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minimal polynomial of a matrix with some unknown entries Question is to prove that :
characteristic and minimal polynomial of $ \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right) $ is $x^3-ax^2-bx-c$.
what i have done so far is :
characteristic polynomial of a matrix $A$ is given by $\det(A-xI)$
in case of $A= \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)$ we have $\det(A-xI)=\det\left( \begin{array}{cccc}
-x & 0 & c \\
1 & -x & b \\
0 & 1 & a-x \end{array} \right)=-(x^3-ax^2-bx-c)$
So, i have got the characteristic polynomial as $x^3-ax^2-bx-c$.
Now, the problem is how do i find minimal polynomial.
As $a,b,c$ are arbitrary, I can not factorize $x^3-ax^2-bx-c$ so as to see which factor gives me minimal polynomial.
I am confused.
please suggest me some hint.
EDIT : This is just after Mr.Will Jagyy's hint :
I have $A= \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)$ then, $A^2= \left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)\left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)=\left( \begin{array}{cccc}
0 & c & ac \\
0 & b & c+ab \\
1 & a & b+a^2 \end{array} \right)$
Now,
$A^2+rA+sI=\left( \begin{array}{cccc}
0 & c & ac \\
0 & b & c+ab \\
1 & a & b+a^2 \end{array} \right)+r\left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)+s\left( \begin{array}{cccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right)=\left( \begin{array}{cccc}
s & c & * \\
r & b+s & * \\
1& * & * \end{array} \right)$
As element of $3^{rd}$ row $1^{st}$ column is $1$ in above matrix, this can never be $0$
i.e., $A^2+rA+sI$ can never be $0$.
Now, $A+rI=\left( \begin{array}{cccc}
0 & 0 & c \\
1 & 0 & b \\
0 & 1 & a \end{array} \right)+r\left( \begin{array}{cccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \end{array} \right)=\left( \begin{array}{cccc}
r & *& * \\
\color{magenta}{1} & * & * \\
* & \color{magenta}{1} & * \end{array} \right)\neq 0$ if $r\neq 0$
Thus, $A^2+rA+sI\neq 0$ and $A+rI\neq 0$ for any $r,s$.
Thus, minimal polynomial for $A$ can not be of order less than $3$.
Thus, minimal polynomial for $A$ has to be $x^3-ax^2-bx-c$.
I have written this just to make sure i have tried in correct way as i can not write this in a comment.
I would be thankful if there is any other way to proceed further.. Thank you :)
|
You don't need to factor anything. Your characteristic polynomial, call it $f(x) = x^3 - a x^2 - b x - c,$ satisfies $f(A) = 0,$ meaning $$ A^3 - a A^2 - b A - c I = 0 $$ as matrices. This is Hamilton-Cayley. Actually, given your question, you should carefully check that with actual matrices. What are $A^2$ and $A^3?$
So, there is a (monic) minimal polynomial, call it $m(x).$ Maybe $m(x) = f(x).$ But maybe $m$ is quadratic or even linear. So, carefully write out the matrix expression $$ A^2 + r A + s I $$ and see if any choice of $r,s$ results in the zero matrix. Then, check $$ A + t I $$ to see if any choice of $t$ results in the zero matrix.
If the minimal polynomial is not quadratic or linear, then it is cubic.
Meanwhile, i collected a bunch of equivalent conditions at Given a matrix, is there always another matrix which commutes with it? which includes an overall answer to your question.
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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|
How to find $x + y + z$?
Q. If $x^{1/3} + y^{1/3} + z^{1/3} = 0$, then
(A) $x + y + z = 3 xyz$
(B) $x + y + z = 0 $
(C) $( x + y + z)^3= 27 xyz$
(D)$ x^3 + y^3 + z^3 = 0$
What I've done:
$(x^{1/3} + y^{1/3} + z^{1/3})^3 = 0^3$
$=> (x^{1/3} + y^{1/3})^3 + z+3(x^{1/3} + y^{1/3})z(x^{1/3} + y^{1/3}+z^{1/3})=0$
$=> (x^{1/3} + y^{1/3})^3 + z=0$
Now what??
|
use this
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$$
so if $a+b+c=0$,then we have
$$a^3+b^3+c^3=3abc$$
then if $$x^{\frac{1}{3}}+y^{\frac{1}{3}}+z^{\frac{1}{3}}=0$$
$$\Longrightarrow x+y+z=3(xyz)^{\frac{1}{3}}$$
so
$$(x+y+z)^3=27xyz$$
\begin{align*}&a^3+b^3+c^3-3abc \\
&=(a^3+3a^2b+3ab^2+b^3+c^3)-(3abc+3a^2b+3ab^2)\\
&=[(a+b)^3+c^3]-3ab(a+b+c)\\
&=(a+b+c)(a^2+b^2+2ab-ac-bc+c^2)-3ab(a+b+c)\\
&=(a+b+c)(a^2+b^2+c^2+2ab-3ab-ac-bc)\\
&=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)
\end{align*}
|
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"url": "https://math.stackexchange.com/questions/507632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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|
If $ k = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$. Then $\lfloor k \rfloor =$ If $ k = \sqrt{n\cdot (n+1)\cdot(n+2)\cdot (n+3)}$, where $n\in \mathbb{N}$. Then $\lfloor k \rfloor = $
$\underline{\bf{My\; Try}}::$ We can write the expression $n\cdot (n+1)\cdot(n+2)\cdot (n+3) = (n^2+3n).(n^2+3n+2)$
$ = (n^2+3n)^2+2\cdot (n^2+3n)+1-1 = (n^2+3n+1)^2-1<(n^2+3n)^2$
Now I Did not Understand How can i solve further
Help Required.
Thanks
|
$$k<\sqrt{n.(n+1).(n+2).(n+3)+1}=n^2+3n+1$$
$$n.(n+1).(n+2).(n+3)=n^4+6n^3+11n^2+6n>(n^2+3n)^2$$
so $\lfloor k \rfloor =n^2+3n $
|
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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|
How find the maximum of $\frac{1}{1+x^2}+\frac{1}{1+y^2}+\frac{z}{1+z^2}$ let $x,y,z$ are postive numbers,and such
$$xy+zx+yz=1$$
find the maxum of
$$\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{z}{1+z^2}$$
my try
note
$$x^2+1=x^2+xy+yz+xz=(x+y)(x+z)$$
$$y^2+1=(y+x)(y+z)$$
$$z^2+1=(z+x)(z+y)$$
$$\dfrac{1}{1+x^2}+\dfrac{1}{1+y^2}+\dfrac{z}{1+z^2}=\dfrac{y+z+x+z+z(x+y)}{(x+y)(y+z)(x+z)}$$
I think we must use
$$8(a+b+c)(ab+bc+ac)\le 9(a+b)(b+c)(a+c)$$
$$LHS\le \dfrac{9}{8}\dfrac{x+y+z+z(x+y+1)}{x+y+z}=\dfrac{9}{8}(1+\dfrac{z(x+y+1)}{x+y+z})$$
then follow I can't works,Thank you
|
Due to symmetry of the formula, you can suppose x=y=t. Hence
$$t^2+2tz=1$$
$$z=\frac{1-t^2}{2t}$$
$$f(t)=\frac{2}{1+t^2}+\frac{2t(1-t^2)}{4t^2+(1-t^2)^2}=2\frac{1+t+t^2-t^3}{(1+t^2)^2}$$
$$f'(t)=2\frac{(-3t^2+2t+1)(1+t^2)-4t(-t^3+t^2+t+1)}{(1+t^2)^3}$$
$$f'(t)=2\frac{t^4-2t^3-6t^2-2t+1}{(1+t^2)^3}$$
$P(t)=t^4-2t^3-6t^2-2t+1=(t+1)(t^3-3t^2-3t+1)=(t+1)^2(t^2-4t+1)$
Hence there are extremum in $t=-1$ (not really as it is a double root), and $t={2\pm\sqrt{3}}$. You can easily verify the maximum is for $2-\sqrt{3}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}$ for $n>1,n\in\Bbb N$ by Induction Proving $\frac{1}{n+1} + \frac{1}{n+2}+\cdots+\frac{1}{2n} > \frac{13}{24}$ for $n>1,n\in\Bbb N$
To solve it I used induction but it is leading me nowhere my attempt was as follows:
Lets assume the inequality is true for $n = k$ then we need to prove that it is true for $k+1$
so we need to prove $\frac1{k+2} + \frac1{k+3}+\cdots+\frac1{2(k+1)} > 13/24$
I don't know where to go from here please help.
|
Using induction we first show this is true for $n=2$:
$\frac{1}{2+1}+\frac{1}{2+2}=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}=\frac{14}{24}\gt\frac{13}{24}$
Therefore it is indeed true for $n=2$.
Now lets assume it is true for some $n=k$, therefore:
$S_k=\frac{1}{k+1}+\frac{1}{k+2}+...+\frac{1}{2k}\gt\frac{13}{24}$
Finally we need to prove that this implies it is also true for $n=k+1$:
$S_{k+1}=\frac{1}{(k+1)+1}+\frac{1}{(k+1)+2}+...+\frac{1}{2(k+1)-2}+\frac{1}{2(k+1)-1}+\frac{1}{2(k+1)}$
$\qquad=\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2(k+1)}$
$\qquad=-\frac{1}{k+1}+\frac{1}{k+1}+\frac{1}{k+2}+\frac{1}{k+3}+...+\frac{1}{2k}+\frac{1}{2k+1}+\frac{1}{2(k+1)}$
$\qquad=-\frac{1}{k+1}+S_k+\frac{1}{2k+1}+\frac{1}{2(k+1)}$
$\qquad=S_k+\frac{1}{2k+1}+\frac{1}{2(k+1)}-\frac{1}{k+1}$
$\qquad=S_k+\frac{1}{2(2k+1)(k+1)}$
$\qquad\gt S_k$
$\therefore S_{k+1}\gt \frac{13}{24}$
|
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|
Help with partial fraction $\int_{4}^{5}\frac{x^3-3x^2-9}{x^3-3x^2}dx$ I can't figure out why I keep getting this answer wrong.
First, simplifying the answer:
$$\int_{4}^{5}1-\frac{9}{x^2(x-3)}dx$$
Setting up the partial fraction:
$$\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-3}$$
Expanding...
$$\frac{A(x^2)(x-3)+B(x)(x-3)+Cx^3}{x^2(x-3)}$$
Further expansion...
$$\frac{A(x^3-3x^2)+B(x^2-3x)+Cx^3}{x^2(x-3)}$$
Grouping...
$$\frac{(A+C)x^3+(B-3A)x^2(-3Bx)}{x^2(x-3)}$$
After solving the system of equations I got the following values:
$$A=1, B=3, C=-1$$
So my new integral is now:
$$\int_{4}^{5}\frac{1}{x}+\frac{3}{x^2}-\frac{1}{x-3}dx$$
Solving I got this:
$$ln|x|+2xln|x^2|-ln|x-3||_{4}^{5}$$
Final answer:
$$ln5+10ln25-ln2-(ln4+8ln16)$$
|
Recall that the original integrand was $$1 - \frac{9}{x^2(x-3)}$$ Did you forget to account for the $1$, which would integrate to $x$, and/or forget that you are subtracting the quotient from $1$?
So you should end with $$\int_{4}^{5} 1 - \Big(\frac{1}{x}+\frac{3}{x^2}-\frac{1}{x-3}\Big)dx$$
Also, use the power rule for integrating $$\int \frac 3{x^2} \,dx = -3x^{-1} = -\frac 3x$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How find this equation $\prod\left(x+\frac{1}{2x}-1\right)=\prod\left(1-\frac{zx}{y}\right)$ let $x,y,z\in(0,1)$, find the pairs of $(x,y,z)$ such
$$\left(x+\dfrac{1}{2x}-1\right)\left(y+\dfrac{1}{2y}-1\right)\left(z+\dfrac{1}{2z}-1\right)=\left(1-\dfrac{xy}{z}\right)\left(1-\dfrac{yz}{x}\right)\left(1-\dfrac{zx}{y}\right)$$
my try:use
$$x+\dfrac{1}{2x}-1\ge 2\sqrt{\dfrac{1}{2}}-1=\sqrt{2}-1$$
so
$$LHS\ge (\sqrt{2}-1)^3$$
so I guess
$$RHS\le (\sqrt{2}-1)^3$$
But I can't prove it.Thank you
|
My approach lacks something at the end.
$$x+\dfrac{1}{2x}-1 = \frac{2x^2}{2x}-\frac{2x}{2x}+\dfrac{1}{2x} = \frac{2x^2 -2x +1}{2x}$$
$$1 - \frac{xy}{z} = \frac{z-xy}{z}$$
$$
\begin{align}
\left(x+\dfrac{1}{2x}-1\right)\left(y+\dfrac{1}{2y}-1\right)\left(z+\dfrac{1}{2z}-1\right)&=\left(1-\dfrac{xy}{z}\right)\left(1-\dfrac{yz}{x}\right)\left(1-\dfrac{zx}{y}\right)\\
\frac{2x^2 -2x +1}{2x}\frac{2y^2 -2y +1}{2y}\frac{2z^2 -2z +1}{2z} &= \frac{z-xy}{z}\frac{x-yz}{x}\frac{y-zx}{y}\\
(2x^2 -2x +1)(2y^2 -2y +1)(2z^2 -2z +1) &= 8(z-xy)(x-yz)(y-zx)\\
\end{align}
$$
Clearly,
$$\dfrac{1}{8} \leq (2x^2 -2x +1)(2y^2 -2y +1)(2z^2 -2z +1) \lt 1$$
My work in progress is to show that $(z-xy)(x-yz)(y-zx) \leq \dfrac{1}{64}$, but I thought I'd put this up as it seems a simpler proof for the LHS.
|
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|
Find the value of $\sin25^° \sin35^° \sin85^°$ Trigo problem :
Find the value of $\sin25^° \sin35^° \sin85^°$.
My approach :
Using $2\sin A\sin B = \cos(A-B) -\cos(A+B)$
$$
\begin{align}
& \phantom{={}}[\cos10^{°} -\cos60^°] \sin85^° \\
& = \frac{1}{2}[2\cos10^°\sin85^° -2\cos60^° \sin85^°] \\
& = \frac{1}{2}[2\cos10^°\sin85^° -2 \frac{1}{2} \sin85^°] \\
& = \frac{1}{2}[2\cos10^°\sin85^° -\sin85^°] \\
& = \frac{1}{2} [\sin 95^° + \sin 75^° -\sin85^°]
\end{align}
$$
After this I am unable to solve further.... please guide... thanks.
|
\begin{align}
& \phantom{={}} \sin25^{°}\sin35^{°}\sin85^{°} \\
& =\sin85^{°}\sin35^{°}\sin25^{°} \\
& =\frac{1}{2}(2\sin85^{°}\sin35^{°})\sin25^{°} \\
& =\frac{1}{2}[ \cos50^{°} -\cos120^{°}] \sin25^{°} \\
& = \frac{1}{4}[ 2\cos50^{°}\sin25^{°} +2\cos60^{°} \sin25^{°}] \\
& = \frac{1}{4}[ 2\cos50^{°}\sin25^{°} +2 \frac{1}{2} \sin25^{°}] \\
& = \frac{1}{4}[ 2\cos50^{°}\sin25^{°} +\sin25^{°}] \\
& = \frac{1}{4} [ \sin75^{°} -\sin 25^{°} +\sin25^{°}] \\
& =\frac{1}{4} \sin75^{°}
\end{align}
|
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|
boolean logic simplify To prove:
$(X+Y)(X'+Z) = XZ + X'Y$
I try to simply $(X+Y)(X'+Z)$ to $XZ + X'Y + YZ$
then I have no idea how to simply further.
Thanks in advance!!
|
$$\begin{align} (X + Y)(X' + Z) & = (X+Y)X' + (X+Y)Z \\ \\ & = \underbrace{X'X}_{= 0} + X'Y + XZ + YZ \\ \\ & = X'Y + XZ + YZ \\ \\ & = X'Y + XZ + \underbrace{(X+ X')}_{= 1}YZ \\ \\ & = X'Y + XZ + XYZ + X'YZ \\ \\ & = X'Y\underbrace{(1 + Z)}_{= 1} + XZ\underbrace{(1 + Y)}_{= 1} \\ \\ &= X'Y + XZ \end{align}$$
|
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|
How many even numbers of four digits can be formed with the digits 0,1,2,3,4,5 and 6 no digit being used more? My attempt to solve this problem is:
First digit cannot be zero, so the number of choices only $6 (1,2,3,4,5,6)$
The last digit can be pick from $0,2,4,6$, so the number of choices only 4
Second digit can be only pick from the rest, so the number of choices only 5
Third digit can be only pick from the rest, so the number of choices only 4
The total number of choices is $6\cdot 4\cdot 5\cdot 4= 480$
So, is my solution true? Or I miss something?
Thanks
|
For an even no. last digit needs to be even = 4 choices
Total 4 digit even numbers = (Total even numbers) - (Total 3 digit even numbers)
Total even numbers: $4\cdot 6\cdot 5\cdot 4 = 480$
Total 3-digit even numbers: $3\cdot 5\cdot 4\cdot 1 = 60$
Total 4-digit even numbers: $480 - 60 = 420$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality....(RMO $1994$...question $8$) If $a$, $b$, $c$ are positive real numbers such that $a+b+c = 1$, prove that
$$(1+a)(1+b)(1+c) ≥ 8(1-a)(1-b)(1-c)\text{.}$$
|
Note that by A.M - G.M we have $$ (b+c)+(c+a) \geq 2 \cdot \sqrt{(b+c)\cdot(c+a)}$$ This says $2c+b+a =1+c\geq 2 \cdot \sqrt{(b+c)\cdot (c+a)}$. Similary get inequalities for $(1+b)$ and $(1+c)$ multiply and get the answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the matrix representation of the following linear transformation $T: M_{2x2} (R) \rightarrow P_{2} (R)$ defined by $T \begin{pmatrix} a & b \\ c & d \end{pmatrix}= (a+b) + (2d)x + bx^2$.
Let $\beta = \left \{ \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}, \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} \right \}$ and $\gamma = {1, x, x^2}$. Compute $[T]_{\beta}^{\gamma}$.
This is what I have:
\begin{align}
[T]_{\beta}^{\gamma} &= \left [ [T\begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}]_{\gamma} \, \, \, [T\begin{pmatrix} 0 & 1 \\ 0 & 0 \end{pmatrix}]_{\gamma} \, \, \, [T\begin{pmatrix} 0 & 0 \\ 1 & 0 \end{pmatrix}]_{\gamma}\, \, \, [T\begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix}]_{\gamma}\right ]
&= \begin{pmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 0 & 2x \\ 0 & x^2 & 0 & 0 \end{pmatrix}
\end{align}
but I'm not sure if it is correct. How can a matrix turn a 2x2 matrix into a polynomial?
|
Let $V$ a vector space and $\vec{v}$. The $\vec{v}$ representation with respect to a basis $B=\left\langle { \vec{\beta}_1, \cdots, \vec{\beta}_n } \right\rangle$ of $V$ is a colmun vector $$Rep_B(\vec{v})=\left( {\begin{array}{*{20}{c}}
{c_1} \\
\vdots \\
{c_n} \\
\end{array}} \right)
$$
such that $\vec{v}=c_1\vec{\beta}_1 + \cdots + c_n\vec{\beta}_n$. Now for to find the transformation matrix you need to find $Rep_B(T(\left( {\begin{array}{*{20}{c}}
{1} & {0} \\
{0} & {0} \\
\end{array}} \right)
))$,
$Rep_B(T(\left( {\begin{array}{*{20}{c}}
{0} & {1} \\
{0} & {0} \\
\end{array}} \right)
))$,
$Rep_B(T(\left( {\begin{array}{*{20}{c}}
{0} & {0} \\
{1} & {0} \\
\end{array}} \right)
))$ and $Rep_B(T(\left( {\begin{array}{*{20}{c}}
{0} & {0} \\
{0} & {1} \\
\end{array}} \right)
))$, where $B=\left\langle 1,x,x^2\right\rangle$, then put these four column vectors to form a matrix. According wiht this, in your matrix you will not have $x^2$ nor $2x$, but you will have $1$ and $2$, i.e. the respective coefficients, do you see why?.
Note that the matrix formed above will not transform from $2\times 2$ matrices to polynomials, it will transform from the representation of $2\times 2$ matrices to representation of polynomials with respect to given bases.
|
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|
How to solve this system of equation? I need to solve the following system of $(x,y)$:
\begin{cases}
3y^3+3x\sqrt{1-x}=5\sqrt{1-x}-2y\\
x^2-y^2\sqrt{1-x}=\sqrt{2y+5}-\sqrt{1-x}
\end{cases}
|
Let's resurrect this old question. Two solutions to
\begin{cases}
3y^3+3xz=5z-2y\\
x^2-y^2z=\sqrt{2y+5}-z\end{cases}
are $(x,\,y)= (-3,\,2)$ for $z=\sqrt{1-x}$ and,
$$x = \text{Real root}(7 - x - 3 x^2 + 7 x^3 - 8 x^4 + 2 x^5 - x^6 + x^7=0)$$
$$y = \text{Real root}(2 + y + 2 y^2 + 4 y^3 - 3 y^5 + y^7=0)$$
for $z=-\sqrt{1-x}$.
Unless your system has special symmetries, I'm afraid the only way to find solutions that algebraic numbers of high degree is to use resultants.
|
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|
Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$. Determine $p$ for one to have $\displaystyle\frac{1}{5}+\frac{1}{6}=\frac{1+1}{5+6}=\frac{2}{11}$, in $\mathbb{Z_p}$.
I know that $p$ must, $13, 43, 61, 101,103$.
|
Try multiplying both sides by $11 \cdot 5 \cdot 6$, to get $$121 = 11 \cdot 6 + 11 \cdot 5 = 2 \cdot 5 \cdot 6 = 60$$ in $\mathbb{Z}/p\mathbb{Z}$.
|
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|
Show that the sequence ${a_n}$ converges where $a_n = \sqrt{1+\sqrt{2+\sqrt{3+\cdots+\sqrt{n}}}}$ for $n\geq 1$. The original question was to determine whether the sequence converges, but I have checked for extremely high values of $n$ and it seems as though it does converge. This lead me to wonder if there was an "easy" method of showing that the sequence is bounded (since it is monotone, it then follows that it converges). Thanks.
|
By repeatedly rationalizing the numerators and using that $a_n\geq1$ for all $n$,
$$
a_{n+1}-a_n=\frac{\sqrt{2+\sqrt{3+\cdots\sqrt{n+1}}}-\sqrt{2+\sqrt{3+\cdots\sqrt{n}}}}{a_{n+1}+a_n}\\
\leq\frac{\sqrt{2+\sqrt{3+\cdots\sqrt{n+1}}}-\sqrt{2+\sqrt{3+\cdots\sqrt{n}}}}{2}\\
=\frac12\,\frac{\sqrt{3+\sqrt{4+\cdots\sqrt{n+1}}}-\sqrt{3+\sqrt{4+\cdots\sqrt{n}}}}{\sqrt{2+\sqrt{3+\cdots\sqrt{n+1}}}+\sqrt{2+\sqrt{3+\cdots\sqrt{n}}}}\\
\leq\frac12\,\frac{\sqrt{3+\sqrt{4+\cdots\sqrt{n+1}}}-\sqrt{3+\sqrt{4+\cdots\sqrt{n}}}}{2\sqrt2}\\
\leq\cdots\\
\leq\frac{\sqrt{n+1}}{2^{n+1}}\leq\frac{n+1}{2^n}
$$
By telescoping, we see that the sequence is Cauchy, i.e.
$$
a_{n+k}-a_n=\sum_{j=1}^ka_{n+j}-a_{n+j-1}\leq\sum_{j=1}^k\frac{n+j+1}{2^{n+j}}\leq\frac3{2^n}.
$$
So the limit $L$ exists.
|
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|
by completing the square find in terms of k the roots of the equation $x^2 + 2kx-7=0$ By completing the square find in terms of $k$ the roots of the equation $$x^2 + 2kx-7=0$$
prove for all real values of $k$, the roots are real
|
$$x^2+2kx-7=x^2+2\cdot x\cdot k+k^2-k^2-7=0$$
$$(x+k)^2=k^2+7$$
$$x+k=\pm\sqrt{k^2+7}$$
$$x=-k\pm\sqrt{k^2+7}$$
because $k^2+7\geq 0$ for all real $k$ all roots are real.
|
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|
The sum of three square roots bounded below by $\sqrt{82}$
Let $a,b,c >0$ and $a+b+c \le 1$. Prove: $$\sqrt{a^2+1/a^2}+\sqrt{b^2+1/b^2}+\sqrt{c^2+1/c^2} \ge \sqrt{82}$$
Progress: I tried to have 3 vectors $(a,1/a)$, $(b,1/b)$ and $(c,1/c)$, played around with the vector inequalities but no success.
|
In the first proof only condition $a+b+c=1$ is used. What if $a+b+c<1$? Also a simpler alternative way to show $\frac{1}{a}+ \frac{1}{b}+\frac{1}{c}\geq 9$ when $a+b+c\leq 1$ is the Cauchy-Schwarz inequality:
$9=\|( \sqrt{a}, \sqrt{b}, \sqrt{c})\cdot(\frac{1}{\sqrt{c}}, \frac{1}{\sqrt{b}}, \frac{1}{\sqrt{c}})\|^2\leq (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$.
|
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|
Dervation of $\sum_{k=0}^n(r+1)^k= \frac{(r+1)^{n+1}-1}{r}$ How can one derive the following identity?
$\sum_{k=0}^n(r+1)^k= \frac{(r+1)^{n+1}-1}{r}$
I have playing around with binomial coefficients and index shiftings but wasn't able to get anywhere.
|
For $r\neq 1$, the sum of the first n terms of a geometric series is:
$a + ar + a r^2 + a r^3 + \cdots + a r^{n-1} = \sum_{k=0}^{n-1} ar^k= a \, \frac{1-r^{n}}{1-r}$,
where a is the first term of the series, and r is the common ratio. We can derive this formula as follows:
\begin{align}
&\text{Let }s = a + ar + ar^2 + ar^3 + \cdots + ar^{n-1}. \\[4pt]
&\text{Then }rs = ar + ar^2 + ar^3 + ar^4 + \cdots + ar^{n} \\[4pt]
&\text{Then }s - rs = a-ar^{n} \\[4pt]
&\text{Then }s(1-r) = a(1-r^{n}),\text{ so }s = a \frac{1-r^{n}}{1-r} \quad \text{(if } r \neq 1 \text{)}.
\end{align}
As n goes to infinity, the absolute value of r must be less than one for the series to converge. The sum then becomes
$a+ar+ar^2+ar^3+ar^4+\cdots = \sum_{k=0}^\infty ar^k = \frac{a}{1-r} \Leftrightarrow |r|<1$
When $a = 1$, this simplifies to: $1 \,+\, r \,+\, r^2 \,+\, r^3 \,+\, \cdots \;=\; \frac{1}{1-r}$, the left-hand side being a geometric series with common ratio r. We can derive this formula:
\begin{align}
&\text{Let }s = 1 + r + r^2 + r^3 + \cdots. \\[4pt]
&\text{Then }rs = r + r^2 + r^3 + \cdots. \\[4pt]
&\text{Then }s - rs = 1,\text{ so }s(1 - r) = 1,\text{ and thus }s = \frac{1}{1-r}.
\end{align}
|
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|
How prove this $a+b\le 1+\sqrt{2}$ let $0<c\le b\le 1\le a$, and such $a^2+b^2+c^2=3$, show that
$a+b\le 1+\sqrt{2}$
My try: let $ c^2=3-(a^2+b^2)\le b$
|
Set $a^2+b^2=3-c^2=k^2$, then we want to maximize
$$a+b=b+\sqrt{k^2-b^2}$$
Since it is increasing in $k$, and $k^2=3-c^2\le 3$, we achieve maximum at $k=\sqrt{3}$, $c=0$ (!)
Set $b=\sqrt{3}\sin\alpha \le 1$. Therefore
$$0\le \sin(\alpha)\le\frac{1}{\sqrt{3}}<\frac{1}{\sqrt{2}}\implies 0^{\circ}\le\alpha<45^{\circ}$$
And we want to maximize
$$a+b=\sqrt{3}(\sin\alpha+\cos\alpha)=\sqrt{6}\sin(\alpha+45^{\circ})$$
Since both $b$ and the function are incresing in $\alpha\in[0,45^{\circ}]$ and decreasing in $c$, we reach maximum at $c=0$, $b=1$, $a=\sqrt{2}$.
|
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|
Correctness of proof of division in ceiling function My professor asked me to present a proof to my fellow students tomorrow that
$$\left\lceil\frac{n}{2^k}\right\rceil = \left\lceil\frac{\left\lceil\frac{n}{2^{k-1}}\right\rceil}{2}\right\rceil$$
I tried to prove it as follows, but I am not sure if my proof is correct.
Proof
The above is equivalent to $$\left\lceil\frac{n}{2^{k+1}}\right\rceil = \left\lceil\frac{\left\lceil\frac{n}{2^k}\right\rceil}{2}\right\rceil$$
Let us say that $x = \left\lceil\frac{n}{2^{k+1}}\right\rceil$ for ease of notation.
It is then clear that $$x-1 \lt \frac{n}{2^{k+1}} \le x$$
We will now prove what is asked using contradiction. Assume that $$\left\lceil\frac{n}{2^{k+1}}\right\rceil \lt \left\lceil\frac{\left\lceil\frac{n}{2^k}\right\rceil}{2}\right\rceil$$
Then there is an integer $a$ such that $$\left\lceil\frac{n}{2^{k+1}}\right\rceil \le a \lt \left\lceil\frac{\left\lceil\frac{n}{2^k}\right\rceil}{2}\right\rceil$$
This is equivalent to $$\left\lceil\frac{n}{2^k}\right\rceil \le 2*a \lt \frac{n}{2^k}$$ which can obviously not be true.
QED
Do you see any errors in my proof? Is it clear? I'm a bit doubtful about the last step, where I multiply $a$ by 2.
|
Your reasoning is a bit shaky in a few places. Let me see if I can help you a bit with the rigor.
To further simplify, let's say $x=\left\lceil\frac{n}{2^{k+1}}\right\rceil$ and $y=\left\lceil\frac{n}{2^k}\right\rceil,$ so we must prove that $x=\left\lceil\frac{y}2\right\rceil$. Since $\frac{n}{2^k}\le y,$ then $\frac{n}{2^{k+1}}\le\frac{y}2,$ and so $x\le\left\lceil\frac{y}2\right\rceil$. Now we can assume that $x\ne\left\lceil\frac{y}2\right\rceil,$ from which it follows that $x<\left\lceil\frac{y}2\right\rceil.$ Then indeed, there is an integer $a$ such that $x\le a<\left\lceil\frac{y}2\right\rceil.$ In fact, $x$ is such an integer! More relevant, though, is the fact that $$\frac{n}{2^{k+1}}\le x<\left\lceil\frac{y}2\right\rceil.$$ Since $x$ is an integer less than $\left\lceil\frac{y}2\right\rceil$, then we have $x<\frac{y}2,$ and so $$2x<y.\tag{$\spadesuit$}$$ On the other hand, $$\frac{n}{2^{k+1}}\le x,$$ so $$\frac{n}{2^k}\le 2x,$$ and since $2x$ is an integer, then $$y=\left\lceil\frac{n}{2^k}\right\rceil\le 2x.\tag{$\clubsuit$}$$ By $(\spadesuit)$ and $(\clubsuit),$ we have our contradiction.
|
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|
17539 decimal to binary not getting the same result I'm trying to convert 17539 to binary. My math says its 110000010010001, but online calculators like this and this say it equals to 100010010000011. Who is making something wrong.
|
Note that
$$\begin{align}
17539 &= \color{red}1\cdot 2^{14}+ \color{red}0\cdot2^{13} + \color{red}0\cdot2^{12} +\color{red}0\cdot2^{11}+ \color{red}1\cdot 2^{10} \\ &+ \color{red}0\cdot2^{9} + \color{red}0\cdot2^{8} + \color{red}1\cdot 2^7 + \color{red}0\cdot2^{6} + \color{red}0\cdot2^{5} \\&+ \color{red}0\cdot2^{4} +\color{red}0\cdot2^{3} +\color{red}0\cdot2^{2} +\color{red}1\cdot 2^1 + \color{red}1\cdot 2^0.
\end{align}
$$
|
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|
How do I get the equation for this parabola in standard form? How do I get the equation for this parabola in standard form?
$ y = f(x)= 2x^2+3x-2$
|
If you mean vertex form, you can complete the square.
$$\begin{align}f(x) &= 2x^2 + 3x - 2 \\
&= 2\left(x^2+\frac{3}{2}x\right) -2 \\
&= 2\left(x^2+\frac{3}{2}x+\frac{9}{16}-\frac{9}{16}\right) -2\\
&= 2\left(x^2 +\frac{3}{2}x+\frac{9}{16}\right) -2\left(\frac{9}{16}\right)-2\\
&= 2\left(x+\frac{3}{4}\right)^2 - \frac{25}{8}\end{align}$$
|
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|
Evaluating $\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx$ I am trying to prove that
\begin{equation}
\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x = \frac{\pi^{3}}{16} - 3\mathrm{G}\log\left(2\right)
\tag{1}
\end{equation}
where $\mathrm{G}$ is Catalan's Constant.
I was able to express it in terms of Euler Sums but it does not seem to be of any use.
\begin{align}
&\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x
\\[3mm] = &\
\frac{1}{16}\sum_{n = 1}^{\infty}
\frac{\psi_{1}\left(1/4 + n\right) -
\psi_{1}\left(3/4 + n\right)}{n} \tag{2}
\end{align}
Here $\psi_{n}\left(z\right)$ denotes the polygamma function.
Can you help me solve this problem $?$.
|
This is a partial solution.
Let us put, for $0\leq t\leq 1$,
$$F(t) = \int_0^1 \frac{\log x \log(1-tx^4)}{1+x^2} dx$$
Then
$$F'(t) = -\int_0^1 \frac{x^4\log x}{(1+x^2)(1-tx^4)} dx = -\int_0^1 \frac{x^4\log x}{1+x^2} \sum_{n=0}^\infty t^nx^{4n} dx$$
$$=-\sum_{n=0}^\infty t^{n} C_{4(n+1)}$$
where $$C_m = \int_0^1 \frac{x^{m}\log x}{1+x^2} dx.$$
One has $C_0 = -G$. Multiplying both sides of the identity
$$x^m = \frac{x^m}{1+x^2} + \frac{x^{m+2}}{1+x^2}$$
by $\log x$ and integrating from $0$ to $1$, one finds the recurrence formula
$$C_m + C_{m+2} = \frac{-1}{(1+m)^2}$$
and therefore
$$C_{m+4} - C_m = \frac{-1}{(3+m)^2} + \frac{1}{(1+m)^2}.$$
Therefore,
$$C_0 = -G$$
$$C_4 = -G +1 - \frac{1}{3^2}$$
$$C_8 = -G + 1 - \frac{1}{3^2} + \frac{1}{5^2} - \frac{1}{7^2}.$$
and so on. (Remark that $C_{4n} \to 0$ by definition of $G$.) Now, remark that $F(0) = 0$, so your integral is
$$F(1) = \int_0^1 F'(t) dt = -\sum_{n=0}^\infty \frac{C_{4(n+1)}}{n+1} = -\sum_{n=1}^\infty \frac{C_{4n}}{n}.$$
Now, it should be a matter of partial summation to transform the sum $-\sum_{n=1}^\infty \frac{C_{4n}}{n}$ into $\pi^3/16 -3G\log 2$ (in a manner similar to this), but I don't see it right away. I'll think about it a bit more later.
|
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|
Finding the Maximum and Minimum Distance by Lagrange's Method of Multipliers Q. Find the maximum and minimum distance of a point from origin such that the point lies in the curve $3x^2+4xy+6y^2=140$
I am unable to solve these three equations simultaneously for $(x,y)$
$2x+\lambda(6x+4y)=0$
$2y+\lambda(4x+12y)=0$
$3x^2+4xy+6y^2=140$
|
Another related approach that can be taken makes use of the symmetry of the geometric figure and the function to be extremized. The conic section, although rotated, still retains symmetry about the origin, that is, if the point $ \ (x,y) \ $ lies on the curve, $ \ (-x,-y) \ $ does as well. The distance-from-the-origin function $ \ s(x,y) \ = \ \sqrt{x^2 \ + \ y^2} \ $ has "radial" symmetry about the origin. Since the points $ \ (x,y) \ $ and $ \ (-x,-y) \ $ are equidistant from the origin, the points on the curves at extremal distances from the origin lie on lines $ \ y \ = \ mx \ $ (or possibly the vertical line $ \ x \ = \ c \ $ ) .
We can thus search for such points by inserting $ \ y \ = \ mx \ $ into the equation for the conic section and using the result in finding critical points of the "distance-squared" function
$$ s^2 \ = \ x^2 \ + \ y^2 \ = \ x^2 \ + \ (mx)^2 \ = \ ( \ 1 \ + \ m^2 \ ) \ x^2 \ \ . $$
From the equation for our rotated ellipse, we thus obtain
$$ 3x^2 \ + \ 4xy \ + \ 6y^2 \ = \ 140 \ \ \Rightarrow \ \ 3x^2 \ + \ 4mx^2 \ + \ 6m^2x^2 \ = \ 140 $$
$$ x^2 \ = \ \frac{140}{6m^2 \ + \ 4m \ + \ 3} \ \ \Rightarrow \ \ s^2 \ = \ \frac{140 \ ( \ 1 \ + \ m^2 \ )}{6m^2 \ + \ 4m \ + \ 3} \ \ . $$
By implicit differentiation, we locate critical values of $ \ m \ $ from
$$ 2s \ \frac{ds}{dm} \ = \ \frac{[ \ (6m^2 \ + \ 4m \ + \ 3) \cdot 140 \cdot 2m \ ] \ - \ [ \ ( \ 1 \ + \ m^2 \ ) \cdot 140 \cdot (12m \ + \ 4) \ ]}{(6m^2 \ + \ 4m \ + \ 3)^2} \ = \ 0 $$
$$ \Rightarrow \ \ [ \ (6m^2 \ + \ 4m \ + \ 3) \ m \ ] \ - \ [ \ ( \ 1 \ + \ m^2 \ ) \cdot (6m \ + \ 2) \ ] \ = \ 0 $$
[we can just investigate the numerator, as the denominator has no real zeroes]
$$ \Rightarrow \ \ 6m^3 \ + \ 4m^2 \ + \ 3m \ - \ 6m^3 \ - \ 2m^2 \ - \ 6m \ - \ 2 \ = \ 0 \ \ \Rightarrow \ \ 2m^2 \ - \ 3m \ - \ 2 \ = \ 0 \ \ . $$
[Note that we can also arrive here by inserting $ \ y \ = \ mx \ $ into the result from the Lagrange equations, $ \ 2x^2 \ + \ 3xy \ - \ 2y^2 \ = \ 0 \ $ . ]
By factoring or applying the quadratic formula, we find the solutions $ \ m \ -\frac{1}{2} \ , \ 2 \ $ , so these are the slopes of the lines through the origin on which the extremal-distance points lie (perhaps unsurprisingly, the lines are perpendicular).
For the question at hand, we don't need to know the coordinates of those points (although they can be determined readily), so we compute the extremal distances as
$$ \mathbf{m \ = \ 2 \ : } \quad s^2 \ = \ \frac{140 \ ( \ 1 \ + \ 2^2 \ )}{6 \cdot 2^2 \ + \ 4 \cdot 2 \ + \ 3} \ = \ \frac{140 \cdot 5}{35} \ = \ 20 \ \ \Rightarrow \ \ s \ = \ 2 \ \sqrt{5} \ \ ; $$
$$ \mathbf{m \ = \ -\frac{1}{2} \ : } \quad s^2 \ = \ \frac{140 \ ( \ 1 \ + \ \left[-\frac{1}{2}\right]^2 \ )}{6 \cdot \left[-\frac{1}{2}\right]^2 \ + \ 4 \cdot \left[-\frac{1}{2}\right] \ + \ 3} \ = \ \frac{140 \cdot \left(\frac{5}{4}\right)}{\left(\frac{5}{2}\right)} \ = \ 70 $$
$$ \Rightarrow \ \ s \ = \ \sqrt{70} \ \ . $$
Here is a graph illustrating the solution:
*The ellipse in question is marked in blue; the points closest to the origin lie on the red line, tangent to the red circle of radius $ \ 2 \sqrt{5} \ $ ; those farther from the origin are on the green line, tangent to the green circle of radius $ \ \sqrt{70} \ $ .
$$ \ \ $$
[And I apologize, but as an American of a not insignificant age, I just couldn't look at that graph without being put in mind of this: ]
|
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|
How to find the determinant of this matrix I have the following matrix:
$
\begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{bmatrix}
$
My approach is to rref the matrix so that i can find the determinant by multiplying along the diagonals.
I attempted to do an rref and ended up with
$
\begin{bmatrix}
1 & a & 1 & 1 \\
1-a & a-1 & 0 & 0 \\
0 & 1-a & a-1 & 1 \\
0 & 0 & 1-a & a-1 \\
\end{bmatrix}
$
I then factored out (1-a) to get this
$
\begin{bmatrix}
1 & a & 1 & 1 \\
1 & -1 & 0 & 0 \\
0 & 1 & -1 & 1 \\
0 & 0 & 1 & -1 \\
\end{bmatrix}
$
which will then make my determinant a multipile of $(1-a)^3$:
$\det(A) = (1-a)^3x$
But now here I don't know what to do. I have a feeling that my approach is wrong. Any help or guide please?
|
To find the determinant you could just do it brute force by "Expansion by Minors" using the first row:
$$\begin{align}
\det \begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{bmatrix} = a\det
\begin{bmatrix}
a & 1 & 1 \\
1 & a & 1 \\
1 & 1 & a \\
\end{bmatrix} \\- 1\det\begin{bmatrix}
1 & 1 & 1 \\
1 & a & 1 \\
1 & 1 & a \\
\end{bmatrix} \\+ 1\det \begin{bmatrix}
1 & a & 1 \\
1 & 1 & 1 \\
1 & 1 & a \\
\end{bmatrix} \\- 1\det\begin{bmatrix}
1 & a & 1 \\
1 & 1 & a \\
1 & 1 & 1 \\
\end{bmatrix}
\end{align}
$$
|
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|
Finding the limit of $x \sin\frac{\pi}{x}$ How can I find the limit of the following
$x\rightarrow\infty$
$x \sin\frac{\pi}{x}$
I did
$\dfrac{\sin\frac{\pi}{x}}{\frac{1}{x}}$
I took the derivative using l hospital and got.
$\dfrac{-1x^{-2} \cos \dfrac{\pi}{x}}{-1x^{-2}}$
Simplying I get
$\cos \frac{\pi}{x}$ but I am stuck.
another problem I have is
$\dfrac{\ln(x)}{\cot x}$ as $x\rightarrow0$
I did
$\dfrac{\dfrac{1}{x}}{-\csc^2(x)}$
But I am unsure how to go on.
|
For problems like this you can expand sine in terms of the following taylor series:
$$ sin (p) = p - \frac{p^3}{3!} + \frac{p^5}{5!} - \frac{p^7}{7!} + ...$$
Substituting $p = \frac{\pi}{x}$, we get:
$sin(\frac{\pi}{x}) = \frac{\pi}{x} - \frac{\pi^3}{x^3 3!} + \frac{\pi^5}{x^5 3!} - ...$
We are evaluating $f(x) = x \space sin(\frac{\pi}{x})$, so if we multiply above by $x$, we see that the first term is $\pi$, the other terms are powers of $\frac{1}{x}$
Hence when you apply the limit, the answer tends to $\pi$
|
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|
Show that $\frac{(1-y^{-1})^2}{(1-y^{-1 -c})(1-y^{-1+c} )}$ is decreasing in $y > 1 $. I am interested in the function
$f(y) = \frac{(1-y^{-1})^2}{(1-y^{-1 -c})(1-y^{-1+c} )},$
for values of $c \in (0,1)$, and $y > 1$, and have been trying to show that the function is decreasing.
I have tried differentiating the function, but this does not yield a particularly amenable formula to decide whether $f'(y) < 0$. Similarly I have considered the ratios $f(y)/f(\tilde y)$, but again have been unable to ascertain an inequality for this.
I'd appreciate any other approaches to the problem.
|
EDIT
An easier proof:
Note
\begin{eqnarray}
f(y) &=& \frac{(1-y^{-1})^2}{(1-y^{-1-c})(1-y^{-1+c})} \\
&=& \frac{(y^{1/2}-y^{-1/2})^2}{(y^{\frac{1+c}{2}}-y^{-\frac{1+c}{2}})(y^{\frac{1-c}{2}}-y^{-\frac{1-c}{2}})} \\
&=& \left(\frac{\sinh x}{\sinh \ (1+c)x}\right)^2,
\end{eqnarray}
where $x = \log\sqrt{y}$. Thus, it's sufficient to show
$$
t(x) =\frac{\sinh x}{\sinh \ (1+c)x}
$$
is decreasing for $x>0$. Note
\begin{eqnarray}
t'(x) &=& \frac{\sinh (1+c)x \cosh x - (1+c)\sinh x \cosh(1+c)x}{\sinh^2 (1+c)x} \\
&=& \cosh x \cosh(1+c)x \frac{\tanh (1+c)x - (1+c)\tanh x}{\sinh^2 (1+c)x}.
\end{eqnarray}
All factors here are strictly positive for $x>0$ except $s(x)\equiv\tanh (1+c)x - (1+c)\tanh x$, which is strictly negative. So see this, note
\begin{eqnarray}
s'(x) = (1+c) \cosh^{-2} (1+c) x - (1+c) \cosh^{-2} x
\end{eqnarray}
and use the fact that $\cosh x$ is strictly increasing for $x>0$.
ORIGINAL PROOF (Adds nothing over one above)
You are interested in the function
$$
f(y) = \frac{(1-y^{-1})^2}{(1-y^{-1-c})(1-y^{-1+c})}
$$
for $y>1$. Note that $f(y)$ is strictly decreasing for $y>1$ iff $x\mapsto \frac{(1-x)^2}{(1-x^{1+c})(1-x^{1-c})}$ is strictly increasing for $0<x<1$ (because $s\mapsto \frac{1}{s}$ is strictly decreasing on $s>0$). Observe
$$
\frac{(1-x)^2}{(1-x^{1+c})(1-x^{1-c})} = \frac{1}{1 - \frac{x^{1+c} + x^{1-c} - 2 x}{(1-x)^2}}.
$$
The right hand side can be written $1/(1-g(x))$ where
$$
g(x) = \frac{x^{1+c} + x^{1-c} - 2 x}{(1-x)^2}.
$$
We want to show that $1/(1-g(x))$ is strictly increasing for $0<x<1$. This will follow if we are able to show $g(x)$ is positive and strictly increasing for $0<x<1$. One can write
$$
g(x) = \frac{x}{(1-x)^2} (x^c + x^{-c} - 2).
$$
The two factors can be rewritten as follows:
\begin{eqnarray}
\frac{x}{(1-x)^2} &=& \frac{1}{(x^{1/2} - x^{-1/2})^2} \\
x^c + x^{-c} - 2 &=& (x^{c/2} - x^{-c/2})^2.
\end{eqnarray}
We end up with the lovely expression for $g(x)$:
$$
g(x) = \left(\frac{x^{c/2} - x^{-c/2}}{x^{1/2} - x^{-1/2}}\right)^2.
$$
It is now obvious $g(x) > 0$.
Now to show $g$ is strictly increasing.
We can again use the "increasing functions of increasing functions are increasing" trick to ignore the square. We also substitute $s=-\frac{1}{2}\log x$. With this transformation, $0<s<+\infty$, and since $s$ is strictly decreasing as a function of $x$, we must now show $h(s)$ is strictly decreasing for $0<s$ where
$$
h(s) = \frac{\sinh(cs)}{\sinh s}.
$$
($h$ is obtained simply by substituting $s=-\frac{1}{2}\log x$ in the $g(x)$ expression above.)
Compute:
\begin{eqnarray}
h'(s) &=& \frac{c \sinh s \cosh (cs) - \sinh(cs) \cosh s}{\sinh^2 s} \\
&=& \cosh s\cosh(cs)\frac{c\tanh s - \tanh(cx) }{\sinh^2 s}.
\end{eqnarray}
The $\cosh$ and $\sinh$ factors are positive for $s>0$. To see that
$$
p(s) = c\tanh s - \tanh(cx)
$$
is negative, note that $p(0)=0$ and $p'(s) = c(\cosh^{-2} s - \cosh^{-2} (cs))$. Since $\cosh s$ is strictly increasing for $s>0$, $p'(s)<0$, so $h(s)$ is indeed strictly decreasing in $s$. We are done.
|
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|
Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$ Prove that there are no positive integers $a,b,c,d$ such that $a^2 + b^2 = c^2$ and $a^2 - b^2 = d^2$?
Could you show me the proof?
|
I'll borrow newzad's caclulations and we'll prove that that last equation can't be a square number. We have:
$$a^2 - b^2 = n^2(k^4 - 6k^2 + 1)$$
Because $n^2$ is a square of some natural number $n$ we need to prove that $k^4 - 6k^2 + 1$ is a square number in order RHS to be also a square. First make a substitution $k^2 = t$ to reduce it to quadratic equation. And because we want it to be a square, set it to $m^2$. So we have:
$$k^4 - 6k^2 + 1 = t^2 - 6t + 1 = m^2$$
$$t^2 - 6t + (1 - m^2) = 0$$
Now solve this quadratic equation with respect to $t$. Because all coefficients are integers it'll have a integer solution, if the discriminant is a square number so we have:
$$D = b^2 - 4ac = (-6)^2 - 4(1-m^2) = 36 - 4 + 4m^2 = 2^2(9 - 1 + m^2) = 2^2(m^2 + 8)$$
Again we have that the discriminant will be a square number if $m^2 + 8$ is a square so we have:
$$m^2 + 8 = l^2$$
$$m^2 - l^2 = -8$$
$$(m-l)(m+l) = -8$$
By checking for factors of $-8$ we get that $m=\pm1$ and $l=\pm3$ is the only integer solutions.
So we can subtitute back in the initial equation and we have:
$$t^2 - 6t + 1-1 = 0$$
$$t^2 - 6t = 0$$
$$t(t-6) = 0$$
$$t_1 = 0 \quad \quad \quad t_2 = 6$$
This implies that:
$$k_{1/2} = 0 \quad \quad \quad k_{3/4} = \pm\sqrt{6}$$
But none of this is possible, because in newzad's caculation we have:
$$\frac{a}{b} + \frac{c}{b} = k$$
Because $a,b,c \in \mathbb{N}$ it's impossible for $k=0$ and also because $a,b,c$ are positive integers we have that $\frac{a}{b}$ and $\frac{c}{b}$ are rational numbers, but we know that a sum of two rational numbers, will never be an irrational numbers, so it impossible $k = \pm \sqrt{6}$
Because there aren't any more possibilities for $k$ we have that it's impossible for $a,b,c,d \in \mathbb{N}$ the following to hold:
$$a^2 + b^2 = c^2 \quad \quad \quad a^2 - b^2 = d^2$$
Q.E.D.
|
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|
Find the limit of $\sum\limits_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)$
$$\lim_{n\rightarrow\infty}\sum_{k=1}^n\left(\sqrt{1+\frac{k}{n^2}}-1\right)$$
Note that $\forall x\ge 0, \sqrt{x}-1\le\sqrt{1+x}-1\le x$
Then
$$\sum_{k=1}^n\left(\sqrt{\frac{k}{n^2}}-1\right)\le S_n\le \sum_{k=1}^n\frac{k}{n^2}=\frac{1}{2}-\frac{1}{2n}$$
How do I evaluate the sum on the left?
|
We have $$S_n=\sum_{1,n} \sqrt{1+k/n^2}-1=\sum_{1,n}\frac{k/n^2}{\sqrt{1+k/n^2}+1}$$
hence
$$\frac{n(n+1)}{2n^2 (\sqrt{1+1/n}+1)}=\sum_{1,n}\frac{k/n^2}{\sqrt{1+n/n^2}+1}\leq S_n\leq \sum_{1,n}\frac{k/n^2}{\sqrt{1}+1}=\frac{n(n+1)}{4n^2}$$
and so $\lim S_n =\frac{1}{4}$ by squeezing.
|
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|
Use the epsilon-delta definition of limits to evaluate the limit $\lim_{x\to1}\frac{x^2-x-2}{2x-3}$ The Problem: Evaluate the given limit and prove your conclusion using only definitions from the epsilon-delta limit definition: $$\lim_{x \rightarrow1}\frac{x^2-x-2}{2x-3}$$
First I evaluated the limit and found the limit of the function to be 2. For every $\varepsilon > 0$, there exists a $\delta$ so that $|x-1|< \delta \Rightarrow \frac{x^2-x-2}{2x-3}-2 < \varepsilon$. After doing some algebra on the epsilon inequality, I end up with $|\frac{x^2-5x+4}{2x-3}| < \varepsilon$. Since I can't prove the limit by defining x in terms of $\varepsilon$ and $x$, my professor has told my class that I should choose a specific value for delta for the purpose of substituting it into the inequality to find an x in terms of epsilon alone.
I let $\delta=1$, and it follows from $|x-1|<1$ that $|x|<2$, and that $|\frac{1}{2x-3}|>1$ (for eliminating the denominator. I think that $|\frac{1}{2x-3}|>1$ has set me back in solving this rigorously, since the expression is greater than, rather than less than 1. How do I proceed?
|
Note that $\frac{x^2-x-2}{2x-3} -2 = \frac{1}{2}(1-\frac{5}{2x-3})(x-1)$, so we have $|\frac{x^2-x-2}{2x-3} -2 | = \frac{1}{2}|1-\frac{5}{2x-3}||x-1|$.
The $x \mapsto |1-\frac{5}{2x-3}|$ part looks like
We see that we need to 'stay 'away' from $x=\frac{3}{2}$. If we choose $1-\frac{1}{5} < x < 1+\frac{1}{5}$, we see that $3-2x > 3-2(1+\frac{1}{5}) = \frac{3}{5}$, and so
$|1-\frac{5}{2x-3}| = |1+\frac{5}{3-2x}| < 1+ 5 (\frac{5}{3}) = \frac{28}{3}$.
Then we have
$|\frac{x^2-x-2}{2x-3} -2 | \le \frac{1}{2}\frac{28}{3}|x-1|$.
Now, in addition to $|x-1| < \frac{1}{5}$, we need another constraint on $|x-1|$ so that the last expression is less than $\epsilon$.
|
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|
If $a,b \in \Bbb R$, prove that $|ab| \le (a^2+b^2)/2$ So far I have the first case when $a=b$:
\begin{align*}
|ab|
&= |b^2|\\
&=|b|^2\\
&=\frac{2|b|^2}2\\
&=\frac{b^2+b^2}2\\
&=\frac{a^2+b^2}2
\end{align*}
Case 2: $a>b$
Case 3 $a<b$
I've been stuck on this problem for a few hours now and don't know how to proceed. Am I approaching it wrong? Should I not be thinking about the cases of $a$ in terms of $b$? Thank you for your help.
|
If $ab\ge0$ then $|ab|=ab\le ab+\dfrac{(a-b)^2}2=\dfrac{a^2+b^2}2$.
If $ab\le0$ then $|ab|=-ab\le-ab+\dfrac{(a+b)^2}2=\dfrac{a^2+b^2}2$.
|
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|
Matrix Equation $A^3-3A=\begin{pmatrix}-7 & -9\\ 3 & 2\end{pmatrix}$ How can I solve in $\mathcal{M}_{2}(\mathbb{Z})$ the equation $$A^3-3A=\begin{pmatrix}-7 & -9\\ 3 & 2\end{pmatrix}?$$
I try to use $$A^2-Tr(A)A+detA\cdot I_2=O_2$$ but I don't still obtain anything.
thanks.
|
Let $t=\operatorname{trace}(A)$ and $d=\det(A)$. By Cayley-Hamilton theorem, $A^2 = tA - dI$ and hence
$$
\pmatrix{-7&-9\\ 3&2} = A^3 - 3A = tA^2 - (d+3)A = (t^2-d-3)A - tdI.\tag{1}
$$
Taking traces on both sides, we get $-5 = (t^2-d-3)t - 2td$. Hence $t$ divides $5$, i.e. $t=1,-1,5,-5$, and we can express $d$ in terms of $t$ as $d = \frac13\left(t^2 + \frac5t\right) - 1$. Since $d$ must be an integer, $(t,d)=(1,1)$ or $(-5,7)$. However, by inspecting the $(2,1)$-th entries on both sides of $(1)$, we see that $(t^2-d-3)\mid3$. Therefore $(t,d)=(1,1)$. So, from $(1)$ we obtain
$$
\pmatrix{-7&-9\\ 3&2} = -3A - I\quad\Longrightarrow\quad A=\pmatrix{2&3\\ -1&-1}.
$$
|
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|
Prove : $\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}$ $a;b;c>0$ such that $a^2+b^2+c^2=\frac{5}{3}$. Prove :
$\frac{1}{a}+\frac{1}{b}-\frac{1}{c}< \frac{1}{abc}$
|
Hint: $(a+b-c)^2 \ge 0 \implies (a^2+b^2+c^2) \ge 2(bc + ca - ab)$
Can you relate both sides to the question at hand?
|
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|
Integrals of trignometric functions Question is to Prove that :
$$\int_0 ^{2\pi} \frac{d\theta}{a+b\sin \theta}=\frac{2\pi}{\sqrt{a^2-b^2}} \text{for}\ a>b>0$$
using residue theory.
What i have done so far is :
I transformed functio of $\theta$ as function of complex entity $z$ with $z=e^{i\theta}$.
Then, we have : $dz=i.e^{i\theta}.d\theta\Rightarrow dz=i.z.d\theta \Rightarrow d\theta =\frac{dz}{iz}$
$$\Rightarrow \int_0 ^{2\pi} \frac{d\theta}{a+b\sin \theta}=\int_{|z|\leq 1} \frac{1}{a+b.\frac{1}{2i}.(\frac{z^2-1}{z})}.\frac{dz}{iz}$$
$$\int_{|z|\leq 1} \frac{1}{a+b.\frac{1}{2i}.(\frac{z^2-1}{z})}.\frac{dz}{iz}= 2 \int_{|z|\leq 1} \frac{1}{bz^2+(2ai)z-b} dz$$
So, now the actual problem comes :
I have to find poles for this and use residue theorem to evaluate the integrals :
$bz^2+(2ai)z-b=0 \Rightarrow z=\frac{-2ai \pm \sqrt{-4a^2+4b^2}}{2b}$. But, $a>b$ so, i would like to see this as
$\frac{-2ai \pm 2i\sqrt{a^2-b^2}}{2b}=\frac{-ai \pm i \sqrt{a^2-b^2}}{b}$
Now, i am not able to decide which of $\frac{-ai + i \sqrt{a^2-b^2}}{b}$ and $\frac{-ai - i \sqrt{a^2-b^2}}{b}$ is in $\{ z\in \mathbb{C} : |z|\leq 1\}$
I tried considering their modulus..
Please help me to understand this problem.
As this is just my third problem in "solving integrals using residue theory", I do not have much experience to tackle this problem...
Once i know what are the poles in $\{ z\in \mathbb{C} : |z|\leq 1\}$ then i know how to use residue theorem and conclude required result.
Thank You.
EDIT :
we need to consider $|\frac{-ai - i \sqrt{a^2-b^2}}{b}|=\big|-i(\frac{a+\sqrt{a^2-b^2}}{b})\big|=\big|\frac{a+\sqrt{a^2-b^2}}{b}\big| \leq |\frac{a}{b}|+|\sqrt{(\frac{a}{b})^2-1}|$
As $a>b$, we have $\frac{a}{b}>1$ SO, I do not have specific upper bound for $\frac{a}{b}$ .. I am confused..
even if i have $|\frac{-ai + i \sqrt{a^2-b^2}}{b}|$ then also i can not see..
|
Consider your roots:
$$z_{\pm} = -i \left [ \frac{a}{b} \pm \sqrt{\left ( \frac{a}{b}\right)^2-1}\right]$$
Note that, when $a>b$, $a/b > 1$. Therefore $|z_+| > 1$. Note also that
$$|z_-| = \frac{a}{b} - \sqrt{\left ( \frac{a}{b}\right)^2-1} = \frac{1}{\frac{a}{b} + \sqrt{\left ( \frac{a}{b}\right)^2-1}} = \frac{1}{|z_+|} < 1$$
The residue at the pole $z=z_-$ is
$$\frac{2}{2 b z_-+i 2 a} = \frac{i}{a-\sqrt{a^2-b^2}-a} = \frac{-i}{\sqrt{a^2-b^2}}$$
The value of the integral is then $2 \pi/\sqrt{a^2-b^2}$, as was to be demonstrated.
|
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|
What is $\int \frac1{1+(a\tan x)^2}dx$? What is $\int \frac1{1+(a\tan x)^2} \mathrm dx$?
This is a difficult integral. If you can, please give a step-by-step solution - I would be delighted.
|
It's not that hard if you take the suggestion from Clayton:
$u=a \tan{x}$; $x = \arctan{(u/a)}$; $dx=a\, du/(u^2+a^2)$. Then the integral is
$$a \int \frac{du}{(u^2+a^2)(u^2+1)} $$
Assume $a \ne 1$. Use partial fractions:
$$\frac{1}{u^2+a^2}-\frac{1}{u^2+1} = \frac{1-a^2}{(u^2+a^2)(u^2+1)}$$
so the integral becomes
$$\frac{a}{1-a^2} \int du \left (\frac{1}{u^2+a^2}-\frac{1}{u^2+1} \right ) = \frac{a}{1-a^2} \left (\frac{1}{a} \arctan{\frac{u}{a}} - \arctan{u} \right )+C$$
Therefore
$$\int \frac{dx}{1+a^2 \tan^2{x}} = \frac{1}{1-a^2} \left (x - a \arctan{(a \tan{x})}\right )+C $$
When $a=1$, however, we get that the integral is
$$\int \frac{dx}{1+\tan^2{x}} = \int dx \cos^2{x} = \frac{x}{2} + \frac12 \sin{x} \cos{x} + C$$
|
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|
Finding element in binary representation of $GF(2^6)$ I got the following task:
Let $F = GF(2^6
)$ be K[x] modulo the primitive polynomial $h(x) = 1 +x
^2 +x
^3 +x
^5 +x
^6$
, and let $\alpha$ be the
class of x. I have a table with the binary representation of the elements of $GF(2^6
) = \{0; 1; \alpha^1; \alpha^2
; ...; \alpha^{62}\}$.
Is there a quicker way to check $\alpha^{52}$ than performing a long division of $x^{52}$ by $h(x)$ ?
It has $2^6$ elements so after $\alpha^{62}$ it will be back to 1 again, but maybe there is some trick to find it faster rather than performing such a long division?
I have also $\alpha^{13}$ so it's $\alpha^{13^4}$, but no idea how to use it
|
You say you have $\alpha^{13}$, so you can substitute suitable values into $\alpha^{13} = b_0 + b_1 \alpha + \ldots + b_5 \alpha^5$. Then $$\begin{eqnarray}\alpha^{26} & = & (\alpha^{13})^2 \\
& = &(b_0 + b_1 \alpha + \ldots + b_5 \alpha^5)^2 \\
& = &b_0 + b_1\alpha^2 + \ldots + b_5 \alpha^{10}
\end{eqnarray}$$
where the cross-terms cancel and so do the squares of the coefficients, because we're in characteristic $2$.
But we have $\alpha^6 = 1 + \alpha^2 + \alpha^3 + \alpha^5$, whence $\alpha^8 = 1 + \alpha$ and $\alpha^{10} = \alpha^2 + \alpha^3$. So
$$\begin{eqnarray}
\alpha^{26} & = & b_0 + b_1\alpha^2 + b_2\alpha^4 + b_3(1 + \alpha^2 + \alpha^3 + \alpha^5) + b_4(1 + \alpha) + b_5 (\alpha^2 + \alpha^3) \\
& = & (b_0+b_3+b_4) + (b_4)\alpha + (b_1+b_3+b_5)\alpha^2 + (b_3+b_5)\alpha^3 + (b_2)\alpha^4 + (b_3)\alpha^5
\end{eqnarray}$$
Repeat for $\alpha^{52}$.
PS Minor nitpick: $2^6-1=63$, not $62$.
|
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|
How is $\sqrt{\frac{\sqrt{3}+2}{4}} = \frac{\sqrt{2}(\sqrt{3}+1)}{4}$ How is $\sqrt{\frac{\sqrt{3}+2}{4}} = \frac{\sqrt{2}(\sqrt{3}+1)}{4}$?
(Prove by using algebraic manipulation not by calculation)
I've tried to come up with something myself but I can't find a solution, I must be missing something.
|
$$\begin{align}\sqrt{\frac{\sqrt{3}+2}{4}} &= \frac{\sqrt{2}(\sqrt{3}+1)}{4}\\
{\frac{\sqrt{3}+2}{4}} &= \frac{{2}(\sqrt{3}+1)^2}{16}\\
{\frac{\sqrt{3}+2}{4}} &= \frac{4+2\sqrt{3}}{8}\\
{\frac{\sqrt{3}+2}{4}} &= \frac{\sqrt{3} + 2}{4}\tag*{$\blacksquare$}\end{align}$$
|
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|
Derivative of trigonometric function How i can find the derivative of this trigonometric function
$csc^4(8x^4-5)$
i tried to do it my self and i got to this
$ 4[csc(8x^4-5)]^3 * [-csc(8x^4-5)cotan(8x^4-5)] $
The answer in the book is $ -128 x^3csc^4(8x^4-5)cot an(8x^4-5)$
|
$$\csc^4(8x^4-5)=\frac{1}{\sin^4(8x^4-5)}=-\frac{(\sin^4(8x^4-5))'}{\sin^8(8x^4-5)}=-\frac{4\sin^3(8x^4-5)\cdot (\sin(8x^4-5))'}{\sin^8(8x^4-5)}$$
$$=-\frac{4\sin^3(8x^4-5)\cdot \cos(8x^4-5)(8x^4-5)'}{\sin^8(8x^4-5)}=-\frac{4\sin^3(8x^4-5)\cdot \cos(8x^4-5)\cdot 32x^3}{\sin^8(8x^4-5)}$$
$$=-\frac{128 x^3\cdot \cos(8x^4-5)}{\sin^5(8x^4-5)}=-128x^3\frac{\cos(8x^4-5)}{\sin(8x^4-5)}\cdot \frac{1}{\sin^4(8x^4-5)}$$
$$=-128x^3\cot(8x^4-5)\csc^4(8x^4-5)$$
|
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|
Error analysis of exponential function By definition: $$ e^x = \lim_{n \rightarrow \infty} ( 1 + \frac{x}{n} ) ^ n$$
I am interesting in calculating the error $$\left | e^x - \left( 1 + \frac{x}{n} \right) ^ n \right|$$ for some fixed $n \in \mathbb{N}$.
What I've done so far:
First note:
$$e^x = \sum_{k = 0}^\infty \frac{x^k}{k!} = 1 + x + \frac{x^2}{2!} + \dots \frac{x^n}{n!} + O\left(\frac{1}{(n+1)^{(n+1)}}\right)$$ and
$$\left( 1 + \frac{x}{n} \right) ^ n = \sum_{k = 0}^n {n \choose k} \frac{x^k}{n^k} = 1 + x + \frac{x^2}{2!}\left(1 - \frac{1}{n} \right) + \dots + \frac{x^n}{n!}\left(1 - \frac{1}{n} \right) \dots \left(1 - \frac{n-1}{n} \right)$$
Subtracting term by term yields $$\frac{x^2}{2!}\left(\frac{1}{n} \right) + \frac{x^3}{3!}\left(\frac{3}{n} - \frac{2}{n^2}\right) + \frac{x^4}{3!}\left(\frac{6}{n} - \frac{11}{n^2} + \frac{6}{n^3}\right) + \dots + O\left (\frac{1}{(n+1)^{(n+1)}}\right)$$
Where can I got from here? I want to be able to calculate the exact difference, and I don't think the expression above is good enough for that task.
Any help would be highly appreciated.
|
I take a simple-minded approach: whenever practical, write the Taylor formula with Lagrange form of remainder. Since $(1+x/n)^n=\exp(n \ln(1+x/n))$, I begin with the logarithm:
$$
\ln(1+x/n) = \frac{x}{n}+\frac{(x/n)^2}{2}\frac{-1}{(1+\xi)^2}
\tag{1}$$
where $\xi$ is between $0$ and $x/n$. Formula (1) is Taylor's formula applied to $t\mapsto \ln(1+t)$.
So,
$$
(1+x/n)^n = e^x \exp\left(-\frac{x^2}{2n} \frac{1}{(1+\xi)^2}\right)
\tag{2}$$
which is already a reasonable error estimate. When $x>0$, it can be brought into the form you seek without too much sweat. Since $\xi>0$ in (2), we have
$$
0<e^x-(1+x/n)^n < e^x(1-e^{-x^2/(2n)})
\tag{3}$$
Use the Mean Value theorem (i.e., the simplest form of the Lagrange remainder formula):
$$
1-e^{-x^2/(2n)} = \frac{x^2}{2n} e^{-\zeta},\quad
0<\zeta<\frac{x^2}{2n}
\tag{4}$$
Hence, the conclusion: for $x>0$,
$$
0<e^x-(1+x/n)^n < \frac{x^2}{2n}e^x
\tag{5}$$
Of course, (5) is not telling us anything new when $x^2>2n$. But it correctly shows that $x^2/n$ is what determines the quality of approximation. This quadratic relation is visible on the spreadsheet below, where I calculated the difference $e^x-(1+x/n)^n$ and colored the cells where the difference is at least $1$.
|
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|
Conjecture $\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}\stackrel?=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi$ $$\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[4]x\ \sqrt[4]{2-x\,\sqrt3}}\stackrel?=\frac{2\,\sqrt2}{3\,\sqrt[8]3}\pi\tag1$$
The equality numerically holds up to at least $10^4$ decimal digits.
Can we prove that the equality is exact?
An equivalent form of this conjecture is
$$I\left(\frac{\sqrt3}2;\ \frac14,\frac14\right)\stackrel?=\frac23,\tag2$$
where $I\left(z;\ a,b\right)$ is the regularized beta function.
Even simpler case:
$$\int_0^1\frac{\mathrm dx}{\sqrt{1-x}\ \sqrt[6]{9-x}\ \sqrt[3]x}\stackrel?=\frac\pi{\sqrt3},\tag3$$
which is equivalent to
$$I\left(\frac19;\ \frac16,\frac13\right)\stackrel?=\frac12.\tag4$$
A related question.
|
Here is another approach for evaluating the integral (3).
Using @achille hui's transformation, we can write
$$
I=\int_0^1\frac{dx}{\sqrt[3]{x}\sqrt{1-x}\sqrt[6]{9-x}}= \frac{4\pi^2 2^{\frac{1}{3}}}{\Gamma\left(\frac{1}{3}\right)^3}\int_0^\frac{1}{\sqrt{2}}\frac{1}{\sqrt{1+x^6}}dx
$$
Using the substitution $x=\sqrt{t}$, we get
$$
I=\frac{\pi^2 2^{\frac{4}{3}}}{\Gamma\left(\frac{1}{3}\right)^3}\int_0^\frac{1}{2}\frac{1}{\sqrt{t+t^4}}dt
$$
We can now transform this integral into a beta integral using the substitution $t=\frac{1-y}{2+y}$. This gives us
\begin{align*}
I
&= \frac{\pi^2 2^{\frac{4}{3}}}{\Gamma\left(\frac{1}{3}\right)^3} \int_0^1\frac{1}{\sqrt{1-y^3}}dy \\
&= \frac{\pi^2 2^{\frac{4}{3}}}{3\Gamma\left(\frac{1}{3}\right)^3}B\left(\frac{1}{3},\frac{1}{2} \right) \\
&= \frac{\pi^2 2^{\frac{4}{3}}}{3\Gamma\left(\frac{1}{3}\right)^3}\left( \frac{\sqrt{3}\Gamma\left(\frac{1}{3}\right)^3}{\pi 2^{\frac{4}{3}}}\right) \\
&= \frac{\pi}{\sqrt{3}}
\end{align*}
|
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|
How to derive this second derivative using the quotient rule? If a given first derivative is: $\ {dy \over dx} = {-48x \over (x^2+12)^2} $
What are the steps using the quotient rule to derive the second derivative: $\ {d^2y \over dx^2} = {-144(4-x^2) \over (x^2+12)^3} $
My Steps:
\begin{align*}
{d^2y \over dx^2}
&= {-48(x^2 +12)^2 - 2(x^2+12)(2x)(-48x)\over (x^2+12)^4} \\
&=-48{(x^2 +12)^2 - 4x^2(x^2+12)\over (x^2+12)^4} \\
&= -48{(x^2 +12)( - 4x^2 +(x^2+12))\over (x^2+12)^4} \\
&= -48{(x^2 +12)( - 4x^2)\over (x^2+12)^3} \\
&= {???}
\end{align*}
|
First, make sure you know what the quotient rule is - which you can derive from the product rule say by taking the derivative of $g(x)/f(x)$.
Then, simplify. Looking at the product rule, expect to have to cancel a factor of $x^2+12$ from the top and bottom.
|
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|
Polar coordinates differential equation I have the following ODE:
$$\dot x=-y(x^2+y^2), \dot y=x(x^2+y^2)$$
I want to sketch the phase portrait (manually) and I want to find the flow $\phi_t$, the orbit $O(x_0)$ and the limit set $\omega(x_0)$
I start by taking polar coordinates and change the system to $\dot r=-r^3\sin\theta, \dot\theta=r^3\cos\theta$
The phase portrait then looks like the one a stable centre, right?
How can I continue to find the flow of the function, i.e the solution of the differential equation?
|
We are given:
$$\dot x=-y(x^2+y^2), \dot y=x(x^2+y^2)$$
Recall, when we are doing polar coordinates, we have
$$x = r \cos \theta, y = r \sin \theta, x^2+y^2 = r^2$$
When we differentiate this, we have:
$$2 x x' + 2 y y' = 2 r r'$$
This gives us:
$$rr' = x(-y(x^2+y^2)) + y(x(x^2+y^2)) = 0 \rightarrow r' = 0$$
To find the angle, we take:
$$\dfrac{r \sin \theta}{r \cos \theta} = \tan \theta = \dfrac{y}{x}$$
Using the quotient rule gives us:
$$\theta' = \dfrac{x y' - y x'}{r^2} = \dfrac{x^2(x^2+y^2)+y^2(x^2+y^2)}{r^2}= \dfrac{r^4}{r^2} = r^2$$
Thus in polar coordinates, the original system becomes
*
*$r' = 0$
*$\theta' = r^2$
Can you now solve this system and draw the phase portrait?
The phase portrait should look like:
|
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|
How find the minimum of the $abc$ let $a,b,c>2$ be prime numbers,such
$$a|b^5+1,b|c^5+1,c|a^5+1$$
Find the minimum of $abc$
My try:
I think use
$$a^{p-1}=1(modp)$$
But I don't,Thank you for your help.
|
Note that $a, b, c$ are pairwise relatively prime and thus pairwise distinct.
If $\min(a, b, c) \geq 7$, WLOG assume that $a=\max(a, b, c) \geq 13$, so $a>b \pm 1$. Now $b^5 \equiv -1 \pmod{a}, b^{10} \equiv 1 \pmod{a}, b \not \equiv \pm 1 \pmod{a}, b^2 \not \equiv 1 \pmod{a}$. Thus the order of $b \pmod{a}$ is $10$, so since by Fermat's little theorem $b^{a-1} \equiv 1 \pmod{a}$, we have $10 \mid a-1$. Thus $a \geq 31$. We get $abc \geq 7(11)(31)=2387$.
If $\min(a, b, c)=5$, WLOG assume that $a=\min(a, b, c)=5$. Then $c \mid 5^5+1=3126=6(521)$ so $c=521$ since $c>a=5$. Then $abc=5(521)b>5(521)(2)=5210$.
If $\min(a, b, c)=3$, WLOG assume that $a=\min(a, b, c)=3$. Then $c \mid 3^5+1=244=4(61)$ so $c=61$ since $c>2$. Now $5 \leq b, 3 \mid b^5+1$ and $b \mid 61^5+1$. We have $5 \nmid 61^5+1$ and $3 \nmid 7^5+1$ clearly so $b \geq 11$ and $abc \geq 3(11)(61)=2013$. Indeed $(a, b, c)=(3, 11, 61)$ works, so the minimum value of $abc$ is $2013$.
|
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|
Calculate $a^3+b^3+c^3+d^3$ for the real roots of $x^4+2x^3-3x^2-3x+2$ $a,b,c,d \in \mathbb{R}$ are the real roots of $x^4+2x^3-3x^2-3x+2$.
Calculate $a^3+b^3+c^3+d^3$.
With approximation i found out, that $a^3+b^3+c^3+d^3 = -17$, but how can I proof that without calculating the roots exactly?
Cheers
|
As biquadratic equation has exactly $4$ roots, all of $a,b,c,d$ are real
Clearly, $abcd\ne0$
and $\displaystyle a^4+2a^3-3a^2-3a+2=0\implies a^3+2a^2-3a-3=-\frac2a$
Similarly, for $b,c,d$
Summing we get $\sum a^3+2\sum a^2-3\sum a-3\sum 1=-2\sum \frac 1a$
Now, we need
$\sum a$
$\displaystyle\sum\frac1a=\frac{\sum abc}{abcd}$
$\displaystyle\sum a^2=(\sum a)^2-2\sum ab$
Vieta's formulas is calling for usage
|
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|
Prove that $C_n < 4n^2$ for all n greater than or equal to 1 $C_1 = 0$, $C_n = C_{\lfloor n/2\rfloor} + n^2$ for all $n \ge 1$
Prove that $C_n < 4n^2$ for all $n \ge 1$.
I don't know how to even approach this. I remember something about inductive proofs...but i really don't understand that, could you please explain that to me?
|
Why not unroll the recursion to get a precise answer?
Let $$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k$$
be the binary representation of $n.$
This yields the exact formula
$$C_n = \sum_{k=0}^{\lfloor \log_2 n \rfloor - 1}
\left(\sum_{j=k}^{\lfloor \log_2 n \rfloor} d_j 2^{j-k}\right)^2$$
where $n\ge 2$ and zero otherwise.
Clearly this is maximized when all digits are equal to one, giving
$$C_n\le \sum_{k=0}^{\lfloor \log_2 n \rfloor - 1}
\left(\sum_{j=k}^{\lfloor \log_2 n \rfloor} 2^{j-k}\right)^2
= \frac{16}{3} 2^{2 \lfloor \log_2 n \rfloor}
- 8 \times 2^{\lfloor \log_2 n \rfloor} + \lfloor \log_2 n \rfloor
+ \frac{8}{3}. $$
For a string of ones we have $n=2^j-1$ for some $j$ with $j-1 = \lfloor \log_2 n \rfloor.$ This gives
$$2^{\lfloor \log_2 n \rfloor} = \frac{1}{2} (n+1)
\quad\text{or}\quad
\lfloor \log_2 n \rfloor = \log_2(n+1) - 1.$$
Hence we have the following bound on the leading term
$$ \frac{16}{3} 2^{2 \lfloor \log_2 n \rfloor}
= \frac{16}{3} 2^{2 \log_2(n+1) - 2}
= \frac{4}{3} (n+1)^2.$$
Furthermore recalling the condition on $n$ we have
$$- 8 \times 2^{\lfloor \log_2 n \rfloor} = -8 \times \frac{1}{2} (n+1).$$
Putting these pieces together we find that
$$C_n\le \frac{4}{3}n^2-\frac{4}{3}n + \lfloor \log_2 n \rfloor
< \frac{4}{3}n^2$$
because we certainly have $4/3 \times n > \log_2 n.$
To conclude, let us briefly consider the lower bound, which occurs when there is a one followed by a string of zeros, giving
$$C_n\ge \sum_{k=0}^{\lfloor \log_2 n \rfloor - 1}
\left(2^{\lfloor \log_2 n \rfloor - k}\right)^2
= \frac{4}{3} 2^{2\lfloor \log_2 n \rfloor} -\frac{4}{3}.$$
In this scenario we have $\lfloor \log_2 n \rfloor = \log_2 n,$ so that the lower bound gives $$C_n\ge \frac{4}{3} n^2 -\frac{4}{3}.$$
Note that the bounds from the two bit patterns are no longer directly comparable because the substitutions for $\lfloor \log_2 n \rfloor$ are different. What we may say, however, is that $$C_n \sim \frac{4}{3} n^2.$$
The coefficient on the leading term of the bounds in $\lfloor \log_2 n \rfloor$ fluctuates between $4/3$ and $16/3$ on each interval where $\lfloor \log_2 n \rfloor$ is constant. However when we reach the end of this interval $\lfloor \log_2 n \rfloor$ is off by almost one from the true value, reducing $16/3$ to $4/3$ when the bounds are expressed in $n.$
This link points to a series of similar calculations.
|
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|
Trig equations solution Solving the following for x:
$$
\frac{3\cos(2x)+5\cos(x)-1}{\sqrt{-\cot(x)}}=0
$$
The solution says that the answer is $x=-\frac{\pi}{3}+2\pi k$ where $k$ is an integer.
I am not sure why there is the minus sign in front. Can someone please help me out?
|
First off, there is a domain issue with the function on the left-hand side of the equation, since the denominator is undefined wherever $ \ \cot x \ \ge \ 0 \ . $ Since the tangent function is positive in the first and third quadrants, so is the cotangent function; also, $ \ \cot x \ = \ 0 \ $ wherever $ \ \cos x \ = \ 0 $ and undefined where $ \ \sin x \ = \ 0 $ . Putting this all together tells us that there are only solutions to the equation for $ \ (\frac{2n-1}{2}) \cdot \pi \ < \ x \ < \ n \cdot \pi \ , $ the second and fourth quadrants without their boundaries.
The numerator is zero for
$$ \ 3 \ \cos(2x) \ + \ 5 \cos x \ - \ 1 \ = \ 3 \ (2 \cos^2 x \ - \ 1 ) + \ 5 \cos x \ - \ 1 $$
$$ = \ 6 \cos^2 x \ + \ 5 \cos x \ - \ 4 \ = \ (3 \ \cos x \ + \ 4 ) \ \cdot \ ( 2 \ \cos x \ - \ 1) \ = \ 0 . $$
The first factor is never zero, but the second one is for $ \ \cos x \ = \ \frac{1}{2} \ \Rightarrow \ x \ = \ \pm \frac{\pi}{3} + 2k\pi \ , $ as Oliver has said. However, the $ + \frac{\pi}{3} + 2k\pi \ $ "family" of solutions is barred since those are in the first quadrant. Thus, the permissible family of solutions is $ - \frac{\pi}{3} + 2k\pi \ . $
|
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How to find $f$ if $f(f(x))=\frac{x+1}{x+2}$ let $f:\mathbb R\to \mathbb R$,and such
$$f(f(x))=\dfrac{x+1}{x+2}$$
Find the $f(x)$
My try
I found $f(x)=\dfrac{1}{x+1}$
because when $f(x)=\dfrac{1}{x+1}$,then
$$f(f(x))=f\left(\dfrac{1}{x+1}\right)=\dfrac{1}{\dfrac{1}{x+1}+1}=\dfrac{x+1}{x+2}$$
so $f(x)=\dfrac{1}{x+1}$ such this condition,But $f(x)$ Have other form? Thank you
|
Suppose $f(x)=\frac{ax+b}{x+c}$. Then
$$
f(f(x))= \frac{(a^2+b)x+b(x+c)}{(a+c)x + b+c^2}=\frac{x+1}{x+2}
$$
yields the equations
$$
\frac{a^2+b}{a+c}=1, \frac{b(a+c)}{a+c}=b=1, \frac{b+c^2}{a+c}=2, \frac{a^2+1}{a+c}=1$$
from which we find
$$c=a^2-a+1$$
and hence
$$a^4-2a^3+a^2-2a=a(a^3-2a^2+a-2)=0$$
This last equation has but two real solutions, $a=0$ and $a=2$, corresponding to the two solutions given by sos440: $$ f(x)=\frac{1}{x+1} \mbox{ and } f(x)=\frac{2x+1}{x+3}.$$
Added: The last equation has $a=\pm i$ as roots, but these yield the degenerate functions $f(x)=i$ and $f(x)=-i$ which do not yield the required $f(f(x))$.
|
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|
Evaluation of $\int^1_0 \cos^2\frac{(m+n)\pi x}{2}\sin^2\frac{(n-m)\pi x}{2}dx$. Just wondering whether the following integration is something special. To be more specific, is it equal to some constant real number, please? I found a integration table involving sine and cosine but it requires that sine and cosine have the same argument, which is clearly not the case here.
$$\int^1_0 \cos^2\frac{(m+n)\pi x}{2}\sin^2\frac{(n-m)\pi x}{2}dx$$
|
First notice that $$\cos\frac{(m+n)\pi x}{2}\sin \frac{(n-m)\pi x}{2} = \frac{1}{2} \left( \sin n\pi x -\sin m\pi x\right)$$
From this, we deduce that the given integral formula is equal to
$$\frac{1}{4}\int_0^1 {\left( \sin n\pi x -\sin m\pi x\right)}^2dx \\= \frac{1}{4}\int_0^1 { \sin^2 n\pi x +\sin^2 m\pi x - \sin n\pi x \sin m \pi x}dx \\ = \frac{1}{4}\int_0^1 { \frac{1-\cos 2n\pi x}{2} +\frac{1-\cos 2m\pi x}{2} + \frac{1}{2}\left( \cos (n+m)\pi x - \cos (n-m)\pi x\right)}dx$$
The calculations left are easy.
|
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|
Function range problem $(sin^2\theta +\sin\theta -1)/(sin^2\theta -\sin\theta+ 2)$ .
It is asked to find the range of this function.
Here i was assumed a variable
$z=sin\theta$ so that i get
$-1<=z<=1 $
and then i obtained a quadratic equation as :-
$(y-1)z^2 -(y+1)z + 2y +1 =0$ from here i need some help or if there's some other way for this problem that might also help. Thanks!
|
Since $z^2 - z + 2 = (z - 1/2)^2 + 7/4$, the given function's domain is $\mathbb{R}$.
$$f(z) = \frac{z^2 + z - 1}{z^2 - z + 2} = 1 + \frac{2z- 3}{z^2 - z + 2}$$
Let us differentiate the function.
$$f'(z) = \frac{(2z - 3)' (z^2 - z + 2) - (2z - 3) (z^2 - z + 2)'}{(z^2 - z + 2)^2} \\= \frac{2z^2 - 2z + 4 - (2z - 3)(2z - 1)}{\text{A positive number}} = - \frac{2z^2 - 6z - 1}{\text{A positive number}}$$
What can you talk about the shape of $f(z)$? When $f(z)$ has local maximum/minimum?
|
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Solve for $\sin^2(x) = 3\cos^2(x)$ I am trying to solve the following equation for x. The textbook's answer and my answer differ and I keep getting the same answer. Could someone please tell me what I'm going wrong? Notice that this is "sin squared x" and 3 * "cos squared x"
$\sin^2x = 3\cos^2x$ //Just rewriting the equation again
$1-\cos^2x = 3\cos^2x$ //Using the Pythagorean identities to substitute in for $\sin^2x$
I then add $\cos^2x$ to both sides yielding: $$1 = 4\cos^2x$$
I then divide by $4$ yielding: $$\frac 1 4 = \cos^2x$$
I then take the square root of both sides yielding: $$\frac 1 2 = \cos x$$
Then, I determine that the places where the $\cos x$ is positive $(1/2)$ is $\pi/3$ and $5\pi/3$
The textbook's answer is $\pi/3$ and $2\pi/3$. However the $\cos(2\pi/3)$ is $-(1/2)$ meaning that I must've solved the equation incorrectly.
Does anyone see my mistake?
Thanks!
|
we know that $\sin^2x=3\cos^2x$ so$$1-\cos^2x=3\cos^2x \Rightarrow 4\cos^2x=1 \Rightarrow \cos^2x=\frac{1}{4} \Rightarrow \cos x=\pm \frac{1}{2}$$
and now we calculate values of $x$
$$\mbox{if} \qquad \cos x=\frac{1}{2},\mbox{ then}\qquad x=\arccos\left(\frac{1}{2}\right)=\frac{\pi}{3},\frac{5\pi}{3}$$
$$\mbox{if} \qquad \cos x=-\frac{1}{2},\mbox{ then}\qquad x=\arccos\left(-\frac{1}{2}\right)=\frac{2\pi}{3},\frac{4\pi}{3}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$ are linearly dependent.
Functions $f$ and $g$ are independent on an interval $D$ if $af(x) + bg(x) = 0$ implies that $a = 0$ and $b = 0$ $\forall x \in D$
let $\alpha$, $\beta$, $\gamma$ be real constants. Prove that
$\sin(x + \alpha)$, $\sin(x + \beta)$ and $\sin(x + \gamma)$
are linearly dependent vectors in $C^0[0, 1]$. Be convincing in your reasoning (argument)
I was researching and found Wronskian. Using the Wronskian for three functions. The determinant of $f$, $g$ and $h$ is $W(f, g, h) = $
$$
\begin{vmatrix}
f & g & h \\
f' & g' & h' \\
f'' & g'' & h'' \\
\end{vmatrix}
$$
If $W(f, g, h) \neq 0$ then $f(x)$, $g(x)$ and $h(x)$ are linearly independent.
If $f(x)$, $g(x)$, and $h(x)$ are linearly dependent then $W(f, g, h) = 0$
My attempt
Let
$f(x) = \sin(x + \alpha)$, $g(x) = \sin(x + \beta)$ and $h(x) = \sin(x + \gamma)$
$W(f, g, h) =$
$$
\begin{vmatrix}
\sin(x + \alpha) & \sin(x + \beta) & \sin(x + \gamma) \\
\cos(x + \alpha) & \cos(x + \beta) & \cos(x + \gamma) \\
-\sin(x + \alpha) & -\sin(x + \beta) & -\sin(x + \gamma) \\
\end{vmatrix}
$$
$= \sin(x + \alpha)[-\sin(x + \gamma)\cos(x + \beta) + \cos(x + \gamma)\sin(x + \beta)] - sin(x + \beta)[-\sin(x + \gamma)\cos(x + \alpha) + \cos(x + \gamma)\sin(x + \alpha)] + \sin(x + \gamma)[-\sin(x+ \beta)\cos(x + \alpha) + \cos(x + \beta)\sin(x + \alpha)]$
$= -\sin(x + \alpha)[\sin((x + \gamma) +(x + \beta))] + \sin(x + \beta)[\sin((x + \gamma) + (x + \alpha))] - \sin(x + \gamma)[\sin((x + \beta) + (x + \alpha))] = 0$
By Wronskian, $f(x)$, $g(x)$ and $h(x)$ are linearly dependent since $W(f, g, h) = 0$
Not sure if this argument is sound?
|
As Mohamed pointed out in his answer, we can use the angle addition formula for $\sin$, viz.:
$\sin (x + \alpha) = \cos \alpha \sin x + \sin \alpha \cos x, \tag{1}$
$\sin (x + \beta) = \cos \beta \sin x + \sin \beta \cos x, \tag{2}$
$\sin (x + \gamma) = \cos \gamma \sin x + \sin \gamma \cos x, \tag{3}$
to see that all three functions lie in the subspace generated by $\sin x$ and $\cos x$, which is of course $\text{Span} \{ \sin x, \cos x \}$, and since $\sin (x + \alpha)$, $\sin (x + \beta)$, $\sin (x + \gamma)$ are thus three vectors in a subspace of dimension two, a linear dependence between them must exist.
Going a step further, we can actually solve for $\sin x$, $\cos x$ in terms of $\sin (x + \alpha)$, $\sin(x + \beta)$ and thus find the linear dependence explicitly. We have, from
(1) and (2),
$\begin{bmatrix} \cos \alpha & \sin \alpha \\ \cos \beta & \sin \beta \end{bmatrix} \begin{pmatrix} \sin x \\ \cos x \end{pmatrix} = \begin{pmatrix} \sin (x + \alpha) \\ \sin(x + \beta) \end{pmatrix}; \tag{4}$
the inverse of
$\begin{bmatrix} \cos \alpha & \sin \alpha \\ \cos \beta & \sin \beta \end{bmatrix} \tag{5}$
is readily seen to be
$(\cos \alpha \sin \beta - \sin \alpha \cos \beta)^{-1}\begin{bmatrix} \sin \beta & -\sin \alpha \\ -\cos \beta & \cos \alpha \end{bmatrix} = (\sin(\beta - \alpha))^{-1}\begin{bmatrix} \sin \beta & -\sin \alpha \\ -\cos \beta & \cos \alpha \end{bmatrix}, \tag{6}$
from which it follows that
$\begin{pmatrix} \sin x \\ \cos x \end{pmatrix} = (\sin(\beta - \alpha))^{-1}\begin{bmatrix} \sin \beta & -\sin \alpha \\ -\cos \beta & \cos \alpha \end{bmatrix}\begin{pmatrix} \sin (x + \alpha) \\ \sin(x + \beta) \end{pmatrix}. \tag{7}$
(7) gives $\sin x$, $\cos x$ in terms of $\sin(x + \alpha)$, $\sin(x + \beta)$; the resulting formulas may then be inserted into (3) to express $\sin(x + \gamma)$ in terms of $\sin(x + \alpha)$, $\sin(x + \beta)$; I leave the necessary algebra to my readers. One must of course remain vigilant to the possibility that $\sin(\beta - \alpha) = 0$; but this
is a special case best handled on its own.
Hope this helps. Cheerio,
and as always,
Fiat Lux!!!
|
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|
Integral $\int_0^1\frac{\ln x}{\left(1+x\right)\left(1+x^{-\left(2+\sqrt3\right)}\right)}dx$ There is a curious known integral:
$$\int_0^1\frac{\ln\left(1+x^{2+\sqrt{3\vphantom{\large3}}}\right)}{1+x}dx=\frac{\pi^2}{12}\left(1-\sqrt{3\vphantom{\large3}}\right)+\ln \left(1+\sqrt{3\vphantom{\large3}}\right)\ln2.$$
If we consider $\alpha=2+\sqrt{3\vphantom{\large3}}$ as a parameter and take a derivative w.r.t. $\alpha$ at this point, we get the following:
$$I=\int_0^1\frac{\ln x}{\left(1+x\right)\left(1+x^{-\left(2+\sqrt{3\vphantom{\large3}}\right)}\right)}dx.$$
Is it possible to express the integral $I$ in a closed form?
|
Here is a partial progress report. I am basically repeating Jim Belk's analysis from the previous answer.
Set $F(a) = \int_{x=0}^1 \frac{\log(1+x^a)}{1+x} dx$. Then
$$F(a) = \int_{x=0}^1 \int_{y=0}^{x^a} \frac{dx dy}{(1+x)(1+y)} = \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)}$$
so
$$F(a) + F(a^{-1}) = \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)} + \int_{0 \leq y \leq x^{1/a} \leq 1} \frac{dx dy}{(1+x)(1+y)}$$
$$= \int_{0 \leq y \leq x^a \leq 1} \frac{dx dy}{(1+x)(1+y)} + \int_{0 \leq y^a \leq x \leq 1} \frac{dx dy}{(1+x)(1+y)} = \int_{0 \leq x,y \leq 1} \frac{dx dy}{(1+x)(1+y)} = (\log 2)^2.$$
(In order to combine the integrals, first switch the names of $x$ and $y$ in the second one.)
So
$$F'(a) - a^{-2} F'(a^{-1})=0.$$
This gives a linear relation between $F'(2 + \sqrt{3})$ and $F'(2-\sqrt{3})$. If we find a second one, we can solve the linear equations and be done.
Notice that
$$F'(a) = \int_{x=0}^1 \frac{x^a \log x dx}{(1+x)(1+x^a)} = \sum_{m,n=0}^{\infty} \int_{x=0}^1 (-1)^{m+n} x^{m+(n+1) a} \log x dx.$$
Integrating by parts, $\int_{x=0}^1 x^b \log x dx = \frac{-1}{(b+1)^2}$. So, ignoring issues of convergence, we should have
$$F'(a) = \sum_{m,n=0}^{\infty} \frac{(-1)^{m+n+1}}{(m+(n+1) a + 1)^2} = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1}}{(m+n a)^2}$$
In the last step, we turned $m+1$ and $n+1$ into $m$ and $n$ to make things pretty. My guess is that the convergence issues can be dealt with for any $a>0$, but I haven't thought much about it.
So
$$F'(a) + F'(a^{-1}) = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \left( \frac{(-1)^{m+n+1}}{(m+n a)^2} +\frac{(-1)^{m+n+1}}{(m+n a^{-1})^2} \right).$$
Putting $a=2 + \sqrt{3}$, this is
$$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} (-1)^{m+n+1} \frac{2 (m^2+4mn+7n^2)}{(m^2+4mn+n^2)^2} $$
$$= 2 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1}}{m^2+4mn+n^2} +12 \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{m+n+1} n^2}{(m^2+4mn+n^2)^2}.$$
Here is where I run out of ideas. The first sum is basically the one at the end of Jim Belk's post, but I have no ideas for the second one.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to prove this inequality $(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$? Question:
If $a,b,c$ are nonnegative real numbers such that $a+b+c=3,$ then
$$(a^2+bc^4)(b^2+ca^4)(c^2+ab^4) \leq 64$$
My try: I found the equality holds only if $(a,b,c)=(2,0,1)$ or all of its permutations.
But I can't prove this inequality it. I would appreciate very much a proof.
This problem comes from:http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&t=562119
|
Without loss of generality, assume $a$ is smallest of $a, b, c$. Also, let $$f(a,b,c)=(a^2+bc^4)(b^2+ca^4)(c^2+ab^4)$$and firstly, if $a\le c\le b$, then$$f(a,b,c)-f(a,c,b)=(b^3-a^3) (c^3 - a^3) (b ^3- c^3) (a b c - 1)<0$$therefore we can assume $a\le b\le c$. Now, we will prove $$f(0,b,a+c)\ge f(a,b,c)$$which is, after full expansion,$$a (-a^6 b^4 c + a^5 b^3 - a^5 c^3 - a^4 b^5 c^5 - a^3 b c^7 - a^2 b^6 + 20 a^2 b^3 c^3 \\+ 15 a b^3 c^4 - a b^2 c^2 - b^7 c^4 + 6 b^3 c^5)+ 6 a^5 b^3 c + 15 a^4 b^3 c^2\ge0$$
and it is obvious that $-a^6 b^4 c + a^5 b^3\ge0$, $- a^5 c^3 - a^4 b^5 c^5 - a^2 b^6 + 20 a^2 b^3 c^3\ge0$ and $6 a^5 b^3 c + 15 a^4 b^3 c^2\ge0$. Therefore it is enough to show$$b c^2 (-a^3 c^5 + 15 a b^2 c^2 - a b - b^6 c^2 + 6 b^2 c^3)\ge0$$and from $-a^3c^5\ge-ab^2c^5$, $-ab\ge-abc^3\ge-b^2c^3$ and $-b^6c^2>-4b^3c^2\ge-4b^2c^3$, it is left to show$$b^2c^2(-a c^3 + 15 a + c)\ge0$$and we can divide it with three cases.
Case 1) $c^3\le15$: $-a c^3 + 15 a + c>a(15-c^3)\ge0$.
Case 2) $15^{1/3}< c\le2.6$: Firstly, $c>2.4$. The equation is decreasing with respect to $a$, therefore we need to show only for maximal value of $a$. If $a>0.3$, then $a+b+c\ge2a+c>0.6+2.4=3$, so maximal value is $0.3$. Also, $c^3\le2.6^3<20$. Therefore, $$-a c^3 + 15 a + c\ge0.3(15-c^3)+c\ge-1.5+2>0$$
Case 3) $2.6<c\le3$: Similarly, it is enough to show for maximal value of $a$ which is $0.2$. Therefore, $$-a c^3 + 15 a + c\ge0.2(15-27)+2.6>0$$
Therefore, we can assume that $0=a\le b\le c$. Now $f(a,b,c)=b^3c^6\le2^6\left(\frac{3\times b+6\times0.5c}{9}\right)^9=64$ and it is proved.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality: $ab^2+bc^2+ca^2 \le 4$, when $a+b+c=3$. Let $a,b,c $ are non-negative real numbers, and $a+b+c=3$.
How to prove inequality
$$
ab^2+bc^2+ca^2\le 4.\tag{*}
$$
In other words, if $a,b,c$ are non-negative real numbers, then how to prove inequality
$$
27(ab^2+bc^2+ca^2)\le 4(a+b+c)^3.\tag {**}
$$
$\color{gray}{\mbox{(Without using "universal" Lagrange multipliers method).}}$
Thanks!
|
Let $a=3x, b=3y, c=3z$ and $f(x,y,z)=x^2y+y^2 z +z^2 x$
Without loss of generality let's assume that $x= \max { (x,y,z)}$
Then $f(x+\frac{z}{2}, y+\frac{z}{2}, 0) - f(x,y,z)=yz(x-y)+\frac{xz}{2}(x-z) + \frac{z^3}{8} + \frac{z^2y}{4} \ge 0$
And $f(x,y,0)= x^2y = 4 \frac{x}{2} \frac{x}{2} y \le 4\left( \frac{x+y}{3} \right)^3=\frac{4}{27}$
From AM-GM inequality.
|
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|
If $x\le c\le y$ and $|x-y|<\delta$, then $|\frac{g(x)-g(y)}{x-y}-g'(c)|<\epsilon$ Alright, so here goes the question:
Consider a function $g$ that is differentiable at some point $c$ on an interval $[a,b]$.
Prove that: $\forall$$\epsilon$ $>0$ $\exists$$\delta$ $> 0$ such that when $a\leq x \leq c \leq y \leq b$, and $|x-y|<\delta$, then $\left|\dfrac{g(x)-g(y)}{x-y} - g'(c)\right| < \epsilon$.
What I have so far is that, because $g$ is differentiable at $c$, and $c$ is between $x$ and $y$, $g$ is differentiable at $c$ on $[x,y]$. By the mean value theorem, $f'(c)=(f(y)-f(x))/(y-x)$, which equals $(f(x)-f(y))/(x-y)$. Thus $(f(x)-f(y))/(x-y) - f'(c) = 0$, which is less than any $\epsilon$.
Is what I'm doing legal? Thank you for your help!
|
By the definition of the derivative, for every $\varepsilon > 0$, there is a $\delta > 0$ such that for all $z \in [a,b]$ with $0 < \lvert z-c\rvert < \delta$, we have
$$\left\lvert\frac{g(z)-g(c)}{z-c} -g'(c)\right\rvert < \varepsilon.\tag{1}$$
Now if $c < y < c+\delta$ and $c-\delta < x < c$, we have by $(1)$
$$\begin{align}
\left\lvert\frac{g(y)-g(c)}{y-c} - g'(c)\right\rvert &< \varepsilon,\tag{2}\\
\left\lvert\frac{g(c)-g(x)}{c-x} - g'(c)\right\rvert &< \varepsilon,\tag{3}
\end{align}$$
and we want to combine $(2)$ and $(3)$ so that we obtain
$$\frac{g(y)-g(x)}{y-x} - g'(c).$$
To get the denominator $y-x$, we can multiply $(2)$ with $\dfrac{y-c}{y-x}$, and $(3)$ with $\dfrac{c-x}{y-x}$. That leads to
$$\begin{align}
\frac{y-c}{y-x}&\left(\frac{g(y)-g(c)}{y-c}-g'(c)\right)
+ \frac{c-x}{y-x}\left(\frac{g(c)-g(x)}{c-x}-g'(c)\right)\\
&= \frac{g(y)-g(c)}{y-x} + \frac{g(c)-g(x)}{y-x} - \left(\frac{y-c}{y-x}+\frac{c-x}{y-x}\right)g'(c)\\
&= \frac{g(y)-g(x)}{y-x} - g'(c),
\end{align}$$
and therefore we find
$$\begin{align}
\left\lvert \frac{g(y)-g(x)}{y-x} - g'(c)\right\rvert &= \left\lvert \frac{y-c}{y-x}\left(\frac{g(y)-g(c)}{y-c}-g'(c)\right)
+ \frac{c-x}{y-x}\left(\frac{g(c)-g(x)}{c-x}-g'(c)\right)\right\rvert\\
&\leqslant \frac{y-c}{y-x}\left\lvert\frac{g(y)-g(c)}{y-c} - g'(c)\right\rvert
+ \frac{c-x}{y-x}\left\lvert\frac{g(c)-g(x)}{c-x} - g'(c)\right\rvert\\
&< \frac{y-c}{y-x}\varepsilon + \frac{c-x}{y-x}\varepsilon\\
&= \varepsilon.
\end{align}$$
So we have the desired inequality if $c-\delta < x < c < y < c+\delta$. If $x = c < y < c+\delta$ or $c-\delta < x < c = y$, the inequality holds by $(1)$, and if $\lvert y-x\rvert < \delta$ with $x\leqslant c \leqslant y$, then we have $c-\delta < x \leqslant c \leqslant y < \delta$, so we have established
$$\left\lvert \frac{g(y)-g(x)}{y-x} - g'(c)\right\rvert < \varepsilon$$
for all $x,y$ with $\lvert y-x\rvert < \delta$, $x\leqslant c \leqslant y$ and $x \neq y$.
|
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|
On the equation $(1-x)^2/x + (1-y)^2/y + (1-z)^2/z + 4 = 0$ The problem is to solve the equation,
$$\frac{(1-x)^2}{x} + \frac{(1-y)^2}{y} + \frac{(1-z)^2}{z} + 4 = 0\tag{1}$$
in the rationals. Treating this as an equation in $z$, easy solutions would involve $z = \pm 1$, $z = \pm x, \pm y$. More complicated ones would be to make the discriminant $D$ of $(1)$ a square,
$$D = -4x^2y^2 + (x+y-2xy+x^2y+xy^2)^2 = t^2\tag{2}$$
with one solution being,
$$ x = a/b$$
$$ y = -b/a\, (p_1/p_2)$$
$$p_1 = 7a^6 + 4a^5b - 14a^4b^2 + 12a^3b^3 - 25a^2b^4 + 8a b^5 - 8b^6$$
$$p_2 = 8a^6 - 8a^5b + 25a^4b^2 - 12a^3b^3 + 14a^2b^4 - 4a b^5 - 7b^6$$
and $a,b$ being the legs of the Pythagorean triple $a^2+b^2 = c^2$.
However, can someone find a polynomial parameterization of small degree to $(1)$?
Edit (a few days later):
Courtesy of Allan MacLeod, a simple parameterization to $(1)$ can be given by using $x=1/y$ and the discriminant $(2)$ greatly simplifies to just solving,
$$y^2+1 = w^2$$
Hence, his first answer below can also be expressed in terms of Pythagorean triples $a^2+b^2 = c^2$ as,
$$x=a/b,\;\; y = b/a,\;\; z = \frac{(a+c)(b-c)}{ab}$$
|
Consider the problem written as
\begin{equation*}
\frac{(1-f)^2}{f}+\frac{(1-g)^2}{g}+\frac{(1-h)^2}{h}+4=0
\end{equation*}
which gives the quadratic in $h$
\begin{equation*}
h^2+\frac{f^2g+f(g-1)^2+g}{f\,g}+1=0
\end{equation*}
As Tito stated, for this to have a rational solution (assuming $f,g$ rational) the discriminant must be a rational square. After simplification,
this means there must be $t \in \mathbb{Q}$ with
\begin{equation*}
t^2=f^2g^4+2f(f^2-1)^2g^3+(f^4-4f^3+4f^2-4f+1)g^2+2f(f-1)^2g+f^2
\end{equation*}
Defining $Y=t\,f$ and $X=g\,f$ gives the quartic
\begin{equation*}
Y^2=X^4+2(f-1)^2X^3+(f^4-4f^3+4f^2-4f+1)X^2+2(f^2-f)^2X+f^4
\end{equation*}
There is an obvious rational point $X=0, Y = f^2$, so the curve is birationally equivalent to an elliptic curve. Using Mordell's method
the elliptic curve is
\begin{equation*}
v^2=u^3+(f^4-4f^3-2f^2-4f+1)u^2+16f^4u
\end{equation*}
with
\begin{equation*}
g=\frac{u(f-1)^2-v}{2f(4f^2-u)}
\end{equation*}
The curve has a point of order $2$ at $u=0$, two points order $4$ when $u=4f^2$ and $4$ points of order $8$ at $u=4f$ and $u=4f^3$. These give undefined solutions
or permutations of $(f,-f,1)$ or $(f,-1/f,1)$. Thus the torsion subgroup is usually isomorphic to $\mathbb{Z}8$. There are $3$ rational points of order $2$ when
$f^2-6f+1=\square$ when the torsion subgroup is $\mathbb{Z}2 \times \mathbb{Z}8$.
Numerical tests show that the elliptic curve often has rank $0$, and so no other solutions. For $f=10$, the rank is $1$ with generator $(-125,-8250)$
which gives $g=-5/28$ and $h= -7/4,-4/7$.
If we set $f=(k^2-1)/2k$, the parametric solution quoted comes from
\begin{equation*}
u=\frac{(k-1)^2(k^6+2k^5-13k^4-4k^3-5k^2-6k-7)^2}{2(2k^6-4k^5+5k^4-12k^3+4k^2+1)^2}
\end{equation*}
Numerical tests suggest that this is often $3$ times a generator point. Thus, it might be possible to find a smaller parametric solution by doing some further algebra.
|
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|
Evaluate $ \sum\limits_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}$ The question was: Evaluate, ${\textstyle {\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}}}.$
And I go, since $\frac{n}{n^{4}+n^{2}+1}\sim\frac{1}{n^{3}}$ and we know that ${\displaystyle \sum_{n=}^{\infty}\frac{1}{n^{3}}}$ converges. so ${\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}}$ is convergent as well.
But I find it hard to calculate the sum. can you give me some hints?
|
$\newcommand{\+}{^{\dagger}}%
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\begin{align}
&\sum_{n = 1}^{\infty}{n \over n^{4} + n^{2} + 1}
=
\sum_{n = 1}^{\infty}{n \over \pars{n^{2} + 1/2}^{2} + 3/4}
\\[3mm]&=
\sum_{n = 1}^{\infty}{n \over \bracks{n^{2} -\pars{-1/2 - \sqrt{3}\,\ic/2}}
\bracks{n^{2} -\pars{-1/2 + \sqrt{3}\,\ic/2}}}
=
\sum_{n = 1}^{\infty}{n \over \pars{n^{2} - \xi^{2}}\pars{n^{2} - {\xi^{*}}^{2}}}
\end{align}
where $\xi^{2} \equiv \pars{-1 - \root{3}\,\ic}/2 = \expo{4\pi\ic/3}$
\begin{align}
&\sum_{n = 1}^{N}{n \over n^{4} + n^{2} + 1}
=
{1 \over \xi^{2} - {\xi^{*}}^{2}}\sum_{n = 1}^{N}\pars{%
{n \over n^{2} - \xi^{2}} - {n \over n^{2} - {\xi^{*}}^{2}}}
=
{1 \over 2\ic\Im\pars{\xi^{2}}}\,2\ic\Im\sum_{n = 1}^{N}
{n \over n^{2} - \xi^{2}}
\\[3mm]&=
-\,{2\root{3} \over 3}\Im\bracks{{1 \over 2}\sum_{n = 1}^{N}
\pars{{1 \over n - \xi} + {1 \over n + \xi}}}
=
-\,{\root{3} \over 3}\Im\sum_{n = 1}^{N}
\pars{{1 \over n + \xi} + {1 \over n - \xi}}
\\[3mm]&=
-\,{\root{3} \over 3}\Im\pars{%
\sum_{n = 1}^{N}{1 \over n + \xi} - \sum_{n = 1}^{N}{1 \over n - \xi^{*}}}
\end{align}
\begin{align}
&\xi = \expo{2\pi\ic/3} = \cos\pars{2\pi \over 3} + \sin\pars{2\pi \over 3}\ic
=
-\,{1 \over 2} + {\root{3} \over 2}\,\ic
\\[3mm]&
\mbox{Notice that}\ \xi^{*} = -\,{1 \over 2} - {\root{3} \over 2}\,\ic
= -1 - \pars{-\,{1 \over 2} + {\root{3} \over 2}\,\ic} = - 1 - \xi
\end{align}
\begin{align}
&\sum_{n = 1}^{\infty}{n \over n^{4} + n^{2} + 1}
=
-\,{\root{3} \over 3}\lim_{N \to \infty}\Im\pars{%
\sum_{n = 1}^{N}{1 \over n + \xi} - \sum_{n = 1}^{N}{1 \over n + 1 + \xi}}
\\[3mm]&=
-\,{\root{3} \over 3}\lim_{N \to \infty}\Im\pars{%
\sum_{n = 1}^{N}{1 \over n + \xi} - \sum_{n = 2}^{N + 1}{1 \over n + \xi}}
\\[3mm]&=
-\,{\root{3} \over 3}\lim_{N \to \infty}\Im\pars{%
{1 \over 1 + \xi } + \sum_{n = 2}^{N}{1 \over n + \xi}
-
\sum_{n = 2}^{N}{1 \over n + \xi} - {1 \over N + 1 + \xi}}
=
-\,{\root{3} \over 3}\Im\pars{1 \over 1 + \xi }
\\[3mm]&=
-\,{\root{3} \over 3}\Im\bracks{1 \over \pars{1 + \root{3}\ic}/2}
=
-\,{\root{3} \over 6}\Im\pars{1 - \root{3}\ic} = {1 \over 2}
\end{align}
$$\color{#0000ff}{\large%
\sum_{n = 1}^{\infty}{n \over n^{4} + n^{2} + 1} = {1 \over 2}}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/571973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 1
}
|
How to prove $\sum_{n=1}^\infty\operatorname{arccot}\frac{\sqrt[2^n]2+\cos\frac\pi{2^n}}{\sin\frac\pi{2^n}}=\operatorname{arccot}\frac{\ln2}\pi$? How can I prove the following identity?
$$\sum_{n=1}^\infty\operatorname{arccot}\frac{\sqrt[2^n]2+\cos\frac\pi{2^n}}{\sin\frac\pi{2^n}}=\operatorname{arccot}\frac{\ln2}\pi$$
|
Rewrite the sum as
$$\sum_{n=1}^{\infty} \arctan{\left [\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}+\cos{\left (\frac{\pi}{2^n}\right)}}\right ]}$$
Now let
$$a_n = \arctan{\left [\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}-\cos{\left (\frac{\pi}{2^n}\right)}}\right ]}$$
Then one may show (nontrivial!) that
$$a_n - a_{n-1} = \arctan{\left [\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}+\cos{\left (\frac{\pi}{2^n}\right)}}\right ]}$$
so that we have a telescoping sum:
$$\sum_{n=1}^{\infty} (a_n-a_{n-1}) = a_{\infty}-a_0$$
where
$$a_{\infty} = \lim_{n\to\infty} a_n $$
Now,
$$\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}-\cos{\left (\frac{\pi}{2^n}\right)}} \sim \frac{\pi/2^n}{1+\frac{\log{2}}{2^n}-1} = \frac{\pi}{\log{2}}$$
Note also that $a_0=0$. therefore the sum is simply
$$\arctan{\frac{\pi}{\log{2}}}$$
which is the stated result.
ADDENDUM
For completeness, I'll outline a few steps to demonstrate how to prove the difference equation above. Start with the relation
$$\arctan{p}-\arctan{q} = \arctan{\frac{p-q}{1+p q}}$$
so we need to work with the following argument of the arctangent:
$$\frac{\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}-\cos{\left (\frac{\pi}{2^n}\right)}} - \frac{\sin{\left (\frac{2\pi}{2^n}\right)}}{2^{2^{1-n}}-\cos{\left (\frac{2\pi}{2^n}\right)}}}{1+\frac{\sin{\left (\frac{\pi}{2^n}\right)}}{2^{2^{-n}}-\cos{\left (\frac{\pi}{2^n}\right)}}\frac{\sin{\left (\frac{2\pi}{2^n}\right)}}{2^{2^{1-n}}-\cos{\left (\frac{2\pi}{2^n}\right)}}}$$
which simplifies somewhat to
$$\frac{-2^{2^{-n}} \sin \left(\pi 2^{1-n}\right)+2^{2^{1-n}} \sin \left(\pi
2^{-n}\right)+\sin \left(\pi 2^{1-n}\right) \cos \left(\pi 2^{-n}\right)-\sin
\left(\pi 2^{-n}\right) \cos \left(\pi 2^{1-n}\right)}{2^{3\ 2^{-n}}+\sin
\left(\pi 2^{1-n}\right) \sin \left(\pi 2^{-n}\right)-2^{2^{-n}} \cos
\left(\pi 2^{1-n}\right)+\cos \left(\pi 2^{-n}\right) \cos \left(\pi
2^{1-n}\right)-2^{2^{1-n}} \cos \left(\pi 2^{-n}\right)}$$
Now you may show that the numerator is equal to
$$\sin \left(\pi 2^{-n}\right) \left(2^{2^{1-n}}-2^{2^{-n}+1} \cos \left(\pi
2^{-n}\right)+1\right)$$
and the denominator is equal to
$$2^{2^{-n}} \left(2^{2^{1-n}}-2^{2^{-n}+1} \cos \left(\pi
2^{-n}\right)+1\right)+\cos \left(\pi 2^{-n}\right)
\left(2^{2^{1-n}}-2^{2^{-n}+1} \cos \left(\pi 2^{-n}\right)+1\right)$$
Cancelling the common factors in the numerator and denominator produces the desired result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/577557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
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"answer_id": 0
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|
Proving for every odd number $x$, $x^2$ is always congruent to $1$ or $9$ modulo $24$ A problem I have been presented with asks the following:
Prove for every odd number $x$, $ x^2$ is always congruent to $1$ or $9$ modulo $24$.
This seems odd and non-intuitive to me. Of course, it must be true other wise they wouldn't be asking for me to prove it.
I know that:
$9$ modulo $24$ $=$ $9$
$1 = 1$
How could every odd number in existence squared be equal to either 1 or 9?
|
Every number is equivalent mod 24 to one of 0, 1, 2, 3 ,4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, or 23.
But our number $n$ is odd, so it is equivalent mod 24 to one of 1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, 23.
Then:
$$\begin{array}{r|r|l}
n & n^2 & n^2\pmod{24} \\\hline
1 & 1 & 1 \\
3 & 9 & 9 \\
5 & 25 & 1 \\
7 & 49 & 1 \\\hline
9 & 81 & 9 \\
11 & 121 & 1\\
13 & 169 & 1 \\
15 & 225 & 9 \\\hline
17 & 289 & 1 \\
19 & 361 & 1 \\
21 & 441 & 9 \\
23 & 529 & 1
\end{array}
$$
So there you go.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/578749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
How find this Fibonacci sequence sum $\sum_{k=0}^{\infty}\frac{1}{F_{2^k}}$ let sequence $\{F_{n}\}$ such $$F_{1}=1,F_{2}=1,F_{m+1}=F_{m}+F_{m-1},m\ge 2$$
Find this value
$$I=\sum_{k=0}^{\infty}\dfrac{1}{F_{2^k}}$$
My try: I know this
$$F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{\sqrt{5}+1}{2}\right)^n-\left(\dfrac{\sqrt{5}-1}{2}\right)^n\right)$$
so
$$\dfrac{1}{F_{2^k}}=\dfrac{\sqrt{5}}{\left(\dfrac{\sqrt{5}+1}{2}\right)^{2^k}-\left(\dfrac{\sqrt{5}-1}{2}\right)^{2^k}}$$
so
we only find this sum
$$I=\sum_{k=0}^{\infty}\dfrac{\sqrt{5}}{\left(\dfrac{\sqrt{5}+1}{2}\right)^{2^k}-\left(\dfrac{\sqrt{5}-1}{2}\right)^{2^k}}$$
But I can't.Thank you
|
let
$$a=\dfrac{1+\sqrt{5}}{2}\Longrightarrow \dfrac{\sqrt{5}-1}{2}=a^{-1}$$
so
$$I=\sum_{n=0}^{\infty}\dfrac{\sqrt{5}}{a^{2^n}-a^{-2^n}}$$
so let $$a^{2^n}=x$$
then
$$\dfrac{1}{a^{2^n}-a^{-2^n}}=\dfrac{x}{x^2-1}=\dfrac{1}{x-1}-\dfrac{1}{x^2-1}=\dfrac{1}{a^{2^n}-1}-\dfrac{1}{a^{2^{n+1}}-1}$$
so
\begin{align*}I&=\sum_{n=0}^{\infty}\dfrac{\sqrt{5}}{a^{2^n}-a^{-2^n}}=\sqrt{5}\sum_{n=0}^{\infty}\left(\dfrac{1}{a^{2^n}-1}-\dfrac{1}{a^{2^{n+1}}-1}\right)\\
&=\lim_{n\to\infty}\left(\dfrac{\sqrt{5}}{a-1}-\dfrac{1}{a^{2^{n+1}}-1}\right)\\
&=\dfrac{\sqrt{5}}{a-1}\\
&=\dfrac{7-\sqrt{5}}{2}
\end{align*}
|
{
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"url": "https://math.stackexchange.com/questions/578920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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|
new $\arctan$ series working for any $x$? Using Log series, we can write:
$$
\frac{1}{2 i} \left( \text{Log} \left( 1 + e^{\text{i2t}} \right) - \text{Log} \left( 1 + e^{\text{-i2t}} \right) \right)
=
\frac {1} {2 i}\left ( \sum _ {k = 1}^{\infty}\frac {(-1)^{k + 1}} {k} e^{\text{i2t}^k} - \sum _ {k = 1}^{\infty}\frac {(-1)^{k + 1}} {k} e^{-\text{i2t}^k} \right)
$$
$$
\frac {1}{2 i} \text {Log} \left( \frac{1 + e^{\text {i2t}}}{1 + e^{-\text {i2t}}} \right)
=
\frac{1}{2 i} \sum_{k = 1}^{\infty}\frac{(-1)^{k + 1}}{k} \left( e^{\text{i2kt}} - e^{-\text{i2kt}} \right)
$$
$$
\frac {1}{2 i} \text {Log} \left( \frac{e^{\text{i2t}} \left( e^{-\text{i2t}}+1 \right)}{1+e^{-\text{i2t}}} \right)
=
\sum_{k = 1}^{\infty} \frac{(-1)^{k + 1}}{k} \left( \frac{e^{\text {i2kt}} - e^{-\text {i2kt}}}{2 i} \right)
$$
$$
\frac{1}{2i} \text{Log} \left( e^{\text{i2t}} \right)
=
\sum_{k = 1}^{\infty}\frac {(-1)^{k + 1}} {k}\text {Sin} (2kt)
$$
$$
t
=
\sum_{k = 1}^{\infty}\frac {(-1)^{k + 1}} {k}\text {Sin} (2kt)
$$
Here is the plot of the sum in red with sucessive aproximations and y=t in blue:
Choosing $t = \text{ArcTan} (x)$, we have:
$$
\text{ArcTan} (x)
=
\sum_{k = 1}^{\infty} \frac {(-1)^{k + 1}} {k} \text{Sin} ( 2k \text{ ArcTan} x )
$$
Using $z=1 + i x=|z| e^{i \text{arcTan}x }$ and its complex conjuguate $z^*=1 - i x=|z| e^{-i \text{arcTan}x }$, with $|z|=\sqrt{1 + x^2}$, we can write:
$$
\text{Sin} ( 2k \text{ Arctan} x)
=
\frac {z^{2 k} - z^{*2 k}} {2i |z| ^{2 k}}
=
\frac {(1 + ix)^{2k} - (1 - ix)^{2k}} {2i (1 + x^2)^k}
=
\frac{\sum_{q=0}^k (-1)^q \left( \begin{array}{c} 2k \\ 2q+1 \end{array} \right) x^{2 q+1}}{\left(1+x^2\right)^k}
$$
Thus:
$$
\text{ArcTan} (x)
=
\sum_{k = 1}^{\infty} \frac {(-1)^{k + 1}} {k}
\frac{\sum_{q=0}^k (-1)^q \left( \begin{array}{c} 2k \\ 2q+1 \end{array} \right) x^{2 q+1}}{\left(1+x^2\right)^k}
$$
Here are the plots of $\text{ArcTan}(x)$ in blue, and successive aproximations to the classic ArcTan series $\text{ArcTan}(x)=\sum_{k = 0}^{\infty} \frac {(-1)^k} {2k + 1} x^{2k + 1}$ in green, working for $x<1$, and the one I gave in red, working for any $x$ as the sum goes to infinity:
is that result known?
|
I do not know whether your series for $\arctan$ is new (it may be), but when evaluating its advantages, you should consider alternatives other than the classical McLaurin series. For example, the identity $\tan 2\theta=\frac{2\tan \theta}{1-\tan^2\theta}$ can be turned into $$\arctan x = 2\arctan \frac{x}{1+\sqrt{1+x^2}}$$
Note that the argument of $\arctan$ on the right is always less than $1$ in absolute value. Hence,
$$\arctan x = 2 \, \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}\left(\frac{x}{1+\sqrt{1+x^2}}\right)^{2k+1},\quad x\in\mathbb R$$
The content of parentheses is an irrational function of $x$ but it only needs to be computed once. Partial sums up to $k=0,1,2,3$ are shown below; the purple curve is the arctangent itself.
(The horizontal window is $[0,4]$, as in your post.)
|
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|
Integral involving logarithm and cosine For $a,b>0$ I would like to compute the integral $$I=\int_0^{2\pi} -\log{\sqrt{a^2+b^2-2ab\cos{t}}}~dt.$$
Numerical computations suggest that $$I=\min\{-\log{a},-\log{b}\}.$$
How can I prove this? I tried to find the antiderivative using Mathematica, but the result looks awfull.
Have you seen such an integral? Some reference would be great!
|
We have
$$I=\dfrac12\int_0^{2\pi} -\log(a^2+b^2-2ab\cos{t})dt=\int_0^{\pi} -\log(a^2+b^2-2ab\cos{t})dt$$
We then have
$$I=-2\pi\log(b) -\int_0^{\pi} \log((a/b)^2+1-2(a/b)\cos{t})dt$$
Let us call $a/b$ as $a$ from now on. Hence, we want to evaluate the integral
$$I(a) = \displaystyle \int_0^{\pi} \ln \left(1-2a \cos(x) + a^2\right) dx$$ Some preliminary results on $I(a)$. Note that we have $$I(a) = \underbrace{\displaystyle \int_0^{\pi} \ln \left(1+2a \cos(x) + a^2\right) dx}_{\spadesuit} = \overbrace{\dfrac12 \displaystyle \int_0^{2\pi} \ln \left(1-2a \cos(x) + a^2\right) dx}^{\clubsuit}$$ $(\spadesuit)$ can be seen by replacing $x \mapsto \pi-x$ and $(\clubsuit)$ can be obtained by splitting the integral from $0$ to $\pi$ and $\pi$ to $2 \pi$ and replacing $x$ by $\pi+x$ in the second integral.
Now let us move on to our computation of $I(a)$.
\begin{align}
I(a^2) & = \int_0^{\pi} \ln \left(1-2a^2 \cos(x) + a^4\right) dx = \dfrac12 \int_0^{2\pi} \ln \left(1-2a^2 \cos(x) + a^4\right) dx\\
& = \dfrac12 \int_0^{2\pi} \ln \left((1+a^2)^2-2a^2(1+ \cos(x))\right) dx = \dfrac12 \int_0^{2\pi} \ln \left((1+a^2)^2-4a^2 \cos^2(x/2)\right) dx\\
& = \dfrac12 \int_0^{2\pi} \ln \left(1+a^2-2a \cos(x/2)\right) dx + \dfrac12 \int_0^{2\pi} \ln \left(1+a^2+2a \cos(x/2)\right) dx
\end{align}
Now replace $x/2=t$ in both integrals above to get
\begin{align}
I(a^2) & = \int_0^{\pi} \ln \left(1+a^2-2a \cos(t)\right) dt + \int_0^{\pi} \ln \left(1+a^2+2a \cos(t)\right) dt = 2I(a)
\end{align}
Now for $a \in [0,1)$, this gives us that $I(a) = 0$. This is because we have $I(0) = 0$ and $$I(a) = \dfrac{I(a^{2^n})}{2^n}$$ Now let $n \to \infty$ and use continuity to conclude that $I(a) = 0$ for $a \in [0,1)$. Now lets get back to our original problem. Consider $a>1$. We have
\begin{align*}
I(1/a) & = \int_0^{\pi} \ln \left(1-\dfrac2{a} \cos(x) + \dfrac1{a^2}\right)dx\\
& = \int_0^{\pi} \ln(1-2a \cos(x) + a^2) dx - 2\int_0^{\pi} \ln(a)dx\\
& = I(a) - 2 \pi \ln(a)\\
& = 0 & \text{(Since $1/a < 1$, we have $I(1/a) = 0$)}
\end{align*}
Hence, we get that
$$I(a) = \begin{cases} 2 \pi \ln(a) & a \geq 1 \\ 0 & a \in [0,1] \end{cases}$$
Adapted from here
|
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|
If the roots of $ax^3 - x^2 + bx - 1$ are real and positive, prove $b \geq \sqrt{3}$ Let
$$p(x) = ax^3 - x^2 + bx - 1$$
such that $a, b \in \mathbb{R}$ and the roots of $p(x)$ are real and positive.
Prove that
$$b \geq \sqrt{3}$$
Let $\alpha, \beta, \gamma \in \mathbb{R^{+}}$ be the roots of $p(x)$
$$\alpha + \beta + \gamma = \sigma_1 = \frac{1}{a}$$
$$\alpha\beta + \beta\gamma + \gamma\alpha = \sigma_2 = \frac{b}{a}$$
$$\alpha\beta\gamma = \sigma_3 = \frac{1}{a}$$
Consider the inequality:
$$\alpha^2 + \beta^2 + \gamma^2 \geq \alpha\beta + \beta\gamma + \gamma\alpha$$
Since the inequality is cyclic, we may assume $\alpha \geq \beta \geq \gamma$. Now, the above is clearly true by the rearrangement inequality.
$$\implies \sigma_1^2 - 2\sigma_2 \geq \sigma_2$$
$$\implies \sigma_1^2 \geq 3\sigma_2$$
$$\implies \frac{1}{a^2} \geq 3\frac{b}{a}$$
$$\implies 1 \geq 3ab \tag{1}$$
Now, from the AM-GM inequality:
$$\alpha + \beta + \gamma \geq 3\sqrt[3]{\alpha\beta\gamma}$$
Since $\alpha + \beta + \gamma = \alpha\beta\gamma$,
$$\alpha^3\beta^3\gamma^3 \geq 27\alpha\beta\gamma$$
$$\implies \alpha\beta\gamma = \frac{1}{a} \geq 3\sqrt{3}$$
$$\implies \frac{1}{3\sqrt{3}} \geq a$$
Putting this in $(1)$, we get:
$$3ab \leq \frac{b}{\sqrt{3}}$$
We know that $3ab \leq 1$, but we can't say that $\frac{b}{\sqrt{3}}\leq 1$ as well. Did I go wrong somewhere? Or is there some way to solve it from here?
|
Everything you have derived is true, but it's not enough to finish the problem. Here is one way to do it:
We have $\sigma_2^2 = \alpha^2\beta^2 + \beta^2\gamma^2 +\alpha^2\gamma^2 + 2(\alpha+\beta+\gamma)(\alpha\beta\gamma) = \alpha^2\beta^2 + \beta^2\gamma^2 +\alpha^2\gamma^2 + 2\sigma_1\sigma_3$. Now we have that $\displaystyle \frac{1}{\alpha^2} + \frac{1}{\beta^2} + \frac{1}{\gamma^2} \geq \frac{1}{\alpha\beta} + \frac{1}{\beta\gamma} +\frac{1}{\alpha\gamma}$, and multiplying this by $(\alpha\beta\gamma)^2$ we get $\alpha^2\beta^2 + \beta^2\gamma^2 +\alpha^2\gamma^2 \geq (\alpha+\beta+\gamma)(\alpha\beta\gamma) = \sigma_1\sigma_3$. Hence we get $\displaystyle \frac{b^2}{a^2} = \sigma_2^2 \geq 3\sigma_1\sigma_3 = \frac{3}{a^2}$, or $b^2 \geq 3$. Therefore $b\geq\sqrt{3}$ as required.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How prove this $\frac{a}{(bc-a^2)^2}+\frac{b}{(ca-b^2)^2}+\frac{c}{(ab-c^2)^2}=0$ let $a,b,c\in \mathbb{R}$, if such
$$\dfrac{1}{bc-a^2}+\dfrac{1}{ca-b^2}+\dfrac{1}{ab-c^2}=0$$
show that
$$\dfrac{a}{(bc-a^2)^2}+\dfrac{b}{(ca-b^2)^2}+\dfrac{c}{(ab-c^2)^2}=0$$
Does this problem has nice methods?
My idea:let
$$(ca-b^2)(ab-c^2)+(bc-a^2)(ab-c^2)+(bc-a^2)(ca-b^2)=0$$
then I can't. Thanks
|
One approach uses factoring. Define rational functions
$$ R_1 := \frac1{b c-a a} + \frac1{c a-b b} + \frac1{a b-c c}, \tag{1}$$
$$ R_2 := \frac{a}{(b c-a a)^2} + \frac{b}{(c a-b b)^2} + \frac{c}{(a b-c c)^2}, \tag{2}$$
and polynomials
$$ P_1:=a+b+c,\qquad \qquad P_2:=a b+b c+c a,\\ P_3:=P_2\!-\!a a\!-\!b b\!-\!c c,\quad
P_4:=a b c P_1\!-\!a a b b\!-\!b b c c\!-\!c c a a,
\tag{3} $$
and product of polynomials
$$ D := (b c-a a)(c a-b b)(a b-c c). \tag{3} $$
The factorization of $\,R_1,R_2\,$ are
$$ R_1 = \frac{P_2P_3}D, \qquad R_2 = \frac{P_1 P_2 P_3 P_4}{D^2}. \tag{4} $$
The result is that $\,R_1 = 0\,$ if $\,P_2P_3=0\,$ and this also implies
that $\,R_2 = 0.$
|
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|
Calculating limit of function To find limit of $\lim_{x\to 0}\frac {\cos(\sin x) - \cos x}{x^4} $.
I differentiated it using L Hospital's rule. I got
$$\frac{-\sin(\sin x)\cos x + \sin x}{4x^3}\text{.}$$ I divided and multiplied by $\sin x$.
Since $\lim_{x\to 0}\frac{\sin x}{x} = 1$, thus I got
$\frac{1-\cos x}{4x^2}$.On applying standard limits, I get answer $\frac18$. But correct answer is $\frac16$. Please help.
|
Without L'Hôpital
The expansions $\sin(x) = x - \frac{x^3}{6} + O(x^5)$ and $\cos(u) = 1 - \frac{u^2}{2} + \frac{u^4}{24} + O(u^6)$ combine joyfully to give
$$
\cos(\sin x) - \cos(x) = \left(1 - \frac{x^2}{2} + \frac{5x^4}{24}\right) - \left(1-\frac{x^2}{2} + \frac{x^4}{24}\right) + O(x^5)
$$
so finally,
$$
\lim_{x\to 0} \frac{\cos(\sin x) - \cos(x)}{x^4} = \frac{5}{24}-\frac{1}{24} = \frac{1}{6}.
$$
|
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|
Can $a^2+b+2$ and $b^2+4a$ both be perfect squares? Are there any positive integers $a$ and $b$ so that $a^2+b+2$ and $b^2+4a$ are both perfect squares?
|
No.
$b^2+4a$ is a perfect square $>b^2$, and $b^2+4a \not =(b+1)^2$ (fails by checking parity), so $b^2+4a \geq (b+2)^2$ so $a \geq b+1$
$a^2+b+2$ is a perfect square $>a^2$, so $a^2+b+2 \geq (a+1)^2$ so $b \geq 2a-1$.
Combining, $b \geq 2a-1 \geq 2(b+1)-1=2b+1>b$, a contradiction.
|
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|
Convex Quadrilateral: $ \dfrac {\tan A + \tan B + \tan C + \tan D}{\tan A \tan B \tan C \tan D} = \cot A + \cot B + \cot C + \cot D $ Problem
Let $ABCD$ be a convex quadrilateral with no right angles. Show that $$ \dfrac {\tan A + \tan B + \tan C + \tan D}{\tan A \tan B \tan C \tan D} = \cot A + \cot B + \cot C + \cot D. $$
Source: Geometry Unbound by Kiran Kedlaya.
Attempt: Well, all we really know about a convex quadrilateral are that $ \angle A + \angle B + \angle C + \angle D = 360^\circ $ and that the polygon is convex. Well, that's obvious. But any starts to a proof of this useful fact would be helpful!
|
Multiply the original identity by $\tan A \tan B \tan C \tan D$
$$
\tan A + \tan B + \tan C + \tan D = \tan B \tan C \tan D + \tan A \tan C \tan D + \tan A \tan B \tan D + \tan A \tan B \tan C.
$$
then find $\tan D$ from it
$$
\tan D = \frac{\tan A \tan B \tan C - \tan A - \tan B - \tan C}{1 - \tan B \tan C - \tan A \tan C - \tan A \tan B}.
$$
On the other hand
$$
\tan D=\tan(2\pi -A - B - C)=-\tan(A + B + C)
$$
So it is remains to show that
$$
\tan(A + B + C)=\frac{\tan A + \tan B + \tan C - \tan A \tan B \tan C}{1 - \tan B \tan C - \tan A \tan C - \tan A \tan B}
$$
which was done here.
|
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|
For how many integers $a$ is $\frac{2^{10} \cdot 3 ^8 \cdot 5^6}{a^4}$ an integer? In Mathleague $11316$ Target #$4$, the question is:
For how many integers $a$ is $$\frac{2^{10} \cdot 3 ^8 \cdot 5^6}{a^4}$$ an integer?
|
Clearly the number
$$ q =\frac{2^{10} 3^8 5^6 }{a^4} = \left( \frac{2^{5} 3^4 5^3 }{a^2} \right)^2
$$
is a perfect square, and so
$$
\sqrt{q} = \frac{2^{5} 3^4 5^3 }{a^2}
$$ is rational. Now in order for this to be an integer, the denominator must divide the numerator. Therefore $a^2$ must divide $2^53^45^3$. Now the greatest perfect square dividing $2^53^45^3$ is $2^4 3^4 5^2$, and so $a^2$ can be any divisor of this number. Therefore $a$ can be any integer divisor of its square root, $2^2 3^2 5$. There are $3(3)(2)=18$ positive divisors of this number, and so there are $36$ signed possibilities for $a$. So the answer is 36.
|
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|
calculation of $\int\frac{1}{\sin^3 x+\cos^3 x}dx$ and $\int\frac{1}{\sin^5 x+\cos^5x}dx$
Solve the following indefinite integrals:
$$
\begin{align}
&(1)\;\;\int\frac{1}{\sin^3 x+\cos^3 x}dx\\
&(2)\;\;\int\frac{1}{\sin^5 x+\cos^5 x}dx
\end{align}
$$
My Attempt for $(1)$:
$$
\begin{align}
I &= \int\frac{1}{\sin^3 x+\cos ^3 x}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^2 x+\cos ^2 x-\sin x \cos x\right)}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x\right)}\;dx\\
&= \frac{1}{3}\int \left(\frac{2}{\left(\sin x+\cos x\right)}+\frac{\left(\sin x+\cos x \right)}{\left(1-\sin x\cos x\right)}\right)\;dx\\
&= \frac{2}{3}\int\frac{1}{\sin x+\cos x}\;dx + \frac{1}{3}\int\frac{\sin x+\cos x}{1-\sin x\cos x}\;dx
\end{align}
$$
Using the identities
$$
\sin x = \frac{2\tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}},\;\cos x = \frac{1-\tan ^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}
$$
we can transform the integral to
$$I = \frac{1}{3}\int\frac{\left(\tan \frac{x}{2}\right)^{'}}{1-\tan^2 \frac{x}{2}+2\tan \frac{x}{2}}\;dx+\frac{2}{3}\int\frac{\left(\sin x- \cos x\right)^{'}}{1+(\sin x-\cos x)^2}\;dx
$$
The integral is easy to calculate from here.
My Attempt for $(2)$:
$$
\begin{align}
J &= \int\frac{1}{\sin^5 x+\cos ^5 x}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(\sin^4 x -\sin^3 x\cos x+\sin^2 x\cos^2 x-\sin x\cos^3 x+\cos^4 x\right)}\;dx\\
&= \int\frac{1}{(\sin x+\cos x)(1-2\sin^2 x\cos^2 x-\sin x\cos x+\sin^2 x\cos^2 x)}\;dx\\
&= \int\frac{1}{\left(\sin x+\cos x\right)\left(1-\sin x\cos x-\left(\sin x\cos x\right)^2\right)}\;dx
\end{align}
$$
How can I solve $(2)$ from this point?
|
I am not sure how you can continue either (the second term in the denominator can be expressed as $1-\sin(2 x)/2 - \sin^2(2x)/4,$ but I am not aware of any double angle formula for $\sin x + \cos x.$ The simplest approach to your integral is to use the feared $u = \tan \frac{x}2$ substitution, which reduces the integral to a rational function integral....
|
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|
Show that $\frac{\sqrt[n]a}{\sqrt[n]{ab}+\sqrt[n]a+1}+\frac{\sqrt[n]b}{\sqrt[n]{bc}+\sqrt[n]b+1}+\frac{\sqrt[n]c}{\sqrt[n]{ac}+\sqrt[n]c+1}=1$ If $$\sqrt[n]{{abc}} = 1,$$
Prove that $$\frac{\sqrt[n]a}{\sqrt[n]{ab}+\sqrt[n]a+1}+\frac{\sqrt[n]b}{\sqrt[n]{bc}+\sqrt[n]b+1}+\frac{\sqrt[n]c}{\sqrt[n]{ac}+\sqrt[n]c+1}=1.$$
|
Let $x = \sqrt[n]a, y = \sqrt[n]b, z = \sqrt[n]c$. Then you have $xyz = 1$ and
$$\frac{x}{xy+x+1}+\frac{y}{yz+y+1}+\frac{z}{zx+z+1} = \frac{x}{xy+x+1}+\frac{xy}{1+xy+x}+\frac{1}{x+1+xy}$$
where we multiplied the second term's numerator and denominator by $x$ and the third term's by $xy$.
|
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|
Trigonometric inequality to prove Prove the following inequality for each $x$:
$$|\sin x+7\cos x| \leq \sqrt{50}$$
My remark: It seems to me that the solution uses the fact that $1^2+7^2=50$.
Thanks in advance!
|
One of the ways is to use the addition formula: $R \sin (a+x) = (R\cos a) \sin x + (R\sin a) \cos x$. We let $R \cos a = 1$ and $R \sin a = 7$. Hence $50 = 1^2 + 7^2 = R^2 (\cos^2 + \sin^2) = R^2$ and $R = \sqrt{50}$.
$|\sin x + 7\cos x| = |\sqrt{50} \sin (a+x)| \leq \sqrt{50}$
|
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|
Finding $\int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\cos(n\theta)}{3+2\cos\theta} \operatorname{d}\theta$ If $n$ is a positive integer find
$$
\int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\cos(n\theta)}{3+2\cos\theta} \operatorname{d}\theta
$$
I know that I have to use contour integral with a circle of radius 1 centered at the origin, but I am having trouble converting the integral into the form $\int_{|z|}$
$$\int_{|z| = 1} \frac{(1+z+1/z)^n\cos(n\theta)}{3+(z+1/z)} \frac{1}{iz} \operatorname{d}z$$
I cant seem to find a way to expand $\cos(n\theta)$ into a function of $z$.
From the above equation, I can get that the poles of is at $z = -1.5 \pm \frac{\sqrt{5}}{2}$ and only the residual of $z = -1.5 + \frac{\sqrt{5}}{2}$ should be included.
|
I don't know if it will be too late for the answer. What you need is to add $i\sin(n\theta)$ to the integral. Clearly
$$ \int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\sin(n\theta)}{3+2\cos\theta} \operatorname{d}\theta=0. $$
Noting for $z=e^{i\theta}$, ones
$$ 1+2\cos\theta=|1+z|^2-1=(1+z)(1+\frac1z)-1=1+z+\frac1z, $$
and
$$3+2\cos\theta=1+(1+z)(1+\frac1z) $$
and hence
\begin{eqnarray}
&&\int_{0}^{2\pi} \frac {(1+2\cos\theta)^n\cos(n\theta)}{3+2\cos\theta} \operatorname{d}\theta\\
&=&\int_{0}^{2\pi} \frac {(1+2\cos\theta)^ne^{ni\theta }}{3+2\cos\theta} \operatorname{d}\theta\\
&=&-i\int_{|z|=1} \frac {(1+z+\frac1z)^{n}z^n}{1+(1+z)(1+\frac1z)}\frac{\operatorname{d}z}{iz}\\
&=&-i\int_{|z|=1} \frac {(z^2+z+1)^{n}}{z^2+3z+1}\operatorname{d}z\\
&=&-i\int_{|z|=1} \frac {(z^2+z+1)^{n}}{(z-z_1)(z-z_2)}\operatorname{d}z\\
&=&-i\cdot 2\pi i \cdot\text{Res}(\frac {(z^2+z+1)^{n}}{z-z_2},z=z_1)\\
&=&2\pi \frac {(z_1^2+z_1+1))^{n}}{z_1-z_2}\\
&=&\frac{2\pi(3-\sqrt5)^{n}}{\sqrt5}.
\end{eqnarray}
Here $z_1,z_2$ are two of $z^2+3z+1=0$,
$$ z_1=\frac{-3+\sqrt5}{2}, z_2=\frac{-3-\sqrt5}{2}. $$
|
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|
Limit $ \lim_{n \to \infty} \sum_{k=1}^n\frac 1 { \sqrt{n^2+k} } $ (1)
$$ \lim_{n \to \infty} \sum_{k=1}^n \frac 1 { \sqrt{n^2+k} } $$
(2)
$$ \lim_{n\to\infty} \frac {1+\sqrt[n]2 + \sqrt[n]3 + ... \sqrt[n]n} {n} $$
The answers should both be 1.. any hints?
|
For the first problem, note that if $1\le k\le n$, then
$$n^2\lt n^2+k\le n^2+n\lt\left(n+\frac{1}{2}\right)^2.$$
Thus
$$\frac{1}{n+\frac{1}{2}}\le \frac{1}{\sqrt{n^2+k}}\lt \frac{1}{n}.$$
Our sum is therefore between $\frac{n}{n+\frac{1}{2}}$ and $1$.
Squeeze.
|
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|
Trigonometric Integral : $\int\frac{1}{\sin x+ 3\cos x}dx$ I would appreciate if somebody could help me with the following problem
Q: How to integrate this integral
$$\int\frac{1}{\sin x+ 3\cos x}dx$$
|
\begin{align*}
\int \frac 1 {\sin x +3\cos x}\ dx &= \int \frac 1 {2\sin \frac x 2\cos \frac x 2 +3\cos^2 \frac x 2 - 3 \sin^2 \frac x 2}\ dx \\
&= \int \frac 1 {\cos^2 \frac x 2} \cdot \frac 1 {3 + 2\tan \frac x 2 - 3 \tan^2 \frac x 2}\ dx
\end{align*}
Switching $y=\tan \frac x 2 \implies\frac {dy}{dx} = \frac 1 2 \frac 1 {\cos^2 \frac x 2}$
\begin{align*}
\int \frac 1 {\sin x +3\cos x}\ dx
&= \int \frac 1 {3 + 2y - 3 y^2}2\ dy \\
&= 6\int \frac 1 {(3y-1-\sqrt{10})(3y-1+\sqrt{10})} \ dy \\
&= \frac 3 {\sqrt{10}}\int \frac 1 {3y-1-\sqrt{10}} - \frac 1 {3y-1+\sqrt{10}} \ dy \\
\end{align*}
I'll leave the rest.
|
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|
$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$ Show that $$\sqrt{2\sqrt{2\sqrt{2\cdots}}}=2$$
$$\sqrt{2}=\mathbf{2}^{1/2}$$
$$\sqrt{2\sqrt{2}}=\mathbf{2}^{1/2+1/2^2}$$
$$\sqrt{2\sqrt{2\sqrt{2}}}=\mathbf{2}^{1/2+1/2^2+1/2^3}$$
Show the limit of $$\mathbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+\dotsb+\frac{1}{2^n}=1$$ when $n\to\infty$
$$\textbf{S}_{n}=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n}$$
$$\Rightarrow \frac{1}{2}\textbf{S}_{n}=\frac{1}{2}(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^n})$$
$$\Rightarrow \frac{1}{2}\textbf{S}_{n}=(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{n+1}})$$
$$\Rightarrow (1)-(2)=\textbf{S}_{n}-\frac{1}{2}\textbf{S}_{n}=\frac{1}{2}-\frac{1}{2^{n+1}}$$
$$\Rightarrow \textbf{S}_{n}(1-\frac{1}{2})=\frac{1}{2}-\frac{1}{2^{n+1}}$$
$$\Rightarrow \frac{1}{2^{n+1}}\rightarrow\textbf{0}\quad\textit{when n}\rightarrow\infty$$
$$\Rightarrow \textbf{S}_{n}\rightarrow\textbf{1}\quad\textit{when n}\rightarrow\infty$$
$$\Rightarrow \lim_{n \to \infty}\textbf{2}^{\textbf{S}_{n}}=2\quad\textit{when n}\rightarrow\infty$$
|
Let $x_1 = \sqrt{2}$ and define $x_{n+1} = \sqrt{2 x_n}$, then it suffices to show
that $\lim_n x_n = 2$. In order to achieve this goal show that the sequence
$x_n$ is monotonically increasing and bounded above (I will leave this for you to do).
Then the limit exists so let $x = \lim x_n$. Then, using $x_{n+1} = \sqrt{2 x_n}$ and the continuity of the square root function, we get that $x = \sqrt{2 x} \implies x=2$.
|
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|
Limit of series $4\left( \frac {1}{8}+\frac {1}{12}\right) +6\left( \frac {1}{24}+\frac {1}{36}\right) +\ldots$ How to find this serie
$4\left( \dfrac {1}{8}+\dfrac {1}{12}\right) +6\left( \dfrac {1}{24}+\dfrac {1}{36}\right) +8\left( \dfrac {1}{64}+\dfrac {1}{96}\right) +\ldots $
I think it's telescopic, isn't it?
|
The $n$th term appears to be $$2(n+1)\left(\frac1{2^{n+1}(n+1)}+ \frac1{2^{n}3(n+1)}\right).$$
Canceling factors of $2(n+1)$, we can rewrite the $n$th term as
$$\left(\frac{1}{2^n} + \frac{1}{2^{n-1}{3}}\right) = \left(\frac{3}{2^n3} + \frac{2}{2^{n}{3}}\right) = \left(\frac{5}{2^n3}\right)$$
So the series is $$\sum_{n=1}^\infty \frac53 \cdot 2^{-n} = \frac53\sum_{n=1}^\infty 2^{-n} = \frac53.$$
|
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|
How find this $I=\iint_{\Sigma}(x^2+y^2+z^2)^{-\frac{3}{2}}(\frac{x^2}{a^4}+\frac{y^2}{b^4}+\frac{z^2}{c^4})^{-\frac{1}{2}}dS$ Find this Surface integral
$$I=\iint_{\Sigma}(x^2+y^2+z^2)^{-\frac{3}{2}}\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)^{-\frac{1}{2}}dS$$
where
$$\Sigma:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1,a>0,b>0,c>0$$
My try: let
$$x=a\sin{\alpha}\cos{\beta},y=b\sin{\alpha}\sin{\beta},z=c\cos{\alpha},\alpha\in[0,\pi],\beta\in[0,2\pi]$$
$$E=x''_{\alpha}+y''_{\alpha}+z''_{\alpha}=a^2\cos^2{\alpha}\cos^2{\beta}+b^2\cos^2{\alpha}\sin^2{\beta}+c^2\sin^2{\alpha}$$
$$G=x''_{\beta}+y''_{\beta}+z''_{\beta}=a^2\sin^2{\alpha}\sin^2{\beta}+b^2\sin^2{\alpha}\cos^2{\beta}$$
$$F=x'_{\alpha}x'_{\beta}+y'_{\alpha}y'_{\beta}+z'_{\alpha}z'_{\beta}=-a^2\sin{\alpha}\cos{\alpha}\sin{\beta}\cos{\beta}+b^2\sin{\alpha}\cos{\alpha}\sin{\beta}\cos{\beta}$$
so
$$EG-F^2=(a^2b^2\cos^2{\alpha}+a^2c^2\sin^2{\alpha}\sin^2{\theta}+b^2c^2\sin{\alpha}\cos^2{\beta})\sin^2{\alpha}$$
then
$$\int_{\Sigma}f(x,y,z)dS=\int_{\Delta}f(a\sin{\alpha}\cos{\beta},b\sin{\alpha}\sin{\beta},c\cos{\alpha})\sqrt{EG-F^2}d\alpha d\beta$$
where $\Delta:0\le\alpha\le \pi,0\le \beta\le 2\pi$
$$f(x,y,z)=(x^2+y^2+z^2)^{-\frac{3}{2}}\left(\dfrac{x^2}{a^4}+\dfrac{y^2}{b^4}+\dfrac{z^2}{c^4}\right)^{-\frac{1}{2}}$$
and follow I fell very ugly,Thank you very much
|
Let $\vec{r} = (x,y,z)$ and $\varphi(\vec{r}) = \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}$, we have
$$\vec{\nabla}\varphi(\vec{r}) = (\frac{2x}{a^2}, \frac{2y}{b^2},\frac{2z}{c^2})
\quad\implies\quad
\begin{cases}
\vec{r}\cdot \vec{\nabla}\varphi(\vec{r}) &= 2 \left(\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2}\right) = 2\varphi(\vec{r})\\
\left|\vec{\nabla}\varphi(\vec{r})\right|^2 &= 4\left(\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}\right)
\end{cases}$$
On the surfaces $\Sigma$, $\varphi(\vec{r}) = 1$, we can rewrite the factor
$$\frac{1}{\sqrt{\frac{x^2}{a^4} + \frac{y^2}{b^4} + \frac{z^2}{c^4}}} d\text{S}
\quad\text{ as }\quad
\vec{r} \cdot \frac{\vec{\nabla}\varphi(\vec{r})}{\left|\vec{\nabla}\varphi(\vec{r})\right|} d\text{S}
$$
But $\;\displaystyle \frac{\vec{\nabla}\varphi(\vec{r})}{\left|\vec{\nabla}\varphi(\vec{r})\right|}$ is nothing but the normal vector $\hat{n}{(\vec{r})}\;$ for the surfaces $\Sigma$. This means
$$I = \int_{\Sigma} \frac{\vec{r}}{|\vec{r}|^3} \cdot \hat{n} d\text{S}
= - \int_{\Sigma} \vec{\nabla}\frac{1}{|\vec{r}|} \cdot \hat{n} d\text{S}
$$
Since $\;\displaystyle \nabla^2 \frac{1}{|\vec{r}|} = 0\;$ for $\vec{r} \ne \vec{0}$,
we can use Divergence Theorem to deform the surface $\Sigma$ to the unit sphere $S^2$ centered at $\vec{0}$ without changing the value of $I$. As a result,
$$I = \int_{S^2} \frac{\vec{r}}{|\vec{r}|^3} \cdot \hat{n} d\text{S} = 4\pi$$
|
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|
Which is bigger, $1+3\sqrt{2}$ or $3\sqrt{3}$? Which is bigger, $a = 1+3\sqrt2$ or $b = 3\sqrt3$?
To find out result, I am doing: $(3\sqrt3)^2-(1+3\sqrt2)^2=8-2\sqrt{18}$.
But how can I find if the value of $8-2\sqrt{18}$ is positive or negative?
Thank you.
|
HINT:
Note that $(3\sqrt3)^2-(1+3\sqrt2)^2=27-1-6\sqrt2-18=8-6\sqrt2$. Next note that $\sqrt2>\frac86$.
|
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|
How find this equation solve this equation
$$\sqrt{\sqrt{3}-\sqrt{\sqrt{3}+x}}=x$$
My try: since
$$\sqrt{3}-x^2=\sqrt{\sqrt{3}+x}$$
then
$$(x^2-\sqrt{3})^2=x+\sqrt{3}$$
|
Let $y = \sqrt{\sqrt3 + x}$, then the equation becomes: $\sqrt3 - y = x^2$, and $y^2 = \sqrt3 + x$.
So:
$$\begin{align*}
y^2 - x & = y + x^2 \\
\Rightarrow (y + x)(y - x - 1) & = 0
\end{align*}$$
And we can go from here.
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Show that the sum of the series Show that the sum of the series is greater than 24
$$\frac{1}{\sqrt{1}+\sqrt{3}}+\frac{1}{\sqrt{5}+\sqrt{7}}+\frac{1}{\sqrt{9}+\sqrt{11}} +\cdots+\frac{1}{\sqrt{9997}+\sqrt{9999}} > 24$$
I see that
$\frac{1}{\sqrt{3}+\sqrt{1}}\cdot \frac{\sqrt{3}-\sqrt{1}}{\sqrt{3}-\sqrt{1}}=\frac{\sqrt{3}-\sqrt{1}}{2}$
In each term, the denominator is 2 so the series becomes $\frac{1}{2}\sum^{2499}_{n=0}[\sqrt{4n+3}-\sqrt{4n+1}]$
|
This was not far from being telescopic. So let us make this telescopic.
$$4S> \frac{2}{\sqrt{1}+\sqrt{3}}+ \frac{2}{\sqrt{3}+\sqrt{5}}+\frac{2}{\sqrt{5}+\sqrt{7}} +...+ \frac{2}{\sqrt{9997}+\sqrt{9999}}+ \frac{2}{\sqrt{9999}+\sqrt{10001}}$$
$$
=\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+\ldots+\sqrt{9999}-\sqrt{9997}+\sqrt{10001}-\sqrt{9999}
$$
$$
=\sqrt{10001}-1>99
$$
|
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|
Find the solution of the equation Find all real solutions of this equation :
$$x=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
|
The function
$$f(x)=\sqrt{2+\sqrt{2-\sqrt{2+x}}}$$
is non-negative and decreasing on its domain, $-2\le x\le2$. Since $f(2)=\sqrt2\lt2$, the equation $x=f(x)$ has exactly one solution. That solution will be a root of the polynomial of degree $8$ that comes from repeated squaring:
$$((x^2-2)^2-2)^2=2+x$$
L.F.'s comment to Nigel Overmars's answer indicates that $x=2\cos(2\pi/9)$ is a solution, so the upshot here is that it is the only solution.
The solution $x=2\cos(2\pi/9)$ can be verified using standard trig identities:
$$\begin{align}
\sqrt{2+2\cos(2\pi/9)}&=2\cos(\pi/9)\\
\sqrt{2-2\cos(\pi/9)}&=2\sin(\pi/18)\\
&=2\cos(\pi/2-\pi/18)\\
&=2\cos(4\pi/9)\\
\sqrt{2+2\cos(4\pi/9)}&=\cos(2\pi/9)
\end{align}$$
|
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|
Probability that $5$ divides $a^2+b^2$ is $\frac9{25}$
Two positive integers $a$ and $b$ are randomly selected with replacement, then prove that the probability of $(a^2 +b^2)/5$ being positive integer is $9/25$.
I found out the pattern of the last digits of $a^2$ , $b^2$ but then the sample space isn't finite so I dropped idea of thinking of finding the a's and b's such that their square's last digit is either 0 or 5. Please help.
|
$1^2 mod 5 \equiv 1$
$2^2 mod 5 \equiv 4$
$3^2 mod 5 \equiv 4$
$4^2 mod 5 \equiv 1$
$5^2 mod 5 \equiv 0$
We want $a^2+b^2 mod 5 \equiv 0$,
That happens for (1,3) (1,2) (4,3) (4,2) (3,1) (2,1) (3,4) (2,4) and (5,5)
where (x,y) represents (a mod 5, b mod 5)
5*5 ways to express (x,y)
So the probability is 9/25
|
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|
Absolute value of cubic polynomial roots lower than 1 Assume we have a cubic polynomial
$ x^3 +bx^2+xc+d=0 $, with $b,c,d$ real numbers.
Let $x_1, x_2, x_3 $ be the roots, either real or complex.
What is the relation of the coefficients $b,c$ and $d$ in order to have the roots inside the unit sphere, that means
$ |x_i| < 1$ for $i=1,2,3$ ?
|
What you're trying to determine is whether the conditions on $b,c,d$ that make $x^3 + bx^2 + cx + d$ a Schur polynomial. As mentioned before, a sufficient (but perhaps unnecessary) condition is that
$$
1>b>c>d
$$
Is true. For the precise conditions, one may apply either the Jury test or Bistritz test
Or, apply the Routh-Hurwitz criterion to
$$
(z-1)^3p\left(\frac{z+1}{z-1}\right) = \\(1+b+c+d)z^3+
(3+b-c-3d)z^2 + (3 - b - c - 3d)z + (1 - b + c - d)
$$
That last option leads to the following statement:
Let $a_3 = 1+b+c+d$, $a_2 = 3+b-c-3d$, $a_1 = 3 - b - c - 3d$, and $a_0 = 1 - b + c - d$. We may state that $x^3 + bx^2 + cx + d$ has its roots in the unit ball if and only if all of the following conditions are satisfied:
*
*$a_i>0$ for each $i$
*$a_2 a_1> a_3 a_0$
|
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|
Trouble solving $\int\sqrt{1-x^2} \, dx$ I am trying to learn how to solve integrals and I've got the hang out of a lot of examples, but I haven't got the slightest idea how to solve this example, this is how far I've got:
$$
\int\sqrt{1 - x^2} \, dx = x\sqrt{1-x^2} - 2\int\frac{x^2}{\sqrt{1-x^2}} \, dx
$$
Can you please help me solve it, and also some tips concerning the integration are welcome.
Thank you
|
Your idea of integrating by parts is good (but you did it wrong):
\begin{align}
\int\sqrt{1 - x^2} \, dx
&= x\sqrt{1-x^2} - \int x\frac{-2x}{2\sqrt{1-x^2}} \, dx \\
&= x\sqrt{1-x^2} + \int\frac{x^2}{\sqrt{1-x^2}}\,dx\\
&= x\sqrt{1-x^2} + \int\frac{x^2-1+1}{\sqrt{1-x^2}}\,dx\\
&= x\sqrt{1-x^2} - \int\sqrt{1-x^2}\,dx+\int\frac{1}{\sqrt{1-x^2}}\,dx\\
&= x\sqrt{1-x^2} + \arcsin x - \int\sqrt{1-x^2}\,dx
\end{align}
Therefore
$$
2\int\sqrt{1 - x^2} \, dx = x\sqrt{1-x^2} + \arcsin x
$$
and
$$
\int\sqrt{1 - x^2} \, dx = \frac{1}{2}x\sqrt{1-x^2} + \frac{1}{2}\arcsin x + c
$$
|
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|
calculating mod 7 Problem: Calculate $10^{10^{10}} \pmod 7$
According to Fermat's little theorem: $a^{p-1}\equiv1 \pmod p$, so $10^6\equiv1 \pmod 7$ and $10^n\equiv4 \pmod 6$, $n$ being any integer, why can we write $10^{10^{10}}\equiv10^4 \pmod 7$?
Similarly, if it's $10^n\equiv1 \pmod 3$, n being any integer, then wouldn't the equation become $10^{10^{10}}\equiv10^1\equiv3\pmod 7$?
I'm confused on the translation phase.
Thanks
|
Let me see if I can make it clear
Starting with $10^6 \equiv 1 \mod 7$, we have
$$ \begin{align}
10^{12} &= \left(10^6\right)^2 \equiv 1^2 = 1 \mod 7\\
10^{18} &= \left(10^6\right)^3 \equiv 1^3 = 1 \mod 7\\
10^{24} &= \left(10^6\right)^4 \equiv 1^4 = 1 \mod 7\\
\cdots&
\end{align} $$
This tells us how to find $10^n \mod 7$. For example, if I want $10^{73} \mod 7$
$$
10^{73} = 10^{72 + 1} = 10^{72}\, 10 = \left(10^6\right)^{12} ~10 = 10 \mod 7 = 3
$$
In general when calculating $a^n \mod p$ where $p$ is a prime, we cast off multiples of $p-1$ from n i.e.
$$
a^n \mod p \equiv a^m \mod p ~~\hbox{where $n \equiv m \mod (p-1)$ }$$
Note: If $p$ is not a prime, then replace $p-1$ by $\phi(p)$
|
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|
equation $x^2+ax+6a = 0$ has integer roots, Then integer values of $a$ is If the equation $x^2+ax+6a = 0$ has integer roots, Then integer values of $a$ is
$\bf{My\; Try}::$ Let $\bf{\alpha,\beta}$ be two roots of given equation $x^2+ax+6a = 0$
So $\bf{\alpha+\beta = -a}$ and $\bf{\alpha \cdot \beta = 6a}$ and $\bf{\alpha,\beta \in \mathbb{Z}}$
So $\bf{\alpha \cdot \beta =-6\alpha -6\beta \Rightarrow 6\alpha+\alpha \cdot \beta +6\beta +36 = 36}$
So $\bf{(\alpha+6)\cdot (\beta+6) = 36 = 6\times 6 = 9\times 4 = 18\times 2 = 36\times 1}$ and many more
But I did not understand how can i calculate for all ordered pairs.
Is there is any other method to solve it
Help Required
Thanks
|
What you have done so far is absolutely correct. However, for the sake of completeness I will repeat the beginning of your solution. Using Vieta's equations, we know that $$\alpha+\beta=-a$$ and $$\alpha\beta=6a$$
By substituting the first equation into the second, we get:
$$ \alpha\beta=-6\alpha-6\beta$$
Which simplifies to:
$$(\alpha+6)(\beta+6)=36$$
Since we are only interested in $a$ a solution $(x,y)$ would be the same as a solution $(y,x)$, hence, we assume without loss of generality that $\alpha \leq \beta$.
$$ $$ Hence, we know that $$(\alpha+6,\beta+6)=(1,36),(2,18),(3,12),(4,9),(6,6),(-36,-1),(-18,-2),(-12,-3),(-9,-4),(-6,-6)$$
Which give us our values for $\alpha$ and $\beta$:
$$(\alpha,\beta)=(-5,30),(-4,12),(-3,6),(-2,3),(0,0),(-42,-7),(-24,-8),(-18,-9),(-15,-10),(-12,-12)$$
Therefore, the possible values for $a$ are:
$$a=-25, -8, -3, -1, 0, 49, 32, 27, 25, 24$$
However, I do not think that this was your problem in solving the question. As indicated in your question, we were unsure how to calculate all ordered pairs. The key to solving problem was realizing that there were only a small number of ways to express 36 as the product of 2 integers. In fact, there were only 10 ways to do this. This is true for most small integers. When in doubt as to the feasibility of testing all cases, it can be useful to calculate the number of factors that the integer has. This can be done by breaking the number down into its prime factorization. For example, $36=2^2 \cdot 3^2$, to find the number of positive factors, we simply find the number of ways to separate $2^2 \cdot 3^2$ into 2 groups. There are 3 possibilities for 2, we can either have $2^0$ or $2^1$ or $2^2$ in the first group, each of which will produce unique factors, similarly, there are 3 possibilities for 3. Hence, the number of positive factors of $36$ is $3 \cdot 3=9$ As such, the total number of factors that $36$ has is $18$ (including negative factors). The number of ways to express 36 as the product of $2$ integers would, under normal circumstances, be half this, as each factor should have a unique other factor with which it can multiply to form $36$, (e.g. $1 \cdot 36=36$), however, 36 is a perfect square and has to repeated factors($6\cdot 6$ and $-6 \cdot -6$). Hence, the number of ways 2 express $36$ as the product of $2$ integers is $\frac{9\cdot 2 -2}{2} +2=10$, which is quite feasible to test.
|
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|
How to prove $(n!)^4\le2^{n^2+n}$? This may sound like a newbie but
question is to show that
$$(n!)^4\le2^{n^2+n} for \quad n=1,2,3...$$
I know it is true for n=1. $(1!)^4\le2^2$
and
assume it is true for $1<m\le n$ for all $\quad m\in N$
we have to show for m=n+1.
$((n+1)!)^2\le^? 2^{(n+1)^2+n+1}$
$((n+1)!)^4=(n!)^4.(n+1)^4\le 2^{n^2+n}.(n+1)^4$
so it is enough to show
$(n+1)^4\le4^{n+1}$
it is not true for n=2 but $(2!)^4\le2^{2^2+2}$ is true
so we can check for $n\ge3$
$(n+1)^4=n^4+4n^3+6n^2+4n+1\le^?4^{n+1}=4(4^n)$
I need to show $4n+1\le4^n$ ,$4n^3\le4^n$,$6n^2\le4^n,n^4\le4^n$ How Can I continue?
|
I would rather write $$2^{n^2+n}=4^{\frac{n(n+1)}2}=4^14^24^3\cdots 4^n$$
Then looking at the factors individually, it suffices to how that for each $n\geqslant 4$, that $n^4\leqslant 4^n$. Observe the last inequality is false for $n=3$; but your inequality is. Taking $\log$s, this is equivalent to showing that $$\frac{\log n}n\leqslant \frac{\log 4}4$$ for every $n\geqslant 4$. The inequality is obviously true for $n=4$, and if $x>4$ (in fact, if $x>e$) $$\frac{1-\log x}{x^2}<0$$ so $\dfrac{\log n}n$ decreases for $n>3$.
|
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|
$3$ never divides $n^2+1$ Problem: Is it true that $3$ never divides $n^2+1$ for every positive integer $n$? Explain.
Explanation: If $n$ is odd, then $n^2+1$ is even. Hence $3$ never divides $n^2+1$, when $n$ is odd.
If $n$ is even, then $n^2+1$ is odd. So $3$ could divide $n^2+1$.
And that is where I am stuck. I try to plug in numbers for $n$ but I want a more general form of showing that $3$ can't divide $n^2+1$ when $n$ is even.
|
Since $n-1,n,n+1$ are three successive integers so one of them must be divisible by 3 hence there product must be divisible by $3$ i.e. $3|(n-1)n(n+1) \implies 3|n^3-n$ , now
$3|n^2+1 \implies 3|n(n^2+1) \implies 3|n^3+n \implies 3|n^3+n-(n^3-n) \implies 3|2n$ , since 3 does
not divide $2$ so $3|n \implies 3|n^2 \implies 3|n^2+1-n^2 \implies 3|1$ , contradiction !
|
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|
Solving $\frac{\partial^2 u}{\partial x^2}+\frac{\partial^2 u}{\partial y^2}=c$ Question:
let $a,b,c$ be positive constants. Find $u=u(x,y)$ if is satisfies the partial differential equation
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c$$
and the boundary condition
$$u=0,\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.$$
my try: I know this is screened Poisson equation
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c$$
I only find this poisson equation one of the solution
$$u(x,y)=-\dfrac{c}{2}\cdot\dfrac{a^2b^2}{a^2+b^2}\left(1-\dfrac{x^2}{a^2}-\dfrac{y^2}{b}\right)$$
because
$$\dfrac{\partial u}{\partial x}=-\dfrac{c}{2}\cdot\dfrac{a^2b^2}{a^2+b^2}\left(-\dfrac{2x}{a^2}\right)$$
$$\dfrac{\partial^2 u}{\partial x^2}=c\cdot\dfrac{b^2}{a^2+b^2}$$
and
$$\dfrac{\partial^2 u}{\partial y^2}=c\cdot\dfrac{a^2}{a^2+b^2}$$
so
$$\dfrac{\partial^2 u}{\partial x^2}+\dfrac{\partial^2 u}{\partial y^2}=c\cdot\left(\dfrac{a^2}{a^2+b^2}+\dfrac{b^2}{a^2+b^2}\right)=c$$
and when
$$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\Longrightarrow u=0$$
and this solution is uniqueness? and How prove it? Thank you
Thank you very much!
|
The shape that your boundary conditions describe is in fact an ellipse. So if you switch to an elliptic coordinate system your pde should separate out nicely.
|
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|
Proving that if $a^2+b^2=c^2$, then $a+b\ge c$. Hello, I'm trying to prove this statement.
Let a,b & c be three positive real numbers and if $a^2+b^2=c^2$ then $a+b\ge c$
Any help, please?
|
Suppose Not, which means that $a + b < c $
Square both sides
$(a + b)^2 < c^2 $
$a^2 + b^2 + 2ab < c^2 $
But by assumption $a^2 + b^2 = c^2$
Therefore,
$a^2 + b^2 + 2ab < a^2 + b^2 $
Which means
$2ab < 0 $ where a & b are positive numbers
Contradiction as no 2 positive multiplication is less than 0
therefore,
$a+b≥c $
|
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|
integer $m$ for which $x^3-mx^2-mx-(m^2+1)=0\Rightarrow m^2+m(x^2+x)-x^3+1=0$ has integer roots Calculation of all integer values of $m$ for which the equation
$p(x)=x^3-mx^2-mx-(m^2+1)$ has integer roots.
$\bf{My\; Try}::$ Given $x^3-mx^2-mx-(m^2+1)=0\Rightarrow m^2+m(x^2+x)-x^3+1=0$
for integer roots(i.e $x\in \mathbb{Z}$), discriminant of quadratic equation must be perfect square.
So $(x^2+x)^2+4(x^3-1)=k^2$
Now How can i solve after that
help Required
Thanks
|
$\frac{-(x^2+x) \pm \sqrt{(x^2+x)^2+4(x^3-1)}}{2} = n$ where $n$ is an integer.
$\Rightarrow 4n^2 + (x^2+x)^2 + 4n(x^2+x) = 4(x^3-1) + (x^2+x)^2$
$\Rightarrow n^2+ n(x^2+x) = x^3-1$
$\Rightarrow n(n+x^2+x) = (x-1)(x^2+x+1)$
Hint:
For what integer values of n and x are the LHS and RHS equal?
|
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|
How find this $\left(\frac{1}{x^2+a^2}\right)^{(n)}$ Prove that
$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=(-1)^{(n)}n!\dfrac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}$$
my try:
since
$$\dfrac{1}{x^2+a^2}=\dfrac{1}{2ai}\left(\dfrac{1}{x-ai}-\dfrac{1}{x+ai}\right),i=\sqrt{-1}$$
so
$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=\dfrac{(-1)^nn!}{2ai}\left(\dfrac{1}{(x-ai)^{n+1}}-\dfrac{1}{(x+ai)^{n+1}}\right)$$
so
let$$x=a\cot{\theta},0<\theta<\pi,$$
then
$$x\pm ai=a(\cos{\theta}\pm i\sin{\theta})/\sin{\theta}$$
so
$$\dfrac{1}{(x\pm ai)^{n+1}}=\dfrac{\sin^{n+1}{\theta}}{a^{n+1}}[\cos{(n+1)\theta}\mp i\sin{(n+1)\theta}]$$
so$$\left(\dfrac{1}{x^2+a^2}\right)^{(n)}=(-1)^{(n)}n!\dfrac{\sin{[(n+1)\cdot \mathrm{arccot}{(x/a)}]}}{a(x^2+a^2)^{(n+1)/2}}$$
Question:
Have other methods?
Because this is important reslut,so I think this have other methods? Thank you
|
First, you can change variables to reduce to the case $a=1$. Second, the integral is the arctangent, so you are asking for the $(n+1)$th derivative of $\arctan(x)$. Maple says:
$$
\left(\frac{d}{dx}\right)^n\arctan x =
\frac{1}{2}\,{2}^{n}
G^{1, 3}_{3, 3}\left({x}^{2}\, \Big\vert\,^{0, 0, 1/2}_{0, (n-1)/2, n/2}\right)
{x}^{1-n}
$$
in terms of the Meijer G-function.
|
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|
Find max: $\frac{1}{a^3+2b^3+6}+\frac{1}{b^3+2c^3+6}+\frac{1}{c^3+2a^3+6}$ For $a,b,c>0$ and $abc=1$. Find max:
$\frac{1}{a^3+2b^3+6}+\frac{1}{b^3+2c^3+6}+\frac{1}{c^3+2a^3+6}$
I used AM-GM for $a^3+2b^3$ but I don't know how to continue ...
|
From the AM-GM Inequality,
$a^3+b^3+1\ge 3ab$,
$b^3+1+1\ge 3b$
$\Rightarrow a^3+2b^3+6\ge 3(ab+b+1)$
Similarly, $b^3+2c^3+6\ge 3(bc+c+1)$,$c^3+2a^3+6\ge 3(ca+a+1)$.
We have $\frac 1{a^3+2b^3+6}+\frac 1{b^3+2c^3+6}+$$\frac 1{c^3+2a^3+6}\le \frac 13(\frac 1{ab+b+1}+\frac 1{bc+c+1}+\frac 1{ca+a+1})$
$=\frac 13(\frac 1{ab+b+1}+\frac {ab}{b+1+ab}+\frac{b}{1+ab+b})$
$=\frac 13$.
The equality occurs when $a=b=c=1$
|
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|
Describing a linear map geometrically I have the following linear map $\mathbb{R}^2\to\mathbb{R}^2:$
$$\begin{pmatrix}x\\y\end{pmatrix}\mapsto \begin{pmatrix}9y-5x\\7y-4x\end{pmatrix}$$
I am asked to describe this geometrically.
I can see what is happening: the upper left quadrant is being squeezed into the upper right quadrant, in between the lines $y=\frac{7}{9}x$ and $y=\frac{4}{5}x$. Likewise the lower right quadrant is scrambled in between the same lines in the lower left quadrant. The other quadrants are stretched in the obvious way.
I can't find the proper terms to describe this well though. A rotation followed by a squeeze? Thanks to anyone who responds.
(I can upload an image if anyone is interested, but the lines are so close that it will not be very clear)
|
Suppose you have the map
$$
T_2
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
9y + x \\
y
\end{pmatrix}.
$$
Then this is called a horizontal shear. There is more information here: Shear mapping. Vertical shear is similarly defined. Now consider three additional maps
$$
T_1
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
-5x \\
y
\end{pmatrix} \\
T_3
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
x \\
-\frac{1}{5}y
\end{pmatrix} \\
T_4
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
x \\
y + \frac{4}{5}x
\end{pmatrix}.
$$
Then the given map is just
$$
T_4T_3T_2T_1
\begin{pmatrix}
x \\
y
\end{pmatrix}
=
\begin{pmatrix}
9y - 5x \\
7y - 4x
\end{pmatrix}.
$$
So the map is a combination of horizontal scaling followed by horizontal shear followed by vertical scaling and finally a vertical shear. Note that you can write transformations like this in terms of simpler transformations in many different ways. So the one I gave is just one possible way.
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/625189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
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|
Inequality $\frac{a + \sqrt{ab} + \sqrt[3]{abc}}{3} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}.$ Someone can to help me with a hint in the following problem:
Show that for any $a,b,c>0$,
$$\frac{a + \sqrt{ab} + \sqrt[3]{abc}}{3} \leq \sqrt[3]{a \cdot \frac{a+b}{2} \cdot \frac{a+b+c}{3}}.$$
I have tried using the Hölder inequality, but can not apply it efficiently.
Thanks!
|
By AM-GM inequality:
$\sqrt{ab}\le\sqrt[3]{ab.\frac{a+b}{2}}$
We prove that: $$\sqrt[3]{\frac{2}{a+b}.\frac{3}{a+b+c}}\left(a+\sqrt[3]{ab.\frac{a+b}{2}}+\sqrt[3]{abc}\right)\le3\sqrt[3]{a}$$
Also by AM-GM: $$\sqrt[3]{\frac{2a}{a+b}.\frac{3a}{a+b+c}}\le\frac{1+\frac{2a}{a+b}+\frac{3a}{a+b+c}}{3}$$
$$\sqrt[3]{\frac{3b}{a+b+c}}\le\frac{2+\frac{3b}{a+b+c}}{3}$$
$$\sqrt[3]{\frac{2b}{a+b}.\frac{3c}{a+b+c}}\le\frac{1+\frac{2b}{a+b}+\frac{3c}{a+b+c}}{3}$$
Sum up term by term, we obtain complete proof!
Equality holds iff $a=b=c$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/627543",
"timestamp": "2023-03-29T00:00:00",
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|
Why is it that $2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2$? Why is it that $$2^{10} + 2^{9} + 2^{8} + \cdots + 2^{3} + 2^2 + 2^1 = 2^{11} - 2?$$
|
Your sum is as follow:
$2(2^9+2^8+\cdots +2+1)$
which can be proved simply is equal to $2 \frac{2^{10}-1}{2-1}$ as desired.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/628501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Let $n=x^2+y^2$; $n=2^{2k}m$ or $n=2^{2k+1}m$ with $m$ odd. Prove that $2^{k}$ divides both $x$ and $y$. Let $n=x^2+y^2$ where x,y are integers, be one of the forms $n=2^{2k}m$ respectively $n=2^{2k+1}m$ with m odd. Prove that $2^{k}$ divides both x and y.
|
Let $(x,y)=2^rs$ where $s$ is odd
and $\displaystyle\frac xX=\frac yY=2^rs\implies (X,Y)=1$
$\displaystyle\implies x^2+y^2=2^{2r}s^2(X^2+Y^2)$
Now as $(X,Y)=1,$
Case $1:$ Either both are odd :
As $(2c+1)^2=4c^2+4c+1\equiv1\pmod4, X^2+Y^2=2\pmod4$ i.e., divisible by $2,$ but not by $4$
Case $2:$ $X,Y$ are of opposite parity
$\implies X^2+y^2\equiv0+1\pmod4$ i.e., odd
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/632930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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|
Prove that $ \lim_{n \rightarrow \infty } \frac{n+6}{n^2-6} = 0 $. My attempt:
We prove that $ \lim\limits_{n \rightarrow \infty } \dfrac{n+6}{n^2-6} =0$.
It is sufficient to show that for any $ \epsilon \in\textbf{R}^+ $, there exists an $ K \in \textbf{R}$ such that for any $ n > K $
$ \displaystyle \left| \frac{n+6}{n^2-6} - 0 \right| < \epsilon $.
Suppose $ n>4 $. Then $ n+6 < 7n $ and $ n^2 -6 > \frac{1}{2} n^2 $. So
$ \displaystyle \left| \frac{n+6}{n^2-6} - 0 \right| = \frac{n+6}{n^2-6 } < \frac{7n}{\frac{1}{2}n^2} = \frac{14}{n} $.
Consider $ K = \max\{4,\displaystyle \frac{14}{\epsilon}\} $ and suppose $ n> K $. Then $ n > \displaystyle \frac{14}{\epsilon} $. This implies that $ \epsilon > \displaystyle \frac{14}{n} $. Therefore
$ \displaystyle \left| \frac{n+6}{n^2-6} - 0 \right| = \frac{n+6}{n^2-6 } < \frac{7n}{\frac{1}{2}n^2} = \frac{14}{n} < \epsilon $.
Thus $ \lim\limits_{n \rightarrow \infty } \displaystyle \frac{n+6}{n^2-6}=0 $.
Is this proof correct? What are some other ways of proving this? Thanks!
|
Your proof is correct. Just to be pedantic, you can define $K$ directly as $14/\varepsilon$, since you precedently assumed that $n>4$.
As you've already noted in the comments, the actual way to solve limits without going through all those $\delta$-$\epsilon$ arguments is to use limits properties, such as the preservation of operations $+,\cdot$ (and, with some attention, $\div$). With these little theorems the proof is very easy: $$\{\dfrac{n+6}{n^2-6}\}_{n\to \infty}=\{\dfrac{1+6/n}{n-6/n}\}_{n\to \infty}=\dfrac{ \lim 1 + \lim 6/n}{\lim n -\lim6/n}=0$$
since all but the left bottom corner terms are bounded.
|
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"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solving $2\cos(x) = \sin(x)$ How would you solve equations of the form $ a \sin (x+b) = \sin (x)$?
Eg. $ 2 \cos(x) = \sin(x) $
I realy have no idea how I would solve this kind of equations.
|
I would do a tan half angle substitution with $t = \tan \frac{x}{2}$ which leads to
$$ \cos(x) = \cos\left( 2 \tan^{-1} t \right) = \frac{1-t^2}{1+t^2} \\
\sin(x) =\sin \left( 2 \tan^{-1} t \right) =\frac{2 t}{1+t^2} $$
which transforms your problem to
$$ \frac{1-t^2}{1+t^2} a \sin(b) + \frac{2 t}{1+t^2} \left(a \cos(b)-1\right) =0 \rightarrow$$
$$ 2 t \left(a \cos b -1\right) + a (1-t^2)\sin b=0 $$
The above has two solutions
$$ \boxed{ t = \dfrac{a\cos b-1 \pm \sqrt{c}}{a \sin b} }$$ where $c=1+a (a-2\cos b) $, and $ \boxed{x= 2 \tan^{-1} t }$.
So with $a=2$ and $b=0$ the result is $x =\pm \pi$
Notes:
This method allows the transformation of any trigonometric expression into a polynomial expression. it is often used in Robotics.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/633672",
"timestamp": "2023-03-29T00:00:00",
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|
What is the integrating factor of $2xy'+x^{2}e^{1-x^2}y=2$ I know how to start solving it, its dividing everything by $2x$, but I can't solve
$$\int \dfrac{x^2e^{1-x^2}}{2x}\,dx$$
|
Starting from where you left off (divide both sides by $2x$), we have
$$\dfrac{dy}{dx} + \dfrac{1}{2}xe^{1 - x^2}y = \dfrac{1}{x}$$
Since (by the standard form of linear equation) $p(x) = \dfrac{1}{2}xe^{1 - x^2}$, the integrating factor is
$$\begin{aligned}
\mu(x) = e^{\int p(x)\,dx} = e^{\int \frac{1}{2}xe^{1 - x^2}\,dx}\\
\end{aligned}$$
We need to evaluate the integral, which occurs as the power of $e$ of the integrating factor. Using substitution method, we see that if $u = 1 - x^2$ and $du = -2x\,dx$, then
$$\begin{aligned}
\dfrac{1}{2}\int xe^{1 - x^2}\,dx &= \dfrac{1}{2} \int -\dfrac{1}{2}e^{u}\,du\\
&= -\dfrac{1}{4} \int e^u\,du\\
&= -\dfrac{1}{4}e^u + \mbox{C}\\
&= -\dfrac{1}{4}e^{1 - x^2} + \mbox{C}
\end{aligned}$$
which answers the part of your question about integration. So for the integrating factor (The constant is neglected since it is arbitrary) becomes
$$\mu(x) = e^{-\frac{1}{4}e^{1 - x^2}}$$
As an extra bonus, I provided you the general solution of the given ordinary differential equation:
$$\begin{aligned}
\mu(x)\dfrac{dy}{dx} + \mu(x) \cdot \dfrac{1}{2} xe^{1 - x^2}y &= \dfrac{\mu(x)}{x}\\
e^{-\frac{1}{4}e^{1 - x^2}}\dfrac{dy}{dx} + e^{-\frac{1}{4}e^{1 - x^2}} \cdot \dfrac{1}{2} xe^{1 - x^2}y &= \dfrac{e^{-\frac{1}{4}e^{1 - x^2}}}{x}\\
\left[e^{-\frac{1}{4}e^{1 - x^2}}y\right]' &= \dfrac{e^{-\frac{1}{4}e^{1 - x^2}}}{x}\\
e^{-\frac{1}{4}e^{1 - x^2}}y &= \int \dfrac{e^{-\frac{1}{4}e^{1 - x^2}}}{x}\,dx\\
y(x) &= \dfrac{\int \dfrac{e^{-\frac{1}{4}e^{1 - x^2}}}{x}\,dx}{e^{-\frac{1}{4}e^{1 - x^2}}}
\end{aligned}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
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.