Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Determine limit: $\lim_{x \to -\infty} (\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}) $ Determine the limit of:
$$\lim_{x \to -\infty} \left(\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}\right) $$
I've tried a few times, most notably the following two versions. I'm looking for a comment on both, since both amount to a wrong answer.
First attem... | Your Second attempt is true, but when you bring $x$ from radicals, take a minus to $x$, because $x<0$.
For your first attempt use this:
$$L = \lim_{x \to -\infty} \left(\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}\right) $$
$$ \begin{split} &= \lim_{x \to -\infty} \left(\left(\sqrt{(x+1)^2-1}\right) - \left(\sqrt{(x-1)^2-1}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $f(g(x)) = 4x^2-8x$ and $f(x)=x^2-4$, then what's the value of $g(x)$? I'm a little stuck with this simple function:
If ${f(g(x)) = 4x^2-8x}$ and ${f(x)=x^2-4}$, then what's the value of ${g(x)?}$
Any tips?
| If $g$ is a polynomial, then note that the degree can't be higher than $1$. Suppose $g(x)=ax+b$. Then
$$f(g(x))=f(ax+b)=a^2x^2+2abx+b^2-4$$
Now we want $a^2=4$, $2ab=-8$, and $b^2-4=0$. Two such solutions are $a=2,b=-2$, or $a=-2,b=2$.
Pluggin those back in, we have
$$f(2x-2)=4x^2-8x+4-4=4x^2-8x$$
and
$$f(-2x+2)=4x^2-8... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
Find a minimum of $x^2+y^2$ under the condition $x^3+3xy+y^3=1$ As in the title, I've tried to find a maximum and mininum of $x^2+y^2$ when $x^3+3xy+y^3=1$ holds. It is not too hard to show that $x^2+y^2$ has no maximum, but I can't find a minimum.
Lagrange multiplier gives a dirty calculation so I can't handle it. Is ... | $$x^3+3xy+y^3-1=(x+y-1)(x^2+y^2-xy+x+y+1)=0$$
So either $y=1-x$ or $(x,y)=(-1,-1)$
Subbing this in gives $x^2+(1-x)^2$ or $2$ to be minimized. Considering the first simple calculus gives us that $\frac{dy}{dx}=2x-2(1-x)=4x-2$. Setting this equal to zero gives $x=y=\frac{1}{2}$ as a possible minimum/maximum. Looking at ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2107048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Closed form for the limit of the iterated sequence $a_{n+1}=\frac{\sqrt{(a_n+b_n)(a_n+c_n)}}{2}$ Is there a general closed form or the integral representation for the limit of the sequence: $$a_{n+1}=\frac{\sqrt{(a_n+b_n)(a_n+c_n)}}{2} \\ b_{n+1}=\frac{\sqrt{(b_n+a_n)(b_n+c_n)}}{2} \\ c_{n+1}=\frac{\sqrt{(c_n+a_n)(c_n+... | I can show that
the sum of the squares of the terms
decreases at each step.
I think this implies convergence,
but I am not completely sure.
$a_{n+1}
=\frac{\sqrt{(a_n+b_n)(a_n+c_n)}}{2} \\
b_{n+1}=
\frac{\sqrt{(b_n+a_n)(b_n+c_n)}}{2} \\
c_{n+1}
=\frac{\sqrt{(c_n+a_n)(c_n+b_n)}}{2}
$
I'm going to play around
and see i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Find the inverse of a lower triangular matrix of ones Find the inverse of the matrix $A=(a_{ij})\in M_n$ where
$$
\begin{cases}
a_{ij}=1, &i\geq j,\\
a_{ij}=0, &i<j.
\end{cases}
$$
The only method for finding inverses that I was taught was by finding the adjugate matrix. So $A^{-1}=\frac{1}{\det A}\operatorname{adj(A)}... | Just to show a different approach.
Consider the matrix $\mathbf E$, having $1$ only on the first subdiagonal
$$
\mathbf{E} = \left\| {\,e_{\,n,\,m} = \left\{ {\begin{array}{*{20}c}
1 & {n = m + 1} \\
0 & {n \ne m + 1} \\
\end{array} } \right.\;} \right\| = \left\| {\,\left( \begin{gathered}
0 \\
n - m - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2111562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
$2\sin 2x-2\cos2x=\frac{\cos x+\cos3x}{\cos x-\sin x}$ Solve the equation :
$$2\sin 2x-2\cos2x=\frac{\cos x+\cos3x}{\cos x-\sin x}$$
my try :
$$\cos x-\sin x \neq 0 \to \cos x \neq \sin x$$
$$x\neq k\pi+\frac{\pi}{4}$$
$$2(\sin 2x -\cos 2x)=\frac{\cos x+\cos3x}{\cos x-\sin x}$$
$$2(2\sin x \cos x -\cos 2x)=\frac{\cos x... | Hint:
$\cos3x+\cos x=2\cos2x\cos x$
$\sin2x-\cos2x=\cos x(\cos x+\sin x)$
$\iff \cos^2x+\cos2x=\sin x\cos x$
Divide both sides by $\cos^2x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Infimum of two variable function on the closed unit disc I am trying to compute
$$\inf_{x,y} \frac{ax-by}{1+x^2+y^2}$$
subject to the constraint $x^2+y^2 \leq 1$. Here $a,b$ are any two fixed, real numbers. I am having trouble computing this using standard derivative techniques, and Wolfram alpha is unable to recognize... | It might be easier to work in polar coordinates $(r,\theta)$. Note that
$$f(x,y)=\frac{ax-by}{1+x^2+y^2}=\frac{r(a\cos(\theta)-b\sin(\theta))}{1+r^2}=g(r,\theta)$$
Then, we have
$$\begin{align}
\frac{\partial}{\partial r}\left(\frac{r(a\cos(\theta)-b\sin(\theta))}{1+r^2}\right)&=\frac{(1-r^2)}{(1+r^2)^2}\,(a\cos(\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Rotating rectangle 90 degrees clockwise I have a rectangle in the cartesian plane defined by the top left and Bottom right as $(3, 5),(5, 3)$
We rotate this around the origin clockwise by 90 degrees, what is the new top left and bottom right point?
Is it: $(3, -3), (5, -5)$
?
This doesnt graphically make sense to me t... | Rotations in the plane by an angle $\theta$ can be calculated using the rotation matrix
$$\begin{bmatrix}\cos \theta& -\sin \theta \\ \sin \theta& \cos\theta\end{bmatrix}.$$
So in the case of a rotation of $\theta=90^\circ$ this matrix is
$$\begin{bmatrix}0& -1 \\ 1& 0\end{bmatrix}.$$
So the new corners of your rectang... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2115569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Integral of a logarithmic function What is the value of integral $$\int_0^1 \log[(1+x)^{1/2} +(1-x)^{1/2}]dx$$ ? I'm applying integration by parts but couldn't find any substitution after couple of steps
| Let $$I = \int^{1}_{0}\ln\bigg(\sqrt{1+x}+\sqrt{1-x}\bigg)\cdot 1dx$$
Integration by parts
$$I = \log\bigg(\sqrt{1+x}+\sqrt{1-x}\bigg)\cdot x\bigg|_{0}^{1}+\frac{1}{2}\int^{1}_{0}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\cdot \frac{1}{\left(\sqrt{1+x}\cdot \sqrt{1-x}\right)}xdx$$
$$I = \frac{\ln2}{2}+\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Limit $\lim_{x\rightarrow +\infty}\sqrt{x}e^{-x}\left(\sum_{k=1}^{\infty}\frac{x^{k}}{k!\sqrt{k}}\right)$ $$\lim_{x\rightarrow +\infty}\sqrt{x}e^{-x}\left(\sum_{k=1}^{\infty}\frac{x^{k}}{k!\sqrt{k}}\right)$$
Any hint will be appreciated.
Note: There is a related question on MathOverflow: Asymptotic expansion of $\sum\l... | (I have overwritten my previous incorrect answer.)
For any positive constant $c > 1$ we have: \begin{align*}S(cx) = e^{-x}\sum\limits_{k \ge cx} \frac{x^k}{k!} &= e^{-x}\frac{x^{cx}}{(cx)!}\left(1 + \frac{x}{(cx+1)} + \frac{x^2}{(cx+1)(cx+2)} + \cdots\right) \\& \le e^{-x}\frac{x^{cx}}{(cx)!}\sum\limits_{k=0}^{\infty}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2117742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Calculate remainder of $12^{34^{56^{78}}}$ when divided by $90$
Calculate remainder of $12^{34^{56^{78}}}$ when divided by $90$
First of all,I'm not sure about the order of calculation of powers!
Secondly I don't know the rules of finding remainder moulus 90
| My solution is rather routine and makes good use of Euler totient theorem and modulus properties.
$$\begin{align*}12^{34^{45^{78}}} \pmod{90} &\equiv 6\left[\frac{12^{34^{45^{78}}}}{6} \pmod{15} \right],\qquad\qquad\gcd(12,90)=6\\
&\equiv 18\left[\frac{12^{34^{45^{78}}}}{18} \pmod{5}\right],\qquad\qquad\gcd(12,15)=3\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Find all triplets of natural numbers $(x,y,z)$ that satisfy this equation: $2x^{2}y^{2}+2y^{2}z^{2}+2x^{2}z^{2}-x^{4} -y^{4}-z^{4}=576$ I've tried
$(x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(x^{2}-z^{2})^{2}-x^{4}-y^{4}-z^{4}=-576$
$(x^{2}-y^{2}-z^{2})(x^{2}-y^{2}+z^{2})+(y^{2}-z^{2}-x^{2})(y^{2}-z^{2}+x^{2})+(x^{2}-z^{2... | HINT:
$2x^{2}y^{2}+2y^{2}z^{2}+2x^{2}z^{2}-x^{4} -y^{4}-z^{4}$
$=(2xy)^2-(x^2+y^2-z^2)^2$
$=\{2xy+(x^2+y^2-z^2)\}\{2xy-(x^2+y^2-z^2)\}$
$=\{(x+y)^2-z^2\}\{z^2-(x-y)^2\}=\cdots$
Observe that $576$ is even
and so are the difference & sum of any two of
$\{x+y+z,x+y-z,z-x+y,z+x-y\}$
So, each of the four multiplicands hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2120651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to show that $\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx=3\pi$
Consider
$$I=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^4}\mathrm dx
\qquad J=\int_{0}^{\infty}{(2x)^4\over (1-x^2+x^4)^3}\mathrm dx$$
I want to show that $I=3\pi$ and that $I=J$.
F... | Consider the transformation $x=1/y$. Then
$$
J=16\int_0^{\infty} dy\frac{1}{y^2} \frac{1/y^4}{(1-1/y^2+1/y^4)^3}=8\int_{-\infty}^{\infty} dy\frac{1}{(y^2-1+1/y^2)^3}=\\8\int_{-\infty}^{\infty} dy\frac{1}{((y-1/y)^2+1)^3}\underbrace{=}_{(\star)}8\int_{-\infty}^{\infty}dz\frac{1}{(z^2+1)^3}=3\pi
$$
and your proof is comp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
5 triangles with the same area inside a pentagon A pentagon ABCDE contains 5 triangles whose areas are each one. The triangles are ABC, BCD, CDE, DEA, and EAB. Find the area of ABCDE?
Is there a theorem for overlapping triangle areas?
| It was asked how to find the area algebraically.
Assume that the pentagon $ABCDE$ is regular and the area $[ABC] = 1$. Let $s$ be its side length. The angle at B is $3\pi/5$. The area of $ABC$ is then $s^2/2 \sin (3\pi/5)$, which must equal 1. The area of the pentagon is $1/4 \times 5 \times s^2 \times \cot (\pi/5)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to i solve this Exponential equation How to solve this exponential equation?
$$7 \cdot 3^{x+1} - 5^{x+2}= 3^{x+4}- 5^{x+3}$$
| Group them according to the base:
\begin{align}
&7\cdot 3^{x+1}-3^{x+4}=5^{x+2}-5^{x+3}\\
\implies&3^{x+1}(7-3^3)=5^{x+2}(1-5)\\
\implies&3^{x+1}\cdot (-20)=5^{x+2}\cdot(-4)\\
\implies&3^{x+1}\cdot 5=5^{x+2}\\
\implies&3^{x+1}=5^{x+1}
\end{align}
Generally, you would now use logarithms, although this case is rather obv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Help me find the following integral by using basic formulas of integral $$\int\frac{\sqrt{4+x^2}+2\sqrt{4-x^2}}{\sqrt{16-x^4}}dx$$
i tried to break this expression into two parts(sum of integrals) the first gives me $\arcsin(x/2)$(by formula) as $16-x^4 = (4-x^2)(4+x^2)$ and square roots of $4+x^2$ are cancelled out(i... | Hint:$$\frac { \sqrt { 4+{ x }^{ 2 } } +2\sqrt { 4-{ x }^{ 2 } } }{ \sqrt { 16-{ x }^{ 4 } } } =\frac { 1 }{ \sqrt { 4-{ x }^{ 2 } } } +\frac { 2 }{ \sqrt { 4+{ x }^{ 2 } } } $$ then substite $x=2\sin { \theta } $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
equation of the plane passing through the intersection of two ellipsoids I want to find the equation of the plane passing the intersection of two intersecting ellipsoids. the intersection of two ellipsoids is always a ellipse. I need to find the equation of the planar surface containing this ellipse.
Does have anyone a... |
Only two similar and translated ellipsoids guarantee planar ellipse section:
For $\lambda > 0$,
\begin{align*}
\frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2}
&= 1 \tag{1} \\
\frac{(x-\alpha)^2}{a^2}+\frac{(y-\beta)^2}{b^2}+\frac{(z-\gamma)^2}{c^2}
&= \lambda^2 \tag{2} \\
\end{align*}
$(1)-(2)$,
$$\frac{2\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Letters order - combinatorics Question: How many words is it able to create from the letters $LYCANTHROPIES$ so $C$ won't stand next to $A$ won't stand next to $N$ and $N$ won't stand next to T.
I thought about $13! - 3\cdot 2 \cdot 12! - 2 \cdot 10!$, when $13!$ is the maximum number of words I can create with no limi... | At first we see that $LYCANTHROPIES$ consists of $13$ pairwise different characters.
The number of all words of length $13$ which can be built from these characters is
\begin{align*}
13!
\end{align*}
Forbidden subwords are the following six words
\begin{align*}
&AC,CA,\\
&AN,NA\\
&NT,TN
\end{align*}
Therefore we h... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128886",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that $17$ is the only prime of the form $p^q +q^ p$ , where $p$ and $q$ are prime
Show that $17$ is the only prime of the form $p^q +q^ p$ , where $p$ and $q$ are prime
My attempt so far is first assume $p$ and $q$ are prime. Now $17=2^3+3^2.$
Now fix $p=2$ and let $q>3$ then $q=3x+1$ or $q=3x+2$, $x \in \mathbb... | If $p,q$ are both odd (or both even) then $p^q+q^p$ is even and $>2$, hence nt prime.
So if $n=p^q+q^p$ is prime we can assume wlog that $q=2$ and $n=p^2+2^p$ with $p$ odd.
Then $2^p\equiv 2\pmod 3$.But if $p$ is not a multiple of $3$, then $p^2\equiv 1\pmod 3$ and so $2^p+p^2$ is a multiple of $3$. As $2^p+p^2$ cannot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2129457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove $a_n = n 3^{n-1}$, recurrence relation. Let $a_0 = 0, a_1 = 1$, and let $a_{n+2} = 6a_{n+1} - 9a_n$ for $n\geq 0.$
Prove $a_n = n 3^{n-1}$ for $n\geq 0.$
attempt: Suppose $a_0 = 0, a_1 = 1$, and let $a_{n+2} = 6a_{n+1} - 9a_n$.
Then notice when $n = 0$, we have $a_2 = 6_{1} - 9a_0 = 6-0 = 6$.
And $a_2 = 2*(3^{2-... | First off let us see if the base case holds.
$a_{0}$=$0*3^n-1$=0 that means $a_{0}$ holds
$a_{1}$=$1*3^0=1$ that means $a_{1}$ holds too
Then to do the inductive step we have to see if n+1 holds.
$a_{n+3}$=$6(n+2)*3^n$ + $2$ - $1$ - $9(n+1)*3^n$
=$2(n+2)*3^n+2$-$(n+1)*3^n+2$=$3^n+2(2(n+2)-n+1)$= $3^{n+2}(n+3)$
That mea... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the maximum and minimum value of $\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$ Find the maximum and minimum value of $$\cos^2x-6\sin(x)\cos(x)+3\sin^2x+2$$.
i simplified and reach to expression as follows :
$5 + 2\sin(x)[\sin(x)-3\cos(x)]$. How do i go from here?
Thanks
| $$
\begin{align}
\frac{d}{dx} \cos^2x-6\sin(x)\cos(x)+3\sin^2x+2 &= -6 \cos(2 x) + 2 \sin(2 x)
\end{align}
$$
as a result of a simple differential calculation; now you'd like the $RHS$, namely $ -6 \cos(2 x) + 2 \sin(2 x)$, to be $0$ as to find the local minima and maxima of the function.
$$
\begin{align}
-6 \cos(2 x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2134658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Differentiate $y = 6 \cdot 3^{2x - 1}$ I'm trying to differentiate $y = 6 \cdot 3^{2x - 1}$, not really sure about my answer. I tried using wolfram alpha, but it doesn't really help me... I feel I'm kind of close, but I just seem to not be able to make the final leap.
So: $\frac{d}{dx}(6 \cdot 3^{2x - 1}) = 6 \frac{d}... | We can clear up the situation by starting from the get go, with the reminder that for real $a, b, c$, we have $a^b\cdot a^c = a^{b+c}$.
$$y = 6 \cdot 3^{2x - 1} = 2\cdot 3\cdot 3^{2x - 1} = 2\cdot {3^1\cdot 3^{2x - 1}} = 2\cdot 3^{1 + 2x - 1} = 2\cdot 3^{2x}$$
Now find $y'$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Find the GCD of ... Find the GCD of:
$y^2-10y+24+6x-9x^2$, $2y^4-18x^2y^2-48xy^2-32y^2$.
My Attempt:
$$
\begin{split}
1^{st} \text{expression} &= y^2-10y+24+6x-9x^2\\
&={(y)^2-2\cdot(y)\cdot5+(5)^2}-(5)^2+24+6x-9x^2\\
&=(y-5)^2-{25-24-6x+9x^2}\\
&=(y-5)^2-{(1)^2-2\cdot(1)\cdot(3x)+(3x)^2}\\
&=(y-5)^2-(1-3x)^2\\
&=(y-5+... | You don't need to factor that quartic expression from scratch; you just have to try out all factors of the quadratic expression.
Use synthetic division to see that
$$
\frac{2y^4-18x^2y^2-48y^2x-32y^2}{y-3x-4} = 2y^2(y+3x+4)
$$
So the GCD is $(y-3x-4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding partial sum of a series I need to find sum of n terms of series: $$-1 +2-3+4-5+6 .... $$ Which is in the form $$\sum_{k=1}^{n}(-1)^{k}k$$
I am not sure how i should approach this
| When $n$ is even, such as $$-1+2-3+4-5+6=\color{red}{-1+1}+1\color{red}{-3+3}+1\color{red}{-5+5}+1=3$$
So inductively the total is $$\frac n2$$
When $n$ is odd, such as $$-1+2-3+4-5+6-7=\color{red}{-1+1}+1\color{red}{-3+3}+1\color{red}{-5+5}+1-7=-4$$
So inductively the total is $$-\frac{n+1}{2}$$
| {
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"url": "https://math.stackexchange.com/questions/2137234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Range of $f(z) = |1-z|+|1+z^2|$, where $z$ is a complex number If $z$ is a complex number such that $|z| = 1.$ then range of $f(z) = |1-z|+|1+z^2|$
Attempt: assuming $z=x+iy$ and $|z| = 1$ so $x^2+y^2$
so $f(x,y) = \sqrt{(1-x)^2+y^2}+\sqrt{(1+x^2-y^2)^2+4x^2y^2}$
could some help me, thanks
| HINT:
WLOG $z=\cos2y+i\sin2y$ where $y$ is real
$$1+z^2=1+\cos4y+i\sin4y=2\cos2y(\cos2y+i\sin2y)$$
$$\implies|1+z^2|=2|\cos2y|$$
$$1-z=1-\cos2y-i\sin2y=2\sin^2y-2i\sin y\cos y=-2i\sin y(\cos y-i\sin y)$$
$$\implies|1-z|=2|\sin y|$$
As $\cos2y=1-2\sin^2y=1-2s^2,$ writing $|\sin y|=s$
Case$\#1:$ For $2s^2\le1\iff0\le s\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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line through origin meets two lines at $P$ and $Q,$ then $(PQ)^2$ A line from the origin meet the lines $\displaystyle \frac{x-2}{1} = \frac{y-1}{-2} = \frac{z+1}{1}$ and $\displaystyle \frac{x-\frac{8}{3}}{2} = \frac{y+3}{-1}=\frac{z-1}{1}$
at points $P$ and $Q$ respectively, then $(PQ)^2$ is
Attempt: assuming equati... | We then have substituting the point $P $ into the equation of the first line as: $$\frac {a\lambda -2}{1} = \frac {b\lambda + 1}{-2} = \frac {c\lambda +1}{1} = k $$ giving us $$ a\lambda = 2+k$$ $$b\lambda =-2k-1$$ $$c\lambda = k-1$$
Similarly substituting $Q $ in the second line gives us: $$a\mu = 2k_1 + \frac {8}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2138421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Infinitely many solutions for PDE $u_x+u_y=2xu$ I want to show that the following PDE of the function $u(x,y)$ has infinitely many solutions:
$$
\left\{
\begin{array}{c}
u_x+u_y=2xu \\
u(x,x)=e^{x^2} \\
\end{array}
\right.
$$
By using the method of characteristics and choosing a curve $\Gamma(r,r,e^{r^2})$ in $u(x(... | Consider the Ansatz $u(x,y) = \exp\left(ax^2 + bxy + cy^2\right)$, then $a,b,c$ must satisfy
\begin{align}
u_x + u_y &= \left( 2ax + by \right) \exp\left( a^2 + bxy + cy^2 \right) + \left( bx + 2cy \right) \exp\left( ax^2 + bxy + cy^2 \right) \\
&=\left( 2a + b \right)x u(x,y) + \left( 2c +b \right) y u(x,y)
\end{align... | {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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The sum of numbers being unexpected squares We have the following result:
If $a$, $b$ and $c$ are pairwise coprime integers, such that
$$\frac 1a+\frac 1b=\frac 1c$$
then $a+b$, $a-c$ and $b-c$ are perfect squares.
What I did.
I tried to prove first that $a+b$ is a perfect square.
Multiplying by $abc$ we get
$$c(a+b... | The equation $\frac1a+ \frac1b = \frac1c$ is equivalent to $ab=c(a+b) \implies (a−c)(b−c)=c^2$.
Since $a,b,c$ are positive integers, ${1\over{a}}<{1\over{c}}$ and ${1\over{b}}<{1\over{c}}$. Hence $a>c$ and $b>c$. For each pair of integers $x,y $ satisfying $xy=c^2$, we get $a−c=x$ , $b−c=y$.
$\gcd(a,b) = 1$ thus if a p... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the minimum real number $\lambda$ so that the following relation holds ($x>y$): $\lambda\frac{x^3-y^3}{(x-y)^3}\geq \frac{x^2-y^2}{(x-y)^2}$
ّFind the minimum real number $\lambda$ so that the following relation holds for arbitrary real numbers $x,y$($x>y$): $$\lambda\frac{x^3-y^3}{(x-y)^3}\geq \frac{x^2-y^2}{(x-... | Your work is interesting and I have a possible solution using that method,
First I'd like to check out this approach.
I would always factor first. I don't know if you did that so I'm working it out.
We are given that $x > y$ so $(x-y) >0 $ and all divisions by $(x-y)$ are defined. Multiplication by $(x-y)$ doesn't chan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How do I find the diameters of the circles in this geometry puzzle? My family and I like to do a daily quiz but this particular question has had us baffled for weeks. Please help. We only have basic mathematical knowledge.
| $\triangle DCE$, $\triangle ECA$ and $\triangle AED$ are right-angled triangles. We therefore have
$$ \begin{align}
AD^2 &= AE^2 + DE^2 \\
&= AC^2 + CE^2 + EC^2 + CD^2 \\
&= AC^2 + (AC - 5)^2 + (AC - 5)^2 + (AC - 9)^2
\end{align}$$
But we also have $AD = 2AC - 9$, so we can solve for $AD$:
$$ \beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2140204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
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"answer_id": 2
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Is there a quicker way to evaluate $\int_{0}^{\infty} \frac{1-x^2}{x^4+3x^2+1}\ dx$? The integral is: $$\int_{0}^{\infty} \frac{1-x^2}{x^4+3x^2+1}\ dx$$
My procedure:
$$4\int_{0}^{\infty} \frac{1-x^2}{(2x^2+3)^2-5}\ dx=4\int_{0}^{\infty} \frac{1-x^2}{(2x^2+3-\sqrt5)(2x^2+3+\sqrt5)}\ dx$$
$$$$Using partial fractions to ... | Substitute $x\mapsto\frac{1-x}{1+x}$ to reveal an integral of an odd function over an interval symmetric about the origin:
$$\begin{align*}
I &= \int_0^\infty \frac{1-x^2}{1+3x^2+x^4} \, dx \\[1ex]
&= 8 \int_{-1}^1 \frac x{4+6x^2+5x^4} \, dx \\[1ex]
&= \boxed{0}
\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Evaluating $\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$ Let $A(z_1),B(z_2),C(z_3)$ are complex numbers satisfying $|z-\sqrt{3}i|=1$ and $3z_1+\sqrt{3}i=2z_2+2z_3$. The question asks to find the value of $$\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$$ I tried to shift the origin to $\sqrt{3}i$ and transformed the original condition in... | Put $t_j = z_j - \sqrt{3}i$. Then $|t_j|=1$ and $3t_1 = 2t_2 + 2t_3$. Hence $\frac{3}{4}t_1 = \frac{1}{2}(t_2 + t_3)$. The points $t_j$ lie on the unit circle, with center $O$ at the origin and form a triangle $ABC$, with $A = t_1$ etc. The midpoint $D$ of the chord $BC$ is $\frac{3}{4}t_1$ and hence lies on $OA$ at a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Find the remainder of $\frac{3^{11}-1}{2}$ divided by $9$ I have the following question:
Find the remainder of $$\frac{3^{11}-1}{2}$$ divided by $9$
I tried to reformat the question:
$$\frac{3^{11} -1 } {2} \times \frac{1}{9} = \frac{3^{11}-1}{18}$$
Since $3^2 = 9$
$$\frac{3^2(3^9) -1}{3^2 \times2}$$
I don't know whe... | $3^{11}$ is divisible by $9$.
So, $3^{11} - 1$, when divided by $9$, will give a remainder of $8$.
Division equation-
$a=bq+r$
$a= 3^{11} - 1$
$b= 9$
$r= 8$
$3^{11} - 1 = 9q + 8$
The $q$ given here will be even since $3^{11} - 1$ is even, $r$ is even and so $9q$ must also be even. Hence, $q$ is of the form $2k$ for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Verify that the following sequence defined recursively admits as closed formula another sequence Verify that the following sequence defined recursively
$\left \{ a_n \right \}_n \left \{ \begin{array}{rcl} a_0 & = & 0 & n = 0 \\ a_1 & = & 1 & n=1 \\ a_n & = & a_{n-1} + a_{n-2} & n \ge 2 \end{array} \right .$
admits a... | Since $\frac{1+\sqrt{5}}{2}$ and $\frac{1-\sqrt{5}}{2}$ are roots of the polynomial $x^2-x-1=0$
You can substitute $p^2=p+1$ and $q^2=q+1$.
So
$$\frac{1}{\sqrt{5}}[p^{n-1}+p^{n-2}-q^{n-1}-q^{n-2}]=\frac{1}{\sqrt{5}}[p^{n-2}(p+1)-q^{n-2}(q+1)]=\frac{1}{\sqrt{5}}[p^n-q^n]$$
| {
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"url": "https://math.stackexchange.com/questions/2149466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove that $\sum\limits_{cyc}\frac{a^2-bd}{b+2c+d}\geq0$
Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\frac{a^2-bd}{b+2c+d}+\frac{b^2-ca}{c+2d+a}+\frac{c^2-db}{d+2a+b}+\frac{d^2-ac}{a+2b+c}\geq0$$
This inequality is a similar to the following inequality of three variables.
Let $a$, $b$ and $c$ be po... | The following quantity is clearly positive
\begin{eqnarray*}
((2(d^3+b^3)+8bd(b+d))(b-d)^2+(2(a^3+c^3)+8ca(c+a))(c-a)^2)+
((a+c)(5(d^2+b^2)(b-d)^2+12bd(b-d)^2)+(b+d)(5(a^2+c^2)(a-c)^2+12ac(a-c)^2))+
2((b^3+d^3)(a-c)^2+(a^3+c^3)(b-d)^2)+
14(bd(a+c)(a-c)^2+ac(b+d)(b-d)^2)
\end{eqnarray*}
After doing some algebra, this is... | {
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"url": "https://math.stackexchange.com/questions/2151579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find a given logarithmic definite integral Find the following integral, where $a$ is a real number bigger than $1$:
$$\int_1^{a^2} \frac{\ln x}{\sqrt x(x + a)}\,\mathrm dx.$$
By using the substitution $t = \sqrt x$, I got this new integral which seems to be easier to solve, but I haven't found any way to do it yet:
$$4... | Let $t=\sqrt{x}$
\begin{equation}
I = \int\limits_{1}^{a^{2}} \frac{\ln x}{\sqrt{x}(x+a)} dx
= 4 \int\limits_{1}^{a} \frac{\ln t}{t^{2}+a} dt
\end{equation}
Integrating by parts, we have
\begin{align}
I_{1} &= \int\limits_{1}^{a} \frac{\ln t}{t^{2}+a} dt \\
&= \frac{\ln t}{\sqrt{a}} \tan^{-1}\left( \frac{t}{\sqrt{a}} \... | {
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"url": "https://math.stackexchange.com/questions/2154470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why do you need two fractions for partial fraction decomposition with repeated factors? For example, suppose my denominator contains $(x - 1)(x - 1)$.
I know I need two fractions, one with $(x - 1)$ and one with $(x - 1)^2$ as the denominator. But I'm looking for a deeper reason as to why. It makes sense when you go th... | You don't actually need two fractions for the squared factor. If for example you wanted to express $\frac{4x}{(x+1)(x-1)^2}$ in partial fractions, you could express it as $$\frac{4x}{(x+1)(x-1)^2}\equiv\frac{a}{x+1}+\frac{bx+c}{(x-1)^2}$$ This gives you $$4x\equiv a(x-1)^2+(bx+c)(x+1)\equiv (a+b)x^2+(-2a+b+c)x+(1+c)$$a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2158948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Proving the Hypergeometric Sequence
Question: How do you prove$$_4F_3\left[\begin{array}{c c}\frac 12n+1,n,-x,-y\\\frac 12n,x+n+1,y+n+1\end{array};-1\right]=\dfrac {\Gamma(x+n+1)\Gamma(y+n+1)}{\Gamma(n+1)\Gamma(x+y+n+1)}\tag{1}$$For $\Re(2x+2y+n+2)>0$
I'm not sure how to prove this. There are a multitude of other sim... | An established identity, useful to transform a $_4F_3 $ hypergeometric sequence with negative unit argument in a $_3F_2$ sequence, is
$$_4F_3 \left[\begin{array}{c c} a, b, c, d \\ a - b +1, a - c+1, a - d+1 \end{array};-1\right] \\ = \dfrac {\Gamma[a - b+1] \Gamma[a - c+1]}{\Gamma[a+1] \Gamma[a - b - c+1]} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2159662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Changing the order of double integrals in polar coordinates. How do i change the order of the integrals of a multiple integral of the following: $\int_0^{2\pi}\int_0^{1+\cos(\theta)}r\text{ }dr\text{ }d\theta$ ?
| For each $r_0$ between $0$ and $2$, the circle $r = r_0$ and the cardioid $r = 1+ \cos\theta$ intersect at most two points between $-\pi$ and $\pi$: $\cos^{-1}(r_0 - 1)$, and $- \cos^{-1}(r_0 - 1)$. When $r_0=0$, both points are the origin; when $r_0 = 2$, both points are $(2,0)$.
(Screenshot from Desmos; you can ad... | {
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"url": "https://math.stackexchange.com/questions/2160658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Polynomial system If there are 3 numbers $x,y,z$ satisfying
$f=x+y+z=3$ , $g=x^2+y^2+z^2=5$ , $h=x^3+y^3+z^3=7$ then prove that they also satisfy
$x^4+y^4+z^4=9$ but not $x^5+y^5+z^5=11$
I dont know how to tackle this to be honest, i have started trying to write $x^4+y^4+z^4-9$ as $r_1\times f + r_2\times g+ r_3\time... | For notational ease, let's define a few variables:
\begin{align*}
j&=x^3y+x^3z+xy^3+xz^3+y^3z+yz^3\\
k&=x^2y^2+x^2z^2+y^2z^2\\
l&=x^2yz+xy^2z+xyz^2.
\end{align*}
Now, let's start multiplying things out:
\begin{align*}
fh&=(x^4+y^4+z^4)+(xy^3+xz^3+yx^3+yz^3+zx^3+zy^3)\\
&=(x^4+y^4+z^4)+j\\
g^2&=(x^4+y^4+z^4)+2(x^2y^2+x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2162199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How to simplify $\frac{75}{8}\times\sqrt{\frac{a^3}{9}-\frac{a^3}{25}}$ How would you go about simplifying the expression $\frac{75}{8}\times\sqrt{\frac{a^3}{9}-\frac{a^3}{25}}$?
| We have:
$$\frac{75}{8}\times\sqrt{\frac{a^3}{9}-\frac{a^3}{25}}$$
Let's deal with the stuff inside the square root first.
$${\frac{a^3}{9}-\frac{a^3}{25}}$$
Finding a common denominator and simplifying:
$${\frac{(a^3)(25)-(a^3)(9)}{225}}$$
$${\frac{25a^3-9a^3}{225}}$$
$${\frac{16a^3}{225}}$$
$${\frac{16}{225}}\times a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Binomial coefficient complex expression I have been trying to find the coefficient of
$a^2x^3$, in the expansion of $(a+x+c)^2(a+x+d)^2$, without success.
I am having trouble expanding the above expression, because I can't find a way to merge them into one.
How can I solve this? Thanks
| Here is a variation to determine the coefficient of $a^2x^3$ without a full expansion of the expression. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in an expression.
We obtain
\begin{align*}
[a^2x^3]&(a+x+c)^2(a+x+d)^2\\
&=[a^2]\left([x^2](a+x+c)^2\right)\left([x^1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Solve an integral $\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$ Solve an integral $$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx$$
Using the third substitution of Euler,
$$\sqrt{x^2-1}=(x-1)t\Rightarrow x=\frac{1+t^2}{1-t^2},dx=\frac{4t}{(1-t^2)^2}$$
we get an integral $$\int\frac{1-t^2}{1+t^4}dt=\frac{1}{2}\int\frac{1-\sqrt 2 t}{t^2-\... | Here's what I tried. Setting $x=\sec \theta$, we get
\begin{align*}
\int \frac{1}{(x^2+1)\sqrt{x^2-1}} \mathrm{d}x &= \int \frac{\sec \theta \tan \theta}{(\sec^2 \theta+1)\sqrt{\sec^2 \theta-1}} \mathrm{d}\theta \\
&= \int \frac{\sec \theta}{\sec^2 \theta+1} \mathrm{d}\theta \\
&= \int \frac{\cos \theta}{\cos^2 \theta+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Sum of series $\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$ Find the radius of convergence and the sum of power series
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}$$
Radius of convergence is $R=1$, and the interval of convergence is $-1<x<1$.
I am having trouble in finding the sum.
Here is what I have tried.
$$\sum... |
How to find the sum of $\displaystyle \,\sum_{n=1}^{+\infty}n(xt)^{2n}\,?$
One may start with the standard finite evaluation:
$$
1+x+x^2+...+x^n=\frac{1-x^{n+1}}{1-x}, \quad |x|<1, \tag1
$$ then by differentiating $(1)$ we have
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2172875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
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Calculus: Partial Derivatives Question Consider the following function.
$H(x,y) = 4 \ln(x^3 + 4y^2)$
(a) Find $f_{xx}(2,3)$.
(b) Find $f_{yy}(2,3)$.
(c) Find $f_{xy}(2,3)$.
I know that I should use second order partial derivatives to solve this problem.
$f_{xx} = (f_x)_x$
$f_{yy} = (f_y)_y$
$f_{xy} = (f_x)_y$
but ... | $H(x,y)=4 ln(x^3+4y^2)$
So $f_{x}(x,y)=\frac{4}{x^3+4y^2}(3x^2)=\frac{12x^2}{x^3+4y^2}$
And $f_{xx}(x,y)=\frac{12(-x^4+8xy^2)}{(x^3+4y^2)^2}$
Plugging in for $x=2, y=3$ will leave you with $f_{xx}(2,3)=96/121$
The trick is to treat $y$ as a constant. Keep that in mind when you partially differentiate.
$f_{y}(x,y)=\frac... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate the sum of series $\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$ Evaluate the sum of series
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}$$
I have tried two methods:
1) using power series
2) using partial sums
but I can't find the sum.
1) Using power series:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{... | $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
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"language": "en",
"url": "https://math.stackexchange.com/questions/2181754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Triangle with given data constructible? So, this is a problem in abstract algebra, not some elementary geometry: Is it possible to construct a triangle $ABC$ given $a$, which is side $BC$, $b$ which is side $AC$ and angle $\beta-\gamma$, where $\beta$ is the angle at point $B$ while $\gamma$ is the angle at point $C$? ... | \begin{eqnarray*}
\sin \beta = \frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{2ca} \\
\sin \gamma = \frac{\sqrt{(a+b+c)(-a+b+c)(a-b+c)(a+b-c)}}{2ab} \\
\cos \beta = \frac{c^2+a^2-b^2}{2ca} \\
\cos \gamma = \frac{a^2+b^2-c^2}{2ab} \\
\end{eqnarray*}
Quick sanity check ...
\begin{eqnarray*}
\cos \alpha = -\cos \beta \cos \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2183377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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18 positive integers satisfying $(x-a)(x-2a)(x-a^2)<0$ and $a>2$ find natural number a
let $a>2$ be a constant. If there are just 18 positive integers satisfying the inequality $(x-a)(x-2a)(x-a^2)<0$ then find the natural number $a$.
zeroes: $a\ \ \ \ \ \ \ 2a\ \ \ \ \ \ \ a^2$
$\ \ -ve\ \ +ve\ \ \ -ve\ \ \ \ +ve... | You already wrote that the zeroes are $a$, $2a$ and $a^2$.
Next, you need to check that the equation is true for $x<a$ and for $2a<x<a^2$ since $a$ is a natural number.
Since you only consider positive integers, you also have $x>0$.
There are $a-1$ positive integers between $0$ and $a$, and $a^2-2a-1$ positive intege... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2187690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Does the equation $U^2+V^2=A^2+sB^2$ with $s$ squarefree have a complete integer solution? I’m looking for a complete solution (parameterization or other) to the equation in the title, i.e.,
$$
U^2+V^2=A^2+sB^2,
$$
where $s$ is squarefree [if necessary]. When $s=1$, the solution is well-known (and easy to derive), so... | For the equation.
$$U^2+W^2=A^2+tB^2$$
You can write such a parameterization.
$$U=2ps(z^2+tq^2+x^2-y^2)+2x((p^2-s^2)y+(p^2+s^2)z)$$
$$W=(p^2-s^2)(z^2+tq^2-x^2+y^2)+2y(2psx+(p^2+s^2)z)$$
$$A=(p^2+s^2)(z^2-tq^2+x^2+y^2)+2z(2psx+(p^2-s^2)y)$$
$$B=2q(2psx+(p^2-s^2)y+(p^2+s^2)z)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Maximum value of expression: $\frac{ab+bc+cd}{a^2+b^2+c^2+d^2}$ What is the maximum value of
$$\frac{ab+bc+cd}{a^2+b^2+c^2+d^2}$$
where $a,b,c$, and $d$ are real numbers?
| Have a look at the Rayleigh quotient of two quadratic forms. This expression is homogeneous of degree zero in vector $(a,b,c,d)$ so you can look for its maximum on unit sphere.
Then if $X=(a,b,c,d)$ is on this sphere and
$$Q = \frac{1}{2}\begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192095",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve this in an efficient way (without calculators)
Solve for $x$, if
$$(x+4)(x+7)(x+8)(x+11)+20=0.$$
Is there an easier way to solve this than trying to multiply all the values together? I've tried multiplying all of them together, and I get an equation of the fourth degree which I find very hard to factor... | $f(x) = (x+4)(x+7)(x+8)(x+11)+20\ $ is symmetric about the line $x = -7.5$
$y = x + 7.5$
$(y-3.5)(y-0.5)(y+0.5)(y+3.5)+20=0$
$(y^2 - 0.5^2)(y^2 -3.5^2)+20=0$
$y^4 - 12.5 y^2 + 23.0625 = 0$
Use the quadratic formula to solve for $y^2.$
The square roots of that give $y.$
And, subtract $7.5$ to get $x.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 0
} |
Prime number decomposition What is the fastest way to decompose the given number to prime numbers without using calculator?
Example : $$3575$$
What I do is :
$$3575 = 3 \times 10^3 + 5 \times 10^2 + 7 \times 10 + 5 = 3\times5^3\times2^3 + 5^2 \times 2^2 \times 5 + 7 \times 5 \times 2 + 5$$
But now I do not know how to... | $$ 3600 - 25 = 60^2 - 5^2 = (60 + 5) (60 - 5) = 65 \cdot 55 $$
Then we see $65 = 5 \cdot 13$ and $55 = 5 \cdot 11$
This is usually called Fermat factorization. Start with the first square larger than the number, see if the difference is a square. If that does not work take the square just larger than that. To save tim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2195496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Using BMCT to prove the sequence $a_{n+1} = \sqrt{2 + a_n}$ is bounded and increasing. My solution:
Proof of (1)
$$< \sqrt{2 + 2} \text{ by I. H.}$$
$$= 2$$ As required. Therefore by PMI $\{a_n\}$ is bounded above by 2
Proof of (2)
$$(a_n - 2)(a_n + 1) < 0$$
Therefore $\{a_n \}$ is strictly increasing.
By BMCT it stat... | Here is a generalization.
My original answer is at the end.
If
$a_{n+1}
=\sqrt{a_n+d^2-d}
$
where $d > 1$,
then once
$a_n < d$
then
$a_n \to d$
linearly.
This problem is the case
$d = 2$.
If $a_n \lt d$
then
$a_{n+1} \lt \sqrt{d+d^2-d}
=d
$,
so that
all subsequent $a_n < d$.
Let $a_n = b_n+d$.
Want to show that
$b_n \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2196026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How can I prove this trigonometric equation with squares of sines? Here is the equation:
$$\sin^2(a+b)+\sin^2(a-b)=1-\cos(2a)\cos(2b)$$
Following from comment help,
$${\left(\sin a \cos b + \cos a \sin b\right)}^2 + {\left(\sin a \cos b - \cos a \sin b\right)}^2$$
$$=\sin^2 a \cos^2b + \cos^2 a \sin^2 b + \sin^2 a \c... | Given that
$$ \cos(u) = \frac{e^{iu} + e^{-iu}}{2} , \sin(u) = \frac{e^{iu} - e^{-iu}}{2i} \text{ , and } e^{2u}+ e^{-2u} = \left(e^{u} - e^{-u}\right)^2 + 2,$$
we prove the identity by
$$\begin{align} 1 - \cos(2a) \cos(2b) & = 1 - \frac{ \left( e^{2ia} + e^{-2ia} \right) \left(e^{2ib} + e^{-2ib} \right)}{4}\\
& = \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2198091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 5
} |
Computing $ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$ $$ \int_0^1 \frac{1 + 3x +5x^3}{\sqrt{x}}\ dx$$
My idea for this was to break each numerator into its own fraction as follows
$$ \int_0^1 \left(\frac{1}{\sqrt{x}} + \frac{3x}{\sqrt{x}} + \frac{5x^3}{\sqrt{x}}\right)dx$$
$$ \int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2})\... | Putting $t=\sqrt{x}$ you have $dt=\frac{1}{2\sqrt{x}}dx$ and the limits stay the same.
$$\int_0^12(1+3t^2+5t^6)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
How to evaluate $\int \frac{2 x^3 - 3 x^2 - 26 x + 38}{x^4 - 2 x^3 - 13 x^2 + 38 x - 12} \, dx$
Evaluate the integral $$\int \frac{2 x^3 - 3 x^2 - 26 x + 38}{x^4 - 2
x^3 - 13 x^2 + 38 x - 12} \, dx$$
I tried to split the integral with partial fractions but I could not find a suitable factorization for the denominato... | Well, I can take a stab at starting it:
$$
\begin{align*}
\int \frac{2 x^3 - 3 x^2 - 26 x + 38}{x^4 - 2
x^3 - 13 x^2 + 38 x - 12} \, dx &= \int \frac{4 x^3 - 6 x^2 - 26 x + 38}{x^4 - 2
x^3 - 13 x^2 + 38 x - 12} \, dx\\ &\qquad+ \int \frac{-2x^{3} + 3x^{2}}{x^4 - 2
x^3 - 13 x^2 + 38 x - 12}\,dx\\
&=\ln\left|x^4 - 2
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Show: $\cos \left( \frac{ 3\pi }{ 8 } \right) = \frac{1}{\sqrt{ 4 + 2 \sqrt{2} }}$ I'm having trouble showing that:
$$\cos\left(\frac{3\pi}{8}\right)=\frac{1}{\sqrt{4+2\sqrt2}}$$
The previous parts of the question required me to find the modulus and argument of $z+i$ where $z=\operatorname{cis{\theta}}$. Hence, I found... | $$
\begin{aligned}
\text{By}\cos ^{2} x &=\frac{1+\cos 2 x}{2}, \textrm{ we have } \\
\cos \left(\frac{3 \pi}{8}\right) &=\sqrt{\frac{1+\cos \frac{3 \pi}{4}}{2}} \\
&=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}} \\
&=\frac{\sqrt{2-\sqrt{2}}}{2}
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2206391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Probability Interview Question - Brain teaser
Player A has a thirty-sided and Player B has a twenty-sided die. They both roll the dies and whoever gets the higher roll wins. If they roll the same amount Player B wins. What is the probability that Player B win?
So I have
$$1 - \left(\frac{1}{3}+\frac{1}{2}\cdot\frac{2... | Let $A$ throw $1..30$.
$B$ has no chance against $21..30$ with probability $\frac{1}{3}$: it wins $0\times\frac{1}{3}$ such throws.
$B$ wins $\frac{2}{3}\times\frac{1}{20}$ equal throws.
$B$ also wins half of the remaining $19$ throws of $A$: $\frac{1}{2}\times\frac{19}{20}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 4
} |
Analyze solutions to a matrix 3x5 matrix with two parameters and find a unique solution I have the linear system (over $\mathbb{R}$) for which I need to find a unique solution:
$$\begin{cases}
4x+8y+7z+3cw = 3b \\
x+2y+2z+cw=b \\
2x+4y+2z+(c-1)w=b
\end{cases}$$
for which the corresponding matrix is:
$$\begin{bmatrix}
4... | So we have system of equations:
$\begin{cases}
4x+8y+7z+3cw=3b
\\
x+2y+2z+cw=b
\\
2x+4y+2z+(c-1)w=b
\end{cases}$
And the matrix of it is:
$\begin{bmatrix}
4&8&7&3c&3b
\\
1&2&2&c&b
\\
2&4&2&c-1&b
\end{bmatrix}$
Using these steps:
$1)\,R1\rightleftharpoons R2$
$2)\,R2=R2-4R1$
$3)\,R3=R3-2R1$
$4)\,R1=R1+R2$
$5)\,R3=R3-2R2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2211436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving equations with floor function and square roots? I was solving a problem to discover n and after I transformed the problem it gave me this equation:
\begin{equation*}
\left\lfloor{\frac{2}{3}\sqrt{10^{2n}-1}}\right\rfloor = \frac{2}{3}(10^{n}-1)
\end{equation*}
So I tried to simplify it by defining:
\begin... | Note that
$$-8(10^n-1)\ \le\ 0\ <\ 4\cdot10^n+5$$
is true for all $n\in\mathbb Z^+$. In other words
$$-8\cdot10^n+4\ \le\ -4\ <\ 4\cdot10^n+1$$
$$\iff\ 4(10^n-1)^2\ \le\ 4(10^{2n}-1)\ <\ (2\cdot10^n+1)^2$$
$$\iff\ 2(10^n-1)\ \le\ 2\sqrt{10^{2n}-1}\ <\ 2\cdot10^n+1$$
$$\iff\ \frac23(10^n-1)\ \le\ \frac23\sqrt{10^{2n}-1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2213807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integration of rational functions How would you integrate a rational function like this:
$$\frac{2x+3}{x^2+2x+10}$$
| $x^{2}+2x+10$ prime $\rightarrow$
$$I=\int\dfrac{2x+3}{x^{2}+2x+10}dx=\int \dfrac{2x+2+1}{x^{2}+2x+10}dx=\int \dfrac{2x+2}{x^{2}+2x+10}dx +\int \dfrac{1}{(x^{2}+2x+1)+9}dx=\ln (x^{2}+2x+10)+J$$
$$J=\int \dfrac{1}{(x+1)^{2}+3^{2}}dx=\int \dfrac{1}{9\left[ \left ( \dfrac{x+1}{3}\right)^{2}+1\right]}dx=\dfrac{1}{3}\int \d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2214614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Evaluating the indefinite integral $\int \frac{x^3}{\sqrt{4x^2 -1}}~dx$.
Find $$\int {x^3\over \sqrt{4x^2 -1}}\,dx.$$
Let
$2x = \sec u$, $2 =\sec (u) \tan(u) u^{'}(x).$ Then
$$
\begin{align*}
\int {x^3\over \sqrt{4x^2 -1}}\,dx &= \frac1{16}\int {\sec^3 u\over \tan u}\tan u \sec u \, du\\
&= \frac1{16}\int {\sec^4... | There is nothing wrong with your approach; you just need to fiddle around with trig identities to see that $\tan(\sec^{-1}2x)=\sqrt{\sec^2(\sec^{-1}2x)-1}=\sqrt{(2x)^2-1}$, etc. But another, possibly easier, way to do the integral is to let $u=4x^2$ so that $du=8x\,dx$ and thus
$$\int{x^3\over\sqrt{4x^2-1}}dx={1\over3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2216699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Proving the existence and number of *real* roots for $x^3 - 3x + 2$ I need to find how many real roots this polynomial has and prove there existence. I was wondering if my logic and thought process was correct.
Determine the number of real roots and prove it for $x^3 - 3x + 2$
First, note that $f'(x) = 3x^2 - 3$ and ... | $x^3 - 3x + 2$
Possible roots are ±1 and ±2:
$f(1) = 1 - 3 + 2 = 0$
$f(2) = 8 - 6 + 2 = 4$
$f(-1) = (-1) - (-3) + 2 = 6$
$f(-2) = (-8) - (-6) + 2 = 0$
$x=1$ and $x=-2$ are the roots so we can factor the expression into $(x-1)(x+2)(x-1)$
We have the equation $(x-1)^2(x+2) = 0$
$(x-1)^2 = 0 \lor x + 2 = 0$
$\sqrt{(x-1)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2217630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Calculting limit $\cos(\sqrt{x+1})-\cos(\sqrt{x})$ as $x\to \infty$ I am not sure how to calculate the limit:
$$\lim_{x\to\infty}\cos(\sqrt{x+1})-\cos(\sqrt{x})$$
I applied trigonometric identity to get:
$$L=-\lim_{x\to \infty} 2\sin\left(\dfrac{1}{2}\dfrac{1}{(\sqrt{1+x}-\sqrt{x})}\right)
\sin\left(\dfrac{1}{2}\dfrac... |
METHODOLOGY $1$: Pre-Calculus Approach
Recalling that $\cos(x)-\cos(y)=-2\sin\left(\frac{x-y}{2}\right)\sin\left(\frac{x+y}{2}\right)$, $|\sin(\theta)|\le |\theta|$, and $|\sin(\theta)|\le 1$, we have
$$\begin{align}
\left|\cos(\sqrt{x+1})-\cos(\sqrt{x})\right|&=2\left|\sin\left(\frac{\sqrt{x+1}-\sqrt{x}}{2}\right)\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2218736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Determining a matrix given the characteristic and minimal polynomial Let $p_a=(x-2)^2(x-7)^4x$ be the characteristic polynomial of the matrix $A$ and $(x-2)^2(x-7)x$ the minimal polynomial. Determine the matrix $A$.
My work: I know the matrix has to be $7x7$ and in its diagonal it must have two $2$, four $7$ and one $0... | The minimal polynomial in this case gives you the information about the relevant Jordan blocks. Since it has $(x-2)^2$ as a factor, you must have one $2 \times 2$ Jordan block associated to the eigenvalue $2$ (and not two $1 \times 1$ Jordan blocks). To see why, note that the minimal polynomial of
$$ \begin{pmatrix} 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2219832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Mathematical Olympiad Treasures Problem 1.11 Let $n$ be a positive integer, prove that:
$3^{3^{n}}(3^{3^{n}}+1) + 3^{3^{n} + 1} - 1$
is not prime
The solution states:
Observe:
$3^{3^{n}}(3^{3^{n}}+1) + 3^{3^{n} + 1} - 1 = a^{3} + b^{3} + c^{3} - 3abc$
for $a = 3^{3^{n-1}}$, $b = 9^{3^{n-1}}$, $c = -1$
After that main ... | $$3^{3^{n}}(3^{3^{n}}+1) + 3^{3^{n} + 1} - 1=\\
3^{2\cdot3^n}+3^{3^n}-1+3^{3^n+1}=\\
9^{3^n}+3^{3^n}-1+3^{3^n+1}\\
(9^{3^{n-1}})^3+(3^{3^{n-1}})^3+(-1)^3-3(9^{3^{n-1}}\cdot 3^{3^{n-1}}\cdot(-1))$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2221424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate the$\int_0^1\int_{\sin^{-1} y}^{\pi/2} \cos x\sqrt{1+\cos^2}\,dxdy$ and $\int_0^2\int_0^1\int_y^1 \sinh(z^2)\,dzdydx$
*
*Evaluate the following integrals:
\begin{gather}
\int_0^1\int_{\sin^{-1} y}^{\pi/2} \cos x\sqrt{1+\cos^2 x}\,dxdy;\\
\int_0^2\int_0^1\int_y^1 \sinh(z^2)\,dzdydx
\end{gather}
*... | Help with number 1?
Hint. You are on the right track. To evaluate
$$
\int \sqrt{2-u^2}du
$$ one may use the change of variable $u=\sqrt{2}\cdot \sin t$ obtaining
$$
\begin{align}
\int \sqrt{2-u^2}du&=2\int \cos^2 t \:dt
\\&=2\int \left(\frac12+\frac{\cos (2t)}2 \right)dt
\\&=\int \left(1+\cos (2t) \right)dt
\\&=t+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2223429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving Trigonometric Equation How to get $ \left(5+\sqrt{6}\right)$ from the expression
$$6\sqrt { 3 } \sin { \left( \frac { 2\pi }{ 3 } -\arcsin { \left( \frac { 5\sqrt { 3 } }{ 9 } \right) } \right) } $$
without using calculator for examination purpose?
| by using identity $$\sin { \left( \alpha -\beta \right) =\sin { \alpha \cos { \beta -\sin { \beta \cos { \alpha } } } } } $$ we get $$6\sqrt { 3 } \sin { \left( \frac { 2\pi }{ 3 } -\arcsin { \left( \frac { 5\sqrt { 3 } }{ 9 } \right) } \right) } =6\sqrt { 3 } \left[ \sin { \frac { 2\pi }{ 3 } \cos { \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2225052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
"Extraneous solution" solves original equation For the given equation:
$$x - 10 = \sqrt{9x}$$
when one simplifies, through the following steps:
\begin{align*}
x^2 - 20x + 100 &= 9x\\
x^2 - 29x + 100 &= 0\\
x &= 25, 4
\end{align*}
we check for extraneous solutions to make sure we have not altered from the set of solutio... | This is because you thought that $\sqrt{36}=\pm 6$ though it is not. Look when we do the checking at $x=4$, we get $x-10=4-10=-6$ whereas $\sqrt{9x}=\sqrt{36}=6$ which shows that $x-10=\sqrt{9x}$ does not hold when $x=4$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I prove that this sequence is monotonic? I have a sequence $(u_n)$ that is defined as:
$u_0 = 2$,
$u_{n+1} =\frac{u_n}{2} + \frac{1}{u_n}$
I have tried to prove that it is monotonic using induction but I wasn't able to succeed.
How can I prove it easily ?
Thank you
| Let us show first that
$$u_n \ge \sqrt{2}$$
This can be done by induction: $u_1 \ge 2$, and if $u_m \ge \sqrt{2}$ then
$$u_{m+1} = \dfrac{u_m}{2}+\dfrac{1}{u_m} \geq \sqrt{2}$$
The last inequality holds because of the following reason. Consider the function
$$f(x) = \dfrac{x}{2} + \dfrac{1}{x}$$
Its derivative:
$$f'(x... | {
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"url": "https://math.stackexchange.com/questions/2226280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $a_n+b_n\sqrt{3}=(2+\sqrt{3})^n$, then what's $\lim_{n\rightarrow\infty}\frac{a_n}{b_n}$? Let $a_n$ and $b_n$ be integers defined in the following way: $$a_n+b_n\sqrt{3}=(2+\sqrt{3})^n.$$ Compute $$\lim_{n\rightarrow\infty}\frac{a_n}{b_n}.$$
I tried expanding using binomial theorem:
$$ (2+\sqrt{3})^n=\binom{n}{0}2+\... | For fun, you can also do this with linear algebra. Let $V = \mathbb{Q}(\sqrt{3})$ as a $\mathbb{Q}$-vector space with basis $\{1,\sqrt{3}\}$. The "multiplication-by-$(2+\sqrt{3})$" map looks like $\begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix},$ so $$\begin{pmatrix} a_n \\ b_n \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 1 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2226394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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What is the probability that if $5$ balls are distributed to $3$ bags (with no bag left empty) that there is exactly one ball in the first bag? A friend of mine gave me the following exercise:
There are $5$ balls and $3$ bags, there are no empty bags (each bag contains at least $1$ ball). What is the probability to hav... | The answer to your question depends on whether the balls are distinct. If the balls are identical and the bags are distinct (which seems to be a reasonable assumption since there is a first bag), your answer is correct.
What is the probability that if five identical balls are distributed to three bags so that there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2231429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Roots of Quadratic lies in $(0, 1)$ Given that the Quadratic equation $f(x)=ax^2-bx+c=0$ has two roots in $(0 \: 1)$
if $a,b,c \in \mathbb{N}$ Find Minimum values of $a$ and $b$
Since $a$ is Natural number graph of parabola will be open upwards.
Now $f(0) \gt 0$ and $f(1) \gt 0$ so we get
$c \gt 0$ and $a+c \gt b$ and ... | By inspection, it looks like $y = f(x) = 4x^2 - 4x + 1$ is a candidate.
Analysis:
*
*$f(0) = c > 0$ because $c \in \mathbb{N}$.
Thus, the parabola is open upwards because it has to cross two points $0 < x_1, x_2 < 1$ on the $x$ axis and passing the point $(0,c)$ in the first quadrant.
*
*Because the parabola... | {
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"timestamp": "2023-03-29T00:00:00",
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$1+(1+2+4)+(4+6+9)+(9+12+16)+.......+(361+380+400)=?$ I came across this question today.
$1+(1+2+4)+(4+6+9)+(9+12+16)+.......+(361+380+400)$=?
Now, my workout
If we simplify the expression it comes to be
$1+7+19+37+.....+1141$
Here we see that from the second term to the first term there is a difference of 6 and then f... | Let :
$$\text{S}= 1+7+19+37+\dots +a_n$$
$$\begin{align}~~~~~~~~~~~\text{-S}=~~~-1-7-19- \dots -a_{n-1}-a_n \end{align}$$
Adding these two :
$$0=1+6+12+18+ \cdots 6(n-1)-a_n$$
This gives $$a_n=1+6+12+18+ \cdots 6(n-1)$$
Now ,
$$a_n=1+6(1+2+3 \cdots (n-1))$$
$$a_n=1+6\cdot \frac{n(n-1)}2=3n^2-3n+1$$
$$S=\sum a_n= \sum \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2233718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given functions $f,g$ find the direct form for $f(x)=g(x)+f(x-1)$ Let $g(x)=1+x+\frac{x(x+1)}{2}$ and let $f(x)=g(x)+f(x-1)$. Find the closed form for $f(x)$ and $x\geq0$ given that $f(0)=0$.
So what we are asked are is a direct form for $$\sum_{n=1}^{k}g(n)$$
I want to show that $$\sum_{n=0}^{k}g(n)=\frac{1}{2}(\sum_{... | Hint: $\;g(k)=1 + \frac{3}{2} k + \frac{1}{2} k^2\,$, then by direct calculation:
$$
f(n) = \sum_{k=1}^{n}g(k) = \sum_{k=1}^{n} 1 + \frac{3}{2} \sum_{k=1}^{n} k + \frac{1}{2} \sum_{k=1}^{n} k^2 = n + \frac{3}{2} \frac{n(n+1)}{2} + \frac{1}{2}\frac{n(n+1)(2n+1)}{6} = \cdots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2234008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trigonometric equation 3 $$2 \arcsin x = \arccos 2x$$
I tried applying $\sin$ to the equation, so it results
$$\sin (2 \arcsin x) = \sin(\arccos 2x)$$
$$2 \sin(\arcsin x) \cdot \cos(\arcsin x) = \sin (\arccos 2x)$$
$$2x \cdot \sqrt{1 - x^2} = \sqrt{1- 4x^2}$$
But this doesn't result in the correct answer $\frac{\sqrt{3... | NOTE: I would suggest doing $\cos(2\arcsin x)=\cos(\arccos 2x)$ instead which seems easier since $\cos(2\arcsin x)=1-2\sin^2(\arcsin x)=1-2x^2$
About your method you're not missing anything
$$(2x)^2(1-x^2)=1-4x^2\\4x^2-4x^4=1-4x^2\\4x^4-8x^2+1=0\\x^2=t\\4t^2-8t+1=0\\t_{1,2}=\frac{8\pm\sqrt{64-16}}{8}=\frac{8\pm4\sqrt{3... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find limit of sequence using integral Given sequence $$a_m = \frac{1}{m^2} \sum \limits_{k = 1}^m \sqrt[3]{(mx + k + 1)\cdot (mx + k)^2}$$
Find its limit using integral.
I thought it may be solved using either Euler-Mclaurin formula or Riemann sum. Unfortunately, the function under the sum sign is pretty uncomfy to ope... | Hint. One may write, as $n \to \infty$,
$$
\begin{align}
a_n = \frac{1}{n^2} \sum \limits_{k = 1}^n \sqrt[3]{(nx + k + 1)\cdot (nx + k)^2}&= \frac{1}{n} \sum \limits_{k = 1}^n \sqrt[3]{\left(x + \frac{k+1}{n}\right)\cdot \left(x + \frac{k}{n}\right)^2}
\end{align}
$$ giving
$$
\frac{1}{n} \sum \limits_{k = 1}^n\sqrt[3]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2236470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If matrix $A$ in $\mathbb{R}^3 $ such that, $A^3 = I$, $\det A = 1$. Is there a such matrix which is not orthogonal, rotation and identity? I tried to use Cayley–Hamilton theorem to learn something about the matrix.
Using the theorem we have:
$p(A)=0 = - A^3 + \text{tr} A\cdot A^2 -\left(\begin{vmatrix}a_{11} && a_{12}... | Since $A^3-I=0$ you get that the possible eigenvalues are $1, e^{2\pi i/3}, e^{4\pi i/3}$.
Moreover, since $A$ is $3 \times 3$ it has 1 or 3 real eigenvalues.
Case 1: All eigenvalues are real. Then, the minimal polynomial of $A$ is $(x-1)^k$ and divides $x^3-1$, therefore it is $x-1$. This shows that $A$ is diagonaliz... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2238320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
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Questions on orthogonal projections on a subspace
Let $u= \begin{pmatrix} 3 \\ 1 \\ 1\end{pmatrix}$ and $S = \left\{\frac{1}{3} \begin{pmatrix} 2 \\ -1 \\ -2\end{pmatrix}, \frac{1}{3}\begin{pmatrix} 1 \\ -2 \\ 2\end{pmatrix} \right\}$.
*
*Find the unique vectors $w \in W$ and $z \in W^\perp$, such that $z = u... | $w=(u\cdot v_1)v_1+(u\cdot v_2)v_2=(3/3)v_1+(3/3)v_2$
After plugging in the vectors and simplifying I got $w=(1,-1,0)$, thus $z=(2,2,1)$.
Projection of $u$ on $W$ is $(2,2,1)$.
$z\cdot v_1=4-2-2=0$
$z\cdot v_2=2-4+2=0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $n$ is an odd prime, and $a$ and $b$ are co-prime, and $a+b$ is not a multiple of $n$, prove that $\frac{a^n+b^n}{a+b}$ is co-prime to $a+b$. If $n$ is an odd prime, and $a$ and $b$ are co-prime, and $a+b$ is not a multiple of $n$, prove that $\frac{a^n+b^n}{a+b}$ is co-prime to $a+b$.
How does even even begin to pr... | Note that the problem does not make sense if $n=2$, as $\dfrac{a^2+b^2}{a+b}$ is not generally an integer. So we assume $n\ne2$, and hence $n$ is odd.
If $\dfrac{a^n+b^n}{a+b}$ and $a+b$ have a common factor then they have a common prime factor $p$. So
$$p\mid a+b\quad\hbox{and}\quad p\mid a^{n-1}-a^{n-2}b+a^{n-3}b^2... | {
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"timestamp": "2023-03-29T00:00:00",
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Least possible polynomial deegree of complex roots What is the least possible deegree of polynomial with real coefficients having roots $2\omega , 2+3\omega , 2+3\omega ^2 , 2-\omega -\omega ^2$
As there are four roots so the deegree should be four but the answer is given as five . how ?
| Such a polynomial cannot have degree $4$, because we have, using $1+\omega+\omega^2=0$,
\begin{align}f(x) & =(x-2\omega)(x-(2+3\omega))(x-(2+3\omega^2))(x-(2-\omega-\omega^2))\\
& =x^4 + 2x^3( - \omega - 2) + 2x^2(4\omega + 5) + x( - 20\omega - 21) + 42\omega,
\end{align}
which does not have real coefficients. Also, n... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\lim_{n\rightarrow \infty}\frac{1+2^2+3^3+4^4+\cdots +n^n}{n^n}$ Finding $$\lim_{n\rightarrow \infty}\frac{1+2^2+3^3+4^4+\cdots +n^n}{n^n}$$
Attempt: $$\lim_{n\rightarrow \infty}\bigg[\frac{1}{n^n}+\frac{2^2}{n^n}+\frac{3^3}{n^n}+\cdots \cdots +\frac{n^n}{n^n}\bigg] = 1$$
because all terms are approaching to z... | By Stolz $$\lim_{n\rightarrow \infty}\frac{1+2^2+3^3+4^4+\cdots +n^n}{n^n}=\lim\limits_{n\rightarrow\infty}\frac{n^n}{n^n-(n-1)^{n-1}}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2243586",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Limit of trigonometric function $\lim_{x\to\pi/3} \frac{1 - 2\cos(x)}{\sin(x - \frac{\pi}{3})}$ I want to compute this limit:
$\displaystyle \lim_{x\to\pi/3} \frac{1 - 2\cos(x)}{\sin(x - \frac{\pi}{3})}$
Using L'Hopital, is easy to get the result, which is $\sqrt{3}$
I tried using linear approximation (making $u = x - ... | This is actually much easier than it looks. By letting $x=u+\frac\pi3$, we get
$$
\lim_{x\to\pi/3} \frac{1-2\cos(x)}{\sin(x-\frac\pi3)} = \lim_{u\to0}\frac{1-\cos(u)+\sqrt{3}\sin(u)}{\sin(u)}
$$
Now, we write the limit on the right as
$$
\lim_{u\to0} \frac{1-\cos(u)}{\sin(u)} + \lim_{u\to0}\frac{\sqrt{3}\sin(u)}{\sin(u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\frac{a}{b} = \frac{c}{d}$ why does $\frac{a+c}{b + d} = \frac{a}{b} = \frac{c}{d}$? Can anyone prove why adding the numerator and denominator of the same ratios result in the same ratio? For example, since $\dfrac{1}{2}=\dfrac{2}{4}$ then $\dfrac{1+2}{2+4}=0.5$.
| We know $ad=bc$, so $ab+bc=ab+ad$. If you factor out this equation you get $b(a+c)=a(b+d)$ and then you get $\frac{a}{b}=\frac{a+c}{b+d}$. Similarly, you can prove the other equality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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What is(are) the value(s) of $\sqrt{i}+\sqrt{-i}$? When I am going to find out the value of $\sqrt{i}+\sqrt{-i}$, I stuck to evaluate $\sqrt{i}\times \sqrt{-i}$.
Progress: $\sqrt{i}+\sqrt{-i}=\sqrt{(\sqrt{i}+\sqrt{-i})^2}=\sqrt{2\times \sqrt{i}\times\sqrt{-i}}$.
Now $\sqrt{i}\times\sqrt{-i}=\sqrt{i\times (-i)}=\sq... | I'm using your method, but I'm restraining myself from writing $\sqrt{z}$ for $z$ complex. Let's use use only $[i^2=-1]$ and $[x^2=y^2\iff x=\pm y]$
Let's have $a^2=i$ and $b^2=-i$, we are searching for the value of $(a+b)$.
$(a+b)^2=a^2+2ab+b^2=i+2ab-i=2ab$
$a^2b^2=(i)(-i)=1\iff ab=\pm 1$
*
*If $ab=1$ then $(a+b)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2251769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Calculate sum of coefficients of polynomial Let $$(x + 1)(x^2 + 2)(x^2 + 3)(x^2 + 4)(x^2 + 5) = \sum_{k=0}^{9} (A_k \cdot x^k)$$
Compute:
*
*$$\displaystyle \sum_{k=0}^{9} A_k$$
*$$\displaystyle \sum_{k=0}^{4} A_{2k}$$
I tried to figure out from Viete's Sums how to rewrite this but I can't find the coefficients for... | Well, since
$$\sum_{k=0}^{9} A_kx^k=(x + 1)(x^2 + 2)(x^2 + 3)(x^2 + 4)(x^2 + 5)$$
we can just set $x=1$ to see
$$\sum_{k=0}^{9} A_k=(1 + 1)(1^2 + 2)(1^2 + 3)(1^2 + 4)(1^2 + 5)=720$$
Try setting $x=-1$; can you now find $\sum_{k=0}^{4} A_{2k}$?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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$\cos^2(\frac{\pi}{101})+\cos^2(\frac{2\pi}{101})+\cos^2(\frac{3\pi}{101})+...+\cos^2(\frac{100\pi}{101})=?$
Find the value: $$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cd... | HINT:
Note that
$$\cos^2(x)=\frac{1+\cos(2x)}{2}$$
Then, the problem boils down to evaluating (See This Answer)
$$\sum_{k=1}^{100}\cos(2k\pi/101)=\text{Re}\left(\sum_{k=1}^{100}\left(e^{i2\pi/101}\right)^k\right)$$
The sum $\sum_{k=1}^{100}\cos(2k\pi/101)$ can also be easily evaluated by multiplying by $\frac{\sin(2\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
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Prove $f(x) =\ \frac{x^5}{5!} +\frac{x^4}{4!} +\frac{x^3}{3!}+\frac{x^2}{2!} +x+1$ has only one root. We have to prove that the equation $\displaystyle \frac{x^5}{5!} +\frac{x^4}{4!} +\frac{x^3}{3!}+\frac{x^2}{2!} +x+1=0$ have exactly one real root .
My sir told me it is just an application of derivative .
But I could... | Let's name the sequence of Taylor polynomials for $e^x,$ so
$$ f_0(x) = 1, $$
$$ f_1(x) = 1 + x, $$
$$ f_2(x) = 1 + x + \frac{x^2}{2}, $$
$$ f_3(x) = 1 + x + \frac{x^2}{2}+ \frac{x^3}{6}, $$
$$ f_4(x) = 1 + x + \frac{x^2}{2}+ \frac{x^3}{6}+ \frac{x^4}{24}, $$
$$ f_5(x) = 1 + x + \frac{x^2}{2}+ \frac{x^3}{6}+ \frac{x^4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2257878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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show that $(n+1)(n+2)...(2n)$ is divisible by $2^n$ but not by $2^{n+1}$ Is this proof correct?
Suppose $2^k$ is the largest power of $2$ in the sequence $n+1, n+2, ... 2n$
Then we can compute the power of 2 in the product as $n/2 + n/2^2 + ... n/2^k = n(1 + 2 + .... 2^{k-1})/2^k = n$.
| You can also proceed by induction.
Let $a_n = (n+1)(n+2)\ldots(2n)$.
Then $a_{n+1}/a_n = (2n+1)(2n+2)/(n+1) = 2(2n+1)$.
Since $2n+1$ is odd, only one $2$ factor is added at each step.
Since $a_1 = 2^1$, you get $a_n = 2^n \times o_n$ where $o_n$ is an odd number
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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show that $\sum_{i = 1}^{2k}\frac{ (-1)^{i+1}}{i} = \sum_{i = k+1}^{2k} \frac{1}{i}$ I have a proof but it does not seem elegant. Is there a more elegant solution? Thanks.
Consider $X = \sum_{i = 1}^{2k}\frac{(-1)^{i+1}}{i} = X_1 + X_2$ where
$X_1 = 1 - \frac{1}{2} + ... + \frac{1}{k-1} - \frac{1}{k}$
$X_2 = \frac{1}{k... | You're right, there's a much simpler way:
$$S=\sum^{2k}_{i=1}\left[\frac{(-1)^{(i+1)}}{i}\right]=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6} \cdot\cdot\cdot +\frac{1}{2k-1}-\frac{1}{2k}$$
$$S=\color{blue}{\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6} \cdot\cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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X and Y intercept basic issue Finding the Intercepts of the Graph of an Equation
Given an equation involving x and y, we find the intercepts of the graph as follows
*
*x-intercepts have the form (x,0); set y = 0 in the equation and solve for x.
*y-intercepts have the form (0,y); set x = 0 in the equation and solve f... | In the first part, $\begin{align}
0^2 & \ne 2^2{{(x + 4)}}
-2^2
\end{align}$ In general, $(a-b)^2\ne a^2-b^2$
Similar problem in the second part. Just evaluate square root 4 (hint it's 2) and avoid the following missteps.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integrate $\int \frac {1}{(x+2)(x+3)} \textrm {dx}$ Integrate $\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$
My Attempt:
$$\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$$
$$\int \dfrac {1}{x+2} \textrm {dx} . \int \dfrac {1}{x+3} \textrm {dx}$$
$$\dfrac {\textrm {log (x+2)}}{1} . \dfrac {\textrm {log (x+3)}}{1} + C$$
$$\textrm ... | The method you need to use is Partial Fractions.
$\frac{1}{(x+3)(x+2)} =\frac{A}{x+2}+\frac{B}{x+3}$
The LCD is (x+2)(x+3.)
$\frac{1}{(x+3)(x+2)} =\frac{A(x+3) + B(x+2)}{(x+2)(x+3)}$
$$\\$$
Then you set both of the numerators equal to each other:
1 = A(x+3) + B(x+2)
$$\\$$
When does x+3 = 0?
x= -3
Plug -3 into x:
1 = $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259958",
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"answer_id": 2
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Relativistic sum with magnitude c
Pick any two vectors (in 3 dimensions) having magnitude equal to c and check whether the relativistic sum of them also has magnitude c. Is u v equal to v u?
| Let $ ||u||_2 = ||v||_2 = c$. Then:
\begin{align}
||u \oplus v||_2 &= \frac{1}{c^2 + u\cdot v}\left\vert\left\vert
c^2(u+v)
+
\frac{u\times(u\times v)}{1 + \sqrt{1-(u\cdot u)/c^2}} \right\vert\right\vert_2 \\
&=
\frac{1}{c^2 + u\cdot v}
||
\underbrace{c^2(u+v) }_a
+
\underbrace{ u\times(u\times v) }_b
||_2 \\
&= \... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0$ , why $cot(2\theta) =\dfrac{{A}-{B}}{C}$ in conic sections? I would like to know why
$$
cot(2\theta) =\dfrac{{A}-{B}}{C}
$$
given $$
Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0
$$
Thank You!
| This is a question that is covered in a typical high school discussion of analytic geometry.
First, we note that we can perform a simple translation of the coordinate system so that the origin is located at the center of symmetry of the conic section; i.e., we may assume that $D = E = 0$. In such a case, we suppose th... | {
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"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do i solve this limit: $\lim_{x\to 0}{x-\sin(\sin(...(\sin x)))\over x^{3}}$ I have the next limit :
$$\large \lim_{x\to 0}{x-\sin(\sin(\overbrace {\cdot \ \cdot \ \cdot }^n(\sin(x))\overbrace {\cdot \ \cdot \ \cdot }^n))\over x^{3}}$$
$\sin(\sin(...(\sin(x))...))$-is n times.
I have no idea. Someone can help me? T... | Let us rewrite
\begin{align}
\frac{x-\sin^{[n]}x}{x^3}&=\frac{x-\sin x+\sin x-\sin\sin x+\ldots+\sin^{[n-1]} x-\sin^{[n]} x}{x^3}=\\
&=\frac{x-\sin x}{x^3}+\frac{\sin x-\sin\sin x}{x^3}+\ldots+\frac{\sin^{[n-1]}x-\sin^{[n]}x}{x^3}
\end{align}
and calculte the limit of each fraction separately.
*
*Since
$$
\sin t=t-... | {
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"answer_count": 3,
"answer_id": 2
} |
Finding the power function of the given test. We have a density $X$ defined as $f(x,\theta)=\theta x^{\theta -1}I_{(0,1)}(x)$.
The hypothesis to test is given as follows:
$H_0:\theta \leq1$ Vs $H_1:\theta >1$
A sample size of two is selected, and the critical region is defined as follows:
$C=\{(x_1,x_2):\frac{3}{4x_1}\... | The power function is
$$\begin{align}\pi(\theta)&=\mathbb{P}_\theta(C)\\
&=\mathbb{P}_\theta\left(X_1X_2\ge\frac{3}{4}\right)\\
&=1-\mathbb{P}_\theta\left(X_1X_2<\frac{3}{4}\right).\\
\end{align}$$
Now
$$\begin{align}\mathbb{P}_\theta\left(X_1X_2<\frac{3}{4}\right)&=\int_0^1{\mathbb{P}_\theta\left(X_1<\frac{3}{4x_2}\,\... | {
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"answer_count": 1,
"answer_id": 0
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Find the exact value of $\frac{\cos( \beta)}{\cos( \beta) -1}$ with $\sin(\beta - \pi) = \frac{1}{3}$
Let $h$ be $$\frac{\cos( \beta)}{\cos( \beta) -1}$$
With $\sin(\beta-\pi)=\frac{1}{3}$ and $\beta \in
{]}\pi,\frac{3\pi}{2}{[}$ determine the exact value of $h(\beta)$.
I tried:
$$\sin(\beta-\pi) = \sin(\beta)\cos(... | Your method is correct, you just forgot one subtlety:
$$\sin^2\beta +\cos^2\beta = 1\implies \cos\beta = \color{red}{\pm}\sqrt{1-\sin^2\beta}$$
We are given that $\pi<\beta<\frac{3\pi}{2}\implies \color{red}{\cos\beta <0}$
Therefore you should instead get $\cos\beta = \color{red}-\frac{2\sqrt{2}}{3}$
Subbing this in w... | {
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"answer_count": 3,
"answer_id": 0
} |
Show that a system has a period solution by finding a trapping region (Poincaré-Bendixson Theorem) \begin{align*}
\dot{x}&=4x+2y-x(x^2+y^2)\\
\dot{y}&=-2x+y-y(x^2+y^2)
\end{align*}
I want to show that this system has at least one periodic solution by constructing a trapping region where the Poincaré-Bendixson theorem c... | There is a small sign error in the trigonometric term in your solution for $\dot{r}$. A complete solution follows the sake future readers.
Problem statement
Is there a periodic solution for the following dynamical system?
$$
%
\begin{align}
%
\dot{x} &= 4 x+2 y - x\left(x^2+y^2\right)\\
%
\dot{y} &= -2 x+y-y \left(x^... | {
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"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Determine the limit to which $\prod_{n=2}^{\infty}\left (1+\frac{(-1)^n}{n}\right )$ converges Background: this is Arfken et al mathematical methods 12.5.4 and the answer is 1.
Using the infinite sin product we need the alternating terms in red to cancel when $\pi$ is plugged into z but I don't know how to do that:
$$\... | Here's a different approach notice that for $n=2k$
we have that the general term is equal to
$$\frac{2k+(-1)^{2k}}{2k}=\frac{2k+1}{2k}$$
And for $n=2k+1$ we have $$\frac{2k+1+(-1)^{2k+1}}{2k+1}=\frac{2k}{2k+1}$$
And the product of those two is exactly $1$,hence writing the partial product
$$P_{2k+2}=\frac{3}{2}\cdot\fr... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Factoring $x^5 - 5x^4 + 1$ The original problem was stated like this:
Prove that the polynomial $x^5 - 5x^4 + 1$ does not have roots of multiplicity 4.
So, factoring the polynomial would answer the question, but I don't know how to do it.
Using the root test, ${1, -1}$ should be roots, but If I divide by $x-1$ or $x+1$... | As suggested in the comments, try this $$\begin{align}(x-a)(x-b)^4&=(x-a)(x^4-4x^3b+6x^2b^2-4xb^3+b^4)\\&=(x^5-(a+4b)x^4+(4ab+6b^2)x^3-(4b^3+6ab^2)x^2+(b^4+4ab^3)x-ab^4)\end{align}$$
For this to match the expression $x^5-5x^4+1$, we need:
$$\begin{align}&a+4b=5\\&b(4a+6b)=0\\&b^2(4b+6a)=0\\&b^3(b+4a)=0\\&ab^4=-1
\end{a... | {
"language": "en",
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"question_score": "3",
"answer_count": 9,
"answer_id": 0
} |
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.