Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to compute $\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$? $$\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx =\frac{\pi}{48\sqrt{3}}$$
I have difficulty to evaluating above integrals.
First I try the substitution $x^4 =t$ or $x^4 +x^2+1 =t$ but it makes integral worse.
Using Mathematica I fou... | A yet another approach:
\begin{align}
I&=\int_0^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx\\
&=\int_0^1 \frac{x^4}{(x^4+ x^2 +1)^3} dx+\int_1^\infty \frac{x^4}{(x^4+ x^2 +1)^3} dx\\
&=\int_0^1\frac{x^4}{(x^4+ x^2 +1)^3} dx+\int_0^1\frac{x^6}{(x^4+ x^2 +1)^3} dx\\
\end{align}
where I used $x \to 1/x$.
Now use $x\to \tan x$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 1
} |
Normal convergence: $\sum_{n = 1}^{\infty} \frac{x}{(1 + (n-1)x)(1 + nx)}$ I want to prove that:
$$\sum_{n = 1}^{\infty} \frac{x}{(1 + (n-1)x)(1 + nx)}$$
is not normally convergent on $[a, \infty)$ for fixed $a>0$.
Let $U_n(x)$ denote the general term.
We have:
$$||U_n||_{\infty} = \sup_{x \ge a} \left( \frac{x}{(1 + (... | Your first inequality is invalid, if you make the (positive) denominator smaller, since the numerator is also positive, you make the fraction larger.
Actually, the series is normally convergent on $[a,+\infty)$ for all $a > 0$, since for $n \geqslant 2$
$$0 < U_n(x) = \frac{x}{(1+(n-1)x)(1+nx)} < \frac{x}{(n-1)xnx} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Derivative of $f(x) = \frac{\cos{(x^2 - 1)}}{2x}$ Find the derivative of the function $$f(x) = \frac{\cos{(x^2 - 1)}}{2x}$$
This is my step-by-step solution: $$f'(x) = \frac{-\sin{(x^2 - 1)}2x - 2\cos{(x^2 -1)}}{4x^2} = \frac{2x\sin{(1 - x^2)} - 2\cos{(1 - x^2)}}{4x^2} = \frac{x\sin{(1 - x^2)} - \cos{(1 - x^2)}}{2x^2} ... | $$f(x) = \frac{\cos{(x^2 - 1)}}{2x}\Longrightarrow$$
$$\frac{df(x)}{dx}\left(\frac{\cos{(x^2 - 1)}}{2x}\right)=$$
$$\frac{1}{2}\left(\frac{df(x)}{dx}\left(\frac{\cos{(x^2 - 1)}}{x}\right)\right)=$$
$$\frac{1}{2}\left(\frac{x\frac{df(x)}{x}(\cos(1-x^2))-\cos(1-x^2)\frac{df(x)}{dx}x}{x^2}\right)=$$
$$\frac{x\frac{df(x)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Do the spaces spanned by the columns of the given matrices coincide? Reviewing linear algebra here.
Let $$A = \begin{pmatrix}
1 & 1 & 0 & 0 \\
1 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 1 & 1
\end{pmatrix} \qquad \text{and} \qquad
B = \begin{pmatrix}
1 & 0 & 0 \\
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1
\end{pmat... | There is a very elementary solution. By applying the column transformations $C_1\rightarrow C_1-C_2$ and $C_3 \rightarrow C_3-C_4$ to $A$ you get $$A^*=
\begin{bmatrix}
0 & 1 & 0 & 0 \\
0 & 1 & 0 & 0 \\
0 & 0 & 1 & 0 \\
0 & 0 & 0 & 1
\end{bmatrix}$$
If you look at the colums of $A^*$ and $B$ they ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1293755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Evaluate the double sum $\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}$ As a follow up of this nice question I am interested in
$$
S_1=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}
$$
Furthermore, I would be also very grateful for a solution to
$$
S_2=\sum_{m=1}^{\i... | Clearly, $S_1$=$S_2$ (this can be shown by reversing the order of summation, as was noted above).
Using
$$
\frac{ 1}{m n\left(m^2-n^2\right)^2}=\frac{ (m+n)^2-(m-n)^2}{4 m^2 n^2\left(m^2-n^2\right)^2}
$$
we get
$$
S_1=\sum_{m=1}^{\infty}\sum_{n=1}^{m-1}\frac{ 1}{m n\left(m^2-n^2\right)^2}=\frac{1}{4}\sum_{m=1}^{\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1294185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 0
} |
Calculating $\int_0^\pi \frac{1}{a+b\sin^2(x)} dx $ How do I calculate this integral? $a \gt b$ is given.
$$\int_0^\pi \frac{1}{a+b\sin^2(x)} dx $$
I am confused since WolframAlpha says one the one hand, that $F(\pi) = F(0) = 0 $ , but with some random values it isn't 0.
What am I missing?
Note that I am not really i... | Suppose we seek to evaluate
$$\frac{1}{2} \int_0^{2\pi} \frac{1}{a+b\sin^2 x} dx.$$
Put $z = \exp(ix)$ so that $dz = i\exp(ix) \; dx$ and hence
$\frac{dz}{iz} = dx$ to obtain
$$\frac{1}{2} \int_{|z|=1}
\frac{1}{a+b(z-1/z)^2/4/(-1)}\frac{dz}{iz}
\\ = \frac{1}{2} \int_{|z|=1}
\frac{4}{4a-b(z-1/z)^2}\frac{dz}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296351",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Show that $\frac {\sin x} {\cos 3x} + \frac {\sin 3x} {\cos 9x} + \frac {\sin 9x} {\cos 27x} = \frac 12 (\tan 27x - \tan x)$ The question asks to prove that -
$$\frac {\sin x} {\cos 3x} + \frac {\sin 3x} {\cos 9x} + \frac {\sin 9x} {\cos 27x} = \frac 12 (\tan 27x - \tan x) $$
I tried combining the first two or the la... | In general, we have
$$\sum_{k=1}^n \dfrac{\sin\left(3^{k-1}x \right)}{\cos\left(3^kx\right)} = \dfrac{\tan\left(3^nx\right)-\tan(x)}2$$
We can prove this using induction. $n=1$ is trivial. All we need to make use of is that
$$\tan(3t) = \dfrac{3\tan(t)-\tan^3(t)}{1-3\tan^2(t)}$$
For the inductive step, we have
$$\sum_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1299957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Proving a differential equation is a circle So, I have solved the differential equation, to find the general solution of:
$$\frac{y^2}{2} = 2x - \frac{x^2}{2} + c$$
I am told that is passes through the point $(4,2)$. Using this information, I found $C$ to be $2$. Then I am asked to prove that the equation forms a circ... | You have,
$$ y^2 +x^2 = 4x+4,$$
what we need to do is group the $x$'s together and complete the square,
$$ y^2 + \left( x^2-4x \right) = 4.$$
So we need to complete the square for the expression $x^2-4x$. The idea is to write this polynomial as a perfect square plus a constant. To do this we first consider the express... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to prove that the ring of all algebraic integers is a Bézout domain? I was told that the ring of all algebraic integers (that is, the complex numbers which are roots of a monic polynomials with integral coefficients) is a Bézout domain. But I have no idea how to prove it. Can anyone help me with this? Thanks in adv... | First, an example:
Consider the algebraic number field $\mathbb{Q}(\sqrt{-5})$. Its class number is two. That means there exist ideals of the ring of integers $\mathbb{Z}[\sqrt{-5}]$ of this field that are not principal, but the square of any ideal is principal. For instance, we have the decomposition
$$6 = 2 \cdot 3 =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Trigonometric relationship in a triangle If in a triangle $ABC$, $$3\sin^2B+4\sin A\sin B+16\sin^2A-8\sin B-20\sin A+9=0$$ find the angles of the triangle. I am unable to manipulate the given expression. Thanks.
| Let $X = \sin A$ and $Y = \sin B$. Then we have
$$16X^2 + 4XY + 3Y^2 - 20X - 8Y + 9 = 0$$
Rearranging the terms to solve for Y, we get
$$3Y^2 + (4X-8)Y + (16X^2 - 20X + 9) = 0$$
The discriminant for this quadratic equation becomes
$$(4X-8)^2 - 4*3*(16X^2-20X+9) = 16X^2 - 64X + 64 - 192X^2 + 240X - 108 = -176X^2 + 176X ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Matrices $\begin{pmatrix} a &b \\ c &d \end{pmatrix}\begin{pmatrix} x\\y \end{pmatrix}=k\begin{pmatrix}x\\y \end{pmatrix}$ If $\begin{pmatrix}
a &b \\
c &d
\end{pmatrix}\begin{pmatrix}
x\\y \end{pmatrix}=k\begin{pmatrix}x\\y \end{pmatrix}$, prove that $k$ satisfies the equation $k^2-(a+d)k+(ad-bc)=0$.
If the roots of... | $Ax=kx$
Roots of the given equations are the eigen-values of $A$.
So sum of roots = trace (A) = a+b
Product of roots = det (A)= ad-bc
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Showing that $x^3+2y^3+4z^3=2xyz$ has no integer solutions except $(0,0,0)$. Let $x,y,z\in \mathbb{Z}$ satisfy the equation:
$$
x^3+2y^3+4z^3=2xyz
$$
How do I prove that the only solution is $x=y=z=0$?
| We have $x^3 +2y^3+4z^3=2xyz$ so x is even and $8x_1^3+2y^3+4z^3=4x_1yz$ $\implies$
$4x_1^3+y^3+2z^3=2x_1yz$ so y is even and $4x_1^3+8y_1^3+2z^3=4x_1y_1z$
$\implies$ $2x_1^3+4y_1^3+z^3=2x_1y_1z$ so z is even and
$2x_1^3+4y_1^3+8z_1^3=4x_1y_1z_1$ $\implies$ $x_1^3+2y_1^3+4z_1^3=2x_1y_1z_1$.
Consequently, the procedure ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
$\mathbb{Q}\left ( \sqrt{2},\sqrt{3},\sqrt{5} \right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3}+\sqrt{5} \right )$ Prove that
$$\mathbb{Q}\left ( \sqrt{2},\sqrt{3},\sqrt{5} \right )=\mathbb{Q}\left ( \sqrt{2}+\sqrt{3}+\sqrt{5} \right )$$
I proved for two elements, ex, $\mathbb{Q}\left ( \sqrt{2},\sqrt{3}\right )=\mathbb{Q}\... | This is a boring but purely algebraic derivation.
For any three distinct positive integers $a,b,c$, let $x = \sqrt{a}+\sqrt{b}+\sqrt{c}$, we have
$$\begin{align}
& (x-\sqrt{a})^2 = (\sqrt{b}+\sqrt{c})^2 = b + c + 2\sqrt{bc}\\
\implies & (x^2 + a - b - c) - 2\sqrt{a}x = 2\sqrt{bc}\\
\implies & (x^2 + a - b - c)^2 - 4\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$ Question:
Given $x + y = 12$, find the minimum value of $\sqrt{x^2+4} + \sqrt{y^2+9}$?
Key:
I use $y = 12 - x$ and substitute into the equation, and derivative it.
which I got this
$$\frac{x}{\sqrt{x^2+4}} + \frac{x-12}{\sqrt{x^2-24x+153}} = f'(x).$$
However, aft... | If you are aware of the rarely seen in action "Minkowski" inequality..then here you go: $\sqrt{x^2+2^2}+\sqrt{y^2+3^2} \geq \sqrt{(x+y)^2+(2+3)^2}=\sqrt{12^2+5^2} = \sqrt{169}=13$, and rest assured that this extrema is achievable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Need a help to find the series formula for A005836 I need a help in finding the series formula for the Sequence $A005836$
$a(1)=0$
$a(2)=1$
$a(3)=3$
$a(4)=4$
$a(5)=9$
so series of above sequence would be
$A(1)=0$
$A(2)=1$
$A(3)=4$
$A(4)=8$
$A(5)=17$
where $$A(n)=a(1)+a(2)+a(3)+........+a(n)$$
the value of $n$ can be up... | I suspect there is a relatively simple recursion with $O(\log n)$ complexity:
$$A(2^n+b) = (3^n-1)2^{n-2}+3^n b+A(b)$$ for $0 \le b \lt 2^b$ and with $A(0)=0$, where the $(3^n-1)2^{n-2}$ represents $A(2^n)$, the $3^n b$ represents the leading ternary digits of the final $b$ terms, and the $A(b)$ representing the other ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1308961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Finding $\lim\limits_{n\to\infty }\frac{1+\frac12+\frac13+\cdots+\frac1n}{1+\frac13+\frac15+\cdots+\frac1{2n+1}}$ I need to compute: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}$.
My Attempt: $\dis... | By the Stolz-Cesaro theorem from https://en.wikipedia.org/wiki/Stolz%E2%80%93Ces%C3%A0ro_theorem,
$$\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}=\lim_{n\to\infty}\frac{\frac{1}{n+1}}{\frac{1}{2n+3}}=2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 0
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How do I rewrite this rational expression? How do I rewrite the rational expression:
$$\frac{x^3+5x^2+3x-10}{x+4}$$
But in the form of:
$$q(x) + \frac{r(x)}{b(x)}$$
| Use Polynomial long division. The only way I know how to format that here would be something like:
$$\begin{split}
x^3 + 5x^2 + 3x - 10 &= (x+4)(x^2) + x^2 + 3x - 10 \\
&= (x+4)(x^2) + (x+4)(x) - x - 10 \\
&= (x+4)(x^2) + (x+4)(x) - (x+4)(1) - 6 \\
&= (x+4)(x^2 + x - 1) - 6
\end{split}$$
So:
$$\frac{x^3 + 5x^2 + 3x - 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1310780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Simplify $\prod_{k=1}^5\tan\frac{k\pi}{11}$ and $\sum_{k=1}^5\tan^2\frac{k\pi}{11}$ My question is:
If $\tan\frac{\pi}{11}\cdot \tan\frac{2\pi}{11}\cdot \tan\frac{3\pi}{11}\cdot \tan\frac{4\pi}{11}\cdot \tan\frac{5\pi}{11} = X$ and $\tan^2\frac{\pi}{11}+\tan^2\frac{2\pi}{11}+\tan^2\frac{3\pi}{11}+\tan^2\frac{4\pi}{11}+... | Like Roots of a polynomial whose coefficients are ratios of binomial coefficients,
$$\tan(2n+1)x=\dfrac{\binom{2n+1}1\tan x-\binom{2n+1}3\tan^3x+\cdots}{1-\binom{2n+1}2\tan^2x+\cdots}$$
If $\tan(2n+1)x=0,(2n+1)x=r\pi$ where $r$ is ay integer
$\implies x=\dfrac{r\pi}{2n+1}$ where $r\equiv-n,-(n-1),\cdots,0,1,\cdots,n\pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
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Determinate set $A\subseteq\mathbb{R}$ so that for every $a\in A$ and every $x\in\mathbb{R}$ the condition $ax^2+x+3\ge0$ is valid Quadratic function is always greater than $0$ if
$$a>0$$ and $$D=0$$
Solving for $D$ we have
$$1-12a=0\Rightarrow a=\frac{1}{12}$$
So, $$a\in[\frac{1}{12},+\infty)$$
Is this the only condit... | $ax^2+x+3\ge 0$ for every $x\in\mathbb R$ if and only if $a\gt 0$ and $D=1-12a\color{red}{\le} 0$, i.e. $a\color{red}{\ge} \frac{1}{12}$.
For the limit, you can have
$$\begin{align}\lim_{x\to\infty}\left(x+1-\sqrt{ax^2+x+3}\right)&=\lim_{x\to\infty}\frac{(x+1-\sqrt{ax^2+x+3})(x+1+\sqrt{ax^2+x+3})}{x+1+\sqrt{ax^2+x+3}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the maximum value of $a^2+b^2+c^2+d^2+e^2$ Find the maximum value of $a^2+b^2+c^2+d^2+e^2$ with the following condition:
i) $a\ge b\ge c\ge d\ge e\ge 0$
ii) $a+b\le m$
iii) $c+d+e \le n$
Where $m,n\ge 0$.
Playing with numbers I found that for $m=n$ we should get the maximum value is $m^2$.
| It seems the following.
Since a set $C$ of all sequences $(a,b,c,d,e)$ satisfying Conditions i-iii is compact, a continuous function $$f(a,b,c,d,e)= a^2+b^2+c^2+d^2+e^2$$ attains its maximum $M$ on the set $C$ at some point $p=(a, b, c, d, e)$. If $b>c$ then a point $p’=(m-c, c, c, d, e)$ also belongs to the set $M$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Generating functions (Counting?) Determine how many ways Brian, Katie, and Charlie can split a $\$50$ dinner bill such that Brian and Katie each pay an odd number of dollars and Charlie pays at least $\$5$.
How can I use generating functions to solve this problem? Or should I use a counting approach?
| The generating function for the number of ways that the three can pay a bill of $n$ dollars is
$$
\begin{align}
&(x+x^3+x^5+\dots)(x+x^3+x^5+\dots)(x^5+x^6+x^7+\dots)\\[3pt]
&=x^7\frac1{(1-x^2)^2}\frac1{1-x}\\
&=x^7\frac1{(1-x)^3}\frac1{(1+x)^2}\\
&=x^7\left[\frac14\frac1{(1-x)^3}+\frac14\frac1{(1-x)^2}+\frac3{16}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
solving difficult complex number proving if $z= x+iy$ where $y \neq 0$ and $1+z^2 \neq 0$, show that the number $w= z/(1+z^2)$ is real only if $|z|=1$
solution :
$$1+z^2 = 1+ x^2 - y^2 +2xyi$$
$$(1+ x^2 - y^2 +2xyi)(1+ x^2 - y^2 -2xyi)=(1+ x^2 - y^2)^2 - (2xyi)^2$$
real component
$$(1+ x^2 - y^2)x - yi(2xyi) = x + x^3... | I tried to solve this with geometry and how the arithmetic operations on complex numbers can be interpreted geometrically.
*
*By dividing two complex numbers, their arguments (angles, $\varphi$)
are subtracted.
*If $w$ should be real, its argument has to be zero.
That means $\varphi(z) = \varphi(1+z^2)$.
Here's a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
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Solving for n in the equation $\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$
Solving for $n$ in the equation $$\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$$
Can anyone show me a numerical method step-by-st... | Let $f(x)=\left ( \frac{1}{2} \right )^{x}+\left ( \frac{1}{4} \right )^{x}+\left ( \frac{3}{4} \right )^{x}-1$. Now note that $f(1)=\frac12$ and $f(2)=-\frac18$. Hence given that $$f'(x)=-4^{-x}(3^x \log\frac43+2^x\log 2+\log4)<0$$ there is one unique root in $(1,2)$. The rest is numerics, this could be approximated t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Express 99 2/3% as a fraction? No calculator My 9-year-old daughter is stuck on this question and normally I can help her, but I am also stuck on this! I have looked everywhere to find out how to do this but to no avail so any help/guidance is appreciated:
The possible answers are:
$1 \frac{29}{300}$
$\frac{269}{300}$... | There are multiple roads that lead to Rome:
First, take out the $\%$. We find $\frac{299}{3}$
$99 \frac{2}{3} = 99 + \frac{2}{3} = \frac{1}{3} \cdot ( 297 + 2) = \frac{1}{3} \cdot 299 = \frac{299}{3}$
or
$99 \frac{2}{3} = 99 + \frac{2}{3} = 3 \cdot \frac{99}{3} + \frac{2}{3} = \frac{297}{3} + \frac{2}{3} = \frac{299}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1318621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "65",
"answer_count": 14,
"answer_id": 6
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How to calculate $\sum_{n=1}^\infty \frac{2}{(2n-1)(2n+1)}$ How can I calculate the sum of the following infinite series: $$\sum_{n=1}^\infty \frac{2}{(2n-1)(2n+1)}$$
| This is how you do it formally:
$$\sum_{i=1}^N\frac{2}{(2n-1)(2n+1)} = \sum_{i=1}^N \left (\frac{1}{2n-1}-\frac{1}{2n+1} \right)= \sum_{i=1}^N \left (\frac{1}{2n-1} \right) - \sum_{i=1}^N \left (\frac{1}{2n+1} \right) = 1 + \sum_{i=2}^{N} \left (\frac{1}{2n-1} \right) - \sum_{i=1}^{N-1}\left (\frac{1}{2n+1} \right) - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
why the convolution of two functions of moderate decrease is again function of moderate decrease? I need to prove that a convolution of 2 functions of moderate decrease is a function of moderate decrease.
I tried to split the integral into two integrals but I couldn't manage to bound any one of them by $\frac{A}{x^2+1}... | Take $z = x - y/2$.
$$ \begin{align*}\int_{-\infty}^{\infty} \frac{1}{1 + x^2} \cdot \frac{1}{1 + (y-x)^2} \mathrm{d}x &= 2 \int_0^\infty \frac{1}{1 + (y/2 + z)^2} \cdot \frac{1}{1 + (y/2 - z)^2} \mathrm{d}z \\
& \leq \frac{2}{1 + (y/2)^2} \int_0^\infty \frac{1}{1 + (y/2 - z)^2} \mathrm{d}z \\
& \leq \frac{2}{1 + (y/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Bounding $\sum_{k=1}^N \frac{1}{1-\frac{1}{2^k}}$ I'm looking for a bound depending on $N$ of $\displaystyle \sum_{k=1}^N \frac{1}{1-\frac{1}{2^k}}$.
The following holds $\displaystyle \sum_{k=1}^N \frac{1}{1-\frac{1}{2^k}} = \sum_{k=1}^N \sum_{i=0}^\infty \frac{1}{2^{ki}}=\sum_{n=0}^\infty \sum_{\substack{q\geq 0 \\ 1... | Note that
$$\frac{1}{1-\frac{1}{2^k}} = 1+ \frac{1}{2^k-1}$$
so
$$N \le \sum_{k=1}^N\frac{1}{1-\frac{1}{2^k}} \le N+ \sum_{k=1}^{+\infty} \frac{1}{2^k-1} \le N+2$$
since $1\le \sum_{k=1}^{+\infty} \frac{1}{2^k-1} \le 2$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Finding a polynomial by divisibility
Let $f(x)$ be a polynomial with integer coefficients. If $f(x)$ is divisible by $x^2+1$ and $f(x)+1$ by $x^3+x^2+1$, what is $f(x)$?
My guess is that the only answer is $f(x)=-x^4-x^3-x-1$, but how can I prove it?
| This is a straightforward application of CRT = Chinese Remainder Theorem.
By hypothesis $\ (x^2\!+1)g = f = -1 + (x^3\!+x^2\!+1)h\ $ for some polynomials $\,g,h,\,$ so
${\rm mod}\ x^2\!+\!1\!:\ x^2\equiv -1\,\Rightarrow\,1\equiv (x^3\!+x^2\!+1)h\equiv -xh\,\Rightarrow\, h \equiv -1/x\equiv x^2/x\equiv \color{#c00}x$
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1322741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate the gcd of two polynomials $\mod 7$ I need to find gcd of $x^4-3x^3-2x+6$ and $x^3-5x^2+6x+7$ in $\mathbb Z/7 \mathbb Z[x]$, the integer polynomials mod $7$. Please any help will be appreciated.
| In general, you use the same method as gcd of numbers: the Euclidean algorithm (detailed below). In this specific (i.e. non-general) case you can take a short-cut: since you see that the constant term in your second polynomial is $0$ mod $7$, we know that $0$ is a root mod $7$. So the second polynomial factors as $(a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $2|x|/(1+x^{2})<1$ true if $|x|<1$? [Solved] I was asked to show that the series
$$\sum_{k\geq 0}a_{k}\left ( \frac{2x}{1+x^{2}} \right )^{k}$$
is convergent for $x\in (-1,1)$, where $\{ a_{k}\}_{k\geq 0}$ is a real bounded sequence.
My main question is, how do I know that $|2x/(1+x^{2})|<1$, that is, $2|x|/(1+x^{... | Looking at positive values only, $\frac{2x}{1 + x^2} <1$ leads to $2x < 1 + x^2$ which leads to $0 < x^2 - 2x + 1 = (x - 1)^2$--this is true for any value of $x \neq 1$--that is all values of $0 \leq x < 1 \cup 1 < x < \infty$. If we consider negative values then we have that $\frac{-2x}{1 + x^2} < 1$ which leads to $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Integral of Absolute Value of $\sin(x)$ For the Integral: $\int |\sin (ax)|$, it is fairly simple to take the Laplace transform of the absolute value of sine, treating it as a periodic function.
$$\mathcal L(|\sin (ax)|) = \frac{\int_0^\frac{\pi} {a} e^{-st} \sin (ax)} {1 - e^{-\frac{ \pi s} {a}}} = \frac{a (1 + e^{-\f... | Suppose $F(x)$ is an antiderivative of $|\sin(x)|$.
On the interval $[n\pi, (n+1)\pi]$, $|\sin(x)| = (-1)^n \sin(x)$, so we must have $F(x) = (-1)^{n+1}\cos(x) + c_n$ there, where $c_n$ is some constant.
Comparing the values at $n\pi$ from both sides, since $\cos(n\pi) = (-1)^n$,
we get
$ 1 + c_{n-1} = -1 + c_n$, i.e. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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High School Trigonometry ( Law of cosine and sine) I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine:
Let $\alpha$, $\beta$ and $\gamma$ be the angles of arbitrary triangle... | From Sine Rule in the triangle, we have $$\frac{\sin\alpha}{a}=\frac{\sin\beta}{b}=\frac{\sin\gamma}{c}=k \space (\text{let any arbitrary constant})$$ Thus by substituting the values of $a=k\sin \alpha$, $b=k\sin \beta$ & $c=k\sin \gamma$ in the given expression, we get $$\frac{b-2a\cos\gamma}{a\sin\gamma}+\frac{c-2b\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 4
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Determining irreducible components of $Z(y^4 - x^6, y^3 - x y^2 - y x^3 + x^4) \subset \mathbb A^2$ I'm trying to determine the irreducible components of the zero set $Z(I)$ for the ideal $I = (y^4 - x^6,\, y^3 - x y^2 - y x^3 + x^4)$, in the affine space $\mathbb A^2$ over an algebraically closed field $k$.
I can fact... | The key fact that we will use to find the irreducible components is that an algebraic subset of a variety is irreducible iff its coordinate ring is an integral domain. Therefore, consider the coordinate ring of the algebraic set $Z(y^2 - x^3)$: it is $\frac{k[x,y]}{(y^2 - x^3)}$, which is an integral domain, hence $Z(y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1327291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sums $a_k=[\frac{2+(-1)^k}{3^k}, (\frac{1}{n}-\frac{1}{n+2}), \frac{1}{4k^2-1},\sum_{l=0}^k{k \choose l}\frac{1}{2^{k+l}}]$
Determien the sums of the following series'.
1:$\sum_{k=0}^\infty \frac{2+(-1)^k}{3^k}$
2:$\sum_{k=0}^\infty (\frac{1}{n}-\frac{1}{n+2})$
3:$\sum_{k=0}^\infty \frac{1}{4k^2-1}$
4:$\sum_{k=0}^\inf... | $1-$
$$\sum_{k=0}^{\infty }\frac{2+(-1)^k}{3^k}=\sum_{k=0}^{\infty }\frac{2}{3^k}+\frac{(-1)^k}{3^n}=2\frac{1}{1-1/3}+\frac{1}{1+1/3}$$
$3-$
$$\sum_{k=0}^{\infty }\frac{1}{4k^2-1}=\sum_{k=0}^{\infty }0.5[\frac{1}{2k-1}-\frac{1}{2k+1}]=0.5[1/1-1/3+1/3-1/5+1/5+....]=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1327732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Calculating fluxintegral out of the surface $1-x^2-y^2$ I am trying to calculate the flux integral of the vector field
$$
\vec{F} = (x,y,1+z)
$$
Out of the surface $z=1-x^2-y^2$, $z\geq 0$
Answer : $\frac{5\pi}{2}$
I begin by defining a vector that traces out the surface and calculate the cross product of its derivate ... | You should consider that $ds$=$\frac{dxdy}{\sigma}$ and thus projecting the whole surface on the XOY plane where $\sigma$=$|\vec n . \vec k|$ and $\vec n$ is the unit vector that is in this case $\vec n$ = $\frac{\vec grad(S)}{||\vec grad(S)||}$ = $\frac{2x\vec i + 2y\vec j + 2z\vec k}{\sqrt{4x^2 +4y^2 +4z^2}}$=$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1327807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing $\sum_{n\geq 0}n\frac{1}{4^n}$ Can I compute the sum
$$
\sum_{n\geq 0}n\frac{1}{4^n}
$$
by use of some trick?
First I thought of a geometrical series?
| The sum can be expanded as follows:
\begin{align*}
\begin{array}{l|cccc}
\text{once}&1/4^1&&&\\
\text{twice}&1/4^2&1/4^2&&\\
\text{3 times}&1/4^3&1/4^3&1/4^3&&\\
\vdots\\
\text{$n$ times}&1/4^n&1/4^n&\cdots&1/4^n\\
\vdots
\end{array}
\end{align*}
Now, instead of summing row-by-row, try summing column-by-column:
\begin{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
evaluate $\int_0^{2\pi} \frac{1}{\cos x + \sin x +2}\, dx $ This is supposed to be a very easy integral, however I cannot get around.
Evaluate:
$$\int_0^{2\pi} \frac{1}{\cos x + \sin x +2}\, dx$$
What I did is:
$$\int_{0}^{2\pi}\frac{dx}{\cos x + \sin x +2} = \int_{0}^{2\pi} \frac{dx}{\left ( \frac{e^{ix}-e^{-ix}}{2i... | Note that: $$\frac{1}{iz \left ( \frac{z-z^{-1}}{2i} + \frac{z+z^{-1}}{2}+2 \right )}$$
is a rational function. Write it as a quotient of polynomials, then factor the denominator and apply partial fractions.
If you change the integral to:
$$\int_{-\pi}^{\pi} \frac{1}{\cos x + \sin x +2}\, dx$$ you can apply the Weiers... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Centroid of a Triangle on a inscribed circle $AB$ is the hypotenuse of the right $\Delta ABC$ and $AB = 1$.
Given that the centroid of the triangle $G$ lies on the incircle of $\Delta ABC$, what is the perimeter of the triangle?
| Assume that $C=(0,0),B=(a,0),A=(0,b)$ and $c=\sqrt{a^2+b^2}=AB$. We have:
$$ G = \left(\frac{a}{3},\frac{b}{3}\right) \tag{1}$$
and the inradius is given by
$$ r = \frac{2\Delta}{a+b+c} = \frac{ab}{a+b+c},\tag{2}$$
so the equation of the incircle is given by:
$$ \left(x-\frac{ab}{a+b+c}\right)^2 + \left(y-\frac{ab}{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329263",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimum value of $\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\frac{24}{5\sqrt{5a+5b}}$ Let $a\ge b\ge c\ge 0$ such that $a+b+c=1$
Find the minimum value of $P=\sqrt{\dfrac{a}{b+c}}+\sqrt{\dfrac{b}{c+a}}+\dfrac{24}{5\sqrt{5a+5b}}$
I found that the minimum value of $P$ is $\dfrac{78}{5\sqrt{15}}$ when $a=b=\dfrac{3}{8};c... | $P=\sqrt{\dfrac{a}{1-a}}+\sqrt{\dfrac{b}{1-b}}+\dfrac{24}{5\sqrt{5a+5b}}$
$m=\sqrt{\dfrac{a}{1-a}},n=\sqrt{\dfrac{b}{1-b}} \implies a=\dfrac{m^2}{m^2+1},b=\dfrac{n^2}{n^2+1},\\a\ge b\ge c \implies a\ge \dfrac{1}{3} \implies m^2\ge \dfrac{1}{2} ,a+b=1-c \ge 1-b \implies n\ge \sqrt{\dfrac{1}{2m^2+1}}\implies mn\ge ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Quick way to solve the system $\displaystyle \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} = \frac{65}{36}$, $xy-x+y=118$. Consider the system
$$\begin{aligned} \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} & = \frac{65}{36}, \\ xy -x +y & = 118. \end{aligned}$$
I have solved... | Consider $$ \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} = \frac{65}{36}$$ and set $a=\left( \frac{3}{2} \right)^{x-y}$. So, you have $a-\frac 1a=\frac{65}{36}$, that is to say $a=\frac{9}{4}$. So $$\left( \frac{3}{2} \right)^{x-y}=\frac{9}{4}=\left( \frac{3}{2} \right)^2$$ so $x-y=2$.
For the r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(... | Let $a^3=2-\sqrt3,b^3=2+\sqrt3$, we have
$$a^3+b^3=4$$
$$a+b=x$$
$$ab=1$$
Consider
\begin{align}
x^3=(a+b)^3&=3ab(a+b)+a^3+b^3\\&=3x+4
\end{align}
Therefore
$$x^3-3x-4=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 5
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Finding $ \lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x} $ I'm kind of stuck on this problem, I could use a hint.
$$
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x}
$$
After some algebra, I get
$$
{\lim_{x\to 2}\frac{x+2 - 2x}{x(x-2)-\sqrt{x+2}+\sqrt{2x}}}
$$
EDIT above should be:
$$
\lim_{x\to 2}\frac{x+2 - 2x}{x(x... | \begin{align}
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x}&=
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x(x-2)}\frac{\sqrt{x+2}+\sqrt{2x}}{\sqrt{x+2}+\sqrt{2x}} \\
&=\lim_{x \to 2} \frac{(x+2)-(2x)}{x(x-2)(\sqrt{x+2}+\sqrt{2x})} \\
&=\lim_{x \to 2} \frac{-(x-2)}{x(x-2)(\sqrt{x+2}+\sqrt{2x})} \\
&=\lim_{x \to 2} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Value of $x$ in $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$ How can we find the value of $x$ in $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$?
Note that $\sin^{-1}$ is the inverse sine function.
I'm asking for the solution $x$ for this equation.
Please work out the solution.
| $\bf{My\; Solution::}$ Given $$\sin^{-1}(x)+\sin^{-1}(1-x) = \cos^{-1}(x)$$
$$\displaystyle \sin^{-1}(1-x) = \cos^{-1}(x)-\sin^{-1}(x)\Rightarrow (1-x) = \sin \left[\cos^{-1}(x)-\sin^{-1}(x)\right]$$
Using $$\sin^{-1}(x)+\cos^{-1}(x) = \frac{\pi}{2}$$
So $$\displaystyle (1-x) = \sin\left[\frac{\pi}{2}-2\sin^{-1}(x)\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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How to find sum of the infinite series $\sum_{n=1}^{\infty} \frac{1}{ n(2n+1)}$ $$\frac{1}{1 \times3} + \frac{1}{2\times5}+\frac{1}{3\times7} + \frac{1}{4\times9}+\cdots $$
How to find sum of this series?
I tried this: its $n$th term will be = $\frac{1}{n}-\frac{2}{2n+1}$; after that I am not able to solve this.
| $$\frac{1}{n(2n+1)}= \frac 1n-\frac{2}{2n+1}=2\left(\frac{1}{2n}-\frac{1}{2n+1}\right)$$
Hence
$$\sum_{n=1}^{\infty} \frac{1}{ n(2n+1)} =2 \sum_{n=1}^{\infty} \left(\frac{1}{2n}-\frac{1}{2n+1}\right)= 2\left(1-\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}\right)=\color{red}{ 2(1-\ln 2)} $$
since $${\displaystyle \ln(1+x)=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
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Product of Matrices I Given the matrix
\begin{align}
A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right)
\end{align}
consider the first few powers of $A^{n}$ for which
\begin{align}
A = \left( \begin{matrix} 1 & 2 \\ 3 & 2 \end{matrix} \right)
\hspace{15mm} A^{2} = \left( \begin{matrix} 7 & 6 \\ 9 & 10 \end{m... | It is often possible to diagonalize a square matrix, that is find a diagonal matrix $D$ and another matrix $P$ such that $$A=PDP^{-1}.$$ This is a nice property because $$A^n=PDP^{-1}PDP^{-1}...PDP^{-1}=PD^nP^{-1}.$$
The $P$ and $P^{-1}$ matrices cancel except on the ends, so we only have to take a power of the diagona... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1336283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The partial fraction decomposition of $\dfrac{x}{x^3-1}$ I was trying to decompose $\dfrac{x}{x^3-1}$ into Partial Fractions. I tried the following:
$$\dfrac{x}{(x-1)(x^2+x+1)}=\dfrac{A}{(x-1)}+\dfrac{B}{(x^2+x+1)}$$
$$\Longrightarrow A(x^2+x+1)+B(x-1)=x$$
Putting $x=1,$
$$A(1+1+1)+B(1-1)=1\Rightarrow A=\dfrac{1}{3}$$
... | $\frac{x}{x^3-1}=\frac{x}{(x-1)(x^2+x+1)}$
$\frac{x}{(x-1)(x^2+x+1)}=\frac{A}{x-1}+\frac{Bx+C}{x^2+x+1}$
$x=A(x^2+x+1)+(Bx+C)(x-1)$
Let $x=1$
$1=A(1^2+1+1)$
$1=3A$
$A=\frac{1}{3}$
So, $x=\frac{1}{3}(x^2+x+1)+(Bx+C)(x-1)$
$x=\frac{1}{3}x^2+\frac{1}{3}x+\frac{1}{3}+Bx^2-Bx+Cx-C$
$x=(\frac{1}{3}+B)x^2+(\frac{1}{3}-B+C)x+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How do I prove this combinatorial identity using inclusion and exclusion principle? $$\binom{n}{m}-\binom{n}{m+1}+\binom{n}{m+2}-\cdots+(-1)^{n-m}\binom{n}{n}=\binom{n-1}{m-1}$$
Note that we can show this with out using inclusion and exclusion principle by using Pascal's Identity i.e. $C(n,k)=C(n-1,k)+C(n-1,k-1)$.
| Note: This answer is inspired by the deleted answer of @diracpaul.
We analyze lattice paths generated from $(1,0)$ steps in $x$-direction and $(0,1)$ steps in $y$-direction. The number of paths from $(0,0)$ to $(m,n-m)$ is $\binom{n}{m}$. This is valid since for paths of length $n$ we can select $m$ steps in $x$-dir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Conditions of a differential equation Consider the differential equation
\begin{align}
2 x^2 y'' + x(x^2 - 1) y' + (2 x^2 - x +1)y = 0 \hspace{5mm} y(0) = 0, y'(0)=1.
\end{align}
A solution readily found is
\begin{align}
y(x) &= B_{0} \, \sqrt{x} (x+1) \, e^{- \frac{x^{2}}{4}} \\
& \hspace{5mm} + B_{1} \, e^{- \frac{3 ... | We develop $y(x)$ in the vicinity of $x=0$
$$2\Gamma\left(\frac{3}{4},-\frac{x^2}{4}\right)+i\Gamma\left(\frac{1}{4},-\frac{x^2}{4}\right)=2\Gamma\left(\frac{3}{4}\right)+i\Gamma\left(\frac{1}{4}\right)-(2i)^{3/2}\sqrt{x}+O(x)$$
$$4e^{\frac{3\pi i}{4}}e^{-\frac{x^2}{4}}\sqrt{2}(x+1)\left(2\Gamma\left(\frac{3}{4},-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Minimization on compact region I need to solve the minimization problem
$$\begin{matrix} \min & x^2 + 2y^2 + 3z^2 \\ subject\;to & x^2 + y^2 + z^2 =1\\ \; & x+y+z=0
\end{matrix}$$
I was trying to verify the first order conditions using lagrange multipliers
$$\left\{\begin{matrix}
2x = 2\lambda_1x + \lambda_2 \\
4y = 2 ... | $x^2+2y^2+3z^2=x^2+y^2+z^2+y^2+2z^2=1+y^2+2z^2, x=-(y+z),(y+z)^2+y^2+z^2=1 \implies y^2+yz+z^2=\dfrac{1}{2}$, now the problem become : find min of $y^2+2z^2$ with $y^2+yz+z^2=\dfrac{1}{2}$,it shoud be easier now.
or we can go further :$y=\dfrac{-z \pm \sqrt{2-3z^2}}{2},f(z)=\left(\dfrac{-z \pm \sqrt{2-3z^2}}{2}\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How can I prove irreducibility of polynomial over a finite field?
I want to prove what $x^{10} +x^3+1$ is irreducible over a field $\mathbb F_{2}$ and $x^5$ + $x^4 +x^3 + x^2 +x -1$ is reducible over $\mathbb F_{3}$.
As far as I know Eisenstein criteria won't help here.
I have heard about a criteria of irreducibility... | Neither polynomial has zeros in the respective prime field, so they do not have linear factors either.
Let's check the first polynomial $p(x)=x^{10}+x^3+1\in\Bbb{F}_2[x]$. The claim is that it is irreducible. A more efficient way of using the cited result is to observe that if $p(x)$ were a product of two or more irred... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 3,
"answer_id": 0
} |
How to get the results of this logarithmic equation? How to solve this for $x$:
$$\log_x(x^3+1)\cdot\log_{x+1}(x)>2$$
I have tried to get the same exponent by getting the second
multiplier to reciprocal and tried to simplify $(x^3+1)$.
| Keep in mind $(\log_a b)(\log_c a) = \log_c b$.
Setting $a = x, b = x^3 + 1$, and $c = x + 1$, we have
$\log_x (x^3 + 1)\cdot \log_{x+1}x = \log_{x + 1}(x^3 + 1)$.
So your inequality is equivalent to
$\log_{x + 1}(x^3 + 1) > 2
\\ x^3 + 1 > (x + 1)^2
\\ x^3 + 1 > x^2 + 2x + 1
\\ x^3 - x^2 - 2x > 0
\\ x(x + 1)(x - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving a trigonometric identity with tangents Prove that: $$\tan^227^\circ +2 \tan27^\circ \tan36^\circ=1$$
any help, I appreciate it.
| The solution is nothing more than some computation and observing $2\cdot27 = 90 - 36.$
\begin{align*}
\tan^2(27^\circ) + 2 \tan(27 ^ \circ) \tan (36^\circ) &= \tan^2(27^\circ) + 2 \tan(27 ^ \circ) \cot (54^\circ) \\[1ex]
&= \tan^2(27^\circ) + 2 \, \frac{\tan(27 ^ \circ)}{ \tan(54^\circ)} \\[1ex]
&= \tan^2(27^\circ) + 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove equality of Fibonacci sequence
Let $u(n)$ — the Fibonacci sequence. Prove that
$$u(1)^3+...+u(n)^3=\frac{ 1 }{ 10 } \left[ u(3n+2)+(-1)^{n+1}6u(n-1)+5 \right].$$
I suppose we need prove that equality by iduction in $n$. It's obviously holds at $n=1$. So let $P(k)$ holds, then also must $P(k + 1)$ holds. What... | Consider the series
\begin{align}
\sum_{r=1}^{n} t^{r} = \frac{t \, (1-t^{n})}{1-t}
\end{align}
and
\begin{align}
F_{n}^{3} &= \frac{1}{5 \sqrt{5}} \, (\alpha^{n} - \beta^{n})^{3} \\
&= \frac{1}{5} \left( F_{3n} - 3 (-1)^{n} \, F_{n} \right),
\end{align}
since $\sqrt{5} \, F_{n} = \alpha^{n} - \beta^{n}$ and $2 \alpha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
solve $|x-6|>|x^2-5x+9|$
solve $|x-6|>|x^2-5x+9|,\ \ x\in \mathbb{R}$
I have done $4$ cases.
$1.)\ x-6>x^2-5x+9\ \ ,\implies x\in \emptyset \\
2.)\ x-6<x^2-5x+9\ \ ,\implies x\in \mathbb{R} \\
3.)\ -(x-6)>x^2-5x+9\ ,\implies 1<x<3\\
4.)\ (x-6)>-(x^2-5x+9),\ \implies x>3\cup x<1 $
I am confused on how I proceed.
Or... | HINT:
As $x^2-5x+9=\dfrac{(2x-5)^2+11}4\ge\dfrac{11}4>0\forall $ real $x$
So, $|x^2-5x+9|=+(x^2-5x+9)\forall $ real $x$
So, the problem reduces to two cases for $|x-6|$
Also, $|x-6|>\dfrac{11}4\implies \pm(x-6)>\dfrac{11}4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Quadratic Absolute Value Equation Problem:
Find all $x$ such that $|x^2+6x+6|=|x^2+4x+9|+|2x-3|$
I can't understand how to get started with this. I thought of squaring both sides of the equation to get rid of the modulus sign, but then couldn't understand how to deal with the following term:$$2|(x^2+4+9)(2x+3)|$$Any ... | You-know-me has already provided a good answer.
But I decided to undelete my answer in the hope that this answer might be of some help.
Since we have $x^2+4x+9=(x+2)^2+5\gt 0$, we have
$$|x^2+4x+9|=x^2+4x+9.$$
Then, since we have $x^2+6x+6=0\iff x=-3\pm\sqrt{3}$ and $2x-3=0\iff x=\frac 32$, separate it into three case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1346966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
One-One Correspondences Adam the ant starts at $(0,0)$. Each minute, he flips a fair coin. If he flips heads, he moves $1$ unit up; if he flips tails, he moves $1$ unit right.
Betty the beetle starts at $(2,4)$. Each minute, she flips a fair coin. If she flips heads, she moves $1$ unit down; if she flips tails, she mov... | Adam and Betty may meet only after three steps from the start, in the points $(2,1),(1,2)$ or $(0,3)$.
After three steps, Adam is in $(2,1)$ with probability $\frac{3}{8}$, in $(1,2)$ with probability $\frac{3}{8}$ and in $(0,3)$ with probability $\frac{1}{8}$.
After three steps, Betty is in $(2,1)$ with probability $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Does the equation $2\cos^2 (x/2) \sin^2 (x/2) = x^2+\frac{1}{x^2}$ have real solution?
Do the equation
$$2\cos^2 (x/2) \sin^2 (x/2) = x^2+\frac{1}{x^2}$$
have any real solutions?
Please help. This is an IITJEE question.
Here $x$ is an acute angle.
I cannot even start to attempt this question. I cannot understand.... | the right hand of the equation, you have $$x^2 + \frac1{x^2} = \left(x-\frac 1x\right)^2 + 2 \ge 2 \tag 1$$and equality occurs for $$x = \pm 1.$$
on the left hand side, we have $$ \frac12 \sin^2 x = 2\cos^2 (x/2) \sin^2 (x/2) \le \frac12 \tag 2$$ equality for $$\sin x = \pm 1.$$
but $(1)$ and $(2)$ are inconsistent, th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Differentiate the Function $ h(z)=\ln\sqrt{\frac{a^2-z^2}{a^2+z^2}}$
Differentiate the function $$h(z)=\ln\sqrt{\frac{a^2-z^2}{a^2+z^2}}$$
My try:
$$h(z) = \frac{1}{2}\ln\left(a^2-z^2\right)-\frac{1}{2}\ln\left(a^2+z^2\right)$$
so
$$h'(z) = \frac{1}{2}\cdot\frac{2a-2z}{a^2-z^2}-\frac{1}{2}\cdot\frac{2a+2z}{a^2+z^2}$$... | Albeit you used logarithmic differentiation, you must never forget the power of substitutions.
Let $$u=\sqrt{\frac{a^2-z^2}{a^2+z^2}}.$$ And, for more simplicity, let $$v=\frac{a^2-z^2}{a^2+z^2}. $$ Then $u=\sqrt{v} $ with $h(z)=\ln(u).$
We find: $${d}h(z)=\frac{du}{u}$$ $$du=\frac{1}{2}v^{-1/2}dv $$ $$dv=\frac{-2z}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to calculate $\int \frac{\sin x}{\tan x+\cos x} \, dx$ How to calculate $$\int \frac{\sin x}{\tan x+\cos x} \, dx\text{ ?}$$ I got to $$\int \frac{-u}{u^2-u-1} \, du$$ while $u=\sin x$ but can I continue from here?
| By replacing $\sin x$ with $u$ we have:
$$I=\int\frac{\sin x}{\tan x+\cos x}\,dx = \int \frac{\sin x\cos x}{1+\sin x-\sin^2 x}\,dx = \int \frac{u}{1+u-u^2}\,du\tag{1}$$
but the roots of $1+u-u^2$ occur at $u=\frac{1\pm\sqrt{5}}{2}$, so that:
$$ \frac{u}{1+u-u^2} = \frac{-\frac{5+\sqrt{5}}{10}}{u-\left(\frac{1+\sqrt{5}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
The number of numbers lying between 1 and 200 which are divisible by either of 2 , 3 or 5? The number of numbers lying between 1 and 200 which are divisible by either of two , three or five?
| A)numbers divisible by 2: $\frac{200}{2} = 100$
B)numbers divisible by 3: $\frac{200}{3} = 66$
C)numbers divisible by 5: $\frac{200}{5} = 40$
counting twice
AB)numbers divisible by 6: $\frac{200}{6} = 33$
AC)numbers divisible by 10: $\frac{200}{10} = 20$
BC)numbers divisible by 15: $\frac{200}{15} = 13$
counting 3 time... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Product of cosines: $ \prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right) $
Evaluate
$$ \prod_{r=1}^{7} \cos \left({\dfrac{r\pi}{15}}\right) $$
I tried trigonometric identities of product of cosines, i.e, $$\cos\text{A}\cdot\cos\text{B} = \dfrac{1}{2}[ \cos(A+B)+\cos(A-B)] $$
but I couldn't find the product.
Any help w... | I think it's worth noting the product is also evaluable just remembering, besides the well known $\displaystyle \cos\frac{\pi}{3}=\frac{1}{2},$ the somewhat nice $$\displaystyle\cos\frac{\pi}{5}=\frac{1+\sqrt{5}}{4}=\frac{\phi}{2}$$ (where $\phi$ is the golden section) and iterating the sum/difference formula for the c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 5,
"answer_id": 1
} |
this inequality $\prod_{cyc} (x^2+x+1)\ge 9\sum_{cyc} xy$ Let $x,y,z\in R$,and $x+y+z=3$
show that:
$$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 9(xy+yz+xz)$$
Things I have tried so far:$$9(xy+yz+xz)\le 3(x+y+z)^2=27$$
so it suffices to prove that
$$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 27$$
then the problem is solved. I stuck in here
| Assume that $x,y,z$ are roots of the polynomial
$$ q(w) = w^3-3w^2+s w-p \tag{1}$$
where $s=xy+xz+yz$ and $p=xyz$.
In order that $x,y,z\in\mathbb{R}$, the discriminant of $q$ must be non-negative, hence:
$$ 54ps + 9s^2 \geq 4s^3+27p^2+108 p.\tag{2}$$
On the other hand, $2x+1,2y+1,2z+1$ are roots of the monic polynomi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How find $I= \int_{x=0}^{ \frac{1}{2} } \int_{y=x}^{1-x} ( \frac{x-y}{x+y})^{2}\, dy\,dx$ In $$I= \int_{x=0}^{ \frac{1}{2} } \int_{y=x}^{1-x} \left( \frac{x-y}{x+y}\right)^{2} \,dy\,dx$$
follow the change of variables on $x= \frac{1}{2} (r-s),y= \frac{1}{2} (r+s)$ and find$I$
My try
$J=\begin{vmatrix} x'_{r} & x'_{s}... | Off to a good start. Draw the picture! The hard part is determining the integration limits. Really, you just need to understand what the lines of constant $r$ and $s$ mean. Basically, they are lines at 45-degree angles to the axes. So, really, imagine that if $r \in [0,1]$, then what are the limits of $s$? Well, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Why does the equation of a circle have to have the same $x^2,y^2$ coefficients? In one of my geometry texts, it tells me they should be the same but not why. I am unsatisfied with this.
Suppose that:
$$ax^2+by^2 + cx + dy + f = 0 \text{ such that } a \neq b$$
is the equation of some circle. Upon completing the square a... | If $a\ne b$ and both of them are non-zero, then your equation above
$$\begin{align*}
a\left(x+\frac{c}{2a}\right)^2 + b\left(y+\frac{d}{2b}\right)^2 &= \frac{c^2}{4a} + \frac{d^2}{4b} - f\\
\left(x+\frac{c}{2a}\right)^2 + \frac ba\left(y+\frac{d}{2b}\right)^2 &= \frac{c^2}{4a^2} + \frac{d^2}{4ab} - \frac fa\\
\left(x+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Probability that team $A$ has more points than team $B$
Seven teams play a soccer tournament in which each team plays every other team exactly once. No ties occur, each team has a $50\%$ chance of winning each game it plays, and the outcomes of the games are independent. In each game, the winner is awarded a point and... | Hints:
*
*Team A has $5$ more games to play, and so does team B.
*The outcome of games is independent. Thus, for team A or B, the probability that the team wins exactly $k$ more games (after their game with each other) has binomial distribution, namely
$$P(\text{exactly $k$ more games won})={5 \choose k}\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Differentiate the Function: $y=\frac{ae^x+b}{ce^x+d}$ $y=\frac{ae^x+b}{ce^x+d}$
$\frac{(ce^x+d)\cdot [ae^x+b]'-[(ae^x+b)\cdot[ce^x+d]'}{(ce^x+d)^2}$
numerator only shown (') indicates find the derivative
$(ce^x+d)\cdot(a[e^x]'+(e^x)[a]')+1)-[(ae^x+b)\cdot (c[e^x]'+e^x[c]')+1]$
$(ce^x+d)\cdot(ae^x+(e^x))+1)-[(ae^x+b)\cd... | Another method would be to use logarithmic differentiation. Your function $$y=\frac{ae^x+b}{ce^x+d} $$ is identical to $$\ln(y)=\ln\bigg(\frac{ae^x+b}{ce^x+d}\bigg)=\ln(ae^x+b)-\ln(ce^x+d). $$ We find the differentials for both variables $x$ and $y$. $$\frac{1}{y}dy=\bigg(\frac{ae^x}{ae^x+b}-\frac{ce^x}{ce^x+d}\bigg)dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Find the roots of this 6th degree polynomial Hey guys I'm reviewing for a test and I'm getting stuck on one part, I can't remember what to do next.
$x^6+16x^3+64$
$(x^3)^2+16x^3+64$
let $x^3=w$
$w^2+16w+64$
$(w+8)^2$ now substitute again
$(x^3+8)^2$
Now what do I do?
| Then you have to solve $x^3+8=0$:
\begin{align*}
x^3+8=0\Longleftrightarrow
x^3=-8
\end{align*}
But
$$
-8=8(-1)=8e^{i\pi}
$$
Thus
$$
x=2e^{ki\pi/3}\;\;\;\;k=0,1,2.
$$
Hence
$$
x^6+16x^3+64=(x^3+8)^3=(x-2)^3(x-2e^{i\pi/3})^3(x-2e^{2i\pi/3})^3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Given that $2\cos(x + 50) = \sin(x + 40)$ show that $\tan x = \frac{1}{3}\tan 40$ Given that:
$$
2\cos(x + 50) = \sin(x + 40)
$$
Show, without using a calculator, that:
$$
\tan x = \frac{1}{3}\tan 40
$$
I've got the majority of it:
$$
2\cos x\cos50-2\sin x\sin50=\sin x\cos40+\cos x\sin40\\
$$
$$
\frac{2\cos50 - \sin40}... | Using the well-known identity $$\cos (90^{\circ} - x) = \sin x$$ since the cosine function is just a $90^{\circ}$ horizontal translation of the sine function.
Taking $x = 40^{\circ}$ gives us $\cos 50^{\circ} = \sin 40^{\circ}$ and letting $x=50^{\circ}$ establishes the second result. Then $$\begin{align}2\cos x\cos50... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to prove $\cos{\frac{\pi}{11}}$ is a root of I want to show that $x=\cos{\frac{\pi}{11}}$ is a solution of equation :
$$8x^2-4x+\frac{1}{x}-4=4\sqrt{\frac{1-x}{2}}$$
Thanks in advance.
| We have to prove that for $x=\cos\frac{\pi}{11}$ we have:
$$ 4x(2x^2-1)-(4x^2-1)=4x\sqrt{\frac{1-x}{2}}\tag{1}$$
or:
$$ 4\cos\frac{\pi}{11}\cos\frac{2\pi}{11}-\frac{\sin\frac{3\pi}{11}}{\sin\frac{\pi}{11}}=4\cos\frac{\pi}{11}\sin\frac{\pi}{22}\tag{2}$$
that is equivalent to:
$$ \sin\frac{4\pi}{11}-\sin\frac{3\pi}{11}=2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding all groups of order $7$ up to isomorphism? I'm learning group theory but I didn't learn any concepts of building groups.
I know that there exists the identity group $\{e\}$, and the group with 2 elements: $\{e,a\}$.
If I try to create a group with 3 elements, let's say: $\{e,a,b\}$ then we would have:
$ea = a, ... | Groups of order $1$, $2$, $3$, $5$ and $7$ are unique up to isomorphism: for $1$ it's obvious, the others are prime numbers.
Let's tackle the $4$ case; one of the elements is the identity, so the group is $\{1,a,b,c\}$. Every group of even order has an element of order $2$ (a very easy case of Cauchy's theorem). Let's ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1361137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
is there a general formula for cases like $\sqrt{2}$ = $\frac {2}{\sqrt{2}}$? I just noticed that $\sqrt{2}$ is equal to $\frac {2}{\sqrt{2}}$:
$\sqrt{2} = 1.414213562$
$\frac {2}{\sqrt{2}} = 1.414213562$
It is confirmed by a hand-calculator.
I tried to proof this as follows:
$\sqrt{2}$ = $\frac {2}{\sqrt{2}}$
$2^{\fra... | It will be true for any positive real $x$ that
$$\frac{x}{\sqrt{x}} = \frac{x}{x^{1/2}} = x^1\cdot x^{-1/2} = x^{1-1/2} = x^{1/2} = \sqrt{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
How many $5$ element sets can be made?
Let $m$ be the number of five-element subsets that can be chosen from the set of the first $14$ natural numbers so that at least two of the five numbers are consecutive. Find the remainder when $m$ is divided by $1000$.
I think the opposite would be to find, the number of sets w... | We wish to calculate the number of subsets of five elements of the set $\{1, 2, 3, \ldots, 14\}$ in which no two of the numbers are consecutive.
Suppose the five non-consecutive numbers selected from the set $\{1, 2, 3, \ldots, 14\}$ are $x_1, x_2, x_3, x_4, x_5$, where $x_1 < x_2 < x_3 < x_4 < x_5$. Let
\begin{align... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
How find $\lim\limits_{n \to \infty } \sqrt{n} \cdot \sqrt[n]{\left( \lim\limits_{n \to \infty } a_n\right)-a_n}$? Let $a_n= \sqrt{1+ \sqrt{2+\cdots+ \sqrt{n} } }$
How find
$\lim\limits_{n \to \infty } \sqrt{n} \cdot \sqrt[n]{\left( \lim\limits_{n \to \infty } a_n\right)-a_n}$ ?
| Call the limit $\lim\limits_{n\to \infty} a_n = a$.
Let us denote: $$a_{k,n} = \sqrt{k+\sqrt{k+1+\sqrt{\cdots + \sqrt{n}}}}$$
Since, $\displaystyle (a_{k,n+1} - a_{k,n}) = \frac{a_{k+1,n+1} - a_{k+1,n}}{a_{k,n+1}+a_{k,n}}$ we have: $
\displaystyle (a_{n+1} - a_n) = \frac{\sqrt{n+1}}{\prod\limits_{k=1}^{n}(a_{k,n+1}+a_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
"answer_id": 0
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Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital... | You have
$$
-\frac{1}{x^2} + \frac{1}{\sin ^2x}
$$
You can't apply L'Hopital's rules to either of the two fractions above since the numerators do not approach $0$ or $\pm\infty$. But you can first add the fractions and then use L'Hopital's rule:
$$
-\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{-\sin^2 x + x^2}{x^2\sin^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
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Bounds for $\frac{x-y}{x+y}$ How can I find upper and lower bounds for $\displaystyle\frac{x-y}{x+y}$? So I do see that
$$\frac{x-y}{x+y} = \frac{1}{x+y}\cdot(x-y) = \frac{x}{x+y} - \frac{y}{x+y} > \frac{1}{x+y} - \frac{1}{x+y} = \frac{0}{x+y} = 0$$
(is it correct?) but I don't get how to find the upper bound.
| Look at pairs $(x,y)$ such that $x = 1-y$. For these pairs, $\frac{x-y}{x+y} = 1-2y$ and we have $\mbox{lim}_{y\rightarrow \pm\infty}(1-2y) = \mp \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of combinations of selecting $r$ numbers from first $n$ natural numbers of which exactly $m$ are consective. Number of combinations of selecting $r$ numbers from first $n$ natural numbers of which exactly $m$ are consective.
Say $g(n,r,m)$ is the number of such combinations. The two cases of $m=r$ and $m=1$ are ... | You can do this using generating functions. If we let $x^k$ represent $k$ consecutive numbers and one gap, the number of ways to make them fill $n+1$ slots (plus $1$ for the extra gap) is
$$
[x^r]\left(1+x+\cdots+x^m\right)^{n-r+1}=[x^r]\left(\frac{1-x^{m+1}}{1-x}\right)^{n-r+1}\;,
$$
where $[x^r]$ denotes extracting t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=3/2$
Prove that $$\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=\frac{3}{2}$$
I thought of rewriting $$\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$ as $$\cos^2(90^{\circ}+ (\theta +30^{\circ... | Notice,
$$\color{blue}{\sin (A+B)\sin(A-B)=\sin^2A-\sin^2 B}$$
$$\color{blue}{\sin (A+B)+\sin(A-B)=2\sin A\cos B}$$
Now, I will follow OP's steps
$$LHS=\cos^2 \theta+\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$
$$=\cos^2 \theta+\cos^2(90^{\circ}+ (\theta +30^{\circ})) + \cos^2(90^{\circ}-(\theta-30^{\ci... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 6
} |
Value of an expression with cube root radical What is the value of the following expression?
$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}$$
| The expression strangely reminds of Cardano's formula for the cubic equation
$$x=\sqrt[3]{-\frac q2-\sqrt{\frac{q^2}4+\frac{p^3}{27}}}+\sqrt[3]{-\frac q2+\sqrt{\frac{q^2}4+\frac{p^3}{27}}}.$$
Identifying, we have
$$q=-76,p=3,$$
and $x$ is a root of $$x^3+3x-76=0.$$
By inspection (trying values close to $\sqrt[3]{76}$)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Maximize the area of a triangle inscribed in a semicircle. http://i.imgur.com/Q5gjaSG.png
Consider the semicircle with radius 1, the diameter is AB. Let C be a point on the semicircle and D the projection of C onto AB. Maximize the area of the triangle BDC.
My attempt so far, I'm new at these problems and have only don... | Another approach is to use $\angle CBD=\phi\in(0,\frac\pi2)$:
\begin{align}
|BC|&=|AB|\cos\phi=2\cos\phi
\\
|BD|&=|BC|\cos\phi=2\cos^2\phi
\\
S_{\triangle BCD}&=
\tfrac12|BD|\cdot|BC|\sin\phi
\\
&=
2\sin\phi\cos^3\phi
\\
S'(\phi)
&=
2\cos^2\phi(\cos^2\phi-3\sin^2\phi)
\\
&=
2\cos^2\phi(1-4\sin^2\phi)
\end{align}
The o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$ Question:
$ \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6° $
I have partially solved this:-
$$ \sin78^\circ-\sin42^\circ +\sin6^\circ-\sin66^\circ $$
$$ 2\cos\left(\frac{78^\circ+42^\circ}{2}\right) \sin\left(\frac{78^\circ-42^\circ}{2}\right) + 2\cos\le... | You have done your question already. Note that $sin18^\circ=\frac{\sqrt{5}-1}{4}$ and $cos36^\circ=\frac{\sqrt{5}+1}{4}$.If you had done mulitiple angles in trigonometry you can easily calculate these.
For $sin18^\circ$ let, $A=18^\circ$. Then $5A=90^\circ$
$2A=90^\circ-3A$
$Sin2A=cos3A$. Use $Sin2A=2sinAcosA$ and $C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
Triple Integral with difficult limits. I don't understand how to count integral bounds from the inequality.
$$\iiint\limits_{V}(x-2)dV $$ where
$$V=\left\{(x, y, z):\frac{(x-2)^2}{9}+\frac{(y+3)^2}{25}+\frac{(z+1)^2}{16} < 1\right\}$$
| As @JohnMa hinted, we infer from the symmetry about $x-2=0$ that the volume integral is zero. If one wishes to forgo using the symmetry argument, then we can proceed to evaluate the volume integral directly.
METHOD 1:
We use Cartesian coordinates and take as the inner integral, the integration over $x$ and write
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Contradiction between integration by partial fractions and substitution Integration by substitution:
$$\int \frac {dx}{x^2-1}$$
Let $x=\sec\theta$ and $dx=\sec\theta\tan\theta \,d\theta$
$$\int \frac {dx}{x^2-1} = \int \frac{\sec\theta\tan\theta \,d\theta}{\sec^2\theta-1} = \int \frac {\sec\theta\tan\theta\, d\theta}{\... | $$
\ln\frac{|x-1|}{\sqrt{|x^2-1|}} = \ln\frac{\sqrt{|x-1|}\sqrt{|x-1|}}{\sqrt{|x-1|}\sqrt{|x+1|}} = \ln\sqrt{\frac{|x-1|}{|x+1|}} = \frac 1 2 \ln\left|\frac{x-1}{x+1}\right|
$$
In response to comments I've made this more complete than it was. Notice that
*
*I take no square roots of anything except non-negative num... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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Non-linear system of equations Solve following system of equations over real numbers:
$$
x-y+z-u=2\\
x^2-y^2+z^2-u^2=6\\
x^3-y^3+z^3-u^3=20\\
x^4-y^4+z^4-u^4=66
$$
This does not seem as hard problem. I have tried what is obvious here, to write $x^2-y^2$ as $(x-y)(x+y)$, $x^3-y^3$ as $(x-y)(x^2+xy+y^2)$ etc. Problem is ... | rewrite the equations:
$x+z-(y+u)=2\\
x^2+z^2-(y^2+u^2)=6\\
x^3+z^3-(y^3+u^3)=20\\
x^4+z^4-(y^4+u^4)=66$
$a=x+z,c=xz,b=y+u,d=yu, \\x^2+z^2=(x+z)^2-2xz=a^2-2c,\\x^3+z^3=(x+z)(x^2+z^2-xz)=(x+z)((x+z)^2-3xz)=a(a^2-3c)\\x^4+z^4=(x^2+z^2)^2-2(xz)^2=((x+z)^2-2xz)^2-2(xz)^2=a^4-4a^2c+2c^2$
you can go from here now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Notation - matrix calculatoin i am self-studying linear algebra, and came across the following statement about singular value decomposition (X = $USV^T$). I am somewhat confused on how to interpret the (|) and (--) symbols. Should i interpret that X is the sum of r matrices, of which the first one is equal to the (scal... | In the equation, $u_1$ is an $n\times 1$ matrix (i.e., a column vector) and $v_1^t$ is a $1 \times n$ matrix, i.e., a row vector. The product shown is an "outer product", i.e., the result is an $n \times n$ matrix. The little bars are indeed meant to indicate the "shape" of the iterm, i.e., that $u_1$ consists of a bun... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Given that $p$ is an odd prime, is the GCD of any two numbers of the form $2^p + 1$ always equal to $3$? I have checked it for some numbers and it appears to be true. Also I am able to reduce it and get the value $3$ for specific primes $p_1$, $p_2$ by using the Euclidean algorithm but I am not able to find a general a... | Let $p_1, p_2$ be any two distinct prime numbers and assume that
$$q \mid2^{p_1} + 1 \, \mbox{ and } \, q \mid 2^{p_2} + 1$$
Then
$$2^{p_1} \equiv -1 \equiv 2^{p_2} \pmod{q}$$
In particular, there exists a power of $2$ which is $-1 \pmod{q}$. Let $a$ be the smallest number for which $2^a \equiv -1 \pmod{q}$ and let ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Finding $\frac {a}{b} + \frac {b}{c} + \frac {c}{a}$ where $a, b, c$ are the roots of a cubic equation, without solving the cubic equation itself Suppose that we have a equation of third degree as follows:
$$
x^3-3x+1=0
$$
Let $a, b, c$ be the roots of the above equation, such that $a < b < c$ holds. How can we find th... | By Vieta's formulas, we have
$$a+b+c=-\frac{0}{1}=0\tag1$$
$$ab+bc+ca=\frac{-3}{1}=-3\tag2$$
$$abc=-\frac{1}{1}=-1\tag3$$
From $(1)(2)(3)$, we have
$$P=a^2b+ab^2+b^2c+bc^2+c^2a+ca^2=(a+b+c)(ab+bc+ca)-3abc=3\tag4$$
Now, set
$$Q=\frac ab+\frac bc+\frac ca,\ \ \ R=\frac ba+\frac cb+\frac ac.$$
Then, we have
$$Q+R=\frac ab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 1
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Tangent line parallel to another line At what point of the parabola $y=x^2-3x-5$ is the tangent line parallel to $3x-y=2$? Find its equation.
I don't know what the slope of the tangent line will be. Is it the negative reciprocal?
| Edit: since the tangent is parallel to the given line: $3x-y=2$ hence the slope of tangent line to the parabola is $\frac{-3}{-1}=3$
Let the equation of the tangent be $y=3x+c$
Now, solving the equation of the tangent line: $y=3x+c$ & the parabola: $y=x^2-3x-5$ by substituting $y=3x+c$ as follows $$3x+c=x^2-3x-5$$ $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
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Find the sum: $\sum_{i=1}^{n}\dfrac{1}{4^i\cdot\cos^2\dfrac{a}{2^i}}$ Find the sum of the following :
$S=\dfrac{1}{4\cos^2\dfrac{a}{2}}+\dfrac{1}{4^2\cos^2\dfrac{a}{2^2}}+...+\dfrac{1}{4^n\cos^2\dfrac{a}{2^n}}$
| $$S=S+\frac{1}{4^n \sin^2\frac{a}{2^n}}-4^{-n}\csc^2 (a/2^n)\\=\sum_{k=1}^{n-1}\frac{1}{4^k\cos^2(a/2^k)}+\frac{1}{4^{n-1}\sin^2{a/2^{n-1}}}-4^{-n}\csc^2 (a/2^n)\\
=\cdots=\frac{1}{\sin^2(a)}-4^{-n}\csc^2 (a/2^n)=\csc^2(a)-4^{-n}\csc^2 (a/2^n)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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$U_n=\int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ . $U_n= \int_{n^2+n+1}^{n^2+1}\frac{\tan^{-1}x}{(x)^{0.5}}dx$ where
Find $\lim_{n\to \infty} U_n$ without finding the integration
I don't know how to start
| Let $f(x)=\frac{\arctan x}{\sqrt x}$ and then
$$ f'(x)=\frac{2x-(x^2+1)\arctan x}{2x\sqrt x\arctan x}<0\text{ for large }x>0. $$
So $f(x)$ is decreasing for large $x>0$. By the Mean Value Theorem, there is $c\in(n^2+1,n^2+n+1)$ such that
\begin{eqnarray}
\int^{n^2+n+1}_{n^2+1}\frac{\arctan x}{\sqrt x}dx&=&n\frac{\arcta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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Cauchy-Schwarz inequality problem The problems:
*Prove that $$\frac{\sin^3 a}{\sin b} + \frac{\cos^3 a}{\cos b} \geqslant \sec (a-b),$$ for all $a,b \in \bigl(0,\frac{\pi}{2}\bigr)$.
*Prove that $$\frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} + \frac{1}{2\sqrt[3]{abc}} \geqslant \frac{(a+b+c+\sqrt[3]{abc})^2}{(a+b... | You already have an answer for (1). For (2), note by Cauchy Schwarz inequality,
$$LHS = \sum_{cyc} \frac{c^2}{c^2(a+b)}+\frac{(\sqrt[3]{abc})^2}{2abc} \ge \frac{(a+b+c+\sqrt[3]{abc})^2}{\sum_{cyc} c^2(a+b) + 2abc}$$
and further the denominator factorises as
$$\sum_{cyc} c^2(a+b) + 2abc = (a+b)(b+c)(c+a)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to find: $\int^{2\pi}_0 (1+\cos(x))\cos(x)(-\sin^2(x)+\cos(x)+\cos^2(x))~dx$? How to find: $$\int^{2\pi}_0 (1+\cos(x))\cos(x)(-\sin^2(x)+\cos(x)+\cos^2(x))~dx$$
I tried multiplying it all out but I just ended up in a real mess and I'm wondering if there is something I'm missing.
| $\displaystyle\int_0^{2\pi}(\cos x+\cos^2x)(\cos 2x+\cos x)dx=\int_0^{2\pi}\big(\cos x+\frac{1}{2}+\frac{1}{2}\cos 2x\big)\big(\cos 2x+\cos x\big)dx$
$\displaystyle\int_0^{2\pi}\big(\cos x\cos 2x+\cos^{2}x+\frac{1}{2}\cos 2x+\frac{1}{2}\cos x+\frac{1}{2}\cos^{2} 2x+\frac{1}{2}\cos 2x\cos x\big)dx$
$\displaystyle\int_0^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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find total integer solutions for $(x-2)(x-10)=3^y$ I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
How many integer solutions ($x$, $y$) are there of the equation $(x-2)(x-10)=3^y$?
(A)1 (B)2 (C)3 (D)4 (E)5
If let $y=0$, we ha... | By the Fundamental Theorem of Arithmetic, $x-2$ and $x-10$ must be powers of $3$, so we may write $$x-2= \pm 3^a $$ and $$x-10 = \pm 3^b.$$ Substracting the latter from the former we get $$8=\pm 3^a\mp3^b \\ 2^3 =3^a(\mp3^{b-a}\pm1)= 3^b (\pm3^{a-b} \mp1),$$ whence, again by FTA, $(a,b)= (0,2),( 2,0)$ and thus $(x,y)= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Identities involving binomial coefficients, floors, and ceilings I found the following four apparent identities:
$$
\begin{align}
\sum_{k=0}^n
2^{-\lfloor\frac{n+k}{2}\rfloor}
{\lfloor\frac{n+k}{2}\rfloor\choose k}
&=
\frac{4}{3}-\frac{1}{3}(-2)^{-n},\\
\sum_{k=0}^n
2^{-\lceil \frac{n+k}{2}\rceil }
{\lflo... | Assuming that $n$ is even, $n=2m$, we have:
$$\begin{eqnarray*} \sum_{k=0}^{n} 2^{-\left\lfloor\frac{n+k}{2}\right\rfloor}\binom{\left\lfloor\frac{n+k}{2}\right\rfloor}{k}&=&\sum_{j=0}^{m}2^{-(m+j)}\binom{m+j}{2j}+\sum_{j=0}^{m-1}2^{-(m+j)}\binom{m+j}{2j+1}\\&=&4^{-m}+2^{-m}\sum_{j=0}^{m-1}2^{-j}\binom{m+j+1}{m-j}\\&=&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Is this:$\sum_{n=1}^{\infty}{(-1)}^{\frac{n(n-1)}{2}}\frac{1}{n}$ a convergent series? Is there someone who can show me how do I evaluate this sum :$$\sum_{n=1}^{\infty}{(-1)}^{\frac{n(n-1)}{2}}\frac{1}{n}$$
Note : In wolfram alpha show this result and in the same time by ratio test it's not a convince way to judg that... | After $ \ n = 1 \ $ , the exponents of (-1) are binomial coefficients which are "double-alternating" between even and odd integers. So the series looks like
$$ 1 \ - \frac{1}{2} \ - \ \frac{1}{3} \ + \ \frac{1}{4} \ + \ \frac{1}{5} \ - \ \frac{1}{6} \ - \ \frac{1}{7} \ + \ \ldots \ \ . $$
As lulu I think properly obje... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Proving the Fibonacci sum $\sum_{n=1}^{\infty}\left(\frac{F_{n+2}}{F_{n+1}}-\frac{F_{n+3}}{F_{n+2}}\right) = \frac{1}{\phi^2}$ and its friends In this article, (eq.92) has,
$$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{F_{n+1}F_{n+2}} = \frac{1}{\phi^2}\tag1$$
and I wondered if this could be generalized to the tribonacci numb... | Let's start off with the case of the Fibonacci sequence. We have
$$\sum_{n = 1}^k \left( \frac{F_{n+2}}{F_{n+1}} -\frac{F_{n+3}}{F_{n+2}} \right) = \left(\frac{F_3}{F_2} - \frac{F_4}{F_3}\right) + \left(\frac{F_4}{F_3} - \frac{F_5}{F_4}\right) + \cdots + \left(\frac{F_{k+2}}{F_{k+1}}- \frac{F_{k+3}}{F_{k+2}}\right)\\
=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Find expression in terms of x Knowing that $$\frac{dy}{dx}= k\cdot x^{\frac{1}{3}}$$ and given that it passes through points $(1,4)$ and $(8,16)$, find an expression for the path in terms of $x$.
I found out that $$y= \frac34 k x^{\frac43}$$ by integrating $$\frac{dy}{dx}.$$
What do I do next?
| We have, $$\frac{dy}{dx}=kx^{\frac{1}{3}}$$ $$\implies dy=kx^{\frac{1}{3}}dx$$ Integrating both the sides w.r.t. $x$ as follows $$\int dy=\int kx^{\frac{1}{3}}dx$$ $$\implies y=k\int x^{\frac{1}{3}}dx=k\left(\frac{x^{\frac{1}{3}+1}}{\frac{1}{3}+1}\right)+C$$ $$\implies y=\frac{3kx^{\frac{4}{3}}}{4}+C$$ Since, the curve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
remainder of $a^2+3a+4$ divided by 7
If the remainder of $a$ is divided by $7$ is $6$, find the remainder when $a^2+3a+4$ is divided by 7
(A)$2$ (B)$3$ (C)$4$ (D)$5$ (E)$6$
if $a = 6$, then $6^2 + 3(6) + 4 = 58$, and $a^2+3a+4 \equiv 2 \pmod 7$
if $a = 13$, then $13^2 + 3(13) + 4 = 212$, and $a^2+3a+4 \equiv 2 \pmod ... | $a = 6 \quad(\mathrm{mod} 7)$
$a^2 = 36 = 1 \quad(\mathrm{mod} 7)$
$3a = 18 = 4\quad (\mathrm{mod} 7)$
$a^2 + 3a + 4 = 1 + 4 + 4 = 9 = 2 \quad(\mathrm{mod} 7)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Geometric problem based on angle bisectors I am not asking a question,i just want to conform,is my method of solving problem correct?
Given a triangle ABC.It is known that AB=4,AC=2,and BC=3.The bisector of angle A intersects the side BC at point K.The straight line passing through point B and being parallel to AC int... | You have a mistake
$$b^2 n + c^2 m =a(d^2 + mn)$$
$$2^2 (2) + 4^2 (1) = 3(d^2 + 2)$$
$$8 + 16 = 3(d^2 + 2)$$
$$24 = 3(d^2 + 2) \Rightarrow d^2 = 6$$
Hence $KM^2 = (2AK)^2 = 4(AK^2) = 24$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $x$ in the triangle
the triangle without point F is drawn on scale, while I made the point F is explained below
So, I have used $\sin, \cos, \tan$ to calculate it
Let $\angle ACB = \theta$, $\angle DFC = \angle BAC = 90^\circ$, and $DF$ is perpendicular to $BC$ (the reason for it is to have same $\sin, \cos, ... | Let, $\angle ACB=\alpha$ then we have $$AC=BC\cos \alpha=(9+3)\cos\alpha=12\cos\alpha$$ $$\implies DC=AD=\frac{AC}{2}=6\cos\alpha$$
in right $\triangle DFC$ we have
$$DF=(DC)\sin\alpha=6\sin\alpha\cos\alpha$$ $$FC=(DC)\cos\alpha=6\cos\alpha\cos\alpha=6\cos^2\alpha$$ $$FE=FC-CE=6\cos^2\alpha-3$$ Now, applying Pythago... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Finding a triangle ABC if $2\prod (\cos \angle A+1)=\sum \cos(\angle A-\angle B)+\sum \cos \angle A+2$ Find $\triangle ABC$ if $\angle B=2\angle C$ and $$2(\cos\angle A+1)(\cos\angle B+1)(\cos\angle C+1)=\cos(\angle A-\angle B)+\cos(\angle B-\angle C)+\cos(\angle C-\angle A)+\cos\angle A+\cos\angle B+\cos\angle C+2$$
| Hint:$$A+B+C=\pi,B=2C\Rightarrow A=\pi-3C$$
$\cos A=-\cos 3C$
$\cos B=\cos 2C$
$\cos(A-B)=-\cos 5C$
$\cos(B-C)=\cos C$
$\cos(C-A)=-\cos 4C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"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.