Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solving an exponential with three different bases $$2^x+4^x=8^x$$ Solve for $x$. I reduced the bases to $2$ and try to use logarithms but could not get passed the logarithm of an expression. When typing into online calculators they say there is no solution but graphing shows an answer.
| Set $y = 2^x$, then you get
\begin{align}
y+y^2= y^3 \ \ \implies \ \ \ y(1+y-y^2) = 0 \ \ \implies \ \ y=0 \ \ \text{ or } \ \ y = \frac{1\pm\sqrt{5}}{2}.
\end{align}
Hence it follows that
\begin{align}
y = 2^x=\frac{1+\sqrt{5}}{2} \ \ \implies \ \ x = \log_2\left(\frac{1+\sqrt{5}}{2}\right).
\end{align}
Note that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Find the natural solutions of $a^3-b^3=999$ I want to find the natural solutions of $a^3-b^3=999$.
I got $a^3-b^3=(a-b)\cdot(a^2+ab+b^2)$, so if we consider the equation in $\mathbb{Z}/3\mathbb{Z}$ we get
$$(a-b)\cdot(a^2+ab+b^2) \equiv0 \text{ mod }3$$
and because $\mathbb{Z}/3\mathbb{Z}$ is a domain, we get
$$a\equ... | By Fermat Little Theorem
$$x^3 \equiv x \pmod{3}$$
Therefore
$$0 \equiv a^3-b^3 \equiv a-b \pmod{3}$$
This shows that $a-b=3k$.
Then
$$3^3 \cdot 37 =a^3-b^3=(a-b)(a^2+ab+b^2)=(a-b)((a-b)^3+3ab)=3k(9k^2+3ab) \Rightarrow \\
3 \cdot 37=k(3k^2+ab)$$
Since $k <3k^2+ab$ the only posibilities are
$$k=1 \\
3k^2+ab=3 \cdot 37... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
What is $\lim_{x\to 2} \frac{\sqrt{x+2}-2}{x-2}$? I tried multiplying by the conjugate which gave:
$$\frac{x-2}{(x-2)\sqrt{x+2}+2x-4}$$
But i'm still gettting $\frac{0}{0}$. According to my textbook the answer should be $\frac{1}{4}$, but how do I get there?
| Let $x=2+h$ and let $h \rightarrow 0$ hence
$$
\frac{\sqrt{4+h}-2}{h}=\frac{2\sqrt{1+\frac{h}{4}}-2}{h}=\frac{2\left(1+\frac{h}{8}-1+o\left(h\right)\right)}{h} \underset{h \rightarrow 0}{\rightarrow}\frac{1}{4}
$$
Hence
$$
\frac{\sqrt{x+2}-2}{x-2} \underset{x \rightarrow 2}{\rightarrow}\frac{1}{4}]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 4
} |
Number of solution is twice $(x,y)$ Problem: Count the number of $2 \times 2$ matrices $A$ with $A^TA=-I$ in $Z_p$ for $p>2$.
Answer: if $p$ is an odd prime, the number of such matrices $A$ is twice the number of solutions $(x,y)$ to the congruence $x^2+y^2 \equiv -1 \pmod p$.
What's the reason behind "twice"?
| Let
\begin{eqnarray*}
A=
\begin{bmatrix}
a &b \\c &d \\
\end{bmatrix} .
\end{eqnarray*}
Then we require
\begin{eqnarray*}
\begin{bmatrix}
a &c \\b &d \\
\end{bmatrix}
\begin{bmatrix}
a &b \\c &d \\
\end{bmatrix} =
\begin{bmatrix}
-1 &0 \\0 &-1 \\
\end{bmatrix} .
\end{eqnarray*}
So
\begin{eqnarray*}
a^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Stretching an ellipse along major or minor axis Consider the ellipse given by:
$$
Ax^2 + Bxy + Cy^2 + Dx + Ey + F =0.
$$
What is the equation of an ellipse which has major and minor axis equal to $p$ times the major and minor axis length of the above ellipse.
My attempt is as follows:
We can remove rotation, increase a... | Referring to the standard results here, the centre is given by
$$(h,k)=
\left(
\frac{2CD-BE}{B^2-4AC}, \frac{2AE-BD}{B^2-4AC}
\right)$$
and the transformed conics is
$$\frac{A+C \color{red}{\pm} \sqrt{(A-C)^{2}+B^{2}}}{2} X^2+
\frac{A+C \color{red}{\mp} \sqrt{(A-C)^{2}+B^{2}}}{2} Y^2+
\frac
{\det
\begin{pmatrix}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Irrational Computation of a Rational Cubic Root. \begin{align}
& x^3+5x-18=0\\
\text{Let } & x=a+b\\
& (a+b)^3+5(a+b)-18 =0\\
& a^3+b^3+(3ab+5)(a+b)-18=0
\end{align}
Taking $a$ and $b$ such that $3ab+5=0$.
We have, $a^3+b^3-18=0$
\begin{align}
& a^3-\frac{5^3}{(3a)^3}-18=0\\
& a^6-18a^3-\frac{125}{27}=0\\
& a^3=9+\sqrt... | $$x^3+5x-18=x^3-2x^2+2x^2-4x+9x-18=(x-2)(x^2+2x+9),$$
which gives $x=2$ because $$x^2+2x+9=(x+1)^2+8>0.$$
We need to prove that $$\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}+\sqrt[3]{9-\sqrt{81+\frac{125}{27}}}=2$$ or
$$9+\sqrt{81+\frac{125}{27}}+9-\sqrt{81+\frac{125}{27}}-2^3+3\cdot2\sqrt[3]{9+\sqrt{81+\frac{125}{27}}}\cdot\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
check if $\sum_{k=1}^{\infty}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k$ converges How to check if $\sum_{k=1}^{\infty}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k$ converges?.
$\begin{align}
\sum_{k=1}^{\infty}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k &=
\lim_{n \to \infty} \sum_{k=1}^{n}{ \left( \frac{7k-2}{8... | HINT
Note that eventually
$$0<\frac{7k-2}{8k-3 \sqrt{k}}\le c<1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Addition of $2$ Events
Let $X$ and $Y$ be independent, each uniformly distributed on $\{1, 2, ..., n\}$. Find $P(X + Y = k)$ for $2 \le k \le 2n$.
\begin{align}P(X + Y = k) &= \sum_{(x,y)\,:\,x+y=k} P(x, y) \\
&= \sum_{(x,y)\,:\,x+y=k} \frac{1}{n^2} \\
&= (k - 1)\frac{1}{n^2} \\
&= \frac{k-1}{n^2} \end{align}
When $... | We have $x+y = k$ where $x,y \in \{ 1, \ldots, n\}$.
We need not always have $k-1$ pairs of $(x,y)$. In particular, if we let $x=1$, we will need to let $y=k-1$, however, this need not always be possible, as it is possible that $n<k-1$.
If $n\geq k-1$ and $k\geq $, there are $k-1$ pairs, this case has been discussed.
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Definite Integral = $\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$ for $0\le a<1$ I am trying to find an expression for the following Definite Integral for the range $0 \leq a <1$:
$$D = \int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta.$$
which gives (Wolfram Alpha)
$$D= \left[
\frac{\si... | Let $z=e^{i\theta}$ and then
$$ \cos\theta=\frac12(z+z^{-1}),\sin\theta=\frac1{2i}(z-z^{-1}),d\theta=\frac{1}{iz}dz. $$
So
\begin{eqnarray}
&&\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta\\
&=&\int_{|z|=1}\frac{[\frac1{2i}(z-z^{-1})]^2}{(1-\frac a2(z+z^{-1}))^3}\frac{1}{iz}dz\\
&=&-\int_{|z|=1}\frac{2 i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
U-substitution of 2x in trigonometric substitution Find
$$\int^{3\sqrt{3}/2}_0\frac{x^3}{(4x^2+9)^{3/2}} \,\mathrm{d}x.$$
The text says to use substitution of $u = 2x$. How did they get $u = 2x$ and not $u = x^3$?
| The hint:
$$\frac{x^3}{\sqrt{(4x^2+9)^3}}=\frac{x^3+\frac{9}{4}x-\frac{9}{4}x}{\sqrt{(4x^2+9)^3}}=\frac{x}{4\sqrt{4x^2+9}}-\frac{9x}{4\sqrt{(4x^2+9)^3}}.$$
Can you end it now?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$ This is a question from Math Olympiad.
If $\{x,y,z\}\subset\Bbb{R}^+$ and if $$x^2 + xy + y^2 = 3 \\ y^2 + yz + z^2 = 1 \\ x^2 + xz + z^2 = 4$$ find the value of $xy+yz+zx$.
I basically do not know how to approach this question. Pleas... | The given equations $(1),(2),(3)$ can be expressed as:
$$\begin{cases} x^3-y^3=3(x-y) \\ y^3-z^3=y-z \\ z^3-x^3=4(z-x)\end{cases} \stackrel{+}{\Rightarrow} x=3z-2y \ \ (4)$$
Plug $(4)$ in $(1)$ and consider it with $(2)$:
$$\begin{cases} 3z^2-3zy+y^2=1 \\ y^2+yz+z^2=1\end{cases} \Rightarrow 2z^2-4zy=0 \Rightarrow z=2y ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 6,
"answer_id": 5
} |
Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can ... | When you multiply two polynomials, the result is the sum of each pair of terms, one from the "left" and one from the "right", multiplied together.
With $(1+x+x^2+x^3+x^4+x^5)^2$, you'll get $1*1$ once (there's only one way to pick "1" from each side). But you'll get $x$ twice, once from taking the left's $1$ and the ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 1
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Kernel/Image expression
Express the kernel of the 1 × 4 matrix A = \begin{bmatrix}1&2&3&4 \end{bmatrix}
as the
image of a 4 × 3 matrix B.
I understand that the kernel of a matrix is solving the system for A$\vec{x}$ = 0, but I have no idea what this question is asking nor how to do it. Are there any kind souls wh... | $Ax=0\Rightarrow [1 \ 2 \ 3 \ 4][x \ y \ z \ v ]^t=0\Rightarrow x+2y+3z+4v=0\Rightarrow x=-2y-3z-4v$
$\Rightarrow \begin{pmatrix}
x \\
y \\
z \\
v
\end{pmatrix}=\begin{pmatrix}
-2y \\
y \\
0 \\
0
\end{pmatrix}+\begin{pmatrix}
-3z \\
0 \\
z \\
0
\end{pmatrix}+\begin{pmatrix}
-4v \\
0 \\
0 \\
v
\end{pmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2644972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Given that $X,Y$ are independent $N(0,1)$ , show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $N(0,\frac{1}{4})$ It is given that $X,Y \overset{\text{i.i.d.}}{\sim} N(0,1)$
Show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}} \overset{\text{i.i.d.}}{\sim} N(0,\frac{1}... | If you have already proved $\frac{X Y}{\sqrt{X^2 + Y^2}} $ and $\frac{X^2 - Y^2}{\sqrt{X^2 + Y^2}}$ are gaussian, as long as that pair of them is jointly gaussian, then you may use their property: $u,\, v \text{ independent} \iff \operatorname{cov}(u,v) = 0$.
$$ \operatorname{cov} \left( \frac{X Y}{\sqrt{X^2 + Y^2}},\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2647442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Calculus sine proof Suppose that $a, b, c$ are non-zero acute angles such that
$$\frac{\sin(a − b)}{\sin(a + b)} + \frac{\sin(b − c)}{\sin(b + c)} + \frac{\sin(c − a)}{\sin(c + a)}= 0$$
Prove that at least two of $a, b, c$ are equal.
I have no idea how to begin.
| It's just trigonometric manipulation.
$$\frac{\sin(a-b)}{\sin(a+b)}+\frac{\sin(b-c)}{\sin(b+c)}=\frac{\sin(a-b)\sin(b+c)+\sin(b-c)\sin(a+b)}{\sin(a+b)\sin(b+c)}$$
We can now use $\sin x\sin y=\frac{1}{2}(\cos(x-y)-\cos(x+y))$ to gwtite the above expression as
$$\frac{\sin(a-b)\sin(b+c)+\sin(b-c)\sin(a+b)}{\sin(a+b)\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How is $x(2x+7)+3$ equal to $(2x+1)(x+3)$? For some reason, $x(2x+7)+3$ seems like it should be equal to $(2x+7)(x+3)$ instead of $(2x+1)(x+3)$. How does $(2x+1)$ factor out of here?
The original equation was $2x^2+7x+3$.
Proof of this: https://www.desmos.com/calculator/2xpcqznkio
Notice how $x(2x+7)+3$ and $(2x+1)(x+3... | It is impossible for $2x^2+7x+3$ to factor into $(2x+7)(x+3)$, because when you expand the brackets, the constant term is $21$ — not $3$.
A way to factor quadratics without the quadratic formula, where the leading coefficient $a \ne 1$ is like this:
$$
\begin{array} {c|c|c}
2x^2 & ? \\ \hline
? & 3 \\
\end{array}
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Is there any way to prove $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $ by induction since $ \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6} $ we have that for each $n\in \Bbb N$ , $ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{n^2} \leq \frac{\pi^2}{6} $
my problem i... | Take a look at this proof of Daniel J. Velleman.
For $n\geq 1$, we define the positive numbers,
$$I_n:=\int_0^{\pi/2}(\cos(x))^{2n}\,dx\quad\text{and}\quad
J_n:=\int_0^{\pi/2}x^2(\cos(x))^{2n}\,dx.$$
By integration by parts we show that
$$I_n=(2n-1)(I_{n-1}-I_n)\implies I_n=\frac{2n-1}{2n}I_{n-1}$$
and
$$I_n=n(2n-1)J... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
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If $x \cos\theta+y\sin\theta=a$ and $x\sin\theta-y\cos\theta=b$, then $\tan\theta=\frac{bx+ay}{ax-by}$. (Math Olympiad) I tried to solve it but I can’t get the answer. Please help me in proving this trig identity:
If
$$x \cos\theta+y\sin\theta=a$$
$$x\sin\theta-y\cos\theta=b$$
then $$\tan\theta=\frac{bx+ay}{a... | We may write the given equations as:
$$\begin{pmatrix}\cos \theta & - \sin \theta \\ \sin \theta & \cos \theta \end{pmatrix} \begin{pmatrix}x \\ - y \end{pmatrix} = \begin{pmatrix}a \\ b \end{pmatrix} $$
and recognise that the first matrix represents a counterclockwise rotation by $\theta$ about the origin.
We ca... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
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Radical with pattern Let $a =111 \ldots 1$, where the digit $1$ appears $2018$ consecutive times.
Let $b = 222 \ldots 2$, where the digit $2$ appears $1009$ consecutive times.
Without using a calculator, evaluate $\sqrt{a − b}$.
| \begin{align}
a &= \frac{10^{2018}-1}{9} \\
b &= \frac{2(10^{1009}-1)}{9} \\
a-b &= \frac{10^{2018}-2\times10^{1009}+1}{9} \\
&= \frac{(10^{1009}-1)^{2}}{9} \\
\end{align}
Can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Secant and Tangent identity i've been stuck on this question too long
$x = \sec A + \tan A$
show $x + \frac{1}{x} = 2\cdot \sec A$
I've been using $\tan^2 \theta + 1 = \sec^2 \theta$
and $\tan\theta = \frac{\sin \theta}{\cos\theta}$
help would be much appreciated
$x=\sec A +\tan A = \frac{1}{\cos A}+\frac{\sin A}{\cos... | Enough to show:
$$\cos{A}\left( x + \frac{1}{x}\right) = 2$$
$$\cos{A} \cdot x + \frac{\cos{A}}{x} = 1 +\sin{A} + \frac{\cos^2{A}}{1 +\sin{A}} = 1 +\sin{A} + \frac{1-\sin^2{A}}{1 +\sin{A}} = 2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$. Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$.
Attempt at a solution:
$$(\sin y = 3(\cos x \cos y - \sin x \sin y) \sin x) \frac{1}{\cos y}$$
$$\tan y = (3 \co... | Starting from marty cohen's answer
$$\tan (y)=\dfrac{3\sin(2 x)}{5-3\cos(2x)}$$ let $t=\tan(x)$ to get
$$\tan (y)=\frac{3 t}{4 t^2+1}=f(t)$$ So
$$f'(t)=\frac{3-12 t^2}{\left(4 t^2+1\right)^2}\qquad \text{and} \qquad f''(t)=\frac{24 t \left(4 t^2-3\right)}{\left(4 t^2+1\right)^3}$$ The first derivative cancels for $t_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Solve $3\sin^2 x - \cos^2 x - 2 =0$
Find all the angles between $0$ and $360^\circ$ that satisfy $$3\sin^2 x - \cos^2 x - 2 =0$$
My attempt -
$3\sin^2 x - (1-\sin^2x) - 2 =0$
$ 3 \sin^2 x + \sin^2 x = 3 $
$4\sin^2 x = 3 $
$ \sin x= \frac{\sqrt{3}}{2} $
I found that $x= 60,120 $
Why is the answer for this $60,12... | The original definition of $\sin \theta$ of a right-angled triangle is just $\sin \theta =\dfrac {\text {opposite side}}{\text {hypotenuse}}$.
Later, because of its in-adequacy in handling angles larger than $90^0$, the definition is expanded to $\sin \theta = \dfrac {y-ordinate}{\text {radius of the unit circle}} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Minimum value of $h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} $ Minimum value of $h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} $
Find the minimum value of $h(\theta)$
$h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} = 5 \sin (\theta + 53.13) + \sqrt{2} $
Minimum value -
$5\sin (\theta + 53.13) + \sq... | Apply Buniakowski inequality: $(3\sin \theta + (-4)\cos \theta)^2 \le (3^2+(-4)^2)(\sin^2 \theta+\cos^2 \theta) = 25\implies 3\sin \theta - 4\cos \theta \ge -5\implies $ min value $ = -5+\sqrt{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Fourier series for $f(x+4) = f(x)$, $f(x)=1$ for $x\in (0,2), f(x)=-1$ for $x \in (-2,0)$ Given that $f(x+4) = f(x)$ and
$f(x) = -1$ if $-2 < x < 0$, and
$f(x)=1$ if $0 < x < 2$
Find the Fourier series.
I tried it out but I get all $0$ for $a_0$, $a_n$ and $b_n$. Can anyone help me out? I can attach the working if you... | Since $f(x)$ is odd, you should get $a_n =0$ for all $n \ge 0$. The $a_n$ coefficients correspond to the $\cos$ terms which represent the even part of $f(x)$. However, you should not get $b_n = 0$. We see that $$\int_{-2}^2 f(x) \sin\left(\frac{n\pi x}{2}\right)dx = 2 \int^2_0 \sin\left(\frac{n\pi x}{2}\right)dx = 2 \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2657960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the length of the tangent to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ which is intercepted between the axes.
Find the length of the tangent to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ which is intercepted between the axes.
$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}\imp... | Any point on $$x^{2/n}+y^{2/n}=a^{2/n}\ \ \ \ (1)$$ can be chosen as $(a\cos^nt,a\sin^nt)$
Differentiating $(1)$ wrt $x,$ $$\dfrac{dy}{dx}=-\dfrac{x^{(2-n)/n}}{y^{(2-n)/n}}$$
$$\dfrac{dy}{dx}_{\text{ at }(a\cos^nt,a\sin^nt)}=-\dfrac{x^{(2-n)/n}}{y^{(2-n)/n}}=-\dfrac{\cos^{2-n}t}{\sin^{2-n}t}$$
So, the equation of the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that if $3\mid(a^2+b^2)$,then $3\mid a$ and $ 3\mid b$ I am trying to prove this by contradiction. So if $3$ doesn't divide $a$ or $3$ doesn't divide $b$, then the remainder is either $1$ or $2$. I am struggling on what to do next. How do I get the remainder of $a^2$ and $b^2$ for these cases?
Any help is greatly... | Assume $a,b$ are not divisible by $3$. Then:
$$a=3m+1 \ \ \text{or} \ \ a=3m+2; \\
b=3n+1 \ \ \text{or} \ \ b=3n+2.$$
Note:
$$(3k+1)^2=3(3k^2+2k)+1; \ \ (3k+2)^2=3(3k^2+4k+1)+1.$$
Then:
$$a^2+b^2 \equiv 2 \ (\mod 3).$$
Now assume only one of them is divisible by $3$. Then:
$$a^2+b^2 \equiv 1 \ (\mod 3).$$
Contradictio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Why doesn't $A^2=I$ imply $A=\pm I$? Im having trouble believing this T/F Question:
if $\mathrm A^2=I$ then $\mathrm A = \pm \mathrm I$
The answer is False but why?
If the matrix is $\mathrm A = \mathrm I,$ say
\begin{bmatrix}1& 0\\
0 & 1 \end{bmatrix}
then $\mathrm A^2$ is also that. And if $\mathrm A = -\mathrm... | for matrices in $\mathbb{M}_2(\mathbb{R})$
$$
\begin{pmatrix} a & b \\ c & d \end{pmatrix}^2 =
\begin{pmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc \end{pmatrix} =
\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}
$$
requires either $b=c=0, a,d= \pm1$ or $a=-d, bc=1-a^2$
the first type of solution gives $4$ solutions which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
If $A$, $B$ and $C$ are the angles of a triangle then find the value of $\Delta$ I'll state the question from my book below:
If $A$, $B$ and $C$ are the angles of a triangle, then find the determinant value of
$$\Delta = \begin{vmatrix}\sin^2A & \cot A & 1 \\ \sin^2B & \cot B & 1 \\ \sin^2C & \cot C & 1\end{vmatrix}... | $$F=\begin{vmatrix} \sin^2B-\sin^2A & \cot B-\cot A \\ \sin^2C-\sin^2A & \cot C-\cot A \end{vmatrix}$$
$$=\begin{vmatrix} \sin^2B-\sin^2A & -\dfrac{\sin(B-A)}{\sin A\sin B} \\ \sin^2C-\sin^2A & -\dfrac{\sin(C-A)}{\sin C\sin A} \end{vmatrix}$$
$$=\dfrac1{\sin B\sin^2A\sin C}\begin{vmatrix}\sin(B-A)\sin(B+A)\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Prove $f(x) = \frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$ Prove $f(x) =\frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$.
My attempt...
Proof
$$\left| f(x) - f(y) \right| = \left| \frac{1}{1+x^2} - \frac{1}{1+y^2}\right| = \frac{\left|x+y\right|}{\left(1+x^2\right)\left(1+y^2\right)}\left|x-y\ri... | Apply mean value theorem in an interval $[x,y], x,y\in \mathbb R$. So, $\exists \xi \in [x,y]: f(x)-f(y)=f'(\xi)(x-y)$. Also we know that $|f'(\xi)|<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve $x^4 - 8x^3 + 21x^2 - 20x + 5 = 0$ given that the sum of two of its roots is $4$ Here's what I tried:
Let the roots be $a$, $b$, $c$ and $d$, $a+b=4$. Then,
$$a + b + c + d = 8 \Longrightarrow 4 + c+ d = 8 \Longrightarrow a+b = c+d = 4$$
$$(a + b)(c + d) + ab + cd = 21$$
$$ab (c + d) + cd (a + b) = 20 \Longrighta... | make the Ansatz $$x^4-8x^3+21x^2-20x+5=(x^2+Ax+B)(x^2+Cx+D)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Using summation by parts to evaluate an alternating sum I want to evaluate
$$
\sum_{k=0}^n (-1)^k \binom{n}{k} k.
$$
I tried summation by parts, i.e. the formula
$$
\sum_{k=0}^n (f(k+1) - f(k))g(k)
= f(n+1)g(n+1) - f(0)g(0) - \sum_{k=0}^n f(k+1) (g(k+1) - g(k))
$$
with $f(k+1) - f(k) = (-1)^k \binom{n}{k}$ and $g(k... | You need $f(k+1) - f(k) = (-1)^k \binom{n}{k}$. But
$$
(-1)^{k+1}\binom{n-1}{k+1} - (-1)^k \binom{n-1}{k}
= (-1)^{k+1} \left( \binom{n-1}{k+1} + \binom{n-1}{k} \right)
= (-1)^{k+1} \binom{n}{k+1}
$$
So
$$
(-1)^{k}\binom{n-1}{k} - (-1)^{k-1} \binom{n-1}{k-1}
= (-1)^{k} \left( \binom{n-1}{k} + \binom{n-1}{k-1} \ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove $\sum\limits_{n=0}^\infty\frac{5^n(3^{5^{n+1}}-5\cdot3^{5^n}+4)}{(729)^{5^n}-(243)^{5^n}-5\cdot3^{5^n}+1}=\frac12$ This problem is taken from Algerian Olympiad and asks to prove that
$$\sum_{n=0}^{\infty} \dfrac{5^n(3^{5^{n+1}} -5\cdot3^{5^n} + 4)}{(729)^{5^n} - (243)^{5^n}-5\cdot3^{5^n}+1} = \frac 12.$$
Noticing... | I think the "$5 \cdot 3^{5^n}$" in the denominator is a typo and it can be proved that$$
\sum_{n = 0}^\infty \frac{5^n (3^{5^{n + 1}} - 5 \cdot 3^{5^n} + 4)}{3^{6 \cdot 5^n} - 3^{5^{n + 1}} - 3^{5^n} + 1} = \frac{1}{2}.
$$
In fact,$$
\sum_{n = 0}^\infty \frac{5^n (3^{5^{n + 1}} - 5 \cdot 3^{5^n} + 4)}{3^{6 \cdot 5^n} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
Find the eigenvalues of block matrix $C$
Find the eigenvalues of $$C = \begin{bmatrix}\begin{array}{c|c} 0 & A\\ \hline A^T & 0\end{array}\end{bmatrix}$$ where $$A = \begin{bmatrix}
0&0&0&1&1&1&1\\0& 1& 1& 0& 1& 0& 1\\0 &1 &1 &1 &0 &1 &0\\1& 0& 1& 0& 0& 1& 1\\1 &0 &1& 1& 1& 0& 0\\1 &1 &0& 0& 1& 1& 0\\1 &1& 0& 1& 0& 0 ... | Something you have written is wrong but can be corrected. We have$$xI-C=\begin{bmatrix}
\begin{array}{c|c}
xI & -A \\
\hline
-A^T & xI
\end{array}
\end{bmatrix}$$therefore$$|xI-C|=|x^2I-A^TA|$$therefore $$x=\pm|\lambda|$$where $\lambda$ is an eigenvalue of $A$ therefore the eigenvalues of $C$ are the $\pm$ singular val... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Sequence and two limits to calculate Let $(a_n)_{n \ge 0}$ with $a_0>0$ and $a_{n+1}=\frac{a_n}{1+\sqrt{1+a_n^2}}$, for every $n \ge 0$. Find $\lim_{n\to \infty} a_n$ and $\lim_{n\to \infty} (1+a_0^2)(1+a_1^2)...(1+a_n^2)$.
I proved $(a_n)_n$ is convergent and I found $\lim_{n\to \infty} a_n=0$ and for the second limit... | If $a_n=\tan\theta$ for some $\theta\in\left(0,\frac{\pi}{2}\right)$, then $a_{n+1}=\tan\frac{\theta}{2}$ (familiarity with the bisection formulas).
By induction $a_n = \tan\frac{\arctan a_0}{2^n}=\tan\frac{\varphi}{2^n}$ and $1+a_n^2=\sec^2\frac{\varphi}{2^n}$.
By the duplication formula for the sine function
$$ \sin(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Checking if a functional $F(x)$ is a norm in $\mathbb{R}^2$ There is a functional given:
$$F(x) = \sqrt{2x_1^2 + 3x_2^2}$$
Of course $x\in\mathbb{R}^2 \rightarrow x = (x_1, x_2)$
It is easy to check that:
1) $\forall x \in \mathbb{R}^2$ $F(x) \ge 0$
2) $F(x) = 0 \iff x = 0$
3) $ F(\lambda x) = | \lambda | F(x)$
I faile... | You can just use the Cauchy-Schwartz inequality:
\begin{align}
F(x+y)^2 &= 2(x_1+y_1)^2+3(x_2+y_2)^2 \\
&= \underbrace{2x_1^2 + 3x_2^2}_{F(x)^2} + \underbrace{2y_1^2+3y_2^2}_{F(x)^2}+2(2x_1y_1 + 3x_2y_2)\\
&\stackrel{CSB}{\le} F(x)^2 + F(y)^2 + 2\sqrt{2x_1^2+3x_2^2}\sqrt{2y_2^2+3y_2^2}\\
&= F(x)^2 + F(y)^2 + 2F(x)F(y)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Factorize $\det\left[\begin{smallmatrix}yz-x^2&zx-y^2&xy-z^2\\zx-y^2&xy-z^2&yz-x^2\\xy-z^2&yz-x^2&zx-y^2\end{smallmatrix}\right]$ using factor theorem Factorize and prove that
$$
\Delta=\begin{vmatrix}
yz-x^2&zx-y^2&xy-z^2\\
zx-y^2&xy-z^2&yz-x^2\\
xy-z^2&yz-x^2&zx-y^2
\end{vmatrix}\\=\frac{1}{4}(x+y+z)^2\Big[(x-y)^2+(... | By taking elementary operation:$$\Delta=\begin{vmatrix}
yz-x^2&zx-y^2&xy-z^2\\
zx-y^2&xy-z^2&yz-x^2\\
xy-z^2&yz-x^2&zx-y^2
\end{vmatrix}=\begin{vmatrix}
xy+yz+zx-x^2-y^2-z^2&zx-y^2&xy-z^2\\
xy+yz+zx-x^2-y^2-z^2&xy-z^2&yz-x^2\\
xy+yz+zx-x^2-y^2-z^2&yz-x^2&zx-y^2
\end{vmatrix}\\=-(x^2+y^2+z^2-xy-yz-zx)\begin{vmatrix}
1&z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$.
Find all real solutions for $x$ in $2(2^x−1)x^2+(2^{x^2}−2)x=2^{x+1}−2$.
I started off by dividing $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2$ by $2$, and I got $(2^x-1)x^2 + (2^{x^2-1}-1)x = 2^x -1$. I tried dividing by $2^x-1$ on both sides wh... | Using Mathematica
eq = 2 (2^x - 1) x^2 + (2^(x^2) - 2) x == 2^(x + 1) - 2;
sol = Solve[eq, x, Reals]
(* {{x -> -1}, {x -> 0}, {x -> 1}} *)
Verifying the solutions
And @@ (eq /. sol)
(* True *)
Showing the solutions on a plot
Plot[{2 (2^x - 1) x^2 + (2^(x^2) - 2) x, 2^(x + 1) - 2},
{x, -2, 2}, PlotLegends -> Place... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Find an equation of the tangent plane to the given function at the given point. $z = \sin(xy)$ at the point $(x,y) = (2,3\pi /4)$
Work I have done so far:
Partial derivatives:
$$f_x(x,y) = \cos(xy)y \implies f_x\left(2,\frac{3\pi}{4}\right) = \frac{3\pi}{4}\cos\frac{6\pi}{4}\\
f_y(x,y) = \cos(xy)x \implies f_y\left(2,\... | On the left-hand side, you should not have $z-0,$ but $z-f(2,3\pi /4).$ Also, $\cos\frac{6\pi}{4} = \cos\frac{3\pi}{2} = 0,$ so you can make your life a lot easier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding $\frac{\partial^6 f}{\partial x^4 \partial y^2}$
Find $$\frac{\partial^6 f}{\partial x^4 \partial y^2}(0,0)$$ Of the $f(x,y)=\frac{1}{1-x^2y}$
$$\frac{1}{1-x^2y}=\sum_{n=1}^{\infty}(x^2y)^n$$ as $|x^2y|<1$
So we have $\frac{1}{1-x^2y}\approx1+x^2y+x^4y^2$
But how should we continue from here?
| Your method is correct:
$$z=\sum_{n=\color{red}{0}}^{\infty} (x^2y)^n=1+x^2y+x^4y^2+x^6y^3+\cdots$$
$$z_{x^4}=4!y^2+4!x^2y^3+\cdots$$
$$z_{x^4y^2}=4!\cdot 2!+4!x^2\cdot 2!y+\cdots$$
$$z_{x^4y^2}(0,0)=4!2!=48.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2686731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solve the equation $\cos^2x+\cos^22x+\cos^23x=1$
Solve the equation: $$\cos^2x+\cos^22x+\cos^23x=1$$
IMO 1962/4
My first attempt in solving the problem is to simplify the equation and express all terms in terms of $\cos x$. Even without an extensive knowledge about trigonometric identities, the problem is solvable.... | Hint:
Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$0=\cos^2x+\cos^22x+\cos^23x-1$$
$$=\cos(3x+x)\cos(3x-x)+\cos^22x$$
$$=\cos2x(\cos4x+\cos2x)$$
Use
http://mathworld.wolfram.com/ProsthaphaeresisFormulas.html
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Verfication of deduction made using the Cauchy-Schwarz inequality Is the following proof correct?
Show that $$16\leq(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$$ for all positive numbers $a,b,c,d$.
Proof. Let $\mathbf{R}^4$ be the inner product space with the inner product defined as in the ... | it is $$\frac{a}{b}+\frac{b}{a}+\frac{c}{a}+\frac{a}{c}+\frac{a}{d}+\frac{d}{a}+\frac{b}{c}+\frac{c}{b}+\frac{d}{c}+\frac{c}{d}+\frac{b}{d}+\frac{d}{b}+4\geq 12+4=16$$
since $$x+\frac{1}{x}\geq 2$$ for $x>0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2689137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Simplifying calculation for searching for a matrix The exercise it the following: I should find all $$A \in M^{}_{22}(\mathbb{R}) $$ with
$$
A
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
=
\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
A
$$
I solved it like the following but I wond... | Perhaps not more elegant, but you can simplify the problem a little further. Since $$\begin{pmatrix}1&1\\0&1\end{pmatrix}=\mathbf{I}_2+\begin{pmatrix}0&1\\0&0\end{pmatrix}$$ you have, using that $\mathbf{A}\mathbf{I}_2=\mathbf{I}_2\mathbf{A}$,
$$\begin{pmatrix}a&b\\c&d\end{pmatrix}\begin{pmatrix}0&1\\0&0\end{pmatrix}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2697632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the the derivative of $y=\sqrt{1-\sin x}; 0A question I'm attempting is:
Find the derivative of $ y = \sqrt {1 - \sin x} ; 0 < x <\pi/2$.
I did this:
$y = \sqrt {1 - \sin x} = \sqrt {\cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin \frac{x}{2}\cos \frac{x}{2}} = \sqrt { (\sin \frac{x}{2}-\cos \frac{x}{2})^2} = \... | Let apply chain rule
$$\left( \sqrt{f(x)} \right)'=\frac{f'(x)}{2\sqrt{f(x)}}=\frac{-\cos x}{2\sqrt {1 - \sin x}}=\frac12\frac{\sin^2 x/2-\cos^2 x/2}{(\cos x/2-\sin x/2)}=\\=\frac12\frac{ (\sin x/2+\cos x/2)(\sin x/2-\cos x/2) }{(\cos x/2 - \sin x/2)}=-\frac12(\sin x/2+\cos x/2)$$
Note indeed that for $0<x<\pi/2$ since... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
An Example of the Chi-Square Test Here is another problem dealing with the Chi-Square Test. I am hoping that somebody can confirm that I did it correctly or tell me where I went wrong.
Thanks,
Bob
Problem:
An urn contains $6$ red marbles and $3$ white ones. Two marbles are
selected at random from the urn, their colors ... | For $(a)$, yes, your reasoning is correct.
For $(b)$, you set up your hypotheses the wrong way around. With $\chi^2$ test, we set our null hypothesis to be that our results are consistent with the expected results. Furthermore, in statistics, we never talk about accepting the null hypothesis; only whether we reject or ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2699573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How does this transformation happen? $$
ax^2+2hxy+by^2 = a\left(x+ \frac{h}{a}y \right)^2 + \frac{ab-h^2}{a}y^2
$$
How do I go from the equation on the LHS to the equation on the right hand side? My study material mentions "completing the square", which I know but I can't understand how it applies to this particular eq... | For completing the square the coefficient of $\,x^2$ should be one ,so factor out $\,a$ then complete the square as normal with the coefficient of $\,y$ being $\,\frac{2hx}{a}$
ie.
$\,ax^2+2hxy+by^2 $
$= a(x^2+\frac{2hxy}{a}+\frac{by^2}{a}) $
$= a(x^2+\frac{2hxy}{a}+\frac{by^2}{a}+(\frac{hy}{a})^2-(\frac{hy}{a})^2)$
$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2700172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Factoring $x^2-y^2-z^2+2yz+x+y-z$ This is a factorization problem on polynomials. I can't find a way to solve it, neither can a math program called MathWay. Help me, please.
$$x^2-y^2-z^2+2yz+x+y-z$$
My book says that the answer is $(x+y-z)(x-y+z+1)$.
| $$\begin{align}x^2-y^2-z^2+2yz+x+y-z&=x^2-(y-z)^2+x+y-z\\&=(x+(y-z))(x-(y-z))+x+y-z\\&=(x+y-z)(x-y+z+1)\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2703737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$ Related to solving $~~2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$
I've tried some arithmetic to find something like
$a^x=b~\Longrightarrow~x=\log_{a}b$
But what I've found is that
$2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$
$2^8\cdot2^{x-5}+3^{x-5}=2^8\cdot2^{3(x-5)}+3^{2(x-5)}$
$256\cdot2^{x-5}+... | As @JohnHughes hinted at in the comments, the trick to solving $2^{x+3}+3^{x−5}=2^{3x−7}+3^{2x−10}$ is in realizing that powers of two are always even, and powers of three are always odd, so a power of two won't ever affect the value of a power of three. (Edit: This is only true when you are multiplying powers of two b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Formula for discriminant of a polynomial of degree 2 in 3 variables Which is the correct way/ method/Formula to find the discriminant of a quadratic equation $f$ in 3 variables?
i.e., a quadratic form in 3 variables.
Also, how to conclude that, whether this $f$ is reducible or irreducible?
Some one knows this please h... | Let us write the general 2nd degree equation under the form:
$$\tag{1}ax^2+2bxy+cy^2+2dxz+2eyz+fz^2.$$
In order for (1) to be expressible under the form of a product of two first degree polynomials:
$$\tag{2}(px+qy+rz)(sx+ty+uz),$$
coefficients $a,b,c,d,e,f$ must satisfy the pair of constraints :
$\tag{3}\begin{cases}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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4 digit numbers divisible by 11 Four digit numbers are formed using the
digits 1,2,3,4 (repetition is allowed). The
number of such four digit numbers divisible
by 11 is-
(1) 22 (2) 36 (3) 44 (4) 52
I know for a number to be divisible by 11 the sum of digits at even places must be equal to the sum of those at odd places... | Let $[abcd]$ be the number. Then $[abcd] \equiv [ab]+[cd] \pmod{99}$ which implies $[abcd] \equiv [ab]+[cd] \pmod{11}$. Because $a,b,c,d \in \{1,2,3,4\}$,
$[abcd] \equiv [(a+c)(b+d)] \pmod{11}$. Hence $[abcd]$ is a multiple of $11$ if and only if $a+c=b+d$.
\begin{array}{|r|r|r|}
\text{sum} &\text{ac and bd events} & \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Visualizing $3^3+4^3+5^3=6^3$ I was wondering how would the visualization of $3^3+4^3+5^3=6^3$ look? I tried some 3D sketches but when calculating none work. Any ideas would be highly appreciated.
| No complete answer but some thoughts:
Subtracting a 5-cube from a 6-cube leaves
$$6^3-5^3=216-5^3=91.$$
Imagine you place the 5-cube into the 6-cube and check what is left over. Lets assume we rotate it so that the 5-square "on ground" is inscribed into the 6-square.
Now we extend the 5-cube by one unit to the top and... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How is this nested radical expression for $\cos(2\pi /7)$ proven? On the Wolfram Mathworld page for the silver constant, the following identity is present, how is it proven?
$$2\cos\Big(\frac{2\pi}{7}\Big) = \frac{2+\sqrt[3]{\,7+7\sqrt[3]{\,7+...}}}{1+\sqrt[3]{\,7+7\sqrt[3]{\,7+...}}}$$
Link: http://mathworld.wolfram.c... | This is the same as
$$ 2\cos\left(\frac{2\pi}{7}\right)-1=\frac{1}{1+\sqrt[3]{7+7\sqrt[3]{7+\cdots}}}. $$
Write $T=2\cos\left(\frac{2\pi}{7}\right)=\zeta+\zeta^{-1}$ where $\zeta=e^{2\pi i/7}$ and $x=\sqrt[3]{7+7\sqrt[3]{7+\cdots}}$. Then observe
$$ \begin{array}{ll} T & = \zeta+\zeta^{-1} \\ T^2 & = \zeta^2+2+\zeta^{-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $2 \cdot \sum\limits_{k = 0}^{n} \binom{2n}{2k} = 2^{2n}$ without induction I am aware of
$$
2^{2n} = (1 + 1)^{2n} =
\sum_{k = 0}^{2n} \binom{2n}{k}
$$
I then tried grouping the terms and using that $\binom{n}{k} = \binom{n}{n-k}$, to obtain
$$
\begin{align}
\sum_{k = 0}^{2n} \binom{2n}{k} &=
\binom{2n}{0} + \bi... | I would also use
$$0 = (1 - 1)^{2n} = \sum_{k=0}^{2n}(-1)^k\binom{2n}{k}.$$
Adding this to the previous sum,
$$2^{2n} = \sum_{k=0}^{2n}\left[(-1)^k\binom{2n}{k} + \binom{2n}{k}\right].$$
When $k$ is odd, the term is $0$, leaving only even $k$ terms (but doubled!). Therefore,
$$2 \sum_{k=0}^{n} \binom{2n}{2k} = 2^{2n},$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the number of ways in which 6 persons out of 5 men and 5 women can be seated at a round table such that 2 men are never together. Find the number of ways in which 6 persons out of 5 men and 5 women can be seated at a round table such that 2 men are never together.
My attempt:
6 people may be 3 men and 3 women, 2 ... | Label the chairs from $1$ to $6$ in increasing order, label the men and the women from $1$ to $5$.
Consider three cases:
1) There is 1 man: choose a man in $5$ ways, place the man at chair $1$, and permute the $5$ women in the remaining chairs in $5!$ ways. Therefore the number of ways is $5\cdot 5!=600$.
2) There are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$
First,I tried to evaluate like this:
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^... | $$since\ I=\int_{0}^{\frac{\pi }{2}}\frac{x^2}{sinx}dx=-2\int_{0}^{\frac{\pi }{2}}xln(tan(\frac{x}{2}))dx=-8\int_{0}^{\frac{\pi }{2}}t \ln(tan(t))dt\\
\\
=-8\int_{0}^{1}\frac{lnt\arctan(t)}{1+t^2}dt=-4\pi \int_{0}^{1}\frac{ln(t)}{1+t^2}dt+8\int_{0}^{1}\frac{ln(t)arctan(t)}{1+t^2}dt\\
\\
=4\pi G-8\int_{1}^{\infty }\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 9,
"answer_id": 6
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Number of five digit numbers that can be formed using the digits 1,2,3,4,5,6,7,8,9 in which one digit appears once and two digits appear twice Find the number of five digit numbers that can be formed using the digits 1,2,3,4,5,6,7,8,9 in which one digit appears once and two digits appear twice (e.g.41174 is one such nu... | Choose two numbers to appear twice and then one number to appear once. Now we have five numbers. Choose two places for the smaller number that appears twice and then the two places for the larger number that appears twice.
$$\binom{9}{2}\binom{7}{1}\binom{5}{2}\binom{3}{2}=7560$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to prove a sequence is strictly increasing and bounded above? For this proof, I'm not sure if I am doing it right. Here is what I have so far. Can anyone please help me out?
Let $\{a_n\}$ be a sequence defined recursively by
$a_1 = \sqrt{6}$
$a_{n+1} = \sqrt{6+5a_n}, n = 1,2,3,...$
Prove that $\{a_n\}$ is strictly ... | It is correct globally, but I think you are not expressing yourself clearly about why the sequence is strictly increasing. The base case is $a_1<a_2$, which is clearly true. Andn, if $a_n<a_{n+1}$, then $6+5a_n<6+5a_{n+1}$ and therefore $\sqrt{6+5a_n}<\sqrt{6+5a_{n+1}}$. But this means that $a_{n+1}<a_{n+2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\sin\left(\arccos\frac12+\arccos\frac{7}{25}\right)$
Evaluate $$\sin\left(\arccos\frac12+\arccos\frac{7}{25}\right)$$
I know that $\arccos\frac12$ is $60^\circ$. I don't know how to continue.
| $$\sin\left(\arccos\frac12+\arccos\frac{7}{25}\right)=$$
$$\sin(x+y) = \sin x \cos y + \cos x \sin y$$
Where $$x=\arccos\frac12$$ and $$y= \arccos\frac{7}{25}$$
Note that $$ \cos x =\frac {1}{2}$$ and $$ \cos y =\frac {7}{25}$$
We can find $$ \sin x= {\sqrt 3}/2 $$ and $$\sin y = 24/{25}$$
upon substitution we get $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2717882",
"timestamp": "2023-03-29T00:00:00",
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Solve the ODE $\cot (x^2+y^2)dy+xdx+ydy=0$ Solve the ODE $\cot (x^2+y^2)dy+xdx+ydy=0$
i am trying to solving integrating combination
since given that
$\cot (x^2+y^2)dy+xdx+ydy=0$
then $\cot (x^2+y^2)dy+d(xy)=0$ is it correct way ? and we can apply integration from here? can any one help me this problem
| Dividing both sides of $\cot (x^2+y^2)dy+xdx+ydy=0$ by $dy$ we get $\cot (x^2+y^2)+x\frac {dx}{dy}+y=0$
Then,
$$x\frac {dx}{dy}=-y-\cot (x^2+y^2) \iff \frac {dx}{dy}=\frac{-y-\cot (x^2+y^2)}{x} \iff \frac {dy}{dx}=-\frac{x}{y+\cot (x^2+y^2)}$$
Now you can integrate this: $\frac {dy}{dx}=-\frac{x}{y+\cot (x^2+y^2)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Find the minimum value of $\frac{a+b}{2} + \frac{2}{ab-b^{2}}$, with $a,b \in \mathbb{R}$, $a>b>0$ Find the minimum value of $$ \frac{a+b}{2} + \frac{2}{ab-b^{2}},$$
where $a,b \in \mathbb{R}$, $a>b>0$.
Attempt : The only method I knew was using partial derivatives.
Let
$$f(a,b) = \frac{a+b}{2} + \frac{2}{ab-b^{2}}... | Using A.M. >= G.M.
$$\cfrac{a-b}{2} + {b} + \cfrac{2}{b(a-b)} >=3\sqrt[3]{\cfrac{a-b}{2}*{b}* \cfrac{2}{b(a-b)}} = 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2720295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to factorize $2x^2-9x+9$ by completing the square? I know that $x^2-bx+c=(x-k)^2=x^2-2kx+k^2$ if it is a complete square. If not we create one by adding and subtracting $\left(\frac{b}{2}\right)^2$
I tried $$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x^2-\frac{9}{2}x+\left(\frac{9/2}{2}\right)^2-\left(\frac{... | Make the polynomial equal to $0$ to find the roots so that you can factorise.
$$2x^2-9x+9=0$$
$$x^2-\frac{9x}{2}+\frac{9}{2}=0$$
$$x^2-\frac{9x}{2}=\frac{-9}{2}$$
To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of b.
$$(\frac{b}{2})^2=(\frac{-9}{4})^2$$
Ad... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If the integers 1-9 are randomly distributed into three sets of 3 integers, what is the probability that at least one set contains only odd integers? The problem https://www.cut-the-knot.org/Probability/ThreeGroupsOfThree.shtml has been stated here,
If the integers 1-9 are randomly distributed into three sets of 3 int... | The number of ways of forming three labeled sets of three numbers from the set $\{1, 2, 3, 4, 5, 6, 7, 8, 9\}$ is
$$\binom{9}{3}\binom{6}{3}\binom{3}{3}$$
since we must select three of the nine numbers to be in the first set, three of the remaining six numbers to be in the second set, and all three of the remaining th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the $\lim\limits_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$ The task is to find $$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}$$
What I've tried is dividing both the numerator and the denominator by $x$, but I just can't calculate it completely.
I know it should be s... | Let $y=-x\to \infty$ then
$$\lim_{x\rightarrow - \infty}\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+b^2}+x}=\lim_{y\rightarrow \infty}\frac{\sqrt{y^2+a^2}-y}{\sqrt{y^2+b^2}-y}$$
and
$$\frac{\sqrt{y^2+a^2}-y}{\sqrt{y^2+b^2}-y}\frac{\sqrt{y^2+a^2}+y}{\sqrt{y^2+a^2}+y}\frac{\sqrt{y^2+b^2}+y}{\sqrt{y^2+b^2}+y}=\frac{a^2}{b^2}\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Is it true that $b^n-a^n < (b-a)nb^{n-1}$ when $0 < a< b$? A Real Analysis textbook says the identity
$$b^n-a^n = (b-a)(b^{n-1}+\cdots+a^{n-1})$$ yields the inequality
$$b^n-a^n < (b-a)nb^{n-1} \text{ when } 0 < a< b.$$
(Note that $n$ is a positive integer)
No matter how I look at it, the inequality seems to be wrong. ... | \begin{align}
b^n-a^n & = (b-a)(b^{n-1}+ b^{n-2}a + b^{n-3}a^2 + b^{n-4}a^3 + b^{n-5} a^4 +\cdots+a^{n-1}) \\[10pt]
& < (b-a)(b^{n-1} + b^{n-2} b + b^{n-3}b^2 + b^{n-4}b^3+ b^{n-5}b^4 + \cdots + b^{n-1}) \\[10pt]
& = (b-a)(b^{n-1} + b^{n-1} + b^{n-1} + b^{n-1} + b^{n-1} + \cdots + b^{n-1}) \\[10pt]
& = (b-a) n b^{n-1}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Calculate the limit: $\lim_{n\to\infty}n^2\left(\left(1+\frac{1}{n}\right)^8-\left(1+\frac{2}{n}\right)^4\right)$ Calculate the limit:
$\lim\limits_{n\to\infty}n^2\left(\left(1+\dfrac{1}{n}\right)^8-\left(1+\dfrac{2}{n}\right)^4\right)$
My first suggestion was that $\lim\limits_{n\to\infty} = 0$. As in both brackets as... | Let $x=1+\frac{1}{n}$ and $y=1+\frac{2}{n}$ then
$$x^8-y^4=(x^2-y)(x^2+y)(x^4+y^2)$$ and
$$x^2-y=\frac{1}{n^2}$$
Thus the limit is
$$(x^2+y)(x^4+y^2) \to 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Recurrence relation from Generating function - how to find recurrence relation?? Given a generating function equation of the form -
$$ f(x) = A.x.f(x)^2 + B.x.f(x) + C $$
for the sequence $ f(x) = \sum_{n=0}^{\infty} a_n.x^n $. How to find the corresponding recursive relation?
This type of generating function is quit... | $$f(x) = x.f(x)^2 + x.f(x) + 1 \implies x.f(x)^2 + (x-1)f(x) + 1 = 0$$
$$f(x) = \frac{(1-x) \pm \sqrt{(x-1)^2 - 4x}}{2x} = \frac{(1-x) \pm \sqrt{1-6x+x^2}}{2x}$$
To get recurrence from this (and taking the negative root):
$$(1-x) - \sum_{n=0}^{\infty} 2xf_nx^n = (1-3x) - \sum_{n=2}^{\infty} 2f_{n-1}x^n = \sqrt{1-6x+x^... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$? I'm trying to find the inverse of $5+4\sqrt[3]{2}+3\sqrt[3]{4}$ in $\mathbb{Z}[\sqrt[3]{2}]$. I know it is a unit, so there is an inverse, but I feel like I may be doing too much work in the wrong direction. Here's what I have so far:
Let $\alpha = 5+4\sqrt[3]{2}+3\sqr... | Alt. hint: let $\,x=5+4\sqrt[3]{2}+3\sqrt[3]{4}\,$, then successively multiplying with $\,\sqrt[3]{2}\,$:
$$
\begin{cases}
\begin{align}
5-x+4\sqrt[3]{2}+3\sqrt[3]{4} &= 0 \\
12+(5-x)\sqrt[3]{2}+4 \sqrt[3]{4} &= 0 \\
8 + 12 \sqrt[3]{2}+(5-x)\sqrt[3]{4} &= 0
\end{align}
\end{cases}
$$
Eliminating $\,\sqrt[3]{2}\,$ and... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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A problem from Makarov Selected Problems in Real Analysis This is a problem from Makarov’s Selected Problems in Real Analysis. Put $S(n) = \sum_{k=0}^{2n}{\frac{k}{k+n^2}}$ Find the limit as n tends to infinity. The answer is $\frac {1}{2} \ln 5$. My solution goes as follows.
Since $y=\frac{x}{x+n^2}$ is increasing on ... | It is true that
$\sum\limits_{k=0}^{2n}{\frac{k}{k^2+n^2}}\lt S(n)\lt \sum\limits_{k=0}^{2n}{\frac{k}{n^2}}$
Both sides can be formed to Riemann integrals:
$2\frac{1}{2n}\sum\limits_{k=0}^{2n}{\frac{2\frac{k}{2n}}{4\big(\frac{k}{2n}\big)^2+1}}\lt S(n)\lt 4\frac{1}{2n} \sum\limits_{k=0}^{2n}{\frac{k}{2n}}$
$\frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving $45x-3795x^3 +95634x^5 - \cdots + 945x^{41}-45x^{43}+x^{45} = N$? We have the following excerpt from John Stillwell's Mathematics and its History:
My question is, what are the hints which give away the fact that this comes from the expansion of $\sin 45 \theta$?
| Perhaps, it is this:
${\sin n \theta = \dbinom{n}{1}\cos^{n-1}\theta\sin \theta- \dbinom{n}{3}\cos^{n-3}\theta \sin^3 \theta + \dbinom n 5\cos^{n-5}\theta\sin ^{5}\theta...}\\= \color{blue}{\displaystyle\sum_{r=0, 2r+1\le n}(-1)^r\dbinom{n}{2r+1}\cos^{n-2r-1}\theta \sin^{2r+1}\theta} $
Proof:
$(\cos \theta+ i\sin \th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Adding $k^2$ to $1^2 + 2^2 + \cdots + (k - 1)^2$. In his book, Calculus Vol. 1, Tom Apostol mentions that adding $k^2$ to the predicate
$$A(k): 1^2 + 2^2 + \cdots + (k - 1)^2 < \frac{k^3}{3}$$
gives the inequality
$$ 1^2 + 2^2 + \cdots + k^2 < \frac{k^3}{3} + k^2.$$
Why does the RHS $$1^2 + 2^2 + \cdots + (k - 1)^2 $$ ... | The notation does not matter. $1^2 + 2^2 + \cdots + (k - 1)^2 + k^2$ and $1^2 + 2^2 + \cdots + k^2$ both stand for the same value (under the appropriate interpretation of $\cdots$). Both stand for $\displaystyle \sum_{j=1}^{k}j^2$.
Example:
$1^2+2^2+3^2 + \cdots + 7^2 = 1^2+2^2+3^2 + \cdots + 6^2 + 7^2 = 1^2+2^2+3^2 +... | {
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"source": "stackexchange",
"question_score": "1",
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Mini-Challenge on a condition Hello I would like to purpose to you an enigma this is the following :
Let $a,b,c$ be positive real number find the condition on $abc$ and $ab+bc+ca$ and $a+b+c$ to have $a^9+b^9+c^9=3$
I have a solution using the following identity :
$$a^9+b^9+c^9= 3a^3b^3c^3−45abc(ab+bc+ca)(a+b+c)^4+ ... | We can use the following identity to solve the problem in a more efficient manner than remembering a gigantic identity:
$$a^3+b^3+c^3= \sigma_1^3-3\sigma_1\sigma_2+3\sigma_3$$
where $\sigma_1=a+b+c, \ \sigma_2=ab+bc+ca$, and $\sigma_3=abc$ are the Elementary Symmetric Polynomials. Next we find the cubed versions of the... | {
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"source": "stackexchange",
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Definite integral question with rotation Find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line.
$y=x^2, x = y^2;$ rotate about $y=1$.
So the first thing that jumps out to me is that we're rotating this around $y=1$. So it seems like using vertical washers mig... | You've done very well. Your work is correct, and indeed, the answer is $\frac{11\pi}{30}$.
Now it's simply a matter of using the power rule for integration, evaluating at $x=1$, and $x=0$, finding a common denominator. I edited your post to include the missing $dx$ in each integral.
$$\pi \int_0^1 (-2x^2 + 2\sqrt{x} ... | {
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Alternate solutions to similar integrals? So, when I started this problem I couldn't see any reasonable trig substitutions that would make it simpler, and I don't know of a better method, so I attempted to solve it using repeated use of integration by parts. Is there a much easier, better, alternate solution to integra... | Consider:$$\int\sin^4(x)\,\mathrm{d}x$$Apply the reduction formula:$$=\frac{3}{4}\cdot\int\sin^2(x)\,\mathrm{d}x\,-\frac{\cos(x)\sin^3(x)}{x}$$$$=\frac{3}{4}\cdot\left(\frac{1}{2}x-\frac{\cos(x)\sin(x)}{x}\right)-\frac{\cos(x)\sin^3(x)}{x}$$$$\boxed{=\frac{\sin(4x)-8\sin(2x)+12x}{32}+C}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find Jordan blocks from minimal polynomial If the characteristic polynomial is $p=(x-λ)^6$ and the minimal polynomial is $p=(x-λ)^4,$ how do we find all the Jordan blocks?
| From your characteristic polynomial, we see that the matrix has size six-by-six (so the sum of the lengths of the Jordan blocks is six) and only one eigenvalue. From the minimal polynomial, we see that one Jordan block must have length four. That leaves two possibilities for the Jordan blocks:
$$\begin{pmatrix}
\lamb... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of improper integral $ \lim_{x\to \infty} \ \int_1^x x\,e^{t^2-x^2} \,dt$ I need to calculate the limit of the following improper integral:
$$ \lim_{x\to \infty} \ \int_1^x x\,e^{t^2-x^2} \,dt$$
My solution is $\infty$, but when I enter it in WolframAlpha the solution is $\frac{1}{2}$, so I guess mine is wrong, b... | It is
$$
\int \limits_1^x x e^{t^2-x^2} dt = \frac{x}{e^{x^2}} \int \limits_1^x e^{t^2} dt = \frac{x}{e^{x^2}} \cdot (G(x) - G(1))
$$
for some differentiable function $G: [1, \infty) \rightarrow \mathbb{R}$ by the fundamental theorem of calculus. Furthermore, $G'(x) = e^{x^2}$, hence $G$ is strictly increasing and $\li... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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If $a,b\in\mathbb N_+$ satisfy $\sqrt{ab},\sqrt{\frac{a^2+b^2}2}\in\mathbb N_+$, is it necessarily true that $a=b$? This question is related to another question where it asks to find pairs of positive integers $(a, b)$ such that
$$
\frac{a + b}{2},\ \sqrt{ab},\ \frac{2ab}{a + b} \in \mathbb{N}_+.
$$
In that question ... | First, it is easy to argue that, since $\sqrt{ab} \in \mathbb N$, there exists a square free integer $k$ and integers $m,n$ such that
$$a=m^2k\\
b=n^2k$$
($k$ is simply the product of the primes which in the prime decomposition of $a$ appear at an odd power).
Then,
$$\sqrt{\frac{a^2+b^2}{2}}=\sqrt{k^2\frac{m^4+n^4}{2... | {
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Is this set of solutions complete? Let $p,q,r$ be distinct primes and $a,b,c\ge 2$ integers.
The equation $$p^a+q^b=r^c$$ has the following solutions :
$$2^4+3^2=5^2$$
$$2^5+7^2=3^4$$
$$2^2+11^2=5^3$$
$$2^7+17^3=71^2$$
$$7^3+13^2=2^9$$
Can we prove without using any unproven conjecture that this list is complete ?
I... | In this answer, I'll assume Explicit abc-Conjecture is true and will show that the set of solution is finite (the upper bound is very big). Since exactly one of $p$, $q$ and $r$ is $2$, we can split the equation into two cases.
i) $r=2$.
The equation is now $p^a+q^b=2^c$. When $a=b=2$, $p^a+q^b=p^2+q^2\equiv 2\pmod 4$,... | {
"language": "en",
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Generalizing $ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5}$ If $a+b+c=0$ as discussed in this, this, and this post, then,
$$ \frac{a^2+b^2+c^2}{2} \times \frac{a^3+b^3+c^3}{3} = \frac{a^5+b^5+c^5}{5}\tag1 $$
$$ \frac{a^3+b^3+c^3}{3} \times \frac{a^4+b^4+c^4}{2} = \frac{a^7+b^7+c^7}{7}\tag... | Define $\;s_k := (a^k+b^k+c^k)/k.\;$ Then
$\;s_{17}=2s_{13}s_4+3s_9s_5s_3,\; s_{19}=8s_8s_7s_4+3s_{11}s_5s_3.$
Not as nice as the ones for $\;s_{11}\;$ and $\;s_{13}\;$ but they are two terms with positive integer coefficients.
| {
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If $x$, $y$, and $z$ are real numbers such that $x+y+z=8$ and $x^2+y^2+z^2=32$, what is the largest possible value of $z$? I tried swapping $z$ from the first equation to the second, and got $$x^2 + x y - 8 x + y^2 - 8 y + 16=0$$ Not sure where to go from there, and if I'm on the right track at all.
| Writing the root-mean square inequality in the form $\;\displaystyle\frac{a^2+b^2}{2} \ge \left(\frac{a+b}{2}\right)^2\;$ which holds for all $\,a,b \in \mathbb{R}\,$ regardless of signs, and using that the largest of $\,x,y,z\,$ must be $\,z \gt 0\,$:
$$
\begin{alignat*}{3}
\frac{x^2+y^2}{2} \ge \left(\frac{x+y}{2}\ri... | {
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"url": "https://math.stackexchange.com/questions/2750329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How do I derive the formula for the reciprocal of a hypotenuse? Given $a^2 + b^2 = c^2$,
Why is it that the equation below yields the reciprocal of the hypotenuse c, ($\frac{1}c$)?
$\sqrt{(\frac{a}{a^2+b^2})^2 + (\frac{b}{a^2+b^2})^2}$
Worked example:
$3^2 + 4^2 = c^2$
$c = 5$
$\sqrt{(\frac{3}{3^2+4^2})^2 + (\fra... | Expand the equation:
$$
\begin{align}
\sqrt{\left(\frac{a}{a^2+b^2}\right)^2+\left(\frac{b}{a^2+b^2}\right)^2}&=\sqrt{\left(\frac{a}{c^2}\right)^2+\left(\frac{b}{c^2}\right)^2}\\
&=\sqrt{\frac{a^2}{c^4}+\frac{b^2}{c^4}}\\
&=\sqrt{\frac{c^2}{c^4}}\\
&=\sqrt{\frac{1}{c^2}}\\
&=\frac{1}{c}
\end{align}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given $3$ equations with parameters $a, b, c$, evaluate $a^2+b^2+c^2$
Given the equations
$$\begin{align}
x&=cy+bz \\
y&=az+cx \\
z&=bx+ay
\end {align}$$
where $x,y,z$ are not all zero, evaluate $a^2+b^2+c^2$.
I don't understand the way to relate $a,b,c$ and $x,y,z$.
Can someone explain this?
| We can rearrange the equations as $-x+cy+bz=0$, $cx-y+az=0$, $bx+ay-z=0$, and so
$$\begin{pmatrix}-1&c&b\\c&-1&a\\b&a&-1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.$$
This has a solution with $x,y,z$ not all zero if and only if the determinant
$$\begin{vmatrix}-1&c&b\\c&-1&a\\b&... | {
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Notation/simplification: $(n-1)(n-3)(n-5)...(3)(1)$ How would one write this in simplified form? I am aware that the answer is $\frac{n!}{2^{n/2}(n/2)!}$. How to arrive at this answer?
| The number $n$ is requiered to be even.
$$1 \cdot 3 \cdot 5 \cdot \ldots \cdot (n-5)(n-3)(n-1) \\[1em]
=\frac{1 \cdot \color{blue}{2} \cdot 3 \cdot \color{blue}{4} \cdot 5 \cdot \color{blue}{6} \cdot \ldots \cdot (n-5) \color{blue}{(n-4)} (n-3) \color{blue}{(n-2)} (n-1) \color{blue}{n} }{\color{blue}{2 \cdot 4 \cdot 6 ... | {
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Solve a trigonometric equation: $|1-2\sin^2 x|=|\cos x|$ I have difficulty in solving this equation:
$$|1-2\sin^2 x|=\lvert\cos x\rvert.$$
What are the indications to solve the exercise correctly?
| Since both sides are absolute values, you can square both sides:
$$\begin{align}
1-4\sin^2x+4\sin^4x&=\cos^2x \\
0&=-3\sin^2x+4\sin^4x \\
&=\sin^2x\left(4\sin^2x-3\right) \\
\sin x&=0,\pm\frac{\sqrt3}2
\end{align}$$
Following a comment by @labbhattacharjee, we note that $\sin(3x)=3\sin(x)-4\sin^3x$, so
$$\begin{align}... | {
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How do I solve the inequality $x^{12}−x^9+x^4−x+1>0$ using intervals? My book has the question $x^{12}−x^{9}+x^{4}−x+1>0$. The solution given gives three cases, when $x \le 0$, when $0 < x \le 1$ and when $x > 1$. How did they get these intervals? What is the method used?
| Hint: When $x=0$ we get $1>0$ and when $x<0$ then $-x^9,-x>0$ so $$x^{12}-x^9+x^4-x+1>0$$
For $$0<x\le 1$$ we get $$x=1$$ then we get $$1>0$$ and we substitute $$x=\frac{1}{y}$$ so we obtain
$$1+y^3(y^5-1)+y^{11}(y-1)>0$$
and for $x>1$ we have $$x^9(x^3-1)+x(x^3-1)+1>0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Without using a calculator, is $\sqrt[8]{8!}$ or $\sqrt[9]{9!}$ greater? Which is greater between $$\sqrt[8]{8!}$$ and $$\sqrt[9]{9!}$$?
I want to know if my proof is correct...
\begin{align}
\sqrt[8]{8!} &< \sqrt[9]{9!} \\
(8!)^{(1/8)} &< (9!)^{(1/9)} \\
(8!)^{(1/8)} - (9!)^{(1/9)} &< 0 \\
(8!)^{(9/72)} - (9!)^{8... | Another way to see this is to convert both sides into two "averages".
*
*"Uniform distribution on $\{\log1,\dots,\log8\}$": $$\frac{\log1 + \dots + \log8}{8}$$
*"Uniform distribution on $\{\log1,\dots,\log9\}$": $$\frac{\log1 + \dots + \log9}{9}$$
Intuitively, the later has a greater "expectation", so the result ... | {
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"source": "stackexchange",
"question_score": "34",
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Definite Integral involving Max function. Here is an integral I was attempting to solve $\int\limits_{0}^{\ln t}{\max{\left(1,x\right)dx}}$ but my answer is not coming to be correct. What is wrong in my attempt?
Attempt 1. Let's call the integral $I$
Case 1. Let $0<t<e$ $\implies$ $\ln t < 1$
This means the given inte... | Correct Attempt 1. Let's call the integral $I$
Case 1. Let $0<t\le e$ $\implies$ $\ln t \le 1$
This means the given integral becomes $\int\limits_{0}^{\ln t}{1 dx}$ because from $(0,1)$ the maximum between $1$ and $x$, is $1$ $$\implies I=\ln t \tag{1}$$
Case 2. Let $e\le t< \infty$ $\implies$ $\ln t \ge 1$
This mean... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Simplifying the derivative of $x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$ $$x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$$
So I get:
$$-x^{\frac{2}{3}} \cdot \frac{1}{3} (6-x) ^{\frac{-2}{3}} + (6-x) ^{\frac{1}{3}} \cdot \frac{2}{3} x ^ {\frac{-1}{3}}$$
How does one go about simplifying this?
I guess I can pull out common... | $$f^3 (x)=x^2 (6-x) $$
by differentiation
$$3f^2 (x)f'(x)=2x (6-x)-x^2=3x (4-x) $$
thus
$$f'(x)=\frac {x (4-x)}{f^2 (x)} $$
$$=\frac {x (4-x)}{x^\frac43 (6-x)^\frac23} $$
$$=\frac {4-x}{(6-x)^\frac23}x^{\frac {-1}{3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2768051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximum power of $2$ which divides $3^{1024}-1$ What is the maximum power of $2$ which completely divides $3^{1024}-1$?
I proceeded thus:
$\phi(2^n)=2^{n-1}$ for all $n\ge1$
$$3^{1024}=3^{2^{10}}\equiv1\pmod {2^{11}}$$
$$3^{1024}-1\equiv0\pmod {2^{11}}$$
Since $\phi(2^{11})=2^{10}$. So, maximum power of $2$ must be $1... | Just to give a different approach, since $3^4=81=1+80=1+5\cdot2^4$, we have
$$3^{1024}-1=(1+80)^{256}-1=256\cdot80+{256\choose2}80^2+{256\choose3}80^3+\cdots\\
=2^8(5\cdot2^4)+(2^7\cdot255)(5^2\cdot2^8)+(2^7\cdot85\cdot254)(5^3\cdot2^{12})+\cdots\\
\equiv5\cdot2^{12}\mod2^{13}$$
all terms after the first having higher ... | {
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"url": "https://math.stackexchange.com/questions/2770741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find polynomial over Z2, that has the least degree, for which the field P=GF(64) is the minimum decomposition field. How much such polynomial are? Find polynomial over Z2, that has the least degree, for which the field P=GF(64) is the minimum decomposition field. How much such polynomials are?
Please, answer in detail!... | This late answer uses computer power to find explicitly the irreductible polynomials of degree six in $\Bbb F_2[X]$.
There exists up to isomorphy only one field with $2^6$ elements, $F=\Bbb F_{64}$, its elements are fixed by the corresponding Frobenius isomorphism, so $F$ contains the roots of
$$x^{2^6}-x\ .$$
sage giv... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove this stronger inequality
Let $a,b,c>0$, and $a+b+c=1$,show that
$$\sum_{cyc}\dfrac{a^3+b^2}{b+c}\ge\dfrac{2}{3}+\dfrac{5+\sqrt{2}}{12}\sum_{cyc}(a-b)^2\tag{1}$$
I have prove $$\sum_{cyc}\dfrac{a^3+b^2}{b+c}\ge\dfrac{2}{3}$$
because Use Holder inequality
$$\sum_{cyc}\dfrac{a^3}{b+c}\cdot\sum(b+c)\sum_{cyc}(1)\... | We'll prove a stronger inequality:
Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=1$. Prove that
$$\sum_{cyc}\frac{a^3+b^2}{b+c}\geq\frac{2}{3}+\frac{2}{3}\sum_{cyc}(a-b)^2.$$
Indeed, by C-S
$$\sum_{cyc}\frac{b^2}{b+c}\geq\frac{(a+b+c)^2}{2(a+b+c}=\frac{1}{2}.$$
Thus, it's enough to prove that
$$\sum_{cyc}\f... | {
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"timestamp": "2023-03-29T00:00:00",
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Piecewise function as initial condition in Heat Equation exercise - Confirm solution I have $u_t = u_{xx}$ for $0<x<1$,
$u(0,t) = 0, \; 0 \leq t < \infty$
$u(1,t) = 0, \; 0 \leq t < \infty$
$u(x,0) = \varphi(x), \; 0 \leq x \leq 1$ where $\varphi(x) = \begin{cases}
\frac{5x}{2}, 0 < x < \frac{2}{3} \\
3-2... | Yes, this is correct, it is just an integration by part.
| {
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Binomial Inequality in which Binomial coefficient is in square root.
$$1\cdot \sqrt{C_{1}}+2\cdot \sqrt{C_{2}}+\cdots \cdots +100\cdot \sqrt{C_{100}}\leq \frac{1}{2}\cdot (2^{100}-1)+\frac{20301}{12}$$
where $\displaystyle C_{r}=\binom{n}{r}$
Try: Using Cauchy Schwarz Inequaity
$$\bigg(1^2+2^2+\cdots \cdots +100^2\bi... | Hint.
$$
2\bigg[\frac{100\cdot 101\cdot 201}{6}\cdot (2^{100}-1)\bigg]^{\frac{1}{2}}\le \frac{100\cdot 101\cdot 201}{6} + (2^{100}-1)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2777024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Matrix satisfying $A-I = A^{-1}$ Recall the (infinitely) continued fraction definition of the golden ratio
\begin{align}
\phi = 1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\frac{1}{1+\cdots}}}}}
\end{align}
This is equivalent to the expression
\begin{align}
\phi = 1+\frac{1}{\phi}
\end{align}
Saying that the inverse ... | $$\begin{align}A-I = A^{-1}\end{align}$$
$$\begin{align}A = I + A^{-1}\end{align}$$
$$\begin{align}A^2 = A + I\end{align}$$
$$\begin{align}A^3 = A^2 + A\end{align}=2A+I$$
$$\begin{align}A^4 = A^3 + A^2\end{align}=3A+2I$$
The pattern is now suggesting,
$$A^n = F_n A+F_{n-1}I$$ where $F_n$ is the Fibonacci's sequence.
| {
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"url": "https://math.stackexchange.com/questions/2778724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Complex Number Series If
$$
z + 1/z = -1
$$
where $z$ is a complex number then value of the sum from r = 1 to r = 99 of $$
( z^r + 1/z^r)^2
$$
is equal to
A) 198
B) 3
C) 99
D) 0
I tried by putting z = x + iy but that gives, |z| = 1
| Define $$f_n(z) = z^n + z^{-n}.$$ Then note $$f_1(z) f_n(z) = (z+z^{-1})(z^n + z^{-n}) = z^{n+1} + z^{-(n+1)} + z^{n-1} + z^{-(n-1)} = f_{n+1}(z) + f_{n-1}(z).$$ In the case where $f_1(z) = -1$, we find $$f_{n+1}(z) = -f_n(z) - f_{n-1}(z).$$ Also note that $$f_0(z) = z^0 + z^{-0} = 1 + 1 = 2.$$ Unfolding the first ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find recurrence for $f(x)$ coefficient in the expression $f(x)\cdot(1+2x+2x^2+x^3)=\frac{1}{1-x^3}$ using generating functions?
Let $f(x)=\sum_{i=0}^{\infty} a_ix^i$. Also $f(x)\cdot(1+2x+2x^2+x^3)=\frac{1}{1-x^3}$. Find numbers $b,c,d$ such that:
$$
a_n={3+n-1 \choose n}-ba_{n-1}-ca_{n-2}-da_{n-3}
$$
for $... | I believe Claude is correct about the statement of the problem, but I will attempt to provide some useful commentary as is.
So if we substitute $\sum_{i=0}^{\infty} a_{i}x^i$ in for $f(x)$, we obtain
$$\sum_{i=0}^{\infty} a_{i}x^i + 2x\sum_{i=0}^{\infty} a_{i}x^i + 2x^2\sum_{i=0}^{\infty} a_{i}x^i + x^3\sum_{i=0}^{\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2780454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the Maximum Likelihood estimator for $r$ I have a set of data
$$X \sim N_2(0,\Sigma) \quad \text{with} \quad \Sigma = \pmatrix{ \sigma^2 & r\sigma^2 \\ r\sigma^2 & \sigma^2}$$
and I am asked to find the Maximum Likelihood estimates of $r$.
Firstly I find the log likelihood of my data which is
$$-n\log(2\pi) - 2... | Write the log-likelihood as
$$
\ell(r,\sigma^2)=-\frac{n}{2}\log(1-r^2)-\frac{1}{2\sigma^2}\frac{1}{1-r^2}\left(A-2rB\right) - n\log\sigma^2,
$$
where
$$
A = \sum_{i}x_i^2+y_i^2, \qquad B\sum_{i}x_iy_i.
$$
Then $\frac{\partial \ell}{\partial r}=0$ implies that
$$
nr\sigma^2+\frac{1}{1-r^2}\left((r^2+1)B - Ar \right) =0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2780662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to rewrite a Sum? How to find the following sum?
$$
\frac{2}{n} \sum_{i=1}^{n} \left(\frac{4i}{n} +1\right)
$$
For example
$$
\sum_{i = 1}^n i^2 = \frac{n(n+1)(2n+1)}{6}.
$$
I need to find the value of which this would converge to and I know that it must be six because the integral from $1$ to $3$ of the function ... | $$\frac{2}{n} \sum_{i=1}^{n} \left(\frac{4i}{n} +1\right) = \frac{2}{n}\left( \sum_{i=1}^{n} \frac{4i}{n} + \sum_{i=1}^{n} 1\right) = \frac{8}{n^2}\sum_{i=1}^{n} i + \frac{2}{n}\cdot n = \frac{8}{n^2}\cdot \frac{n(n+1)}{2} + 2 = 4 \left(1+\frac{1}{n} \right) + 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.
Find $a\in\mathbb{R}$ such that the image of $f(x)=\frac {x^2+ax+1}{x^2+x+1}$ is included in $[0,2]$.
My attempt:
We have: $f'(x)=-\frac{(a-1)(x^2-1)}{(x^2+x+1)^2}\implies x = 1$ and $x = -1$ points of extrema.
then ... | You're doing well. The derivative is
$$
f'(x)=-\frac{(a-1)(x^2-1)}{(x^2+x+1)^2}=(1-a)\frac{(x^2-1)}{(x^2+x+1)^2}
$$
Leaving aside $1-a$, the fraction is negative for $|x|<1$.
If $1-a>0$, the function has a maximum at $-1$ and a minimum at $1$. The situation is reversed for $1-a<0$. For $a=1$ the function is constant an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2783982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Suppose $a,b,c\geq 0$ and $ab+bc+ca=1$, prove that $a\sqrt{a^2+1}+b\sqrt{b^2+1}+c\sqrt{c^2+1}\geq 2$. Suppose $a,b,c\geq 0$ and $ab+bc+ca=1$, prove that
$$a\sqrt{a^2+1}+b\sqrt{b^2+1}+c\sqrt{c^2+1}\geq 2.$$
In my opinion, I do this problem by th following:
$$\sum_{cyc}a\sqrt{a^2+1}=\sum_{cyc}a\sqrt{(a+b)(c+a)},$$
but I... | I think your idea works, let's proceed:
Introduce the following substitution: $$a + b = u^2$$
$$b + c = v^2$$
$$c + a = w^2$$
Then we have:
$$2a = u^2 + w^2 - v^2$$
$$2b = v^2 + u^2 - w^2$$
$$2c = w^2 + v^2 - u^2$$
$ab + bc + ca = 1$ becomes:
$$4 = 2u^2v^2 + 2v^2w^2 + 2w^2u^2 - u^4 - v^4 - w^4$$
So the inequality becom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2785890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$. Show that $\binom{2n}{n}\geq2^n$ for all $n\in\mathbb{N}_0$.
First, observe that $\binom{2n}{n} = \frac{(2n)!}{n!(2n-n)!}=\frac{(2n)!}{(n!)^2}=\frac{1\cdot 2\cdot 3\cdot...\cdot 2n-1\cdot 2n}{(n!)^2}=\frac{(1\cdot 3\cdot 5\cdot ... \cdot 2n-1)\cdot(2\cdot 4\... | From this identity and inequality, the proof becomes obvious
$$
\binom{2n}n = \sum_{k=0}^n {\binom nk}^2\ge \sum_{k=0}^n \binom nk.
$$
This identity is by comparison of coefficients of $x^n$ in
$$
(1+x)^{2n}=(1+x)^n (1+x)^n.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2787651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
This function has two different integrals? $f(x)=∫\frac{1}{x^2}dx$
Integrating by u-substitution:
$u=x^2$
$du=2dx$
$\frac{1}{2}du = dx$
$∫\frac{1}{x^2}dx=$ $∫\frac{1}{u}\times\frac{1}{2}du$
$\frac{1}{2}$∫ $\frac{1}{u}du$
$=\frac{1}{2}ln u+c$
$=\frac{1}{2}ln x^2+c$
$=lnx+c$
Another way:
$∫\frac{1}{x^2}dx=∫x^{-2}dx $
... | Your mistake is in changing the $dx$: it should be $du = 2x \, dx$, or $dx = \frac{1}{2}u^{-1/2} \, du$, which gives
$$ \int \frac{dx}{x^2} = \int \frac{1}{u} \frac{1}{2u^{1/2}} \, du = \frac{1}{2} \int u^{-3/2} \, du = -u^{-1/2}+C = -\frac{1}{x}+C. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2788929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.