Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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$\lim_{n\to \infty} \frac{3}{n}(1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+...+\sqrt{\frac{n}{4n-3}})$ fine the limits :
$$\lim_{n\to \infty} \frac{3}{n}(1+\sqrt{\frac{n}{n+3}}+\sqrt{\frac{n}{n+6}}+...+\sqrt{\frac{n}{4n-3}})$$
My Try :
$$\lim_{n\to \infty} \frac{3}{n}(1+\frac{1}{\sqrt{1+\frac{3}{n}}}+\frac{1}{\sqrt{1+... | $$\frac3n\sum_{k=0}^{n-1}\sqrt{\frac n{n+3k}}=\frac3n\sum_{k=0}^{n-1}\sqrt{\frac1{1+\frac{3k}n}}\xrightarrow[n\to\infty]{}3\int_0^1\frac1{\sqrt{1+3x}}\,dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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express $a$ in terms of $b$ and $c$
Given that $$c=\frac{\sqrt{a+3b}}{a-3b}$$ express $a$ in terms of $b$ and $c$
My attempt,
\begin{align}c^2(a^2-6ab+9b^2)&=a+3b\\
c^2a^2+(-6bc^2-1)a+9b^2c^2-3b&=0\\
a&=\frac{-(-6bc^2-1)\pm \sqrt{(-6bc^2-1)^2-4c^2(9b^2c^2-3b)}}{2c^2}\\
a&=\frac{6bc^2+1\pm \sqrt{24bc^2+1}}{2c^2}\end{a... | A slightly different take on Yves Daoust's approach:
Let $\sqrt{a+3b}=u\ge0$, so that $a=u^2-3b$. Then
$$c={u\over u^2-6b}\implies cu^2-u-6bc=0$$
The quadratic formula gives
$$u={1\pm\sqrt{1+24bc^2}\over2c}$$
as the formal solutions. It is convenient to rewrite them as
$$u={1+\sqrt{1+24bc^3}\over2c}\qquad\text{and}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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logarithm proof I'm trying to prove the following inequalities:
\begin{align}
\frac{b+c}{b} \geq \frac{\log(\frac{a}{b})}{\log(\frac{a+c}{b+c})}, c \in (0,1)
\end{align}
I know that $\log(\frac{a}{b}) > \log(\frac{a+c}{b+c})$ for $ c > 0$, but I'm stuck as to how to proceed with the proof. This is not a homework and is... | Let $a,b,c \in \mathbb{R}^+$ and $a>b$.
Assume
$$\left(\frac{a+c}{b+c}\right)^{b+c}>\left(\frac{a}{b}\right)^{b}$$
Then
$$\log\left(\left(\frac{a+c}{b+c}\right)^{b+c}\right)>\log\left(\left(\frac{a}{b}\right)^{b}\right)$$(as both sides are positive and $\log$ is an increasing function)
$$\implies (b+c)\log\left(\frac{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to evaluate the closed form for $\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right)=\pi\cdot F(a,b)?$ Proposed:
$$\int_{0}^{\pi/2}{\mathrm dx\over \sin^2(x)}\ln\left({a+b\sin^2(x)\over a-b\sin^2(x)}\right)=\pi\cdot F(a,b)\tag1$$
Where $a\ge b$
Examples:
Where $F(1,1)= \s... | Let $u=\cot{x}$. Then $du = dx/\sin^2{x}$ and $\sin^2{x}=1/(1+u^2)$, and the limits become $\infty$ and $0$, so the integral becomes
$$ \int_0^{\infty} \log{\left( \frac{a+b/(1+u^2)}{a-b/(1+u^2)} \right)} \, du = \int_0^{\infty} \log{\left( \frac{a+b+au^2}{a-b+au^2} \right)} du $$
One can now integrate this by parts t... | {
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"timestamp": "2023-03-29T00:00:00",
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How to find the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which goes through $(3,1)$ and has the minimal area?
How to find the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which goes through $(3,1)$ and has the minimal area? Ellipse area is given as $\pi ab$.
My approach is to use Lagrange method where the constrain... | We have the general equation of an ellipse:
$$ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 $$
whose area is:
$$ f(a,b) = \pi a b$$
We want to minimize the area $f(a,b)$ subject to the constraint that the ellipse passes through the point $(3,1)$, that is:
$$ g(a,b) = \frac{9}{a^2} + \frac{1}{b^2} = 1 $$
Applying the method of... | {
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"question_score": "1",
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There exists a $2 \times 2$ matrix $R$ such that $r = R v$ for all 2-dimensional vectors $v$. Find $R$. For a vector $v$, let $r$ be the reflection of $v$ over the line
$$x = t \begin{pmatrix} 2 \\ -1 \end{pmatrix}.$$
There exists a $2 \times 2$ matrix $R$ such that
$$r = R v$$
for all 2-dimensional vectors $v$. Find $... | If $u$ is the projection of $v$ onto $w$, the reflection of $v$ over $w$ is given by $2u-v$. See the diagram below.
Hence $$v'=2u-v=2\begin{pmatrix}\frac{4}{5}&-\frac{2}{5}\\-\frac{2}{5}&\frac{1}{5}\end{pmatrix}v-\begin{pmatrix}1&0\\0&1\end{pmatrix}v = \begin{pmatrix}\frac{3}{5}&-\frac{4}{5}\\-\frac45&-\frac35\end{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2288500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Find $\sin(A)$ and $\cos(A)$ given $\cos^4(A) - \sin^4(A) = \frac{1}{2}$ and $A$ is located in the second quadrant.
Question: Find $\sin(A)$ and $\cos(A)$, given $\cos^4(A)-\sin^4(A)=\frac{1}{2}$ and $A$ is located in the second quadrant.
Using the fundamental trigonometric identity, I was able to find that:
• $\cos^... | Hint
$$\left( \cos(A)+ \sin(A) \right)^2 = 1+2 \sin(A) \cos(A)=\frac{1}{2} \\
\left( \cos(A)- \sin(A) \right)^2 = 1-2 \sin(A) \cos(A)=\frac{3}{2} $$
Take the square roots, and pay attention to the quadrant and the fact that $\cos^4(A) >\sin^4(A)$ to decide is the terms are positive or negative.
Alternate simpler soluti... | {
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"timestamp": "2023-03-29T00:00:00",
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From $a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$ to $a_n$, Case 2
Find and prove by induction an explicit formula for $a_n$ if $a_1=1$ and, for $n \geq 1$,
$$P_n: a_{n+1}= \frac{3a_n}{(2n+2)(2n+3)}$$
Checking the pattern:
$$a_1=1 $$
$$a_2= \frac{3}{4 \cdot 5}$$
$$a_3= \frac{3^2} { 4 \cdot 5 \cdot 6 \cdot 7}$$
$$a_4= \frac{... | $a_{n+1} = \frac{3a_n}{(2n+2)(2n+3)} = \frac{3\cdot 3!\cdot 3^{n-1}}{(2n+2)(2n+3)(2(n-1)+3)!} = \frac{3!3^n}{(2n+3)!}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291696",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that for every integer $n$, $n^3 - n$ is divisible by 3 using modular arithmetic Problem:
Show that for every integer $n$, $n^3 - n$ is divisible by 3 using modular arithmetic
I was also given a hint:
$$n \equiv 0 \pmod3\\n \equiv 1 \pmod3\\n \equiv 2 \pmod3$$
But I'm still not sure how that relates to the questi... | Using the hint is to try the three cases:
Case 1: $n \equiv 0 \mod 3$
Remember if $a \equiv b \mod n$ then $a^m \equiv b^m \mod n$ [$*$]
So $n^3 \equiv 0^3 \equiv 0 \mod 3$
Remember if $a \equiv c \mod n$ and $b \equiv d \mod n$ then $a+b \equiv c + d \mod n$ [$**$]
So $n^3 - n\equiv 0 - 0 \equiv 0 \mod n$.
Case 2: $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the simplified form of $\frac {1}{\cos x + \sin x}$ Find the simplified form of $\dfrac {1}{\cos x + \sin x}$.
a). $\dfrac {\sin (\dfrac {\pi}{4} +x)}{\sqrt {2}}$
b). $\dfrac {\csc (\dfrac {\pi}{4} + x)}{\sqrt {2}}$
c). $\dfrac {\sin (\dfrac {\pi}{4} + x)}{2}$
d). $\dfrac {\csc (\dfrac {\pi}{4} + x)}{2}$
My Attemp... |
This is the graph of the function $\sin x+\cos x$.
Notice that itself looks like a wave.
So, we should be able to deduce that(since only the amplitude and phase seem to have changed),
$$\sin x+\cos x=A\sin( x+\phi)$$
Expanding using the identity for $\sin(A+B)$,
$$\sin x+\cos x=A\sin( x)\cos\phi+A\cos(x)\sin\phi$$
Com... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Is $f(x) = x^{10}-x^5+1$ solvable by radicals?
Is $f(x) = x^{10}-x^5+1$ solvable by radicals?
So far I've showed that $f$ is irreducible because if we let $y=x^5$ then $f(y)=y^2-y+1$ which is irreducible because it has a negative discriminant. I also know that $f$ has no real roots so I've concluded that $Gal(L_f/\ma... | Yes, of course.
Let $x+\frac{1}{x}=a$.
Hence,
$$x^{10}-x^5+1=x^{10}+x^7-x^7+x^6-x^5-x^6+1=$$
$$=(x^2-x+1)(x^7(x+1)-x^5-(x^3-1)(x+1))=$$
$$=(x^2-x+1)(x^8+x^7-x^5-x^4-x^3+x+1)=$$
$$=(x^2-x+1)x^4\left(x^4+\frac{1}{x^4}+x^3+\frac{1}{x^3}-x-\frac{1}{x}-1\right)=$$
$$=(x^2-x+1)x^4(a^4-4a^2+2+a^3-3a-a-1)=$$
$$=(x^2-x+1)x^4(a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that if $(x_n) \rightarrow x$ then $(\sqrt{x_n}) \rightarrow \sqrt{x}$. Let $x_n \ge 0$ for all $n \in \mathbf{N}$ and $x>0$, show that if $(x_n) \rightarrow x$ then $(\sqrt{x_n}) \rightarrow \sqrt{x}$.
My textbook does the following proof:
Let $\epsilon >0$, we must find an $N$ such that $n \ge N$ implies $|\sqrt... | Because $\sqrt{x_n}+\sqrt{x}\geq \sqrt x$ and thus $$\frac{1}{\sqrt{x_n}+\sqrt x}\leq \frac{1}{\sqrt x}.$$
When you'll see continuous function, such a proof is easier using the continuity of $x\longmapsto \sqrt x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Limit of function calculations I must solve limit of next function:
$$\lim_{x\to \infty}\frac{2x^3+x-2}{3x^3-x^2-x+1}$$
Does my calculations are proper? If not where is my mistake?
$$=\lim_{x\to \infty}\frac{x^3\left(2+\frac{1}{x^2}-\frac{2}{x^3}\right)}{x^3\left(3-\frac{1}{x}-\frac{1}{x^2}+\frac{1}{x^3}\right)} \\
\ =... | You are correct, if you have the ratio of two polynomial of the same degree $n$ then
$$\lim_{x\to +\infty}\frac{a_nx^n+a_{n-1}x^{n-1}+\dots +a_0}{b_nx^n+b_{n-1}x^{n-1}+\dots +b_0}=\lim_{x\to +\infty}\frac{x^n(a_n+\frac{a_{n-1}}{x}+\dots +\frac{a_0}{x^n})}{x^n(b_n+\frac{b_{n-1}}{x}+\dots +\frac{b_0}{x^n})}\\=\lim_{x\to ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding solutions to $2^x+17=y^2$
Find all positive integer solutions $(x,y)$ of the following equation:
$$2^x+17=y^2.$$
If $x = 2k$, then we can rewrite the equation as $(y - 2^k)(y + 2^k) = 17$, so the factors must be $1$ and $17$, and we must have $x = 6, y = 9$.
However, this approach doesn't work when $x$ is ... | It looks like I need to spell out the details for insipidintegrator.
If $x$ is even, the prime $17$ is the product of $y+2^{x/2}$ and $y-2^{x/2}.$ Averaging, we find $y=9$ whence $x=6.$
If $x$ is odd, write $y-2^{x/2}=\frac{17}{y+2^{x/2}}$. Letting $x=2n+1$, we have
$\Big|\frac{y}{2^n}-\sqrt{2}\Big|=\frac{17}{2^n(y+2^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Degenerate eigenvalues problem for a 4x4 system In summary, my question is whether or not I'm allowed to have the zero vector as my generalised eigenvector. I'm given the system $$x'=\begin{bmatrix} 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ -2 & 2 & -3 & 1 \\ 2 & -2 & 1 & -3 \end{bmatrix}x$$ I also found two eigenvalues: 0 & -... | Let's name your matrix A.
The matrix $(A + 2E) = \begin{pmatrix} 2 & 0 & 1 & 0 \\ 0 & 2 & 0 & 1 \\ -2 & 2 & -1 & 1 \\ 2 & -2 & 1 & -1 \end{pmatrix}$ has two linearly independent ordinary eigenvectors with eigenvalue 0: $\begin{pmatrix} -1 \\ 0 \\ 2 \\ 0 \end{pmatrix}$ and $\begin{pmatrix} 0 \\ -1 \\ 0 \\ 2 \end{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculating $\sum_{k=1}^\infty \frac{k^2}{2^k}=\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots+\frac{k^2}{2^k}+\cdots$ I want to know the value of $$\sum_{k=1}^\infty \frac{k^2}{2^k}=\frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots+\frac{k^2}{2^k}+\cdots$$
I added up to $k=5... | If we start with the power series
$$ \sum_{k=0}^{\infty}x^k=\frac{1}{1-x} $$
(valid for $|x|<1$) and differentiate then multiply by $x$, we get
$$ \sum_{k=1}^{\infty}kx^k=\frac{x}{(1-x)^2}$$
If we once again differentiate then multiply by $x$, the result is
$$ \sum_{k=1}^{\infty}k^2x^k=\frac{x(x+1)}{(1-x)^3}$$
and sett... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to find $\lim\limits_ {n\to\infty}n^5\int_n^{n+2}\frac{{x}^2}{{ {2+x^7}}}\ dx$? How to find $$\displaystyle\lim_ {n\to\infty}n^5\int_n^{n+2}\dfrac{{x}^2}{{ {2+x^7}}}\ dx$$Can I use Mean Value Theorem? Someone suggested I should use Lagrange but I don't know how it would help.
| $$
\begin{align}
2n^5\frac{n^2}{2+(n+2)^7}&\le n^5\int_n^{n+2}\frac{x^2}{2+x^7}\,\mathrm{d}x\le2n^5\frac{(n+2)^2}{2+n^7}\\[12pt]
\frac2{\frac2{n^7}+\left(1+\frac2n\right)^7}&\le n^5\int_n^{n+2}\frac{x^2}{2+x^7}\,\mathrm{d}x\le\frac{2\left(1+\frac2n\right)^2}{\frac2{n^7}+1}
\end{align}
$$
Apply the Squeeze Theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find the range of $f(x) = 3x^4 - 16x^3 + 18x^2 + 5$ without applying differential calculus
Find the range of $f(x) = 3x^4 - 16x^3 + 18x^2 + 5$ without applying differential calculus.
I tried to express $$f(x)=3x^4-16x^3+18x^2+5=A(ax^2+bx+c)^2+B(ax^2+bx+c)+C $$ which is a quadratic in $ax^2+bx+c$ which itself is quadr... | The range is $[k,+\infty)$ where $k$ is the minimum value such that the inequality
$$
3x^4-16x^3+18x^2+5\ge k
$$
is true for any $x \in \mathbb{R}$ and this is the minimum value $k$ such that the equation
$$
3x^4-16x^3+18x^2+5- k=0
$$
has a double solution, that is the discriminant of $3x^4-16x^3+18x^2+5- k$ is null.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2302781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Improper integral $\int \limits_{2}^{4}\frac{\sqrt{(16 - x^2)^5}}{(x^2 - 9x + 20)^3}dx$ I can't figure out how to solve (say whether it converges or diverges) the following improper integral:
$$
\int \limits_{2}^{4}\frac{\sqrt{(16 - x^2)^5}}{(x^2 - 9x + 20)^3}dx
$$
I've tried to simplify this and got:
$$
\int \limits_{... | An alternate substitution is $x = 4 \, \sin(t)$ which leads to, with some difficulty,
\begin{align}
\int \frac{(16 - x^2)^{5/2}}{(x^2 - 9 x + 20)^3} \, dx = \frac{81}{2} \, \frac{x-6}{(x-5)^2} \, \sqrt{16 - x^2} - 63 \, \tan^{-1}\left(\frac{16 - 5x}{3 \, \sqrt{16 - x^2}}\right) - \sin^{-1}\left(\frac{x}{4}\right) + c_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2305911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Existence of square root matrix $B \in \mathbb{C}^{2\times 2}$ for any $A \in \mathbb{C}^{2\times 2}$, where $A^2\neq 0$ I am trying to prove that for any $A \in \mathbb{C}^{2\times 2}$ with $A^2\neq 0$, there exists $B \in \mathbb{C}^{2\times 2}$ with $BB=A$.
I have tried the approach of a general matrix A andB with ... | We have the equations
\begin{eqnarray*}
a^2+bc= A \\
b(a+d)=B \\
c(a+d)=C \\
bc+d^2=D
\end{eqnarray*}
Multiply the first equation by $(a+d)^2$ and use the second & third we have
\begin{eqnarray*}
a^2(a+d)^2 +BC=A(a+d)^2 \\
d= -a +\sqrt{\frac{BC}{A-a^2}}.
\end{eqnarray*}
Now subtract the first & the fourth
\begin{eqna... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Area between $r=4\sin(\theta)$ and $r=2$ I'm trying to find the area between $r=4\sin(\theta)$ and $r=2$.
I found the points of intersections to be $\pi/6,5\pi/6$. Which implies the area is $$A=\frac{1}{2}\int_{\pi/6}^{5\pi/6}(4\sin(\theta))^2-2^2d\theta.$$
Is this correct? Or did I find the area for the following reg... | The desired red region is just the area of a circle with radius $2$ minus the area of the blue region:
$$
\pi(2)^2 - A
= 4\pi - \frac{1}{2}\int_{\pi/6}^{5\pi/6}(4\sin(\theta))^2-2^2d\theta
= 4\pi - \left( 2\sqrt 3 + \frac{4\pi}{3} \right)
= \frac{8\pi}{3} - 2\sqrt 3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Find a basis for orthogonal complement in R⁴
How do I approach part 2? I found the projection of 1. to be (6,-2,2,-2) but what do I do now?
| For vector $\mathbf v = (x_1, x_2, x_3,x_4)$, the dot products of $\mathbf v$ with the two given vectors respectively are zero.
$$\begin{align*}
\begin{bmatrix}1&2&3&4\\2&5&0&1\end{bmatrix}
\begin{bmatrix}x_1\\x_2\\x_3\\x_4\end{bmatrix} &= \begin{bmatrix}0\\0\end{bmatrix}\\
\begin{bmatrix}1&2&3&4\\0&1&-6&-7\end{bmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Calculus Spivak Chapter 2 problem 16(c) The question asks to prove that if $\frac mn \lt \sqrt{2}$, then there is another rational number $\frac {m'}{n'}$ with $\frac mn \lt \frac {m'}{n'} \lt \sqrt{2}$.
Intuitively, it's clear that such a number exists, but I don't understand the solution to this problem. It states: l... | I actually have a slightly different answer to the above, which I think is closer to the book as it relies directly on parts (a) and (b).
If anyone spots anything wrong I'd appreciate if you comment below and point out any mistakes:
We have proven in part (a) that:
$\frac{m^2}{n^2} < 2 \implies \frac{(m + 2n)^2}{(m + n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Find $\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})}$ without L'Hopital I tried:
$$\lim_{x\rightarrow \frac{\pi}{4}}\frac{\cos(2x)}{\sin(x-\frac{\pi}{4})} =
\frac{\frac{\cos(2x)}{x-\frac{\pi}{4}}}{\frac{\sin(x-\frac{\pi}{4})}{x-\frac{\pi}{4}}}$$ and $$\begin{align}\frac{\cos(2x)}{x-\frac{\pi}... | Just as an alternate approach:
I said: $\frac{cos(2\theta)}{sin\theta*cos(\frac{\pi}{4})-cos\theta*(\frac{\pi}{4})}$ =
$\frac{cos(2\theta)}{1/\sqrt(2)*sin\theta-cos\theta}$ Noting that $cos(2\theta)=cos^2\theta-sin^2\theta$ and $sin\theta-cos\theta=-(cos\theta-sin\theta)$
So, we have
$\frac{cos^2\theta-sin^2\theta}{-1/... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Can the following trigonometric equation be transformed into the other? Can $$16\sec^2(x)\tan^4(x)+88\sec^4(x)\tan^2(x)+16\sec^6(x)$$ be proven equal to
$$24\sec^6(x)-8\sec^4(x)+96\sec^4(x)\tan^2(x)-16\sec^2(x)\tan^2(x)$$
I have made about six attempts, but I keep getting stuck. I thought I'd ask, maybe someone el... | Sometimes the easiest thing to do is convert everything into sines and cosines.
\begin{array}{l}
16 \sec^2(x) \tan^4(x) + 88 \sec^4(x) \tan^2(x) + 16 \sec^6(x) \\
=\dfrac{16\sin^4(x)}{\cos^6(x)}+\dfrac{88\sin^2(x)}{\cos^6(x)}
+\dfrac{16}{\cos^6(x)} \\
=\dfrac{16\sin^4(x) + 88 \sin^2(x) + 16}{\cos^6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that if $\frac1a+\frac1b+\frac1c = a+b+c$, then $\frac1{3+a}+\frac1{3+c}+\frac1{3+c} \leq\frac34$
Show that if $a,b,c$ are positive reals, and
$\frac1a+\frac1b+\frac1c = a+b+c$, then
$$\frac1{3+a}+\frac1{3+b}+\frac1{3+c} \leq\frac34$$
The corresponding problem replacing the $3$s with $2$ is shown here:
How... | For the case $k≥3$, $(k\in\mathbb Z)$
Given equality implies $ab+bc+ca=abc(a+b+c)$
We know that $(x+y+z)^2≥3(xy+yz+zx)$
put $x=ab,y=bc,z=ca\Rightarrow (ab+bc+ca)^2\geq3abc(a+b+c)=3(ab+bc+ca)$
$\Rightarrow abc(a+b+c)=ab+bc+ca\geq3$ and $a+b+c\geq3$ (obvious)
$\dfrac {1}{k+a}+\dfrac {1}{k+b}+\dfrac {1}{k+c}≤\dfrac {3}{k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2312672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
How can I find maximum and minimum modulus of a complex number? I have this problem. Let be given complex number $z$ such that
$$|z+1|+ 4 |z-1|=25.$$
Find the greastest and the least of the modulus of $z$.
I tried with minimum.
Put $A(-1,0)$, $B(1,0)$ and $M(x,y)$ present of $z$.
We have $O(0,0)$ is the midpoint of th... | For maximum $|z|$, we have
\begin{align}
|5z|&=|(z+1)+4(z-1)+3|\\
&\le|z+1|+4|z-1|+|3|\\
&\le25+3\\
|z|&\le \frac{28}{5}
\end{align}
with the equality holds if and only if $\displaystyle z=\frac{28}{5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314488",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve for $x$ in $\cos(2 \sin ^{-1}(- x)) = 0$
Solve $$\cos(2 \sin ^{-1}(- x)) = 0$$
I get the answer $\frac{-1}{ \sqrt2}$
By solving like this
\begin{align}2\sin ^{-1} (-x )&= \cos ^{-1} 0\\
2\sin^{-1}(- x) &= \frac\pi2\\
\sin^{-1}(- x) &= \frac\pi4\\
-x &= \sin\left(\frac\pi4\right)\end{align}
Thus $x =\frac{-1}{\... | You need to solve $\cos \left(2 \arcsin(-x) \right) = 0$. Let $y = 2 \arcsin(-x)$ then $\cos y = 0$ so $y = \pi/2 \pm n\pi$. Then,
$$
2 \arcsin(-x) = \frac{pi}{2} \pm n\pi
$$
which implies
$$
x = -\sin \left( \frac{\pi}{4} \pm \frac{n\pi}{2} \right)
$$
Can you simplify this?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find this limit. Compute the value of the limit :
$$
\lim_{x\to\infty}{\frac{1-\cos x\cos2x\cos3x}{\sin^2x}}
$$
I've tried simplifying the expression to
$$
\lim_{x\to\infty}\frac{-8\cos^6x+10\cos^4x-3\cos^2x+1}{\sin^2x}
$$
But I don't know what to do after this.
| Using the identities $\displaystyle \sin^2(x)=\frac{1-\cos(2x)}{2}$, $\displaystyle \cos(4x)=2\cos^2(2x)-1$, and $\displaystyle \cos(x)\cos(3x)=\frac12(\cos(2x)+\cos(4x))$, we obtain
$$\begin{align}
\frac{1-\cos(x)\cos(2x)\cos(3x)}{\sin^2(x)}&=\frac{1-\frac{\cos(2x)\left(\cos(2x)+\overbrace{(2\cos^2(2x)-1)}^{=\cos(4x)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
equation of ellipse after projection If I have the intersection of $x+z=1$ and $$x^2 +y^2 +z^2=1$$ which is a circle in $O'xyz$. Then I do a projection of this circle on the $O'xy$ plane, it'll be an ellipse. How can I then find the equation of this ellipse?
| If $(x,y,z)$ is a point on the circle, then $(x,y,z)$ satisfy both $x+z=1$ and $x^2+y^2+z^2=1$. It's equations are
$$\begin{cases} z=1-x \\ x^2+y^2+(1-x)^2=1\end{cases}$$
Its projection has $z$-coordinate equals $0$ and keep the $x$ and $y$-coordinates. So the equation of the projection on the $xy$-plane is
\begin{alig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the following Determinant of $12$th degree polynomial Evaluate $$\Delta=\begin{vmatrix}
\frac{1}{(a+x)^2} & \frac{1}{(b+x)^2} & \frac{1}{(c+x^2)}\\
\frac{1}{(a+y)^2} & \frac{1}{(b+y)^2} & \frac{1}{(c+y)^2}\\
\frac{1}{(a+z)^2} & \frac{1}{(b+z)^2} & \frac{1}{(c+z)^2}\\
\end{vmatrix}$$
My Try: I have taken all... | It's obvious that we have a factor $(a-b)(b-c)(c-a)(x-y)(y-z)(z-x)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$ Find the equation of the line through the point $(\frac{1}{2},2)$ and tangent to the parabola $y=\frac{-x^2}{2}+2$ and secant to the curve $y=\sqrt{4-x^2}$
Let the required... | Let $\left(t,-\frac{t^2}{2}+2\right)$ be a tangent point.
Since $\left(-\frac{x^2}{2}+2\right)'=-x$, we get an equation of the tangent line:
$$y+\frac{t^2}{2}-2=-t(x-t).$$
Now, substitute $x=\frac{1}{2}$ and $y=2$, find a values of $t$ (I got $t=0$ or $t=1$) and choose a value, which you need.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Simplification of Trigo expression
Simplify $$\frac{\tan^2 x+\cos^2 x}{\sin x+ \sec x}$$
My attempt,
$$=\frac{(\cos^2x+\frac{\sin^2x}{\cos^2x})}{\frac{\sin x \cos x+1}{\cos x}} $$
$$=\frac{\cos^4 x+\sin^2 x}{\cos^2 x}\cdot \frac{\cos x}{\sin x \cos x+1}$$
$$=\frac{\cos^4 x+\sin^2 x}{\cos x(\sin x \cos x+1)}$$
I'm st... | Hint
Use $\tan^2x=\sec^2x -1$
$$\frac{\tan^2 x+\cos^2 x}{\sin x+ \sec x}=\frac{(\sec^2 x-1)+\cos^2 x}{\sin x+ \sec x}=\frac{\sec^2 x-\sin^2 x}{\sin x+ \sec x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$ If $x^3 - 5x^2+ x=0$ then find the value of $\sqrt {x} + \dfrac {1}{\sqrt {x}}$
My Attempt:
$$x^3 - 5x^2 + x=0$$
$$x(x^2 - 5x + 1)=0$$
Either,
$x=0$
And,
$$x^2-5x+1=0$$
??
| $x^3-5x^2+x$ gives $x=0$ or $x^2+1=5x$.
For $x\leq0$ the needed value does not exist.
For $x>0$ we have $\sqrt{x}+\frac{1}{\sqrt{x}}>0$.
Thus,
$$x+\frac{1}{x}=5$$
or $$\left(\sqrt{x}+\frac{1}{\sqrt{x}}\right)^2=7,$$
which gives $\sqrt{x}+\frac{1}{\sqrt{x}}=\sqrt7.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2329371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Calculate $\tan^2{\frac{\pi}{5}}+\tan^2{\frac{2\pi}{5}}$ without a calculator The question is to find the exact value of:
$$\tan^2{\left(\frac{\pi}{5}\right)}+\tan^2{\left(\frac{2\pi}{5}\right)}$$
without using a calculator.
I know that it is possible to find the exact values of $\tan{\left(\frac{\pi}{5}\right)}$ and $... | Let $\tan\left(\frac{\pi}{5}\right)=x$
$$\tan\left(\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}+\frac{\pi}{5}\right)=0$$
$$\implies S_1-S_3+S_5=0$$
($S_k$ represents sum of tangents taken $k$ at a time)
$$\implies 5x-10x^3+x^5=0$$
Now the roots of this equation are $\tan\left(\frac{\pi}{5}\right),\tan\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 3
} |
Find $\arctan(\frac{1}{3})+\arctan(\frac{1}{9})+\arctan(\frac{7}{19})$ Firstly used this formula $$ \begin{align} \arctan(\alpha)+\arctan(\beta)
& =\arctan(\frac{1-xy}{x+y}),\quad x\gt0,y\gt0 \\ &=\arctan(\frac{1-\frac{1}{3}\frac{1}{9}}{\frac{1}{3}+\frac{1}{9}}) \\ &=\arctan(2) \end{align}$$ So it is $\arctan(2)+\arct... | Just calculate:
$$\tan\left(\arctan\frac{1}{3}+\arctan\frac{1}{9}+\arctan\frac{7}{19}\right)=$$
$$=\frac{\frac{\frac{1}{3}+\frac{1}{9}}{1-\frac{1}{3}\cdot\frac{1}{9}}+\frac{7}{19}}{1-\frac{\frac{1}{3}+\frac{1}{9}}{1-\frac{1}{3}\cdot\frac{1}{9}}\cdot\frac{7}{19}}=1,$$
which gives the answer: $45^{\circ}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving trigonometric identity $\frac{\sin(A)}{1+ \cos(A )}+\frac{1+ \cos(A )}{\sin(A)}=2 \csc(A)$
$$
\frac{\sin(A)}{1+\cos(A)}+\frac{1+\cos(A)}{\sin(A)}=2\csc(A)
$$
\begin{align}
\mathrm{L.H.S}&= \frac{\sin^2A+(1+\cos^2(A))}{\sin(A)(1+\cos(A))} \\[6px]
&= \frac{\sin^2A+2\sin(A)\cos(A)+\cos^2(A)+1}{\sin(A)(1+\cos(A))... | You make several mistakes, the main one being
$$
(a+b)^2=a^2+b^2
$$
The mistake is $(1+\cos(A))^2=1+\cos^2(A)$, whereas it should be
$$
(1+\cos(A))^2=1+2\cos(A)+\cos^2(A)
$$
Note that
$$
\frac{a}{b}+\frac{b}{a}=\frac{a^2+b^2}{ab}
$$
where $a=\sin(A)$ and $b=1+\cos(A)$.
In the second step you also arbitrarily insert a t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 3
} |
Finding the volume of the solid generated by revolving the given curve. The objective is to find the volume of the solid generated by revolving the curve $y=\dfrac{a^3}{a^2+x^2}$ about its asymptote.
Observing the given function yields that $y\ne0$, hence $y=0$ is the asymptote to the given curve. Thus, the volume of t... | Your integral is equal $\frac{\pi^2}{2}( \frac{1}{a^2})^{\frac{3}{2}}a^6$ according to wolfram alpha. Your algebra must be wrong somewhere. I recommend trying trigonometric substitutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Find $a^2+b^2+2(a+b)$ minimum if $ab=2$ Let $a,b\in R$,and such
$$ab=2$$
Find the minimum of the $a^2+b^2+2(a+b)$.
I have used $a=\dfrac{2}{b}$, then
$$a^2+b^2+2(a+b)=\dfrac{4}{b^2}+b^2+\dfrac{4}{b}+2b=f'(b)$$
Let $$f'(b)=0,\,b=-\sqrt{2}$$
So $$a^2+b^2+2(a+b)\ge 4-4\sqrt{2}$$
I wanted to know if there is other way to ... | For $a=b=-\sqrt2$ we get a value $4-4\sqrt2$.
We'll prove that it's a minimal value.
Indeed, let $a+b=2k\sqrt{ab}$.
Hence, $|k|=\left|\frac{a+b}{2\sqrt{ab}}\right|\geq1$ and we need to prove that
$$a^2+b^2+2(a+b)\geq4-4\sqrt2$$ or
$$a^2+b^2+\sqrt{2ab}(a+b)\geq(2-2\sqrt2)ab$$ or
$$(a+b)^2+\sqrt{2ab}(a+b)\geq(4-2\sqrt2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2341443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate an indefinite integral Find the value of
$$\int{\frac{x^2e^x}{(x+2)^2}} dx$$
My Attempt: I tried to arrange the numerator as follows:
$$ e^xx^2 = e^x(x+2-2)^2 $$ but that didn't help.
Any guidance on this problem will be very helpful.
| Another method:
\begin{align}
\int \frac{x^2 \, e^{x}}{(x+2)^2} \, dx &= - \int x^2 \, e^{x} \, \frac{d}{dx} \left(\frac{1}{x+2}\right) \, dx \\
&= - \left[ \frac{x^2 \, e^{x}}{x + 2} \right] + \int x(x+2) \, e^{x} \cdot \frac{1}{x+2} \, dx \\
&= - \frac{x^2 \, e^{x}}{x + 2} + \int x \, e^{x} \, dx \\
&= - \frac{x^2 \,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Hyperbolas: Deriving $\frac{x^2}{a^2} + \frac{y^2}{a^2 - c^2} = 1$ from $\sqrt{(x + c)^2 + y^2} - \sqrt{(x - c)^2 + y^2} = \pm2a$ My textbook's section on Hyperbolas states the following:
If the foci are $F_1(-c, 0)$ and $F_2(c, 0)$ and the constant difference is $2a$, then a point $(x, y)$ lies on the hyperbola if an... | You have\begin{multline*}\sqrt{(x+c)^2+y^2}-\sqrt{(x-c)^2+y^2}=\pm2a\Longleftrightarrow\\\Longleftrightarrow(x+c)^2+y^2+(x-c)^2+y^2-2\sqrt{(x+c)^2+y^2}\sqrt{(x-c)^2+y^2}=4a^2.\end{multline*}This is the same thing as saying that$$\sqrt{(x+c)^2+y^2}\sqrt{(x-c)^2+y^2}=-2a^2+c^2+x^2+y^2.$$Squaring both sides, one gets$$\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the largest constant $k$ such that $\frac{kabc}{a+b+c}\leq(a+b)^2+(a+b+4c)^2$
Find the largest constant $k$ such that $$\frac{kabc}{a+b+c}\leq(a+b)^2+(a+b+4c)^2$$
My attempt,
By A.M-G.M, $$(a+b)^2+(a+b+4c)^2=(a+b)^2+(a+2c+b+2c)^2$$
$$\geq (2\sqrt{ab})^2+(2\sqrt{2ac}+2\sqrt{2bc})^2$$
$$=4ab+8ac+8bc+16c\sqrt{ab}$... | One more way...
Noting that replacing $a, b$ with $\frac{a+b}2, \frac{a+b}2$ leaves RHS unchanged but increases the LHS, we have to only check for the case $a=b$. Further as the inequality is homogeneous in $a, b, c$; WLOG we may set $a=1$. Hence we need only look for the minimum of the univariate
$$f(c) = (4+(2+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2343592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Hoffman and Kunze, Linear algebra sec 3.5 exercise 9
Let $V$ be the vector space of all $2\times 2$ matrices over the field of real numbers and let $$B=\begin{pmatrix}2&-2\\-1&1\end{pmatrix}.$$
Let $W$ be the subspace of $V$ consisting of all $A$ such that $AB=0.$ Let $f$ be a linear functional on $V$ which is in th... | There is nothing wrong with your solution.
Perhaps the author had the following approach in mind for which $C$ could be useful. Also it helps to determine the functional completely. Let us consider the following standard basis of $V$:
$$\mathcal{B}=\{E_1,E_2,E_3,E_4\}=\left\{\begin{bmatrix}1&0\\0&0\end{bmatrix}, \begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $\frac {\sin A + \tan A}{\cos A}=9$, find the value of $\sin A$. If $\dfrac {\sin A + \tan A}{\cos A}=9$, find the value of $\sin A$.
My Attempt:
$$\dfrac {\sin A+\tan A}{\cos A}=9$$
$$\dfrac {\sin A+ \dfrac {\sin A}{\cos A}}{\cos A}=9$$
$$\dfrac {\sin A.\cos A+\sin A}{\cos^2 A}=9$$
$$\dfrac {\sin A(1+\cos A)}{\cos^... | Hint: using the tangent half-angle formulas, let $\,t=\tan(A/2)\,$, then the equation becomes:
$$
\frac{2t}{1+t^2} + \frac{2t}{1-t^2}=9 \,\frac{1-t^2}{1+t^2} \;\;\iff\;\; 9 t^4 - 18 t^2 - 4 t + 9 = 0
$$
The quartic has $2$ real roots which can be solved in radicals, but the calculations are not pretty.
[ EDIT ] Onc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2344532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the Roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$ Once I came across the following problem: find the roots of $(x+1)(x+3)(x+5)(x+7) + 15 = 0$.
Here it is how I proceeded:
\begin{align*}
(x+1)(x+3)(x+5)(x+7) + 15 & = [(x+1)(x+7)][(x+3)(x+5)] + 15\\
& = (x^2 + 8x + 7)(x^2 + 8x + 15) + 15\\
& = (x^2 + 8x + 7)[(x^2 + 8x +... | HINT.-Looking about integer solutions for $f(x)=(x+1)(x+3)(x+5)(x+7) + 15 = 0$ possible values should be even and negative so the only candidates are $-2,-4$ and $-6$. We verified that $-2$ and $-6$ are roots. The other two roots are solutions of $$\frac{x^4+16x^3+86x^2+176x+120}{x^2+8x+12}=x^2+8x+10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 2
} |
Closed form of function composition Given that $f(x)=\dfrac{x+6}{x+2}$, find $f^{n}(x)$ where $f^{n}(x)$ indicates the $n$th iteration of the function.
I first tried to find a pattern but there didn't seem to be an obvious one:
$$f(x) = \dfrac{x+6}{x+2}$$
$$f^2(x) = \dfrac{7x + 18}{3x + 10}$$
$$f^3(x) = \dfrac{25x + 7... | \begin{align}
a_{n+1} &= a_n + b_n \\
b_{n+1} &= 6a_n + 2b_n\\
c_{n+1} &= c_n + d_n \\
d_{n+1} &= 6c_n + 2d_n
\end{align}
Let's write it in matrix form:
$$\begin{bmatrix} a_{n+1} & c_{n+1}\\ b_{n+1} & d_{n+1} \end{bmatrix}= \begin{bmatrix} 1 & 1 \\6 & 2 \\ \end{bmatrix}\begin{bmatrix} a_{n} & c_{n}\\ b_{n} & d_{n} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solving equations with complex numbers I have to solve:
$Z^3+\bar{\omega^7} = 0$ and $Z^5\omega^{11}=1$
From the second equation, I got $Z^5=\omega$ and from the first I got $Z^3=-\omega^2$. I plugged in omega from the first result into $Z^3=-\omega^2$, giving me $Z=0$ or $Z^7=-1$, finally giving me 8 solutions:$0,-1,-... | Let's assume that $\omega$ is a root of unity, but not necessarily a cube root of unity, and see what happens.
It's clear that $Z\not=0$. The equation $Z^3+\overline\omega^7=0$ can be rewritten as $Z^3\omega^7=-1$, which, by squaring both sides, implies $Z^6\omega^{14}=1$. Combining with the other equation, $Z^5\omeg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2346016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Riccati D.E., vertical asymptotes
For the D.E.
$$y'=x^2+y^2$$
show that the solution with $y(0) = 0$ has a vertical asymptote at some point $x_0$. Try to find upper and lower bounds for $x_0$:
$$y'=x^2+y^2$$
$$x\in \left [ a,b \right ]$$
$$b> a> 0$$
$$a^2+y^2\leq x^2+y^2\leq b^2+y^2$$
$$a^2+y^2\leq y'\leq b^2+y^... | 1. $x_0$ exists
First note that $y'''(x)$ is increasing$^{[1]}$. It is also easy to see that $y'(0)=y''(0)=0$ but $y'''(0)=2$$^{[2]}$, so by Taylor's theorem$^{[3]}$,
$$
y(x)=\frac{x^3}{6}y'''(c)\ge \frac{x^3}{3},\qquad (*)
$$
for all $x>0$ such that $y$ is defined. Choose one such $x=\epsilon>0$. Then if $x>\epsilo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sum_{x=0}^{n}(-1)^x\binom{n}{n-x} (n+1-x)^n=n!$ I figure out these thing when "playing" with numbers:
$$3^2-2.2^2+1^2=2=2!$$
$$4^3-3.3^3+3.2^3-1^3=6=3!$$
$$5^4-4.4^4+6.3^4-4.2^4+1^4=24=4!$$
So I go to the conjecture that:
$$\binom{n}{n}(n+1)^n-\binom{n}{n-1}n^n+\binom{n}{n-2}(n-1)^n-...=n!$$
or
$$\sum_{x=0... | First of all note that $$\sum_{k=0}^{n}\dbinom{n}{n-k}\left(-1\right)^{k}\left(n-k+1\right)^{n}=\sum_{k=0}^{n}\dbinom{n}{k}\left(-1\right)^{k}\left(n-k+1\right)^{n}$$ then from the special case of the Melzak's identity: $$\sum_{k=0}^{n}\left(-1\right)^{k}\dbinom{n}{k}\frac{f\left(y-k\right)}{x+k},=\frac{f\left(x+y\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 6,
"answer_id": 4
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Other ways to evaluate the integral $\int_{-\infty}^{\infty} \frac{1}{1+x^{2}} \, dx$?
$$\int_{-\infty}^{\infty}\frac{1}{x^2+1}\,dx=\pi $$
I can do it with the substitution $x= \tan u$ or complex analysis. Are there any other ways to evaluate this?
| You can use partial fractions:
$$
\begin{align}
\int_{-\infty}^\infty \frac{dx}{1+x^2}
& = \int_{-\infty}^\infty \frac{1}{2} \left( \frac{1}{1+ix} + \frac{1}{1-ix} \right) dx \\
& = \frac{1}{2i} \bigg[\log(1+ix) - \log(1-ix)\bigg]_{-\infty}^\infty \\
& = \frac{1}{2i} \left[ \lim_{x\to\infty} \log\left( \frac{1+ix}{1-ix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2349224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 1
} |
let $f$ be a differentiable function. Compute $\frac{d}{dx}g(2)$, where $g(x) = \frac{f(2x)}{x}$. let $f$ be a differentiable function and $$\lim_{x\to 4}\dfrac{f(x)+7}{x-4}=\dfrac{-3}{2}.$$
Define $g(x)=\dfrac{f(2x)}{x}$. I want to know the derivative
$$\dfrac{d}{dx}g(2)=?$$
I know that :
$$\dfrac{d}{dx}g(2)=\dfrac{... | HINT
\begin{align*}
\frac{dg}{dx}(2) & = \lim_{x\rightarrow 2} = \frac{g(x) - g(2)}{x - 2} = \lim_{x\rightarrow 2}\frac{\displaystyle\frac{f(2x)}{x} - \frac{f(4)}{2}}{x - 2} = \lim_{x\rightarrow 2}\frac{2f(2x)-xf(4)}{2x(x-2)}\\
& = \lim_{u\rightarrow 4}\frac{2f(u) - u\displaystyle\frac{f(4)}{2}}{u\left(\displaystyle\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2351494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove that number of non-isomorphic ordered tree with 'n' vertices is nth catalan number. according to wikipedia $C$n is the number of non-isomorphic ordered trees with n vertices. But I can't seem to be able to prove this result. How do we do that?
where $n$th catalan number is:
$$
C_n = \frac 1{n+1} \binom{2n}{n}
$$
| We have from basic principles for the species of ordered trees the
species equation
$$\mathcal{T} = \mathcal{Z} +
\mathcal{Z} \mathfrak{S}_{\ge 1}(\mathcal{T}).$$
This yields the functional equation for the generating function $T(z)$
$$T(z) = z + z \frac{T(z)}{1-T(z)}$$
which is
$$T(z) (1-T(z)) = z (1-T(z)) + z T(z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2351676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculate Derivative of a map
Consider the maps from $R^2 \to R^2$ such that $F(u, v) = (e^{u + v}, e^{u - v})$ and $G(x, y) = (xy, x^2 - y^2)$. Calculate $D(F \circ G)(1, 1)$ by directly composing.
I got
$F \circ G = (e^{x^2 +xy - y^2}, e^{y^2 + xy - x^2})$
But how do I get the derivative matrix?
| Remember that the derivative of a vector field is its Jacobian. If $F:\mathbb{R}^2\to\mathbb{R}^2$ is differentiable function with $F(x,y)=(f_1(x,y),f_2(x,y))$ where both $f_i$ are differentiable, then we have: $$D_{(x,y)}F = \left(
\begin{array}{cc}
\frac{\partial}{\partial x}f_1(x,y) & \frac{\partial}{\partial y}f_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $a + \frac{1}{a} = -1$, then the value of $(1-a+a^2)(1+a-a^2)$ is?
If $a + \frac{1}{a} = -1$ then the value of $(1-a+a^2)(1+a-a^2)$ is?
Ans. 4
What I have tried:
\begin{align}
a + \frac{1}{a} &= -1 \\
\implies a^2 + 1 &= -a \tag 1 \\
\end{align}
which means
\begin{align}
(1-a+a^2)(1+a-a^2) &=(-2a)(-2a^2) \\
&=4a^... | Without solving the quadratic:
$$ a^3 = - a^2 - a = a + 1 - a = 1 $$
which was found by using the equation $a^2 + a + 1 = 0$ twice. This means that $a$ is a non-real cube root of unity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2353620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
If $ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$ then what is the value of $a^2-ax$ is equal to
If $$ a=\frac{\sqrt{x+2} + \sqrt {x-2}}{\sqrt{x+2} -\sqrt{x-2}}$$ then
the value of $a^2-ax$ is equal to:
a)2 b)1 c)0 d)-1
Ans. (d)
My attempt:
Rationalizing $a$ we get,
$ x+ \sqrt {x^2-4}$
$a^2=(x+\sqrt{... | $$\dfrac a1=\dfrac{\sqrt{x+2}+\sqrt{x-2}}{\sqrt{x+2}-\sqrt{x-2}}$$
calling for Componendo and Dividendo
$$\dfrac{a+1}{a-1}=\dfrac{\sqrt{x+2}}{\sqrt{x-2}}$$
Squaring we get $$\dfrac{a^2+1+2a}{a^2+1-2a}=\dfrac{x+2}{x-2}$$
Again apply componendo and dividendo, $$\dfrac{a^2+1}{2a}=\dfrac x2$$
Now simplify
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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What is the number of triples (a, b, c) of positive integers such that the product $a.b.c=1000$ , and $a \leq b \leq c $?
What is the number of triples (a, b, c) of positive integers such that the product $a.b.c=1000$, and $a \leq b \leq c$?
My try:
The prime factorization of $1000$ is $2^3\cdot 5^3$
$a\cdot b \cdo... | Your computation of $N=10$ is correct and $100$ is the number of ordered triples that have product $1000$. You have failed to account for the condition that $a \le b \le c$. All of the unordered triples that have three distinct elements have shown up six times, so you have overcounted. Those that have two or three e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
proof that $n^4+22n^3+71n^2+218n+384$ is divisible by $24$ Let $n$ be a positive integer number.
How do we show that the irreducible polynomial $n^4+22n^3+71n^2+218n+384$ is divisible by $24$ for all $n$?
| Another way to do it is to note:
$n^4+22n^3+71n^2+218n+384 \equiv $
$n^4 - 2n^3 - n^2 + 2n \mod 8$.
If $n \equiv 0, \pm 1, \pm 2, \pm3, 4 \mod 8$ then $n^4 - 2n^3 - n^2 + 2n\equiv$
$0, 1\mp 2 - 1 \pm 1, 16 \mp 16 -4 \pm 4, 81 \mp 54 - 9 \pm 6, 4^4 \mp 2*4^3 - 16 + 8 \equiv$
$0, 0, 72 \mp 48, 0 \equiv 0 \mod 8$.
So $8$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2356982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
How to calculate $\lim_{x\to 0^+} \frac{x^x- (\sin x)^x}{x^3}$ As I asked, I don't know how to deal with $x^x- (\sin x)^x$.
Please give me a hint. Thanks!
| $$\lim_{x\rightarrow0^+}\frac{\ln{x}-\ln{\sin{x}}}{x^2}=\lim_{x\rightarrow0^+}\left(\frac{\ln\left(1+\frac{x}{\sin{x}}-1\right)}{\frac{x}{\sin{x}}-1}\cdot\frac{\frac{x}{\sin{x}}-1}{x^2}\right)=$$
$$=\lim_{x\rightarrow0}\left(\frac{x-\sin{x}}{x^3}\cdot\frac{x}{\sin{x}}\right)=\lim_{x\rightarrow0}\frac{1-\cos{x}}{3x^2}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How does the trigonometric identity $1 + \cot^2\theta = \csc^2\theta$ derive from the identity $\sin^2\theta + \cos^2\theta = 1$? I would like to understand how would the original identity of $$ \sin^2 \theta + \cos^2 \theta = 1$$ derives into
$$ 1 + \cot^2 \theta = \csc^2 \theta $$
This is my working:
a) $$ \frac{\... | The same way; you start with $\sin^2\theta + \cos^2\theta = 1$ and divide both sides by $\cos^2\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Find the number of points of differentiability for the following function. If
$$f(x)=\begin{cases}
\cos x^3&;x\lt0\\
\sin x^3 - |x^3-1|&;x\ge0
\end{cases}$$
then find the number of points where $g(x)=f(|x|) \text { is non differentiable.}$
| The question is asking about $f(|x|)$, by symmetry, since $g(x)$ is not differentiable at $x=1$, it is not differentiable at $x=-1$ as well.
To show that it is not differentiable at $x=1$:
If it is differentiable at $x=1$, then the following limit exists.
\begin{align}\lim_{x \rightarrow 1} \frac{f(x)-f(1)}{x-1}&= \li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2360233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Lengthy integration problem $$ \int\frac{x^3+3x+2}{(x^2+1)^2(x+1)} \, dx$$
I managed to solve the problem using partial fraction decomposition.
But that approach is pretty long as it creates five variables. Is there any other shorter method to solve this problem(other than partial fractions)?
I also tried trigonometri... | There are much better ways to find the coefficients in partial fractions than solving five equations in five variables.
Writing your function as
$$ \frac{x^3 + 3 x + 2}{(x^2+1)^2 (x+1)} = \frac{Q(x)}{(x^2+1)^2} + \frac{E}{x+1} $$
multiply both sides by $x+1$ and substitute $x=-1$. We get
$$ \frac{-2}{2^2} = 0 + E$$
so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Find number of real solutions of $3x^5+2x^4+x^3+2x^2-x-2=0$
Find a number of real roots of $$f(x)=3x^5+2x^4+x^3+2x^2-x-2$$
I tried using differentiation:
$$f'(x)=15x^4+8x^3+3x^2+4x-1=0$$ and I found number of real roots of $f'(x)=0$ by drawing graphs of $g(x)=-15x^4$ and $h(x)=8x^3+3x^2+4x-1$ and obviously from graph... | $f'(x)=15x^4+8x^3+3x^2+4x-1>0$ for $x>\frac{1}{2}$ and $f\left(\frac{1}{2}\right)<0$.
Hence, since $\lim\limits_{x\rightarrow+\infty}f(x)=+\infty$, we see that there is unique root for $x>\frac{1}{2}$.
Now, prove that $f(x)<0$ for all $x\leq\frac{1}{2}$.
For example, for $0\leq x\leq\frac{1}{2}$ we have
$$3x^5+2x^4+x^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Evaluate $\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$
Evaluate
$$\lim_{ x\to \infty} \left( \tan^{-1}\left(\frac{1+x}{4+x}\right)-\frac{\pi}{4}\right)x$$
I assumed $x=\frac{1}{y}$ we get
$$L=\lim_{y \to 0}\frac{\left( \tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4... | Well, we can start as you, by setting $y=\frac{1}{x}$. Now, our limits transforms to:
$$L=\lim_{y\to0}\frac{\tan^{-1}\left(\frac{1+y}{1+4y}\right)-\frac{\pi}{4}}{y}$$
Now, let $f:\mathbb{R}\to\mathbb{R}$ with
$$f(x)=\tan\left(\frac{1+x}{1+4x}\right)$$
Note that $f(0)=\tan^{-1}(1)=\frac{\pi}{4}$. So, we have:
$$L=\lim_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2365914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 3
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Unable to reach the desired answer in trigonometry. The question is:
If $\sin x + \sin y = \sqrt3 (\cos y - \cos x)$
show that $\sin 3x + \sin 3y= 0 $
This is what I have tried:
*
*Squaring of the first equation (Result: Failure)
*Tried to use the $\sin(3x)$ identity but got stuck in the middle steps because I co... | I was trying to find out how the condition was conceived.
$$\sin3x+\sin3y=0\implies\sin3x=\sin(-3y)$$
$\implies3x=180^\circ n+(-1)^n(-3y)$ where $n$ is any integer
$\iff x=60^\circ n+(-1)^{n+1}y$
If $n$ is even $=2m$(say), $x=120^\circ m-y$
$$\implies x+y\equiv\begin{cases}0 &\mbox{if }3\mid m\\120^\circ& \mbox{if } n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2368879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Symbolic Notation for $\theta "=" \arcsin(-.5)$? I'm teaching PreCalculus and the following issue has always bugged me.
Problem: Solve $\sin\theta = -.5$ for $0 \le \theta \le 2\pi$.
Solution:
\begin{align*}
\sin\theta &= -.5\\
\theta &= \arcsin(-.5) = -\frac\pi6
\end{align*}
But to get this into our desired domain, ou... | The way I always explained it to my students was "First, you find a solution, then you find the solution."
Let $\theta_0 = \arcsin(-\frac 12)$.
The range of $\arcsin$ is $-\frac{\pi}{2} \le \theta_0 \le \frac{\pi}{2}$. Since, on the unit circle, $\sin \theta = y=-\frac 12$, we see quickly that
$\theta_0 = -\frac{\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2370728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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$x^5 + y^2 = z^3$ While waiting for my döner at lunch the other day, I noticed my order number was $343 = 7^3$ (surely not the total for that day), which reminded me of how $3^5 = 243$, so that $$7^3 = 3^5 + 100 = 3^5 + 10^2.$$
Naturally, I started wondering about nontrivial integer solutions to $$x^5 + y^2 = z^3 \tag{... | There is a beautiful connection between $a^5+b^3=c^2$ and the icosahedron. Consider the unscaled icosahedral equation,
$$\color{blue}{12^3u v(u^2 + 11 u v - v^2)^5}+(u^4 - 228 u^3 v + 494 u^2 v^2 + 228 u v^3 + v^4)^3 = (u^6 + 522 u^5 v - 10005 u^4 v^2 - 10005 u^2 v^4 - 522 u v^5 + v^6)^2\tag1$$
By scaling $u=12x^5$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2373028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Minimum of the given expression
For all real numbers $a$ and $b$ find the minimum of the following expression.
$$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2$$
I tried expressing the entire expression in terms of a single function of $a$ and $b$. For example, if the entire expression reduces to $(a-2b)^2+(a-2b)+5$ then its mini... | Let $a=\frac{17}{15}$ and $b=\frac{4}{5}$.
Hence, we get a value $\frac{2}{15}$.
Thus, it remains to prove that
$$(a-b)^2 + (2-a-b)^2 + (2a-3b)^2\geq\frac{2}{15}$$ or
$$10(3a-3b-1)^2+3(5b-4)^2\geq0$$
Done!
I got my solution by the following way.
We need to find a maximal $k$ for which the following inequality is true ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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how to prove $\frac{b^2}{b_1^2}=\frac{ac}{a_{1c_1}}$? If the Ratio of the roots of $ax^2+bx+c=0$ be equal to the ratio of the roots of $a_1x^2+b_1x+c_1=0$, then how one prove that $\frac{b^2}{b^2_1}=\frac{ac}{a_1 c_1}$?
| Hint :
let $\alpha$ and $\beta$ be the roots of $ax^2+bx+c=0$ & let $\gamma$ and $\delta$ be the roots of $a_1 x^2+b_1 x+c_1 =0$.
The ratio of their roots are equal if
\begin{eqnarray*}
\frac{\alpha}{\beta} = \frac{\gamma}{\delta}.
\end{eqnarray*}
Further hint : $\color{red}{\alpha+\beta=-\frac{b}{a}}$ & $\alpha \beta... | {
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"url": "https://math.stackexchange.com/questions/2375736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that the equation $x^3+y^3+z^3-(x^2z+y^2x+z^2y)=2$ has no solution in natural numbers I asked myself which primes $p$ can be written as $p=x^3+y^3+z^3-(x^2z+y^2x+z^2y)$ with $x,y,z \in \mathbb{N}$.
But for $p \neq 2$ we have the solution $x=y=\frac{p-1}{2}$ and $z=\frac{p+1}{2}$. So the only prime for which I ca... | EDIT: You're right stackExchangeUser; my proof doesn't work. With a similar tack, we can still salvage this:
\begin{align*}
&x^3 + y^3 + z^3 - (x^2 z + y^2 x + z^2 y) \\
= ~ &(x + y + z)^3 - 4(x^2 z + y^2 x + z^2 y) - 3(x^2y + y^2 z + z^2 x) - 6xyz
\end{align*}
So, we are solving,
$$(x + y + z)^3 = 2 + 4(x^2 z + y^2 x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2375828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Finding the minimum value of $\cot^2A + \cot^2B+ \cot^2C$ where $A$, $B$ and $C$ are angles of a triangle. The question is:
If $A+B+C= \pi$, where $A>0$, $B>0$, $C>0$, then find the minimum value of $$\cot^2A+\cot^2B +\cot^2C.$$
My solution:
$(\cot A + \cot B + \cot C)^2\ge0$ // square of a real number
$\implies ... | it is equivalent to
$$\frac{1}{\sin(A)^2}+\frac{1}{\sin(B)^2}+\frac{1}{\sin(C)^2}\geq 4$$
with $$\sin(A)=\frac{a}{2R}$$ etc and $$S=\sqrt{s(s-a)(s-b)(s-c)}$$ and $$S=\frac{abc}{4R}$$ we get
$$b^2c^2+c^2a^2+a^2b^2-(-a+b+c)(a-b+c)(a+b-c)(a+b+c)\geq 0$$
and this is equivalent to
$$a^4+b^4+c^4\geq a^2b^2+b^2c^2+c^2a^2$$
wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Recurrence relation $a_n = 11a_{n-1} - 40a_{n-2} + 48a_{n-3} + n2^n$ I didn't do a lot of maths in my career, and we asked me to solve the following recurrence relation:
$$a_{n} = 11a_{n-1} - 40a_{n-2} + 48a_{n-3} + n2^n$$
with
$a_0 = 2$, $a_1 = 3$ and $a_2 = 1$
What is the procedure to solve such relation? So far, I ... | Just to offer another approach, generating functions can be used as well:
$\begin{align}
G(x) &= \sum_{n=0}^{\infty} a_n x^n \\
G(x) &= 2x^0 + 3x^1 + x^2 + \sum_{n=3}^{\infty}(11a_{n-1} - 40a_{n-2} + 48a_{n-3} + n2^n)x^n \\
G(x) &= 2 + 3x + x^2 + 11\sum_{n=3}^{\infty}a_{n-1}x^n - 40\sum_{n=3}^{\infty}a_{n-2}x^n + 48\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Express a formula in terms of trigonometric expressions I am studying Kerr Black holes using Hobson's General relativity an introduction for physicists book.
In order to find circular radius for photons, two conditions need to be satisfied:
$$r_c=3\mu\frac{b-a}{b+a}$$ and
$$(b+a)^3=27\mu^2(b-a)$$
According to the book... | \begin{align}
r_c&=3\mu\frac{b-a}{b+a} \tag{1}\label{1}
\\
(b+a)^3&=27\mu^2(b-a) \tag{2}\label{2}
\end{align}
To get the expression for b from \eqref{1},
\begin{align}
b-a&=\frac{r_c}{3\mu}(b+a)
,
\end{align}
combined with \eqref{2},
\begin{align}
(b+a)^3&=27\mu^2\frac{r_c}{3\mu}(b+a)
,\\
(b+a)^2&=9\mu{r_c}
,\\
b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Need help integrating $\int e^x(\frac{x+2}{x+4})^2 dx $ I have simplified the problem a bit using integration by parts, with $u = e^x(x+2)^2$ and $v = 1/(x+4)^2$ but I'm then stuck with how to integrate this:
$$\int\frac{e^x(x^2+4x+8)}{x+4}dx. $$
I've considered substituting $t = e^x$, but this doesn't seem to make the... | \begin{align*}\int e^x\left(\frac{x+2}{x+4}\right)^2\,\mathrm dx&=\int\frac1{(x+4)^2}e^x(x+2)^2\,\mathrm dx\\&=-\frac{e^x(x+2)^2}{x+4}+\int e^x(x+2)\,\mathrm dx\\&=-\frac{e^x(x+2)^2}{x+4}+e^x(x+2)-\int e^x\,\mathrm dx\\&=-\frac{e^x(x+2)^2}{x+4}+e^x(x+1)\\&=\frac{xe^x}{x+4}.\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2376729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the maximum positive integer that divides $n^7+n^6-n^5-n^4$
Find the maximum positive integer that divides all the numbers of the form $$n^7+n^6-n^5-n^4 \ \ \ \mbox{with} \ n\in\mathbb{N}-\left\{0\right\}.$$
My attempt
I can factor the polynomial
$n^7+n^6-n^5-n^4=n^4(n-1)(n+1)^2\ \ \ \forall n\in\mathbb{N}.$
If... | Without using the link I gave you (which is overkilling the problem, by the way), you can see that $2^4$ divides $$f(n):=n^7+n^6-n^5-n^4=(n-1)\,n^4\,(n+1)^2$$
by considering the case $n$ is odd and the case $n$ is even. Since $n-1$, $n$, and $n+1$ are consecutive integers, $3$ must divide $f(n)$. That is, $2^4\cdot 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Definite integral for a 4 degree function
The integral is:
$$\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$$
I used an approach that involved substitution of x by $a\tan\theta$. No luck :\ . Help?
| $\displaystyle\int_0^a \frac{x^4}{(x^2+a^2)^4}dx$
Where do we get with the substitution you have suggested?
$x = a\tan\theta\\
dx = a\sec^2\theta\\
\displaystyle\int_0^{\frac \pi 4} \frac{(a^4\tan^4\theta)(a\sec^2\theta)}{(a^2\tan^2\theta+a^2)^4}d\theta\\
$
Looks promising:
Keep simplifying
$\displaystyle\int_0^{\frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2377946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Prove this inequality $2(a+b+c)\ge\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}$ For $a,b,c$ are positive real numbers satisfy $a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$. Prove that $$2\left(a+b+c\right)\ge\sqrt{a^2+3}+\sqrt{b^2+3}+\sqrt{c^2+3}$$
We have:$a+b+c=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge \frac{9}{a+b+c}\Leftrigh... | Since the function $f(t) := \sqrt{1+t}$ is concave in $[-1, +\infty)$, we have that
$$
f(t) \leq f(3) + f'(3) (t-3)
\qquad \forall t\geq -1,
$$
i.e.
$$
f(t) \leq 2 + \frac{1}{4}(t-3) = \frac{5}{4} + \frac{1}{4} t
\qquad \forall t\geq -1.
$$
Using this inequality we have that
$$
\sqrt{a^2+3} = a \sqrt{1+ 3/a^2} \leq
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2378941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Inequality $\frac{x_1^2}{x_1^2+x_2x_3}+\frac{x_2^2}{x_2^2+x_3x_4}+\cdots+\frac{x_{n-1}^2}{x_{n-1}^2+x_nx_1}+\frac{x_n^2}{x_n^2+x_1x_2}\le n-1$
Show that for all $n\ge 2$
$$\frac{x_1^2}{x_1^2+x_2x_3}+\frac{x_2^2}{x_2^2+x_3x_4}+\cdots+\frac{x_{n-1}^2}{x_{n-1}^2+x_nx_1}+\frac{x_n^2}{x_n^2+x_1x_2}\le n-1$$
where $x_i$... | Let $\frac{x_2x_3}{x_1^2}=\frac{a_1}{a_2}$,... and similar, where $a_i>0$ and $a_{n+1}=a_1$.
Thus, we need to prove that:
$$\sum_{i=1}^n\frac{1}{1+\frac{a_i}{a_{i+1}}}\leq n-1$$ or
$$\sum_{i=1}^n\left(\frac{1}{1+\frac{a_i}{a_{i+1}}}-1\right)\leq-1$$ or
$$\sum_{i=1}^n\frac{a_i}{a_i+a_{i+1}}\geq1,$$
which is true because... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $ x \in \left(0,\frac{\pi}{2}\right)$. Then value of $x$ in $ \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
If $\displaystyle x \in \left(0,\frac{\pi}{2}\right)$ then find a value of $x$ in $\displaystyle \frac{3}{\sqrt{2}}\sec x-\sqrt{2}\csc x = 1$
$\bf{Attempt:}$ From $$\frac{3}{\sqrt{2}\cos x}-\frac{\sqrt{2}}{\s... | I think it's better to make the following.
Let $x=\frac{\pi}{4}+t$, where $t\in\left(-\frac{\pi}{4},\frac{\pi}{4}\right)$.
Hence, we need to solve that
$$3\sin{x}-2\cos{x}=\sqrt2\sin{x}\cos{x}$$ or
$$3(\sin{t}+\cos{t})-2(\cos{t}-\sin{t})=\cos^2t-\sin^2t$$ or
$$\sin{t}(5+\sin{t})+(1-\cos{t})\cos{t}=0$$ or
$$\sin\frac{t}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2379567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
If $0^\circ\leqslant x<360^\circ$, what is the maximum number of solutions to the equation $\sin x = a$ where a is a real number? I tried solving the question, but I kept getting $5$ solutions. My book only has $4$ choices: $0$, $1$, $2$, or $3$ solutions. My solutions were $0^\circ$, $90^\circ$, $150^\circ$, $180^\cir... | Clearly, we need $-1\le a\le1$ for at least one real solution
If $\sin x_1=\sin x_2$
Using Prosthaphaeresis Formulas, $$\sin x_1-\sin x_2=2\sin\dfrac{x_1-x_2}2\cos\dfrac{x_1+x_2}2.$$
If $\sin\dfrac{x_1-x_2}2=0\implies\dfrac{x_1-x_2}2=m180^\circ\iff x_1\equiv x_2\pmod{360^\circ}.$
If $\cos\dfrac{x_1+x_2}2=0\implies\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Showing the Fourier sine series converges
The Fourier sine series for $f(x) = x$, $-2 < x < 2$ is
$$f(x) = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}\sin \frac{n\pi x}{2}$$
For each $x$ in the interval to what does the Fourier since series for $f(x)$ converge, can we prove pointwise convergence, converg... | *
*Your proof of convergence in $L^2$ is correct.
*Let us prove the pointwise convergenge in the open interval $(-2,2)$.
Consider, for $-2 < x < 2$,
$$ F(x) = \frac{4}{\pi}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}e^{ \frac{n\pi x}{2}i}$$
If $x\in (-2,2)$, then we have that $\sum_{n=1}^{\infty}(-1)^{n+1}e^{ \frac{n\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Element of certain order in special linear space. What would be the conditions (if any) on the trace of an element in $SL(2,p)$ in order for it to have order 5 ? (assuming $p= \pm1 mod 10$)
For example, any traceless element in $SL(2,p)$ has order 4 (straightforward proof).
Any suggestion or comment is tremendously va... | Suppose $A$ is an element of order $5$ in $\DeclareMathOperator{\SL}{SL} \SL_2(\mathbb{F}_p)$ and let $m(x)$ be its minimal polynomial. If $m$ has degree $1$, then $A$ is a scalar matrix hence must be of the form
$$
\begin{pmatrix}
c & 0\\
0 & c
\end{pmatrix}
$$
for some $c$. But then $c^2 = \det(A) = 1$, so $A$ has ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2380745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Pythagorean Triple: $\text{Area} = 2 \cdot \text{perimeter}$ Find the unique primitive Pythagorean triple whose area is equal to twice the perimeter.
So far I set the sides of the triangle to be $a, b,~\text{and}~c$ where $a$ and $b$ are the legs of the triangle and c is the hypotenuse.
I came up with 2 equations whic... | Rewrite the first equation as $c = \frac{ab}{4} - a - b$. Square it to get $$c^2 = a^2 + b^2 + \frac{a^2b^2}{16} - \frac{a^2b}{2} - \frac{ab^2}{2} + 2ab$$
Now using the other equation, we see that
$$\frac{a^2b^2}{16} - \frac{a^2b}{2} - \frac{ab^2}{2} + 2ab = 0$$
Since $a,b > 0$ divide by $ab$ and multiply by $16$ to ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Formula for consecutive residue of primitive modulo n. \begin{align*}
3^0 \equiv 1\mod 7\\
3^1 \equiv 3\mod 7\\
3^2 \equiv 2\mod 7\\
3^3 \equiv 6\mod 7\\
3^4 \equiv 4\mod 7\\
3^5 \equiv 5\mod 7\\
3^6 \equiv 1\mod 7\\
3^7 \equiv 3\mod 7\\
\end{align*}
Now just focusing on 1, 3, 2, 6, 4, 5, 1....
How to devise a formula ... | Fermat says $3^{6k+r}$mod$7=3^r$mod$7,0\le r \le 5$. Up to $r=5$ the calculation is very simple, no?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2381921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find minimum value that the trigonometric expression may take For $x\in\left(0, \frac{\pi}{2}\right)$ find a minimal value, which the expression
$$\sec x+\csc x+\sec^{2}x+\csc^{2}x$$
can take.
My attempt:
I followed the trigonometrical approach and obtained
$$\sec x+\csc x+\sec^{2}x+\csc^{2}x=\sqrt{\left(2\csc 2x+1\r... | Let $\sin{x}=a$ and $\cos{x}=b$.
Hence, $a^2+b^2=1$ and by AM-GM we obtain:
$$\sec x+\csc x+\sec^{2}x+\csc^{2}x=$$
$$=\frac{a+b}{ab}+\frac{1}{a^2b^2}\geq\frac{2\sqrt2}{\sqrt{a^2+b^2}}+\frac{4}{(a^2+b^2)^2}=4+2\sqrt2.$$
The equality occurs for $a=b=\frac{1}{\sqrt2}$, which says that we got a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2382235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Finding the value of a given trigonometric series.
Find the value of $\tan^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+\tan^2\dfrac{4\pi}{16}+\tan^2\dfrac{5\pi}{16}+\tan^2\dfrac{6\pi}{16}+\tan^2\dfrac{7\pi}{16}.$
My attempts:
I converted the given series to a simpler form:
$\tan^2\dfrac{\pi}{16}+... | $$\tan^2\dfrac{\pi}{16}+\tan^2\dfrac{2\pi}{16}+\tan^2\dfrac{3\pi}{16}+\tan^2\dfrac{4\pi}{16}+\tan^2\dfrac{5\pi}{16}+\tan^2\dfrac{6\pi}{16}+\tan^2\dfrac{7\pi}{16}=$$
$$=\tan^2\dfrac{\pi}{16}+\cot^2\dfrac{\pi}{16}+\tan^2\dfrac{3\pi}{16}+\cot^2\dfrac{3\pi}{16}+\tan^2\dfrac{\pi}{8}+\cot^2\dfrac{\pi}{8}+1=$$
$$=\left(\tan\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$?
Why should we have $\sin^2(x) = \frac{1-\cos(2x)}{2}$ knowing that $\sin^2(x) = 1 - \cos^2(x)$?
Logically, can you not subtract $\cos^2(x)$ to the other side from this Pythagorean identity $\sin^2(x)+\cos^2(x)=1?$
When I ... | Both formulas are true, however, both are useful in different contexts (applications).
*
*You use $\sin^2(x) = \frac{1-\cos(2x)}{2}$ for integrating $\sin^2(x)$.
*You use $\sin^2(x) = 1 - \cos^2(x)$, for example, when solving $\sin^2(x) = 2\cos(x)$.
Note that it is just in some way more "natural" to write $\sin^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Find $\cos2\theta+\cos2\phi$, given $\sin\theta + \sin\phi = a$ and $\cos\theta+\cos\phi = b$
If
$$\sin\theta + \sin\phi = a \quad\text{and}\quad \cos\theta+\cos\phi = b$$
then find the value of $$\cos2\theta+\cos2\phi$$
My attempt:
Squaring both sides of the second given equation:
$$\cos^2\theta+ \cos^2\phi + 2... | HINT: use that $$\cos(2\theta)+\cos(2\phi)=\cos(\theta-\phi)\cos(\theta+\phi)$$
and $$\sin(\theta)+\sin(\phi)=2\cos\left(\frac{\theta-\phi}{2}\right)\sin\left(\frac{\theta+\phi}{2}\right)$$
and
$$\cos(\theta)+\cos(\phi)=2\cos\left(\frac{\theta-\phi}{2}\right)\cos\left(\frac{\theta+\phi}{2}\right)$$
so another idea, and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2383791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Polynomial $ax^2 + (b+c)x + (d+e)$ Let $a, b, c, d$ be real number such that polynomial $ax^2 + (b+c)x + (d+e)$ has real roots greater than $1$. Prove that polynomial $ax^4+bx^3+cx^2+dx+e$ has at least one real root.
Is my work correct ?
Let $r$ be real root of $ax^2+(c+b)x+(e+d)$, so $ar^2+cr+e=(br+d)(-1)$.
Let $P(x)... | Assume the roots are $r_1,r_2$. Then:
$$a(x-r_1)^2(x-r_2)^2=ax^2+(-ar_1-ar_2)x+ar_1r_2=0.$$
Hence the second equation:
$$f(x)=ax^4-ar_1x^3-ar_2x^2+(ar_1r_2-e)x+e=0.$$
Note:
$$f(r_1)=-er_1+e$$
$$f(0)=e$$
Now IVT is applicable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Partial fractions and linear vs quadratic factors I was watching some videos on partial fraction decompistion and I got confused on one of the examples:
Say for example you have $$\frac{x+4}{x^2(x^2 +3)^2}.$$
The partial fraction equation of this is apparently:
$$\frac{A}{x} + \frac{B}{x^2} + \frac{Cx+E}{x^2 +3} ... | Hint: $$\frac{ax+b}{x^2} = \frac{a}{x}+\frac{b}{x^2}.$$ Therefore, $$\frac{A}{x}+\frac{ax+b}{x^2}=\frac{A'}{x}+\frac{b}{x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2384930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Sum of series of fractions I am trying to find the $f$ formula that returns the sum of the series created by fractions that have constant nominator and shifting by one denominator.
Here are some examples:
$$f(3) = \frac{3}{1} + \frac{3}{2} + \frac{3}{3} = 5.5$$
or
$$f(4) = \frac{4}{1} + \frac{4}{2} + \frac{4}{3} + \fr... | The numbers
$$1+\frac12+\frac13+\cdots+\frac1n$$
are called the harmonic numbers and often denoted $H_n$. There is
no simple closed formula for $H_n$, but $H_n$ is approximately $\ln n+\gamma$ for large $n$, where $\gamma$ is Euler's constant.
You are considering $nH_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385514",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the real and imaginary part of z let $z=$ $$ \left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta} \right)^n$$
Rationalizing the denominator:
$$\frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta - i\cos\theta}\cdot\left( \frac{1 + \sin\theta + i\cos\theta}{1 + \sin\theta + i\cos\theta}\right) =... | HINT: Express the fraction as $r e^{i\theta}$ and compute $r^n e^{i n\theta}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 0
} |
Is there a geometric method to show $\sin x \sim x- \frac{x^3}{6}$ I found a geometric method to show $$\text{when}\; x\to 0 \space , \space \cos x\sim 1-\frac{x^2}{2}$$ like below :
Suppose $R_{circle}=1 \to \overline{AB} =2$ in $\Delta AMB$ we have $$\overline{AM}^2=\overline{AB}\cdot\overline{AH} \tag{*}$$and
$$\o... | Not a complete answer, I am afraid. Approximating the arc $AB$ as a line segment leads to the correct approximation of $\cos(x) \sim 1 - \frac{x^2}{2}$ but leads to the inaccurate result for the corresponding sine as $ \sin(x) \sim x - \frac{x^3}{8}$. Nevertheless, I post my approach, since it has not been covered in t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2389537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 6,
"answer_id": 4
} |
Finding a solution to a system Let $i$ be of the form $i=2^a3^b5^c$, where $a,b,c\ge 0$ are integers.
Consider numbers $x_{i,4},x_{i,6}$ where $x_{i,6}$ is defined when $2$ or $3$ divides $i$, $x_{i,4}$ is defined only when $2$ divides $i$.
The constraints are
*
*If $2$ divides $i$, then $$x_{i, 4} + x_{i,6} \le \... | Actually this system have no solution, first $x_2,y_2,x_3,y_3,x_5,y_5 \in \mathbb{R} \geq 0$.
And $x_2+y_2=1 = \sum \limits_{k=1}^{\infty} \frac{1}{2^k}$
And $x_3+y_3 = \frac{1}{2} = \sum \limits_{k=1}^{\infty} \frac{1}{3^k}$
And $x_5 +y_5 = \frac{1}{4} = \sum \limits_{k=1}^{\infty} \frac{1}{5^k}$.
So $\sum x_{i,4} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2390917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to change the appearence of the correct answer of $\cos55^\circ\cdot\cos65^\circ\cdot\cos175^\circ$ I represented the problem in the following view and solved it: $$\begin{align}-\sin35^\circ\cdot\sin25^\circ\cdot\sin85^\circ\cdot\sin45^\circ&=A\cdot\sin45^\circ\\ -\frac{1}{2}(\cos20^\circ-\cos70^\circ)\cdot\frac{1... | As $\cos175^\circ=\cos(180^\circ-5^\circ)=-\cos5^\circ,$
Like prove that : cosx.cos(x-60).cos(x+60)= (1/4)cos3x
$$4\cos(60^\circ-5^\circ)\cos5^\circ\cos(60^\circ+5^\circ)=\cos(3\cdot5^\circ)$$
Now use $15=60-45$ or $=45-30$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2391564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $\frac {a_{n+1}}{a_n} \ge 1 -\frac {1}{n} -\frac {1}{n^2}$ then $\sum\limits_na_n$ diverges
Let $(a_n)_{n \ge 1}$ be a sequence of positive real numbers such that, for every $n\ge1$,
$$\frac {a_{n+1}}{a_n} \ge 1 -\frac {1}{n} -\frac {1}{n^2} \tag 2$$ Prove that $x_n=a_1 + a_2 + .. + a_n$ diverges.
It is clear t... | It's easy to show that, for every $n\ge3$,
$$ 1 -\frac {1}{n} -\frac {1}{n^2}
\ge \frac {n-2}{n-1}$$ It follows that, for every $n\ge3$,
$$\frac{a_{n+1}}{a_n}\ge \frac {n-2}{n-1}$$
Thus,
$$\frac {a_4}{a_3} \ge \frac 1 2\qquad
\frac {a_5}{a_4} \ge \frac 2 3\qquad
\ldots\qquad
\frac {a_{n-1}}{a_{n-2}} \ge \frac {n-4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2392220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Find $x\in \Bbb Z$ such that $x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Tried (without sucess) two different approaches: (a) finding $x^3$ by raising the right expression to power 3, but was not able to find something useful in the result t... | $$x=\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}\\x^3=2+\sqrt5+2-\sqrt5+3\sqrt[3]{2+\sqrt{5}}\cdot\sqrt[3]{2-\sqrt{5}}(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}})\\x^3=4+3\cdot(-1)\cdot(x)$$so $$x^3+3x-4=0 \\(x-1)(x^2+x+4)\to\\ x=1,x^2+x+4=0 ,\Delta <0\\x=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2395060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 7,
"answer_id": 0
} |
Taylor expansion of $\cos^2(\frac{iz}{2})$
Expand $\cos^2(\frac{iz}{2})$ around $a=0$
We know that $$\cos t=\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{{2n!}}$$
So $$\cos^2t=[\sum_{n=0}^{\infty}(-1)^n\frac{t^{2n}}{{2n!}}]^2=\sum_{n=0}^{\infty}(-1)^{2n}\frac{t^{4n}}{{4n^2!}}$$
We have $t=\frac{iz}{2}$
$$\sum_{n=0}^{\infty}... | $$ \cos^2(\frac{iz}{2}) = \frac{1}{2}(1+\cos(2*\frac{iz}{2})) = \frac{1}{2}(1+\cos(iz)) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} (-1)^n\frac{(iz)^{2n}}{(2n)!}) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} (-1)^n (i)^{2n}\frac{(z)^{2n}}{(2n)!}) = \frac{1}{2}(1+\sum_{n=0}^{+\infty} (-1)^{2n}\frac{(z)^{2n}}{(2n)!}) = \frac{1}{2}(1+\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
If $A$ is diagonalizable, find $\alpha$ and $\beta$
Let $A$ be a $5 \times 5$ matrix whose characteristic polynomial is given by
$$p_A(\lambda)=(λ + 2)^2 (λ − 2)^3$$
If $A$ is diagonalizable, find $\alpha$ and $\beta$ such that
$$A^{-1} = \alpha A + \beta I$$
I am unable to find the inverse of $5\times 5$ m... | The fact that A is diagonalizable means that there exist an invertible matrix, P, such that $PAP^{1}= \begin{bmatrix}2 & 0 & 0 & 0 & 0 \\ 0 & 2 & 0 & 0 & 0 \\ 0 & 0 & -2 & 0 & 0 \\ 0 & 0 & 0 & -2 & 0 \\ 0 & 0 & 0 & 0 & -2 \end{bmatrix}$. So $(PAP^{1})^{-1}= PA^{-1}P^{-1}= \begin{bmatrix}\frac{1}{2} & 0 & 0 & 0 & 0 \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int_{-\infty}^{\infty}\frac{1}{(x^2+4)^5}dx$ $$\int_{-\infty}^{\infty}\frac{1}{(x^2+4)^5}dx$$
I am trying to use residue.
We first need to find the singularities, $x^2+4=0\iff x=\pm 2i$
Just $2i$ is in the positive part of $i$ so we take the limit
$lim_{z\to 2i}\frac{1}{(z^2+4)^5}$ but the limit is $0$
| For any $a>0$ we have
$$ \int_{-\infty}^{+\infty}\frac{dx}{x^2+a}=\frac{\pi}{\sqrt{a}} $$
and by applying $\frac{d^4}{da^4}$ to both sides we get:
$$ 24\int_{-\infty}^{+\infty}\frac{dx}{(x^2+a)^5}=\frac{105 \pi}{16 a^4\sqrt{a}} $$
so by rearranging and evaluating at $a=4$ we get:
$$ \int_{-\infty}^{+\infty}\frac{dx}{(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2396943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Simplify $\sqrt{6-\sqrt{20}}$ My first try was to set the whole expression equal to $a$ and square both sides. $$\sqrt{6-\sqrt{20}}=a \Longleftrightarrow a^2=6-\sqrt{20}=6-\sqrt{4\cdot5}=6-2\sqrt{5}.$$
Multiplying by conjugate I get $$a^2=\frac{(6-2\sqrt{5})(6+2\sqrt{5})}{6+2\sqrt{5}}=\frac{16}{2+\sqrt{5}}.$$
But I sti... | In this particular problem, you can pretty much guess the answer.
$$\sqrt{6-\sqrt{20}}=\sqrt{6-2\sqrt{5}}$$
Now, suppose that the $-2\sqrt{5}$ was the middle term of a perfect square trinomial, where $x = \sqrt{5}$. In other words, that middle term is $-2x$.
What would the first and last term look like? Obviously it wo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2400336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
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.