Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Rearrangement inequality and minimal value of $\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}$ For $x \in \left(0, \dfrac{\pi}{2}\right)$, is the minimum value of $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x} = 1$? So considering ($\dfrac{1}{\cos x}$, $\dfrac{1}{\sin x}$) and ($\sin^3x$, $\cos^3x$), is it right to ... | Your proof is right.
Also, by C-S we obtain:
$$\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}=\frac{\sin^4x}{\sin{x}\cos x} +\frac{\cos^4x}{\sin x\cos{x}}\geq\frac{(\sin^2x+\cos^2x)^2}{2\sin{x}\cos{x}}=\frac{1}{\sin2x}\geq1.$$
The equality occurs for $x=\frac{\pi}{4},$ which says that $1$ is a minimal value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Evaluate $\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}$. Problem
Evaluate $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}.$$
Solution
Notice that $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^... | $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}=\lim\limits_{t \to 0}\frac{(1+\sqrt{2}t)^{\sqrt{2}}-(1-\sqrt{2}t)^{\sqrt{2}}}{t}\\
=2\sqrt2\lim\limits_{s \to 0}\frac{(1+s)^{\sqrt{2}}-(1-s)^{\sqrt{2}}}{2s}$$
is obviously the derivative of $2\sqrt2r^{\sqrt 2}$ taken at $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2800617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving recurrence relation of this form by iteration for closed form The recurrence relation is of the form:
$$ G(n) = 5G(n-1) +\frac{4^n}{4^2},\quad G(1)=3$$
My answer always differs from that on wolfram Alpha where is my mistake??
My steps are
$ G(n) = 5^2G(n-2) +\frac{5\cdot 4^n}{4^3}+\frac{4^n}{4^2} $
$G(n)= 5^3G(... | Following your approach we have:
$$
\begin{align*}
G(n) &= 5G(n-1) +4^{n-2}=5\left(5G(n-2) +4^{n-3}\right)+4^{n-2}\\&=
5^2G(n-2) +5\cdot 4^{n-3}+4^{n-2}\\
&=5^{n-1}G(1) +5^{n-2}\cdot 4^{0}+\dots+5^1\cdot 4^{n-3}+5^0\cdot4^{n-2}\\
&=5^{n-1}G(1)+5^{n-1}-4^{n-1}=4\cdot 5^{n-1}-4^{n-1}\end{align*},$$
where at the end we us... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2801807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Trouble proving the trigonometric identity $\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$ I have become stuck while solving a trig identity. It is:
$$\frac{1-2\sin(x)}{\sec(x)}=\frac{\cos(3x)}{1+2\sin(x)}$$
I have simplified the left side as far as I can:
\begin{align}
\frac{1-2\sin(x)}{\sec(x)}
&=\frac{1-2\s... | First, in case you don't know how, I'll expand $\cos{3x}$.
$$\require{cancel}\begin{aligned}\cos{3x}&=\cos\left(x+\left(x+x\right)\right)\\&=\cos x\cos\left(x+x\right)-\sin x\sin\left(x+x\right)\\&=\cos{x}\left(\cos^{2}x-\sin^{2}x\right)-\sin{x}\left(2\sin{x}\cos{x}\right)\\&=\cos^3 x-\sin^{2}{x}\cos{x}-2\sin^{2}{x}\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2803617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
The value of $\sum_{1\leq l< m How to solve this summation ? Also, I'm not sure what does $1\leq l< m <n$ supposed to imply in the development of summation form.
$$\sum_{1\leq l< m <n}^{} \frac{1}{5^l3^m2^n}$$
This one is from the Galois-Noether Contest in 2018:
Galois-Contest
| Outline:
You have
$$
\sum_{1\leq \ell<m<n} \frac{1}{5^\ell3^m2^n}
= \sum_{n=1}^\infty\sum_{m=1}^{n-1}\sum_{\ell=1}^{m-1} \frac{1}{5^\ell3^m2^n}
= \sum_{n=1}^\infty\frac{1}{2^n}\sum_{m=1}^{n-1}\frac{1}{3^m}\sum_{\ell=1}^{m-1} \frac{1}{5^\ell} \tag{1}
$$
Since $$
\sum_{\ell=1}^{m-1} \frac{1}{5^\ell} = \frac{1}{5}\cdot\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2806034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Prove $\sec^2\frac{\pi}{7}+\sec^2\frac{2\pi}{7}+\sec^2\frac{3\pi}{7}=24$ using the roots of a polynomial I have to
prove $\sec^2\frac{\pi}{7}+\sec^2\frac{2\pi}{7}+\sec^2\frac{3\pi}{7}=24$ by using the roots of the polynomial $x^3-21x^2+35x-7=0$
I tried to factor the polynomial but it didn't work and later found it ... | let $t=\tan(\theta)$, we have
\begin{eqnarray*}
\tan(7 \theta) =\frac{ 7t-35t^3+21t^5-t^7}{1-21t^2+35t^4-7t^6}.
\end{eqnarray*}
Set $\tan(7 \theta) =0$ then the polynomial
\begin{eqnarray*}
7t-35t^3+21t^5-t^7=0
\end{eqnarray*}
has roots $t=0, \tan( \pi/7), \cdots ,\tan( 6 \pi/7)$. So
\begin{eqnarray*}
x^3-21x^2+35x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2807082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to integrate the product of two or more polynomials raised to some powers, not necessarily integral This question is inspired by my own answer to a question which I tried to answer and got stuck at one point.
The question was:
HI DARLING.
USE MY ATM CARD, TAKE ANY AMOUNT OUT, GO SHOPPING AND TAKE YOUR FRIENDS FO... | Noticing that
$ 3 x^3-x^2+2 x-4=\frac{3}{2}(2 x-3)\left(x^2-3 x+2\right)+\frac{5}{2}\left(5 x^2-7 x+2\right),$
we have $$
I=\frac{3}{2} \underbrace{\int_0^1(2 x-3) \sqrt{x^2-3 x+2} d x}_J+\frac{5}{2} \underbrace{\int_0^1 \frac{5 x^2-7 x+2}{\sqrt{x^2-3 x+2}} d x}_K
$$$$
\begin{aligned}
J =\frac{3}{2} \int_0^1 \sqrt{x^2-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Finding The Zeros Of $\frac{z^2\sin z}{\cos z -1}$ $$f(z)=\frac{z^2\sin z}{\cos z -1}$$
$$\frac{z^2\sin z}{\cos z -1}=0\iff z^2\sin z=0$$ so $z=\pi k$ and $z=0$ are zeros, to find the order we must derive $\frac{z^2\sin z}{\cos z -1}$?
| Since
$\sin(2x)=2\sin(x)\cos(x)$
and
$\cos(2x)=\cos^2(x)-\sin^2(x)
=1-2\sin^2(x)$
so
$\cos(2x)-1
=-2\sin^2(x)
$,
$f(2z)
=\frac{4z^2\sin 2z}{\cos 2z -1}
=\frac{4z^22\sin(z)\cos(z)}{-2\sin^2(z)}
=\frac{-2z^2\cos(z)}{\sin(z)}
=-2\frac{z}{\sin(z)}z\cos(z)
$.
The zeros are
$z=0$
(taking the limit
since it is undefined there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2814928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find minimal polynomial of element Find minimal polynomial of element $3-2\sqrt[3]{2}-\sqrt[3]{4}$ $\in {\displaystyle \mathbb {Q}(\sqrt[3]{2}) } $ over the field $ {\displaystyle \mathbb {Q} } $.
Thank you for any help.
| Here's a systematic way to find the minimal polynomial, which works in general setting. We know that $\{1,\sqrt[3]{2},\sqrt[3]{4}\}$ is a basis of $\mathbb{Q}(\sqrt[3]{2})$ as a vector space over $\mathbb{Q}$. Now consider the following relations:
$$(3-2\sqrt[3]{2}-\sqrt[3]{4})\cdot1 = 3-2\sqrt[3]{2}-\sqrt[3]{4}$$
$$(3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2816857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
solving a linear diophantine equation in two variables I'm a bit confused in applying the Euclidean algorithm to solve the following equation:
$$11x + 34 = 13y + 35$$
$$x,y\in\Bbb Z; \qquad0 < x,y$$
How can I go about solving it, if there are solutions?
| $11x - 13 y = 1$
$$ \gcd( 13, 11 ) = ??? $$
$$ \frac{ 13 }{ 11 } = 1 + \frac{ 2 }{ 11 } $$
$$ \frac{ 11 }{ 2 } = 5 + \frac{ 1 }{ 2 } $$
$$ \frac{ 2 }{ 1 } = 2 + \frac{ 0 }{ 1 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccc}
& & 1 & & 5 & & 2 & \\
\frac{ 0 }{ 1 } & \frac{ 1 }{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2817508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve $\lim_{x\to1}=\frac{x^2+x-2}{1-\sqrt{x}}$? let $f(x)=\dfrac{x^2+x-2}{1-\sqrt{x}}$
How do I solve this limit?
$$\lim_{x\to1}f(x)$$
I can replace the function with its content
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}$$
Then rationalizing the denominator
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}\cdot\dfrac{1... | $$\lim_{x\to1}\dfrac{(x^2+x-2)(1+\sqrt{x})}{1-x}=\\
\lim_{x\to1}(1+\sqrt{x})\lim_{x\to1}\dfrac{(x^2+x-2)}{1-x}\\
=2\lim_{x\to1}\frac{(x+2)(x-1)}{1-x}=-6
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Finding reason for removal of the logarithm in ODE I'm given an ODE
$$y' = \frac{x + 2y + 1}{2x + 3},~y(5/2) = 1/4.$$
To solve this I'm doing the following:
*
*Solve system of equations
$$2x_0 + 2y_0 = 1,$$
$$2x_0 = 3,$$
this gives me solutions $x_0 = \frac{3}{2}$, $y_0 = -\frac{1}{4}$. Then, I'm transforming varia... | Since the initial condition is given for $x=\frac52$ we are assuming the solution for the domain $x>-\frac32\implies \frac{x + 3/2}{4}>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Problem related to probability from Sheldon Ross' book The stumbled upon following problem from Sheldon Ross's book:
Seven balls are randomly withdrawn from an urn that contains 12 red, 16 blue and 18 green balls. What is probability that either exactly 3 red balls or exactly 3 blue balls are withdrawn?
In this pdf, ... | I'd interpret the question as that exactly one of the two events (3 red or 3 blue) occurs but not both ("either or" strongly hints at that). This amounts to $P(3 \text{ red}) + P(3 \text{blue}) - 2P(3 \text{ red and } 3 \text{ blue })$ which is
$${{\binom{12}{3}\binom{16+18}{4}}\over{\binom{12+16+18}{7}}} + {{\binom{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Area of Shaded Region Inside Quadrilateral I recently saw a similar problem online but found the area in a completely different way.
Problem:
There is a unit square, i.e. a square with side length equal to $\text 1$. Two lines are placed inside the square. Lets call them $\color{red}{\text {Line 1}}$ and $\color{blue}{... | There is a simpler approach. Consider an extension of the diagram you drew:
Note that the desired area "A" is the sum of the areas of the circular sector and the obtuse triangle. As shown in the diagram, the angle subtended by an arc from the vertex on the circumference is half of the angle measured from the center ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2821290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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How many triangles have sides $a$, $a+1$, $a+2$ and angles $x$ and $2x$?
How many triangles can I draw if its arms are $a$, $a+1$, and $a+2$, and two angles are $x$ and $2x$?
I think there are only two solutions to this problem. A triangle with arms $1$, $2$, $3$ and $2$, $3$, $4$. But I am not sure.
| This can be done with $a = 1, \frac{1}{2}(1 + \sqrt{13})$ and $4$ only. Not for $a = 2$. It's a simple matter of putting a set of equations into a symbolic calculator and letting it chew through them.
If the angles are $x, y, z$. Then the Cosine Law says, for instance,
$$ a^2 = (a+1)^2 + (a+2)^2 - 2(a+1)(a+2) \cos x $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Doubt in change of basis matrix Let $A=\{(1,0,5) (4, 5,5)(1,1,4)\}; B= \{(1,3,2)(-2,-1,1)(1,2,3) \}$. To find change of basis matrix.
$\boxed{ M_{A\to B} = M_{A\to e}M_{e\to B} } ---(1)\\= \begin{pmatrix}
1 & 4 & 1 \\
0 & 5 & 1 \\
5 & 5 & 4
\end{pmatrix}^{-1} \begin {pmatrix}
1 & -2 & 1 \\
3 & -1 & 2 \\
2 & 1 & 3
\... | We need
$$M_{A\to B} = M_{e\to B}M_{A\to e}$$
that is
$$\begin {pmatrix}
1 & -2 & 1 \\
3 & -1 & 2 \\
2 & 1 & 3
\end {pmatrix}^{-1}\begin{pmatrix}
1 & 4 & 1 \\
0 & 5 & 1 \\
5 & 5 & 4
\end{pmatrix} $$
indeed the first matrix takes a vector in basis $A$ to the standard basis and the second takes a vector in the standar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2825886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show that all eigenvalues of this pentadiagonal matrix are double degenerate I am trying to show in general that the following pentadiagonal matrix $\mathbf{M}$ has double degenerate eigenvalues,
\begin{equation}
\mathbf{M} =
\left[ \begin{array}{cccccccc}
4 & 0 & a & 0 & 0 & 0 & 0 & 0 \\
0 & 3 & 0 & a & 0 & ... | Shortly speaking, such a matrix can be essentially split into two submatrices formed by odd and even rows and columns, and those two matrices are obtained from one another by conjugating by $\mathbf{J}$. To be more detailed: consider the matrix
\begin{equation}
\mathbf{S} =
\left[ \begin{array}{cccccccc}
1 & 0 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving $\cos(3x) = \cos(2x)$ I'm struggling with solving given trigonometric equation
$$\cos(3x) = \cos(2x)$$
Let's take a look at the trigonometric identities we can use:
$$\cos(2x) = 2\cos^2-1$$
and
$$\cos(3x) = 4\cos^3(x) -3\cos(x)$$
Plugging into the equation and we have that
$$4\cos^3(x) -3\cos(x) = 2\cos^2(x)-1... | Hint
You can use sum-product equivalence. Which is:
$$\cos(A)-\cos(B)=-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$$
so,
$$\cos(3x)-\cos(2x)=0\to-2\sin\left(\frac{3x+2x}{2}\right)\sin\left(\frac{3x-2x}{2}\right)=0$$
$$\sin\left(\frac{5x}{2}\right)\sin\left(\frac{x}{2}\right)=0$$
so,
$$\sin\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 1
} |
Finding limit using Euler number $$\lim_{x\rightarrow \infty }\left ( 1+\frac{3}{x+2} \right )^{3x-6}$$
I've tried to factor and simplfy the expression. I got:
$${\left ( 1+\frac{3}{x+2} \right )^{\frac{1}{x+2}}}^{3({x^2-4})}$$
I set $x$ to $1/t$ I get:
$${\left ( 1+\frac{3}{\frac{1}{t}+2} \right )^{\frac{1}{\frac{1}{t... | Let $y = \left(1+\dfrac{3}{x+2}\right)^{3x-6} $. Then
$$\ln y=(3x-6)\ln\left(1+\dfrac{3}{x+2}\right)\text{.}$$
On the left-hand side, by continuity of $\ln$,
$$\lim_{x \to \infty}\ln y = \ln \left(\lim_{x \to \infty}y\right)\text{.}$$
On the right-hand side,
$$\lim_{x \to \infty}(3x-6)\ln[1+3/(x+2)]=\lim_{x \to \inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2826327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Showing that $4b^2+4b = a^2+a$ has no non-zero integer solutions? The problem is
Show that $4b^2+4b = a^2+a$ has no integer solutions where none of $a, b$ are zero.
I have a solution but I think there must be some better ways:
My Solution:
$$4b^2+4b = a^2+a$$
$$(2b+a)(2b-a)+4b-a= 0$$
Now letting $x = 2b + a$ and $y =... | Hmmm...
Show that $4b^2 + 4b = a^2 + a$ has no integer solutions where none of $x, y$ are zero.
Where do $x$ and $y$ come from? Is the following what was meant?
Show that $4b^2 + 4b = a^2 + a$ has no integer solutions where none of $a, b$ are zero.
I'll proceed along the lines of the latter. Clearly $a$ must be eve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2831439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Prove that $m^2(r^2-a^2)+2am(b-c)+2bc-b^2-c^2+r^2<0$ If the circle $(x-a)^2+(y-b)^2=r^2$ and the line $y=mx+c$ do not meet:
Prove that $m^2(r^2-a^2)+2am(b-c)+2bc-b^2-c^2+r^2<0$
These are the steps I have thus taken (although they may be wrong/useless):
*
*Rearranged the circle equation for y:
$y=\sqrt{r^2-x^2+2ax-a^... | Substitute the equation for the line into the equation for the circle
\begin{eqnarray*}
(x-a)^2+(mx+c-b)^2= r^2 \\
(m^2+1)x^2 -2x(a+m(b-c))+a^2+(c-b)^2-r^2=0.
\end{eqnarray*}
This is a quadratic in $x$ and for no solutions to exist we require the disciminant to be negative so
\begin{eqnarray*}
(a+m(b-c))^2-(m^2+1)(a^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2831778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
For the differential equation given by $(x^2-y^2)dx+3xydy=0$ the general solution is $(x^2+2y^2)^3 =cx^2$
For the differential equation given by $$(x^2-y^2)dx+3xy \ dy=0$$ the general soultion is $$(x^2+2y^2)^3 =cx^2$$ Check this solution by differentiating and rewriting to get the original function.
I solved for $\f... | Calling
$$
f(x,y) = (x^2 + 2 y^2)^3 -c x^2 = 0
$$
we have
$$
df=f_x dx+f_y dy = 0\to y' = -\frac{f_x}{f_y} = \frac{x \left(c-3 \left(x^2+2 y^2\right)^2\right)}{6 y \left(x^2+2 y^2\right)^2}
$$
now substituting $c = \frac{\left(x^2+2 y^2\right)^3}{x^2}$ and simplifying we obtain
$$
y' = \frac{y}{3 x}-\frac{x}{3 y}
$$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2832007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Maximizing by setting Derivative equal to 0. Stuck (struggling) I'm having immense difficulty doing this question.
Want to maximize (respect to x):
$f(x)$=$(x-x_0)(x-x_1)(x-x_2)(x-x_3)$ where
*
*$x_1-x_0=h$
*$x_2-x_1=h$
*$x_3-x_2=h$.
Want to get the value of x in terms of $x_1$ and $h$.
First I take the derivativ... | let $a = \frac {x_1+x_2}{2}$ then
$x_1 = a - \frac {h}{2}\\
x_0 = a - 3\frac {h}{2}\\
x_2 = a + \frac {h}{2}\\
x_3 = a + 3\frac {h}{2}$
This will take advantage of some of the symmetry of our fuction
$f(x) = (x-a - \frac {3h}{2})(x-a + \frac {3h}{2})(x-a - \frac {h}{2})(x-a + \frac {h}{2})\\
((x-a)^2 - \frac {9h^2}{4})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2833216",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Doubt with an inequality involving integrals I have to prove this inequality about integrals. I did it but I'm not sure if my arguments were correct. Take a look:
Prove:
$$0\leq\int_{-1}^1\frac{1-x^2}{x^4+1}dx\leq\int_{-1}^1\frac{2+x^4}{x^4+1}dx$$
My try is this:
Let $f(x)=\frac{1-x^2}{x^4+1}$ with $f:[-1,1]\to\mat... | Alt. hint: $\displaystyle\;\;\int_{-1}^1\frac{-x^2}{x^4+1}dx \lt 0 \lt \int_{-1}^1\frac{x^4}{x^4+1}dx\,$, and $\displaystyle\;\int_{-1}^1\frac{1}{x^4+1}dx \lt \int_{-1}^1\frac{2}{x^4+1}dx\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using algebraic identities to prove a number is not prime $$3^{3^n}(3^{3^n}+1)+3^{3^n +1}-1$$
I want the prove that the number is not prime. I used the identity $a^3+b^3+c^3-3abc $. I couldn't simplify to the state where the factors could be observed.
| The other approach does show a quite large factor of the number; good to know how to approach this.
The expression $$ 3^{3^n} 3^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 $$
first becomes
$$ 9^{3^n} + 3^{3^n} + 3^{3^n + 1} - 1 $$
The first two terms really are cubes, since the exponents are divisible by 3. If we look for $a^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2836683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determine Minimum Value.
Find the minimum value of
$$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}$$
for $x>0$.
When $x=1$, $$\frac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=6$$
I tried to plot some points on a graph and I observed that the minimum value is $6$.
Any hints would be sufficient.
Thanks... | $$\dfrac{(x+1/x)^6-(x^6+1/x^6)-2}{(x+1/x)^3+(x^3+1/x^3)}=\dfrac{(x+1/x)^6-(x^3+1/x^3)^2}{(x+1/x)^3+(x^3+1/x^3)}=(x+1/x)^3-(x^3+1/x^3)$$ Therefore your function reduces to $3(x+1/x)$ which has its minimum when $x=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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expected sum of the numbers that appear on two dice, each biased so that a 3 comes up twice as often as each other number? This question is already posted here,but i want to check my approach.
Question
What is the expected sum of the numbers that appear on
two dice, each biased so that a $3$ comes up twice as often
as... | Your approach is correct. To save a bit of the arithmetic and the detour through calculating the probabilities, you could also argue that the problem statement amounts to the result being uniformly randomly drawn from $\{1,2,3,3,4,5,6\}$, so the expected result for one die is the mean of that set,
$$
\frac{1+2+3+3+4+5+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2839494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Minimum $e$ where $a,b,c,d,e$ are reals such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$ I have a question about this 1978 USAMO problem:
Given that $a,b,c,d,e$ are real numbers such that $a+b+c+d+e=8$ and $a^2+b^2+c^2+d^2+e^2=16$, find the maximum value that $e$ can attain.
I had the following solution:
Let $a+... | Intuitive approach:
For a given $$(a+b+c+d=k)$$
We want
$$Min(a^2+b^2+c^2+d^2)$$
In order to maximize $e$
This is true only when $a=b=c=d$
From that we have
$$4x+e=8,\; 4x^2+e^2=16$$
Giving $x=\frac{6}{5}$ and $e=\frac{16}{5}$
Leading to the maximum value
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
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Probability for repeated incidents . I had problems on understanding the following problem :
A die is continuously thrown until 4 is obtained. What is the probability that the dice is thrown an even number of times?
Here , a solution states that probability is :
$$\frac{4}{6}\times \frac{2}{6}+\frac{4}{6}\times \fra... | *
*See, that the probabilty of throwing $4$ in $k$-th throw ($k\geq 1$)is equal to:
$$p(k)=\frac{1}{6}\left(\frac{5}{6}\right)^{k-1}$$
So we have to sum up all the probabilities for even $k$:
$$P=\sum_{n=1}^{\infty}p(2n)=\sum_{n=1}^{\infty}\frac{5}{6}\left(\frac{5}{36}\right)^{2n-2}=\sum_{n=0}^{\infty}\frac{5}{36}\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2841799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate integral $\int_0^\pi \sin^4\left(x+\sin 3x\right)dx$ Evaluate $\int_0^\pi \sin^4\left(x+\sin 3x\right)dx$
My working let I=$\int_0^\pi \sin^4\left(x+\sin 3x\right)dx$
$=\int_0^\pi \frac18\left(\cos (4x+4\sin 3x)-4\cos(2x+2\sin 3x)+3\right)dx$
$=\frac{3\pi}{8}+\frac18\int_0^\pi\cos (4x+4\sin 3x)dx-\frac12\int_0... | Before we start, let us look at a related family of integrals.
For any integer $n$ and $\lambda \in \mathbb{R}$, let $J_n(\lambda)$ be the integral
$$J_n(\lambda) \stackrel{def}{=} \int_{-\pi}^{\pi} e^{in(x+\lambda\sin(3x))} dx$$
It is easy to see $J_0(\lambda) = 2\pi$ independent of $\lambda$. Furthermore, $J_n(\lambd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842118",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
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If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ are $7$ and $1/7$ , then the value of $c$ is?
I cross multiplied the equation and tried to find it's discriminant, but I don't think it gets me anywhere. A little hint would be appre... | Hint: suppose $\,c, x \gt 0\,$, then $\,u = x + \dfrac{c}{x} \ge 2 \sqrt{c}\,$ by AM-GM with equality iff $\,x = \sqrt{c}\,$, and:
$$
f(x)=\frac{x^2-3x+c}{x^2+3x+c} \color{red}{\cdot \frac{\;\;\cfrac{1}{x}\;\;}{\cfrac{1}{x}}}=\dfrac{x+ \cfrac{c}{x}-3}{x+ \cfrac{c}{x}+3} = \dfrac{u-3}{u+3}=1 - \cfrac{6}{u+3} \ge 1 - \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2842481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Solving system of polynomial equations stemming from a constrained optimization problem Constrained optimization (maximization and minimization) problem with two equality constraints.
$$\begin{array}{ll} \text{extremize} & x y z\\ \text{subject to} & x^2 + y^2 + z^2 = 9\\ & x + y - z = 3\end{array}$$
The Lagrangian, $\... | You can reduce the problem in the beginning to:
$$\text{optimize} \ xy(x+y-3) \ \text{subject to} \\
\ x^2+y^2+(x+y-3)^2=9.$$
Hence:
$$L(x,y,\lambda)=x^2y+xy^2-3xy+\lambda(9-x^2-y^2-(x+y-3)^2).$$
FOC:
$$\begin{cases}
L_x=2xy+y^2-3y-2x\lambda-2\lambda(x+y-3)=0\\
L_y=x^2+2xy-3x-2y\lambda-2\lambda(x+y-3)=0\\
L_{\lambda}=9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2843277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$a+b+c=1.$ Show that $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$
Assume $a,b,c>0$ and $a+b+c=1.$ Show that
$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq
\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$
Here's what I tried:
$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+b}+\frac{a+b+c}... | Hint
Note $t\mapsto \frac1t$ is convex and $(1-a,1-b,1-c)\succ (\frac12+\frac a2, \frac12+\frac b2, \frac12+\frac c2)$, so it’s Karamata’s Inequality.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find rational numbers $a,b,c$ satisfying $(2^{1/3}-1)^{1/3} = a^{1/3}+b^{1/3}+c^{1/3}$.
$\textbf{Problem} $ Find rational numbers $a,b,c$ satisfying
\begin{align*}
(2^\frac13-1)^\frac13 = a^\frac13+b^\frac13+c^\frac13
\end{align*}
My Attempt: I try to $2^\frac13-1 = (a^\frac13+b^\frac13+c^\frac13)^3$ and compare wi... | Set $t:=\sqrt[3]{2}$. Then, $t^3=2$ and $t^3-1=1$, whence
$$t-1=\frac{t^3-1}{t^2+t+1}=\frac{1}{t^2+t+1}=\frac{3}{3t^2+3t+3}=\frac{3}{t^3+3t^2+3t+1}=\frac{3}{(t+1)^3}\,.$$
Therefore,
$$\sqrt[3]{t-1}=\frac{\sqrt[3]{3}}{t+1}\,.$$
Now, $t^3+1=3$, so
$$\sqrt[3]{t-1}=\frac{\sqrt[3]{3}\left(t^2-t+1\right)}{t^3+1}=\frac{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2847624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Asymptotes of an implicit curve Per the method described at How to find asymptotes of implicit function? , I proceeded to find the asymptotes of
$$
x^3 + 3x^2y - 4y^3 - x + y + 3 = 0
$$
Whilst, it correctly generates the asymptote $$ y=x $$ , the remaining asymptote(s):-
$$ 2y + x = ±1 $$ can't be deduced.
Instead ano... | I'd try to give another answer. Recall from wikipedia and the most common definition of asymptotic line, to find an asymptotic line of $f(x)$ as $x\to\infty$, we need to compute
(1) $a=\lim_{x\to\infty}\frac{f(x)}{x}$
(2) $b=\lim_{x\to\infty}(f(x) - ax)$.
Then the asymptotic line would be $y=ax+b$.
-
By implicit fu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
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Minimizing in 3 variables
Find the least possible value of the fraction $\dfrac{a^2+b^2+c^2}{ab+bc}$, where $a,b,c > 0$.
My try:
$a^2+b^2+c^2 = (a+c)^2 - 2ac + b^2$,
$= (a+c)/b +b/(a+c) -2ac/b(a+c)$
AM > GM
$3\sqrt[3]{-2ac/b(a+c)}$
And I cant somehow move on.
| You can do the following:
$$\frac{a^2+b^2+c^2}{b(a+c)}\geq \frac{\frac{1}{2}(a+c)^2+b^2}{b(a+c)}=\frac{1}{2}\frac{a+c}{b}+\frac{b}{a+c}$$
where at the first step we used $2(a^2+c^2)\geq (a+c)^2$
If you now set $\frac{a+c}{b}=x$ then you just have to minimize $\frac{x}{2}+\frac{1}{x}$ over the positive real numbers.
But... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2853202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Inverse of a circulant matrix with a specific pattern I'm trying to invert the following circulant matrix:
$$\begin{bmatrix}1 & -1/4 & 0&0 &0&\cdots&0&-1/4\\ -1/4 & 1 & -1/4 & 0&0&\cdots&0&0\\0 & -1/4 & 1 & -1/4&0&\cdots&0&0\\\vdots&\vdots&\vdots&\vdots&\vdots&\ddots&\vdots&\vdots
\\-1/4 & 0 & 0 & 0 & 0&\cdots&-1/4 & ... | The polynomial $g$ is defined as $g(z) = a_0 + a_1 z + \cdots + a_s x^s$. Thus, $s=-1$.
Let $B=A^{-1}$:
\begin{align*}
&B= \begin{bmatrix}b_0 & b_1 & b_2&b_3 &b_4&\cdots &b_{n-2}&b_{n-1}\\ b_{n-1} & b_0 & b_1 & b_2&b_3&\cdots & b_{n-3}&b_{n-2}\\b_{n-2} & b_{n-1} & b_0 & b_1&b_2&\cdots&b_{n-4}&b_{n-3}\\\vdots&\vdots&\v... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2855429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the area of the figure that is bound by the lines: $y=|x^2-1|$ and $y=3|x|-3$?
Find the area of the figure that is bound by the lines:
$y=|x^2-1|$ and $y=3|x|-3$
I tried: $$A=\int_{-1}^{1}(|x^2-1|-3|x|+3)dx$$
Is it correct? Please, help for the right solution.
I dont know, how can I evaluate $\int|x|dx.$
| You should draw the graphs of the two given curves carefully before solving this type of problems, and to do that you have two find the points at which they intersect each other. Let's do that
First write the equation of two given curves explicitly-
$
y=|x^2-1|=
\begin{cases}
x^2-1, & \text{if $x\in(-\infty,-1]\cup[1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding prime factors of large expression (without calculator) Show that: $5^3(5^3(253)+3)+1 = 19 \times 251 \times 829$.
I tried setting $n=5$, so that $253 = 23 \times 11 = (4n+3)(2n+1)$ and going from there, but the resulting polynomial in $n$ was $8n^6 + 10n^7 + 3n^6 + 3n^3 + 1$, which turns out to be irreducible o... | My eye is drawn to the $253$ in the expression being close to the factor $251$ and the fact that $5^3$ is very close to half of $251$. Given the known result I would write
$$5^3(5^3(253)+3)+1=5^3(5^3\cdot (251+2)+3)+1\\=5^6\cdot 251+5^3\cdot253+1\\=(5^6+5^3+1)251$$
then just compute $5^6+5^3+1=15625+125+1=15751$ and d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Calculate $\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$
Calculate:$$\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n}$$
Even though I know how to handle limits like this, I would be interested in other ways to approach to tasks similar to this. My own solution will be at the bottom.
My own solution
$$\lim_{n\to\i... | Considering that $\frac{4}{n^2}<< \frac{1}{n}$ for big $n$ we have
$$
\lim_{n\to\infty} \sqrt{5n^2+4}~-~\sqrt{5n^2+n} \equiv \lim_{h\to 0}\frac{\sqrt{5+4h^2}-\sqrt{5+h}}{h} = \lim_{h\to 0}\frac{\sqrt{5+o(h^2)}-\sqrt{5+h}}{h}
$$
now remember the derivative definition
$$
\lim_{h\to 0}\frac{\sqrt{x+h}-\sqrt x}{h} = \frac ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2859281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Define a sequence $(a_n)$ as follows $a_1 = 1$ and for $n\geq 1,\;a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$ Define a sequence $(a_n)$ as follows $a_1 = 1$ and for $n\geq 1$
$$a_{n+1} = (-1)^n\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)$$
then what will be it's $\limsup (a_n)$ and $\limin... | I think the following will help the readers to understand the solution:
We know $$\frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\ge\sqrt{|a_n| \times \frac {2}{|a_n|}}=\sqrt 2$$
$$\implies \frac{1}{2} \left(|a_n| + \frac {2}{|a_n|}\right)\ge\sqrt 2$$
Given that $~a_1 = 1~$ and for $n\geq 1$
$$a_{n+1} = (-1)^n\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve $\sqrt{49-x^2}-\sqrt{25-x^2}=3$? I recognize the two difference of squares: $49-x^2$ and $25-x^2$.
I squared the equation to get:
${49-x^2}-2(\sqrt{(49-x^2)(25-x^2)})+{25-x^2}=9$
However, I can't quite figure out how to remove the root in the middle. Any help is appreciated.
| Set $\;\;t^2=25-x^2,\;$ the equation becomes $$\begin{align}\sqrt{24+t^2}-\sqrt{t^2}&=3\\ \sqrt{24+t^2}&=|t|+3\\24+t^2&=t^2+6|t|+9\\|t|&=\frac{5}{2}
\end{align}$$
Then $x^2=25-\frac{25}{4}=3\cdot \frac{25}{4},$ from where $x=\pm \frac{5}{2}\sqrt 3.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2866088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
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Maximum minus minimum of $c$ where $a+b+c=2$ and $a^2+b^2+c^2=12$ Let $a,b,$ and $c$ be real numbers such that
$a+b+c=2 \text{ and } a^2+b^2+c^2=12.$
What is the difference between the maximum and minimum possible values of $c$?
$\text{(A) }2\qquad \text{ (B) }\frac{10}{3}\qquad \text{ (C) }4 \qquad \text{ (D) }\frac{... | Solution
Notice that $$a+b=2-c\tag1,$$and $$a^2+b^2=12-c^2.\tag2$$
Since $$ 2ab \leq a^2+b^2,$$Hence$$(a+b)^2=a^2+b^2+2ab \leq 2(a^2+b^2).\tag3$$Put $(1),(2)$ into $(3)$. We obtain $$(2-c)^2 \leq 2(12-c^2),$$namely $$3c^2-4c-20=(c+2)(3c-10) \leq 0.$$
As a result, $$-2\leq c \leq \frac{10}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2867661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 5
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Finding 4-up numbers From Mathcounts Target 2018:
*A 4-up number is defined as a positive integer that is divisible by neither 2 nor 3 and does not have 2 or 3 as any of its digits. How many numbers from 400 - 600 inclusive are 4-up numbers?
So I tried to find the complement, which is a number that is either d... | HINT.-You have to find the numbers of three digits $abc$ with the restrictions$$\begin{cases}a=4,5\\b=0,1,4,5,6,7,8,9\\c=1,5,7,9\end{cases}$$ Then you have in a first step $2\cdot8\cdot4=64$ possibilities from which as second and final step you have to discard those such that $a+b+c\equiv 0\pmod3$. For this you can cal... | {
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"url": "https://math.stackexchange.com/questions/2868540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Computing the determinant of a $4\times4$ matrix
Compute the determinant of \begin{vmatrix} 2 & -1 & 1 & 3 \\ 1 & 2 & 0 & 1 \\ 2 & 1 & 1 & -1 \\ -1 & 1 & 2 & -2 \end{vmatrix}
My try:
$$\begin{vmatrix} 2 & -1 & 1 & 3 \\ 1 & 2 & 0 & 1 \\ 2 & 1 & 1 & -1 \\ -1 & 1 & 2 & -2 \end{vmatrix}_{R_2\rightarrow2R_2-R_1\\R_3\right... | You multiplied two rows by two, so your answer (184) is 4 times greater than the actual determinant. This video may help.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2870183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Ratio of angles in a triangle came across this question in a math contest and can't quite figure out an approach to the question.
Triangle $ABC$ has $AB=AC\neq BC$ and $∠BAC ≤ 90º$. $P$ lies on $AC$, and $Q$ lies on $AB$ such that $AP = PQ = QB = B$C. Find the ratio of $∠ACB$ to $∠APQ$.
I have tried to establish some s... | Let us set $AP=PQ=QB=BC=1$, the unit.
Let us denote by $x$ the angle in $A$. Then the sine theorem in $\Delta APQ$ gives
$$
AQ = AP\cdot\frac{\sin(180^\circ-2x)}{\sin x}
=AP\cdot\frac {\sin(2x)}{\sin x}
=AP\cdot 2\cos x
=2\cos x
\ .
$$
The sides of $\Delta ABC$ are thus $BC=1$ and $1+2\cos x$, and the last again. The s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
I tried really hard but the most I could get i... | Alternatively, the Vieta's formula's for $x^2 + 3x -1 = 0$:
$$x_1+x_2=-3, x_1x_2=-1 \Rightarrow x_1^2+x_2^2=(x_1+x_2)^2-2x_1x_2=9+2=11.$$
Plugging $x_1,x_2$ to $x^4 + ax^2 + bx + c = 0$ and subtracting:
$$\begin{cases}x_1^4 + ax_1^2 + bx_1 + c = 0\\ x_2^4 + ax_2^2 + bx_2 + c = 0\end{cases} \Rightarrow \\
(x_1-x_2)(x_1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2873902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
$Z$ ~ $N(0, 1)$. Find $\operatorname{Var}(Z)$ $Z$ ~ $N(0, 1)$. Find $\operatorname{Var}(Z)$
We know:
$E(Z) = 0$
$\operatorname{Var}(Z) = E(Z^2) - E(Z)^2 = E(Z^2) - 0^2 = E(Z^2)$
Thus, we need:
$E(Z^2) = \int_{-\infty}^{\infty} z^2 \frac{1}{\sqrt{2\pi}}e^{-1/2(z^2)}dz$
Using int by parts, $u = z, du = dz$, then $dv = z ... | The standard method for evaluating $\int_{-\infty}^{\infty}e^{-\frac{1}{2}z^2}dz$ isn't an obvious one, but one worth remembering. It involves squaring it and turning it into a double integral.
First note that
$$\int_{-\infty}^{\infty}e^{-\frac{1}{2}z^2}dz=2\int_{0}^{\infty}e^{-\frac{1}{2}z^2}dz$$
as $e^{-\frac{1}{2}z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the quotient and the remainder of $(n^6-7)/(n^2+1)$
Given that $n$ belong to $\mathbb{N}$.
Find the quotent and the remainder of $(n^6-7)/(n^2+1)$.
So I tried to divide them up and got a negative expression $(-n^4-7)$.
How to continue?
Or what can be done differently?
How to find the quotent and the remainder?
| $$ \left( x^{6} - 7 \right) $$
$$ \left( x^{2} + 1 \right) $$
$$ \left( x^{6} - 7 \right) = \left( x^{2} + 1 \right) \cdot \color{magenta}{ \left( x^{4} - x^{2} + 1 \right) } + \left( -8 \right) $$
$$ \left( x^{2} + 1 \right) = \left( -8 \right) \cdot \color{magenta}{ \left( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Finding the dimension of subspace $S=\{ \left[\begin{smallmatrix} a & b\\ c & d\\ \end{smallmatrix}\right] \mid c = a + b, d=a \}.$
Let the subspace $S$ from $M_{2\times2}$: $$S=\left\{ \begin{bmatrix}
a & b\\
c & d\\
\end{bmatrix} \mid c = a + b\textrm{ and }d=a \right\}.$$ What is the dimension of $S$?
According to... | The subspace $S$ consist of matrices of the form
$
\begin{pmatrix}
a & b \\
a+b& a \\
\end{pmatrix}
$. Every such matrix can be written as a linear combination
$$
\begin{pmatrix}
a & b \\
a+b& a \\
\end{pmatrix}=
a\begin{pmatrix}
1 & 0 \\
1 & 1 \\
\end{pmatrix}+
b\begin{pmatrix}
0 & 1 \\
1 & 0 \\
\end{pma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate: $u =\int_0^\infty\frac{dx}{x^4 +7x^2+1}$ Evaluate:
$u =\displaystyle\int_0^\infty\dfrac{dx}{x^4 +7x^2+1}$
Attempt:
$$u = \int_0^\infty \dfrac{dx}{\left(x^2+ \left(\dfrac{7 - \sqrt {45}}{2}\right)\right)\left(x^2+ \left(\dfrac{7 + \sqrt {45}}{2}\right)\right)}$$
$$u = \int_0^\infty \dfrac{dx}{(x^2+a^2)(x^2... | $$\dfrac{7 - \sqrt {45}}{2}=\left(\dfrac{3 - \sqrt {5}}{2}\right)^2$$ and ...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
How to decompose $X_p = X_r +X_n$? From an original matrix $A$, I have solved for $X_p$, $C(A^T)$, and $N(A)$. The last step is the problem below.
How should I approach this? Thank you for taking the time.
Decompose $X_p = X_r + X_n$, where $X_r \in C(A^T)$ and $X_n \in N(A)$.
$$
X_p = \begin{pmatrix}
3 \\
... | Since I am breaking my Xp into orthogonal projections, I would just need to solve for my scalars at this point (X below).AX=B solving for X. So my final solution would be my C(AT) and N(A) multiplied by my scalar (x).
A = \begin{pmatrix}
16,5,9,-1 \\
2,11,7,-3 \\
3,10,6,3 \\
13,8,12,1 \\
\end{pmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2880876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Intersection of 3 planes along a line I have three planes:
\begin{align*}
\pi_1: x+y+z&=2\\
\pi_2: x+ay+2z&=3\\
\pi_3: x+a^2y+4z&=3+a
\end{align*}
I want to determine a such that the three planes intersect along a line. I do this by setting up the system of equations:
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
x+ay+2z&... | You mean subtract $(a+1)$ times the second row from the third row.
After that the system become
$$x+y+z=2$$
$$(a-1)y+z=1$$
$$(2-a)z=0$$
If $a=2$, then we have $y+z=1$ and $x=1$ which is a line.
If $a \ne 2$, then $z=0$, hence we have $(a-1)y=1$ and $x+y=2$, to be consistent, clearly $a \ne 1$, and we can solve for $y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve $ \binom{a}{2} + \binom{b}{2} = \binom{c}{2} $ with $a,b,c \in \mathbb{Z}$ I am trying to solve the Diophantine equation:
$$ \binom{a}{2} + \binom{b}{2} = \binom{c}{2} $$
Here's what it looks like if you expand, it's variant of the Pythagorean triples:
$$ a \times (a-1) + b \times (b-1) = c \times (c-1) $$
I was ... | $$ (2a-1)^2 + ( 2b-1)^2 = 1 + (2c-1)^2 $$
Taking any $c,$
if the right hand side is not twice a prime, there will be nontrivial $a,b,$ while factoring the right hand side gives a method for finding them other than raw search. So that is one approach.
Parameterizing all uses the modular group $SL_2 \mathbb Z$: take in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883759",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 1
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Express $z = \dfrac{3i}{\sqrt{2-i}} +1$ in the form $a + bi$, where $a, b \in\Bbb R$.
Express $$z = \frac{3i}{\sqrt{2-i}} +1$$ in the form $a + bi$, where $a, b \in\Bbb R$.
I figure for this one I multiply by the conjugate of $\sqrt{2 +1}$? But I’m still struggle to achieve the form $a+bi$.
| You're gonna need to convert $\sqrt {2-i}$ to the form $\alpha + i \beta$.
To do that, convert to polar
$$ 2-i=r(\cos \theta + i \sin \theta)$$
So that
$$\sqrt{2-i}=\sqrt r \cdot (\cos \theta + i \sin \theta)^{1/2} = \sqrt r \cdot \left(\cos \frac{\theta}{2}+i \sin \frac{\theta}{2}\right) $$
Some of the ingredients a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Partial fraction of $\frac{2s+12}{ (s^2 + 5s + 6)(s+1)} $ then inverse transform it
Find the inverse Laplace transform of
$$\mathcal{L}^{-1}\left(\frac{2s+12}{ (s^2 + 5s + 6)(s+1)}\right)$$
I recognise I need to use partial fractions to solve it and that is where I got stuck. Here’s my working,
After factoring the... | $A + B + C = 0\quad[1]$
$ 4A + 3B + 5C = 2\quad[2]$
$3A + 2B + 6C = 12\quad[3]$
$ $
$[2]-3[1]$
$A+2C=2\quad[4]$
$ $
$[3]-2[1]$
$A+4C=12\quad[5]$
$ $
$[5]-[4]$
$2C=10$
$ $
$\therefore C=5$
$\therefore A=-8$
$\therefore B=3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is there a smarter way to differentiate the function $f(x) = \sin^{-1} \frac{2x}{1+x^2}$?
Given $f(x) = \sin^{-1} \frac{2x}{1+x^2}$,
Prove that $$f'(x) = \begin{cases}\phantom{-}\frac{2}{1+x^2},\,|x|<1 \\\\ -\frac{2}{1+x^2},\,|x|>1 \end{cases}$$
Obviously the standard approach would be to use the chain rule and simpl... | Method$\#1:$
By Chain rule,
$$\dfrac{d\arcsin\dfrac{2x}{1+x^2}}{dx}=\dfrac1{\sqrt{1-\left(\dfrac{2x}{1+x^2}\right)^2}}\left(\dfrac2{1+x^2}-\dfrac{4x^2}{(1+x^2)^2}\right)$$
$$=\dfrac{2(1+x^2)(1-x^2)}{|1-x^2|(1+x^2)^2}=?$$
Method$\#2:$
Let $\arctan x=y\implies x=\tan y$
Using principal values $$\arcsin\dfrac{2x}{1+x^2}=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve $8\sin x=\frac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$
Solve
$$8\sin x=\dfrac{\sqrt{3}}{\cos x}+\frac{1}{\sin x}$$
My approach is as follow
$8 \sin x-\frac{1}{\sin x}=\frac{\sqrt{3}}{\cos x}$
On squaring we get
$64 \sin^2 x+\frac{1}{\sin^2 x}-16=\frac{3}{\cos^2 x}$
$(64\sin^4 x-16\sin^2 x+1)(1-\sin^2 x)=3 \sin^2 ... | Hint to solve the equation you got
If you make $t=\sin^{2}x$ then you get $-64t^3+80t^2-20t+1=0.$ It has no integer solutions. So we consider $z=1/t.$ Then we have
$$z^3-20z^2+80z-64=0.$$ It easy to see that $4$ is a root. So we have
$$z^3-20z^2+80z-64=(z-4)(z^2-16z+16).$$ Solve the quadratic equation and finally use ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Find the maximum value of $a+b$ The question:
Find the maximum possible value of $a+b$ if $a$ and $b$ are different non-negative real numbers that fulfill
$$a+\sqrt{b} = b + \sqrt{a}$$
Without loss of generality let us assume that $a\gt b$. I rearrange the equation to get $$a - \sqrt{a} = b - \sqrt{b}$$
If $f(x)= x... | For $a\ne b,$
$$\sqrt a+\sqrt b=1$$
WLOG $a=\cos^4t,b=\sin^4t$
$$a+b=1-2\sin^2t\cos^2t=1-\dfrac{\sin^22t}2$$
Now $0\le\sin^22t\le1$
Or $a+b=1-\dfrac{1-\cos4t}4=\dfrac{3+\cos4t}4$
Now $-1\le\cos4t\le1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2889732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
How can I factorize this: "$X^3 + X^2 + X - 3$" I am going to elementary school & I am living in one of those deprived areas of Africa.
I can solve mathematical questions like this:
$$X^3 + X^2 + X +1 = X^2(X+1)+(X+1) = (X+1)(X^2+1)$$
Or even
\begin{align}X^2 − 2X + X^2 - X + 1 &= (X^2 - 2X + 1) + (X^2 - X) \\
&= (X - ... | Surely $1$ is a root of $x^3+x^2+x-3$ therefore $x-1$ is a factor of it. We have $$x^3+x^2+x-3=(x-1)(x^2+x+1+x+1+1)=(x-1)(x^2+2x+3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2890772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve the system $x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$
Find all real numbers $x,\ y,\ z$ that satisfy $$x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$$
First natural move would be rewriting the system as:
$$x^{2}+y^{2}+2xy=4z-1$$
$$x^{2}+z^{2}+2xz=4y-1$$
$$z^{2}+y^{2}+2zy=4x-1$$
Thus,
$$2x^{2}+2... | Solution. $\blacktriangleleft$ Subtract every pair of the equation, we would obtain something like
\begin{align*}
y-z &= (x+y) - (x+z) \\ &= \sqrt {4z-1} - \sqrt {4y-1} \\&= \frac {4z-1 - 4y+1} {\sqrt {4z-1} + \sqrt {4y-1}} \\&= 4\frac {z-y} {\sqrt {4z-1} + \sqrt {4y-1}},
\end{align*}
then $y -z$ must be zero, otherwis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Sum of roots of equation $x^4 - 2x^2 \sin^2(\displaystyle {\pi x}/2) +1 =0$ is My try:
$$x^4-2x^2\sin^2(\frac{\pi x}{2})+1=0\\x^4+1=2x^2\left (1-\cos^2\left(\frac{\pi x}{2}\right)\right )\\(x^2-1)^2=-2x^2\cos^2\left(\frac{\pi x}{2}\right)\\(x^2-1)^2+2x^2\cos^2\left(\frac{\pi x}{2}\right)=0\\x^2-1=0\,\text{and}\, 2x^2\c... | Solving for $x^2$ we have
$$
x^2 = \frac 12\left(2\sin^2\left(\frac{x \pi}{2}\right)\pm\sqrt{4\sin^4\left(\frac{x \pi}{2}\right)-4}\right)
$$
now we know that $-1\le \sin \left(\frac{x \pi}{2}\right)\le 1$ so the only real possible solution is for $x = \pm 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Prove $\ \frac{z-1}{z+1} $ is imaginary no' iff $\ |z| = 1 $ Let $\ z \not = -1$ be a complex number. Prove $\ \frac{z-1}{z+1} $ is imaginary number iff $\ |z| = 1 $
Assuming $\ |z| = 1 \Rightarrow \sqrt{a^2+b^2} = 1 \Rightarrow a^2+b^2 = 1 $ and so
$$\ \frac{z-1}{z+1} = \frac{a+bi-1}{a+bi+1} = \frac{a-1+bi}{a+1+bi} \c... | In the last step of your proof, before you use the assumption that $(a^2+b^2)=1$, you've deduced
$$\frac{z-1}{z+1}=\frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a}$$
Now, by assumption for the other direction, you have
$$\frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a}=xi$$
as $\frac{z-1}{z+1}$ is assumed to be imaginary, i.e. $\frac{z-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
(Should be) simple algebra exercise I have the following coupled equations
\begin{equation}
\tag{1}
a_{1}\left(T-T_{m}\right)+b_{1}M^{2}+cP^{2}=0
\end{equation}
\begin{equation}
\tag{2}
a_{2}\left(T-\theta_{2}\right)+b_{2}P^{2}+cM^{2}=0
\end{equation}
\begin{equation}
\tag{3}
\theta_{2}=\frac{ca_{1}}{a_{2}b_{1}}\left(T... |
$\displaystyle
M^{2}\left(b_{1}b_{2}-c^{2}\right)=a_{2}c\left(T-T_{e}\right)+a_{1}b_{2}\left(T_{m}-T\right)+\frac{a_{1}c^{2}}{b_{1}}\left(T_{e}-T_{m}\right)
$
$$
\begin{align}
M^{2}\left(b_{1}b_{2}-c^{2}\right) &= a_{2}c\left(T-T_{e}\right)+a_{1}b_{2}\left(T_{m} \color{red}{-T_e+T_e} -T\right)+\frac{a_{1}c^{2}}{b_{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Unable to prove that $\sqrt{i} + \sqrt{-i}$ is a real number. I did this :-
$$
\sqrt{i} = \sqrt{\frac{1}{2}.2i} = \sqrt{\frac{1}{2}.(1 + 2i - 1)} = \sqrt{\frac{1}{2}.(1 + 2i + i^2)} = \sqrt{\frac{1}{2}.(1+i)^2} = \frac{1}{\sqrt{2}}(1+i) \\= \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \mbox{ ---------
1} \\ \mbox{Also, }... | Using the the polar representation for complex numbers, there holds that
$$ \sqrt{i}=e^{i\frac{\pi}{4}}~~~\mbox{and}~~~\sqrt{-i}=e^{-i\frac{\pi}{4}}$$
so that
$$ \sqrt{i}+\sqrt{-i}=e^{i\frac{\pi}{4}}+e^{-i\frac{\pi}{4}}=2\cos\left(\frac{\pi}{4}\right)=\sqrt{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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find the range $(x_{1}+x_{2}+\cdots+x_{2009})^2=4(x_{1}x_{2}+x_{2}x_{3}+\cdots+x_{2009}x_{1})^2$ let $x_{i}\ge 0(i=1,2,3,\cdots,2009$,and such $$(x_{1}+x_{2}+\cdots+x_{2009})^2=4(x_{1}x_{2}+x_{2}x_{3}+\cdots+x_{2009}x_{1})=4$$
find the range $\sum_{i=1}^{2009}x^2_{i}$
I guess the range is $[1.5,2]$
because I think wh... | I think you are right!
We'll prove that
$$(x_1+x_2+...+x_{2009})^2\geq4(x_1x_2+x_2x_3+...+x_{2009}x_1)$$ for all non-negatives $x_i$.
Indeed, let $x_2=\max\limits_{i}\{x_i\}$.
Thus, $$x_2x_{2009}\geq x_1x_{2009}$$ and by AM-GM we obtain:
$$4(x_1x_2+x_2x_3+...+x_{2009}x_1)\leq4(x_1+x_3+...+x_{2009})(x_2+x_4+...+x_{2008}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899143",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Using $\epsilon$-$\delta$ approach, prove that $\lim_{(x,y)\rightarrow (1,3)}\frac{x}{y}=\frac{1}{3} $ How to prove that $$ \lim_{(x,y)\rightarrow (1,3)}\frac{x}{y}=\frac{1}{3}$$
By $\epsilon$-$\delta$ definition.
What i did is this:
Let $\epsilon$ be grater than zero; i need to find a $\delta(\epsilon)>0$ such that $|... | More generally: assume $\delta < 3$. Since you are working on an open circle of centre $(1,3)$ and radius $\delta$, you can say that $\lvert y \rvert > 3-\delta$. This implies $$\frac{4\delta}{3\lvert y \rvert} < \frac{4\delta}{3(3-\delta)},$$
and then you can set $\delta := \frac{9\varepsilon}{4+3\varepsilon}$
to find... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
how to factor $f(x)=x^4-7x^2+1$ How do you factor this equation? I have tried to use factoring method where the expanded b term factors add to B and multiply to get C.
| It's easy to see that $x^4-7x^2+1$ has no rational roots. So if it factors (over the integers), it can only do so as the product of two quadratics. Since the lead and constant coefficients are both $1$ and there are no terms of odd degree, the factorization must be of the form $(x^2+bx+\sigma)(x^2-bx+\sigma)$ with $\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$
If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $$\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$$
Here's how I tried it ... | Note that $$\cos(a-b) = \cos(a)\cos(b)+\sin(a)\sin(b) = \Re(x\bar{y})$$
So $$\label{eq1}\cos(a-b)+\cos(b-c)+\cos(c-a) = \Re(x\bar{y}+y\bar{z}+z\bar{x})$$
With that in mind, note that just like $x$, $y$ and $z$, the $xyz$ complex number is of modulus 1. So if we go back to the $x+y+z=xyz$ identity, we see that $$|x+y+z|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
Area between the curve $y$, the $x$-axis and the lines
Find the total area between the curve $y=x^2-4x+3$, the $x$-axis and the lines $x=0$ and $x=3$.
I have drawn the graph and concluded that:
$$\int_0^1 0 - (x^2-4x+3)\, dx + \int_1^3 0 - (x^2-4x+3)\, dx = -\frac{-26}{3}$$ Which differs from the answer key: $4$. Wha... | $\int_0^1 (x^2-4x+3)\mathbb dx-\int_1^3(x^2-4x+3)\mathbb dx=[(\frac{x^3}3-2x^2+3x)]_0^1-[\frac{x^3}3-2x^2+3x]_1^3=\frac43-[0-\frac43]=\frac83$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Query on $\int \frac{dx}{\cos x \cos (2x+a)}$ .
$$\int \frac{dx}{\cos x \cos (2x+a)}$$
MY APPROACH:
I tried using partialization of fraction as follows :
$$ -\frac{1}{\cos a} \int \frac{dx}{\cos x} + \frac{1}{\cos \frac{\pi-2a}{4}} \int \frac{dx}{\cos(2x+a)}$$
$$=-\frac{1}{\cos a} \cdot\ln|\sec x+\tan x| +\frac{1}{\c... | Hint: Use that $$\cos(x)\cos(2x+a)=\frac{1}{2}\left(\cos(x+a)+\cos(3x+a)\right)$$
Then use that
$$\cos(x)=\frac{1-t^2}{1+t^2}$$ and $$dx=\frac{2tdt}{1+t^2}$$ the so-called Weierstrass substitution.
I will post you the solution for your work!
$$\sec (a) \left(-\sqrt{2} \sqrt{\sin (a)-1} \tan
^{-1}\left(\frac{\sqrt{\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Calculate $\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$
Calculate
$$\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$$
My Attempt:
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\l... | You are correct. This is another way to proceed: if $x\to 0^+$ then by using Taylor expansions we obatin
$$f(x):=\log_{\sin(x)}{(\cos (x))}=\frac{\log(\cos(x))}{\log(\sin(x))}=\frac{\log(1-\frac{x^2}{2}+o(x^2))}{\log (x+o(x^2))}\sim-\frac{x^2}{2\log(x)},$$
and it follows that for any $a>0$,
$$\lim_{x\to{0^+}}\frac{\log... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Prove that $3^{16} -33$ and $3^{15} +5$ is divisible by 4 by means of binomial theorem This is a question that I found in a textbook:
Given that $p=q+1$, $p$ and $q$ are integers, then show that
$p^{2n} - 2nq-1$ is divisible by $q^2$ given that $n$ is a positive integer.
By taking a suitable value of $n$, $p$ and $q$,... | By the Euclid's lemma: $3(3^{15}-11)=4\cdot [120+1120+...]$ implies that $3$ must divide either $4$ or the sum. It does not divide $4$, hence the conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2902603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How can I solve $\int_0^1\frac{\arctan(x^2)}{1+x^2}\,\mathrm dx$? This integral appears very similar to $\int\frac{\arctan x}{1+x^2}\,\mathrm dx$, but this question cannot be solved through the same simple substitution of $u=\arctan x$. WolframAlpha cannot find a symbolic solution to this problem, and this Quora answer... | $\begin{align}
\displaystyle\int_0^1 \dfrac{\arctan(x^2)}{1+x^2}dx&=\Big[\arctan x\arctan(x^2)\Big]_0^1-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\
&=\dfrac{\pi^2}{16}-\int_0^1 \dfrac{2x\arctan x}{1+x^4}dx\\
\end{align}$
Since,
$\displaystyle \arctan x=\int_0^1 \dfrac{x}{1+t^2x^2}dt$
then,
$\begin{align}
\displaystyle K&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2903039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 4,
"answer_id": 0
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Show that $\iint_S (x^2+y^2) dA = 9 \pi /4$ In exam it was asked to show that
$$\iint_S x^2+y^2 dA = 9 \pi /4$$ for $$S = {\{(x, y, z) | x>0, y>0,3>z>0, z^2 = 3(x^2 + y^2)}\}$$
I have tried many times but I don't get the $9 \pi /4$.
$$\begin{align}
\iint_S\sqrt{1+f_x^2+f_y^2}\,dA &=\int_0^{\sqrt3}\int_0^{2\pi} r^2\... | I am going try to reproduce your answer (at least how I think that you got there), so we can see where it went wrong.
Let
$$ x=r\cos\theta, \quad y = r\sin\theta. $$
Then
$$ z^2 = 3(x^2+y^2) = 3r^2 \quad \Rightarrow \quad z = r\sqrt{3} $$
where the final implication follows since both $z$ and $r$ are non-negative.
Henc... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Find $\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$ Find $$\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$$ , without using squeeze theorem.
I have done the solution as below using squeeze theorem ...
$$Let \left[\left({x \over x^2+1}+{x ... | Hint :
Take, $\displaystyle a_{x,n}=\frac{x^2}{x^2+n}$. Then use Cauchy's First limit theorem.
As $a_{x,n} \to 1$ when $x\to \infty$, so $\displaystyle \frac{1}{x}\sum_{n=1}^{x}a_{x,n} \to 1$ when $x\to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Express polynomial as a product of real quadratic polynomials with no real roots I want to express
$$P(x) = x^4 + x^2 + 1$$
as as a product of real quadratic polynomials with no real roots.
I know that: $$(x^2 - bx + a^2)(x^2 +bx + a^2)$$
$$ =x^2 + (2a^2 - b^2)x^2 + a^4 $$
I thought that $b^2 = 2a^2 - 1$ in this case ... | Note that
$$x^4+x^2+1 = (x^2-x+1)(x^2+x+1)$$
and
$$x^4-x^2+1 = (x^2-\sqrt{3}x+1)(x^2+\sqrt{3}x+1)$$
Considering the answer given by your textbook, it seems that your textbook wanted you to factorize $x^4-x^2+1$ into quadratic factors. If you haven't misread the problem, it's worth notifying the authors of this error/ty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
The number of natural number $n\leq 50$ such that
$\displaystyle \sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
Try: Let $\displaystyle x=\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}$
So $\displaystyle x... | Using the formula for finding the roots of the general cubic equation we obtain the numerical criterion:
$$
\sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+}\ldots}}}\in \mathbb{N} \Longleftrightarrow
\sqrt[3]{\dfrac{n}{2}+\sqrt{\dfrac{n^2}{4}-\dfrac{1}{27}}}+\sqrt[3]{\dfrac{n}{2}-\sqrt{\dfrac{n^2}{4}-\dfrac{1}{27}}}
\in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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Calculate $A^5 - 27A^3 + 65A^2$, where $A$ is the matrix defined below. If $A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$, find $A^5 - 27A^3 + 65A^2$
$$A=\begin{bmatrix} 0 & 0 & 1 \\
3 & 1 & 0 \\
-2&1&4\end{bmatrix}$$
Let $\lambda$ be its eigenvalue, then
$$(A-\lambda I) = \begin{bmatrix} 0-\lambd... | I don't see a way to find the answer directly but you can certainly simplify your calculations; since $A$ commutes with itself we may factor the polynomial however we want. Write
$$
A^5-27A^3+65A^2=A^2(A^3-27A+65)=A^2\left[(A^3-5A^2+6A-5)+(5A^2-33A+70)\right].
$$
Now by Cayley Hamilton you only need to compute $A^2(5A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Proof Verification of a Limit in Real Analysis Is this following proof correct?
The limit
\begin{align*}
\lim_{n\to\infty}\frac{t^{n+1}}{(n+1)!}
\end{align*}
tends to zero if for every $\epsilon>0$, there exists $N_\epsilon$ such that
\begin{align*}
n\geq N_\epsilon\implies \left|\frac{t^{n+1}}{(n+1)!}\right|<\epsilon... | The proof looks good (actually very good) to me. The only improvement I can suggest is to turn it around a little, so that no cleverness is invoked.
Proof: Given $t>0$ and $\epsilon > 0$ arbitrary, put:
$$\begin{align*}
N &= \lfloor t \rfloor + 1 & (\text{$N$ is the smallest natural number strictly greater than 1})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2911399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof by induction, $1/2 + ... + n/2^n < 2$ So I'm having trouble with proving this homework question by induction.
$$
\frac{1}{2^1} + \frac{2}{2^2} + ... +\frac{n-1}{2^{n-1}} + \frac{n}{2^n} <2
$$
I know how to prove that the series converges to 2 (using things like the ratio method), but actually using induction is ... | You can try to do the following: write
$$ \frac{1}{2} + \frac{2}{2^2} + \cdots + \frac{k}{2^k} + \frac{k+1}{2^{k+1}} = $$
$$ \frac{1}{2} + \frac{1}{2^2} + \cdots + \frac{1}{2^k} + \frac{1}{2^{k+1}} + $$
$$ \frac{1}{2}\left(\frac{1}{2} + \frac{2}{2^2} + \cdots + \frac{k}{2^k}\right).$$
Use that the first summand is bo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Prove that $\frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x^2+y^2+z^2) $
Let $x, y, z\in [a,b]$ and $u, v, w\in [a, b] $, where $0 <a <b $, s.t. $x^2+y^2+z^2=u^2+v^2+w^2$. Show that $$\frac {x^3}{u}+\frac {y^3}{v} + \frac {z^3}{w}\leq \frac {a^4+b^4}{ab (a^2+b^2)}(x^2+y^2+z^2) .$$
... | The inequality is homogeneous, i.e. multiplying all variables with the same factor will not change it. This allows us to demand that the multiplied variables (for which we use the same variable letters as before) satisfy $x^2 + y^2 + z^2 = u^2 + v^2 + w^2 = 3$ and in turn $a \le 1$ and $\sqrt{3-a^2} \ge b \ge 1$. So ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of ways to pay (generating functions) I've just started learning generating functions.
*
*Let $a_n$ be the number of ways in which you can pay $n$ dollars using 1 and 2 dollar bills. Find the generating function for $(a_0, a_1, a_2, \ldots)$ and general term $a_n$.
*Prove that the number of ways to pay $n$ d... | The generating function for problem $1$ is
$$
\begin{align}
\frac1{1-x}\frac1{1-x^2}
&=\frac12\frac1{1-x}\left(\frac1{1-x}+\frac1{1+x}\right)\\
&=\frac12\left(\color{#C00}{\frac1{(1-x)^2}}+\color{#090}{\frac1{1-x^2}}\right)\\
&=\sum_{k=0}^\infty\frac{\color{#C00}{k+1}+\color{#090}{[2\mid k]}}2\,x^k\\
&=\sum_{k=0}^\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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How to evaluate $1+\frac{2^2}{3!}+\frac{3^2}{5!}+\frac{4^2}{7!}+\cdots$? I learnt that $\displaystyle \sum_{n=0}^{\infty} \frac{n+1}{(2n+1)!} = \frac{e}{2}$.
I am wondering what the closed form for $\displaystyle \sum_{n=0}^{\infty} \frac{(n+1)^2}{(2n+1)!}$ is.
I tried using the fact that $ 1+3+5+\cdots+(2n-1) = n^2$,... |
$$\sum_{n=0, \text{odd}}^{\infty} \frac{(\frac{n+1}{2})^2}{n!}$$
$$=\frac{1}{4} \sum_{n=0, \text{odd}}^{\infty} \frac{n^2+2n+1}{n!}$$
$$=\frac{1}{4} \left( \sum_{n=0,\text{odd}}^{\infty} \frac{n}{(n-1)!}+2\sum_{n=0,\text{odd}}^{\infty} \frac{1}{(n-1)!}+\sum_{n=0,\text{odd}}^{\infty} \frac{1}{n!} \right)$$
Now note:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2917988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Getting a closed form from $\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1$ I need to get a closed form from
$$
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} 1
$$
Starting from the most outer summation, I got
$$
\sum_{k=1}^{j} 1 = j
$$
But now I don't know how to proceed with:
$$
\sum_{i=1}^{n-1} \sum_{j=i+1}^{n... | Knowing that
\begin{align}
\sum_{k=1}^{n} (1) &= n \\
\sum_{k=1}^{n} k &= \frac{n \, (n+1)}{2} \\
\sum_{k=1}^{n} k^2 &= \frac{n(n+1)(2n+1)}{6}
\end{align}
then
\begin{align}
S &= \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} \sum_{k=1}^{j} (1) \\
&= \sum_{i=1}^{n-1} \sum_{j=i+1}^{n} j \\
&= \sum_{i=1}^{n-1} \left(\sum_{j=1}^{n} j ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2918464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Principal part of Laurent Series for $\frac{1}{(1-\cosh(z))^2}$ in this exercise I am asked to provide the principal part of the Laurent series of
$$\frac{1}{(1-\cosh(z))^2}$$
And i am kinda struggling with fonding a solution or even a pattern towards one
Thanks in advance to everyone keen to help
| $$(1-\cosh z)^2=3-4\cosh z+\cosh2z$$
then
\begin{align}
\dfrac{1}{(1-\cosh z)^2}
&= \dfrac{1}{3-4\cosh z+\cosh2z} \\
&= \dfrac{1}{3-4 \left(1+\frac{1}{2!}z^2+\frac{1}{4!}z^4+\frac{1}{6!}z^6+\cdots\right) + \left(1+\frac{4}{2!}z^2+\frac{16}{4!}z^4+\frac{64}{6!}z^6+\cdots\right)} \\
&= \dfrac{1}{\frac{1}{4}z^4+\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\int_0^a\frac{x^4}{(a^2+x^2)^4}dx$? I have to evaluate
$$\int_0^a\frac{x^4}{(a^2+x^2)^4}\,{\rm d}x$$
I tried to substitute $x=a\tan\theta$ which then simplifies to
$$\frac1{a^5}\int_0^\frac\pi4\sin^4\theta\cos^2\theta\, {\rm d}\theta$$.
Now,its quite hectic to solve this. Is there any other method o... | This is a rather standard rational function integral, and we only have to decompose it into partial fractions.
To have lighter calculations, we'll set $T=a^2+x^2$. Then
$$x^4=T^2-2a^2x^2-a^4=T^2-2a^2T+a^4$$
which yields the decomposition
\begin{align}
\frac{x^4}{(a^2+x^2)^4}&=\frac{T^2-2a^2T+a^4}{T^4}=\frac1{T^2}-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 3
} |
Solving the factorial equation $(n + 4)! = 90(n + 2)!$ Solve the equation below:
$(n+4)!
= 90
(n+2)!$
I did this:
$(n+4)(n+3)(n+2)!
= 90
(n+2)!$
$n^2+7n+12+90=0$
$n^2+7n+102=0$
Is there anymore to this?
| When you take the $90$ to the other side you need to subtract it. You'll get
$n^2 +7n +12 -90 = 0,$
$n^2 +7n -78 = 0,$
$(n - 6)(n+13) = 0,$
So $n = 6$ or $n = -13$.
Clearly $n = -13 $ doesn't make sense, so we conclude:
$n = 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2921667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Show that $f:M_f \rightarrow A_f$ is bijective when $f(x)=\frac{x-1}{x+1}$ Problem
Show that $f:M_f \rightarrow A_f$ is bijective when $f(x)=\frac{x-1}{x+1}$
Edit
when $M_f \in \mathbb{R}\setminus \{-1\}$ and $A_f \subset \mathbb{R}$
Attempt to solve
$f:M_f\rightarrow A_f$ is bijective when it is injective and surj... | Observe that
$$f(x) = \frac{x - 1}{x + 1} = 1 - \frac{2}{x + 1}$$
Thus, the range of $f$ is $\mathbb{R} - \{1\}$.
To ensure that $f: \mathbb{R} - \{-1\} \to A_f$ is surjective, we must define $A_f = \mathbb{R} - \{1\}$.
For $y \neq 1$,
\begin{align*}
f\left(\frac{1 + y}{1 - y}\right) & = \frac{\frac{1 + y}{1 - y} - 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Cat / mouse probability question There exist 7 doors numbered in order from 1 to 7 (going from left to right). A mouse is initially placed at center door 4. The mouse can only move 1 door at a time to either adjacent door and does so, but is twice as likely to move to a lower numbered door than to a higher numbered d... | I'll take a shot at this........
1) The eventual fateful outcome occurs when the total number of moves, lower versus higher, differs by 3.
So, a way to look at this as an $E(x) = n\cdot p$ type problem is to sum the $(n\cdot p)$s for all possible outcomes.
This will be:
$3(\frac{1}{3})+5(\frac{2}{9})+7(\frac{4}{27})+9(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924838",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
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Value of 'a' for which $f(x)=\frac{x^3}{3} +\frac{x^2}{2} +ax+b $ is injective Let $f(x)=\frac{x^3}{3} +\frac{x^2}{2} +ax+b \forall x \in R$.
Find the least value of a for which $f(x)$ is injective function.
My approach , for $f(x)$ to be injective either f(x) should be increasing or decreasing function.'
$Y=f'(x)=... | If you want $f$ is injective, then $f$ should be increasing since leading part is $x^3$. So $f'(x)=x^2+x+a\geq 0$ for all $x$, so the discriminant $D\leq 0$ and that is iff $1-4a\leq 0$ so $a\geq {1\over 4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Ask a about hard integral of $\int_{0}^{\infty} \log x \log (\frac{a^2}{x^2}+1) \log(\frac{b^2}{x^2}+1)dx$ I want to evaluate the integral:
$$I(a,b)=\int_{0}^{\infty} \log x \log (\frac{a^2}{x^2}+1) \log(\frac{b^2}{x^2}+1)dx$$
Attempt:$$\frac{\partial ^2I}{\partial a\partial b}=4ab\int_{0}^{\infty}\frac{\log x}{(a^2+x^... | This derivation is slightly different from Maxim's, because I'm not fluent in Meijer G. The beginning is the same, and assume $0<b<a.$ Then
$$ I(a,b)=b\,\int_0^\infty (\log{b} + \log{x} ) \log{(1+ \big(\frac{a/b}{x}\big)^2 ) }\log{(1+1/x^2)} dx $$
Let $r=b/a \le 1$ so that some series manipulations are permissible. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Generating function problem, and distributing candy to kids. Here is the problem:
Determine how many ways I can distribute $80$ candies to $3$ kids, such that:
$\bullet$ The first kid receives an arbitrary number of candies (possibly $0$).
$\bullet$ The second kid receives an even positive number of candies.
$\bullet$ ... | If the option to receive $k$ candies is worth $x^k$ the options of the first child sum to $1+x+x^2+\ldots$, the options of the second child sum to $x^2+x^4+x^6+\ldots$, and the options of the third child sum to $1+x^2+x^5$. In such a situation the function
$$f(x)=(1+x+x^2+\ldots)(x^2+x^4+x^6+\ldots)(1+x^2+x^5)$$
(your... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Using power series to evaluate a limit If we want to evaluate the limit of $$\lim_{x\to 0} \frac{\ln(1-8x)+8x+32x^2}{3x^2}$$
I would imagine we need to use the power series expansion of the function $${\ln(1-8x)}$$
and then determine what cancels out.
The correct answer that I found using a calculator is $${-512/9}$$
| $$\lim_{x\to 0} \frac{\ln(1-8x)+8x+32x^2}{3x^2}=\lim_{x\to 0} \frac{\ln(1-8x)}{3x^2}+\frac{8}{3x}+\frac{32}{3}$$
$$\ln(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}-\cdots-\frac{x^n}{n}$$
$$\ln(1-8x)=-8x-\frac{64x^2}{2}-\frac{512x^3}{3}-\cdots-\frac{8^nx^n}{n}$$
$$\frac{\ln(1-8x)}{3x^2}=-\frac{8}{3x}-\frac{32}{3}-\frac{512x}{9}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2925677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Laurent expansion of $f(x)=\frac{1}{(z-1)(z-2)^2}$
I am trying to find the Laurent expansion of $f(x)=\frac{1}{(z-1)(z-2)^2}$ when $0<|z-1|<1$.
I thought that if
\begin{align}
f(z)&=\frac{1}{z-1}-\frac{1}{z-2}+\frac{1}{(z-2)^2} \\
&=\frac{1}{z-1}+\frac{1}{2-z}+\frac{d}{dz}\left(\frac{1}{2-z}\right) \\
&=\frac{1}{z-... | *
*In line 2, $-\frac{d}{dz}$ should be $+\frac{d}{dz}$.
*Your final sum has many powers duplicated and these could be collected into single terms. For example when $n=5$ the first term is $(z-1)^5$ and when $n=6$ the second term is $6(z-1)^5$.
*Easier method: take $\frac1{z-1}$ as a multiplicative factor at the be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2929743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Classifying singular points of $\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}$
I am trying to classify the singular points of the function $$f(z)=\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}.$$
My attempt:
$$f(z)=\frac{\sin(z^2)}{z^3-\frac{\pi}{4}z^2}=\frac{\sin(z^2)}{z^2\left(z-\frac{\pi}{4}\right)}.$$
Hence the singular points ... | Note that $$\sin(z^2)=z^2-\frac{z^6}{3!}+\frac{z^{10}}{5!}-\cdots$$ so $$f(z)=\frac{z^2-\frac{z^6}{3!}+\frac{z^{10}}{5!}-\cdots}{z^2\left(z-\frac{\pi}{4}\right)}=\frac{1-\frac{z^4}{3!}+\frac{z^8}{5!}-\cdots}{z-\frac\pi4}$$ so $z=0$ is a removable singularity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the relationship of the length of triangle's sides.
Denote the three sides of $\triangle ABC$ to be $a,b,c$. And they satisfy $$a^2+b+|\sqrt{c-1}-2|=10a+2\sqrt{b-4}-22 $$
Now determine what kind of triangle $\triangle ABC$ is.
A.Isosceles triangle which its leg and base is not equal.
B.equilateral triangle
... | Noodling:
$a^2$ and $10a$ in equations imply that maybe I should attempt to complete the square.
So I get
$a^2 - 10a +25+b +|\sqrt{c-1} - 2|=(a-5)^2 + b+ |\sqrt{c-1} -2|= 2\sqrt{b-4} + 3$
And, well this seems a bit weird but that is an even number in front of the $\sqrt{b-4}$ and we have $\sqrt{b-4}$ and $b$ variables ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Integrate $\int \sqrt{\frac{1+x}{1-x}}dx$ using $x = \cos(u) $ I have to integrate $\int \sqrt{\frac{1+x}{1-x}}$ using $x = \cos(u)\Rightarrow dx = -\sin(u) du$
My attempts:
$$\int \sqrt{\frac{1+x}{1-x}}dx=\int \sqrt{\frac{(1+x)(1-x)}{(1-x)(1-x)}}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx=\int \frac{\sqrt{1-x^2}}{1-x}dx$$
$$\i... | $$\frac{1+\cos u}{1-\cos u}=2\cot^2\left(\frac{u}{2}\right).$$
So your integral is
$$\int\sqrt{\frac{1+x}{1-x}}\, dx=-\sqrt{2}\int \sin u \cot\left(\frac{u}{2}\right) \, du=-2\sqrt{2}\int \cos^2 \left(\frac{u}{2}\right)\, du$$
Here we are using $\sin u =2\sin \left(\frac{u}{2}\right)\cos \left(\frac{u}{2}\right)$. Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Finding value of $\int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx$ without contour Integration
Finding value of $\displaystyle \int^{2\pi}_{0}\frac{\sin^2(x)}{a-b\cos x}dx\;\;, a>b>0$ without contour Integration
Try: Let $\displaystyle I =\int^{2\pi}_{0}\frac{\sin^2 x}{a-b\cos x}dx=2\int^{\pi}_{0}\frac{\sin^2 x}{a-b\cos... | From $a>b$ then $\left|\dfrac{b}{a}\cos x\right|<1$ with the expansion
\begin{align}
\int_{0}^{2\pi}\frac{\sin^2(x)}{a-b\cos x}\ dx
&= \dfrac{1}{a}\int_{0}^{2\pi}\sin^2(x)\sum_{n\geq0}\left(\dfrac{b}{a}\cos x\right)^n\ dx \\
&= \dfrac{1}{a}\sum_{n\geq0}\left(\dfrac{b}{a}\right)^n\int_{0}^{2\pi}\sin^2x\cos^nx\ dx \\
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding the domain and the range of $f(x) = \ln (1-2\cos x).$ Finding the domain and the range of $f(x) = \ln (1-2\cos x).$
My Attempt:
I arrived at $x < \cos^{-1}(1/2),$ then how can I complete?
| the domain of $\ln (z)$ is $z \gt 0$. So we need $1- 2\cos(x) \gt 0$.
so $$-2\cos(x) > -1$$ $$2\cos(x) \lt 1$$ $$\cos(x) < \frac{1}{2}$$
if we graph $\cos(x)$ we see that $\cos(x) \ge \frac{1}{2}$ for $x \in \big[ \frac{-\pi}{3}, \frac{\pi}{3} \big]$ but we need to exclude this to keep $\cos(x) \lt \frac{1}{2}$, so we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2939508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"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.