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 attempt
$$L = \lim_{x \to -\infty} \left(\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}\right) $$
I thought it'd be nice to get rid of as much ugliness as possible by moving $x$ out of the roots:
$$
\begin{split}
L &= \lim_{x \to -\infty} \left(x\left(\sqrt{1 + \frac{2}{x}}\right)
- x\left(\sqrt{1 - \frac{2}{x}}\right)\right)\\
&= \lim_{x \to -\infty} \left(x\left(\sqrt{1 + \frac{2}{x}} - \sqrt{1 - \frac{2}{x}}\right)\right)
\end{split}
$$
Seems simple enough, since $\frac{1}{x}$ should be an (infinitely) small number I thought it could be discarded with regard to $\sqrt1$, yielding:
$$L = \lim_{x \to -\infty} \left(x\left(\sqrt{1} - \sqrt{1}\right)\right) = 0$$
Wrong, Second attempt
Multply both sides with the conjugate:
$$
\begin{split}
L &= \lim_{x \to -\infty} \sqrt{x^2 +2x} - \sqrt{x^2 - 2x}) \\
&= \lim_{x \to -\infty} \frac{4x}{(\sqrt{x^2 +2x} + \sqrt{x^2 - 2x})} \\
&= \lim_{x \to -\infty} \frac{4x}{x\sqrt{1 + \frac{2}{x}}
+ x\sqrt{1 - \frac{2}{x}}} \\
&= \lim_{x \to -\infty} \frac{4}{\sqrt{1 + \frac{2}{x}}
+ \sqrt{1 - \frac{2}{x}}}\\
&= \frac{4}{\sqrt{1} + \sqrt{1}} = \frac{4}{2} = 2
\end{split}
$$
Wrong as well... The answer should be $-2$, but I don't get how to 'get the negative' in...
|
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:
$$L = \lim_{x \to -\infty} \left(\sqrt{x^2 +2x} - \sqrt{x^2 - 2x}\right) $$
$$ \begin{split} &= \lim_{x \to -\infty} \left(\left(\sqrt{(x+1)^2-1}\right) - \left(\sqrt{(x-1)^2-1}\right)\right)\\ &= \lim_{x \to -\infty} |x+1| - |x-1| \end{split} $$
$$=\lim_{x \to -\infty}\,-x-1+x-1=-2$$
|
{
"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
$$f(-2x+2)=4x^2-8x+4-4=4x^2-8x.$$
|
{
"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 there any elegant way to find it? Thanks for any help.
p.s. Sorry. I make a typo in the $xy$-coefficient.
|
$$x^3+3xy+y^3-1=(x+y-1)(x^2+y^2-xy+x+y+1)=0$$
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 the second derivative $\frac{d^2y}{dx^2}=4$ shows it is a minimum. Evaluating it gives $x^2+y^2=\frac{1}{2}$ which is less than the value at $(x,y)=(-1,-1)$ so the minimum value is $\frac{1}{2}$ at $(x,y)=\left(\frac{1}{2},\frac{1}{2}\right)$.
|
{
"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: $$a_{n+1}=\frac{\sqrt{(a_n+b_n)(a_n+c_n)}}{2} \\ b_{n+1}=\frac{\sqrt{(b_n+a_n)(b_n+c_n)}}{2} \\ c_{n+1}=\frac{\sqrt{(c_n+a_n)(c_n+b_n)}}{2}$$
in terms of $a_0,b_0,c_0$?
$$L(a_0,b_0,c_0)=\lim_{n \to \infty}a_n=\lim_{n \to \infty}b_n=\lim_{n \to \infty}c_n$$
For the most simple case $a_0=b_0$ we have some interesting closed forms in terms of inverse hyperbolic or trigomonetric functions:
$$L(1,1,\sqrt{2})=\frac{1}{\ln(1+\sqrt{2})}$$
$$L(1,1,1/\sqrt{2})=\frac{2 \sqrt{2}}{\pi}$$
$$L(1,1,2)=\frac{\sqrt{3}}{\ln(2+\sqrt{3})}$$
$$L(1,1,1/2)=\frac{3 \sqrt{3}}{2 \pi }$$
$$L(1,1,3)=\frac{\sqrt{2}}{\ln(1+\sqrt{2})}$$
$$L(1,1,1/3)=\frac{2 \sqrt{2}}{3 \arcsin \left( \frac{2 \sqrt{2}}{3} \right)}$$
$$L(1,1,\sin 1)=\frac{2 \cos 1}{\pi -2}$$
$$L(1,1,\sin 1/2)=\frac{2 \cos 1/2}{\pi -1}$$
$$L(1,1,\cosh 1)=\sinh 1$$
$$L(1,1,\cosh 2)=\frac{\sinh 2}{2}$$
These values are obtained by Wolfram Alpha and Inverse Symbolic Calculator and checked with Mathematica.
This result seems familiar to me, but I'm pretty sure I don't know how to obtain a general closed form or even integral representation.
This question is closely related, but the sequence is very different.
Update:
It seems likely that the closed form for the special case $a_0=b_0=1$ is:
$$L(1,1,x)=\frac{\sinh \left(\cosh ^{-1}\left(x\right)\right)}{\cosh ^{-1}\left(x\right)}$$
However, the proof would be nice as well as the more general case.
Thanks to Sangchul Lee I now see that this sequence is exactly Carlson's transformation. Change:
$$a^2=A,\quad b^2=B,\quad c^2=C$$
$$A_{n+1}=\frac{1}{4} (A_n+\sqrt{A_nB_n}+\sqrt{B_nC_n}+\sqrt{C_nA_n})$$
$$B_{n+1}=\frac{1}{4} (B_n+\sqrt{A_nB_n}+\sqrt{B_nC_n}+\sqrt{C_nA_n})$$
$$C_{n+1}=\frac{1}{4} (C_n+\sqrt{A_nB_n}+\sqrt{B_nC_n}+\sqrt{C_nA_n})$$
Accoding to Wikipedia and references wherein, we have:
$$R_F(A_{n+1},B_{n+1},C_{n+1})=R_F(A_n,B_n,C_n)$$
$$R_F(A,B,C)=\frac{1}{2} \int_0^\infty \frac{dt}{\sqrt{(t+A)(t+B)(t+C)}}$$
Which is exactly the 'closed form' I wanted.
|
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 if anything interesting happens,
with a goal of proving convergence,
maybe.
Multiplying, we get
$a_{n+1}b_{n+1}c_{n+1}
=(a_n+b_n)(a_n+c_n)(b_n+c_n)/8
=(a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2)/8
$
(omitting the "_n")
so that
$\begin{array}\\
\dfrac{a_{n+1}b_{n+1}c_{n+1}}{abc}
&=(\frac{a}{c} + \frac{a}{b} + \frac{b}{c} + 2 + \frac{c}{b} + \frac{b}{a} + \frac{c}{a})/8\\
&=(\frac{a}{c} + \frac{c}{a} + \frac{a}{b} + \frac{b}{a} + \frac{b}{c}+ \frac{c}{b} + 2 )/8\\
&=(\frac{a}{c} -2+ \frac{c}{a} + \frac{a}{b}-2 + \frac{b}{a} + \frac{b}{c}-2+ \frac{c}{b} + 8 )/8\\
&=(g(\dfrac{a}{c}) + g(\dfrac{a}{b}) + g(\dfrac{b}{c}))/8+1
\qquad\text{where } g(x) = (\sqrt{x}-\dfrac1{\sqrt{x}})^2=x+\dfrac1{x}-2\\
\end{array}
$
Note that
$g'(x)
=1-\dfrac1{x^2}
$.
Don't know what to do with this,
so I'll try something else.
$\begin{array}\\
a_{n+1}-a
&=\frac{\sqrt{(a+b)(a+c)}}{2}-a\\
&=a(\frac{\sqrt{(1+b/a)(1+c/a)}}{2}-1)\\
\end{array}
$
Again nothing.
Let's try this.
Since
$(a+b+c)^2
\le 3(a^2+b^2+c^2)
$
with equality iff
$a=b=c$,
$\begin{array}\\
a_{n+1}^2+b_{n+1}^2+c_{n+1}^2
&=((a+b)(a+c)+(a+b)(b+c)+(a+c)(b+c))/4\\
&=(a^2+ab+ac+bc+ab+ac+b^2+bc+ab+ac+bc+c^2)/4\\
&=(a^2+b^2+c^2+3ab+3ac+3bc)/4\\
&=(\frac32(a^2+b^2+c^2+2ab+2ac+2bc)-\frac12(a^2+b^2+c^2))/4\\
&=(\frac32(a+b+c)^2-\frac12(a^2+b^2+c^2))/4\\
&\le(\frac32 3(a^2+b^2+c^2)-\frac12(a^2+b^2+c^2))/4\\
&=a^2+b^2+c^2\\
\end{array}
$
Therefore,
if at least two of
$a, b, c$ are distinct,
the sum of their squares
decreases.
This is not a proof of convergence,
but a start.
Let's try a variation on
this last result.
$\begin{array}\\
a_{n+1}^2+b_{n+1}^2+c_{n+1}^2
&=((a+b)(a+c)+(a+b)(b+c)+(a+c)(b+c))/4\\
&=(a^2+ab+ac+bc+ab+ac+b^2+bc+ab+ac+bc+c^2)/4\\
&=(a^2+b^2+c^2+3ab+3ac+3bc)/4\\
\end{array}
$
so
$\begin{array}\\
a_{n+1}^2+b_{n+1}^2+c_{n+1}^2
-(a^2+b^2+c^2)
&=(-3a^2-3b^2-3c^2+3ab+3ac+3bc)/4\\
&=-3(a^2+b^2+c^2-ab-ac-bc)/4\\
&=-3(2a^2+2b^2+2c^2-2ab-2ac-2bc)/8\\
&=-3((a-b)^2+(a-c)^2+(b-c)^2)/8\\
\end{array}
$
This shows precisely
how the sum of squares
decreases at each step.
I think that
this implies convergence.
|
{
"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 $A^{-1}=\frac{1}{\det A}\operatorname{adj(A)}$
$$A=\begin{pmatrix}
1 & 0 &0 &\ldots &0\\
1 & 1 & 0 &\ldots &0\\
1 & 1 & 1 &\ldots &0\\
\vdots &\vdots & \vdots &\ddots & \vdots\\
1 & 1 &1 & \ldots &1 \end{pmatrix}$$
This is a triangular matrix so $\det A=1.$
To find the adjugate I first need to find the cofactor matirx$(C)$ of $A$.
$$C=\begin{pmatrix}
1&-1&0&0&\ldots &0\\
0 & 1& -1 &0&\ldots & 0\\
0 & 0 & 1 & -1&\ldots &0\\
0 & 0 & 0 &1 &\ldots &0\\
\vdots &\vdots &\vdots &\vdots&\ddots & \vdots\\
0 & 0 &0 &0 &\ldots &1\end{pmatrix}$$
$$C^T=\operatorname{adj}(A)=\begin{pmatrix}
1 & 0 & 0 & 0&\ldots &0\\
-1 & 1& 0 & 0 &\ldots &0\\
0&-1&1&0&\ldots &0\\
0& 0 &-1 &1 &\ldots&0\\
\vdots &\vdots &\vdots &\vdots &\ddots&\vdots\\
0&0&0&0&\ldots &1\end{pmatrix}=A^{-1}$$
Is this correct? Also, can I leave it like that or should I somehow write it more formally?
|
Just to show a different approach.
Consider the matrix $\mathbf E$, having $1$ only on the first subdiagonal
$$
\mathbf{E} = \left\| {\,e_{\,n,\,m} = \left\{ {\begin{array}{*{20}c}
1 & {n = m + 1} \\
0 & {n \ne m + 1} \\
\end{array} } \right.\;} \right\| = \left\| {\,\left( \begin{gathered}
0 \\
n - m - 1 \\
\end{gathered} \right)\;} \right\|
$$
Multiply it by $\mathbf A$ , and it is easily seen that $$\mathbf E \, \mathbf A=\mathbf A-\mathbf I \quad \Rightarrow \quad \left( \mathbf I -\mathbf E \right)\,\mathbf A=\mathbf I$$
In another way, consider that the powers of $\mathbf E$ are readily found and have a simple formulation
$$
\begin{gathered}
\mathbf{E}^{\,\mathbf{2}} = \left\| {\,\sum\limits_{0\, \leqslant \,k\,\left( { \leqslant \,n - 1} \right)} {\left( \begin{gathered}
0 \\
n - k - 1 \\
\end{gathered} \right)\left( \begin{gathered}
0 \\
k - m - 1 \\
\end{gathered} \right)} \;} \right\| = \left\| {\,\left( \begin{gathered}
0 \\
n - m - 2 \\
\end{gathered} \right)\;} \right\| = \left\| {\,\left\{ {\begin{array}{*{20}c}
1 & {n = m + 2} \\
0 & {n \ne m + 2} \\
\end{array} } \right.\;} \right\| \hfill \\
\quad \vdots \hfill \\
\mathbf{E}^{\,\mathbf{q}} = \mathbf{E}^{\,\mathbf{q} - \mathbf{1}} \,\mathbf{E} = \left\| {\,\sum\limits_{0\, \leqslant \,k\,\left( { \leqslant \,n - 1} \right)} {\left( \begin{gathered}
0 \\
n - k - \left( {q - 1} \right) \\
\end{gathered} \right)\left( \begin{gathered}
0 \\
k - m - 1 \\
\end{gathered} \right)} \;} \right\| = \hfill \\
= \left\| {\,\left( \begin{gathered}
0 \\
n - m - q \\
\end{gathered} \right)\;} \right\|\quad \left| {\;0 \leqslant \text{integer}\;q} \right. \hfill \\
\end{gathered}
$$
Therefore
$$
\mathbf{A} = \sum\limits_{0\, \leqslant \,j} {\mathbf{E}^{\,\mathbf{j}} } = \frac{\mathbf{I}}
{{\mathbf{I} - \mathbf{E}}}
$$
To connect to the answer of Jean Marie, note that
$$
\left( {\mathbf{I} - \mathbf{E}} \right)\left\| {\,\begin{array}{*{20}c}
{x_{\,0} } \\
{x_{\,1} } \\
\vdots \\
{x_{\,n} } \\
\end{array} \;} \right\| = \left\| {\,\begin{array}{*{20}c}
{x_{\,0} ( - 0)} \\
{x_{\,1} - x_{\,0} } \\
\vdots \\
{x_{\,n} - x_{\,n - 1} } \\
\end{array} \;} \right\| = \left\| {\,\begin{array}{*{20}c}
{\nabla x_{\,0} \;\left| {x_{\, - 1} = 0} \right.} \\
{\nabla x_{\,1} } \\
\vdots \\
{\nabla x_{\,n} } \\
\end{array} \;} \right\|
$$
|
{
"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 :
$$\cos x-\sin x \neq 0 \to \cos x \neq \sin x$$
$$x\neq k\pi+\frac{\pi}{4}$$
$$2(\sin 2x -\cos 2x)=\frac{\cos x+\cos3x}{\cos x-\sin x}$$
$$2(2\sin x \cos x -\cos 2x)=\frac{\cos x+\cos3x}{\cos x-\sin x}$$
$$2(2\sin x \cos x -(1-2\sin^2x)=\frac{\cos x+\cos3x}{\cos x-\sin x}$$
then ?
|
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 $a,b$ as constants. I have plugged in $\pi$ and $e$ as constants into WA and gotten a result, but this obviously is not sufficient when trying to achieve a general result.
I suspect the answer is (assuming $a,b>0$, for example), that this infimum occurs at $x= \frac{-a}{\sqrt{a^2+b^2}}$ and $y= \frac{b}{\sqrt{a^2+b^2}}$, but I cannot rigorously show it, or get WA to give me a general answer.
|
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
$$\begin{align}
\frac{\partial}{\partial r}\left(\frac{r(a\cos(\theta)-b\sin(\theta))}{1+r^2}\right)&=\frac{(1-r^2)}{(1+r^2)^2}\,(a\cos(\theta)-b\sin(\theta)) \tag 1\\\\
\frac{\partial}{\partial \theta}\left(\frac{r(a\cos(\theta)-b\sin(\theta))}{1+r^2}\right)&=-\frac{r}{1+r^2}(a\sin(\theta)+b\cos(\theta))\tag 2
\end{align}$$
Setting $(1)$ to zero, yields $r=1$ or $\tan(\theta)=a/b$. Setting $(2)$ to zero, yields $r=0$ or $\tan(\theta)=-b/a$.
Hence, the possible local extrema are $r=1$, $\tan(\theta)=-b/a$ or $r=0$ and $\tan(\theta)=a/b$.
Note that $g(0,\theta)=0$, while $g(1,\arctan(-b/a))=\pm \frac12 \sqrt{a^2+b^2}$. Therefore, the minimum value of $g(r,\theta)$ is $-\frac12 \sqrt{a^2+b^2}$ which occurs when $r=1$, $\cos(\theta)=-a/\sqrt{a^2+b^2}$, and $\sin(\theta)=b/\sqrt{a^2+b^2}$.
Transforming to Cartesian coordinates, the minimum of $f(x,y)$ is $-\frac12 \sqrt{a^2+b^2}$, which occurs when $x=-a/\sqrt{a^2+b^2}$ and $y=b/\sqrt{a^2+b^2}$ as conjectured!
|
{
"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 though?
|
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 rectangle are
$$ \begin{bmatrix}0& -1 \\ 1& 0\end{bmatrix} \begin{bmatrix}3 \\ 5\end{bmatrix} = \begin{bmatrix}-5\\ 3\end{bmatrix},
\qquad\begin{bmatrix}0& -1 \\ 1& 0\end{bmatrix} \begin{bmatrix}5 \\ 3\end{bmatrix} = \begin{bmatrix}-3\\ 5\end{bmatrix}, $$
$$ \begin{bmatrix}0& -1 \\ 1& 0\end{bmatrix} \begin{bmatrix}3 \\ 3\end{bmatrix} = \begin{bmatrix}-3\\ 3\end{bmatrix},
\qquad\begin{bmatrix}0& -1 \\ 1& 0\end{bmatrix} \begin{bmatrix}5 \\ 5\end{bmatrix} = \begin{bmatrix}-5\\ 5\end{bmatrix}. $$
So the cooardinate of the top left corner is $(-5,5)$ and the coordinate of the bottom right corner is $(-3,3)$
Keep in mind: Due to the rotation of the rectangle itself, the new top left corner is not the image of the old top left coordinate.
|
{
"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
$$I = \log\bigg(\sqrt{1+x}+\sqrt{1-x}\bigg)\cdot x\bigg|_{0}^{1}+\frac{1}{2}\int^{1}_{0}\frac{\sqrt{1+x}-\sqrt{1-x}}{\sqrt{1+x}+\sqrt{1-x}}\cdot \frac{1}{\left(\sqrt{1+x}\cdot \sqrt{1-x}\right)}xdx$$
$$I = \frac{\ln2}{2}+\frac{1}{2}\int^{1}_{0}\frac{x^2}{(1+\sqrt{1-x^2})\cdot \sqrt{1-x^2}}dx$$
Put $x= \sin \theta$ Then $dx = \cos \theta d\theta$
$$I = \frac{\ln2}{2}+\frac{1}{2}\int^{\frac{\pi}{2}}_{0}\frac{\sin^2 \theta\cdot \cos \theta}{(1+\cos \theta)\cdot \cos \theta}d\theta$$
$$I = \frac{\ln2}{2}+\frac{1}{2}\int^{\frac{\pi}{2}}_{0}(1-\cos \theta)d\theta =\frac{\ln2}{2}+\frac{\pi}{4}-\frac{1}{2}.$$
|
{
"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
}
|
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 $\sum\limits_{n=1}^{\infty} \frac{x^{2n+1}}{n!{\sqrt{n}}}$. The MO question references this question and it also links to some other posts containing similar expressions.
|
(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}\frac{1}{c^k} < \frac{c}{c-1}e^{-x}x^{cx}\frac{e^{cx}}{(cx)^{cx}} = \frac{ce^{-((c\log c) -c +1)x}}{c-1}\end{align*}
where, we used the inequality, $x! > e^{-x}x^{x}$ for sufficiently large $x$. Also note that $(c\log c) - c + 1 > 0$ for all $c > 0$
Again, for sufficiently large $x$: \begin{align*}T(x/c) = e^{-x}\sum\limits_{0 \le k < x/c} \frac{x^k}{k!} = e^{-x}\sum\limits_{0 \le k < x/c} \frac{c^k(x/c)^k}{k!} & \le e^{-x}\frac{(x/c)^{x/c}}{(x/c)!}\sum\limits_{0 \le k < x/c} c^k \\& < e^{-x}e^{x/c}\frac{xc^{x/c}}{c} = \frac{x}{c}e^{-\left(1 - \frac{1}{c} - \frac{\log c}{c}\right)x}\end{align*} where, we note that $(1 - \frac{1}{c} - \frac{\log c}{c}) > 0$.
Hence, we have for any $c>1$, $$\displaystyle e^{-x}\sum\limits_{x/c \le k \le cx} \frac{x^k}{k!} = 1 - S(cx) - T(x/c) \quad \underbrace{\longrightarrow}_{ x \to +\infty} \quad 1 \tag{1}$$
Now, we have the simple estimates, $$\frac{e^{-x}}{\sqrt{cx}}\sum\limits_{x/c \le k \le cx} \frac{x^k}{k!} < e^{-x}\sum\limits_{x/c \le k \le cx} \frac{x^k}{k!\sqrt{k}} < \frac{e^{-x}}{\sqrt{x/c}}\sum\limits_{x/c \le k \le cx} \frac{x^k}{k!} \tag{2}$$
and $$e^{-x}\sum\limits_{k \not\in [x/c,cx]} \frac{x^k}{k!\sqrt{k}} < T(x/c) + S(cx) \tag{3}$$ where, both the terms in the upper bound decays exponentially.
Hence, from $(1),(2)$ and $(3)$ we have that the required limit: $$\displaystyle \lim\limits_{x \to \infty} \sqrt{x}e^{-x}\sum\limits_{k=1}^{\infty} \frac{x^k}{k!\sqrt{k}} \in \left(\frac{1}{\sqrt{c}},\sqrt{c}\right)$$ for arbitrary $c > 1$, i.e., the required limit is $1$ (as pointed out by user tired in the comments).
I suppose for any function $\alpha(x)$ with atmost polynomial growth and continuous at $x = 1$ we can slightly modify the above argument and show, $$\lim\limits_{x \to \infty} e^{-x}\sum\limits_{k=1}^{\infty} \alpha\left(\frac{k}{x}\right)\frac{x^k}{k!} = \alpha(1)$$
|
{
"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.
$$\begin{align*}12^{34^{45^{78}}} \pmod{90} &\equiv 6\left[\frac{12^{34^{45^{78}}}}{6} \pmod{15} \right],\qquad\qquad\gcd(12,90)=6\\
&\equiv 18\left[\frac{12^{34^{45^{78}}}}{18} \pmod{5}\right],\qquad\qquad\gcd(12,15)=3\\
&\equiv 18\left[\frac{2^{34^{45^{78}}\pmod 4}}{3} \pmod{5}\right],\qquad\qquad\phi(5)=5-1=4\\
&\equiv 18\left[\frac{2^0}{3} \pmod{5}\right]\\
&\equiv 18\left[\frac{6}{3} \pmod{5}\right]\\
&\equiv 18\left[2\right]\equiv 36
\end{align*}$$
|
{
"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
$(x^{2}-y^{2})^{2}+(y^{2}-z^{2})^{2}+(x^{2}-z^{2})^{2}-x^{4}-y^{4}-z^{4}=-576$
$(x^{2}-y^{2}-z^{2})(x^{2}-y^{2}+z^{2})+(y^{2}-z^{2}-x^{2})(y^{2}-z^{2}+x^{2})+(x^{2}-z^{2}-y^{2})(x^{2}-z^{2}+y^{2})=-576$
I wanted to factor 576 somehow but I cant put the left side of the equation under a common multiplier, please help.
|
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 have to be even.
$$\dfrac{x+y+z}2\cdot\dfrac{x+y-z}2\cdot\dfrac{z+x-y}2\cdot\dfrac{z-x+y}2=\dfrac{576}{2^4}=36$$
|
{
"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 $I=J$.
First, we noticed that $x^4+x^2+1=(x^2-x+1)(x^2+x+1)$
So it gives us an idea to try and factorise $x^4-x^2+1$ but cannot find any factors.
Integrate $I$ (We try some substitutions to see where it will get us to)
$x=\sqrt{u}$ then $dx={2\over \sqrt{u}}du$
$$I=16\cdot{1\over 2}\int_{0}^{\infty}{u^{3/2}\over (1-u+u^2)^4}\mathrm du$$
$u=\tan(y)$ then $du=\sec^2(y)dy$
$$I=8\int_{0}^{\pi/2}{\tan^{3/2}(y)\over (1-\tan(y)+\tan^2(y))^4}{\mathrm dy\over \cos^2(y)}$$
then simplified down to
$$I=128\int_{0}^{\pi/2}{\cos^6(y)\tan^{3/2}(y)\over (2-\sin(2y))^4}\mathrm dy$$
we further simplified down to
$$I={128\over 2^{3/2}}\int_{0}^{\pi/2}{\cos^3(y)\sin^{3/2}(2y)\over (2-\sin(2y))^4}\mathrm dy$$
Not so sure what is the next step.
|
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 complete (for the next to last equality $(\star)$ we applied Glasser's Master theorem which is not difficult to proof for this special case)
Furthermore applying the same substitution $x=1/y$ again, it is a matter of straightforward algebra that
$$
\Delta=J-I=-\Delta
$$
so
$$
\Delta=0 \,\,\text{or}\,\,I=J=3\pi
$$
QED
|
{
"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 $1/4 \times 5 \times s^2 \times \cot (\pi/5)$ (see for example the Wikipedia article on regular polygons, https://en.wikipedia.org/wiki/Regular_polygon#Area). Therefore we have
$$ {[ABCDE] \over [ABC]} = {1/4 \times 5s^2 \times \cot (\pi/5) \over s^2/2 \times \sin(3\pi/5)} = {5 \over 2} {\cot (\pi/5) \over \sin(3\pi/5) } $$
So we need to evaluate
$$x = {\cot (\pi/5) \over \sin(3\pi/5)}$$. Recalling the definition of the cotangent, we get
$$x = {\cos(\pi/5) \over \sin(\pi/5) \sin(2\pi/5)} $$
Now using the double angle formula, $\sin(2\pi/5) = 2 \sin(\pi/5) \cos(\pi/5)$ and so we have
$$x = {\cos \pi/5 \over 2 \sin^2 \pi/5 \cos \pi/5} = {1 \over 2 \sin^2 \pi/5} $$
and knowing that $$ \sin {\pi \over 5} = {1 \over 4} \sqrt{10 - 2\sqrt{5}}$$ (this is a standard fact one can look up) gives
$$ x= {8 \over (10 - 2 \sqrt{5})} = {8(10 + 2\sqrt{5}) \over (10+2\sqrt{5})(10-2\sqrt{5})} = {80 + 16 \sqrt{5} \over 80} = {5 + \sqrt{5} \over 5}$$
Plugging this back in we get, as desired,
$${[ABCDE] \over[ABC]} = {5 \over 2} {5 + \sqrt{5} \over 5} = {5 + \sqrt{5} \over 2}. $$
|
{
"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 obvious.
|
{
"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 took square root of 4 out of integral in denominator and got $1-(x/2)^2$ and in the second part i've got $$2\int \dfrac{dx}{(4+x)^{1/2}}$$ and what to do next
|
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 any idea in this matter?
thanks in advance for your any response
|
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*}
$(1)-(2)$,
$$\frac{2\alpha x}{a^2}+\frac{2\beta y}{b^2}+\frac{2\gamma z}{c^2}=
1-\lambda^2+\frac{\alpha^2}{a^2}+\frac{\beta^2}{b^2}+\frac{\gamma^2}{c^2}$$
which is the radical plane of the two ellipsoids.
The centre of section is:
$$
\frac
{1-\lambda^2+\dfrac{\alpha^2}{a^2}+\dfrac{\beta^2}{b^2}+\dfrac{\gamma^2}{c^2}}{2\left(
\dfrac{\alpha^2}{a^2}+\dfrac{\beta^2}{b^2}+\dfrac{\gamma^2}{c^2}
\right)}
\begin{pmatrix} \alpha \\ \beta \\ \gamma \end{pmatrix}
$$
Note briefly:
No intersections if $\, \dfrac{\alpha^2}{a^2}+\dfrac{\beta^2}{b^2}+\dfrac{\gamma^2}{c^2}>(1+\lambda)^2$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2128445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
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 limitations, $2 \cdot 12!$ is the number of words when two letters stand next to each other (there are 3 options for this situation) and finally the situation when all the "forbidden" letters stand next to each other ($2 \cdot 10!$).
But the answer is $13! - 3 \cdot 2 \cdot 12! + 2 \cdot 2 \cdot 11! + 2 \cdot 2 \cdot 11! - 2 \cdot 10!$
Any idea what am I missing?
|
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 have to subtract all words which contain (at least) one of these six forbidden words from $13!$
\begin{align*}
-3\cdot 2\cdot 12!\tag{1}
\end{align*}
Since some words contain more than one forbidden word, these are counted more than once in (1) and we have to compensate for it.
We consider two different cases. Words which contain two forbidden subwords consisting of four different characters and words which contain two forbidden subwords consisting of three different characters which do overlap. They are
\begin{align*}
&AC,NT&ANT\\
&AC,TN&CAN\\
&CA,NT&NAC\\
&CA,TN&TNA
\end{align*}
So, we have to add the number of these words giving
\begin{align*}
+2\cdot 2\cdot 11!+2\cdot 2\cdot 11!\tag{2}
\end{align*}
In (2) we count the words which contain three forbidden words too often and we have to compensate for them. Since each character occurs exactly once, the following two forbidden triplets are to consider
\begin{align*}
CANT, TNAC
\end{align*}
So, we have to subtract
\begin{align*}
-2*10!
\end{align*}
and we finally obtain
\begin{align*}
13!-3\cdot 2\cdot 12!+2\cdot 2\cdot 11!+2\cdot 2\cdot 11!-2\cdot 10!
\end{align*}
Note: This answer follows the inclusion-exclusion principle.
|
{
"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
}
|
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 $q=3x+2$, $x \in \mathbb{Z}$
then $2^{3x+1}+(3x+1)^2=9x^2+6x+2(2^{3x})+1$
Now I'm not really sure what to do from here my goal was to show its not prime from both cases of $q$ but I think I"m missing the algebra necessary to show that.
|
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 be equal to $3$, the only valid remaining case is $p=3$, $n=2^3+3^2=17$.
|
{
"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
}
|
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 $a_2 = 2*(3^{2-1}) = 6.$
Suppose as inductive hypothesis this is true for all $n \geq 0.$
Then we need to show $a_{n+1} = (n+1) 3^{(n+1)-1}= (n+1)*3^n$.
Then $a_{n+1} = 6a_{n}-9a_{n-1}$ by the recurrence relation.
So $a_{n+1} = 6a_{n}-9a_{n-1} = 6(n 3^{n-1})- 9a_{n-1} = 3[2(n 3^{n-1})] - 3a_{n-1}]= $ I am stuck after this, I want to show $a_{n+1} = (n+1)*3^n$.
Can someone please help ? Thank you
|
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 means that the inductive step holds, since if we subsitiute n+3 into our formula we get $a_{n+3}$ =$3^{n+2}(n+3)$
|
{
"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.
$$
\begin{align}
-6 \cos(2 x) + 2 \sin(2 x) &= 0 \iff \\
x &= \pi n + \arctan \big( \frac{1}{3} (-1 - \sqrt{10} \big) \bigvee \\
x &= \pi n + \arctan \big( \frac{1}{3} (-1 + \sqrt{10} \big) \\
\forall &n \in \mathbb{Z}
\end{align}
$$
From here you can find the absolute minima and maxima by several methods.
|
{
"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
}
|
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: $\frac{d}{dx}(6 \cdot 3^{2x - 1}) = 6 \frac{d}{dx}(3^{2x - 1}) = 6 \cdot 3^{2x - 1} \ln (3) \cdot 2 = 12 \cdot 3^{2x - 1} \ln (3)$
Now, according to the book, the answer would be: $4 \cdot 3^{2x}\ln (3)$.
What seems obvious is just dividing by $3$, but then I get in to trouble trying to apply it to $\frac{3^{2x - 1}}{3} = 3^{2x - 1} * 3^{-1} = 3^{2x - 2}$...
So how to proceed now?
|
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:
$$
\begin{split}
1^{st} \text{expression} &= y^2-10y+24+6x-9x^2\\
&={(y)^2-2\cdot(y)\cdot5+(5)^2}-(5)^2+24+6x-9x^2\\
&=(y-5)^2-{25-24-6x+9x^2}\\
&=(y-5)^2-{(1)^2-2\cdot(1)\cdot(3x)+(3x)^2}\\
&=(y-5)^2-(1-3x)^2\\
&=(y-5+1-3x)(y-5-1+3x)\\
&=(y-3x-4)(y+3x-6).
\end{split}
$$
But, I could not factorize the $2^{nd}$ expression. Please help.
|
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 $$-\frac{n+1}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2137234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
}
|
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 $2s^2\le1\iff0\le s\le\dfrac1{\sqrt2}\ \ \ \ (1)$
$$|1+z^2|+|1-z|=2(1-2s^2+s)=\dfrac{9-(4s-1)^2}4$$
By $(1),-1\le4s-1\le2\sqrt2-1\implies1\le(4s-1)^2\le(2\sqrt2-1)^2$
Can you take it from here?
Case$\#2:$ For $2s^2>1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2137925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
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 equation of line is $\displaystyle \frac{x-0}{a} = \frac{y-0}{b} = \frac{z-0}{c}$
in parametric form $\displaystyle \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = \lambda$
so coordinate of $P(a\lambda,b\lambda,c\lambda)$
in parametric form $\displaystyle \frac{x}{a} = \frac{y}{b} = \frac{z}{c} = \mu$
so coordinate of $Q(a\mu,b\mu,c\mu)$
so $(PQ)^2 = (a^2+b^2+c^2)(\lambda-\mu)^2$
wan,t be able to go further, could some help me, thanks
|
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: $$a\mu = 2k_1 + \frac {8}{3} $$ $$b\mu = -k_1-3$$ $$c\mu = k_1 +1$$
We thus get from these equations $$\frac {2+k}{\lambda} = \frac {2k_1 + \frac {8}{3}}{\mu} $$ $$\frac {-2k-1}{\lambda} = \frac {-k_1-3}{\mu} $$ $$\frac {k-1}{\lambda} = \frac {k_1+1}{\mu} $$
Now try to express $\lambda $ in terms of $\mu $ and thus both the points $P $ and $Q $ will be expressed in the same parameter $\mu $. Then find the square of the distance. Hope it helps.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2138421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
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 $u(x(r,s),y(r,s))$, I get the characteristic curve $(x(r,s),y(r,s),z(r,s))=(s+r,s+r,e^{(s+r)^2})$
I notice: $x(r,s)=y(r,s)$ $\implies$ $u(x,y)=z(r(x,y),s(x,y))=e^{x^2}=e^{y^2}$. However, this solution that I find is unique. I seem to be missing something to show that there are infinite solutions. Any tips?
|
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}
Thus $2a + b = 2$ and $2c + b = 0$, so $a = 1 - \frac{b}{2}$ and $ c =- \frac{b}{2}$.
This also gives $a + b + c = 1 - \frac{b}{2} + b - \frac{b}{2} = 1$, or
\begin{align}
u(x,x) = \exp\left(x^2 \right)
\end{align}
So for all $b \in \mathbb{R}$ another solution is
\begin{align}
u(x,y) = \exp\left(x^2 \right) \exp\left(- \frac{b}{2}x^2 + b xy - \frac{b}{2} y^2\right) = \exp\left(x^2 \right) \exp\left(- \frac{b}{2} \left(x - y \right)^2\right)
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2139416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
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
$$c(a+b)=ab.$$
Since $a$ and $c$ are coprime, we get by Gauss' lemma that
$$a\mid a+b.$$
Symmetrically,
$$b\mid a+b.$$
Since $a$ and $b$ are coprime,
$$ab\mid a+b.$$
But we obviously have $a+b\mid ab$, so
$$a+b=ab$$
and $c=1$.
I think this would imply $a=b=2$ which is absurd since $a$ and $b$ are coprime...
The question. Is this going somewhere? What else could I do?
|
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 prime $p$ divides both $x,y$ we have $p$ dividing both $a,b$; thus $\gcd(x,y) = 1$. And since $xy=c^2$ we have individually $x = z_1^2$ and $y= z_2^2$.
$a−c=x$ , $b−c=y$. Thus $a+b = 2c +x+y = (z_1+z_2)^2$. Hence $a+b$ is a prefect square.
In addition we have also proved that $a-c$ and $b-c$ are perfect squares.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2139616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
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 $x,y$($x>y$): $$\lambda\frac{x^3-y^3}{(x-y)^3}\geq \frac{x^2-y^2}{(x-y)^2}$$
I simplified the relation into the following: $(\lambda-1)x^2+\lambda xy+(\lambda +1)y^2\geq0$ , but can't go further...
|
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 change the inequality.
$$ \lambda \frac{x^3 - y^3}{(x-y)^3} \geq \frac{x^2 - y^2}{(x-y)^2}$$
$$ \lambda \frac{(x - y)(x^2 + xy + y^2}{(x-y)^3} \geq \frac{(x-y)(x+y)}{(x-y)^2}$$
$$ \lambda (x^2 + xy + y^2) \geq (x-y)(x+y)$$
$$ (\lambda - 1) x^2 + \lambda xy + (\lambda + 1)y^2 \geq 0$$
Right, we agree here. :-)
So here's a technique used in finding limits of functions of two variables:
I got down to a solution with this, below; needs completed calculation and check as this is first draft rough work.
Let $y = kx$. Since $x > y$ given in hypothesis, if $x$ is positive k < 1 (negative is fine) and if $x$ is negative, $k < -1$.
This covers all cases except $x = 0$. If $x = 0$, we calculate separately.
If $x = 0 $ your equation simplifies to
$$(\lambda + 1)y^2 \geq 0$$
$y < 0 $ given $y < x$ but $$y^2 > 0$$. Therefore $(\lambda + 1) \geq 0$ and $\lambda \geq -1$
For y = kx,
$$ (\lambda - 1) x^2 + \lambda xy + (\lambda + 1)y^2 \geq 0$$
$$ (\lambda - 1) x^2 + \lambda x(kx) + (\lambda + 1)(kx)^2 \geq 0$$
$$ \lambda x^2 - x^2 + k \lambda x^2 + \lambda k^2 x^2 + k^2 x^2 \geq 0$$
Separating out the terms with $\lambda$
$$ \lambda x^2 + k \lambda x^2 + \lambda k^2 x^2 \geq x^2 - k^2 x^2$$
$$\lambda x^2(1 + k + k^2) \geq x^2(1 - k^2)$$
We dealt with $x = 0$ above. For $x \neq 0$ we can divide
$$\lambda \geq \frac{(1 - k^2)}{(1 + k + k^2)}$$
Minimize with respect to k subject to the restrictions above [if x is positive k < 1 (negative is fine) and if x is negative, $k<−1$]
$$\frac{(-2k)(1 + k + k^2) -(1-k^2)(1 + 2k)}{(1 + k + k^2)^2} = 0$$
Denominator is never zero
$$(-2k)(1 + k + k^2) -(1-k^2)(1 + 2k) = 0$$
$$ -2k -2k^2 - 2k^3 -1 - 2k + k^2 + 2k^3 = 0$$
$$ -k^2 -4k -1 = 0$$
$$ k = \frac{-4 \pm \sqrt{12}}{2} = -2 \pm \sqrt{3}$$
$$\lambda \geq \frac{(1 - k^2)}{(1 + k + k^2)}$$
Work it out but remember the restrictions on k above.
Again rough work, not proofread, correct as needed. thanks.
Note that if $x = 0$ and $y < 0$ (allowable by definition above) then using either the simplified form or the original form we get $\lambda = -1$.
This disproves some of the answers given.
My calculation above for $k$ when $y = kx , y < x$, may allow even lower value of $\lambda$. I leave it to you to take further for now.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2139957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
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 $AD$:
$$ \begin{align}
(2AC - 9)^2 &= AC^2 + (AC - 5)^2 + (AC - 5)^2 + (AC - 9)^2 \\
4AC^2 - 36 AC + 81 &= AC^2 + 2AC^2 - 20AC + 50 + AC^2 - 18AC + 81 \\
2AC &= 50 \\
AD &= 41
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2140204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 2
}
|
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:
$$4\int_{0}^{\infty} \frac{1-x^2}{(2x^2+3)^2-5}\ dx=4\int_{0}^{\infty} \frac{1-x^2}{(2x^2+3-\sqrt5)(2x^2+3+\sqrt5)}\ dx$$
$$$$Using partial fractions to separate the integral, we obtain:
$$\frac1{\sqrt5}\int_{0}^{\infty} \left(\frac{5-\sqrt5}{2x^2+3-\sqrt5}-\frac{5+\sqrt5}{2x^2+3+\sqrt5}\right)\ dx$$
$$=\frac{5-\sqrt5}{\sqrt5}\int_{0}^{\infty} \frac1{2x^2+3-\sqrt5}\ dx-\frac{5+\sqrt5}{\sqrt5}\int_{0}^{\infty}\frac1{2x^2+3+\sqrt5}\ dx$$
$$$$We need write the fractions in the form $\frac1{u^2+1}$:
$$=\frac{5-\sqrt5}{\sqrt5}\int_{0}^{\infty} \frac1{(3-\sqrt5)\left(\frac{2x^2}{3-\sqrt5}+1\right)}\ dx-\frac{5+\sqrt5}{\sqrt5}\int_{0}^{\infty}\frac1{(3+\sqrt5)\left(\frac{2x^2}{3+\sqrt5}+1\right)}\ dx$$
$$=\frac{5-\sqrt5}{\sqrt5(3-\sqrt5)}\int_{0}^{\infty} \frac1{\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)^2+1}\ dx-\frac{5+\sqrt5}{\sqrt5(3+\sqrt5)}\int_{0}^{\infty}\frac1{\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)^2+1}\ dx$$
$$\frac{5-\sqrt5}{\sqrt{10(3-\sqrt5)}}\int_{0}^{\infty} \frac{\sqrt2}{\sqrt{3-\sqrt5}\left(\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)^2+1\right)}\ dx-\frac{5+\sqrt5}{\sqrt{10(3+\sqrt5)}}\int_{0}^{\infty}\frac{\sqrt2}{\sqrt{3+\sqrt5}\left(\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)^2+1\right)}\ dx$$
$$$$Evaluating the indefinite integrals, we obtain:
$$\arctan\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)-\arctan\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)+C$$
$$$$Evaluating the improper integral:
$$\lim_{z \to \infty}\left[\arctan\left(\frac{\sqrt2x}{\sqrt{3-\sqrt5}}\right)-\arctan\left(\frac{\sqrt2x}{\sqrt{3+\sqrt5}}\right)\right]_0^z$$
$$=\lim_{z \to \infty}\left(\arctan\left(\frac{\sqrt2z}{\sqrt{3-\sqrt5}}\right)-\arctan\left(\frac{\sqrt2z}{\sqrt{3+\sqrt5}}\right)\right)$$
$$=\lim_{z \to \infty}\left(\arctan\left(\infty\right)-\arctan\left(\infty\right)\right)=\frac{\pi}2-\frac{\pi}2=0$$
The way that I evalauted the integral is pretty long and has a lot of calculations. Is there an easier way?
|
Substitute $x\mapsto\frac{1-x}{1+x}$ to reveal an integral of an odd function over an interval symmetric about the origin:
$$\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*}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2143667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Evaluating $\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$ Let $A(z_1),B(z_2),C(z_3)$ are complex numbers satisfying $|z-\sqrt{3}i|=1$ and $3z_1+\sqrt{3}i=2z_2+2z_3$. The question asks to find the value of $$\frac{z_2-\sqrt{3}i}{z_3-\sqrt{3}i}$$ I tried to shift the origin to $\sqrt{3}i$ and transformed the original condition in this system but it does not help me. Any ideas how to proceed? Thanks.
|
Put $t_j = z_j - \sqrt{3}i$. Then $|t_j|=1$ and $3t_1 = 2t_2 + 2t_3$. Hence $\frac{3}{4}t_1 = \frac{1}{2}(t_2 + t_3)$. The points $t_j$ lie on the unit circle, with center $O$ at the origin and form a triangle $ABC$, with $A = t_1$ etc. The midpoint $D$ of the chord $BC$ is $\frac{3}{4}t_1$ and hence lies on $OA$ at a distance $\frac{3}{4}$. This means that in the right angled triangle $ODB$, $OB = 1, OD = \frac{3}{4}$. Thus if the angle $BOC$ is $2\theta$, we have $\cos\theta = \frac{3}{4}$. Now $\frac{t_2}{t_3} = \cos 2\theta + i \sin 2\theta$ and $\cos 2\theta = 2\cos^2 \theta -1 = \frac{1}{8}$ and $\sin 2\theta = \pm\sqrt{1-\cos^2 2\theta} = \pm\frac{\sqrt{63}}{8}$. Thus the required ratio is
$$\frac{1}{8} \pm 3\frac{\sqrt{7}}{8} i$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2145406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Find the remainder of $\frac{3^{11}-1}{2}$ divided by $9$ I have the following question:
Find the remainder of $$\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 where to go next. Anyway, this is one of my many attempts to solve this question, and most of them ends with a complicated solution. I don't want to use modular arithmetic for this question. A hint or anything will help me.
|
$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 some natural number $k$.
$3^{11} - 1 = 18k + 8$
Divide both sides by $2.$
$3^{11} - 1 = 18k + 8$
$(3^{11} - 1)/2 = 9k + 4$
Hence, the remainder is $4$ when $(3^{11} - 1)/2$ is divided by $9$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2146798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
}
|
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 as closed formula this sequence:
$\left \{ b_n \right \}_n = \frac{1}{\sqrt{5}} \cdot \left [ \left ( \frac{1 + \sqrt{5}}{2} \right )^n - \left ( \frac{1 - \sqrt{5}}{2} \right )^n \right ] \quad \forall n \ge 0$
I have verified the base step with $n=2$, i.e. I have to verify the following statement:
$P(2) : a_2 = b_2 \iff 1 = 1$
but, I encounter some problems in the resolution of the inductive step:
I have set as true the statement:
$\begin{array}{lcl}P(n-2) : a_{n-2} & = & b_{n-2} \\ & = & \frac{1}{\sqrt{5}} \cdot \left [ \left ( \frac{1 + \sqrt{5}}{2} \right )^{n - 2} - \left ( \frac{1 - \sqrt{5}}{2} \right )^{n-2} \right ] \end{array}$
and from the truth of this last one I check the truth of
$\begin{array}{lcl}P(n-1) : a_{n-1} & = & b_{n-1} \\ & = & \frac{1}{\sqrt{5}} \cdot \left [ \left ( \frac{1 + \sqrt{5}}{2} \right )^{n - 1} - \left ( \frac{1 - \sqrt{5}}{2} \right )^{n-1} \right ] \end{array}$
and from the truth of this last one I check the truth of
$\begin{array}{lcl}P(n) : a_{n} & = & b_{n} \\ & = & \frac{1}{\sqrt{5}} \cdot \left [ \left ( \frac{1 + \sqrt{5}}{2} \right )^{n} - \left ( \frac{1 - \sqrt{5}}{2} \right )^{n} \right ] \end{array}$
therefore,
$\begin{array}{lcl} a_{n} & = & a_{n-1} + a_{n-2} \\ & = & \frac{1}{\sqrt{5}} \cdot \left [ \left ( \frac{1 + \sqrt{5}}{2} \right )^{n-1} - \left ( \frac{1 - \sqrt{5}}{2} \right )^{n-1} \right ] + \frac{1}{\sqrt{5}} \cdot \left [ \left ( \frac{1 + \sqrt{5}}{2} \right )^{n-2} - \left ( \frac{1 - \sqrt{5}}{2} \right )^{n-2} \right ]\end{array}$
I have also to assign: $p = \frac{1+\sqrt{5}}{2}, q=\frac{1-\sqrt{5}}{2}$ obtaining
$\begin{array}{lcl} a_n & = & \frac{1}{\sqrt{5}} (p^{n-1} - q^{n-1} + p^{n-2} - q^{n-2}) \\ & = & \frac{1}{\sqrt{5}} ( p^np^{-1} - q^nq^{-1} + p^np^{-2} - q^nq^{-2}) \\ & = & \frac{1}{\sqrt{5}} [p^n(p^{-1} + p^{-2}) - q^n(q^{-1} + q^{-2})]\end{array}$
I don't think if it is the right way, if I go to substitute appears a big expression. So this is my way and I stop here.
Please, can you tell me how to resolve this exercise? Many Thanks!
|
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]$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2149466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $\sum\limits_{cyc}\frac{a^2-bd}{b+2c+d}\geq0$
Let $a$, $b$, $c$ and $d$ be positive numbers. Prove that:
$$\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$$
This inequality is a similar to the following inequality of three variables.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^2-bc}{b+c+2a}+\frac{b^2-ca}{c+a+2b}+\frac{c^2-ab}{a+b+2c}\geq0,$$
which we can prove by the following reasoning.
$$\sum\limits_{cyc}\frac{a^2-bc}{b+c+2a}=\frac{1}{2}\sum\limits_{cyc}\frac{(a-b)(a+c)-(c-a)(a+b)}{b+c+2a}=$$
$$=\frac{1}{2}\sum_{cyc}(a-b)\left(\frac{a+c}{b+c+2a}-\frac{b+c}{c+a+2b}\right)=\frac{1}{2}\sum_{cyc}\frac{(a-b)^2}{(b+c+2a)(c+a+2b)}\geq0,$$
but this idea does not help for the starting inequality.
We can make the following. By Holder
$$\sum_{cyc}\frac{a^2}{b+2c+d}=\sum_{cyc}\frac{a^3}{ab+2ac+ad}\geq\frac{(a+b+c+d)^3}{4\sum\limits_{cyc}(ab+2ac+ad)}=\frac{(a+b+c+d)^3}{8\sum\limits_{cyc}(ab+ac)}$$
and
$$\sum_{cyc}\frac{bd}{b+2c+d}=2(a+b+c+d)\left(\frac{bd}{(b+2c+d)(d+2a+b)}+\frac{ac}{(a+2b+c)(c+2d+a)}\right).$$
Thus, it remains to prove that
$$\frac{(a+b+c+d)^2}{16\sum\limits_{cyc}(ab+ac)}\geq\frac{bd}{(b+2c+d)(d+2a+b)}+\frac{ac}{(a+2b+c)(c+2d+a)}$$
and I don't see, what is the rest.
|
The following quantity is clearly positive
\begin{eqnarray*}
((2(d^3+b^3)+8bd(b+d))(b-d)^2+(2(a^3+c^3)+8ca(c+a))(c-a)^2)+
((a+c)(5(d^2+b^2)(b-d)^2+12bd(b-d)^2)+(b+d)(5(a^2+c^2)(a-c)^2+12ac(a-c)^2))+
2((b^3+d^3)(a-c)^2+(a^3+c^3)(b-d)^2)+
14(bd(a+c)(a-c)^2+ac(b+d)(b-d)^2)
\end{eqnarray*}
After doing some algebra, this is the same as
\begin{eqnarray*}
(a^2-bd)(c+2d+a)(d+2a+b)(a+2b+c)+(b^2-ca)(d+2a+b)(a+2b+c)(b+2c+d)+(c^2-db)(a+2b+c)(b+2c+d)(c+2d+a)+(d^2-ac)(b+2c+d)(c+2d+a)(d+2a+b) \geq 0
\end{eqnarray*}
Now divide by $(c+2d+a)(d+2a+b)(a+2b+c)(b+2c+d)$ and the result follows.
(a^2-b*d)*(c+2*d+a)*(d+2*a+b)*(a+2*b+c)+(b^2-c*a)*(d+2*a+b)*(a+2*b+c)*(b+2*c+d)+(c^2-d*b)*(a+2*b+c)*(b+2*c+d)*(c+2*d+a)+(d^2-a*c)*(b+2*c+d)*(c+2*d+a)*(d+2*a+b)-(((2*(d^3+b^3)+8*b*d*(b+d))*(b-d)^2+(2*(a^3+c^3)+8*c*a*(c+a))*(c-a)^2)+((a+c)*(5*(d^2+b^2)*(b-d)^2+12*b*d*(b-d)^2)+(b+d)*(5*(a^2+c^2)*(a-c)^2+12*a*c*(a-c)^2))+2*((b^3+d^3)*(a-c)^2+(a^3+c^3)*(b-d)^2)+14*(b*d*(a+c)*(a-c)^2+a*c*(b+d)*(b-d)^2));
The above computer algebra can be copied & pasted into reduce & serves as justification for the above claim.
The above solution does not use any well known theorems or tricks. I am sure that more elegant solutions exist & We would be interested to see them.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2151579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
}
|
Find a given logarithmic definite integral Find the following integral, where $a$ is a real number bigger than $1$:
$$\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\int_1^a \frac{\ln t}{t^2 + a}\,\mathrm dt.$$
Thank you in advance!
|
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}} \right) \Big|_{1}^{a}
\, - \frac{1}{\sqrt{a}} \int\limits_{1}^{a} \frac{1}{t} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) dt \\
&= \frac{\ln a}{\sqrt{a}} \tan^{-1}(\sqrt{a})
\, - \frac{1}{\sqrt{a}} \int\limits_{1}^{a} \frac{1}{t} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) dt
\end{align}
Let $y=t/ \sqrt{a}$
\begin{align}
I_{2} &= \int\limits_{1}^{a} \frac{1}{t} \tan^{-1}\left( \frac{t}{\sqrt{a}} \right) dt
= \int\limits_{1/\sqrt{a}}^{\sqrt{a}} \frac{1}{y} \tan^{-1}(y) dy \\
\tag{a}
&= \frac{i}{2} \left[ \int\limits_{1/\sqrt{a}}^{\sqrt{a}} \frac{\ln (1-iy)}{y} dy
\, - \int\limits_{1/\sqrt{a}}^{\sqrt{a}} \frac{\ln (1+iy)}{y} dy \right] \\
\tag{b}
&= \frac{i}{2} \left[ \mathrm{Li}_{2}(-iy) - \mathrm{Li}_{2}(iy) \right] \Big|_{1/\sqrt{a}}^{\sqrt{a}} \\
&= \frac{i}{2} \left( \left[ \mathrm{Li}_{2}(-i\sqrt{a}) + \mathrm{Li}_{2}\left(\frac{i}{\sqrt{a}}\right) \right]
- \left[ \mathrm{Li}_{2}(i\sqrt{a}) + \mathrm{Li}_{2}\left(\frac{-i}{\sqrt{a}}\right) \right] \right) \\
\tag{c}
&= \frac{i}{2} \left( \left[ -\frac{\pi ^{2}}{6} - \frac{1}{2} \ln ^{2}(i\sqrt{a}) \right]
- \left[ -\frac{\pi ^{2}}{6} - \frac{1}{2} \ln ^{2}(-i\sqrt{a}) \right] \right) \\
\tag{d}
&= \frac{\pi}{4} \ln a
\end{align}
a. $\tan^{-1}(y) = \frac{i}{2} [\ln (1-iy) - \ln (1+iy)]$
b. Dilogarithm function
\begin{equation}
\mathrm{Li}_{2}(z) = -\int_{0}^{z} \frac{\ln (1-x)}{x} dx
\end{equation}
c. Use the identity
\begin{equation}
\mathrm{Li}_{2}(z) + \mathrm{Li}_{2}(1/z) = -\frac{\pi ^{2}}{6} - \frac{1}{2} \ln ^{2}(-z)
\end{equation}
d. $\ln (\pm iz) = \ln z \pm i\pi /2$
Now we have
\begin{equation}
I = 4I_{1} = \frac{4}{\sqrt{a}} (\ln a) \tan^{-1}(\sqrt{a}) \, - \frac{\pi}{\sqrt{a}} \ln a
\end{equation}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2154470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
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 through and get a common denominator that it will all work out in the end, but I just want a good explanation for it.
|
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)$$and you can then solve the equations:
1)$\ \ a+b=0$
2)$\ -2a+b+c=4$
3)$\ c=0$
for $a,b$ and $c$ to get$$\frac{4x}{(x+1)(x-1)^2}\equiv \frac{-\frac{4}{3}}{x+1}+\frac{\frac{4}{3}x}{(x-1)^2}$$The point is that you have three unknowns ($a,b$and $c$) and three equations, which will generally give a solution.
Were you to assume$$\frac{4x}{(x+1)(x-1)^2}\equiv\frac{a}{x+1}+\frac{b}{(x-1)^2}$$in this case you would get$$4x\equiv a(x-1)^2+b(x+1)\equiv(a+b)x^2+(-2a+b)x+(1+b)$$
You then have to solve
1)$\ a+b=0$
2)$\ -2a+b=4$
3)$\ 1+b=0$
This is three equations in two variables and has no solution.
Trying $$\frac{4x}{(x+1)(x-1)^2}\equiv\frac{a}{x+1}+\frac{b}{(x-1)^2}+c$$ would not help because then after clearing the denominator you finish up with a cubic, so you still have one more equation than you have variables.
Assuming $$\frac{4x}{(x+1)(x-1)^2}\equiv\frac{a}{x+1}+\frac{b}{(x-1)}+\frac{c}{(x-1)^2}$$is usually the most convenient.
If you want to split a rational function into a sum of reciprocals of polynomials then this last form is necessary, but only possible if the denominator of the function you wish to split (in its lowest terms) is a product of linear factors.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2158948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Proving the Hypergeometric Sequence
Question: How do you prove$$_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}$$For $\Re(2x+2y+n+2)>0$
I'm not sure how to prove this. There are a multitude of other similar formulas. Some of which$$\begin{align*}_4F_3\left[\begin{array}{c c}\frac 12n+1,n,n,-x\\\frac 12n,x+n+1,1\end{array};-1\right] & =\dfrac {\Gamma(x+n+1)}{\Gamma(n+1)\Gamma(x+1)}\end{align*}\tag{2}$$For $\Re(2x-n+2)>0$$$_3F_2\left[\begin{array}{c c}\frac 12n+1,n,-x\\\frac 12n,x+n+1\end{array};-1\right]=\dfrac {\Gamma(x+n+1)\Gamma\left(\frac 12n+\frac 12\right)}{\Gamma(n+1)\Gamma\left(x+\frac 12n+\frac 12\right)}\tag3$$
For $\Re(x)>-\frac 12$. I know that the general Hypergeometric Sequence can be written as$$_pF_q\left[\begin{array}{c c}\alpha_1,\alpha_2,\ldots,\alpha_p\\\beta_1,\beta_2,\ldots,\beta_p\end{array};x\right]=\sum\limits_{k=0}^\infty\dfrac {(\alpha_1)_k(\alpha_2)_k\cdots(\alpha_p)_k}{(\beta_1)_k(\beta_2)_k\cdots(\beta_p)_k}\dfrac {x^k}{k!}\tag4$$So $(1)$ becomes$$_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]=\sum\limits_{k=0}^{\infty}\dfrac {\left(\frac 12n+1\right)_k(n)_k(-x)_k(-y)_k}{\left(\frac 12n\right)_k(x+n+1)_k(y+n+1)_k}\dfrac {(-1)^k}{k!}\tag{5}$$However, how do you manipulate the RHS of $(5)$ into the RHS of $(1)$. I think that there could be a sort of elementary transformation involved, but if so, I'm not sure what.
|
An established identity, useful to transform a $_4F_3 $ hypergeometric sequence with negative unit argument in a $_3F_2$ sequence, is
$$_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]} \, _3F_2 \left[\begin{array}{c c} a/2 - d+1, b, c \\ a/2 + 1, a - d+1 \end{array};-1\right]$$
So, rewriting your series as
$$_4F_3 \left[\begin{array}{c c} n, -x, -y, \frac 12n+1\\ n+x+1, n+y+1, \frac 12n \end{array};-1\right] \\$$
we obtain that it is equal to
$$\dfrac {\Gamma(n+x+1)\Gamma(n+y+1)}{\Gamma(n+1)\Gamma(n+x+y+1)} \\ _3F_2 \left[\begin{array}{c c} 0, -x, -y \\ n/2 + 1, n/2 \end{array};-1\right] $$
and since the $_3F_2$ sequence, as a result of the zero among its terms, is equal to $1$, we get your identity.
The same method can be used to generate the second identity. Rewriting it as
$$\begin{align*}_4F_3\left[\begin{array}{c c} n, -x, n,\dfrac 12n+1 \\ n+x+1, 1, \dfrac 12n \end{array};-1\right] & \end{align*}$$
we get
$$ =\dfrac {\Gamma(n+x+1) \Gamma (1)}{\Gamma(n+1)\Gamma(x+1)} \, _3F_2 \left[\begin{array}{c c} 0, -x, n \\ n/2 + 1, n/2 \end{array};-1\right] $$
where again the $_3F_2$ sequence is equal to $1$, giving your identity.
Lastly, another known identity can be used to express in closed form a $_3F_2$ hypergeometric function with negative unit argument:
$$_3F_2 \left[\begin{array}{c c} a, \dfrac a2+1,c \\ \dfrac a2,a-c+1 \end{array};-1 \right]=\dfrac {\sqrt {\pi} \,\, \Gamma(a-c+1)}{2^n \, \Gamma \left(\frac a2+1 \right)\Gamma \left(\frac{a+1}{2}-c \right)}$$
This can be used to prove the third identity of the OP. Rewriting it as follows we get
$$_3F_2 \left[\begin{array}{c c} n, \frac 12n+1,-x \\ \dfrac 12n,n+x+1 \end{array};-1 \right]=\dfrac {\sqrt {\pi} \,\, \Gamma(n+x+1)}{2^n \Gamma \left(\frac n2+1 \right) \Gamma \left(\frac{n+1}{2}+x \right)}$$
Now since it is known that, for given $z $,
$$\dfrac {\Gamma (z)}{\Gamma (2z)}= \dfrac {\sqrt {\pi}}{2^{2z-1}\, \Gamma \left(z+\frac{1}{2} \right)}$$
setting $z=(n+1)/2 \,\,$ we have
$$\dfrac {\Gamma (\frac 12n+ \frac 12 )}{\Gamma (n+1)}= \dfrac {\sqrt {\pi}}{2^{n} \Gamma \left(\frac {1}{2}n+1 \right)}$$
Substituting in the equation above we obtain
$$_3F_2 \left[\begin{array}{c c} n, \frac 12n+1,-x \\ \frac 12n,n+x+1 \end{array};-1 \right]=\dfrac {\Gamma(n+x+1) \Gamma \left(\frac 12n + \frac 12 \right)}{\Gamma(n+1) \Gamma \left(\frac 12n + x+ \frac 12 \right)}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2159662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
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 adjust the circle radius to see the points.)
So
$$
\int_0^{2\pi} \int_0^{1+\cos\theta}r\,dr\,d\theta
= \int_0^2 \int_{-\cos^{-1}(r-1)}^{\cos^{-1}(r-1)}r\,d\theta\,dr
= 2\int_0^2 r\cos^{-1}(r-1) \,dr
$$
To evaluate this, first do integration by parts with $u = \cos^{-1}(r-1)$ and $dv = 2r$:
\begin{align*}
2\int_0^2 r\cos^{-1}(r-1) \,dr
&= \left.r^2 \cos^{-1}(r-1)\right|^2_0 + \int_0^2 \frac{r^2}{\sqrt{1-(r-1)^2}}\,dr \\
&= 2^2 (0) - 0^2 (-\pi) + \int_0^2 \frac{r^2}{\sqrt{1-(r-1)^2}}\,dr \\
\end{align*}
Now substitute $s=r-1$, $ds=dr$, so this is equal to:
$$
\int_0^2 \frac{r^2}{\sqrt{1-(r-1)^2}}\,dr
= \int_{-1}^{1} \frac{(s+1)^2}{\sqrt{1-s^2}}\,ds
$$
Now substitute $s = \sin\theta$:
\begin{align*}
\int_{-1}^{1} \frac{(s+1)^2}{\sqrt{1-s^2}}\,ds
&= \int_{-\pi/2}^{\pi/2}\frac{(\sin\theta+1)^2}{\cos\theta}\,\cos\theta\,d\theta
\\&= \int_{-\pi/2}^{\pi/2}(1 + 2\sin\theta + \cos^2\theta)\,d\theta
\\&= \pi + 0 + \int_{-\pi/2}^{\pi/2} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta
\\&= \pi + \frac{\pi}{2} + 0 = \frac{3\pi}{2}
\end{align*}
On the other hand, without switching the order,
\begin{align*}
\int_0^{2\pi} \int_0^{1+\cos\theta}r\,dr\,d\theta
&= \frac{1}{2}\int_0^{2\pi}(1 + \cos\theta)^2\,d\theta
= \frac{1}{2}\int_0^{2\pi}\left(1 + 2\cos\theta+ \cos^2\theta\right)\,d\theta
\\&= \pi + 0 + \frac12\int_0^{2\pi} \left(\frac{1 + \cos(2\theta)}{2}\right)\,d\theta
\\&= \pi + \frac{\pi}{2} + 0 = \frac{3\pi}{2}
\end{align*}
Given the choice, I prefer the second integral.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2160658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
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 $r_1\times f + r_2\times g+ r_3\times h$ where $r_1,r_2,r_3$ in $C[x,y,z]$ but then i realised that i cant possibly do that by hand.This was an exercise on my first computational algebra course so we havent really learnt anything much yet. I would like a hint if possible.
|
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^2z^2+y^2z^2)\\
&=(x^4+y^4+z^4)+2k\\
f^2g&=((x^2+y^2+z^2)+2(xy+xz+yz))(x^2+y^2+z^2)\\
&=(x^4+y^4+z^4)+2(x^2y^2+x^2z^2+y^2z^2)\\
&\qquad\qquad+2(x^3y+x^3z+x^2yz+xy^3+xy^2z+y^3z+xyz^2+xz^3+yz^3)\\
&=(x^4+y^4+z^4)+2j+2k+2l\\
f^4&=(x^4+y^4+z^4)+4(xy^3+xz^3+yx^3+yz^3+zx^3+zy^3)+6(x^2y^2+x^2z^2+y^2z^2)\\
&\qquad\qquad+12(x^3yz+xy^3z+xyz^3)\\
&=(x^4+y^4+z^4)+4j+6k+12l
\end{align*}
Combining, we have
\begin{align*}
21&=fh=(x^4+y^4+z^4)+j\\
25&=g^2=(x^4+y^4+z^4)+2k\\
45&=f^2g=(x^4+y^4+z^4)+2j+2k+2l\\
81&=f^4=(x^4+y^4+z^4)+4j+6k+12l
\end{align*}
This is now a system of equations where we can use linear algebra to solve. For example, subtracting $6$ times the third from the fourth gives
$$
-189=-5(x^4+y^4+z^4)-8j-6k
$$
Now, adding $3$ times the second to this gives
$$
-114=-2(x^4+y^4+z^4)-8j
$$
Finally, adding $8$ times the first to this gives
$$
54=6(x^4+y^4+z^4)
$$
In other words, $x^4+y^4+z^4=9$.
|
{
"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
}
|
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:
$${\frac{(a^3)(25)-(a^3)(9)}{225}}$$
$${\frac{25a^3-9a^3}{225}}$$
$${\frac{16a^3}{225}}$$
$${\frac{16}{225}}\times a^3$$
We can take square root of these things.
$$\sqrt{\frac{16}{225}}= \frac{\sqrt{16}}{\sqrt{225}}=\frac{4}{15}$$
$$\sqrt{a^3}=(a^3)^{(1/2)}=a^{(3/2)}$$
So,
$$\sqrt{\frac{a^3}{9}-\frac{a^3}{25}} = \frac{4a^{3/2}}{15}$$
Now multiply by $\frac{75}{8}$.
$$\frac{75}{8}\times\frac{4a^{3/2}}{15} = \frac{5}{2}\times\frac{1a^{3/2}}{1} = \frac{5a^{3/2}}{2}$$
|
{
"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
}
|
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](a+x+d)^2\right)\\
&\qquad +[a^2]\left([x^1](a+x+c)^2\right)\left([x^2](a+x+d)^2\right)\tag{1}\\
&=[a^2]\left(2(a+d)\right)+[a^2]\left(2(a+c)\right)\tag{2}\\
&=0
\end{align*}
Comment:
*
*In (1) we observe that in order to obtain the coefficient of $x^3$ one factor has to contribute $x$ and the other factor has to contribute $x^2$. There are no other possibilities to obtain a term with $x^3$.
*In (2) we select the coefficient of $x$ resp. $x^2$ according to the binomial formula
\begin{align*}
(a+x+u)^2=(a+u)^2+2(a+u)x+x^2
\end{align*}
Since there is no term with a factor $a^2$ the result is $0$.
|
{
"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
}
|
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 $$\int\frac{1-t^2}{1+t^4}dt=\frac{1}{2}\int\frac{1-\sqrt 2 t}{t^2-\sqrt 2 t+1}dt+\frac{1}{2}\int\frac{1+\sqrt 2t}{t^2+\sqrt 2t+1}dt$$
Using substitutions $$u=t^2-\sqrt 2t+1$$ and $$v=t^2+\sqrt 2t+1$$
we get $$\int\frac{1-t^2}{1+t^4}dt=-\frac{\sqrt 2}{4}\ln|u|+\frac{\sqrt 2}{4}\ln|v|=-\frac{\sqrt 2}{4}\ln|t^2-\sqrt 2t+1|+\frac{\sqrt 2}{4}\ln|t^2+\sqrt 2t+1|$$
From $$x=\frac{1+t^2}{1-t^2}\Rightarrow t=\sqrt{\frac{x-1}{x+1}}\Rightarrow$$
$$\int\frac{1}{(1+x^2)\sqrt{x^2-1}}dx=$$$$-\frac{\sqrt 2}{4}\ln\left|\frac{x-1}{x+1}-\sqrt 2\sqrt{\frac{x-1}{x+1}}+1\right|+\frac{\sqrt 2}{4}\ln\left|\frac{x-1}{x+1}+\sqrt 2\sqrt{\frac{x-1}{x+1}}+1\right|+c$$
Is there another, quicker method to solve this integral, rather than Euler substitution.
|
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+1} \mathrm{d}\theta \\
&= \int \frac{1}{2-\sin^2 \theta} \mathrm{d}(\sin \theta)\\
&= \frac{1}{2\sqrt 2} \ln \left|\frac{\sqrt 2 + \sin \theta}{\sqrt 2 - \sin \theta}\right|+C
\end{align*}
Since, $\sec \theta = x \implies \sin \theta = \frac{\sqrt {x^2-1}}{x}$, we get
$$\int \frac{1}{(x^2+1)\sqrt{x^2-1}} \mathrm{d}x = \frac{1}{2\sqrt 2} \ln \left|\frac{\sqrt 2x + \sqrt{x^2-1}}{\sqrt 2x - \sqrt{x^2-1}}\right| +C$$
|
{
"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
}
|
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.
$$\sum_{n=0}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=\sum_{n=1}^{+\infty}(n+1)x^{2n+1}\int_0^1t^{2n}dt=\int_0^1\left(\sum_{n=1}^{+\infty}(n+1)x^{2n+1}t^{2n}\right)dt$$
$$=x\int_0^1\left(\sum_{n=1}^{+\infty}(n+1)(xt)^{2n}\right)dt$$
$$\sum_{n=1}^{+\infty}(n+1)(xt)^{2n}=\sum_{n=1}^{+\infty}n(xt)^{2n}+\sum_{n=1}^{+\infty}(xt)^{2n}$$
$$\sum_{n=1}^{+\infty}(xt)^{2n}=\frac{(xt)^2}{1-(xt)^2}$$
How to find the sum of $$\sum_{n=1}^{+\infty}n(xt)^{2n}?$$
EDIT:
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=\left(\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}\right)'$$
$$\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}=\frac{1}{2}x^2\sum_{n=1}^{+\infty}\frac{x^{2n}}{2n+1}$$
$$=\frac{1}{2}x^2\sum_{n=1}^{+\infty}x^{2n}\int_0^1t^{2n}dt=\frac{1}{2}x^2\int_0^1\left(\sum_{n=1}^{+\infty}(xt)^{2n}\right)dt$$
$$=\frac{1}{2}x^2\int_0^1\frac{(xt)^2}{1-(xt)^2}=\frac{1}{2}x^4\int_0^1\frac{t^2}{1-(xt)^2}dt$$
$$=\frac{1}{2}x^4\cdot\frac{1}{x^3}(-x-\frac{1}{2}\ln|x-1|+\frac{1}{2}\ln|x+1|)\Rightarrow$$
$$\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}=\frac{1}{2}x\left(-x-\frac{1}{2}\ln|x-1|+\frac{1}{2}\ln|x+1|\right)\Rightarrow$$
$$\left(\sum_{n=1}^{+\infty}\frac{x^{2n+2}}{2(2n+1)}\right)'=-x-\frac{1}{4}\left(\ln|x-1|+\frac{x}{x-1}\right)+\frac{1}{4}\left(\ln|x+1|+\frac{x}{x+1}\right)\Rightarrow$$
Finally,
$$\sum_{n=1}^{+\infty}\frac{n+1}{2n+1}x^{2n+1}=-x-\frac{1}{4}\left(\ln|x-1|+\frac{x}{x-1}\right)+\frac{1}{4}\left(\ln|x+1|+\frac{x}{x+1}\right)$$
Question: Is this correct?
|
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
$$
1+2x+3x^2+...+nx^{n-1}=\frac{1-x^{n+1}}{(1-x)^2}+\frac{-(n+1)x^{n}}{1-x}, \quad |x|<1, \tag2
$$ and by making $n \to +\infty$ in $(2)$, using $|x|<1$, multiplying by $x$, gives
$$
\sum_{n=1}^\infty nx^n=\frac{x}{(1-x)^2},\qquad |x|<1. \tag3
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2172875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
}
|
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 I do not know how to use these notations and get the solutions. I would appreciate the help.
|
$H(x,y)=4 ln(x^3+4y^2)$
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.
$f_{y}(x,y)=\frac{32y}{x^3+4y^2}$ and $f_{yy}(x,y)=\frac{32(x^3-4y^2)}{(x^3+4y^2)^2}$ (treat $x$ as a constant in this case)
Plugging in for $x=2, y=3$ will leave you with $f_{yy}(2,3)=-56/121$
For $f_{xy} = (f_x)_y$, we partially differentiate $f_{x}(=\frac{12x^2}{x^3+4y^2})$ with respect to $y$.
So $f_{xy}=\frac{\partial}{\partial y}(\frac{12x^2}{x^3+4y^2})=-\frac{96x^2y}{\left(x^3+4y^2\right)^2}$
Plugging in for $x=2, y=3$ will leave you with $f_{xy}(2,3)=-72/121$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2174010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
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:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}x^{k(4k^2-1)}$$
$$f(x)=\sum_{k=1}^{+\infty}(-1)^{k-1}\frac{1}{k(4k^2-1)}x^{k(4k^2-1)}$$
After derivation:
$$f'(x)=\sum_{k=1}^{+\infty}(-1)^{k-1}x^{4k^3-k-1}$$
The problem here is that:
$$\sum_{k=1}^{+\infty}(-1)^{k-1}x^{4k^3-k-1}=x^2-x^{29}+x^{104}-...$$
Is it possible to find the closed form for the last series?
2) Using partial sums:
$$S_n=\sum_{k=0}^{n}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
Now, using the formula:
$$S_n+a_{n+1}=a_0+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}\Rightarrow$$
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
$$S_n=\frac{1}{3}-\frac{1}{30}+...+(-1)^{n}\frac{1}{(n+1)(4(n+1)^2-1)}$$
$$\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}=T_n=-\frac{1}{30}+...+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
$$T_n=S_n-\frac{1}{3}+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
Going back to the formula $$S_n+a_{n+1}=a_0+\sum_{k=1}^{n+1}(-1)^{k}\frac{1}{(k+1)(4(k+1)^2-1)}$$
we have that $S_n$ cancels, so we can't determine partial sums using this method?
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+T_n$$
$$S_n+(-1)^{n+1}\frac{1}{(k+2)(4(k+2)^2-1)}=\frac{1}{3}+S_n-\frac{1}{3}+(-1)^{n+1}\frac{1}{(n+2)(4(n+2)^2-1)}$$
Question: How to find the sum of this series?
|
$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
This is a variant of $\texttt{@Markus Scheuer}$ 'telescopic' fine answer:
\begin{align}
\sum_{k = 1}^{\infty}{\pars{-1}^{k - 1} \over k\pars{4k^{2} - 1}} & =
-2\sum_{k = 1}^{\infty}
{\ic^{2k} \over \pars{2k}\bracks{\pars{2k}^{2} - 1}} =
-2\,\Re\sum_{k = 2}^{\infty}{\ic^{k} \over k\pars{k^{2} - 1}}
\\[5mm] & =
-\,\Re\pars{\sum_{k = 2}^{\infty}{\ic^{k} \over k - 1} -
2\sum_{k = 2}^{\infty}{\ic^{k} \over k} +
\sum_{k = 2}^{\infty}{\ic^{k} \over k + 1}}
\\[5mm] & =\require{cancel}
-\,\Re\bracks{\cancel{\ic\sum_{k = 1}^{\infty}{\ic^{k} \over k}} +
2\ic - \color{#f00}{2\sum_{k = 1}^{\infty}{\ic^{k} \over k}} -
\color{#f00}{\ic}\pars{\color{#f00}{-\ic} + {1 \over 2} +
\cancel{\sum_{k = 1}^{\infty}{\ic^{k} \over k}}}}
\\[5mm] & =
2\,\Re\sum_{k = 1}^{\infty}{\ic^{k} \over k} + 1 =
-2\,\Re\ln\pars{1 - \ic} + 1 = \bbx{\ds{1 - \ln\pars{2}}}
\end{align}
|
{
"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 $C$? Please help, I have no idea how I could even start, I just know that I need some polynomial on something inside triangle which would not be constructible....
|
\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 \gamma + \sin \beta \sin \gamma = \cdots = \frac{b^2+c^2-a^2}{2bc}.
\end{eqnarray*}
Now to the problem in hand ...
\begin{eqnarray*}
\cos (\beta- \gamma)= \cos \beta \cos \gamma + \sin \beta \sin \gamma = \cdots = \frac{a^2 b^2+a^2c^2-b^4+2b^2c^2-c^4}{2a^2bc}
\end{eqnarray*}
Now a little bit of algeabra & we have the following quartic for $c$
\begin{eqnarray*}
c^4-c^2(2b^2+a^2)+2a^2bc \cos (\beta - \gamma)+b^4-a^2b^2=0.
\end{eqnarray*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2183377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
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: $a\ \ \ \ \ \ \ 2a\ \ \ \ \ \ \ a^2$
$\ \ -ve\ \ +ve\ \ \ -ve\ \ \ \ +ve$
Set of solutions:$\{ 3,4,5,\cdots, a-1\}U\{2a+1,2a+2,\cdots,a^2-1\}$
No. of elements =18
$a-3+a^2-2a-1=18$
$a^2-a-22=0$
But this equation has no solution in natural numbers.
What fault have I made in my procedure?
|
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 integers between $2a$ and $a^2$.
In total, we have
$$a^2-2a-1+a-1=a^2-a-2$$
positive integers satisfying your equation. Finally, we solve $a^2-a-2=18$ and find out that $a=5$.
|
{
"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
}
|
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 we can assume $s \ne 1$.
Any references would be appreciated.
|
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
}
|
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
$$Q = \frac{1}{2}\begin{pmatrix} 0 & 1 & 0 & 0 \\ 1 & 0 & 1 & 0 \\ 0 & 1 & 0 & 1 \\ 0 & 0 & 1 & 0\end{pmatrix}$$
You have
$$\frac{ab+bc+cd}{a^2+b^2+c^2+d^2} = \frac{X^T Q X}{X^TX}=X^T Q X$$
This is maximum for eigen-vector of $Q$ associated to its maximum eigen-value (which seems to be half of golden ratio $\frac{(1+\sqrt{5})}{4}$)
|
{
"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 factorise. I'm hoping there is an easier way to solve this question. Any help is appreciated.Thanks :).
The equation I get after multiplying is
$$x^4+30x^3+325x^2+1500x+2484=0,$$
and its roots are
$$-6, \quad -9, \quad \frac{-15\pm\sqrt{41}}2 .$$
|
$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 effectively get rid of "$+$".
|
$$ 3600 - 25 = 60^2 - 5^2 = (60 + 5) (60 - 5) = 65 \cdot 55 $$
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 time, we can rule out some squares since the difference cannot be $3 \pmod 4$ or $2 \pmod 4.$ In this case, $61^2 - 3575 = 146 \equiv 2 \pmod 4 $ cannot be a square. We would jump to $62^2 - 3575,$ then $64^2 - 3575.$ However, the first time worked.
If I gave you $9991$ the next larger square would be $10000$ and $10000 - 9991 = 9,$ so $9991 = (100 +3)(100-3) = 103 \cdot 97$
|
{
"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 states that if a sequence is bounded above and strictly increasing, then it converges. Therefore this sequence by BMCT converges.
Is this correct? Is there anything else I should say or add?
|
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
$b_n \to 0$.
Since
$0 < a_n < d$,
$-d < b_n < 0$
or
$d^2-d < b_n+d^2 < d^2$
or
$\sqrt{d^2-d} < \sqrt{b_n+d^2} < d
$.
Then
$b_{n+1}+d
=\sqrt{b_n+d+d^2-d}
=\sqrt{b_n+d^2}
$
so that
$\begin{array}\\
b_{n+1}
&=\sqrt{b_n+d^2}-d\\
&=(\sqrt{b_n+d^2}-d)\dfrac{\sqrt{b_n+d^2}+d}{\sqrt{b_n+d^2}+d}\\
&=\dfrac{b_n}{\sqrt{b_n+d^2}+d}\\
\text{so}\\
|b_{n+1}|
&=\dfrac{|b_n|}{|\sqrt{b_n+d^2}+d|}\\
&\lt\dfrac{|b_n|}{\sqrt{d^2-d}+d}\\
\end{array}
$
so $b_n \to 0$
linearly.
Note that
this also shows that
$|b_{n+1}|
\gt \dfrac{|b_n|}{2d}
$,
so the convergence
is at most linear.
Since $b_n \to 0$,
$\dfrac{b_{n+1}}{b_n}
\to \dfrac1{2d}
$,
so this is the
exact rate of convergence.
If $a_n \lt 2$
then
$a_{n+1} \lt \sqrt{4}
= 2$,
so the sequence is bounded.
Let $a_n = b_n+2$.
Want to show that
$b_n \to 0$.
Since
$0 < a_n < 2$,
$-2 < b_n < 0$
or
$2 < b_n+4 < 4$
or
$\sqrt{2} < \sqrt{b_n+4} < 2$.
$b_{n+1}+2
=\sqrt{b_n+4}
$
or
$\begin{array}\\
b_{n+1}
&=\sqrt{b_n+4}-2\\
&=(\sqrt{b_n+4}-2)\dfrac{\sqrt{b_n+4}+2}{\sqrt{b_n+4}+2}\\
&=\dfrac{b_n}{\sqrt{b_n+4}+2}\\
\text{so}\\
|b_{n+1}|
&=\dfrac{|b_n|}{|\sqrt{b_n+4}+2|}\\
&\lt\dfrac{|b_n|}{2+\sqrt{2}}\\
\end{array}
$
so $b_n \to 0$
linearly.
Note that
this also shows that
$|b_{n+1}|
\gt \dfrac{|b_n|}{4}
$,
so the convergence
is at most linear.
Since $b_n \to 0$,
$\dfrac{b_{n+1}}{b_n}
\to \dfrac14
$.
|
{
"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,
$${\left(\sin a \cos b + \cos a \sin b\right)}^2 + {\left(\sin a \cos b - \cos a \sin b\right)}^2$$
$$=\sin^2 a \cos^2b + \cos^2 a \sin^2 b + \sin^2 a \cos^2 b + \cos^2 a \sin^2 b$$
I am stuck here, how do I proceed from here?
Edit: from answers I understand how to prove,but how to prove from where I am stuck?
|
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
$$\begin{align} 1 - \cos(2a) \cos(2b) & = 1 - \frac{ \left( e^{2ia} + e^{-2ia} \right) \left(e^{2ib} + e^{-2ib} \right)}{4}\\
& = \frac{e^{2i(a + b)} + e^{-2i(a + b)} + e^{2i(a - b)} + e^{-2i(a - b)} - 4}{-4}\\
& = \frac{ \left( e^{i(a + b)} - e^{-i(a + b)} \right)^2 + 2 + \left( e^{i(a - b)} - e^{-i(a - b)} \right)^2 + 2 - 4}{4i^2} \\
& = \sin^2(a+b) + \sin^2(a - b)\end{align} $$
|
{
"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
$$ \int_0^1 \left(\frac{1}{\sqrt{x}} + \frac{3x}{\sqrt{x}} + \frac{5x^3}{\sqrt{x}}\right)dx$$
$$ \int_0^1 (x^{-1/2} + 3x^{1/2} +5x^{5/2})\ dx $$
$$ \int_0^1 2x^{1/2} + 2x^{3/2} + \frac{10}{7}x^{7/2} $$
Not really sure where to go from there. Should I sub 1 in for the x values and let that be the answer?
|
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 denominator. I think I could do it if I used complex roots but that seems very complicated.
I wasn't too sure on how to proceed from here.
|
Well, I can take a stab at starting it:
$$
\begin{align*}
\int \frac{2 x^3 - 3 x^2 - 26 x + 38}{x^4 - 2
x^3 - 13 x^2 + 38 x - 12} \, dx &= \int \frac{4 x^3 - 6 x^2 - 26 x + 38}{x^4 - 2
x^3 - 13 x^2 + 38 x - 12} \, dx\\ &\qquad+ \int \frac{-2x^{3} + 3x^{2}}{x^4 - 2
x^3 - 13 x^2 + 38 x - 12}\,dx\\
&=\ln\left|x^4 - 2
x^3 - 13 x^2 + 38 x - 12\right|\\ &\qquad+ \int\frac{-2x^{3} + 3x^{2}}{x^4 - 2
x^3 - 13 x^2 + 38 x - 12}\,dx.
\end{align*}
$$
I don't see what you can do from there, however.
|
{
"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 the modulus to be $2\cos{\left(\frac{\pi}{4}-\frac{\theta}{2}\right)}$ units and that the argument would be $\operatorname{arg}(z+i)=\frac{\pi}{4}+\frac{\theta}{2}$.
Now, the next step that I took was that I replaced every theta with $\frac{3\pi}{8}$ in the polar form of the complex number $z+i$. So now it would look like this:
$$z+i=\left[2\cos{\left(\frac{\pi}{8}\right)}\right]\operatorname{cis}{\left(\frac{3\pi}{8}\right)}$$
Then, I expanded the $\operatorname{cis}{\left(\frac{3\pi}{8}\right)}$ part to become $\cos{\left(\frac{3\pi}{8}\right)}+i\sin{\left({\frac{3\pi}{8}}\right)}$. So now I've got the $\cos\left({\frac{3\pi}{8}}\right)$ part but I don't really know what to do next. I've tried to split the angle up so that there would be two angles so I can use an identity, however, it would end up with a difficult fraction instead. So if the rest of the answer or a hint would be given to finish the question, that would be great!!
Thanks!!
|
$$
\begin{aligned}
\text{By}\cos ^{2} x &=\frac{1+\cos 2 x}{2}, \textrm{ we have } \\
\cos \left(\frac{3 \pi}{8}\right) &=\sqrt{\frac{1+\cos \frac{3 \pi}{4}}{2}} \\
&=\sqrt{\frac{1-\frac{\sqrt{2}}{2}}{2}} \\
&=\frac{\sqrt{2-\sqrt{2}}}{2}
\end{aligned}
$$
|
{
"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
$$1 - \left(\frac{1}{3}+\frac{1}{2}\cdot\frac{2}{3}-\frac{1}{30}\right) = \frac{11}{30}.$$
There is $100\%$ chance of winning when A roll from 20 - 30 but for the rest $2/3$ there is only $50\%$ chance of winning, in addition, there is $1/30$ of chance that player A could roll the same number as B
The answer for this question should be $0.35$, I am not sure where I did wrong.
I just figured maybe I should do this instead
$1 - (\frac{1}{3}+(\frac{1}{2}-\frac{1}{30})\cdot\frac{2}{3}) = \frac{16}{45}$ is this right ?
|
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:
$$\begin{bmatrix}
4&8&7&3c&3b \\
1&2&2&c&b \\
2&4&2&c-1&b
\end{bmatrix}
$$
After the following elementary operations:
$R_2 \to R_1; R_2 \to R_2-4R_1, R_3 \to R_3-2R1;R_3 \to R_3-2R_2; R_2 \to -R_2$
I managed to get to the reduced form:
$$\begin{bmatrix}
1&2&2&c&b \\
0&0&1&c&b \\
0&0&0&c-1&b
\end{bmatrix}
$$
At this point I can see which conditions need to hold for the matrix:
1) to not have any solution: $c=1, b \neq 0$
2) to have infinitely many solutions: $c=1, b=0$ or $c \neq 1, b=0$ or $c \neq 1, b \neq 0$
3) $\mathbf{However}$ I don't see any way how there'd be a unique solution to the system because we have more variables than equations and the column of $w$ variable cannot be a column of zeroes.
|
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:
$1)\,R1\rightleftharpoons R2$
$2)\,R2=R2-4R1$
$3)\,R3=R3-2R1$
$4)\,R1=R1+R2$
$5)\,R3=R3-2R2$
$6)\,R2=(-1)\times R2$
I got:
$\begin{bmatrix}
1&2&1&0&0
\\
0&0&1&c&b
\\
0&0&0&c-1&b
\end{bmatrix}$
$\Rightarrow\,
\begin{cases}
x+2y+z=0
\\
z+cw=b
\\
(c-1)w=b
\\
y\quad is\,free
\end{cases}$
Then I started solving them:
$1)\quad (c-1)w=b \\ \hspace{9mm}w=\frac{b}{c-1}$
$2)\quad z+cw=b \\
\hspace{9mm}z+c\times\frac{b}{c-1}=b \\
\hspace{9mm}z+\frac{cb}{c-1}=b \\
\hspace{9mm}z\times (c-1)+cb=b\times (c-1) \\
\hspace{9mm}z(c-1)+cb=cb-b \\
\hspace{9mm}z(c-1)=cb-b-cb \\
\hspace{9mm}z(c-1)=-b \\
\hspace{9mm}z=-\frac{b}{c-1} $
$3)\quad x+2y+z=0 \\
\hspace{9mm}x=-2y-z \\
\hspace{9mm}x=-2y-(-b) \\
\hspace{9mm}x=b-2y$
So the unique solution is:
$\begin{cases}
x=b-2y
\\
y=y
\\
z=-\frac{b}{c-1}
\\
w=\frac{b}{c-1}\hspace{18mm},\,c-1\neq 0
\end{cases}$
|
{
"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{equation*}
k = 10^{n}-1
\end{equation*}
and was left with:
\begin{equation*}
\left\lfloor{\frac{2}{3}\sqrt{k(k+2)}}\right\rfloor = \frac{2}{3}k
\end{equation*}
But I can't get past that. Can anyone help me?
|
Note that
$$-8(10^n-1)\ \le\ 0\ <\ 4\cdot10^n+5$$
is true for all $n\in\mathbb Z^+$. In other words
$$-8\cdot10^n+4\ \le\ -4\ <\ 4\cdot10^n+1$$
$$\iff\ 4(10^n-1)^2\ \le\ 4(10^{2n}-1)\ <\ (2\cdot10^n+1)^2$$
$$\iff\ 2(10^n-1)\ \le\ 2\sqrt{10^{2n}-1}\ <\ 2\cdot10^n+1$$
$$\iff\ \frac23(10^n-1)\ \le\ \frac23\sqrt{10^{2n}-1}\ <\ \frac{2\cdot10^n+1}3=\frac23(10^n-1)+1$$
$$\iff\ \left\lfloor\frac23\sqrt{10^{2n}-1}\right\rfloor\ =\ \frac23(10^n-1)$$
since $10^n-1$ is a multiple of 9 and so is divisible by 3. Hence your equation is true for all non-negative integers $n$.
|
{
"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 $\rightarrow$
$$I=\int\dfrac{2x+3}{x^{2}+2x+10}dx=\int \dfrac{2x+2+1}{x^{2}+2x+10}dx=\int \dfrac{2x+2}{x^{2}+2x+10}dx +\int \dfrac{1}{(x^{2}+2x+1)+9}dx=\ln (x^{2}+2x+10)+J$$
$$J=\int \dfrac{1}{(x+1)^{2}+3^{2}}dx=\int \dfrac{1}{9\left[ \left ( \dfrac{x+1}{3}\right)^{2}+1\right]}dx=\dfrac{1}{3}\int \dfrac{\dfrac{1}{3}}{\left( \dfrac{x+1}{3}\right)^{2}+1}dx=\dfrac{1}{3}\arctan \left( \dfrac{x+1}{3}\right)$$
$$I=\ln (x^{2}+2x+10)+\dfrac{1}{3}\arctan \left( \dfrac{x+1}{3}\right)+C$$
|
{
"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
$$
\begin{align*}
\int {x^3\over \sqrt{4x^2 -1}}\,dx &= \frac1{16}\int {\sec^3 u\over \tan u}\tan u \sec u \, du\\
&= \frac1{16}\int {\sec^4 u} \, du\\
&= \frac1{16}\left(\tan u + {\tan^3 u\over 3} \right) + C\\
&= \frac1{16}\left(\tan (\sec^{-1} 2x) + {\tan^3 (\sec^{-1} 2x)\over 3} \right) + C.
\end{align*}$$
Given answer : $$\dfrac{\left(2x^2+1\right)\sqrt{4x^2-1}}{24}+C$$
Why is my answer incorrect ?
|
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
$$\int{x^3\over\sqrt{4x^2-1}}dx={1\over32}\int{u\over\sqrt{u-1}}du={1\over32}\int\left(\sqrt{u-1}+{1\over\sqrt{u-1}}\right)du\\
={1\over48}(u-1)^{3/2}+{1\over16}(u-1)^{1/2}+C=\sqrt{u-1}\left(u+2\over48\right)+C\\
=\sqrt{4x^2-1}\left(4x^2+2\over48\right)+C=\sqrt{4x^2-1}\left(2x^2+1\over24\right)+C$$
|
{
"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 so
$f'(x) > 0$ for $x \in (-\infty, -1) \cup (1, \infty)$ and since $f'$ is strictly increasing on those intervals, there can be at most one root in each of them.
$f'(x) < 0$ for $x \in (-1,1)$ and since $f'$ is strictly decreasing on this interval it can have at most one root.
Now examine $f(-3) = -16$ and $f(-1) = 4$. By the Intermediate Value Theorem (IVT) $f(c) = 0$ for some $c \in (-3, 1)$ and so $f$ has a root on the interval $(-\infty, 1)$.
Again examine $f(-1) = 4$ and $f(1) = 0$. We cannot say anything about $f$ having a root on the interval $(-1, 1)$.
Likewise examine $f(1) = 0$ and $f(3) = 16$. Again, we cannot say anything about $f$ having a root on $(1, \infty)$.
However, $f(1) = 1 - 3 + 2 = 0$ is clearly a root. And by factorizing the polynomial we get $f(x) = (x+2)(x-1)^2$. Indeed, $1$ is a root with a multiplicity of two.
Hence, $f(x)$ has two real roots.
Additional Comments
Also, do we say two real roots (because of the multiplicity), or three real roots, or do we say two distinct real roots?
While I realize factoring the polynomial gives me the answer I believe the purpose of the question was to do the former analysis, which when the polynomial isn't easily factorized, can provide a lot of insight. That is why I did it all
|
$x^3 - 3x + 2$
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 $(x-1)^2(x+2) = 0$
$(x-1)^2 = 0 \lor x + 2 = 0$
$\sqrt{(x-1)^2} = ±\sqrt{0} \lor x = -2$
$x-1 = ±0 \lor x = -2$
$x = 1±0 \lor x = -2$
$x = 1+0 \lor x = 1-0 \lor x = -2$
$x = 1 \lor x = 1 \lor x = -2$
The fundamental theorem of algebra says every polynomial of degree n has n roots. Since this polynomial is cubic, it has three roots. $x^3 - 3x + 2$ does have three real roots. It's just that two of them happened to be equal.
|
{
"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:
$$L=-\lim_{x\to \infty} 2\sin\left(\dfrac{1}{2}\dfrac{1}{(\sqrt{1+x}-\sqrt{x})}\right)
\sin\left(\dfrac{1}{2}\dfrac{1}{(\sqrt{1+x}+\sqrt{x})}\right)$$
Not sure how to proceed using trigonometry.
Also, is there a method using L'Hospital for $\infty-\infty$ form (not for this question obviously!)?
|
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
$$\begin{align}
\left|\cos(\sqrt{x+1})-\cos(\sqrt{x})\right|&=2\left|\sin\left(\frac{\sqrt{x+1}-\sqrt{x}}{2}\right)\sin\left(\frac{\sqrt{x+1}+\sqrt{x}}{2}\right)\right|\\\\
&=2\left|\underbrace{\sin\left(\frac{1}{2\left(\sqrt{x+1}+\sqrt{x}\right)}\right)}_{\text{bounded by its argument in absolute value}}\,\,\,\underbrace{\sin\left(\frac{\sqrt{x+1}+\sqrt{x}}{2}\right)}_{\text{bounded by}\,1 \,\text{in absolute value}}\right|\\\\
&\le \frac{1}{\sqrt{x+1}+\sqrt{x}}\\\\
&\le \frac{1}{2\sqrt{x}}\to 0
\end{align}$$
as $x\to \infty$.
METHODOLOGY $2$: Calculus Approach
Alternatively, we can use the mean value theorem applied to $\cos(\sqrt{x})$. Then, there exists a number $\xi\in (x,x+1)$ such that
$$\cos(\sqrt{x+1})-\cos(\sqrt{x})=-\frac{\sin(\sqrt{\xi})}{2\sqrt{\xi}}\to 0$$
as $x\to \infty$.
|
{
"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 $0$, so:
\begin{bmatrix}{}
2& & & & & & \\
& 2& & & & &\\
& & 7 & & & &\\
& & & 7 & & &\\
& & & & 7& & \\
& & & & & 7 &\\
& & & & & & 0\\ \end{bmatrix}
I don't know how to follow, what information gives me the minimal polynomial?
|
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
$$ \begin{pmatrix} 2 & 0 \\ 0 & 2 \end{pmatrix} $$
is $(x - 2)$ while the minimal polynomial of
$$ \begin{pmatrix} 2 & 1 \\ 0 & 2 \end{pmatrix} $$
is $(x - 2)^2$.
Similarly, since the minimal polynomial has $(x-7)$ as a factor, al the Jordan blocks associated to the eigenvalue $7$ must be $1 \times 1$. Hence, $A$ is similar to the matrix
$$ \begin{pmatrix} 2 & 1 & 0 & 0 & 0 & 0 & 0 \\
0 & 2 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 7 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & 7 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 7 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 7 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}. $$
|
{
"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 insight the solution follows quite simply.
How are you supposed to find what $a$ and $b$ are. It makes sense to guess that $c = -1$ but finding $a$ and $b$ seems much harder, how do you guess it?
|
$$3^{3^{n}}(3^{3^{n}}+1) + 3^{3^{n} + 1} - 1=\\
3^{2\cdot3^n}+3^{3^n}-1+3^{3^n+1}=\\
9^{3^n}+3^{3^n}-1+3^{3^n+1}\\
(9^{3^{n-1}})^3+(3^{3^{n-1}})^3+(-1)^3-3(9^{3^{n-1}}\cdot 3^{3^{n-1}}\cdot(-1))$$
|
{
"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}
*
*I tried doing integration by parts, got nowhere as nothing seems to reduce. I tried converting $\cos(x)^2$ into $1-\sin(x)^2$ and u-sub, but ended up with $\sqrt{2-u^2}$. Again stuck.
*Know the hyperbolic equivalents $\sinh x = \frac{e^x-e^{-x}}{2}$, and evaluate the integral with $x=z^2$.
Help with number 1? Also maybe links to how to deal with these types of integrals?
|
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
$$
\begin{align}
\int \sqrt{2-u^2}du&=2\int \cos^2 t \:dt
\\&=2\int \left(\frac12+\frac{\cos (2t)}2 \right)dt
\\&=\int \left(1+\cos (2t) \right)dt
\\&=t+\frac{\sin (2t)}2
\\&=t+\sin t \cdot \cos t
\\&=\arcsin\left(\frac{u}{\sqrt{2}}\right)+\frac{1}{2}u \sqrt{2-u^2}.
\end{align}
$$ Thus
$$
\begin{align}
\int_{\arcsin y}^{\pi/2}\cos x \cdot \sqrt{1+\cos^2 x}\:dx&=\int_{\arcsin y}^{\pi/2}\cos x \cdot \sqrt{2-\sin^2 x}\:dx
\\\\&=\int_{y}^{1}\sqrt{2-u^2 }\:du
\\\\&=\frac12+\frac{\pi}4-\frac12\cdot y \sqrt{2-y^2}-\arcsin\left(\frac{y}{\sqrt{2}}\right)
\end{align}
$$ Can you finish it?
|
{
"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 $$6\sqrt { 3 } \sin { \left( \frac { 2\pi }{ 3 } -\arcsin { \left( \frac { 5\sqrt { 3 } }{ 9 } \right) } \right) } =6\sqrt { 3 } \left[ \sin { \frac { 2\pi }{ 3 } \cos { \left( \arcsin { \left( \frac { 5\sqrt { 3 } }{ 9 } \right) } \right) -\cos { \frac { 2\pi }{ 3 } \sin { \left( \arcsin { \left( \frac { 5\sqrt { 3 } }{ 9 } \right) } \right) } } } } \right] =\\ =6\sqrt { 3 } \left[ \frac { \sqrt { 3 } }{ 2 } \sqrt { 1-{ \left( \frac { 5\sqrt { 3 } }{ 9 } \right) }^{ 2 } } +\frac { 1 }{ 2 } \left( \frac { 5\sqrt { 3 } }{ 9 } \right) \right] =6\sqrt { 3 } \left[ \frac { \sqrt { 2 } }{ 6 } +\frac { 5\sqrt { 3 } }{ 18 } \right] =\color{blue}{\sqrt { 6 } +5}$$
|
{
"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 solutions of the original equation while simplifying.
Hence $25$:
\begin{align*}
25-10 &= \sqrt{3 \times 3 \times 5 \times 5}\\
15 &= 15
\end{align*}
Then $4$:
\begin{align*}
4-10 &= \sqrt{36}\\
-6 &= \sqrt{36}
\end{align*}
Since $-6$ is a square root of $36$,
$-6 = -6$. Solutions: $25$, $4$.
However, when I checked this through graphing, it appears that only $25$ is a solution and $4$ is considered extraneous. Why is this so? Square roots accounts for both a positive and a negative value and so the statement $-6 = \sqrt{36}$ should be true.
Looking at the graph of the equation, you could take the positive or the negative of the square root, ending up with two different graphs.
Why did we accept $25$ as a solution but reject $4$?
Any help/explanation would be much appreciated.
|
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 $x=4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2225752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
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:
$$f'(x) = \dfrac{1}{2} - \dfrac{1}{x^2}$$
Here we can note that if $x \ge \sqrt{2}$, then $f'(x) > 0$, which means that $f(x)$ is increasing in $[\sqrt{2}; +\infty)$. So, $f(x) \ge f(\sqrt{2}) = \dfrac{\sqrt{2}}{2} + \dfrac{1}{\sqrt{2}} = \sqrt{2}$ for $x \ge \sqrt{2}$.
So, we have just proven that $u_{m+1} = f(u_m)\ge \sqrt{2}$ (inductive step: if we assume that $u_m \ge \sqrt{2}$ then $u_{m+1} \ge \sqrt{2}$ as well).
Now note that if $u_m \ge \sqrt{2}$ then
$$\dfrac{u_m}{2} \ge \dfrac{1}{u_m} \Rightarrow u_m \ge \dfrac{u_m}{2} +\dfrac{1}{u_m} = u_{m+1}$$
This implies that $u_{m}$ is decreasing monotonically.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2226280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
}
|
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:
$$ (2+\sqrt{3})^n=\binom{n}{0}2+\binom{n}{1}2^2\sqrt{3}+\binom{n}{2}2^3\cdot 3+\ldots+\binom{n}{n-1}2\sqrt{3}^{n-1}+\binom{n}{n}\sqrt{3}^{n}$$
and I think we have that: $$ a_n=\binom{n}{0}2+\binom{n}{2}2^3\cdot 3+\ldots+\binom{n}{n}\sqrt{3}^{n}$$
$$ b_n=\binom{n}{1}2^2\sqrt{3}+\ldots+\binom{n}{n-1}2\sqrt{3}^{n-1}$$
But, frankly, I do not know what to do next.
|
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 $$\begin{pmatrix} a_n \\ b_n \end{pmatrix} = \begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}^n \begin{pmatrix} 1 \\ 0 \end{pmatrix}.$$ In the limit, the largest eigenvalue of $\begin{pmatrix} 2 & 3 \\ 1 & 2 \end{pmatrix}$ dominates, so $\begin{pmatrix} a_n \\ b_n \end{pmatrix}$ will be close to a multiple of the eigenvector $\begin{pmatrix} \sqrt{3} \\ 1 \end{pmatrix}$; i.e. $\frac{a_n}{b_n} \rightarrow \sqrt{3}.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2226394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
}
|
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 have exactly $1$ ball in the first bag?
My reasoning was as follows:
Since there are no empty bags we can assume that we already have $3$ balls in place and we only have $2$ left. So I will denote 1 1 0 as 'there is a ball in the first bag and in the second bag, but not in the third'.
The favorable outcomes are:
0 1 1
0 2 0
0 0 2
The unfavorable outcomes are:
1 0 1
2 0 0
1 1 0
The probability of having exactly $1$ ball in the first bag is favorable outcomes/ (favorable outcomes + unfavorable outcomes) which is $3/6$ or $1/2$. But my friend keeps telling me that the answer is slightly different. Please let me know if there is a flaw in my reasoning. Thanks!
|
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 are no empty bags that exactly one ball is in the first bag?
Assume the bags are distinct. Let $x_k$ be the number of balls placed in the $k$th bag. Then the number of possible outcomes is the number of solutions in the positive integers of the equation
$$x_1 + x_2 + x_3 = 5$$
A particular solution corresponds to the placement of two addition signs in the four spaces between successive ones in a row of five ones. For instance,
$$1 1 + 1 + 1 1$$
corresponds to $x_1 = 2$, $x_2 = 1$, and $x_3 = 2$, while
$$1 1 1 + 1 + 1$$
corresponds to $x_1 = 3$, $x_2 = 1$, and $x_3 = 1$. Thus, the number of possible outcomes is the number of ways we can select two of the four spaces between successive ones to be filled with addition signs, which is
$$\binom{4}{2} = 6$$
as you found. Since an outcome is favorable if exactly one ball is placed in the first bag, the number of favorable outcomes is the number of solutions of he equation
$$1 + x_2 + x_3 = 5$$
in the positive integers, which is equivalent to the number of solutions of the equation
$$x_2 + x_3 = 4$$
in the positive integers. The number of such solutions is the number of ways we can insert one addition sign in the three spaces between successive ones in a row of four ones, which is
$$\binom{3}{1} = 3$$
as you found. Hence, if the five balls are identical, the probability that exactly one ball is placed in the first bag is
$$\frac{\binom{3}{1}}{\binom{4}{2}} = \frac{3}{6} = \frac{1}{2}$$
What is the probability that if five different balls are distributed to three bags so that are no empty bags that exactly one ball is in the first bag?
Assume the bags are distinct. There are $3^5$ ways to distribute five different balls to three bags since there are three choices for each of the five balls. However, this includes distributions in which some of the bags are empty. The number of ways we can distribute five balls to only two of the bags is
$$\binom{3}{1}2^5$$
since there are three ways to choose which bag will be empty and two remaining choices for each of the five balls. However, subtracting this from $3^5$ removes those distributions in which exactly one bag is used twice, once for each of the ways we could have selected one of the other two bags to be empty.
We only want to subtract these cases once, so we need to add them back. There are $\binom{3}{2}$ ways to select which two bags will be empty and one way to place all the balls in the remaining bag. Thus, by the Inclusion-Exclusion Princple, the number of possible outcomes is
$$3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5 = 243 - 3 \cdot 32 + 3 \cdot 1 = 150$$
It remains to count favorable outcomes. There are five ways to select which ball is placed in the first bag. If there were no further restrictions, there would be $2^4$ ways to distribute the remaining balls to the other two bags. However, two of these outcomes place all of the remaining balls in one bag, which would leave an empty bag. Hence, the number of favorable outcomes is
$$\binom{5}{1}\left[2^4 - \binom{2}{1}1^4\right] = 5(16 - 2) = 5 \cdot 14 = 70$$
Thus, if the balls are distinct, the probability is
$$\frac{\binom{5}{1}\left[2^4 - \binom{2}{1}1^4\right]}{3^5 - \binom{3}{1}2^5 + \binom{3}{2}1^5} = \frac{70}{150} = \frac{7}{15}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2231429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
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 since roots are in $(0 \: 1)$ product of roots also lies in $(0 \: 1)$
we get
$$\frac{c}{a} \lt 1$$ or $$a \gt c$$
Also Discriminant $$b^2 \gt 4ac \gt 4c^2$$
now how can we find minimum values of $a$ and $b$
|
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 is open upwards, $f(1)$ must be greater than zero:
$$ f(1) = a - b + c > 0 \implies b < a + c. \quad (1)$$
*
*The product of two roots:
$$x_1\cdot x_2 < 1 \implies c/a < 1 \implies a > c. \quad (2)$$
*
*Finally, to have two real roots:
$$ b^2 - 4ac \ge 0 \implies b^2 \ge 4ac. \quad (3)$$
We will tabulate all possible values that satisfy the three conditions above.
A. Case $c = 1:$
*
*According to $(2)$: $\,a = 2, 3, 4, 5, 6, 7, 8, \ldots$
*
*For $a=2:$
*
*$(1)$ says $b < 3 \implies b = 1, 2.$
*$(3)$ says $b^2 \ge 8 \implies b = 3, 4, 5, \ldots$
*Thus, no values of $b$ satisfy both conditions.
*For $a=3:$
*
*$(1)$ says $b < 4 \implies b = 1, 2, 3.$
*$(3)$ says $b^2 \ge 12 \implies b = 4, 5, 6, \ldots$
*Thus, no values of $b$ satisfy both conditions.
*For $a=4:$
*
*$(1)$ says $b < 5 \implies b = 1, 2, 3, 4.$
*$(3)$ says $b^2 \ge 16 \implies b = 4, 5, 6, \ldots$
*Thus, only $b = 4$ satisfies both conditions.
*The first solution is $a=4,\; b =4\; c=1$
*For $a=5:$
*
*$(1)$ says $b < 6 \implies b = 1, 2, 3, 4, 5.$
*$(3)$ says $b^2 \ge 20 \implies b = 5, 6, \ldots$
*Thus, only $b = 5$ satisfies both conditions.
*The second solution is $a=5,\; b =5\; c=1$
*For $a=6:$
*
*$(1)$ says $b < 7 \implies b = 1, 2, 3, 4, 5, 6.$
*$(3)$ says $b^2 \ge 24 \implies b = 5, 6, \ldots$
*Thus, only $b = 6$ satisfies both conditions.
*The second solution is $a=6,\; b =6\; c=1$
In general, we get a set of solutions $a = b = 4, 5, 6, \ldots$ for $c=1$. The resulting parabola is given by:
$$ y = f(x) = ax^2 -ax + 1 = a (x-1/2)^2 + 1 - a/4, \quad a = 4, 5, 6,\ldots $$
B. Case $c = 2:$
*
*According to (2), $a = 3, 4, 5, 6, 7, 8, \ldots$
*
*For $a=3:$
*
*$(1)$ says $b < 5 \implies b = 1, 2, 3, 4.$
*$(3)$ says $b^2 \ge 24 \implies b = 5, 6, 7, \ldots$
*Thus, no value of $b$ satisfies both conditions.
*For $a=4:$
*
*$(1)$ says $b < 6 \implies b = 1, 2, 3, 4, 5.$
*$(3)$ says $b^2 \ge 32 \implies b = 6, 7, 8, \ldots$
*Thus, no value of $b$ satisfies both conditions.
We can stop here because the next values of $a$ and $b$ to check will exceed the minimum values found in case A.
Thus, the solution with minimum values of $a$ and $b$ is:
$$ y = 4x^2 - 4x +1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2233259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
$1+(1+2+4)+(4+6+9)+(9+12+16)+.......+(361+380+400)=?$ 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 from the third to the second term the difference is 12 and so on...
So we notice that the increase from one term to the other is in form of multiples of 6.ie 6+,12+.18+ and so on....
After that I could not proceed.
|
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 ,
$$a_n=1+6(1+2+3 \cdots (n-1))$$
$$a_n=1+6\cdot \frac{n(n-1)}2=3n^2-3n+1$$
$$S=\sum a_n= \sum \big(3n^2-3n+1\big) =3\sum n^2 - 3\sum n + \sum 1$$
Use $$\sum n^2 =\frac{n(n+1)(2n+1)}{6} $$
$$\sum n =\frac{n(n+1)}{2}$$
$$\sum 1 =n$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2233718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
}
|
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 $$\sum_{n=0}^{k}g(n)=\frac{1}{2}(\sum_{n=0}^{k+1}n+\sum_{n=0}^{k+1}{n^2})=\dfrac{(k+1)(k+2)(k+3)}{6}$$
edit 1: $f,g$ are defined on the integers.
|
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",
"answer_count": 2,
"answer_id": 1
}
|
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 $\frac{\sqrt{3} -1}{2}$
What am I missing ?
|
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
$$(2x)^2(1-x^2)=1-4x^2\\4x^2-4x^4=1-4x^2\\4x^4-8x^2+1=0\\x^2=t\\4t^2-8t+1=0\\t_{1,2}=\frac{8\pm\sqrt{64-16}}{8}=\frac{8\pm4\sqrt{3}}{8}=\frac{2\pm\sqrt{3}}{2}$$
Now we can eliminate the $\frac{2+\sqrt3}{2}$ since then $x>1$ so the square root
of $1-x^2$ is not defined.
$$t=\frac{2-\sqrt3}{2}=\frac{4-2\sqrt3}{4}=\frac{(\sqrt 3-1)^2}{4}\\x^2=\frac{(\sqrt 3-1)^2}{4}\\x=\pm\frac{\sqrt 3-1}{2}$$
We can rule out the negative solution since then $2x\sqrt{1-x^2}$ is negative while $=\sqrt{1-4x^2}$ is positive.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2234225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
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 operate with.
|
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
$$
\frac{1}{n} \sum \limits_{k = 1}^n\sqrt[3]{\left(x + \frac{k}{n}\right)^3}\le a_n \le \frac{1}{n} \sum \limits_{k = 1}^n\sqrt[3]{\left(x + \frac{k+1}{n}\right)^3}
$$ or
$$
\frac{1}{n} \sum \limits_{k = 1}^n\left(x + \frac{k}{n}\right)\le a_n \le \frac{1}{n} \sum \limits_{k = 1}^n\left(x + \frac{k+1}{n}\right)
$$ that is $$
\color{blue}{\frac{1}{n} \sum \limits_{k = 1}^n\left(x + \frac{k}{n}\right)}\le a_n \le \color{blue}{\frac{1}{n} \sum \limits_{k = 1}^n\left(x + \frac{k}{n}\right)}+\frac{1}{n} \sum \limits_{k = 1}^n\frac{1}{n}
$$ Can you finish it?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2236470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
}
|
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:
$p(A)=0 = - A^3 + \text{tr} A\cdot A^2 -\left(\begin{vmatrix}a_{11} && a_{12} \\ a_{21} && a_{22} \end{vmatrix} + \begin{vmatrix}a_{22} && a_{23} \\ a_{32} && a_{33} \end{vmatrix} + \begin{vmatrix}a_{11} && a_{13} \\ a_{31} && a_{33} \end{vmatrix}\right) A + \det A$.
Since $-A^3 + \det A \cdot I = 0$, then $A = \frac{\left(\begin{vmatrix}a_{11} && a_{12} \\ a_{21} && a_{22} \end{vmatrix} + \begin{vmatrix}a_{22} && a_{23} \\ a_{32} && a_{33} \end{vmatrix} + \begin{vmatrix}a_{11} && a_{13} \\ a_{31} && a_{33} \end{vmatrix}\right)}{\text{tr} A}I$. With this in mind I cannot obtain even rotation matrix to say nothing about something else. Any hints?
|
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 diagonalizable and hence
$$A=PIP^{-1}=I$$.
Case 2: $A$ has two complex eigenvalues. Then, $1, e^{2\pi i/3}, e^{4\pi i/3}$ are all eigenvalues, with eigenvectors $u,v, \bar{v}$.
You have $A=PDP^{-1}$ where
$$P=(u v \bar{V})$$ and $D=diag (1, e^{2\pi i/3}, e^{4\pi i/3})$
Now, write $$v=v_1+i v_2$$ Show that
$$A= (u \;v_2 \;v_2) \begin{bmatrix}1 &0 & 0\\
0& \cos(2\pi /3 )& \sin(2\pi /3)\\
0& -\sin(2\pi /3 )& \cos(2\pi /3)
\end{bmatrix}(u \;v_2 \;v_2) ^{-1}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2238320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
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 $z = u - w$.
*Find the orthogonal projection of $u$ on $W$.
*Verify that $z$ is in $W^\perp$.
This is what I have so far:
*
*$w = (u \cdot v_1)v_1 + (u \cdot v_2)v_2 = (3/3)v_1 + (3/3)v_2$, and after plugging in the vectors, I got $w = \begin{pmatrix} 3 \\ -3 \\ 0\end{pmatrix}$, thus $z = \begin{pmatrix} 0 \\ 4 \\ 1\end{pmatrix}$.
*projection of $u$ on $W$ is $\begin{pmatrix} 0 \\ 4 \\ 1\end{pmatrix}$
*When I tried verifying that $z$ is in $W^\perp$, I was not getting $0$, so I'm not sure where I went wrong.
|
$w=(u\cdot v_1)v_1+(u\cdot v_2)v_2=(3/3)v_1+(3/3)v_2$
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",
"url": "https://math.stackexchange.com/questions/2239564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
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 prove this? I have proved that if $a+b$ is a multiple of $n$, then so is $a^n+b^n$, using FLT, but how to progress next eludes me.
|
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
$$p\mid a+b\quad\hbox{and}\quad p\mid a^{n-1}-a^{n-2}b+a^{n-3}b^2-\cdots+b^{n-1}\ .$$
This gives $b\equiv-a\pmod p$ and so the second expression above can be written
$$p\mid na^{n-1}\ .$$
Now $p\mid a+b$ and $n\not\mid a+b$, so $p\ne n$, so $p\mid a$ and hence $p\mid b$. Since $a,b$ are coprime this is a contradiction.
Hence $\dfrac{a^n+b^n}{a+b}$ and $a+b$ are coprime.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2241281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
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, no scalar multiple of it has real coefficients.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2243482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
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 zero except last terms
but answer is not $1$ , could some help me to solve it , thanks
|
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",
"answer_count": 2,
"answer_id": 1
}
|
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 $u = x - \displaystyle \frac{\pi}{3}$)
$\displaystyle \lim_{u\to 0} \frac{1 - 2\cos(u + \frac{\pi}{3})}{\sin(u)} = \lim_{u\to 0} \frac{1 - \cos(u) + \sqrt{3}\sin(u)}{u} \approx \lim_{u\to 0} \frac{1 - 1 + \sqrt{3}u}{u} = \sqrt{3}$
But it bothers me using that sort of linear approximation, I want to get the result in a more formal way. I have tried using double angle properties
$$\cos(2x) = \cos^2(x) - \sin^2(x)$$
$$\sin(2x) = 2\sin(x)\cos(x)$$
But I reach to a point of nowhere, I cannot come up with a way of simplifying expressions to get the results:
$\displaystyle\lim_{x\to\pi/3} 2\frac{3\sin^2(\frac{x}{2}) - \cos^2(\frac{x}{2})}{\sqrt{3}\sin(x) - \cos(x)}$
Is there a way of computing this limit without approximations and without L'Hopital?
|
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
$$
\lim_{u\to0} \frac{1-\cos(u)}{\sin(u)} + \lim_{u\to0}\frac{\sqrt{3}\sin(u)}{\sin(u)}=\lim_{u\to0} \frac{1-\cos(u)}{\sin(u)} + \sqrt{3}
$$
Multiplying the remaining limit by $\frac{1+\cos(u)}{1+\cos(u)}$ gives
$$\begin{align}
\lim_{u\to0} \frac{1-\cos^2(u)}{\sin(u)(1+\cos(u))} + \sqrt{3}&=\lim_{u\to0} \frac{\sin^2(u)}{\sin(u)(1+\cos(u))} + \sqrt{3}\\
&=\lim_{u\to0} \frac{\sin(u)}{1+\cos(u)} + \sqrt{3}\\
&=\sqrt{3}
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2250967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
}
|
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",
"answer_count": 5,
"answer_id": 3
}
|
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 $\sqrt{i}\times\sqrt{-i}=\sqrt{i\times (-i)}=\sqrt{-i\times i}=\sqrt{-1\times i^2}=\sqrt{-1\times -1}=1$
But again, $\sqrt{i}\times\sqrt{-i}=\sqrt{i}\times \sqrt{i}\sqrt{-1}=\sqrt{i}\times \sqrt{i}\times i=i\times i=i^2=-1$.
Which one is correct and what is the logic behind it?
and Finally what are the values of $\sqrt{i}+\sqrt{-i}$
|
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 $(a+b)^2=2$ and $(a+b)=\pm\sqrt{2}$
*If $ab=-1$ then $(a+b)^2=-2$ and $(a+b)=\pm i\sqrt{2}$
We found $4$ possible values, and this is normal, remember that in $\mathbb C$ each complex has $2$ possible roots so $\sqrt{z_1}+\sqrt{z_2}$ would have $4$ possible values (if $z_1\neq z_2$ of course).
Or simply notice that $z=a+b$ is solution of $z^4=(a+b)^4=(2ab)^2=4a^2b^2=4$ and $z^4-4=0$ has $4$ different roots.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2251769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
}
|
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 all powers of $x$. All I know is $A_9 = 1, A_0 = 1\cdot 2 \cdot 3 \cdot 4 \cdot 5$.
|
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",
"url": "https://math.stackexchange.com/questions/2252038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
$\cos^2(\frac{\pi}{101})+\cos^2(\frac{2\pi}{101})+\cos^2(\frac{3\pi}{101})+...+\cos^2(\frac{100\pi}{101})=?$
Find the value: $$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots+\cos^2\left(\frac{99\pi}{101}\right)+\cos^2\left(\frac{100\pi}{101}\right)$$
My attempt:I've tried it by considering the sum
$$\sin^2\left(\frac{\pi}{101}\right)+\sin^2\left(\frac{2\pi}{101}\right)+\sin^2\left(\frac{3\pi}{101}\right)+\sin^2\left(\frac{4\pi}{101}\right)+\sin^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots+\sin^2\left(\frac{99\pi}{101}\right)+\sin^2\left(\frac{100\pi}{101}\right)$$
along with
$$\cos^2\left(\frac{\pi}{101}\right)+\cos^2\left(\frac{2\pi}{101}\right)+\cos^2\left(\frac{3\pi}{101}\right)+\cos^2\left(\frac{4\pi}{101}\right)+\cos^2\left(\frac{5\pi}{101}\right)+\cdots \\ \cdots +\cos^2\left(\frac{99\pi}{101}\right)+\cos^2\left(\frac{100\pi}{101}\right)$$ which gives $ 100$ as resultant but failed to separate the sum of
$$\sin^2\left(\frac{\pi}{101}\right)+\sin^2\left(\frac{2\pi}{101}\right)+\sin^2\left(\frac{3\pi}{101}\right)+\sin^2\left(\frac{4\pi}{101}\right)+\sin^2\left(\frac{5\pi}{101}\right)+\cdots\\ \dots+\sin^2\left(\frac{99\pi}{101}\right)+\sin^2\left(\frac{100\pi}{101}\right)$$ at last.
I tried the next approach by using de Movire's theorem but failed to separate the real and imaginary part.
I've invested a great amount of time in the so it would be better if someone please come up with an answer.
|
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 $\frac{\sin(2\pi/101)}{\sin(2\pi/101)}$ and creative telescoping.
The final answer is $\frac12 (100-1)=\frac{99}{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2254887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
}
|
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 not understand what he mean by that . Can anybody please explain me .
|
Let's name the sequence of Taylor polynomials for $e^x,$ so
$$ f_0(x) = 1, $$
$$ f_1(x) = 1 + x, $$
$$ f_2(x) = 1 + x + \frac{x^2}{2}, $$
$$ f_3(x) = 1 + x + \frac{x^2}{2}+ \frac{x^3}{6}, $$
$$ f_4(x) = 1 + x + \frac{x^2}{2}+ \frac{x^3}{6}+ \frac{x^4}{24}, $$
$$ f_5(x) = 1 + x + \frac{x^2}{2}+ \frac{x^3}{6}+ \frac{x^4}{24}+ \frac{x^5}{120}, $$
$$ f_n(x) = 1 + x + \frac{x^2}{2}+ \frac{x^3}{6}+ \frac{x^4}{24}+ \frac{x^5}{120} + \cdots + \frac{x^n}{n!}. $$
The simple result is that, for $n$ even, we always have $f_n$ positive, no real roots. For $n$ odd, exactly one real root. This is induction, with the important relationships
$$ f_n'(x) = f_{n-1}(x), $$
$$ f_n(x) = f_n'(x) + \frac{x^n}{n!}. $$
Induction, beginning with even: $f_{2n}(x)$ has a single minimum at the point $t$ where $f_{2n-1}(x)$ has its root. This root is never zero. However, there
$$ f_{2n}(t) = f_{2n-1}(t) + \frac{t^{2n}}{(2n)!} = \frac{t^{2n}}{(2n)!} > 0$$
We also need to point out that this means that $f_{2n+1}$ is always increasing, and, being a polynomial of odd degree, has exactly one real root. The cases $2n, 2n+1$ between them complete the induction step.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2257878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
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",
"answer_count": 4,
"answer_id": 2
}
|
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
$X_1 = 1 - \frac{1}{2} + ... + \frac{1}{k-1} - \frac{1}{k}$
$X_2 = \frac{1}{k+1} - \frac{1}{k+2} + ... + \frac{1}{2k-1} - \frac{1}{2k}$
$k$ is even.
Every negative term $d$ in $X_2$ is of the form $\frac{1}{2^my}$ where $y$ is odd and could be $1$. Also $y \le k$
For every such $d$, we will have the following terms
$\frac{1}{y} - \frac{1}{2y} - ... - \frac{1}{2^{m-1}y}$ in $X_1$.
Adding $d$ to the above, we get
$\frac{1}{y} - \frac{1}{2y} - ... - \frac{1}{2^{m-1}y} - \frac{1}{2^my} = \frac{1}{2^my}$
So every negative term $d$ in $X_2$ is now replaced with a term of the same magnitude but with a positive sign.
Hence $X = \frac{1}{k+1} + \frac{1}{k+2} + ... + \frac{1}{2k-1} + \frac{1}{2k}$
|
You're right, there's a much simpler way:
$$S=\sum^{2k}_{i=1}\left[\frac{(-1)^{(i+1)}}{i}\right]=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6} \cdot\cdot\cdot +\frac{1}{2k-1}-\frac{1}{2k}$$
$$S=\color{blue}{\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6} \cdot\cdot\cdot +\frac{1}{2k-1}+\frac{1}{2k}\right]}-2\color{red}{\left[\frac{1}{2}+\frac{1}{4}+\cdot\cdot\cdot\frac{1}{2k}\right]}$$
$$S=\sum^{2k}_{i=1}\frac{1}{i}-2\sum^k_{i=1}\frac{1}{2i}=\sum^{2k}_{i=1}\frac{1}{i}-\sum^k_{i=1}\frac{2}{2i}=\color{purple}{\sum^{2k}_{i=1}\frac{1}{i}-\sum^k_{i=1}\frac{1}{i}}=\sum^{2k}_{i=k+1}\frac{1}{i}$$
Hence proved.
Note: $$\sum^{2k}_{i=1}\frac{1}{i}=\sum^{k}_{i=1}\frac{1}{i}+\sum^{2k}_{i=k+1}\frac{1}{i}$$
|
{
"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
}
|
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 for y
\begin{align}
y & = 2\sqrt{{x + 4}}
-2
\end{align}
Find the X and Y intercepts by setting each variable to 0. I believe I have X right but not Y, please explain where I am messing this up, should be a basic algebra folly.
X intercept: (I believe I have this right, though I should have just subtracted out the -2 early on)
\begin{align}
y & = 2\sqrt{{x + 4}}
-2
\end{align}
\begin{align}
0 & = 2\sqrt{{x + 4}}
-2
\end{align}
\begin{align}
0^2 & = 2^2{{(x + 4)}}
-2^2
\end{align}
\begin{align}
0 & = 4{{x + 16}}
-4
\end{align}
\begin{align}
0 & = 4{{x + 12}}
\end{align}
\begin{align}
-12 & = 4{{x}}
\end{align}
\begin{align}
-3 & = {{x}}
\end{align}
Thus the X intercept is (-3, 0)
Y intercept: I know this is wrong, textbook lists the answer as 2
\begin{align}
y & = 2\sqrt{{x + 4}}
-2
\end{align}
\begin{align}
y & = 2\sqrt{{0 + 4}}
-2
\end{align}
\begin{align}
y & = 2\sqrt{{4}}
-2
\end{align}
(Square both sides to remove the radix?)
\begin{align}
y^2 & = 4*4
-4
\end{align}
\begin{align}
y^2 & = 12
\end{align}
\begin{align}
y & = \sqrt12
\end{align}
Actual answer is 0, 2 for y intercept. So what's my mistake?
|
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.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2259767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Integrate $\int \frac {1}{(x+2)(x+3)} \textrm {dx}$ Integrate $\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$
My Attempt:
$$\int \dfrac {1}{(x+2)(x+3)} \textrm {dx}$$
$$\int \dfrac {1}{x+2} \textrm {dx} . \int \dfrac {1}{x+3} \textrm {dx}$$
$$\dfrac {\textrm {log (x+2)}}{1} . \dfrac {\textrm {log (x+3)}}{1} + C$$
$$\textrm {log} (x+2) . \textrm {log} (x+3) + C$$
Is this correct? Or, How do I proceed the other way?
|
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 = $\require{cancel} \cancel{A(-3+3)}$ + B(-3+2)
$\implies$
1 = -1B
$\implies$
-1 = B
$$\\$$
When does x+2 = 0?
x= -2
Plug -2 into x (and -1 into B because we solved for B above):
1 = A(-2+3) + $\require{cancel} \cancel{-1(-2+2)}$
$\implies$
1 = 1A
$\implies$
1 = A
$$\\$$
Now plug in A= 1 and B= -1 into: $\frac{A}{x+2}+\frac{B}{x+3}$.
Thus your integral is $\int (\frac{1}{x+2}+\frac{-1}{x+3})\,dx$.
When you integrate you get:
$\boxed{\ln \left|{x+2}| - \ln|x+3\right|+C}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2259958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
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 \\
&= \frac{1}{c^2 + u\cdot v} \sqrt{ a\cdot a + b\cdot b + 2(a\cdot b) } \\
\end{align}
Expanding these dot products:
\begin{align}
a\cdot a
&=
||c^2(u+v)||_2^2 \\
&=
c^4[(u\cdot u) + (v\cdot v) + 2(u\cdot v)] \\
&= 2c^4[c^2 + (u\cdot v)]\\
b\cdot b
&=
||u\times(u \times v)||_2 \\
&=
||u||_2^2 ||u\times v||_2^2 - \underbrace{[u\cdot (u\times v)]^2}_0 \\
&=
c^2[||u||_2^2 ||v||_2^2 - (u\cdot v)^2] \\
&= c^2[c^4 - (u\cdot v)^2] \\
a\cdot b
&=
c^2[(u+v)\cdot (u\times(u\times v))] \\
&= c^2[ \underbrace{u \cdot (u\times(u\times v))}_0 + v \cdot (u\times(u\times v)) ] \\
&= c^2[(u\times v)\cdot (v\times u)] \\
&= -||u\times v||_2^2 \\
&= -[c^4 - (u\cdot v)^2]
\end{align}
Plugging back in:
\begin{align}
||u \oplus v||_2
&=
\frac{1}{c^2 + u\cdot v} \sqrt{
2c^4[c^2 + (u\cdot v)]
+
c^2[c^4 - (u\cdot v)^2]
-2[c^4 - (u\cdot v)^2]c^2
} \\
&=
\frac{c}{c^2 + u\cdot v}
\sqrt{ c^4 + 2(u\cdot v) c^2 + (u\cdot v)^2 } \\
&=
\frac{c}{c^2 + u\cdot v}
[c^2 + (u\cdot v)] \\
&= c
\end{align}
as expected.
As for the second part, when the norms are $c$, $u\oplus v = v\oplus u$, since $u$ and $v$ were arbitrary.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2260059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
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 that after some rotation of the conic section by some counterclockwise angle $\theta$, the equation $$Ax^2 + Bxy + Cy^2 + F = 0$$ becomes $$A'u^2 + C' v^2 + F' = 0$$ for suitable constants $A', C', F'$ that depend on $A, B, C, F$ through $\theta$. Such a rotation has the matrix form $$\begin{bmatrix}x \\ y\end{bmatrix} = \begin{bmatrix}\cos \theta & \sin \theta \\ -\sin \theta & \cos \theta \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix}.$$ Written explicitly, we have $$\begin{align*} x &= u \cos \theta + v \sin \theta, \\ y &= u \sin \theta - v \cos \theta, \end{align*}$$ hence
$$\begin{align*}
-F &= A(u \cos \theta + v \sin \theta)^2 + B(u \cos \theta + v \sin \theta)(u \sin \theta - v \cos \theta) + C(u \sin \theta - v \cos \theta)^2 \\
&= (A \cos^2 \theta + B \cos \theta \sin \theta + C \sin^2 \theta) u^2 \\
&\quad + (2A \cos \theta \sin \theta + B(\sin^2 \theta - \cos^2\theta) - 2C \sin \theta \cos \theta) uv \\
&\quad + (A \sin^2 \theta - B \cos \theta \sin \theta + C \cos^2 \theta) v^2.
\end{align*}$$
We want to choose $\theta$ such that the coefficient of the $uv$ term will be zero; i.e., $$(A-C) \sin 2\theta - B \cos 2\theta = 0,$$ or simply, $$\cot 2\theta = \frac{A-C}{B}.$$ This is the origin of the desired relationship.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2261387",
"timestamp": "2023-03-29T00:00:00",
"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? Thank you!
|
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
$$
\sin t=t-\frac{t^3}{6}+o(t^3)
$$
we have the first fraction just as
$$
\frac{x-\sin x}{x^3}=\frac{\frac{x^3}{6}-o(x^3)}{x^3}=\frac16+o(x)\to\frac16.
$$
*Similarly with notation $y=\sin^{[k-1]}x$ the $k$-th fraction is
$$
\frac{\sin^{[k-1]}x-\sin^{[k]}x}{x^3}=\frac{y-\sin y}{x^3}=\frac{y-\sin y}{y^3}\cdot\frac{y^3}{x^3}=\left(\frac16+o(y)\right)\cdot\left(\frac{y}{x}\right)^3\to\frac16\cdot 1^3=\frac16
$$
because
$$
\frac{y}{x}=\frac{\sin^{[k-1]}x}{x}=\frac{\sin^{[k-1]}x}{\sin^{[k-2]}x}\cdot\frac{\sin^{[k-2]}x}{\sin^{[k-3]}x}\cdot\ldots\cdot\frac{\sin x}{x}\to1\cdot 1\cdot\ldots\cdot 1=1.
$$
*Finally the limit is
$$
\underbrace{\frac16+\frac16+\ldots+\frac16}_{n\text{ times}}=\frac{n}{6}.
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2262015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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:
$C=\{(x_1,x_2):\frac{3}{4x_1}\leq x_2\}$
I tried writing the power function, which for a general $\theta$ as follows:
$P(\frac{3}{4X_1}\leq X_2)=P(X_1X_2\geq\frac{3}{4} )$
Now since $X_1,X_2$ both are randoms we have to fix one of them, hence:
$P(X_1X_2\geq\frac{3}{4})=P(X_1\geq\frac{3}{4X_2})=\int_0^1 P(X_1\geq\frac{3}{4X_2}|X_2=x_2)f_{X_2}(x_2)dx_2$
Am I going in the right direction ?
Evaluating this, gives a $log$ term and hence the value of the integral is coming out to be $\infty$.
What am I doing wrong ?
|
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
$$\begin{align}\mathbb{P}_\theta\left(X_1X_2<\frac{3}{4}\right)&=\int_0^1{\mathbb{P}_\theta\left(X_1<\frac{3}{4x_2}\,\Bigg|\, X_2=x_2\right)\,f_\theta(x_2)\,dx_2}\\
&=\int_0^1\mathbb{P}_\theta\left(X_1<\frac{3}{4x_2}\right)\,f_\theta(x_2)\,dx_2\ \ \text{ (because }X_1,X_2\text{ are independent})\\
&=\int_0^1F_\theta\left(\frac{3}{4x_2}\right)\,f_\theta(x_2)\,dx_2
\end{align}$$
where $F_\theta$ is the CDF of $X$, found by integrating the density function $$f_\theta(x) = \theta x^{\theta-1}[0<x<1],$$
i.e.,
$$\begin{align}F_\theta(x)&=\mathbb{P}_\theta(X\le x)\\ \\
&= \int_{-\infty}^x f_\theta(x')\,dx'\\ \\
&=
\begin{cases}
0, & \text{if }x\le 0 \\
x^\theta, & \text{if } 0<x<1\\
1, & \text{if }x\ge 1
\end{cases}\\ \\
&=x^\theta[0<x<1]+1[x\ge 1].\end{align} $$
(For convenience, I'm using Iverson brackets for the indicator functions.)
Thus,
$$\begin{align}\mathbb{P}_\theta\left(X_1X_2<\frac{3}{4}\right)&=\int_0^1F_\theta\left(\frac{3}{4x}\right)\,f_\theta(x)\,dx\\
&=\int_0^1\left\{ \left(\frac{3}{4x}\right)^\theta\left[0<\frac{3}{4x}<1\right]\ \ +\ \ 1\left[\frac{3}{4x}\ge 1\right]\right\}\,\theta x^{\theta-1}\bigg[0<x<1\bigg]\, dx\\
&=\theta\left(\frac{3}{4}\right)^\theta\int_0^1 x^{-1}\left[0<\frac{3}{4x}<1\right]\bigg[0<x<1\bigg]\,dx\ \ \\
&\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad\quad +\ \ \theta \int_0^1x^{\theta-1}\left[\frac{3}{4x}\ge 1\right]\bigg[0<x<1\bigg]\,\, dx\\
&=\theta\left(\frac{3}{4}\right)^\theta\int_{\frac{3}{4}}^1 x^{-1}dx \ + \ \theta\int_0^{\frac{3}{4}}x^{\theta-1}dx\\
&=\theta\left(\frac{3}{4}\right)^\theta\log\frac{4}{3}\ + \ \left(\frac{3}{4}\right)^\theta\\
&=\left(\frac{3}{4}\right)^\theta(\theta\log\frac{4}{3}+1).
\end{align}$$
So the power function is
$$\pi(\theta) = 1 - \left(\frac{3}{4}\right)^\theta(\theta\log\frac{4}{3}+1).
$$
Here's a plot:
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2262117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
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:
$$\sin(\beta-\pi) = \sin(\beta)\cos(\pi)-\cos(\beta)\sin(\pi) = -\sin(\beta)$$
$$\\$$
$$-\sin(\beta) = \frac{1}{3} \Leftrightarrow \\
\sin(\beta) = - \frac{1}{3}$$
$$\\$$
$$\cos \beta = \sqrt{1-\sin^2(\beta)} = \sqrt{1-(-\frac{1}{3})^2} = \sqrt{1-\frac{1}{9}} = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3} = \frac{2\sqrt{2}}{3}$$
And so $h(\beta)$ is :
$$\frac{\frac{2\sqrt{2}}{3}}{\frac{2\sqrt{2}}{3}-1} = \\
\frac{\sqrt{2}}{2\sqrt{2}-3}$$
But my book says the solution is $6\sqrt{2}-8$. What went wrong?
|
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 we get: $$h(\beta) = \frac{-\frac{2\sqrt{2}}{3}}{-\frac{2\sqrt{2}}{3}-1} =\frac{2\sqrt{2}}{2\sqrt{2}+3} =6\sqrt{2}-8\quad\text{as required}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2266507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"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 can be applied.
So far I've converted the system to polar coordinates and got:
\begin{align*}
\dot{r}&=-\frac{1}{2}r(-5+2r^2-3\cos(2\theta))\\
\dot{\theta}&=-r(2+3\cos(\theta)\sin(\theta)).
\end{align*}
Where I'm lost now is constructing the trapping region where $\dot{r}<0$ on the outside and $\dot{r}>0$ on the inside. Graphing this system using streamplot the region is visually clear, but I'm having trouble finding a closed form solution.
|
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?
$$
%
\begin{align}
%
\dot{x} &= 4 x+2 y - x\left(x^2+y^2\right)\\
%
\dot{y} &= -2 x+y-y \left(x^2+y^2\right)
%
\tag{1}
\end{align}
%
$$
Solution method
Use the theorem of Poincare and Bendixson to identify a trapping region, here the gray annulus where the sign of the radial time derivative can change.
The invariant region must
*
*Be closed and bounded,
*Not contain any critical points.
Solution
Identify critical points
At what points
$
\left[
\begin{array}{c}
x \\
y \\
\end{array}
\right]
$ does
$
\left[
\begin{array}{c}
\dot{x} \\
\dot{y} \\
\end{array}
\right]
=
\left[
\begin{array}{c}
0 \\
0 \\
\end{array}
\right]
$?
The only critical point is the origin.
Switch to polar coordinates
The workhorse formula is
$$
%
\begin{align}
%
x &= r \cos \theta, \\
%
y &= r \sin \theta.
%
\end{align}
%
$$
With $r^{2} = x^{2} + y^{2}$, use implicit differentiation to find
$$
r\dot{r} = x \dot{x} + y \dot{y}
\tag{2}
$$
Compute $\dot{r}$
Substituting into $(2)$ using $(1)$, and noting $\cos^{2} \theta = \frac{1}{2} \left( 1 + \cos 2\theta \right)$,
$$
%
\begin{align}
%
r \dot{r} &= x \dot{x} + y \dot{y} \\
%
&= r^{2} - r^{4} \color{blue}{+} 3x^{2} \\
%
&= r^{2} + \frac{3}{2} r^{2} \left( 1 \color{blue}{+} \cos 2\theta \right) - r^{4}
%
\end{align}
%
$$
Therefore
$$
\dot{r} = -r^{3} + \frac{r}{2} \left( 5 \color{blue}{+} 3 \cos 2\theta \right)
\tag{3}
$$
Classify $\dot{r}$
Look for regions where the flow is outward $\dot{r}>0$, and regions where the flow is inward $\dot{r}<0$.
(Note the interesting comment by @Evgeny.)
Classify the problem by examining the limiting cases of $\cos 2\theta$ at $\pm 1$
Outward flow: $\cos 2 \theta \ge -1$
$$
%
\begin{align}
%
\dot{r}_{in} &= -r^{3} + \frac{r}{2} \left( 5 + 3 (-1) \right) \\
%
&= r(1-r^{2})
%
\end{align}
%
$$
When $r<1$, $\dot{r}>0$, and the flow is outward.
Inward flow: $\cos 2 \theta \le 1$
$$
%
\begin{align}
%
\dot{r}_{in} &= -r^{3} + \frac{r}{2} \left( 5 + 3 (-1) \right) \\
%
&= r(4-r^{2})
%
\end{align}
%
$$
When $r>2$, $\dot{r}<0$, and the flow is inward.
Trapping region
The region between the two zones is the annulus centered at the origin with inner and outer radii
$$
%
\begin{align}
%
r_{in} &= 1 \\
%
r_{out} &= 2
%
\end{align}
%
\tag{4}
$$
There are no critical points. There region is closed and bounded. Therefore, a periodic solution exists.
Visualization
The vector field $\left[
\begin{array}{c}
\dot{x} \\
\dot{y} \\
\end{array}
\right]$ in $(1)$ is plotted against the gray trapping region in $(4)$. The red, dashed lines are nullclines which intercept at the critical point.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2267695",
"timestamp": "2023-03-29T00:00:00",
"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:
$$\frac{\sin(z)}{z(1-z^2/\pi^2)}=\prod_{\color{red}{n=2}}^{\infty}(1-\frac{z^2}{n^2\pi^2})\overset{z=\pi}{=}\color{red}{(1-\frac{1}{2})}(1+\frac{1}{2})\color{red}{(1+\frac{1}{3})}(1-\frac{1}{3})\color{red}{(1-\frac{1}{4})}(1+\frac{1}{4})\dots$$
Alternate answers using other values for z like $\pi i/2$ or using other infinite series like cos are welcome.
|
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
$$P_{2k+2}=\frac{3}{2}\cdot\frac{2}{3}\cdot\frac{5}{4}\cdot\frac{4}{5}\cdots\frac{2k+1}{2k}\cdot\frac{2k}{2k+1}\cdot\frac{2k+3}{2k+2}=\frac{2k+3}{2k+2}$$
and $$P_{2k+1}=1$$
So we have that $P_n\to1$ as $n\to\infty$ so the product converges to $1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2269144",
"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 $x+1$ $\frac {x^5 - 5x^4 + 1}{x-1}$, or $\frac {x^5 - 5x^4 + 1}{x-1}$ I get a remainder, so those are not roots?!
|
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:
$$\begin{align}&a+4b=5\\&b(4a+6b)=0\\&b^2(4b+6a)=0\\&b^3(b+4a)=0\\&ab^4=-1
\end{align}$$
From the fifth equation, $b\neq 0$. Then from the fourth equation, $b=-4a$. Substitution into the first yields $a-16a=5\implies a=-\frac13\implies b=\frac43$. But substitution into the second equation gives $4a+6b=-\frac43+\frac{24}3\neq0$. Similarly, the third equation fails.
So the expression $x^5-5x^4+1$ cannot be factored into this form.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2269342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"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.