Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Nested... binomials coefficients? Can I have a proof that this number exists?
The number:
$$\binom{1}{\binom{2}{\binom{3}{\binom{4}{\vdots}}}}$$
If the number exists, then what is the closed form of that number?
| Note that ${i\choose i+1} = 0$.
So $A_2 ={1 \choose 2} = 0$,
$A_3 ={1\choose{2\choose 3}} = {1 \choose 0} = 1$
Now $A_4={1\choose{2\choose {3\choose 4}}} = {1\choose{2\choose 0}} = {1\choose{1}} = 1$
Lastly
$$A_5={1\choose{2\choose {3\choose {4\choose 5}}}} = {1\choose{2\choose {3 \choose 0}}} = {1\choose{2\choose 1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Showing that hyperbolic trigonometric functions parameterize the unit hyperbola I know that the same way circular trigonometry is defined over the circle $ x^2 + y^2 = 1 $, hyperbolic trigonometry is defined over the hyperbola $ x^2 - y^2 = 1 $.
What I don't know is how deduced the formulas
$$ \sinh x = \frac {e^x - e^... | I'll prove $\sinh(x)$ = $\frac {e^x - e^{-x}} {2}$ and leave the proof of $\cosh(x)$ as an exercise for you. So, $\sinh(x) = x + \frac{x^3}{3!} + \frac{x^5}{5!} + ...+\infty$
And we know that, $$e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + ...$$
And we also know that $$e^{-x} = 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Application of diagonalization of matrix - Markov chains Problem: Suppose the employment situation in a country evolves in the following manner: from all the people that are unemployed in some year, $1/16$ of them finds a job next year. Furthermore, from all the people that were employed in some year, $1/8$ of them los... | The method is correct.
The eigenvector corresponding to $\lambda_1=1$ should be
$$v_1 = \begin{pmatrix} 1/2 \\ 1 \end{pmatrix}$$
The signs were wrong in the matrix $\lambda_1I_2-A$.
The signs were also wrong for the other eigenvector. You must have made the same mistake.
| {
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"url": "https://math.stackexchange.com/questions/1365841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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proof by induction $2^n \leq 2^{n+1}-2^{n−1}-1$ My question is prove by induction for all $n\in\mathbb{N}$, $2^n \leq 2 ^{n+1}-2^{n−1}-1$
My proof
$1+2+3+4+....+2^n \leq 2^{n+1}-2^{n−1}-1$
Assume $n=1$,$1 ≤ 2$
Induction step
Assume statement is true for $n=k$, show true for $n=k + 1$
$1+2+3+4+....+2^k+2^k+1 ≤ 2 ^{k+1... | Alternatively, we could prove it directly without induction:
\begin{align*}
2^{n + 1} - 2^{n - 1} - 1
&= (2^n + 2^n) - 2^{n - 1} - 1 \\
&= 2^n + (2^{n - 1} + 2^{n - 1}) - 2^{n - 1} - 1 \\
&= 2^n + 2^{n - 1} - 1 \\
&\geq 2^n + 2^{1 - 1} - 1 &\text{since } n \in \mathbb N \implies n \geq 1 \\
&= 2^n
\end{align*}
as desir... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving a system of five polynomials I am trying to solve the following system of equations for tuple $\left(a,b,c,d,t\right) \in \mathbb{R}^{4} \times [0,1]$, with parameter $\ell\in\mathbb{R}$.
$$
\begin{eqnarray}
a\frac{t^{2}}{2} - bt + 1 = 0 \qquad (1)\\
a \frac{t^{3}}{6} - b \frac{t^{2}}{2} + t - \ell = 0 \qquad (... | Puting $t=0$ in Eq(1) and $t=1$ in Eq(3) shows us that both $0$ and $1$ are not possible values for t. So, dividing by $t$ and $t-1$ should be fine.
Solving Eq(1) and Eq(2) for $a$ and $b$ in terms of $l$ and $t$ we get
\begin{eqnarray*}
a &=& \frac{6(t-2l)}{t^3} \\
b &=& \frac{2(2t-3l)}{t^2} \\
at - b &=& \frac{6t-12l... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Linear transformation, transformation in base1 I have the following problem, I know how to solve it (knowing the answer) however I do not understand what I'm doing
We have the following base in $\mathbb{R}^4$, called $b_1$,$b_2$,$b_3$,$b_4$
$$B = \left\{
\vphantom{\begin{array}{c}1\\1\\1\\1\\1\end{arr... | You don't need to use inverse matrix (you need it if you change base). Let $f\colon V\to V$ be a linear transformation ($V$ for vector space). If $B=\{e_1, \ldots, e_n\}$ is a basis of $V$, and
$$
f(e_1) = a_{11}e_1 + a_{21}e_2 + \ldots + a_{n1}e_n\\
\cdots\\
f(e_n) = a_{n1}e_1 + a_{n1}e_2 + \ldots + a_{nn}e_n\\
$$
the... | {
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Simplifying $ 2\cos^2(x)\sin^2(x) + \cos^4(x) + \sin^4(x)$ who can simplify the following term in a simplest way?
$$ 2\cos^2(x)\sin^2(x) + \cos^4(x) + \sin^4(x)$$
(The answer is 1). Thanks for any suggestions.
| HINT :
$$2ab+a^2+b^2=(a+b)^2$$
with
$$\cos^2 (x)+\sin^2(x)=1.$$
| {
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Integral of square of variable in expression in square root in denominator I'm having trouble with following integral
$$
\int \frac{1}{\sqrt{1+4x^2+4y^2}} dy
$$
that according to WolframAlpha is
$$
\frac{1}{2} \ln( \sqrt{1+4x^2+4y^2} + 2y) + c
$$
which I can verify but I can't figure out what substitution to apply or w... | Notice, the standard formula $$\color{red}{\int\frac{dx}{\sqrt{a^2+x^2}}=\ln(x+\sqrt{a^2+x^2})+c}$$ For derivation, let $x=a\tan \theta \implies dx=a\sec^2\theta d\theta $ $$\int\frac{dx}{\sqrt{a^2+x^2}}=\int\frac{a\sec^2 \theta d\theta}{\sqrt{a^2+(a\tan\theta)^2}}=$$ $$=\int\frac{\sec^2 \theta d\theta}{\sec\theta}=\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370838",
"timestamp": "2023-03-29T00:00:00",
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Wanted : for more formulas to find the area of a triangle? I know some formulas to find a triangle's area, like the ones below.
*
*Is there any reference containing most triangle area formulas?
*If you know more, please add them as an answer
$$s=\sqrt{p(p-a)(p-b)(p-c)} ,p=\frac{a+b+c}{2}\\s=\frac{h_a*a}{2}\\s=... | *
*$s=pr$ where $p=\frac{a+b+c}{2}$ and $r$ is the radius of the inscribed circle.
*$s=\sqrt{r\cdot r_a\cdot r_b\cdot r_c}$ where $r_a,r_b,r_c$ are the exradii of excircles.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
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Given $\tan A + \tan B = 3x$ and $\tan A \tan B = 2x^2$, find $\tan A - \tan B$ Given $$\tan A + \tan B = 3x$$ and $$\tan A \tan B = 2x^{2},$$ how can one find $\tan A - \tan B$? I have tried substitution, but failed to find the answer.
Edit: Can this problem be solved using the formulas for sums and differences of tan... | $tanA+tanB=3x$
Squaring both sides,we get
$tan^2A+tan^2B+2tanA tanB=9x^2$
Subtract $4tanA tanB$ from both sides,
$tan^2A+tan^2B-2tanA tanB=9x^2-4tanA tanB$
$(tan A-tan B)^2=9x^2-4tanA tanB$
Put $tanA tanB$ in above equation,
$(tan A-tan B)^2=9x^2-8x^2=x^2$
$tan A-tan B=\pm x$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$6$ eigenvalues of a $4\times4$-matrix? I am struggling with determining the eigenvalues of the following (symmetric) matrix:
$$ A =\begin{pmatrix}
2 & 1 & 0 & 0 \\
1 & 2 & 0 & 0 \\
0 & 0 & 2 & 1 \\
0 & 0 & 1 & 2
\end{pmatrix} $$
What I did : $$ \operatorname{char}_X(A) =\det\begin{pmatrix}
2 -X & 1 & 0 & 0 \\
1 & 2 -... | The asserted equality of two determinants is incorrect. Let $M$ be the matrix whose determinant appears first; let $N$ be the matrix whose determinant appears after the "equals" sign.
You multiplied the second row by $2-X$. That alters the determinant. Then you added $-1$ times the first row to the second, and that ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Infinite limit of trigonometric series The value of $\displaystyle\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx))$ is
(A) $\sin^4x$
(B) $\sin^2x$
(C) $\cos^2x$
(D) does not exist
My attempt:
$$\lim_{n\to\infty}(\sin^4x+\frac{1}{4}\sin^4(2x)+\cdots+\frac{1}{4^n}\sin^4(2^nx))=$$
$$=(\sin... | Hint: Note that $\sin^2 \theta - \sin^4\theta = \sin^2\theta(1-\sin^2\theta) = \sin^2\theta\cos^2\theta = \dfrac{1}{4}\sin^2 2\theta$.
Hence, $\sin^4\theta = \sin^2\theta - \dfrac{1}{4}\sin^2 2\theta$.
Applying that here gives us $\dfrac{1}{4^k}\sin^4(2^k x) = \dfrac{1}{4^k}\sin^2(2^k x) - \dfrac{1}{4^{k+1}}\sin^2(2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Definite integral question Let $ f(x)$ be a quadratic equation with $f'(3)=3$. If $I=\int_{0}^{\frac{\pi}{3}}t \times \tan(t)dt $ and the value of integral$\int_{3-\pi}^{{3+\pi}}f(x) \times \tan(\frac{x-3}{3})dx $ is equal to $kI$.Then find k.
My attempt:
Put $\frac{x-3}{3}=p$ in below integral
$$\int_{3-\pi}^{{3+\pi}}... | Put $\frac{x-3}{3}=p$ in below integral
$\int_{3-\pi}^{{3+\pi}}f(x) \times \tan\left(\frac{x-3}{3}\right)dx=\int_{\frac{-\pi}{3}}^{\frac{\pi}{3}}f(3p+3) \times \tan(p)\times 3\ dp $
Let $f(x)=Ax^2+Bx+C\Rightarrow f'(x)=2Ax+B\Rightarrow f'(3)=6A+B=3$(given in the question)
$f(3p+3)=A(3p+3)^2+B(3p+3)+C=9Ap^2+18Ap+9A+3Bp+... | {
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"timestamp": "2023-03-29T00:00:00",
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evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
Evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
How to evalute this equation without using calculator?
| We have that $\pm\sqrt{2}\pm\sqrt{3}$ are the roots of the polynomial $(x^2-5)^2-24$, hence:
$$ x^4-10\,x^2+1 = \prod_{\xi_i\in Z}(x-\xi) $$
and $1\pm\sqrt{2}\pm\sqrt{3}$ are the roots of the polynomial:
$$ (x-1)^4-10(x-1)^2+1 = x^4-4x^3-4x^2+16x-8. $$
By Viète's theorem, the sum of the roots of a polynomial $p(x)$ rai... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of $dy/dx$ at $x=8$
Given that variables $xy=40$, find $dy/dx$ at $x=8$.
I used $40/8$ to get $y=5$.
So why is the answer $-5/8$ and not $5/8$?
| We have that
$$xy=40 \implies y = \frac{40}{x}. $$
Remember that
$$\frac{d}{dx}\frac{1}{x} = - \frac{1}{x^2}$$
so
$$\frac{dy}{dx} = \frac{d}{dx}\frac{40}{x} = 40\frac{d}{dx}\frac{1}{x} =- \frac{40}{x^2}. $$
Therefore we have, when $x = 8$, that
$$\frac{dy}{dx} = -\frac{40}{8^2} = -\frac{40}{64} = -\frac58. $$
| {
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Show that $B$ is invertible if $B=A^2-2A+2I$ and $A^3=2I$ If $A$ is $40\times 40$ matrix such that $A^3=2I$ show that $B$ is invertible where $B=A^2-2A+2I$.
I tried to evaluate $B(A-I)$ , $B(A+I)$ , $B(A-2I)$ ... but I couldn't find anything.
| One of the easiest ways to prove $B$ is invertible is the following;
$$B=A^2-2A+2I$$
Using that $A^3=2I$, we can write
$$B=A^2-2A+2I= A^2-2A + A^3 = A(A^2+A-2I)=A(A+2I)(A-I)$$
Now let us prove $A$, $A+2I$ and $A-I$ are invertible.
*
*Supose there is $v$ such that $Av=0$, then we have $0=A^20=A^3v=2Iv=2v$. So we ha... | {
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Finding the equation of a circle. A circle of radius $2$ lies in the first quadrant touching both axis. Find the equation of the circle centered at $(6,5)$ and touching the above circle externally.
Let me share how I answered this question with your suggestions.
Since the radius of the first circle is $2$ and touches b... | Equation of circle of radius 2 and touching the axes is $(x-2)^2+(y-2)^2=4$ and its center is $(2,2)$.
Center of required circle is (6,5).Radius of required circle is half of distance between (6,5) and (2,2) because these two circles touch externally.Therefore radius of required circle$=r=5-2$
Now use $(x-h)^2+(y-k)^2=... | {
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Does $\frac{x+y}{2}>\frac{a+b}{2}$ hold? $a$ and $b$ are two real positive numbers. Given that $x=\sqrt{ab}$ and $y=\sqrt{\frac{a^2+b^2}{2}}$, which one has a higher value, $\frac{x+y}{2}$ or $\frac{a+b}{2}$?
We know that $y=\sqrt{\frac{a^2+b^2}{2}}>\frac{a+b}{2}>x=\sqrt{ab}$ by inequality, and at this point I'm stuck.... | we will prove that $$\frac{x+y}{2}\le \frac{a+b}{2}$$ after squaring is this equivalent to $$2\sqrt{ab}\sqrt{\frac{a^2+b^2}{2}}\le \frac{(a+b)^2}{2}$$ squaring again we get $$ab\left(\frac{a^2+b^2}{2}\right)\le \frac{(a+b)^4}{16}$$ and this is equivalent to $$0\le (a-b)^4$$ and this is true.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Regularity and invertibility of two parameterized matrices?
$$
C=
\begin{bmatrix}
1+a & 2 & 3 & 4 & 5 \\
1 & 2+a & 3 & 4 & 5 \\
1 & 2 & 3+a & 4 & 5 \\
1 & 2 & 3 & 4+a & 5 \\
1 & 2 & 3 & 4 & 5+a \\
\end{bmatrix}
$$
For which values of $a$ is the matrix $C$ regular?
$$
A=
\begin{bmatrix}
3a & a \\
-a & 1 \\
\end... | Actually, there is an easier way to find the values for the matrix $C$ for which it is invertible:
Look at the following representation
$$
\begin{pmatrix}
1+a & 2 & 3 & 4 & 5 \\
1 & 2+a & 3 & 4 & 5 \\
1 & 2 & 3+a & 4 & 5 \\
1 & 2 & 3 & 4+a & 5 \\
1 & 2 & 3 & 4 & 5+a \\
\end{pmatrix}=
\begin{pmatrix}
1 & 2 & 3 & 4 ... | {
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"timestamp": "2023-03-29T00:00:00",
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Finding $P$ such that $P^TAP$ is a diagonal matrix
Let $$A = \left(\begin{array}{cc} 2&3 \\ 3&4 \end{array}\right) \in
M_2(\mathbb{C})$$
Find $P$ such that $P^TAP = D$ where $D$ is a diagonal matrix.
So here's the solution:
$$A = \left(\begin{array}{cc|cc} 2&3&1&0\\ 3&4&0&1 \end{array}\right) \sim \left(\begin{ar... | Generally, in the process of diagonalization, it is easiest to approach via calculating the eigenvalues and corresponding eigenvectors to form an orthonormal eigenbasis.
Such an orthogonal matrix is guaranteed to exist by the Spectral Theorem since our matrix, $A$, is a real symmetric matrix.
step 1: calculate eigenval... | {
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How many pairs of $(x, y)$ satisfied this equation I need help to solve in $\mathbb{Z}$ the following equation
$$yx^{2}+xy^{2}=30$$ I tried to solve it by factor $30$ to $5\times 6$ and I get those two pairs $(2, 3) \& (3, 2) $... is their any other pairs of $(x, y) $? ..
| Factor it as $xy(x+y)=30$. Write $30=2\times3\times5$. Therefore $y$ has to be a factor of $30$.
So $y=\pm1,\pm2,\pm3,\pm5,\pm6,\pm10,\pm15,\pm30$.
However, if $y\geq6$ and $x\geq1$ then $xy(x+y)\geq42$. If $y\geq6$ and $x\leq-1$ then we must also have $x\leq-7$ since otherwise one factor is negative and two are posi... | {
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How does this algebraic trick regarding partial fraction works? Suppose I have to evaluate the integral $$\int \frac{x}{(x-1)(2x+1)(x+3)} \, dx $$
I write it as $$\frac{a_1}{x-1} +\frac{a_2}{2x+1} +\frac{a_3}{x+3}$$
where $a_1$, $a_2$, $a_3$ are constants. once I have foud relationaships between them I solve for the $... | We have $$\frac{x}{(x-1)(2x+1)(x+3)}=\frac{a_1}{x-1}+\frac{a_2}{2x+1}+\frac{a_3}{x+3}.$$
Then, we get $$(x-1) \times \frac{x}{(x-1)(2x+1)(x+3)}=\frac{x}{(2x+1)(x+3)}=a_1+\frac{a_2(x-1)}{2x+1}+\frac{a_3(x-1)}{x+3}. $$ If we take $x=1,$ then we find
$$\frac{1}{3 \times 4}=\frac{1}{12}=a_1.$$ We proceed similarly, we get... | {
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"timestamp": "2023-03-29T00:00:00",
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Solving $x^2\equiv a\pmod{p}$ where $p\equiv5\pmod{8}$ and $a$ is a quadratic residue The exercise asks me to show that one of the values $x = a^{(p+3)/8}$ and $x = 2a \cdot (4a)^{(p-5)/8}$ is a solution to $x^2\equiv a\pmod{p}$, where $p\equiv5\pmod{8}$ is a prime and $a$ is a quadratic residue. My reasoning was that ... | Recall that $p \equiv 5 \mod 8$, and thus $p \not\equiv \pm 1 \mod 8$. Thus, the second supplement to the law of Quadratic Reciprocity shows that $2$ is a quadratic non-residue modulo $p$. In other words, the Legendre symbol $\left(\dfrac{2}{p}\right) = -1$. But Legendre's formula says that $\left(\dfrac{a}{p}\right) \... | {
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Evaluation of $ \int\sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}}\cdot \frac{\sec x}{\sqrt{1+2\sec x}}dx$
Evaluation of $\displaystyle \int\sqrt{\frac{\csc x-\cot x}{\csc x+\cot x}}\cdot \frac{\sec x}{\sqrt{1+2\sec x}}dx$
$\bf{My\; Solution::}$ We can write it as $$\displaystyle \int \sqrt{\frac{1-\cos x}{1+\cos x}}\cdo... | Everything is Ok, for matching the answer See here, $$\displaystyle -\int \frac{1}{t}\times \frac{1}{\sqrt{t^2-1}}dt =\csc^{-1}(t)+\mathcal{C} = \csc^{-1}\left(1+\cos x\right)+\mathcal{C}\\
= \csc^{-1}\left(2\cos^2 \frac{x}{2}\right)+\mathcal{C}$$
Now use, $$\csc^{-1}x=\sin^{-1}(1/x)$$ You will get answer as you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1387431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Show that $\sqrt{6+4\sqrt{2}}-\sqrt{2}$ is rational using the rational zeros theorem Let $r=\sqrt{6+4\sqrt{2}}-\sqrt{2}$, then $r+\sqrt{2}=\sqrt{6+4\sqrt{2}}$.
Squaring both sides, we get $$r^2+2r\sqrt{2}+2=6+4\sqrt{2}$$ which is the same as $r^2-4=2\sqrt{2}$.
Squaring both sides again, we get $r^4-8r^2+16=8$ or $$r^4-... | Check your second line, $r^2+2r\sqrt{2} + 2 = 6 + 4\sqrt{2} \ne r^2-4 -2\sqrt{2}$.
It does, however, equal $ (r-2) (r+2 \sqrt{2}+2) = 0$ which would imply that $r = 2$ or $r= -2(1+\sqrt{2})$.
Thus $2$ is the only rational root of your polynomial and $\sqrt{6+4\sqrt{2}} -\sqrt{2}= 2 $ (since $r =\sqrt{6+4\sqrt{2}} -\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1388369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Reflection formula for Hurwitz Zeta function? In doing some calculus with Mathematica today, I found that
$$\zeta\left(3,\frac{1}{4}\right) - \zeta\left(3,\frac{3}{4}\right) = 2\pi^3$$
by numerically evaluating both sides. Here, $\zeta(x,y)$ denotes the Hurwitz Zeta function defined by
$$\zeta(x,y) = \sum_{n=0}^\infty ... | First define the $\beta(x)$ function by
\begin{align}
\beta(x) = \frac{1}{2} \, \left[ \psi\left(\frac{x+1}{2}\right) - \psi\left(\frac{x}{2}\right) \right] = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{x+k}
\end{align}
for which after two derivatives with respect to x and setting $x = 1/2$ the result becomes
\begin{align}
16 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1390716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving $\frac{1}{1\cdot3} + \frac{1}{2\cdot4} + \cdots + \frac{1}{n\cdot(n+2)} = \frac{3}{4} - \frac{(2n+3)}{2(n+1)(n+2)}$ by induction for $n\geq 1$ I'm having an issue solving this problem using induction. If possible, could someone add in a very brief explanation of how they did it so it's easier for me to understa... | First, check the formula for $n=1$. So:
$$\frac1{1\cdot 3}=\frac34-\frac{2\cdot1+3}{2(1+1)(2+1)}$$
Since this is true, we have shown the so called base case.
Now substitute in the formula $n$ by $n+1$ to get the statement that you have to show:
$$\frac1{1\cdot 3}+\frac1{2\cdot 4}+\cdots+\frac1{(n+1)(n+3)}=\frac34-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1391185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
How do I integrate $\frac {\sin^3x}{\cos^2x}$ How do I integrate $\frac {\sin^{3}x}{\cos^{2}x}$. I have tried to convert to $\tan$, but I could not reach to conclusion. Any help will be appreciated.
Thanks.
| $$\int { \frac { \sin ^{ 3 }{ x } }{ \cos ^{ 2 }{ x } } dx=\int { \frac { \sin { x } \cdot \sin ^{ 2 }{ x } }{ \cos ^{ 2 }{ x } } dx=-\int { \frac { 1-\cos ^{ 2 }{ x } }{ \cos ^{ 2 }{ x } } d\cos { x } } } } =\\ =\int { d\left( \cos { x } \right) } -\int { \frac { 1 }{ \cos ^{ 2 }{ x } } d\left( \cos { x } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
maximum value of $a+b+c$ given $a^2+b^2+c^2=48$? How can i get maximum value of this $a+b+c$ given $a^2+b^2+c^2=48$ by not using AM,GM and lagrange multipliers .
| By the arithmetic-quadratic mean
inequality,
$\frac{a+b+c}{3}
\le \sqrt{\frac{a^2+b^2+c^2}{3}}
= \sqrt{\frac{48}{3}}
=\sqrt{16}
=4
$
with equality iff
$a=b=c$.
Therefore
$a+b+c = 12$
and
$a=b=c = 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1393590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Why am I getting two answers for 8th root of continued fraction Find value of $x$:
$x=\sqrt[8]{2207-\frac{1}{2207-\frac{1}{2207-....and\,so\, on}}}$
On solving ,we have $x^8=2207-\frac{1}{x^8}$
$x^8+\frac{1}{x^8}=2207$
$x^4+\frac{1}{x^4}=47$
$x^2+\frac{1}{x^2}=7$
$x+\frac{1}{x}=3$
$x=\frac{3+\sqrt{5}}{2}\,,\frac{3-\sqr... | You want the value which is greater than $1$ - clearly by estimating.
From $x^8+\frac 1{x^8}=2207$ you know that if $x$ satisfies that equation so will $\frac 1x$. Put $\frac 1x$ in the original equation and modify the eighth root accordingly and you will see where the alternative answer comes from and that this is cle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1396410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Let $a^n = a^{n - 1} + a^{n -2}$. Show that for any $A, B$, $F(n) = Aa^n + Bb^n$ satisfies Fibonacci recurrence relation. $$\begin{align*}
F(n) &= Aa^n + Bb^n\\
&= A(a^{n-1}+a^{n-2}) + B(b^{n-1}+b^{n-2}) \\
&= Aa^{n -1} + Aa^{n-2} + Bb^{n -1} + Bb^{n-2}\\
&= a^{n -1} (A + A^{a-1}) + b^{n - 2} (B + bB)
\end{align*}$$
Us... | Let me try rewriting the question and giving an answer:
Restated question
Suppose that $a$ and $b$ are the two solutions to $x^2 - x - 1 = 0$, so that we have $a^2 = a + 1$ (and similarly for $b$) and hence also
\begin{align}
a^n &= a^{n-1} + a^{n-2}& \text{ (**) }
\end{align}
for $n \ge 2$, and similarly for $b$.
Sh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Convergence of improper integral $\int_{0}^{\frac{\pi}{6}}\dfrac{x}{\sqrt{1-2\sin x}}dx$ I'm trying to determine whether the following improper integral is convergent or divergent.
$$
\int_{0}^{\pi/6}\frac{x}{\sqrt{1-2\sin x}}dx
$$
At first, I substituted $t=\dfrac{\pi}{2} - x $
and then I used $1-\dfrac{1}{2}x^2 \le ... | Since we are only interested in whether the improper integral converges or diverges, let us write out the integrand in terms of their power series and see what we get:
$$ 2 \sin x = 2 \cdot \sum_{n=0}^{\infty} (-1)^n \frac{x^{(2n+1)}}{(2n+1)!} = 2x - \frac{x^3}{3}+\frac{x^5}{60}+ O(x^7)$$
Then we have:
$$ 1 - 2 \sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1397735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$U_n=\int_{0}^{1}x^n(2-x)^ndx,V_n=\int_{0}^{1}x^n(1-x)^ndx,n\in N$ For $U_n=\int_{0}^{1}x^n(2-x)^ndx,V_n=\int_{0}^{1}x^n(1-x)^ndx,n\in N$,which of the following statement is/are true
$(A)\ U_n=2^nV_n\hspace{1cm}(B)\ U_n=2^{-n}V_n\hspace{1cm}(C)\ U_n=2^{2n}V_n\hspace{1cm}(D)\ U_n=2^{-2n}V_n$
$V_n $does not change its fo... | Given $$\displaystyle U_{n} = \int_{0}^{1}x^n\cdot (2-x)^ndx$$ and $$\displaystyle V_{n} = \int_{0}^{1}x^n\cdot (1-x)^ndx$$
Now Let $x=2y$ in $U_{n}\;,$ and $dx = 2dy$ and changing Limit, We get $\displaystyle $
$$\displaystyle U_{n}=2\int_{0}^{\frac{1}{2}}2^n\cdot y^n\cdot 2^n\cdot (1-y)^n dy $$
$$\displaystyle = 2^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1399364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove this identity of ceiling function? My book writes down this identity of least integer function:
$$\lceil x\rceil +\left\lceil x + \frac{1}{n}\right \rceil + \left\lceil x + \frac{2}{n}\right \rceil + \cdots +\left\lceil x + \frac{n -1}{n}\right \rceil = \lceil nx\rceil + n-1 $$.
It didn't deduce it, howev... | Another way of doing it is $$f(x)=\lceil x\rceil+\left\lceil x+\frac1n\right\rceil+\dots +\left\lceil x+1-\frac1n\right\rceil - \left\lceil nx\right\rceil$$
We have $$
\begin{align}
f\left(x+\frac{1}{n}\right)=&\left\lceil x+\frac1n\right\rceil+\left\lceil x+\frac2n\right\rceil+\dots +\left\lceil x+1\right\rceil - \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find $\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}$ Find:
$$\lim\limits_{x\to0}\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}$$
I used L'Hospital's rule, but after second application it is still not possible to determine the limit. When applying Taylor series, I get wrong result ($\frac{-1}{6}$). What met... | L'Hospital's rule is not the α and ω of limits computation. First remove the square roots in the numerator:
$$\frac{\sqrt{1+\tan x}-\sqrt{1+x}}{\sin^2x}=\frac{\tan x-x}{(\sqrt{1+\tan x}+\sqrt{1+x})\sin^2x}$$
Now use equivalents:
*
*$\tan x-x=x+\dfrac{x^3}3+o(x^3)-x$, hence $\;\tan x-x\sim_0 \dfrac{x^3}3$
*$\sqrt{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1400657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
3Finding minimum $f(k)$ ($n$: fixed natural number and $k=0,1,2,\cdots,n-1$ I would appreciate if somebody could help me with the following problem
Q. Finding minimum $f(k)$ where $n \in \mathbb N$ and $k = 0,1,...,n-1$.
$$f(k)={2k+1 \choose k} \times {2n-2k-1\choose n-k}$$
I tried and find
$n=6$ then $f(2),f(3)$ has ... | It’s convenient to let $m=n-1$ and define
$$g_m(k)=\binom{2k+1}k\binom{2m+1-2k}{m+1-k}$$
for $k=0,\ldots,m$; we wish to choose $k$ so as to minimize $g_m(k)$.
Let $\ell=m-k$; then
$$g_m(k)=\binom{2k+1}k\binom{2m+1-2k}{m+1-k}=\binom{2m+1-2\ell}{m+1-\ell}\binom{2\ell+1}\ell=g_m(\ell)=g_m(m-k)\;.$$
That is, $g_m$ is symme... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of trignometric limit I want to find the following limit without L'Hospital.
$ \lim_{x \to \frac{3π}{4}} \frac{1+(\tan x)^{\frac13}}{1-2(\cos x)^2}$
Maybe I should try to get rid of the radical.
| We will use the fact that $(a^3+b^3)=(a+b)(a^2-ab+b^2)$. This will help us rationalize the numerator. With $a=1$and $b=(\tan x)^{1/3}$ we can write
\begin{align*}
\frac{1+(\tan x)^{1/3}}{1-2\cos^2x}& = \frac{1+(\tan x)^{1/3}}{1-2\cos^2x} \cdot \left[\color{blue}{\frac{1-(\tan x)^{1/3}+(\tan x)^{2/3}}{1-(\tan x)^{1/3}+(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1402984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Investigate the limiting behavior at the origin on the three lines $x = 0 , y = 0$ and $y = x$.
Investigate the limiting behavior at the origin on the three lines $x = 0 , y = 0$ and $y = x$.
For the function $f :\mathbb R^2 \to \mathbb R$ given by
$$f(x,y)=\begin{Bmatrix} 0, \quad \{x,y\} = {0,0} \\ \frac{x^2}{x^2+... | You have the right idea but note that approaching along the straight line $x=0$, it is not always true that $\lim_{x,y \to 0} f(0,y) = \lim_{x\to 0} f(x,y)$. The notation is incorrect but you did arrive at the right conclusion (the limit does not exist due to differing values as we change paths).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1403454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove the Identity $\pi=2- \sum_{1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}} $ By considering the fact that $f(\pi/2)=1$, prove the identity
$\pi=2- \sum_{1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}} $
This question was is a subsection in a chapter on Fourier series, can I use my understanding of Fourier series to prove the identit... | I'll play around and see what happens.
We want
$\pi=2- \sum_{1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}}
$
$\begin{array}\\
S
&=\sum_{m=1}^{∞} \frac{(-1)^m}{m^2-\frac{1}{4}}\\
&=\sum_{m=1}^{∞} \frac{(-1)^m}{m^2}\frac1{1-\frac{1}{4m^2}}\\
&=\sum_{m=1}^{∞} \frac{(-1)^m}{m^2}\sum_{j=0}^{\infty}\frac{1}{(4m^{2})^j}\\
&=4\sum_{m=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1404381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How could I find the sum of this infinite series by hand? $$\sum_{n=1}^{\infty}\frac{(7n+32)3^n}{n(n+2)4^n}$$
Thank you!
| Let the $n(\ge1)$th term $T_n=\dfrac{(7n+32)3^n}{n(n+2)4^n}=A\dfrac{\left(\dfrac34\right)^n}n+B\dfrac{\left(\dfrac34\right)^{n+2}}{n+2}$ where $A,B$ are arbitrary constants
$\implies\dfrac{(7n+32)}{n(n+2)}=\dfrac An+ \dfrac{B\left(\dfrac34\right)^2}{n+2}$
$\implies7n+32=A(n+2)+\dfrac{9Bn}{16}$
Comparing the constants... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1405416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Consecutive sets of consecutive numbers which add to the same total I'm looking at examples of numbers that can be written as the sum of integers from $j$ to $k$ and from $k+1$ to $l$. For example $15$ which can be written as $4+5+6$ or $7+8$. Or $27 = 2+3+4+5+6+7 = 8+9+10$. I have been able to find a few numbers wh... | Suppose,
$$X=(a+1)+(a+2)+\dots+b=(b+1)+(b+2)+\dots+c$$
which implies,
$$\Leftrightarrow \frac{b(b+1)}{2}-\frac{a(a+1)}{2}=\frac{c(c+1)}{2}-\frac{b(b+1)}{2}$$
$$2b(b+1)=a(a+1)+c(c+1)$$
or the special Pythagorean triple,
$$(2a-2c)^2+(2a+2c+2)^2=(4b+2)^2$$
where $a,b,c\in\mathbb Z$. We need to find integer solutions of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1408710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Simplifying $\sum_{i=0}^n i^k\binom{n}{2i+1}$ What is the formula for
\begin{eqnarray}\sum_{i=0}^n i^k\binom{n}{2i+1}?\end{eqnarray}
I tried to use the identity
$$
\sum_{i=0}^ni(i-1)\cdots(i-p)\binom{n}{2i+1}=(n-p-2)(n-p-3)\cdots(n-2p-2)2^{n-2p-3}
$$
but got into a mess. Any idea?
| Suppose we seek to evaluate
$$S(n, k) = \sum_{q=0}^{n}
q^k {n\choose 2q+1}.$$
Introduce
$$q^k =
\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{\exp(qz)}{z^{k+1}} \; dz.$$
Observe that with $k\ge 1$ we also get the correct value for $q=0.$
We obtain for the sum
$$\frac{k!}{2\pi i}
\int_{|z|=\epsilon} \frac{1}{z^{k+1}}
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1411602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Show that $ \int_0^2 e^{x^2-x} dx \in [2e^{-1/4},2e^2] $
Show that $ \int_0^2 e^{x^2-x} dx \in [2e^{-1/4},2e^2] $
If $f(x)\leq g(x)$ for $x\in[a,b]$ then $\int^b_af(x)dx\leq \int^b_ag(x)dx$
if $x\in [0,2]$ then $x^2-x\leq x$, so
$$0 \leq \int_0^2 e^{x^2-x} dx\leq \int_0^2 e^{x} dx = e^2-1 <2e^2$$
Which integral sh... | Since $x^2-x=\left(x-\frac{1}{2}\right)^2-\frac{1}{4}$ we have, for $x\in[0,2]$,
$$-\frac{1}{4}\le x^2-x \le 2$$
Then
\begin{align*}
\int_0^2e^{-\frac{1}{4}}\,dx&\le\int_0^2e^{x^2-x}\,dx\le\int_0^2e^2\,dx\\
2e^{-\frac{1}{4}}&\le\int_0^2e^{x^2-x}\,dx\le 2e^2
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1414027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Paramaterization of paraboloid and plane. Consider the paraboloid $z=x^2+y^2$. The plane $2x-4y+z-6=0$ cuts the paraboloid, its intersection being a curve. Find "the natural" parameterization of this curve.
I have set each equation equal to each other by solving for $z$, completed the square to reach $(x-1)^2+(y+2)^2... | Plugging $z=x^2+y^2$ into $2x-4y+z-6=0$ we get
\begin{align*}
2x-4y+x^2+y^2-6&=0\\
x^2+2x+\color{red}{1}+y^2-4y+\color{red}{4}&=6+\color{red}{1+4}\\
(x+1)^2+(y-2)^2&=11
\end{align*}
Then, a parameterization for the curve is
$$\begin{cases}x&=\sqrt{11}\cos t-1 \\ y&=\sqrt{11}\sin t+2\\z&=2\sqrt{11}(2\sin t -\cos t)+16\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1415473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a,b,c,d,e,f$ are non negative real numbers such that $a+b+c+d+e+f=1$, then find maximum value of $ab+bc+cd+de+ef$ $(a+b+c+d+e+f)^2=$ sum of square of each number (X)+ $2($ sum of product of two numbers (Y) $)$
$ab+bc+cd+de+ef \le Y$ since all are positive.
Therefore $1\ge X+(ab+bc+cd+de+ef)$
Edit: From AM GM inequa... | $$a+b+c+d+e+f=1\implies (a+c+e)(b+d+f)\le \frac14 \\ \implies (ab+bc+cd+de+ef)+(ad+af+cf+be)\le \frac14$$
We can show the first bracket can attain full value by say $a=b=\frac12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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integration by parts of $25\, (1-\sin^{2}x)$ I need help solving this integration of parts problem. I've tried a few different solutions and keep getting the wrong answer. This question is in regards to this problem take the integral by parts of:
$$\displaystyle \int (5-5\sin x)(5+5\sin x)dx$$
So first I multiply, and... | Consider
\begin{align}
I &= \int (5 - 5 \sin x) \cdot (5 + 5 \sin x) \, dx \\
&= 25 \, \int (1-\sin x) \cdot (1 + \sin x) \, dx = 25 \, J
\end{align}
Let $dv = 1-\sin x$ then $v = x + \cos x$, $u=1+\sin x$, $du = \cos x$ and
\begin{align}
J &= (x+\cos x)(1 + \sin x) - \int (x + \cos x) \, \cos x \, dx \\
&= (x + \cos ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1416582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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$a,b,c\in \Bbb Z$ and $a\cdot b\cdot c$ is a root of $ax^2+bx+c$. I was curious if there are quadratic equations where $a,b,c\in \Bbb Z$ and $a\cdot b\cdot c$ is a root of $ax^2+bx+c$.
So trivially if $c=0$, $a$ and $b$ can be arbitrary, and if either $a$ or $b$ is zero, this implies that $c=0$, and the other arbitrary... | Here's an alternate approach you might consider (+1 on yours, which works just fine).
Now, if $a,b,c$ satisfy the desired property and $c\ne 0,$ then by Rational Root Theorem we have that $abc$ is an integer factor of $c,$ from which we conclude that either
*
*$c=0,$ in which case any $a,b$ will do (as you observed)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to integrate following indefinite integal? The integral is
$$
\int\frac{x-\sin x}{1-\cos x} \,dx
$$
However, the only guess I have is that the denominator is the derivative of the numerator. Probably the integration by substitution will work here?
| $$\begin{align*}
\int\frac{x-\sin x}{1-\cos x}dx
&= \int \frac{(x-\sin x)(1+\cos x)}{(1-\cos x)(1+\cos x)}\ dx \\
&= \int \frac{x + x\cos x - \sin x -\sin x \cos x}{1-\cos^2 x}\ dx\\
&= \int (x\csc ^2 x + x \cot x \csc x - \csc x - \cot x)\ dx\\
&= -\int x\ d(\cot x) - \int x\ d(\csc x) - \int \csc x\ dx - \int \cot x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1422473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$a,b$ be two positive integers , where $b>2 $ , then is it possible that $2^b-1 \mid 2^a+1$? If $a,b$ be two positive integers , where $b>2 $ , then is it possible that $2^b-1\mid2^a+1$ ? I have figured out that if $2^b-1\mid 2^a+1$, then $2^b-1\mid 2^{2a}-1$ , so $b\mid2a$ and also $a >b$ ; but nothing else. Please he... | If $b$ is odd, then $b\mid 2a$ implies $b\mid a$, then $2^b-1\mid 2^a-1$ and hence $2^b-1\nmid 2^a+1$ (as $2^a+1$ is between $2^a-1$ and $2^a-1+(2^b-1)$).
If $b$ is even, $b=2c$ say, then $c\mid a$, hence $2^c-1\mid 2^a-1$ and $2^c-1\mid 2^b-1$. As $c>1$ this shows $d:=\gcd(2^a-1,2^b-1)>1$ (and of course odd) and so $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1424124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $f(z)=\int_0^1 \frac{dt}{t-z}$ where $z\in \mathbb{C}-[0,1]$.
Evaluate $\displaystyle f(z)=\int_0^1 \frac{dt}{t-z}$ where $z\in \mathbb{C}-[0,1]$. Here, $0\leq t \leq 1$.
My Try:
Can we use the same integration rules here as in real integration? I mean does $$ f(z)=\int_0^1 \frac{dt}{t-z}=\left[\ln|t-z|\vpha... | One simple mistake is you have $\ln|t-z|$ where you must have intended $\ln|t|-\ln|z|$.
Trigger warning: If the phrase "multiple-valued function" causes you pain, then stop reading before this present sentence.
Notice that if $z=x+iy$ and $x$ and $y$ are real then
$$
e^z = e^x e^{iy} = e^x (\cos y+ i\sin y)
$$
and, sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1425863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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A simple double inequality with roots Prove or disprove for integer $n$ greater than $1$:
$$\sqrt{1} + \sqrt{2} + ...+ \sqrt{n-1} < \frac{2n\sqrt{n}}{3} < \sqrt{1} + \sqrt{2} + ...+ \sqrt{n} $$
I think I know a solution, but I am looking for different approaches.
EDIT: Inspired by Prove that $1^2 + 2^2 + ..... + (n-1)... | $\sqrt{x}$ is a concave increasing function on $\mathbb{R}^+$, hence by the Hermite-Hadamard inequality:
$$\frac{1}{2}\cdot\frac{1}{1}+\frac{1}{\sqrt{2}}+\ldots+\frac{1}{\sqrt{n-1}}+\frac{1}{2\sqrt{n}}\leq \int_{1}^{n}\sqrt{x}\,dx = \frac{2}{3}\left(n\sqrt{n}-1\right)\tag{1}$$
as well as:
$$ \frac{(2n+1)\sqrt{2n+1}-1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1426367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is the polynomial $6x^4+3x^3+6x^2+2x+5\in GF(7)[x]$ irreducible? Is the polynomial $6x^4+3x^3+6x^2+2x+5\in GF(7)[x]$ irreducible?
What is the best/simplest/elementary way to approach this? Any solutions or hints are greatly appreciated.
| No, it is not. If we set $p(x)=6x^4+3x^3+6x^2+2x+5$, over $\mathbb{F}_7$ we have:
$$ p(x+2) = -\left(x^4-2x^3-2x+1\right) $$
that is a palyndromic polynomial, from which:
$$\frac{p(x+2)}{x^2} = -\left(x^2+\frac{1}{x^2}\right)+2\left(x+\frac{1}{x}\right) = -\left(x+\frac{1}{x}\right)^2+2\left(x+\frac{1}{x}\right)+2$$
an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1429519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculate product of geometric progressions Let us $p_k(z)=\sum_{i=0}^{k-1}z^{i}.$ For any $m$ find coefficient of the term $z^m$ in the polynomial $\prod_{j=1}^n p_{2j}(z).$
Here are some attempts
$$p_2=1+z$$
$$p_4=1+z+z^2+z^3$$
$$p_6=1+z+z^2+z^3+z^4+z^5$$
$$\vdots$$
$$p_{2n}=1+z+z^2+z^3+\ldots+z^{2n-1}$$
Coefficie... | $$\prod_{j=1}^{n}p_{2j}(z)=\prod_{j=1}^{n}\frac{1-z^{2j}}{1-z}=\sum_{r\geq 0}\binom{r+n-1}{n-1}z^r \prod_{j=1}^{n}(1-z^{2j})$$
so the coefficient of $z^m$ in the RHS is given by a weigthed sum (weigths are given by binomial coefficients) of partition numbers. Have a look at this related question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1430009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
In how many ways can a selection be done of $5$ letters?
In how many ways can a selection be done of $5$ letters out of $5 A's, 4B's, 3C's, 2D's $ and $1 E$.
$ a) 60 \\
b) 75 \\
\color{green}{c) 71} \\
d.) \text{none of these} $
Number of ways of selecting $5$ different letters = $\dbinom{5}{5} = 1$ way
Number of... | Ok so we are breaking this down by case:
Case 1: none of our letters are the same and we pick 5 distinct letters
Case 2a: exactly 2 of the letters are the same and the other 3 are distinct.
Case 2b: 2 of the letters are the same, 2 others are the same, and the fifth is distinct.
Case 3a: exactly 3 of the letters are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1431903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Consider the following system and find the values of b for which the system has a solution So I have this system:
$$\left\{\begin{array}{c}
x_1 &− x_2 &+ 2x_3 &= 2 \\
x_1 &+ 2x_2 &− x_3 &= 2 \\
x_1 &+ x_2 & &= 2 \\
x_1 & & +x_3 &= α
\end{array}\right.$$
And we are asked to find the values of $\alpha$ for ... | HINT: use Rouché–Capelli theorem
A system of linear equations with $\,n\,$ variables has a solution if and only if the rank of its coefficient matrix $\,A\,$ is equal to the rank of its augmented matrix $\,\left[\,A \mid b\,\right].\,$
Let us rewrite your system in matrix form $\,\mathbf A\,\vec{\boldsymbol{x}} = \v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1432632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Permutation of multiset
How many 8-permutation are there of the letters of the word
'ADDRESSES'?
My textbook suggests that we should divide the situation into cases where a different letter is removed. In other words, for the multiset $\{1 \cdot A, 2 \cdot D, 1 \cdot R, 2 \cdot E, 3 \cdot S\}$, we count the number ... | There are nine letters in ADDRESSES, but we have eight positions to fill. Thus, we must omit one of the letters. If we omit the A, we can fill two of the eight positions in $\binom{8}{2}$ ways. We can then fill one of the reamining six positions with an R in $\binom{6}{1}$ ways. We can fill two of the remaining pos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1438216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
check if large number $(9^{81}+6)$ is divisible by $11$ I would like to know if there is a mathematical way to check whether number $9^{81}+6$ is divisible by $11$, without actually calculating the whole number.
| $$9^1 \equiv -2 \bmod{11}$$ $$9^2 \equiv 4 \bmod{11}$$ $$9^6 \equiv 64 \equiv -2 \bmod{11}$$ $$9^{30} \equiv -32 \equiv 1 \bmod{11}$$ $$9^{81} \equiv 9^{30} 9^{30} 9^{6} 9^{6} 9^{6} 9^2 9^1 \equiv (1)(1)(-2)(-2)(-2)(4)(-2) \equiv 64 \equiv 9 \bmod{11}$$ $$9^{81} + 6 \equiv 15 \equiv 4 \bmod{11}.$$
EDIT: I accidentally ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1439772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Taylor expansion $\frac{y-(y+1) \log (y+1)}{y^2 (y+1)}$ I would like to use Taylor expansion around $y=0$ for that expression but i got stuck
$$\frac{y-(y+1) \log (y+1)}{y^2 (y+1)}$$
Taylor expansion around $y=0$
$$\log(1+y)=y-\frac{y^2}{2}+\frac{y^3}{3}+o\left(y^3\right) $$
$$\displaystyle {1 \over y^2 + y^3}= {1 \ov... | No, your expansion is not right.
Although you wrote that $\frac{1}{y^2 + y^3} = y^{-2}(1 - y + y^2/2 + \cdots)$, you seem to have put all of that into the denominator. It shouldn't be there.
Instead, you should have
$$ (-y^2/2 + y^3/6 + \cdots ) y^{-2} (1 - y + y^2/2 + \cdots),$$
from which it is relatively easy to col... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1442479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Polynomial Divisibilty Test I recently came across a question in a book and I was wondering how to go about solving this. I just need a hint about how I could approach it.
I have to show that $x^{6n+2} - x^{6n+1} + 1$ is always divisible by $x^2 - x + 1$ where $n = \{1,2,3,4,\cdots\}$
| Let $\alpha,\beta$ be two (complex) roots of $x^2-x+1$. Then $x^2-x+1=(x-\alpha)(x-\beta)$. Note that $x^2-x+1$ divides $x^3+1=(x+1)(x^2-x+1)$, and hence also $x^6-1=(x^3-1)(x^3+1)$. So both $\alpha,\beta$ satisfy $x^6=1$. Thus we have $\alpha^{6n+2}-\alpha^{6n+1}+1=(\alpha^6)^n\cdot\alpha^2-(\alpha^6)^n\cdot\alpha+1=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
figuring out an integer function $f(1) = 1\\
f(2) = 2\\
f(3) = 6\\
f(4) = 20\\
f(5) = 70\\
f(6) = 252\\
f(7) = 924\\
f(8) = 3432\\
f(9) = 12870$
Then what is $f(n)$ (where $n > 0$)?
I though about many many possibilities but still cannot figure out the expression.
| You could approach this kind of question in this way. You'd like to find what is the relation between $f_n$ and $f_{n+1}$ (you believe that there is some) so that's why most of the time it's a good idea to analyze expressions like $\frac{f_{n+1}}{f_n}$ or $f_{n+1}-f_n$. Let's look at $\frac{f_{n+1}}{f_n}$. You may not... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Inequality related AM-GM It looks easy but I cannot find the solution.
Let $x,y,z$ be positive reals and satisfy $x^2+y^2+z^2=2$, prove that
$$ \left(x^3+\sqrt{3}xyz\right)\left( y^3-\sqrt{3}xyz\right)\leqslant 1. $$
I invented this inequality about a month ago. Essentially it can be proved by brutal force ( homog... | If $y^2-\sqrt3xz<0$ then our inequality is obviously true.
But for $y^2-\sqrt3xz\geq0$ by AM-GM we obtain:
$$(x^3+\sqrt3xyz)(y^3-\sqrt3xyz)=xy(x^2+\sqrt3yz)(y^2-\sqrt3xz)\leq$$
$$\leq\frac{1}{27}(x^2+y^2+xy+\sqrt3yz-\sqrt3xz)^3=$$
$$=\frac{1}{216}(2x^2+2y^2+2xy+2\sqrt3yz-2\sqrt3xz)^3=$$
$$=\frac{1}{216}(6-x^2-y^2-3z^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1443803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving the simultaneous equations $x + y - z = 12$ and $x^2 + y^2 - z^2 = 12$ I found the following question online (it was a past question on the BMO):
find all positive integer solutions $x,y,z$ solving the simultaneous equations
$ x + y - z = 12 \ $ and $\ x^2 + y^2 - z^2 = 12 $ .
Normally these questions have a... | I'll try a generalization.
From
$x+y-z = n
$
and
$x^2+y^2-z^2 = m
$,
$z = x+y-n$
so
$z^2
=(x+y)^2-2n(x+y)+n^2
=x^2+2xy+y^2-2n(x+y)+n^2
$.
Therefore
$m
=x^2+y^2-(x^2+2xy+y^2-2n(x+y)+n^2)
=-2xy+2n(x+y)-n^2
$
so
$xy-n(x+y)
=-(m+n^2)/2
$
or
$(n^2-m)/2
=xy-n(x+y)+n^2
=(x-n)(y-n)
$.
The integer solutions
thus depend on the
f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Non-associative: set with a binary operation, but has inverses and identity I've been thinking about an example of some set with a binary operation which would satisfy all axioms of groups except for associativity. I'm new to Group Theory, so I would appreciate your knowledgeable insight.
My example:
$S={\mathbb{C} - \... | The object you are describing is called a loop. If you want a high powered example that fits in one mouthful, try the multiplicative group loop of non-zero octonions. As a more elementary alternative, I thought I'd give you a finite loop to ponder on (although you may need to work a bit to verify all the properties). I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Errors in partial fractions decomposition I decompose the fraction $\frac{x^2}{(x^2+1)^3}$ into $$\frac{x^2}{(x^2+1)^3}=\frac{Ax+B}{(x^2+1)^2}+\frac{Cx+D}{(x^2+1)^3}.$$
From here I found: $A=C=0,$ and $B=-D=1$
Now $$\frac{x^2}{(x^2+1)^3}=\frac{1}{(x^2+1)^2}-\frac{1}{(x^2+1)^3}$$
but the right side is not equal to the l... | The first expression you got it right, I do the second for you:
$\dfrac{x^5+1}{x^4(x^2+1)}=\dfrac{x^5}{x^4(x^2+1)}+\dfrac{1}{x^4(x^2+1)}=\dfrac{x}{x^2+1}+\dfrac{1+x^2-x^2}{x^4(x^2+1)}=\dfrac{x}{x^2+1}+\dfrac{1+x^2}{x^4(x^2+1)}-\dfrac{x^2}{x^4(x^2+1)}=\dfrac{x}{x^2+1}+\dfrac{1}{x^4}-\dfrac{1}{x^2(x^2+1)}=\dfrac{x}{x^2+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1444713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\sqrt{\frac{n+15}{n+1}}\in\mathbb Q$, then $n=17$ How can one show that :
If $\sqrt{\frac{n+15}{n+1}}\in\mathbb Q$ so $n=17$
I tried using the fact that any number $a\in\mathbb Q$ so $a=\frac{x}{y}$ such that $\gcd(x,y)=1$
So $\frac{n+15}{n+1}=\frac{x^2}{y^2}$
But here I'm stuck.
| If we let $m=n+1$, the equation
$$
\frac{m+14}{m}=\frac{x^2}{y^2}
$$
is equivalent to
$$
m=\frac{14y^2}{(x-y)(x+y)}
$$
we can assume that $x\ge0$ and $y\gt0$ and $(x,y)=1$. Thus,
$$
(x-y,x+y)\mid2\quad\text{and}\quad(x-y,y)=1\quad\text{and}\quad(x+y,y)=1
$$
If $(x-y,x+y)=2$, then, since $(x-y,y)=1$, $y$ must be odd and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Nilpotent matrix and relation between its powers and dimension of kernels Given a 4x4 matrix $T$ over $\mathbb{R}$ such that $T^4 = 0 $, $k_i = \textsf{dim} Ker(T^i)$, I need to check which of the following sequences, $$k_1\leq k_2 \leq k_3 \leq k_4,$$ is NOT possible :
$ 1)\; 1\leq 3 \leq 4 \leq 4$
$2) \; 2\leq3\leq4... | Consider the possible lists invariant factors
of $T$. Since $T$ is $4 \times 4$, then its characteristic polynomial is $T^4$. The invariant factors must divide the characteristic polynomial, hence are all of the form $T^i$. The characteristic polynomial is the product of all invariant factors, so the list of invaria... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1447591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 0
} |
Prove by induction $\sum \frac {1}{2^n} < 1$ Prove by induction $\sum \frac {1}{2^n} < 1$
Well supposing the base case has been shown to be true, I start with the induction step:
Suppose true for n = k:
$$ \frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} < 1$$
Want to show this is true for:
$$ \frac{1}{2} + \frac{1}{4} +... | For $n=k+1$:
$$\frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k} + \frac{1}{2^{k+1}} = \frac{1}{2} + \frac{1}{2}\left(\frac{1}{2} + \frac{1}{4} + ....\frac{1}{2^k}\right) < \frac{1}{2} + \frac{1}{2}\times 1 =1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1448626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
What are the methods of dividing numbers to get weird values like $16\over 17$ without a calculator? I tried estimating it to somewhere near $16\over 20$, but it's a far stretch from getting the actual $16\over 17$. How can one do so? Conventionally, I think for numbers such as $50\over 17$, or for any large numbers, w... | Please don't upvote me or downvote me as I have obviously copied and edited Mr. Hardy's answer. I agree with Mr Hardy that long division is the answer, but there is a small trick that makes subtraction a bit easier
Long division:
$$
\begin{array}{cccccccccc}
& & & 0 & . & 9 & 4 & 1 & 1 & 7 & 6 \\ \\
17 & ) & 1 & 5 & .... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 2
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Prove that $2^n>n^4$ for all $n\geq 17$ I'm always a bit fuzzy on how to solve induction problems involving inequalities. I've managed to get somewhere though, but it looks like I have to go down four levels of induction to prove.
This is what I have so far:
Base ($n=17$):
$$2^{17}>17^4\Rightarrow131072>83521$$
Step:
A... | all that is required for the induction step is $$ (n+1)^4 < 2 n^4 $$
for large enough $n,$ so
$$ \frac{(n+1)^4}{ 2 \, n^4} < 1. $$
This is the same as
$$ \left( \frac{n+1}{n} \right)^4 = \left( 1 + \frac{1}{n} \right)^4 < 2 $$
This does not work for $n \leq 5,$ indeed
$(6/5)^4 = 1296/ 625,$ but it does work for $n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1449320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How many digits will $5^{4^{3^2}}$ produce? According to Wolfram Alpha, $5^{4^{3^2}}$ evaluates to an integer with $183\,231$ digits. How does one find out how many digits such a calculation will produce?
| A number $n \in \mathbf N^+$ has $\lfloor\log_{10}n\rfloor+1$ digits. For $n = 5^{4^{3^2}}$, we have
\begin{align*}
\log_{10} 5^{4^{3^2}} &= 4^9 \cdot \log_{10} 5\\
&= 262\,144 \cdot \log_{10} 5\\
&\approx 183\,230.8
\end{align*}
So $n$ has $183\,231$ digits.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1450806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integrating $x^5 \arcsin x$ Integrating $$\int_0^1 x^5 (\sin^{-1}x) \, dx$$
the answer is $\dfrac{11 \pi}{192}$.
I did substitute $x=\sin^2 u$ and obtained $$2\int_0^{\pi/2}( \sin^{12}x)( \cos x) \, dx$$ but got the incorrect answer. Why isn't this substitution working?
$$\int_0^{\pi/2}( \sin^m x)( \cos^n x)=\frac{(m-1... | We may use the substitution $x=\sin(\theta)$ followed by integration by parts to get:
$$\begin{eqnarray*} I = \int_{0}^{\pi/2}\theta\cos(\theta)\sin(\theta)^5\,d\theta &=& \frac{\pi}{12}-\frac{1}{6}\int_{0}^{\pi/2}\sin(\theta)^6\,d\theta\\&=&\frac{\pi}{12}-\frac{1}{6}\int_{0}^{\pi/2}\cos(\theta)^6\,d\theta\tag{1}\end{e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1451159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving $\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$ I've been trying to solve this over and over without L'Hopital but keep on failing:
$$\lim_{x\to0}\frac{1-\sqrt{\cos x}}{x^2}$$
My first attempt involved rationalizing:
$$\frac{1-\sqrt{\cos x}}{x^2} \cdot \frac{1+\sqrt{\cos x}}{1+\sqrt{\cos x}} = \frac{1-\cos x}{x\cdot... | May be you can use the fact that $$\sqrt{1+x}=1+\frac{x}{2}+o(x)$$ and $$\cos(x)=1-\frac{x^2}{2}+o(x^2).$$ Therefore
$$...=\lim_{x\to 0}\frac{x^2}{4x^2}=\frac{1}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Definite integral of cosine times a quotient of two quartic polynomials If $f(x)=\frac{\left(x^2+x+1\right)^2+\left(x^4+x^2+1\right)}{\left(x^2+x+1\right)^2-\left(x^4+x^2+1\right)}$,then find $\int_{-\pi}^{\pi}\cos x\cdot f(x)dx$
Let $I=\int_{-\pi}^{\pi}\cos x\cdot f(x)dx=\int_{-\pi}^{\pi}\cos x\cdot\frac{\left(x^2+x+1... | The function $f$ simplifies to $f(x)=x+1/x$, so
$$
\int_{-\pi}^\pi \cos(x)f(x)\,dx
$$
diverges.
Addition
This addition was asked for in a comment. The integrals
$$
\int_{-\pi}^0 (x+1/x)\cos x\,dx\quad\text{and}\quad \int_0^{\pi}(x+1/x)\cos x\,dx
$$
both diverges. To see this, we note that $x+1/x\approx 1/x$ and $\cos x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1453734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to solve $\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$ Solved I stuck in this limit, I tried to solve it and gets 1/2 as result. Yet, I was wrong because I forgot a square. Please need help!
$\lim _{x\to \infty \:}\left(\sqrt{x+\sqrt{1+\sqrt{x}}}-\sqrt{x}\right)$
Note: it's $+\infty$
Tha... | $$\begin{align}\sqrt{x+\sqrt{1+\sqrt x}}-\sqrt x&=\frac{\left(\sqrt{x+\sqrt{1+\sqrt x}}-\sqrt x\right)\left(\sqrt{x+\sqrt{1+\sqrt x}}+\sqrt x\right)}{\sqrt{x+\sqrt{1+\sqrt x}}+\sqrt x}\\
&\le \frac{\left(\sqrt{x+\sqrt{1+\sqrt x}}-\sqrt x\right)\left(\sqrt{x+\sqrt{1+\sqrt x}}+\sqrt x\right)}{1+\sqrt x}\\
&=\frac{\sqrt{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1455623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$
Show that $\arctan\frac{1}{2}+\arctan\frac{1}{3}=\frac{\pi}{4}$.
Attempt:
I've tried proving it but it's not equating to $\frac{\pi}{4}$. Please someone should help try to prove it. Is anything wrong with the equation? If there is, please let me know.
| Let $$\displaystyle \tan^{-1}\left(\frac{1}{2}\right)=\alpha\Rightarrow \tan \alpha = \frac{1}{2}$$ and $$\displaystyle \tan^{-1}\left(\frac{1}{3}\right)=\beta\Rightarrow \tan \beta = \frac{1}{3}$$
So $$\displaystyle \tan(\alpha+\beta) = \frac{\tan \alpha+\tan \beta}{1-\tan \alpha\cdot \tan \beta} = \frac{\frac{1}{2}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1456746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
How to simplify $\sin^4 x+\cos^4 x$ using trigonometrical identities? $\sin^{4}x+\cos^{4}x$
I should rewrite this expression into a new form to plot the function.
\begin{align}
& = (\sin^2x)(\sin^2x) - (\cos^2x)(\cos^2x) \\
& = (\sin^2x)^2 - (\cos^2x)^2 \\
& = (\sin^2x - \cos^2x)(\sin^2x + \cos^2x) \\
& = (\sin^2x - \c... | Note that $a^2 + b^2 = (a+b)^2 - 2ab$
$$(\sin^2 x)^2 + (\cos^2 x)^2 = (\sin^2 x + \cos^2 x)^2 - 2\sin^2 x\cos^2 x =(\sin^2 x + \cos^2 x)^2 - 2(\sin x\cos x)^2 = \\ 1 -\frac{ \sin^2 2x}{2}$$
Note the following results:
$$ \sin^2 x + \cos^2 x = 1$$
$$ \sin x \cos x = \frac{\sin 2x}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1458305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Convert $r^2= 9 \cos 2 \theta$ into a Cartesian equation This is how I tried so far...
$r^2= 9 \cos 2 ( \theta)$
$\cos (2 \theta) = \cos ^2 (\theta) - \sin^2 (\theta)$ and
$r^2= x^2 + y^2$
so, it will become
$x^2 + y^2 = 9 [\cos^2 (\theta) - \sin^2 (\theta) ]$
| Hint Polar and Cartesian coordinates are related by the equations
$$x = r \cos \theta, \quad y = r \sin \theta,$$ so a Pythagorean trigonometric identity gives $$x^2 + y^2 = (r \cos \theta)^2 + (r \sin \theta)^2 = r^2 (\cos^2 \theta + \sin^2 \theta) = r^2 .$$ Now, simply rearrange and substitute.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1459152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is the radius of the circle through $(-1,1)$ and touching the lines $x\pm y=2?$ What is the radius of the circle through $(-1,1)$ and touching the lines $x\pm y=2?$
The lines $x+y=2$ and $x-y=2$ are perpendicular to each other and the circle is touching both the lines,these lines are tangents to the circle.Let po... | Let $(h,k)$ be the center of the required circle.
The perpendicular distance from $(h,k)$ to $x+y-2=0$
=The perpendicular distance from $(h,k)$ to $x-y-2=0$
=The distance between $(h,k)$ to $(-1,1)$.
The first two equality gives $|\frac{h+k-2}{\sqrt 2}|=|\frac{h-k-2}{\sqrt 2}|$, which will imply either $k=0$ or $h=2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1465571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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An Inequality for sides and diagonal of convex quadrilateral from AMM Let $\square ABCD$ be a convex quadrilateral. If the diagonals $AC$ and $BD$ have mid-points $E$ and $F$ respectively, show that:
$$\overline{AB} + \overline{BC} +\overline{CD} + \overline{DA} \ge \overline{AC}+\overline{BD}+2\overline{EF}$$
where, $... | The simplest special case is when $\overline{ABCD}$ is a square with side length $a$. In this case, the diagonals are of equal length and given by $a\sqrt{2}$.
$4a \ge 2a\sqrt{2},\quad 4 \ge 2\sqrt{2},\quad$the conjecture holds.
The next simplest case is when $\overline{ABCD}$ is a rectangle of side lengths $a$ and $b$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1466077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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If line through point $P(a,2)$ meets the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ at A and D and meets the coordinate axis at B and C If line through point $P(a,2)$ meets the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ at $A$ and $D$ and meets the coordinate axes at $B$ and $C$ so that $PA$, $PB$, $PC$, $PD$ are in geometri... | WLOG we can take $a>0$, $A$ nearer to $P$ than $D$ and notice that the only case we must consider is when $B$ is the intersection with $x$-axis and $C$ is the intersection with $y$-axis, for otherwise those four segments cannot form a geometric progression. If $b$ is the $x$ coordinate of $B$, the equation of line $PB$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1470586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
if $5\nmid a$ or $5\nmid b$, then $5\nmid a^2-2b^2$. I have a homework as follow:
if $5\nmid a$ or $5\nmid b$, then $5\nmid a^2-2b^2$.
Please help to prove it.
EDIT: MY ATTEMPT
Suppose that $5\mid a^2-2b^2$, then $a^2-2b^2=5n$,where $n\in Z$,
then $a^2-2b^2=(a+\sqrt2b)(a-\sqrt2b)=5n$,
Since 5 is a prime number, we ge... | This is probably what Bernard intended.
The non-zero quadratic residues mod $5$ are $1$ and $4$. Plugging these into $a^2-2b^2$ gives
$$
\begin{array}{c|ccc}
&1&4&a^2\\\hline
1&4&2\\
4&3&1\\
b^2
\end{array}
$$
Since none of the entries are $0$, we get that $a^2-2b^2\not\equiv0\pmod5$ if neither $a$ nor $b$ are $0\pmod5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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In the triangle $ABC$,if median through $A$ is inclined at $45^\circ$ with the side $BC$ and $C=30^\circ$,then $B$ can be In the triangle $ABC$,if median through $A$ is inclined at $45^\circ$ with the side $BC$ and $C=30^\circ$,then $B$ can be equal to
$(A)15^\circ\hspace{1cm}(B)75^\circ\hspace{1cm}(C)115^\circ\hspace{... | If you have access to Geogebra you can easily find the answer yourself:
Now we you only need to prove it. :)
Edit: It seems that the inclination of the median is not clear, but making the two choices on the same figure brings an immediate solution:
In triangle $AED$ we have $\frac{EC}{DC} = \frac{\sin 45^\circ}{\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1471974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Approximation of non-analytic function I have a function which is of the form
\begin{equation}
f(x) = \frac{1 - x^{1/2} + x - x^{3/2} + \ldots}{1+x^{1/2} - x + x^{3/2} - \ldots}.
\end{equation}
Intuitively, I would assume that for small $x$, it holds
\begin{equation}
f(x) \approx \frac{1-x^{1/2}}{1+x^{1/2}}
\end{equati... | Note, that the numerator, is
$$ \sum_{k=0}^\infty (-x^{1/2})^k = \frac{1}{1 + x^{1/2}}, \quad |x| < 1 $$
and the denominator equals
$$ 2 -\sum_{k=0}^\infty (-x^{1/2})^k = 2 - \frac 1{1 + x^{1/2}} $$
Hence
\begin{align*}
f(x) &= \frac{\frac 1{1 + x^{1/2}}}{\frac{2(1 + x^{1/2}) - 1}{1 + x^{1/2}}}\\
&= \frac 1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Linear autonomous systems in the plane: When do phase curves rotate clockwise? For a linear autonomous system in the plane
$$ \mathbf{\dot{x}} = \begin{pmatrix} a & b\\ c & d \end{pmatrix}\mathbf{x} \qquad (a,b,c,d \in \mathbb{R})$$
with determinant $D = ad - bc$ and trace $T = a + d$ we have the characteristic polynom... | Inspired by amd's help I did the math:
The original system
$$ \mathbf{\dot{x}} = \begin{pmatrix} a & b\\ c & d \end{pmatrix}\mathbf{x}, \qquad D = ad - bc, \qquad T = a + d $$
has the eigenvalues
$$ \lambda_{\pm} = \frac{T \pm i\sqrt{|\Delta|}}{2}, \qquad |\Delta| = 4D - T^2 > 0 $$
with corresponding eigenvectors
$$ v_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1478460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$2=1$ Paradoxes repository I really like to use paradoxes in my math classes, in order to awaken the interest of my students. Concretely, these last weeks I am proposing paradoxes that achieve the conclusion that 2=1. After one week, I explain the solution in the blackboard and I propose a new one. For example, I poste... | Here is a simple one:
$$
x=y\\
x^2=xy\\
2x^2=x^2+xy\\
2x^2-2xy=x^2-xy\\
2(x^2-xy)=1(x^2-xy)\\
2=1
$$
The error is quite obviously division by zero (from the 5th to 6th step).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 15,
"answer_id": 2
} |
If $x^{x^4} = 4$, what is the value of $x^{x^2} + x^{x^8}$? If $x^{x^4} = 4$, what is the value of $x^{x^2} + x^{x^8}$ ?
I can find by trial and error, that $x=\sqrt 2$. But, what is the general process to answer questions like this?
| Taking the logarithm base 2 of each side, we have
$\log_2 (x^{x^4}) = x^4\log_2 (x)= \log_2 4 = 2$
Multiplying each side by four we have
$4x^4\log_2 (x) = x^4\log_2 (x^4) = 8$
Relabeling $x^4$ as $u$, we have
$u\log_2(u) = 8$
Utilizing the Lambert W Function, we have that
$u = e^{W(8\ln 2)}=e^{W(4\ln 4)}=4$
Replacing b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1480675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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A trigonometry equation: $3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$
$$3 \sin^2 \theta + 5 \sin \theta \cos \theta - 2\cos^2 \theta = 0$$
What are the steps to solve this equation for $ \theta $?
Because, I am always unable to deal with the product $\sin \theta \cos \theta$.
| Use the linearisation formulae (inverse of the duplication formulae):
\begin{align*}
&=3\frac{1-\cos2\theta}2+\frac52\sin2\theta-2\frac{1+\cos2\theta}2\\
&=\frac12-\frac52(\cos2\theta-\sin2\theta)=\frac12-\frac{5\sqrt 2}2\cos\Bigl(2\theta-\frac\pi4\Bigr),
\end{align*}
whence
\begin{align*}
3\sin^2\theta+&5\sin\theta\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Integral of $x\arctan x/(1+x^2)^{3/2}$ Find
$$\int \frac{x\arctan x}{(1+x^2)^{3/2}}\operatorname{dx} $$
I tried many ways to do it but I can't resolve it. What are the steps to solve this exercise?
| One way is to just integrate by parts,
$$
\begin{split}
\int \frac{x}{(1+x^2)^{3/2}}\arctan x&=-\frac{1}{\sqrt{1+x^2}}\arctan x+\int\frac{1}{(1+x^2)^{3/2}}\,dx\\
&=-\frac{1}{\sqrt{1+x^2}}\arctan x+\frac{x}{\sqrt{1+x^2}}.
\end{split}
$$
Here, the last step might seem mystery. I know it by heart, but one way to conclude ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1481719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find the partial sums of $4+44+444+\cdots$
Find the sum to $n$ terms of $4+44+444+\cdots $
My attempts:
*
*Used successive difference method
*Used $4+(4+40)+(4+40+400)\dots$ method
Failed to get a formula for partial sum in both ways. What am I missing?
| We know that
$$ \sum_{k=0}^n {\frac{1}{x^k}} =
\frac{\frac{1}{x^{n+1}}-1}{\frac 1x - 1} =
\frac{x^{n+1}-1}{x^n(x-1)}$$.
Taking the derivative of both sides results in
$$\displaystyle \sum_{k=1}^n \frac{k}{x^{k+1}} =
\frac{x^{n+1}-(n+1)x^n+n}{x^{n+1}(x - 1)^2}$$.
Multiplying both sides by $x^{n+1}$, we get
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding all positive integer solutions for $x+y=xyz-1$ How do I manually solve $x+y=xyz-1$ assuming that $x, y$ and $z$ are positive integers? I was able to guess all possible solutions, but I do not know how to show that these are the only ones:
$x=1, y=1, z=3$
$x=1, y=2, z=2$
$x=2, y=1, z=2$
$x=2, y=3, z=1$
$x=3, y=2... | We have
$$z=\frac{x+y+1}{xy}=\frac 1y+\frac 1x+\frac{1}{xy}\le 1+1+1$$$$\Rightarrow z=1,2,3$$
Also,
$$zxy-x-y=1\iff z^2xy-zx-zy+1=z+1\iff (zx-1)(zy-1)=z+1$$
So, for $z=1$, we have $(x-1)(y-1)=2$.
For $z=2$, we have $(2x-1)(2y-1)=3$.
For $z=3$, we have $(3x-1)(3y-1)=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
When is $2^n +3^n + 6^n$ a perfect square? Find all $n$ for which $2^n+3^n+6^n$ is a perfect square.
I do not have a specific idea how to solve this one
| If $2^n+3^n+6^n=x^2$ for some positive integer $x$, then we know from a comment by user236182 that $n=2k$ for some positive integer $k$. Hence,
$$\left(x-2^k\right)\left(x+2^k\right)=3^{2k}\left(1+2^{2k}\right)\,.$$
Since $x$ is obviously odd, $\gcd\left(x-2^k,x+2^k\right)=1$, so $3^{2k}$ divides either $x-2^k$ or $x+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1484545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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How do I evaluate this without using taylor expansion :$\lim_{x \to \infty}x^2\log(\frac {x+1}{x})-x\ $?
How do I evaluate this without using Taylor expansion?
$$\lim_{x \to \infty}x^2\log\left(\frac {x+1}{x}\right)-x$$
Note: I used Taylor expansion at $z=0$ and I have got $\frac{-1}{2}$
Thank you for any help
| Another option is to use the following (very convenient) inequality:
$$
\frac{2x}{2+x} \leq \ln(1+x) \leq \frac{x}{\sqrt{1+x}}
$$
which holds for $x \geq 0$. In your case, you have $\ln(1+\frac{1}{x})$, leading to
$$
x^2\frac{\frac{2}{x}}{2+\frac{1}{x}} -x \leq x^2\ln\!\left(1+\frac{1}{x}\right) -x \leq x^2\frac{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1485850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding the roots of $x^3+ax+a=0$ If the roots $x_1,x_2,x_3$ of the equation $x^3+ax+a=0$ satisfy $$\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+\frac{x_3^2}{x_1}=-8$$ then the roots of the equation are?
I wrote the Vieta formulas down but then I'm stuck.Please give your suggestions.
| $\frac{x_1^2}{x_2}+\frac{x_2^2}{x_3}+\frac{x_3^2}{x_1}=-8 \iff \frac{x_1^3}{x_1x_2}+\frac{x_2^3}{x_2x_3}+\frac{x_3^3}{x_1x_2}=-8 \iff \frac{-ax_1-a}{x_1x_2}+\frac{-ax_2-a}{x_2x_3}+\frac{-ax_3-a}{x_1x_3}=-8 \iff \frac{a}{x_1x_2}+\frac{a}{x_2x_3} + \frac{a}{x_1x_3} + \frac{a}{x_1}+ \frac{a}{x_2}+ \frac{a}{x_3}=8 (1)$
Bec... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Limits without L'Hopitals Rule ( as I calculate it?) Prove that:
$\lim z \to \infty \left (z^2 +\sqrt{z^{4}+2z^{3}}-2\sqrt{z^{4}+z^{3}}\right )=\frac{-1}{4}$
| Let $\displaystyle z = \frac{1}{y}\;,$ So when $z\rightarrow \infty\;,$ Then $y\rightarrow 0$
So limit convert into $$\displaystyle \lim_{y\rightarrow 0}\frac{1+\sqrt{1+2y}-2\sqrt{1+y}}{y^2} = \lim_{y\rightarrow 0}\frac{1+(1+2y)^{\frac{1}{2}}-2(1+y)^{\frac{1}{2}}}{y^2}$$
Now Using $$\displaystyle \bullet (1+t)^n = 1+nt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1487860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Find all factored pairs of (a,b) such that...
Determine all possible ordered pairs (a, b) such that
*
*$a − b = 1$
*$2a^2 + ab − 3b^2 = 22$
I've gone as far as factoring the left side of the second equation:
*
*$(a-b)(2a + 3b) = 22$
*$(1)(2a + 3b) = 22$
*$2a + 3b = 22$
I understand that a different (and arguab... | $$
\begin{cases}
a − b = 1 \\
2a^2 + ab − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
2a^2 + ab − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
2(1 + b)^2 + (1 + b)b − 3b^2 = 22
\end{cases}\Longleftrightarrow
$$
$$
\begin{cases}
a = 1 + b \\
2b^2+4b+2 + b+b^2 − ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1491415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
find the area of the region lying inside the circle $r=6$ and inside the cardioid $r=4-3\sin \theta$. Well, I drew a graph to visualise it and I found the interceptions $\theta=\arcsin \left(-\frac{2}{3}\right)$. From the graph, by symmetry, I found that the area of region from $\theta$ to $\pi/2$ and from $\pi/2$ to $... | You want to compute the area integral
$$\int_A {1} dA$$
where $A$ is the intersection of the interior of the circle and the cardioid i.e.
\begin{align*}
A &= \{(x, y) : r = \sqrt{x^2 + y^2} \leq 6\} \cap \{(x, y) : r = \sqrt{x^2 + y^2} \leq 4 - 3 \sin \theta \}\\
&= \{(x, y) : r \leq \min(6, 4 - 3 \sin \theta) \}
\end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1493968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
multiplication over gf(16) Can some one show me how to do multiplication over gf(16) step by step
I found this example online, http://userpages.umbc.edu/~rcampbel/Math413Spr05/Notes/12-13_Finite_Fields.html#An_Example.
An Example:
$x^4 = x+1$
$x^5 = x^2+x$
$(x^2+x+1) (x^3+x^2+1)$
$= x^5 + 2x^4 + 2x^3 + 2x^2 + x + 1$
$=... | By distributivity, all you actually In the general casehave to know is how to multiply $1, x,x^2,x^3$ with each other, i.e. express $x^4,x^5,x^6$. In the general case, here is how it goes:
\begin{align*}
x^4&=\color{red}{x+1} &x^5&=x\cdot x^4=x(x+1)=\color{red}{x^2+x}\\
x^6&=x\cdot x^5=\color{red}{x^3+x^2}&x^7&=x\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1497091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Write the expression in Euler's formula
$a=2+2i$ , $b=5e^{i\frac{\pi}{3}}$
such that $$\frac{b^5}{a^3}=Re^{\theta i}$$
find: $R$ and $\theta$
$$R=\sqrt{2^2+2^2}=\sqrt{8}$$,
$$tan^{-1}=\frac{2}{2}\rightarrow \theta=\frac{\pi}{4}$$
$$a=\sqrt{8}e^{\frac{\pi}{4}i}, a^3=({\sqrt{8}e^{\frac{\pi}{4}i}})^3=(\sqrt{8})^3e^{\fr... | $$\frac{\left(5e^{\frac{\pi}{3}i}\right)^5}{\left(2+2i\right)^3}=$$
$$\frac{\left(5e^{\frac{\pi}{3}i}\right)^5}{\left(|2+2i|e^{\arg(2+2i)i}\right)^3}=$$
$$\frac{\left(5e^{\frac{\pi}{3}i}\right)^5}{\left(\sqrt{2^2+2^2}e^{\tan^{-1}\left(\frac{2}{2}\right)i}\right)^3}=$$
$$\frac{\left(5e^{\frac{\pi}{3}i}\right)^5}{\left(\... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value of $(1-z)\left(1+\frac{z}{2}\right)\left(1-\frac{z}{3}\right)\left(1+\frac{z}{4}\right)\cdots$. Show that: $$(1-z)\left(1+\frac{z}{2}\right)\left(1-\frac{z}{3}\right)\left(1+\frac{z}{4}\right)\cdots = \dfrac{\sqrt{\pi}}{\Gamma\left(1+\dfrac{z}{2}\right)\Gamma\left(\dfrac{1}{2}-\dfrac{z}{2}\right)}.$$
By ... | We can make use of Euler's product for the gamma function, Whittaker and Watson, page 237, §12.11, also (31) of MathWorld:
$$
\frac{1}{\Gamma(z)} =
\lim_{n\rightarrow\infty}
\frac{z \, (1+ z) \cdots (n-1 + z)}{1 \cdot 2 \cdots (n - 1)} n^{-z}.
\qquad (1)
$$
According to (1), we have
$$
\begin{align}
\frac{1}{\Gamma(1 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Transforming the quadratic form $x_1x_2 + x_2x_3 + x_3x_1$ to its normal form. I need to transform $x_1x_2 + x_2x_3 + x_3x_1$ to its normal form described by the inertia theorem. How can I find the transformation? I am allowed to complete squares and perform rotations to find the linear change of coordinates. I know ho... | The matrix of the quadratic form for the given basis $x_1, x_2, x_3$ is
$$
A = \frac 12\begin{pmatrix}
0 & 1 & 1 \\
1 & 0 & 1 \\
1 & 1 & 0
\end{pmatrix}.
$$
For simplicity, I will diagonalize $2A$ first. Eigenvalues of $2A$ can be found by solving
\begin{align}
\lambda^3 - 3 \lambda - 2 & = 0 \\
(\lambda + 1)^2(\lambda... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1501408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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