq a_2\leq a_3\leq a_4\leq a_5$. Define the function $$f(x,y,z)=\cfrac{(a_1(x+y+z)^3+a_2(x)^3+a_3(x+y)^3+a_4(y+z)^3+a_5(z)^3 )^2}{(a_1(x+y+z)^2+a_2(x)^2+a_3(x+y)^2+a_4(y+z)^2+a_5(z)^2 )^3}.$$
Show that for all $x,y,z\in \mathbb R$ we have $f(x,y,z)\leq \cfrac{1}{a_1+a_2+... | For convenience, to simplify, take the sixth root of $f$ to give $$g(x,y,z)=\frac{(a_1(x+y+z)^3+a_2x^3+a_3(x+y)^3+a_4(y+z)^3+a_5z^3)^{1/3}}{(a_1(x+y+z)^2+a_2x^2+a_3(x+y)^2+a_4(y+z)^2+a_5z^2)^{1/2}}$$ which is the ratio of a weighted 3-norm to its 2-norm. Notice also that the function is homogeneous in $x,y,z$, so one c... | {
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Evaluating $\iint_R\big(x^2+y^2\big)\,dA$
Evaluate the following double integral:
$$\iint_R\big(x^2+y^2\big)\,dA,$$
where $R$ is the region given by plane $x^2+y^2\leq a^2$.
My attempts:
\begin{align}
\iint_{R}\big(x^2+y^2\big)\,dA
&=\int_{-a}^{a}\int_{-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}\big(x^2+y^2\big)\,dy\,dx\\... | Hint: Try $x=a\sin(t)$, but really polar coordinates would be the move here.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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What are the non-negative integers of this form? What are $( \frac{x^2-1-my}{x}, \frac{y^2-1-nx}{y} ) \in \mathbb{Z}^2_{\ge 0}$ for $(m,n) \in \mathbb{Z}^2_{\ge 0}$, $(x,y) \in \mathbb{R}^2_{\ge 1}$ and $mx+ny=xy$?
Some investigations:
*
*According to this answer of Ivan Neretin, assuming $x,y$ integers, the ... | A pair $(p,q) = ( \frac{x^2-1-my}{x}, \frac{y^2-1-nx}{y} )$ is achievable if and only if $p^2+q^2+4$ is the sum of two squares.
Proof. In one direction, we assume that the following three polynomials have a common zero:
\begin{split}
f(x,y)&:=x^2-1-my-px,\\
g(x,y)&:=y^2-1-nx-qy, \\
h(x,y)&:=xy-mx-ny.
\end{split} The... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3380735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\lim_{x \rightarrow 0}\frac{\tan x- x-\frac{x^3}{3}}{\sin^5x}$ How do I ind the limit of
$$\lim_{x \rightarrow 0}\frac{\tan x- x-\frac{x^3}{3}}{\sin^5x}$$
by L'Hopital's Rule?
Using Desmos, I get the answer that this limit evaluates to $\frac{3}{15}$, but I can't get that answer.
This is what I've done so fa... | As noted by А.Г., you can not split the limit unless both exist.
You can use L'Hospital right away and see how splittings are done:
$$\lim_{x \rightarrow 0}\frac{\tan x- x-\frac{x^3}{3}}{\sin^5x}\stackrel{LR}=
\lim_{x \rightarrow 0}\frac{\sec ^2x- 1-x^2}{5\sin^4x\cdot \cos x}=\\
\lim_{x \rightarrow 0}\frac{1-\cos^2x-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3382866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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I made an inequality to solve myself, but someone pointed out my solution is wrong. I made the following inequality for myself to solve, but my friend found out an mistake:
If $a,b,c\in\mathbb{R^{+}}$ and $abc=1$, prove that $$\frac{a}{b+1}+\frac{b}{c+1}+\frac{c}{a+1}\ge\frac32$$
I tried to substitute $a=\frac xy, b=... | The friend he told in the problem was me, yeah. And I will post my solution by using the Muirhead Inequality (and also AM-GM Inequality).
$$\dfrac{a}{b+1}+\dfrac{b}{c+1}+\dfrac{c}{a+1}\ge\dfrac{3}{2} \\ \Updownarrow \\ a\left(a+1\right)\left(c+1\right)+b\left(b+1\right)\left(a+1\right)+c\left(c+1\right)\left(b+1\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3383742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\left ( 1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}<2$ Prove that $$\left ( 1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}<2 \tag{1} $$ $\forall$ $n \gt 1$
I tried using Induction:
For th... | Since $f(x)=x^{\frac{1}{n}}$ is a concave function for $n>1$, by Jensen we obtain:
$$\left (1+\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}+\left ( 1-\frac{n^{\frac{1}{n}}}{n} \right )^\frac{1}{n}\leq2\left (\frac{1+\frac{n^{\frac{1}{n}}}{n}+ 1-\frac{n^{\frac{1}{n}}}{n}}{2}\right )^\frac{1}{n}=2.$$
The equality occurs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3383865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
In how many ways can 10 blankets be given to 3 beggars such that each recieves at least one blanket? The question was to find the number of ways in which 10 identical blankets can be given to 3 beggars such that each receives at least 1 blanket. So I thought about trying the multinomial theorem...this is the first time... | First give one blanket to each recipient.
Now distribute the remaining $7$ blankets to $3$ recipients without constraints: $36$
If one person gets $7$ (additional) and the others get none, there are $3$ ways to do that. And likewise for the following distributions:
*
*$700$, $3$ ways
*$610$, $6$ ways
*$520$, $6$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3385830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Show that $x^4-20200y^2=1$ has no solution in postive integers Show that $x^4-20200y^2=1$ has no solution in postive integers.
This topic is a question for the Chinese middle school students' mathematics competition today, so I think this problem has a simple solution.
| We have to solve
$$
x^4-20200y^2=1\tag 1
$$
Observe that $20200=6^2+142^2$. Then (1) becomes
$$
x^4=(6y)^2+(142y)^2+1.\tag 2
$$
Clearly $x=2k+1$ is odd. Hence $x^2=8T+1$,$T=\frac{k(k+1)}{2}$. Hence $x^4=8T_1+1=16T(4T+1)+1\Rightarrow T_1=2T(4T+1)=\frac{4T(4T+1)}{2}$. In general holds the following
THEOREM.
If $x$ is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 2,
"answer_id": 1
} |
Find the range of values of $k$ for which $kx^2 + 8x + k <6$ for all real values of $k$
Find the range of values of $k$ for which $kx^2 + 8x + k <6 $ for all real values of $k$.
I'm unsure if the discriminant must be greater than zero or less than zero.
My working steps: \begin{align}b^2 - 4ac = (8)^2 - 4(-2)(17-k) &... | In order for a quadratic function to always be less than a constant, the leading coefficient must be negative. We can then attempt to complete the square in such a way that the maximum value of the function is $6$, then apply the appropriate inequalities to $k$.
\begin{align}
kx^2+8x+k&=k(x^2+\frac8kx)+k\\
&=k(x^2+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3390858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
For $a,b,c \in \mathbb{R}$ and $|a| \geq |b+c|, |b| \geq |c+a|, |c| \geq |a+b|$ prove then $a+b+c = 0$
For $a,b,c \in \mathbb{R}$ and $|a| \geq |b+c|, |b| \geq |c+a|, |c|
\geq |a+b|$ prove then $a+b+c = 0$
Solution:
$$
|a| + |b| + |c| \geq |b+c| + |c+a| + |a+b| \geq 0
$$
and
$$0 \leq |a+b+c|\leq |a+b| + |c| \leq |a... | hint: Square each of the $| ..|$ inequality and consider $f(a) = a^2 + (2b+2c)a + (b+c)^2$. Show that $f(a) \ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3392565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find limit of sequence $\lim_{n\to\infty} \frac{n}{\sqrt{n^2+n+1}} $ I need help to find the limit:
$$\lim_{n\to\infty} \frac{n}{\sqrt{n^2+n+1}} $$
I've tried this:
$$\lim_{n\to\infty} \frac{\sqrt{n^2}}{\sqrt{n^2+n+1}} =\lim_{n\to\infty} \sqrt{\frac{n^2}{n^2+n+1}} = \lim_{n\to\infty} \sqrt{\frac{1}{1+\frac{1}{n}+\frac{... | Since
$n^2
\lt n^2+n+1
\lt (n+1)^2
=n^2+2n+1
$,
$n
\lt \sqrt{n^2+n+1}
\lt n+1
$
so
$1
\gt \dfrac{n}{\sqrt{n^2+n+1}}
\gt \dfrac{n}{n+1}
= 1-\dfrac1{n+1}
$.
Therefore
$\lim_{n\to\infty} \dfrac{n}{\sqrt{n^2+n+1}}
=1
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3393876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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.