Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that $36|n^{12k-6}-n^6$ Prove that if gcd(n,6)=1 and k>0, then $36|n^{12k-6}-n^6$.
Idea: I want to show that $2|n^{12k-6}-n^6$ and $3|n^{12k-6}-n^6$.
For the first part, since gcd(n,6)=1, n must be an odd number, so it is easy to show that $n^{12k-6}-n^6$ is divisible by 2. I am stuck by the second part and i think that I need to use Fermat's little theorem somehow. Anyone can give some hint?
| We show that $4$ divides our difference, and that $9$ does.
Recall (or prove) that if $n$ is odd then $n^2\equiv 1\pmod{8}$. It follows that $n^{12k-6}\equiv 1\pmod{8}$, and $n^6\equiv 1\pmod{8}$, so in fact our difference is divisible by $8$, and therefore by $4$.
For divisibility by $9$, note that $n^2\equiv 1\pmod{3}$. So $n^6$ has the shape $(1+3t)^3$, which is $1+3(3t)+3(9t^2)+27t^3$, and is therefore congruent to $1$ modulo $9$. For the same reason, $n^{6(2k-1)}$ is congruent to $0$ modulo $9$, and therefore our difference is divisible by $9$.
Alternately, for divisibility by $9$, use Euler's Theorem. We have $\varphi(9)=6$, and therefore $n^6\equiv 1\pmod{9}$, and also $n^{6(2k-1)}\equiv 1\pmod{9}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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how does row vectors span a row space? the column space is very easy to understand, for example, we have:
$$\begin{pmatrix}
1& 0& 0 \\
0& 1& 0\\
0& 0& 1
\end{pmatrix}\times \begin{pmatrix}
a \\ b\\ c
\end{pmatrix}$$
if we let $V_1=\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}, V_2=\begin{pmatrix} 0\\ 1\\ 0 \end{pmatrix},V_3=\begin{pmatrix} 0\\ 0\\ 1 \end{pmatrix}$, so the product of above will be$$ \begin{pmatrix} V_1& V_2& V_3 \end{pmatrix}\times\begin{pmatrix}a \\ b\\ c\end{pmatrix}$$, so any vector in $R^3$ can be seen as $aV_1+bV_2+cV_3$
but for row space, I don't know how can I do the same thing, for example,if we set $V_1=\begin{pmatrix} 1& 0& 0 \end{pmatrix}, V_2=\begin{pmatrix} 0& 1& 0 \end{pmatrix},V_3=\begin{pmatrix} 0& 0& 1 \end{pmatrix}$, then it becomes $$ \begin{pmatrix} V_1\\ V_2\\ V_3 \end{pmatrix}\times\begin{pmatrix}a \\ b\\ c\end{pmatrix}$$, but we cannot even multiply these two matrices,can we? so how can I use row vectors to span $R^3$ in the form as $aV_1+bV_2+cV_3$ where $V_1,V_2,V_3$ are row vectors?
Another question is, are $\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}$ the same as $\begin{pmatrix} 1& 0& 0 \end{pmatrix}$ as they all in $R_3$
| I cannot comment since I don't have enough reps for that.
What about $(a\quad b\quad c)\times \begin{pmatrix} V_1\\ V_2\\ V_3 \end{pmatrix}$?
This is 1 x 3 matrix times 3 x 3 which gives you a 1 x 3 matrix.
Post multiplication of a matrix by a vector gives a linear combination of the columns of the matrix.
Pre multiplication of a matrix by a vector gives a linear combination of the rows of the matrix.
In answer to your second question:
$\begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}$ is not the same as $\begin{pmatrix} 1& 0& 0 \end{pmatrix}$. A vector or point, by default (unless stated otherwise) is assumed to be a column vector. So, $\begin{pmatrix} 1& 0& 0 \end{pmatrix} = \begin{pmatrix} 1\\ 0\\ 0 \end{pmatrix}^t$.
The operations you can perform with a row vector are different from those that can be performed by a column vector. Examples of such different operations are exactly the two examples (pre and post multiplication) you have provided in your OP.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integration of $\int \frac{x^2+20}{(x \sin x+5 \cos x)^2}dx$ How do we integrate
$$\int \frac{x^2+20}{(x \sin x+5 \cos x)^2}dx$$
Could someone give me some hint for this question?
| $\bf{Solution:}$ Using $$\displaystyle (x\cdot \sin x+5\cdot \cos x) = \sqrt{x^2+5^2}\left\{\frac{x}{\sqrt{x^2+5^2}}\cdot \sin x+\frac{5}{\sqrt{x^2+5^2}}\cdot \cos x\right\}$$
$\displaystyle = \sqrt{x^2+25}\cdot \cos\left(x-\phi\right)\;,$ where
$\displaystyle \sin \phi = \frac{x}{\sqrt{x^2+25}}$ and $\displaystyle \cos \phi = \frac{5}{\sqrt{x^2+25}}$ and $\displaystyle \tan \phi = \frac{x}{5}\Rightarrow \phi = \tan^{-1}\left(\frac{x}{5}\right)$
So Integral is $$\displaystyle = \int \sec^2(x-\phi)\cdot \left(\frac{x^2+20}{x^2+25}\right)dx$$
Now Let $$\displaystyle (x-\phi) = y\Rightarrow \left(x-\tan^{-1}\left(\frac{x}{5}\right)\right)=y\;,$$ Then $\displaystyle \left(\frac{x^2+20}{x^2+5^2}\right)dx = dy$
So Integral is $$\displaystyle \int \sec^2(y)dy = \tan y +\mathcal{C} = \tan\left(x-\tan^{-1}\left(\frac{x}{5}\right)\right)+\mathcal{C}$$
So $$\displaystyle \int \frac{x^2+20}{(x\cdot \sin x+5\cdot \cos x)^2}dx = \frac{5\cdot \tan x-x}{5+x\cdot \tan x}+\mathcal{C} = \frac{5\sin x-x\cos x}{5\cos x+x\sin x}+\mathcal{C}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $x+y+z=0$, prove that $\frac{x^2}{2x^2+yz}+\frac{y^2}{2y^2+zx}+\frac{z^2}{2z^2+xy}=1$ A problem in my homework had asked me:
When $x+y+z=0$, evaluate$$\frac{x^2}{2x^2+yz}+\frac{y^2}{2y^2+zx}+\frac{z^2}{2z^2+xy}$$
Without too much difficulty, one can see that the value should be $1$ using $(x,y,z)=(1,0,-1)$.
I decided to use $x=-y-z$, which turned out not to be as difficult as initially thought. However, would someone care to enlighten me to some other methods of doing this?
| The first term is $$\frac{x^2}{2x^2-xy-y^2}=\frac{x^2}{(2x+y)(x-y)}=\frac{x^2}{(x-z)(x-y)}$$
Do the same transformation on each of the fractions and add them up.
So the whole expression is $$\frac{x^2}{(x-z)(x-y)}+\frac{y^2}{(y-z)(y-x)}+\frac{z^2}{(z-y)(z-x)}=...=1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Doubt in solving $\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}+\cot^{-1}\frac{1}{x}=\pi$ Find the value of $x$ if $$\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}+\cot^{-1}\frac{1}{x}=\pi$$ First i tried to calculate the value of
$$\sec^{-1}\sqrt{5}+\csc^{-1}\frac{\sqrt{10}}{3}=\sin^{-1}\frac{2}{\sqrt{5}}+\sin^{-1}\frac{3}{\sqrt{10}}$$ Letting
$$\theta=\sin^{-1}\frac{2}{\sqrt{5}}+\sin^{-1}\frac{3}{\sqrt{10}}$$ taking $\sin$ both sides and using $sin(A+B)=sinAcosB+cosAsinB$ we get
$$\sin\theta=\sin\left(\sin^{-1}\frac{2}{\sqrt{5}}\right)\cos\left(\sin^{-1}\frac{3}{\sqrt{10}}\right)+\cos\left(\sin^{-1}\frac{2}{\sqrt{5}}\right)\sin\left(\sin^{-1}\frac{3}{\sqrt{10}}\right)$$ so
$$\sin\theta=\frac{5}{\sqrt{50}}=\frac{1}{\sqrt{2}}$$
Now my doubt is will $\theta=\frac{\pi}{4}$ or $\theta=\frac{3\pi}{4}$ ?
My book has taken $\theta=\frac{3\pi}{4}$
| The function $\sin^{-1}:[-1,1]\to[-\frac{\pi}2,\frac{\pi}2]$ is increasing. So
$$\frac{\sqrt3}{2}<\frac2{\sqrt5}<1\quad\text{and}\quad\frac{\sqrt3}{2}<\frac3{\sqrt{10}}<1\qquad\implies \qquad\frac{2\pi}{3}<\theta<\pi$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding the coefficient of expansion Question:
Find the coefficient of $x^{11}$ in the expansion of:$$(1+x^2)^4(1+x^3)^7(1+x^4)^{12}$$
The traditional way of doing this, as far as I know, is to first find the coefficient of each term that has $x^{11}$, and then sum it. However, with three individual terms in multiplication, this would be a very tedious job. Is there a shorter way to approach such problems?
| To calculate the coefficient by hand is somewhat cumbersome. Here is a notation which helps to make it better manageable. It is convenient to use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ of a series.
We obtain by successively applying the binomial theorem
\begin{align*}
[x^{11}]&(1+x^2)^4(1+x^3)^7(1+x^4)^{12}\\
&=[x^{11}]\sum_{j=0}^{12}\binom{12}{j}x^{4j}(1+x^2)^4(1+x^3)^7\tag{1}\\
&=\sum_{j=0}^{2}\binom{12}{j}[x^{11-4j}](1+x^2)^4(1+x^3)^7\tag{2}\\
&=\sum_{j=0}^{2}\binom{12}{j}[x^{11-4j}]\sum_{k=0}^7\binom{7}{k}x^{3k}(1+x^2)^4\tag{3}\\
&=\sum_{j=0}^{2}\sum_{k=0}^3\binom{12}{j}\binom{7}{k}[x^{11-4j-3k}](1+x^2)^4\qquad\qquad\quad 4j+3k\leq 11\tag{4}\\
&=\sum_{j=0}^{2}\sum_{k=0}^3\binom{12}{j}\binom{7}{k}[x^{11-4j-3k}]
\sum_{l=0}^{4}\binom{4}{l}x^{2l}\qquad\qquad 4j+3k\leq 11\tag{5}\\
&=\sum_{j=0}^{2}\sum_{k=0}^3\sum_{l=0}^{4}
\binom{12}{j}\binom{7}{k}\binom{4}{l}\qquad\qquad\qquad\qquad 4j+3k+2l= 11\tag{6}\\
\end{align*}
Comment:
*
*In (1) we start expanding the binomial with the greatest power $x^4$ which keeps the number of summands small as we will see in the next step.
*In (2) we use the linearity of the coefficient of operator and apply the rule $[x^{m-n}]=[x^m]x^n$. Note that since the power of $x$ is non negative, the upper limit of the index $j$ is now $2$.
*In (3) we expand the binomial $(1+x^3)^7$.
*In (4) we do a similar job as in (2) and keep in mind, that the power of $x$ is non-negative by stating $4j+3k\leq 11$.
*In (5) we expand the last binomial
*In (6) we apply the coefficient of operator the last time and obtain a summation formula for the coefficient of $x^{11}$.
The final step is to do some bookkeeping and look according to (6) for all triples $(j,k,l)$ with
\begin{align*}
\begin{matrix}
0\leq j \leq 2\\
0\leq k \leq 3\\
0\leq l \leq 4\\
\end{matrix}
\qquad\quad\text{and}\quad\qquad 4j+3k+2l= 11
\end{align*}
We find
\begin{array}{rrrrr}
j&k&l&\binom{12}{j}\binom{7}{k}\binom{4}{l}&\sum\\
\hline
2&1&0&462&462\\
1&1&2&504&966\\
0&1&4&7&973\\
0&3&1&140&\color{blue}{1113}\\
\end{array}
resulting finally in
\begin{align*}
[x^{11}](1+x^2)^4(1+x^3)^7(1+x^4)^{12}=1113
\end{align*}
in accordance with the result of Wolfram Alpha.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \cdots + \frac{1}{{{a_n}}} < 2$
If ${{a}_{1}},{{a}_{2}},\ldots ,{{a}_{n}}$ are distinct odd natural
numbers not divisible by any prime greater than 5, then show that
$\frac{1}{{{a_1}}} + \frac{1}{{{a_2}}} + \cdots + \frac{1}{{{a_n}}}
< 2$.
First I have noted that $a_i$s are in the form $3^a5^b$ where $a,b$ are non-negative integers.Now let ${a_1} < {a_2} < \cdots < {a_n}$. As the inequality involves 2, I thought of the geometric series $1,\frac{1}{2},\frac{1}{{{2^2}}}, \cdots $ and I thought that each $\frac{1}{{{a_i}}}$ would be less than or equal to $\frac{1}{{{2^{i - 1}}}}$ and this works for some $a_i$s such as $\frac{1}{{{a_1}}} = 1 \leqslant 1, \frac{1}{{{a_2}}} = \frac{1}{3} < \frac{1}{2},\frac{1}{{{a_3}}} = \frac{1}{5} < \frac{1}{{{2^2}}}, \frac{1}{{{a_4}}} = \frac{1}{9} < \frac{1}{{{2^3}}}$. But this do not work for $i\ge 5$. I would appreciate any suggestion or idea on this problem.
| $$\sum_{i=1}^n \frac{1}{a_n}<\sum_{i=0}^\infty\sum_{j=0}^\infty\frac{1}{3^i5^j}=\sum_{i=0}^\infty\frac{1}{3^i\left(1-\frac{1}{5}\right)}$$
$$=\frac{5}{4}\sum_{i=0}^{\infty}\frac{1}{3^i}=\frac{5}{4}\cdot \frac{1}{1-\frac{1}{3}}=\frac{15}{8}<2$$
So, in fact, you have a stronger statement: $\sum_{i=1}^n \frac{1}{a_n}<\frac{15}{8}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Where can I read about this 'rule'? I was trying to solve an equation, I got nowhere, and the solution used a 'rule' that I have never seen before, $a^TX\,b = b^TX\,a$.
What is this rule?
Where can I read about it?
Note: $a$ and $b$ are vectors and $X$ is a matrix.
| a^t is for transpose, right?
$
\begin{pmatrix}
a_1 & a_2 \\
\end{pmatrix}
$$
\begin{pmatrix}
x_{11} & x_{12} \\
x_{21} & x_{22} \\
\end{pmatrix}
$$
\begin{pmatrix}
b_1 \\ b_2 \\
\end{pmatrix}
$ = $a_1b_1x_{11} + a_2b_1x_{21} + a_1b_2x_{12} + a_2b_2x_{22} $
but
$
\begin{pmatrix}
b_1 & b_2 \\
\end{pmatrix}
$$
\begin{pmatrix}
x_{11} & x_{12} \\
x_{21} & x_{22} \\
\end{pmatrix}
$$
\begin{pmatrix}
a_1 \\ a_2 \\
\end{pmatrix}
$ = $a_1b_1x_{11} + a_1b_2x_{21} + a_2b_1x_{12} + a_2b_2x_{22} $
So, if X is symmetrical, ($x_{12} = x_{21}$), then this rule is true just by doing multiplication.
| {
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Find all the Zeros and their multiplicities of $f(x)=x^5 +4x^4 +4x^3 -x^2-4x +1$ over $\Bbb Z_5$.
Find all the Zeros and their multiplicities of $f(x)=x^5 +4x^4 +4x^3 -x^2-4x +1$ over $\Bbb Z_5$.
Firstly,I've found the zeros of $f(x)$,just by simply substituting the elements of $\Bbb Z_5=\{0,1,2,3,4\}$ in $f(x)$.I get $f(1)=0$ and $f(3)=0$.So,by factor theorem $(x-1)$ and $(x-3)$ are factors of $f(x)$.
On dividing $f(x)$ by $(x-1)$, I get $x^4 +5x^3 +9x^2+8x +4$ as quotient and $5$ as remainder,which is equal to zero in $\Bbb Z_5$.So,$f(x)=(x-1)(x^4 +5x^3 +9x^2+8x +4)$,thus,multiplicity of $x=1$ is $1$.
On applying the same procedure for $(x-3)$.I get $f(x)=(x-3)(x^4 +7x^3+25x^2 +74x+218)$,thus,multiplicity of $x=3$ is $1$.
But this is Incorrect.I don't know where I am committing mistake.
Is my approach correct?
If there is any general method for finding the quotient and remainder,when some arbitrary polynomial $T(x)$ is divided by another arbitrary polynomial $g(x)$ in some finite field,Please tell me.
Any suggestions are heartily welcome!!
Thank you!
| Start by rewriting $\bar{f}(x)=x^5-x^4-x^3-x^2+x+1$. Divide by $x-1$ to get
$$\bar{f}(x)=(x-1)(x^4-x^2-2x-1)$$
Now divide $x^4-x^2-2x-1$ by $x-3$ to get
$$\bar{f}(x)=(x-1)(x-3)(x^3+3x^2+3x+2)$$
If we consider now $g(x)=x^3+3x^2+3x+2$ we notice that $g(3)=0$. So $3$ is a double root of $f$. Let's divide $g$ by $x-3$ we then get
$$\bar{f}(x)=(x-1)(x-3)^2(x^2+x+1)$$
And we have now completely factored $f$ over $\Bbb{Z}_5$.
We could have as well computed $f'(x)=5x^4-4x^3-3x^2-2x+1$ whose reduction $\pmod 5$ is $x^3+2x^2-2x+1$ and notice that $f'(1)=2$ while $f'(3)=0$ to conclude that $1$ is a simple root and $3$ is a double root.
| {
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Simplifying radicals inside radicals: $\sqrt{24+8\sqrt{5}}$ Simplify: $\sqrt{24+8\sqrt{5}}$
I removed the common factor 4 out of the square root to obtain $2\sqrt{6+2\sqrt{5}}$, but the answer key says it is $2+2\sqrt{5}$.
Am I missing out on some general rule here?
| The nested radical:
$$
\sqrt{6+2\sqrt{5}}=\sqrt{6+\sqrt{20}}
$$
can be denested using the identity (that you can easely verify).
$$
\sqrt{a\pm \sqrt{b}}=\sqrt{\dfrac{a+ \sqrt{a^2-b}}{2}}\pm\sqrt{\dfrac{a- \sqrt{a^2-b}}{2}}
$$
that works if $a^2-b$ is a perfect square. In this case $a^2-b=16$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to find the projw(x) Sorry about formatting, new to this.
Subspace w =
\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} such that $x_1 = x_3$ of R3.
x = \begin{pmatrix}-2\\1\\3\end{pmatrix}
Question is to find $proj_w (x)$.
| We'll assume it's the projection with respect to the standard inner product. The basis for $w$ is $$\left\{\begin{bmatrix}1\\0\\1\end{bmatrix},\begin{bmatrix}0\\1\\0\end{bmatrix}\right\}.$$
Therefore, omitting details, you have a least squares problem for the system $$\begin{bmatrix}1&0\\0&1\\1&0\end{bmatrix}\begin{bmatrix}\alpha\\ \beta\end{bmatrix}=\begin{bmatrix}-2\\1\\3\end{bmatrix}.$$ Call that system $Ax=b$. The least squares solution is $$x=(A^T A)^{-1}A^Tb$$ or $$\begin{bmatrix}\alpha \\ \beta \end{bmatrix}=\begin{bmatrix}2&0\\0&1\end{bmatrix}^{-1}\begin{bmatrix}1&0&1\\0&1&0\end{bmatrix}\begin{bmatrix}-2\\1\\3\end{bmatrix}=\begin{bmatrix}1/2\\1\end{bmatrix}.$$ These are the coefficients of the project, so the projection itself is $$\begin{bmatrix}1&0\\0&1\\1&0\end{bmatrix}\begin{bmatrix}1/2\\1\end{bmatrix}=\begin{bmatrix}1/2\\1\\1/2\end{bmatrix}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $a,b,c,d > 0$ and $a+b+c+d=1$ , find the maximum value of $M=abc+abd+acd+bcd-abcd$
If $a,b,c,d > 0$ and $a+b+c+d=1$ , find the maximum value of $$M=abc+abd+acd+bcd-abcd$$
I don't have any idea how to simplify this equation. I have tried for $a,b,c>0$ and $a+b+c=1$ and I got $ab+ac+bc-abc=(1-a)(1-b)(1-c)<= 8/27$.
Thanks.
| The following solution is similar to my answers
in An inequality $16(ab + ac + ad + bc + bd + cd) \le 5(a + b + c + d) + 16(abc + abd + acd + bcd)$
and An inequality $2(ab+ac+ad+bc+bd+cd)\le (a+b+c+d) +2(abc+abd+acd+bcd)$
Let $x = a + b, \, y = c + d$.
We have $x + y = 1$.
We have
\begin{align*}
M &= ab y + cd x - ab cd\\
&= xy - (x - ab)(y - cd)\\
&\le xy - (x - x^2/4)(y - y^2/4) \tag{1}\\
&= -\frac{1}{16}x^2y^2 + \frac{1}{4}xy(x + y)\\
&= -\frac{1}{16}x^2y^2 + \frac{1}{4}xy\\
&= -\frac{1}{16}(2 - xy)^2 + \frac14 \\
&\le -\frac{1}{16}(2 - 1/4)^2 + \frac14 \tag{2}\\
&= \frac{15}{256}
\end{align*}
where we have used $ab \le (a + b)^2/4 = x^2/4 \le x$ and $cd \le (c + d)^2/4 = y^2/4 \le y$ in (1), and
$xy \le (x + y)^2/4 = 1/4$ in (2).
Also, when $a = b = c = d = 1/4$,
we have $a + b + c + d = 1$ and $M = 15/256$.
Thus, the maximum of $M$ is $15/256$.
| {
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$f$ is an odd, $2\pi$ periodic function. Write explicitly Let $f$ be an odd $2\pi$ periodic function defined on $(0,\pi)$ by $f(x) = \frac12 (\pi - x)$.
I am asked to "write $f$ explicitly". what does this mean? how to do that? thanks for help.
| I assume the question asks for an explicit formula for all the real numbers $(-\infty,\infty)$.
It is an odd function, meaning $f(x) = -f(-x)$ for all $x$, therefore $f(0) = -f(0) = 0$. Therefore we have its values for $(-\pi,\pi)$. Because of the periodicity, this automatically gives us the values for the function for all $x\neq \pi n$ where $n\in N$.
From periodicity and oddness we get $f(-\pi) = f(\pi) = -f(-\pi) = 0$. So we have the values set for a complete $2\pi$ period $(-\pi,\pi]$.
To write a truly explicit formula for $f(x)$ we have to give an explicit formula representing the annoying periodicity, meaning we'd want to deal only with $x -2\pi\lfloor \frac{x} {2\pi} \rfloor$. Note $x + 2\pi n -2\pi\lfloor \frac{x + 2\pi n} {2\pi} \rfloor = x + 2\pi n -2\pi(\lfloor \frac{x +n} {2\pi} \rfloor + n) = x -2\pi\lfloor \frac{x} {2\pi} \rfloor$.
This leaves us with the following function:
$$f(x)=
\begin{cases}
\hfill \frac{1}{2}\left(\pi-\left(x -2\pi\lfloor \frac{x} {2\pi} \rfloor\right)\right) \hfill & \text{ if } x\neq2n\pi \\
\hfill 0 \hfill & \text{ if } x=2n\pi \\
\end {cases}
$$
Where $n$ is an integer. Note that $f(nπ)$ still equals $0$.
This function would also work:
$$f(x)=
\begin{cases}
\hfill -\arctan \left(\tan \left(\frac{x+\pi }{2}\right)\right) \hfill & \text{ if } x\neq2n\pi \\
\hfill 0 \hfill & \text{ if } x=2n\pi \\
\end {cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1750299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$ Solve equation $8\cos x - 6\sin x = 5$ where $0 \le x \le 360$.
I am not asked to use any form, so I am going to use $k\cos(x-\alpha)$.
$8\cos x - 6\sin x = k\cos(x-\alpha)$
$$=k(\cos x\cos\alpha + \sin x\sin\alpha)$$
$$=k\cos\alpha\cos x + k\sin\alpha\sin x$$
equating coefficients:
$k\cos\alpha = 8$
$k\sin\alpha = 6$, or is it $k\sin\alpha = -6$? I find this really confusing
$k = \sqrt{8^2 + 6^2} = 10$
$\alpha$ is in the 1st quadrant where both sin and cos are positive.
$\alpha = \arctan\frac{6}8 = 36.8^{\circ}$
$\therefore10\cos(x - 36.8) = 5$
I have a maximum and minimum value of 10
From here I do not know how to finish solving for x.
| So, $\cos(x-\alpha)=\dfrac5{10}=\cos60^\circ$
$\implies x-\alpha=360^\circ n\pm60^\circ$ where $n$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1755112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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} |
Is it true that $log(i) = \frac\pi2i$ ? If so, are both of these legitimate proofs? They seem too beautiful not to be... Sorry if this is a naive question. I have not yet taken any upper level math courses involving complex numbers. However, in preparation for those courses, together with utilizing the knowledge that mathematicians such as Euler would just play around with infinite series, led me to this result. This seems too beautiful not to be true. If it is true, could someone explain the general framework from which this is derived?
Statement: $$log(i) = \frac\pi2i$$
Proof One: By Euler's Identity $e^{ix} = cos{x} + isin{x}$
So that $$e^{\frac\pi2i} = cos\frac\pi2 + isin\frac\pi2 = i$$
But $$e^{log(i)} = i$$
Thus $$e^{log(i)} = e^{\frac\pi2i}$$
So that $$log(i) = \frac\pi2i$$
Proof Two: Using Power Series
$$log(i) = log(\frac{2i}2) = log\left(\frac{(1+i)(1+i)}{(1+i)(1-i)}\right) = log\left(\frac{1+i}{1-i}\right) = log(1+i) - log(1-i)$$
But $$ log(1+i) = i - \frac{i^2}2 + \frac{i^3}3 - \frac{i^4}4 + ...$$
And $$-log(1-i) = i + \frac{i^2}2 + \frac{i^3}3 + \frac{i^4}4 + ...$$
Thus $$log(1+i) - log(1-i) = 2\{i + \frac{i^3}3 + \frac{i^5}5 + \frac{i^7}7 + ...\} = 2i\{1 - \frac13 + \frac15 - \frac17 + ...\} = 2i\frac\pi4 = \frac\pi2i$$
Therefore $$log(i) = \frac\pi2i$$
Added June 4, 2016: I found a delightful identity for logarithms of complex quantites in general as opposed to the case $a+bi = i$. Since the proof is essentially a generalization of my proof that $log(i) = \frac{\pi}{2}i$ as well as being aesthetically pleasing I thought it was worth posting.
Statement: $$log\left(a+bi\right) = \frac{1}{2}log\left(a^{2}+b^{2}\right) + itan^{-1}\frac{b}{a}$$
Proof: $$log\left(a+bi\right) = log\left({\sqrt{a^{2}+b^{2}}}{\frac{a+bi} {\sqrt{a^{2}+b^{2}}}}\right) = \frac{1}{2}log\left(a^{2}+b^{2}\right) + log\left({\frac{a+bi}{\sqrt{a^{2}+b^{2}}}}\right)$$
Now, $$log\left({\frac{a+bi}{\sqrt{a^{2}+b^{2}}}}\right) = log\left({\frac{1+i\frac{b}{a}}{\sqrt{1+\frac{b^{2}}{a^2}}}}\right) = log\left({1+i\frac{b}{a}}\right) - \frac{1}{2}log\left(1+\frac{b^{2}}{a^2}\right)$$
But, $$log\left(1+i\frac{b}{a}\right) = \left(i\frac{b}{a}\right)-\frac{1}{2}{\left(i\frac{b}{a}\right)}^{2} + \frac{1}{3}{\left(i\frac{b}{a}\right)}^{3} - \frac{1}{4}{\left(i\frac{b}{a}\right)}^{4} +... =
i\left( \left(\frac{b}{a}\right)-\frac{1}{3}{\left(\frac{b}{a}\right)}^{3} + \frac{1}{5}{\left(\frac{b}{a}\right)}^{5} - \frac{1}{7}{\left(\frac{b}{a}\right)}^{7} +... \right) +
\\ \frac{1}{2}\left(\frac{b}{a}\right)-\frac{1}{4}{\left(\frac{b}{a}\right)}^{4} + \frac{1}{6}{\left(\frac{b}{a}\right)}^{6} - \frac{1}{8}{\left(\frac{b}{a}\right)}^{8} +...$$
And,$$ \frac{1}{2}log\left(1+\frac{b^2}{a^2}\right) = \frac{1}{2}\left(\left({\frac{b^{2}}{a^{2}}}\right) - \frac{1}{2}\left({\frac{b^{2}}{a^{2}}}\right)^{2} + \frac{1}{3}\left({\frac{b^{2}}{a^{2}}}\right)^{3} - \frac{1}{4}\left({\frac{b^{2}}{a^{2}}}\right)^{4} +...\right) = \\ \frac{1}{2}\left(\frac{b}{a}\right)^{2}-\frac{1}{4}{\left(\frac{b}{a}\right)}^{4} + \frac{1}{6}{\left(\frac{b}{a}\right)}^{6} - \frac{1}{8}{\left(\frac{b}{a}\right)}^{8} +...$$
So that, $$log\left({1+i\frac{b}{a}}\right) - \frac{1}{2}log\left(1+\frac{b^{2}}{a^2}\right) = i\left( \left(\frac{b}{a}\right)-\frac{1}{3}{\left(\frac{b}{a}\right)}^{3} + \frac{1}{5}{\left(\frac{b}{a}\right)}^{5} - \frac{1}{7}{\left(\frac{b}{a}\right)}^{7} +... \right) = itan^{-1}{\frac{b}{a}}$$
Therefore, $$log(a+bi) = \frac{1}{2}log(a^{2}+b^{2}) + itan^{-1}\frac{b}{a}$$
Example: Let a=0 and b=1 and we have
$$log(i) = itan^{-1}\infty = \frac{\pi}{2}i + 2\pi mi$$
| From Euler's identity $$e^{i\pi/2} = cos(\pi/2) + i sin(\pi/2) = i
$$
Take the natural log of both sides.
The left $\ln(e^{i\pi/2}) = i\pi/2$ and the right is simply $\ln i$.
Since both are equal, $i\pi/2 = \ln i.$
So, it is not arbitrary log but ln. Otherwise, your identity is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1756321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solve the system of equations: $a+b+c=2$, $a^2+b^2+c^2=6$, $a^3+b^3+c^3=8$ If we have \begin{cases} a+b+c=2 \\ a^2+b^2+c^2=6 \\ a^3+b^3+c^3=8\end{cases} then what is the value of $a,b,c$?
| Here I am giving Solution of yours first question, Next time when you post question,
plz show your try.
Let $x=a,x=b\;,x=c$ be the roots of the equation $(x-a)(x-b)(x-c) =0$
So $$x^3-(a+b+c)x^2+(ab+bc+ca)x-abc=0.............(1)$$
Now here $a+b+c=2$ and $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$
So we get $ab+bc+ca = -1$ and $$a^3+b^3+c^3-3abc=(a+b+c)\left[a^2+b^2+c^2-ab-bc-ca\right]$$
So we get $8-3abc=2[6+1]\Rightarrow abc=-2$
Put all value in equation $(1)\;,$ We get $x^3-2x^2-x+2=0$
So we get $$x^3-2x^2-x+2=(x+1)(x-1)(x-2) =0$$
So we get $x=1\;,-1\;,2$
So we get $$a,b,c\in \left\{-1,1,2\right\}$$, bcz Given equation are symmetrical in $a,b,c$
| {
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"url": "https://math.stackexchange.com/questions/1759141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all integral solutions for the Diophantine Equations $x^4 - x^2y^2 + y^4 = z^2$ and $x^4 + x^2y^2 + y^4 = z^2$.
Find all integral solutions for the Diophantine Equations $$x^4 - x^2y^2 + y^4 = z^2$$ and $$x^4 + x^2y^2 + y^4 = z^2$$
I basically think that to solve these equations we need to use the fact that all Pythagoras Triplets are like $(k \cdot 2mn, k \cdot (m^2 - n^2), k \cdot (m^2 + n^2))$. The above equations can be modified a little bit to make all the terms perfect squares. Then the first equation would be $$(x^2 - y^2)^2 + (xy)^2 = z^2$$ and the second one would be $$z^2 + (xy)^2 = (x^2 + y^2)^2$$
I have found no other clues. Please help me proceed.
Thanks.
| Updated
We can try to find solutions ourself.
Equation 1.
$\quad x^4+x^2y^2+y^4 = z^2.$
Let
$$\gcd(x,y) = m,\quad x=mX,\quad y=mY,\quad z=m^2Z,$$
then
$$X^4+X^2Y^2+Y^2=Z^2,\quad\gcd(X,Y)=1.$$
$$(X^2+XY+Y^2)(X^2-XY+Y^2)=Z^2,\quad\gcd(X,Y)=1.$$
Multipliers in the left part are the same parity. If they are even, then both $X$ and $Y$ are even and coprime, but this conditions are incompatible. So the multipliers are odd, and if they can have only odd common divider, which are the divider of expressions
$$3(X^2+XY+Y^2)-(X^2-XY+Y^2) = 2(X+Y)^2$$
and
$$3(X^2-XY+Y^2)-(X^2+XY+Y^2) = 2(X-Y)^2,$$
so numbers $X+Y$ and $X-Y$ share a common odd divider. And this odd divider must divide both their sum $2X$ and difference $2Y$, when $X$ and $Y$ are coprime. This contradiction proves that
$$\gcd(X^2-XY+Y^2,X^2+XY+Y^2)=1.$$
Product of coprime positive factors is the square of an integer. Consequently, each of them is the square of coprime integers:
$$\begin{cases}
X^2+XY+Y^2=U^2,\\
X^2-XY+Y^2=V^2,\\
\gcd(X,Y)=1,\\
\gcd(U,V)=1.
\end{cases}$$
So
$$\begin{cases}
X^2+Y^2=\dfrac{U^2+V^2}2,\\
2XY= U^2-V^2.
\end{cases}$$
Right part of the second equality is even, so $U$ and $V$ has the same parity and are odd. Then, $XY$ is even, when $X$ and $Y$ are coprime.
The task is symmetric in the variables $X$ and $Y$, let $X$ is odd and $Y$ is even:
$$X\cdot\frac Y2 = \dfrac{U+V}2\frac{U-V}2.$$
Note than $\dfrac{U+V}2$ and $\dfrac{U-V}2$ are coprime, because any their common divider also divides their sum $U$ and difference $V$.
The last equation has а solution
$$X=ac,\quad\dfrac Y2=bd,\quad \dfrac{U+V}2=ad,\quad\dfrac{U-V}2=bc,$$
so
$$X=ac,\quad Y=2bd,\quad U=ad+bc,\quad V=ad-bc,\quad Z=\pm(a^2d^2-b^2c^2),$$
where
$$\gcd(ac,2bd)=\gcd(ad,bc)=1.$$
Thus, $a$ and $c$ are odd and $a$, $b$, $c$, $d$ are pairwise coprime numbers.
That numbers must satisfy to equation
$$(ac)^2+4(bd)^2=(ad)^2+(bc)^2,$$
or
$$a^2(c^2-d^2)=b^2(c^2-4d^2).$$
$$\dots$$
Equation 2.
If
$$3(xy)^2+z^2=(x^2+y^2)^2,$$
then
$$
\begin{cases}
xy=2m-1,\\
z=\dfrac{3x^2y^2-1}2,\\
x^2+y^2=\dfrac{3x^2y^2+1}2.
\end{cases}
$$
So
$$(xy+1)(3xy+1)=2(x+y)^2.$$
Then you can use the fact that equation
$$mn=p^2$$
has solutions
$$m=a^2c,\quad n=b^2c,\quad p=abc,$$
for $$m\in\left\{\dfrac{xy+1}2,xy+1\right\}$$
| {
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"url": "https://math.stackexchange.com/questions/1759340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Prove when $abc=1$: $ \frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$
Question: Prove the following inequality which holds for all positive reals $a$, $b$ and $c$ such that $abc=1$:
$$ \frac{a}{2+bc} + \frac{b}{2+ca}+\frac{c}{2+ab} \geq 1$$
My thoughts were turning the right hand side to $abc$ as $abc=1$ however I think this will make the proving even more harder. I also attempted applying the Cauchy-Schwarz inequality and Hölder's inequality but to no avail.
Could someone please show me how to prove it using the inequalities above or another method.
| Substitute $$\frac{x}{y} \ \textrm{for}\ \ a,\quad \frac{y}{z} \ \textrm{for}\ \ b\quad \textrm{and} \quad \frac{z}{x} \ \textrm{for} \ \ c$$ This leaves us with the following and the condition $abc = 1$ is eliminated.
$$
\frac{\frac{x}{y}} {2 + \frac{y}{x}} +
\frac{\frac{y}{z}} {2 + \frac{z}{y}} +
\frac{\frac{z}{x}} {2 + \frac{x}{z}} =
\frac{x^2}{y^2 + 2xy} +
\frac{y^2}{z^2 + 2yz} +
\frac{z^2}{x^2 + 2zx}
$$
By Titu's Lemma,
$$
\frac{x^2}{y^2 + 2xy} +
\frac{y^2}{z^2 + 2yz} +
\frac{z^2}{x^2 + 2zx} \geqslant
\frac{(x + y + z)^2}{\sum_{cyc}{x^2} + 2xy} = \frac{(x + y + z)^2}{(x + y + z)^2} = 1
$$
Hence Proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1765037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
} |
Prove the inequality $\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a}\geq 4$ $a, b, c, d$ are positive reals.
How would I prove the inequality
$$\frac{a+c}{a+b}+\frac{b+d}{b+c}+\frac{c+a}{c+d}+\frac{d+b}{d+a} \geq 4$$
I have tried using the rearrangement inequality with $a\leq b\leq c\leq d$
But it doesn't seem to work well.
Any hints please?
| $$\frac{\color{blue}{a+c}}{a+b}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{blue}{c+a}}{c+d}+\frac{\color{red}{d+b}}{d+a}=\frac{\color{blue}{a+c}}{a+b}+\frac{\color{blue}{a+c}}{c+d}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{red}{b+d}}{d+a}$$
$$\frac{\color{blue}{a+c}}{a+b}+\frac{\color{blue}{a+c}}{c+d}+\frac{\color{red}{b+d}}{b+c}+\frac{\color{red}{b+d}}{d+a}=(\color{blue}{a+c})\color{green}{\left(\frac{1}{a+b}+\frac{1}{c+d}\right)}+(\color{red}{b+d})\color{brown}{\left(\frac{1}{b+c}+\frac{1}{d+a}\right)}$$
WLOG $$\color{brown}{\frac{1}{b+c}+\frac{1}{d+a}} \ge \color{green}{\frac{1}{a+b}+\frac{1}{c+d}} $$
Then
$$(\color{blue}{a+c})\color{green}{\left(\frac{1}{a+b}+\frac{1}{c+d}\right)}+(\color{red}{b+d})\color{brown}{\left(\frac{1}{b+c}+\frac{1}{d+a}\right)} \ge (\color{blue}{a+c}+\color{red}{b+d})\color{green}{\left(\frac{1}{a+b}+\frac{1}{c+d}\right)}$$ $$= (\color{turquoise}{a+b}+\color{orange}{c+d})\left(\color{turquoise}{\frac{1}{a+b}}+\color{orange}{\frac{1}{c+d}}\right) \ge (1+1)^2=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1766378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\frac{1+\sin A}{\cos A}+\frac{\cos B}{1-\sin B}=\frac{2\sin A-2\sin B}{\sin(A-B)+\cos A-\cos B}$ Show that
$$\frac{1+\sin A}{\cos A}+\frac{\cos B}{1-\sin B}=\frac{2\sin A-2\sin B}{\sin(A-B)+\cos A-\cos B}$$
$$\frac{1+\sin A}{\cos A}+\frac{\cos B}{1-\sin B}\\=\frac{\cos A}{1-\sin A}+\frac{\cos B}{1-\sin B} \\
=\frac{\cos A+\cos B-\sin B\cos A-\cos B\sin A}{(1-\sin A)(1-\sin B)}= \ \\
\frac{\cos A+\cos B-\sin(A+B)}{1-\sin A-\sin B+\sin A\sin B}$$
I am stuck here, I could not solve further. Please help.
| The $LHS$ can be written as $\dfrac {1+\sin A}{\cos A} +\dfrac{1+\sin B}{\cos B}$
$=\dfrac{\cos A+\cos B+\sin A\cos B+\cos A\sin B}{\cos A\cos B}$
Multiply Neumaretor and Denominator by $\sin(A-B)+\cos A-\cos B=(\cos A+\sin A\cos B)-(\cos B+\cos A\sin B)$
$=\dfrac{(\cos A+\sin A\cos B)^2-(\cos B+\cos A\sin B)^2}{\cos A\cos B[...]}$
$=\dfrac{\cos^2 A+\sin^2 A\cos^2B+2\sin A\cos A\cos B-\cos^2B-\cos^2A\sin^2B-2\cos A\cos B\sin B}{\cos A\cos B[...]}$
Note that $\sin^2A\cos^2B-\cos^2A\sin^2B=(1-\cos^2A)\cos^2B-\cos^2A(1-\cos^2B)=\cos^2B-\cos^2A$
So LHS becomes
$\dfrac{2\sin A \cos A\cos B-2\cos A\cos B\sin B}{\cos A\cos B[...]}$
$=\dfrac{2\sin A-2\sin B}{\sin(A-B)+\cos A-\cos B}=RHS$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $a^{ab}b+b^{bc}c+c^{ca}a \geqslant \sqrt[6]{5}$ $a,b,c >0$, and $a+b+c=3$, prove
$$ a^{ab}b+b^{bc}c+c^{ca}a \geqslant \sqrt[6]{5}$$
I try to substitute $c=3-a-b$ to reduce the number of variables, but cannot further proceed to solve the problem. I made an Excel spreadsheet and test 100 pairs of $(a,b,c)$, it seems that the inequality is correct.
I cannot even find where the equality occurs. Please help. This is a very unconventional problem
| Hint: for $x>0$ and $y>0$ and $z>0$ we have :
$$x+y+z\geq \left(\frac{x+1}{y+1}+\frac{y+1}{x+1}+\frac{x+1}{z+1}+\frac{z+1}{x+1}+\frac{z+1}{y+1}+\frac{y+1}{z+1}-1\right)^{\frac{1}{6}}\geq (5)^{\frac{1}{6}}$$
Method:
If we put $x+y+z=\lambda$
with $x=a^{ab}a$
$y=b^{bc}b$
$z=c^{ac}c$
And $a+b+c=3$
The question is : when is the maximum reached in the following expression?
$$\frac{x+1}{y+1}+\frac{y+1}{x+1}+\frac{x+1}{z+1}+\frac{z+1}{x+1}+\frac{z+1}{y+1}+\frac{y+1}{z+1}-1$$
The maximum is reached when $x=y=0$ and $z=\lambda$
Following this we have this inequality :
$$\lambda\geq \left(1+2(\lambda+1+\frac{1}{1+\lambda})\right)^{1/6}$$
It occurs for $\lambda\simeq 1.36897$
Now the idea is to repeat the same reasoning with the following expression :
$$\frac{x+2}{y+2}+\frac{y+2}{x+2}+\frac{x+2}{z+2}+\frac{z+2}{x+2}+\frac{z+2}{y+2}+\frac{y+2}{z+2}-1$$
We obtain this :
$$\lambda\geq \left(1+2(\frac{2+\lambda}{2}+\frac{2}{2+\lambda})\right)^{\frac{1}{6}}$$
Its occurs for $\lambda\simeq 1.32985$
Now the idea is to take the following expression :
$$\frac{x+n}{y+n}+\frac{y+n}{x+n}+\frac{x+n}{z+n}+\frac{z+n}{x+n}+\frac{z+n}{y+n}+\frac{y+n}{z+n}-1$$
And with the same reasoning we obtain :
$$\lambda\geq \left(1+2(\frac{n+\lambda}{n}+\frac{n}{n+\lambda})\right)^{\frac{1}{6}}$$
We take the limit :
$$\lambda\geq \lim\limits_{n \to \infty}\left(1+2(\frac{n+\lambda}{n}+\frac{n}{n+\lambda})\right)^{\frac{1}{6}}$$
And we obtain $$\lambda \geq (5)^{\frac{1}{6}}$$
If I'm wrong tell me quickly .
| {
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"timestamp": "2023-03-29T00:00:00",
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Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
$x,y,z >0$, prove
$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$
Note:
Often Stack Exchange asked to show some work before answering the question. This inequality was used as a proposal problem for National TST of an Asian country a few years back. However, upon receiving the official solution, the committee decided to drop this problem immediately. They don't believe that any students can solve this problem in 3 hour time frame.
Update 1: In this forum, somebody said that BW is the only solution for this problem, which to the best of my knowledge is wrong. This problem is listed as "coffin problems" in my country. The official solution is very elementary and elegant.
Update 2: Although there are some solutions (or partial solution) based on numerical method, I am more interested in the approach with "pencil and papers." I think the approach by Peter Scholze in here may help.
Update 3: Michael has tried to apply Peter Scholze's method but not found the solution yet.
Update 4: Symbolic expanding with computer is employed and verify the inequality. However, detail solution that not involved computer has not been found. Whoever can solve this inequality using high school math knowledge will be considered as the "King of Inequality".
| I have finally found a solution . In fact we start to study the 2 variables version of this inequality we have :
$$\frac{a^4}{8a^3+5b^3}+\frac{b^4}{8b^3+5a^3}\geq \frac{a+b}{13}$$
Proof:
We have with $x=\frac{a}{b}$ :
$$\frac{x^4}{8x^3+5}+\frac{1}{8+5x^3}\geq \frac{1+x}{13}$$
Or
$$\frac{5}{13}(x - 1)^2 (x + 1) (x^2 + x + 1) (5 x^2 - 8 x + 5)\geq 0$$
So we have (if we permute the variables $a,b,c$ and add the three inequalities ) :
$$\sum_{cyc}\frac{a^4}{8a^3+5b^3}+\sum_{cyc}\frac{a^4}{8a^3+5c^3}\geq \frac{a+b+c}{6.5}$$
If we have $\sum_{cyc}\frac{a^4}{8a^3+5b^3}\geq\sum_{cyc}\frac{a^4}{8a^3+5c^3}$
We have :
$$\sum_{cyc}\frac{a^4}{8a^3+5b^3}\geq \frac{a+b+c}{13}$$
But also
$$\frac{(a-\epsilon)^4}{8(a-\epsilon)^3+5b^3}+\frac{(b)^4}{8(b)^3+5(c+\epsilon)^3}+\frac{(c+\epsilon)^4}{8(c+\epsilon)^3+5(a-\epsilon)^3}\geq \frac{a+b+c}{13}$$
If we put $a\geq c $ and $\epsilon=a-c$
We finally obtain :
$$\sum_{cyc}\frac{a^4}{8a^3+5c^3}\geq \frac{a+b+c}{13}$$
If we have $\sum_{cyc}\frac{a^4}{8a^3+5b^3}\leq\sum_{cyc}\frac{a^4}{8a^3+5c^3}$
The proof is the same as above .
So all the cases are present so it's proved !
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "124",
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Evaluating the rational integral $\int \frac{x^2+3}{x^6(x^2+1)}dx $
Evaluate $$\int \frac{x^2+3}{x^6(x^2+1)}dx .$$
I am unable to break into partial fractions so I don't think it is the way to go. Neither is $x=\tan \theta$ substitution. Please give some hints. Thanks.
| Below is a nice general way of computing the partial fraction that is a generalization of Heaviside's cover up method to nonlinear denominator factors.
$$\begin{eqnarray} \frac{x^2+3}{(x^2+1)x^6} &=&\ \ \frac{ax+b}{x^2+1}+ \frac{f(x)}{x^6}\quad\text{so clearing denominators yields}\\
x^2+3\ &=&\ \ (ax+b)x^6\! + (x^2+1) f(x)\quad\text{so evaluaing at $x=i$ then $x = -i$} \\
2\ &=& -\!ai - b \\
2\ &=& \ \ \ \ ai - b\quad\text{so subtracting the prior two equations yields}\\
0\ &=& -\!2ai\end{eqnarray}$$
Therefore $\ a=0\ $ so $\ 2 = -b\ $ so $\ b = -2\ $ and, finally, we can compute $\,f(x)\,$ by substituting these $\,a,b\,$ values into the first equation. Then the integral of the decomposition is easy.
See here for further discussion of this method - which often greatly simplifies calculations.
| {
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If $x^2+ax-3x-(a+2)=0\;,$ Then $ \min\left(\frac{a^2+1}{a^2+2}\right)$
If $x^2+ax-3x-(a+2)=0\;,$ Then $\displaystyle \min\left(\frac{a^2+1}{a^2+2}\right)$
$\bf{My\; Try::}$ Given $x^2+ax-3x-(a+2)=0\Leftrightarrow ax-a = -(x^2-3x-2)$
So we get $$a=\frac{x^2-3x-2}{1-x} = \frac{x^2-2x+1+1-x-4}{1-x} = \left[1-x-\frac{4}{1-x}+1\right]$$
Now $$f(a) = \frac{a^2+1}{a^2+2} = \frac{a^2+2-1}{a^2+2} = 1-\frac{1}{a^2+2}$$
So $$f(x) = 1-\frac{1}{\left[(1-x)-\frac{4}{1-x}+1\right]^2+2}$$
Now put $1-x=t\;,$ Then we get $$f(t) =1- \frac{1}{\left(t-\frac{4}{t}+1\right)^2+2}$$
Now How can I maximize $\displaystyle \frac{1}{\left[(1-x)-\frac{4}{1-x}+1\right]^2+2, }\;,$ Help Required, Thanks
| Your function will be maximum when the denominator will be minimum.
So, your work will be to calculate $$\frac{d}{dx}\left[\left\{(1-x)-\frac{4}{1-x}+1\right\}^2+1\right]=0$$
It will come down to $$2\left\{(1-x)-\frac{4}{1-x}+1\right\}\left\{-1-\frac{4}{(1-x)^2}\right\}=0$$
This will give rise to $2$ cases.
Case-$1$:
$$-1-\frac{4}{(1-x)^2}=0 \Rightarrow x \not \in \mathbb{R}$$
Hence $-1-\frac{4}{(1-x)^2}\not =0$ for any real $x$.
Case-$2$:
$$(1-x)-\frac{4}{1-x}+1=0$$
$$(1-x)^2+(1-x)-4=0$$
So $$1-x=\frac{-1\pm \sqrt{1+16}}{2}=\frac{\sqrt{17}\pm 1}{2}$$
Or $$x=1-\frac{\sqrt{17}\pm 1}{2}$$
This will give $2$ values of $x$.
Compute $\frac{d^2}{dx^2}\left[\left\{(1-x)-\frac{4}{1-x}+1\right\}^2+1\right]$ and check for which value of $x$, the expression comes out to be positive.
That is where the function is minimised, or, your actual function is maximised.
Hope this helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality $\sum\limits_{cyc}\frac{a^3}{13a^2+5b^2}\geq\frac{a+b+c}{18}$
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{13a^2+5b^2}+\frac{b^3}{13b^2+5c^2}+\frac{c^3}{13c^2+5a^2}\geq\frac{a+b+c}{18}$$
This inequality is strengthening of the following Vasile Cirtoaje's one, which he created in 2005.
Let $a$, $b$ and $c$ be positive numbers. Prove that:
$$\frac{a^3}{2a^2+b^2}+\frac{b^3}{2b^2+c^2}+\frac{c^3}{2c^2+a^2}\geq\frac{a+b+c}{3}.$$
My proof of this inequality you can see here: https://artofproblemsolving.com/community/c6h22937p427220
But this way does not help for the starting inequality.
A big problem we have around the point $(a,b,c)=(0.785, 1.25, 1.861)$ because the difference between the LHS and the RHS in this point is $0.0000158...$.
I tried also to use Cauchy-Schwarz, but without success.
Also, I think the BW (see here https://math.stackexchange.com/tags/buffalo-way/info I tryed!) does not help.
| I am not a reputable source, but I think I can prove the following theorem: $$\sum_{cyc}\frac{a_{1}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}+k_{2}}$$
PROOF
We will need firstly the following
Lemma 1.
$$\sum_{k=1}^{n}a_{k}^{\alpha+1}\geq\sum_{cyc}a_{1}a_{2}^{\alpha}$$
Proof.
Applying the Rearrangement inequality on $a_{1},a_{2},...,a_{n}$ and $a_{1}^{\alpha},a_{2}^{\alpha},...,a_{n}^{\alpha}$, we have that $\sum_{cyc}a_{1}a_{2}^{\alpha}$ is maximized when $a_{1},a_{2},...,a_{n}$ and $a_{1}^{\alpha},a_{2}^{\alpha},...,a_{n}^{\alpha}$ are similarly sorted.
Therefore, we can affirm that
$$\sum_{k=1}^{n}a_{k}^{\alpha+1}\geq\sum_{cyc}a_{1}a_{2}^{\alpha}$$
Other hand, we need the following
Lemma 2.
$$\sum_{cyc}\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}=\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
Proof.
We establish that
$$\frac{a_{1}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}+\frac{n}{m}=\frac{a_{1}}{k_{1}}$$
Operating, we get that
$$\frac{a_{1}^{\alpha+1}m+n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)}{m\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)}=\frac{a_{1}}{k_{1}}$$
$$k_{1}\left(a_{1}^{\alpha+1}m+n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)\right)=a_{1}m\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)$$
$$k_{1}a_{1}^{\alpha+1}m+k_{1}n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)=a_{1}mk_{1}a_{1}^{\alpha}+a_{1}mk_{2}a_{2}^{\alpha}$$
$$k_{1}n\left(k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}\right)=a_{1}mk_{2}a_{2}^{\alpha}$$
$$\frac{n}{m}=\left(\frac{k_{2}}{k_{1}}\right)\frac{a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}$$
Therefore, we have that
$$\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}=\frac{a_{1}}{k_{1}}$$
And subsequently, repeating the process for each variable, we get that
$$\sum_{cyc}\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{1}a_{2}^{\alpha}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}=\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
Now, we are ready to prove the theorem.
Applying Lemma 1, we derive that
$$\left(\frac{k_{2}}{k_{1}}\right)\sum_{k=1}^{n}a_{k}^{\alpha+1}\geq\left(\frac{k_{2}}{k_{1}}\right)\sum_{cyc}a_{1}a_{2}^{\alpha}$$
Therefore, substituting in the expression of Lemma 2 and operating, we have that
$$\sum_{cyc}\frac{a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
$$\sum_{cyc}\frac{\left(\frac{k_{1}}{k_{1}}\right)a_{1}^{\alpha+1}+\left(\frac{k_{2}}{k_{1}}\right)a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
$$\sum_{cyc}\frac{\left(\frac{k_{1}+k_{2}}{k_{1}}\right)a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}}$$
$$\sum_{cyc}\frac{a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{\left(\frac{k_{1}+k_{2}}{k_{1}}\right)k_{1}}$$
$$\sum_{cyc}\frac{a_{k}^{\alpha+1}}{k_{1}a_{1}^{\alpha}+k_{2}a_{2}^{\alpha}}\geq\frac{\sum_{k=1}^{n}a_{k}}{k_{1}+k_{2}}$$
As we wanted to prove.
The particular case proposed follows from the direct application of the theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
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Suppose that $E[X^n] = 3n$. Find $E[e^X]$... Suppose that $E[X^n] = 3n$. Find $E[e^X]$.
Hint from my professor: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} +···$
Not quite sure how to solve this problem, wouldn't $e^x$ go on exponentially. Any help is really appreciated.
| Warning: What appears below is at best vacuously true --- i.e. true but uninformatitve, since, as several people have pointed out, there is not actually any probability distribution whose $n$th moment, for every positive integer $n$, is $3n$.
\begin{align}
\operatorname{E}\left(e^X\right) & = \operatorname{E}\left( 1 + X + \frac{X^2} 2 + \frac{X^3} 6 + \frac{X^4}{24} + \cdots \right) \\[10pt]
& = 1 + \operatorname{E}(X) + \frac{\operatorname{E}(X^2)} 2 + \frac{\operatorname{E}(X^3)} 6 + \frac{\operatorname{E}(X^4)}{24} + \cdots \\[10pt]
& = 1 + 3 + \frac 6 2 + \frac 9 6 + \frac{12}{24} + \cdots+ \frac{3n}{n!} + \cdots \\[10pt]
& = 1 + \frac 3{0!} + \frac 3 {1!} + \frac 3 {2!} \cdots + \frac 3 {(n-1)!} + \cdots \\[10pt]
& = 1 + 3\left( \frac 1 {0!} + \frac 1 {1!} + \frac 1 {2!} + \frac 1 {3!} + \cdots \right) \\[10pt]
& = 1 + 3e.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove this geometry inequality (1) with $2(DF+EF)\ge BC$ There is a picture which hope to illustrate the configuration:
$\triangle ABC$ is such that $\angle A=\dfrac{2\pi}{3}$, and $F$ is the midpoint of $BC$, and $D,E$ lie on $AB,AC$ respectively, such that $DE ||BC$.
Show that:
$$2(DF+EF)\ge BC$$
maybe use cosine theorem,let $\dfrac{BD}{AB}=k,AB=c,BC=a,AC=b$, and in $\Delta ABC$ we use cosine theorem we have
$$a^2=b^2+c^2+bc\tag{1}$$
and $$DF^2=BD^2+BF^2-2BD\cdot BF\cos{\angle B}
=(kc)^2+\dfrac{a^2}{4}-2kc\cdot\dfrac{a}{2}\dfrac{a^2+c^2-b^2}{2ac}=k^2c^2+\dfrac{a^2}{4}-\dfrac{k(a^2+c^2-b^2)}{2}$$
the same as
$$EF^2=\dfrac{a^2}{4}+(kb)^2-\dfrac{k(a^2+b^2-c^2)}{2}$$
it is equivalent
$$\sqrt{4k^2c^2+a^2-2k(a^2+c^2-b^2)}+\sqrt{4k^2b^2+a^2-2k(a^2+b^2-c^2)}\ge a,0\le k\le 1\tag{2}$$
I have tried some inequality (AM-GM,Cauchy-Schwarz,$\cdots$) to prove $(2)$, but didn't see the solution.
| This is not a complete solution, just some simplification.
Square both sides of (2) and using (1) to get rid of $a^2$, we obtain:
$$2\sqrt{c^2+b^2(1-2k)^2+bc(1-2k)}\sqrt{b^2+c^2(1-2k)^2+bc(1-2k)}+((b^2+c^2)(1-2k)^2+2bc(1-2k))\ge bc\tag{3}$$
Let $\kappa=1-2k$, thus $-1\le \kappa \le 1$. Then (3) becomes
$$2\sqrt{c^2+b^2\kappa^2+bc\kappa}\sqrt{b^2+c^2\kappa^2+bc\kappa}+((b^2+c^2)\kappa^2+2bc\kappa)\ge bc\tag{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$ The question
Prove that:
$$\prod_{n=2}^∞ \left( 1 - \frac{1}{n^4} \right) = \frac{e^π - e^{-π}}{8π}$$
What I've tried
Knowing that:
$$\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right)$$
evaluating at $z=i$ gives
$$ \frac{e^π - e^{-π}}{2i} = \sin(πi) = πi \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right)$$
so:
$$ \prod_{n=1}^∞ \left( 1 + \frac{1}{n^2} \right) = \frac{e^π - e^{-π}}{2π}$$
I'm stucked up and don't know how to continue, any help?
| Note that $$\prod_{n=2}^{\infty} \left(1-\frac{1}{n^{2}}\right) \to \frac{1}{2}$$
This is because $$A_{n} =\prod_{k=2}^{n}\left(1-\frac{1}{n^2}\right) = \prod_{k=2}^{n} \frac{(k-1)(k+1)}{k^2} = \frac{n+1}{2n} \to \frac{1}{2}$$
We have used $\displaystyle \left(1-\frac{1}{n^4}\right) = \left(1+\frac{1}{n^2}\right) \cdot \left(1-\frac{1}{n^{2}}\right)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$
Prove that $2^x = 3 \cdot 9^m+5$ has no positive integer solutions for $m \geq 2$.
I noticed that $x \equiv 5 \bmod 6$ and thus $2^x \equiv 4 \bmod 7$, but that doesn't seem to help me since $3 \cdot 9^4 +5 \equiv 4 \bmod 7$. Pretty much any other mod I use doesn't seem to work so I think proof by contradiction or something may work better.
| Suppose that $x$ is a solution, then:
$ 2^{x} = 3.9^{m} + 5 = 3.9^{m} + 3 + 2$, so:
$$2(2^{x-1} - 1) = 3( 9^{m} + 1) \Rightarrow (3.(9^{m} + 1)) | (2^{x-1} - 1)$$
For $m\geq 2$. But $2^{x-1} - 1$ is odd and $3.(9^{m} + 1)$ is pair.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$
Evaluation of $$\int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$$
$\bf{My\; Try::}$ Let $$I = \int_{0}^{1}\frac{1}{\sqrt{1+x}+\sqrt{1-x}+2}dx$$
Put $x=\cos 2 \theta\;,$ Then $dx = -2\sin 2 \theta d\theta$ and Changing Limit, We get
$$I = \int_{0}^{\frac{\pi}{4}}\frac{2\sin 2 \theta}{\sqrt{2}\cos \theta+\sqrt{2}\sin \theta+2}d\theta$$
So $$I = \int_{0}^{\frac{\pi}{4}}\frac{\sin 2 \theta}{\cos\left(\theta-\frac{\pi}{4}\right)+1}d\theta$$
Now Put $\displaystyle \theta-\frac{\pi}{4}=t\;,$ Then $d\theta = dt$ and changing limits
So $$I = \int_{-\frac{\pi}{4}}^{0}\frac{\cos 2t}{\cos t+1}dt=2\int_{-\frac{\pi}{4}}^{0}\frac{2\cos^2 t-2+1}{\cos t+1}dt$$
after that we can solve easily,
My question is can we solve it without Using Trig substution,
Plz explain me
Thanks
| I'm not sure this is the most straight forward way, but here it is anyway.
Start with the substitution $\sqrt{1-x}\mapsto x $. The integral transforms into
$$I=\int_0^1 \frac{2 x}{2+x+\sqrt{2-x^2}}dx.$$
Next, rationalize the denominator by multiplying the numerator and the denominator by $2+x-\sqrt{2-x^2}.$
This turns out to work very well, because $\,\,(2+x)^2-(2-x^2)=2(1+x)^2.$
So we get that the integral equals
$$I=\int_0^1 \frac{x(2+x-\sqrt{2-x^2})}{(1+x)^2}dx.$$
Finally, integrate by parts (note that $\displaystyle \frac1{(1+x)^2}=-\frac{d}{dx} \frac1{1+x}$):
$$I=-\frac{x(2+x-\sqrt{2-x^2})}{1+x}\Bigg{|}_0^1+\int_0^1 2\left(1+\frac{x-1}{\sqrt{2-x^2}}\right)dx
\\\\=-1+2-2\left(\sqrt{2-x^2}+\sin^{-1}\frac{x}{\sqrt{2}}\right)\Bigg{|}_0^1
\\\\=2\sqrt{2}-1-\frac{\pi}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1783593",
"timestamp": "2023-03-29T00:00:00",
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Find all positive integers $n$ such that $n^2+n+43$ becomes a perfect square Find all positive integers $n$ such that $n^2+n+43$ becomes a perfect square.
Since $n^2+n+43$ is odd,if it's a perfect square it can be written as: $8k+1$,then:
$$n^2+n+43=8k+1\Rightarrow\ n^2+n+42=8k\ \Rightarrow\ n(n+1)\equiv6\pmod8$$
So $n$ becomes $2,5,...$ but $5$ doesn't work!!
| If $n^2+n+43$ is a square, so it is
$$4n^2+4n+172 = (2n+1)^2 + 171. $$
If $171=a^2-(2n+1)^2$, then $171=(a-2n-1)(a+2n+1).$ Since
$$ 171 = 3^2\cdot 19 $$
the only ways for writing $171$ as a product of two positive integers with the same parity are:
$$ 1\cdot 171,\qquad 3\cdot 57,\qquad 9\cdot 19. $$
For instance, if $a-2n-1=9$ and $a+2n+1=19$, then $(19-9)=4n+2$ and $n=2$.
In the same way we get that the whole set of integer solutions is given by:
$$ n\in\color{red}{\{-43,-14,-3,2,13,42\}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Using $\epsilon-\delta$ proof to prove continuity
Use an $\epsilon-\delta$ proof to show that
$f : R \setminus \left \{ \frac{-3}{2} \right \} \rightarrow R$ , $$f(x) = \frac{3x^2-2x-5}{2x+3}$$
is continuous at $x = -1$
Hello there. Can anyone here help me with this? I know i need to show that $|x-l|<\delta$ implies $|f(x) - f(l)| < \epsilon$ but I don't know how to do it for this specific example.
| Let $\epsilon>0$. We want to pick a $\delta$ such that $|x-(-1)|=|x+1|<\delta\implies|f(x)-f(-1)|=|f(x)|<\epsilon$.
Note that
$$
|f(x)|=\bigg|\frac{3x^{2}-2x-5}{2x+3}\bigg|=\bigg|\frac{(3x-5)(x+1)}{2x+3}\bigg|=\bigg|\frac{3x-5}{2x+3}\bigg||x+1|.
$$
We are free top choose $\delta$ as we wish, and, in doing so, we will have a handle on making the $|x+1|$ term above as small as we want. To help us minimize $\big|\frac{3x-5}{2x+3}\big|$, let's see what happens if we would ensure $\delta <\frac{1}{3}$.
If $|x+1|<\frac{1}{3}$, then $-\frac{1}{3}<x+1<\frac{1}{3}$. This implies that
$$
-1<3x+3<1\implies-9<3x-5<-7<9\implies|3x-5|<9
$$
and
$$
-\frac{2}{3}<2x+2<\frac{2}{3}\implies\frac{1}{3}<2x+3<\frac{5}{3}\implies\frac{1}{3}<|2x+3|.
$$
Thus,
$$
\bigg|\frac{3x-5}{2x+3}\bigg|<\frac{9}{1/3}=27.
$$
So, if $\delta<\frac{1}{3}$, it follows that $|f(x)|<27|x+1|$. So, if in addition to this, $\delta<\frac{\epsilon}{27}$, we get that $|f(x)|<27\cdot\frac{\epsilon}{27}=\epsilon$.
Hence, we could choose $\delta=\min\{\frac{1}{3},\frac{\epsilon}{27}\}$. Then, if $|x+1|<\delta$, it follows that $|f(x)|<\epsilon$, as desired.
| {
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Show that $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7} -\frac{1}{8}-\frac{1}{9}-\frac{1}{10} -\cdots $ converge $$1-\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}-\dfrac{1}{7}
-\dfrac{1}{8}-\dfrac{1}{9}-\dfrac{1}{10} ... $$
I added parentheses for each sub-sequence with the same sing.
so i got :
$$1-(\dfrac{1}{2}+\dfrac{1}{3})+(\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6})-(\dfrac{1}{7}
+\dfrac{1}{8}+\dfrac{1}{9}+\dfrac{1}{10}) ... $$
I want to show that the new sequence is a leibniz sequence and by that conclude that is converge.
I managed to show that each pair of parentheses is greater than:
$$\dfrac{2}{n+1} $$
Cant find a way to proceed.
Thanks for helping.
| Compare the absolute value of each homogeneously signed sum with the sum containing an equal number of copies of the first term. For example, the all-negative terms from $-1/7$ to $-1/10$ are compared with four copies $-1/7$. The absolute values of the sums then satisfy
$(1/7+1/8+1/8+1/9+1/10)<4/7$
In general the cluster sums have absolute values less than
$n/({n(n-1)}/2+1)$
which goes to zero as $n$ increases without bound, so the series passes the test for convergence.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$
Prove that $$1^8+2^8+\cdots+99^8 \equiv 1^4+2^4+\cdots+99^4 \pmod{25}.$$
Attempt:
We can easily show that an eighth power can be expressed as a fourth power since $x^8 = (x^2)^4$. Conversely, by Fermat's Little Theorem, $x^{\phi(25)} = x^{20} \equiv 1 \pmod{25}$ if $\gcd(x,25) = 1$ and thus $x^4 = x \cdot x^3 \equiv x^{21} \cdot x^3 \equiv x^{24} \equiv (x^3)^8 \pmod{25}$. $\square$
I am not sure if the above proves the result, but it does show that if we have an $8$th power of an integer modulo $25$ and it is relatively prime to $25$, then it is equal to a fourth power of an integer modulo $25$ and vice-versa. Does that therefore mean they are equivalent?
| If $\gcd(x, 25) \neq 1$, then $x$ is a multiple of $5$, so $x^4$ and $x^8$ are multiples of $25$ and don't contribute to the sum. Thus, we will focus on the cases where $\gcd(x, 25)=1$.
Also, $x \equiv x+25 \pmod{25}$, so we really have the following:
$$4(1^4+2^4+3^4+[...]+24^4) \equiv 4(1^8+2^8+3^8+[...]+24^8) \pmod{25}$$
Now, your work shows that there is a bijective mapping between the fourth powers of the group of units $25$ and the eighth powers of the group of unts of $25$, so for any fourth power on the left side, we can find an eighth power on the right side and then subtract the two from both sides. We can do this repeatedly until we've subtracted all of the powers and gotten to $0 \equiv 0 \pmod{25}$, concluding the proof.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Thoughts on this limit ? $\lim_\limits{x\to 3}\dfrac{\frac{1}{3}-\frac{1}{x}}{x-3}$ $$\lim_\limits{x\to 3}\dfrac{\frac{1}{3}-\frac{1}{x}}{x-3}$$
Normally, one multiplies by $\frac{3x}{3x}$ to eliminate the complex fraction. Here, that will not eliminate the problematic denominator term of $(x-3)$. Neither will subtracting the numerator via LCD. Or will it?! Wait a sec...
| $$\lim_{x\to3}\frac{\frac{1}{3}-\frac{1}{x}}{x-3}=\lim_{x\to3}\frac{\frac{x}{3x}-\frac{3}{3x}}{x-3}=\lim_{x\to3}\frac{\frac{x-3}{3x}}{x-3}=\lim_{x\to3}\frac{x-3}{3x(x-3)}=\lim_{x\to3}\frac{1}{3x}=\frac{1}{3\cdot3}=\frac{1}{9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Orthogonal projection question
Consider the (orthogonal) projection $T: \mathbb{R}^3 \to \mathbb{R}^3$ onto the plane $x - y + z = 0$.
*
*(a) Find the standard matrix $[T]_S$ for $T$.
*(b) Find a new basis $B$ so that $[T]_B$ is rather simple.
How do I do this question?
I created an orthogonal basis and projected $(x,y,z)$.
My result was
$[T]_s=\begin{bmatrix}11/18 & 7/18 & -2/9\\
7/18 & 11/18 & 2/9\\
-2/9 & 2/9 & 4/9
\end{bmatrix}$
and what about part b? Do I make the new basis equal to the orthogonal one I found when solving part a? and is $[T]_B$ a $3 \times 3$ unit matrix?
| Your matrix is wrong. An easy sanity check is that a matrix representing an orthogonal projection should be symmetric with trace equal to the dimension of the space you are projecting onto (in your case, two).
A unit normal vector to your plane is $v = \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix}$ and so the projections of the standard basis vectors are given by
$$ T(e_1) = e_1 - \left<e_1, v \right>v = \begin{pmatrix} 1 \\ 0 \\ 0 \end{pmatrix} - \frac{1}{\sqrt{3}} \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{2}{3} \\ \frac{1}{3} \\ -\frac{1}{3} \end{pmatrix}, \\
T(e_2) = e_2 - \left<e_2, v \right>v = \begin{pmatrix} 0 \\ 1 \\ 0 \end{pmatrix} + \frac{1}{3} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} \frac{1}{3} \\ \frac{2}{3} \\ \frac{1}{3} \end{pmatrix}, \\
T(e_3) = e_3 - \left<e_3, v \right>v = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} - \frac{1}{\sqrt{3}} \frac{1}{\sqrt{3}} \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} = \begin{pmatrix} -\frac{1}{3} \\ \frac{1}{3} \\ \frac{2}{3} \end{pmatrix} $$
and so
$$ [T]_S = \begin{pmatrix} [Te_1]_S & [Te_2]_S & [Te_3]_S \end{pmatrix} = \begin{pmatrix} \frac{2}{3} & \frac{1}{3} & -\frac{1}{3} \\ \frac{1}{3} & \frac{2}{3} & \frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3} & \frac{2}{3} \end{pmatrix}. $$
If you pick an orthonormal basis $(v_1,v_2)$ for the plane $x - y + z = 0$ and take $v_3 = v$, then with respect to the new basis, the matrix will become
$$ [T]_{(v_1,v_2,v_3)} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0 \end{pmatrix}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Why is $5\tan(54^\circ) = \sqrt{25 + 10\sqrt{5}}$ and $\tan\left(\frac{\pi}{5}\right) = \sqrt{5 - 2\sqrt{5}}$? On the Wikipedia Page about Pentagons, I noticed a statement in their work saying that $\sqrt{25+10\sqrt{5}}=5\tan(54^{\circ})$ and $\sqrt{5-2\sqrt{5}}=\tan(\frac {\pi}{5})$
My question is: How would you justify that? My goal is to simplify $\sqrt{25+10\sqrt{5}}$ and $\sqrt{5-2\sqrt{5}}$ without knowledge on the detested radical!
| As $54\cdot5\equiv90\pmod{180},$
if $\tan5x=\infty,5x=180^\circ n+90^\circ$ where $n$ is any integer
$x=36^\circ n+18^\circ$ where $n\equiv0,\pm1,\pm2\pmod5$
Using $\tan5x$ expansion, the roots of $$5t^4-10t^2+1=0$$ are $\tan(36^\circ n+18^\circ)$ where $n\equiv0,\pm1,2\pmod5$
as $n=2\implies\tan(36^\circ n+18^\circ)=?$
$$t^2=\dfrac{10\pm4\sqrt5}{2\cdot5}=\dfrac{5\pm2\sqrt5}5$$
As $\tan54^\circ>\tan18^\circ>0,\tan^254^\circ>\tan^218^\circ,$
$\tan^254^\circ=\dfrac{5+2\sqrt5}5=\dfrac{25+10\sqrt5}{5^2}$
and $\tan^218^\circ=\dfrac{25-10\sqrt5}{5^2}$
For $\tan36^\circ,$ can you use the same idea?
| {
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"timestamp": "2023-03-29T00:00:00",
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Do you write plus/minus if a variable squares equals the square root of a number? For example, if I have $x^2 = \sqrt{49}$. I know that $7$ is the number, but as my final answer, do I write that $x = +\sqrt{7}$ and $-\sqrt{7}$ or just $x = \sqrt{7}$?
| It depends on which field you expect/require the answer to be in:
If the answer is allowed to be a complex number, then the equation $x^2 = \sqrt{49}$ can be rewritten as $x^2 = 7$ OR $x^2 = -7$, which collectively yield four values: $x = +\sqrt{7}$, $x = -\sqrt{7}$, $x = +i\sqrt{7}$, and $x = -i\sqrt{7}$.
If the answer is required to be a real number, then the equation $x^2 = \sqrt{49}$ can be rewritten as $x^2 = 7$ OR $x^2 = -7$, which, since $x^2 = -7$ has no real solution, yields only two values: $x = +\sqrt{7}$ and $x = -\sqrt{7}$.
If the answer is required to be a rational number, or a number in an extension of the rationals which does not contain $\sqrt{7}$, then of course there is no solution.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Sum of real values of $x$ satisfying the equation $(x^2-5x+5)^{x^2+4x-60}=1$ I have this equation from this paper (Q.63)
Find the sum of all real values of $x$ satisfying the equation-$(x^2-5x+5)^{x^2+4x-60}=1$.
My attempt-
$(x^2-5x+5)^{x^2+4x-60}=(x^2-5x+5)^{(x-6)(x+10)}$
So, $x=6$ and $x=-10$ makes the power $0$ and hence the equation becomes one. So, the sum of the real values of $x$ satisfying the equation is $6+(-10)=-4$. But answer is $+3$.
Where have I gone wrong?
| First of all, the equation in your link is : $(x^2-5x+5)^{x^2+4x-60}=1$.
Simply use $\ln$ in both sides :
$ \ln(x^2-5x+5)^{x^2+4x-60} = \ln1 \Leftrightarrow (x^2 + 4x - 60)\ln (x^2 - 5x + 5) = 0$.
That means that : $x^2 + 4x - 60 = 0$ or $(x^2 - 5x + 5) = 1$.
Of course, the functions inside of $\ln$ must be bigger than zero, do not forget to add that in your final answer after you've done these.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to show that any rectangle in ellipse must be oriented parallel to axes? A problem which is often given as an exercise for students learning about calculus and finding extrema, is to find maximal possible area of a rectangle inside an ellipse. Such question was asked, for example, here: Find the area of largest rectangle that can be inscribed in an ellipse (A similar problem in three dimensions is also often asked: Dimensions of a box of maximum volume inside an ellipsoid.)
The solution usually starts by stating that we the rectangle must be oriented in such way that the sides are parallel to the ellipse, which gives a simple expression for the area.
Even though the fact that any rectangle inscribed in an ellipse must be oriented in this way seems intuitively clear, I would like to see an argument showing that this is indeed the case. (I have posted as an answer my attempt using analytic geometry.)
Of course, there is one obvious exception. Rectangle inscribed in a circle can be rotated in any direction we want. So we should assume that the semiaxes of the ellipse have different lengths.
Here is a picture illustrating the situation (shamelessly stolen from this post):
| Suppose we have a rectangle which has all four vertices on the ellipse. If $C$ is the center of the rectangle, then all vertices of the rectangle are in the same distance from $C$. If we choose coordinate system with origin in the point $C$, we get the equation
$$x^2+y^2=r^2.$$
We also want to express the condition that the vertices belong to the ellipse. So far we have only chosen the origin of our Cartesian coordinate system to be the point $C$. We can also choose the axes to be parallel to the axes to the ellipse. If we denote by $(x_0,y_0)$ the center of the ellipse, we get the equation
$$\frac{(x-x_0)^2}{a^2}+\frac{(y-y_0)^2}{b^2}=1.$$
Since rectangle is centrally symmetric w.r.t. the point $C=(0,0)$, the vertices also lie on an ellipse which is centrally symmetric to the ellipse described above. This si precisely the ellipse
$$\frac{(x+x_0)^2}{a^2}+\frac{(y+y_0)^2}{b^2}=1.$$
So we got the following three equations:
\begin{align}
x^2+y^2&=r^2 \tag{1}\\
b^2(x-x_0)^2+a^2(y-y_0)^2&=a^2b^2 \tag{2}\\
b^2(x+x_0)^2+a^2(y+y_0)^2&=a^2b^2 \tag{3}
\end{align}
We are asking for the conditions on $x_0$, $y_0$ and $r$ such that there are at least four solutions.
Notice that by subtracting the equations $(2)$ and $(3)$ we get
$$b^2xx_0+a^2yy_0=0.\tag{4}$$
Let us first show that $x_0=y_0=0$. Assume that this is not true. For example, let us assume $x_0\ne0$. Then we get
$$x=\frac{a^2y_0}{b^2x_0}\cdot y. \tag{5}$$
By plugging $(5)$ into $(1)$ we get
$$y^2\left(1+\frac{a^2y_0}{b^2x_0}\right)^2=r^2 \tag{6}$$
which gives us two possibilities for $y$. Combining this with $(5)$ we get two possibilities for the point $(x,y)$.
So we see that if $x_0\ne0$, then there are at most two solutions of the above system. The case $y_0\ne0$ is symmetric. In order to get a rectangle we need at four points, so we necessarily have1
$$x_0=y_0=0.\tag{7}$$
In the other words, we have found out that the center of the rectangle and the center of the ellipse coincide.
This condition significantly simplifies the original system to
\begin{align}
x^2+y^2&=r^2 \\
b^2x^2+a^2y^2&=a^2b^2
\end{align}
From the first equation we get $y^2=r^2-x^2$. By plugging this value into the second equation we get
\begin{align}
b^2x^2+a^2(r^2-x^2)&=a^2b^2\\
(b^2-a^2)x^2&=a^2(b^2-r^2)
\end{align}
Since we do not consider the case $b=a$, which corresponds to the ellipse being circle, we get
$$x^2=\frac{a^2(b^2-r^2)}{b^2-a^2}.$$
Assuming the RHS is positive, this gives us two possible values $$x=\pm\sqrt{\frac{a^2(b^2-r^2)}{b^2-a^2}}.$$
From these values we can get $$y=\pm\sqrt{r^2-x^2}$$
(again assuming that $r^2-x^2>0$).
Clearly, we got a rectangle with the sides parallel to the axes. (The vertices of rectangle have the form $(x_1,y_1)$, $(-x_1,y_1)$, $(-x_1,y_1)$, $(-x_1,-y_1)$.)
In fact, using the above equations we can also find the conditions on $a$, $b$ and $r$ when the intersection exists. W.l.o.g. we can assume $0<a<b$. (The other case is symmetric.) If we want $a^2(b^2-r^2)/(b^2-a^2)$ to be positive, we need $r<b$.
We also need
$$0<r^2-x^2 = r^2- \frac{a^2(b^2-r^2)}{b^2-a^2} = \frac{a^2(b^2-r^2)}{b^2-a^2} = \frac{b^2(r^2-a^2)}{b^2-a^2}$$
which gives us $a<r$.
So we get a solution only if
$$a<r<b$$
as expected.
1Another possibility how to view the case $(x_0,y_0)\ne(0,0)$, i.e. the case when the center of the rectangle is different from the center of the ellipse, is that in such case we can only get a line segment - which could be considered as a "rectangle having zero width".
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Finding Modular Multiplicative Inverses (Quickly!) as part of an upcoming number theory exam I will need to find the modular multiplicative inverse of every element of ${Z_n}$ (the ones that exist anyway) very quickly. The only way I know is using the Extended Euclidean Algorithm. Is there no way that is faster? The question isn't worth many marks but will take an inordinate amount of time.
| Solve $ax \equiv 1 \pmod{n}$
I would tackle these problems keeping this trick in mind and applying the basic tactic of making $a$ smaller by 'multiplying to the modulus'.
Example 1: Solve $12x \equiv 1 \pmod{19}$.
$\; 12x \equiv 1 \pmod{19} \; \text{ iff}$
$\; -7x \equiv 1 \pmod{19}$
and
$\;-7 \cdot (3) \equiv -21 \equiv -2 \pmod{19}$
$\;-2 \cdot (9) \equiv -18 \equiv 1 \pmod{19}$
and therefore the solution is given by $x \equiv 3 \cdot 9 \equiv 8 \pmod{19}$.
Example 2: Solve $67x \equiv 1 \pmod{97}$.
$\; 67x \equiv 1 \pmod{97} \; \text{ iff}$
$\; -30x \equiv 1 \pmod{97}$
and
$\;-30 \cdot (3) \equiv -90 \equiv 7 \pmod{97}$
$\;7 \cdot (14) \equiv 98 \equiv 1 \pmod{97}$
and therefore the solution is given by $x \equiv 3 \cdot 14 \equiv 42 \pmod{19}$.
Example 3: Solve $7x \equiv 1 \pmod{23}$.
$\;7 \cdot (4) \equiv 5 \pmod{23}$
$\;5 \cdot (5) \equiv 2 \pmod{23}$
$\;2 \cdot (12) \equiv 1 \pmod{23}$
and therefore the solution is given by $x \equiv 4 \cdot 5 \cdot 12 \equiv 10 \pmod{23}$.
Example 4: Solve $6x \equiv 1 \pmod{10}$.
$\; 6x \equiv 1 \pmod{10} \; \text{ iff}$
$\; -4x \equiv 1 \pmod{10}$
and
$\;-4 \cdot (2) \equiv 2 \pmod{10}$
$\;2 \cdot (5) \equiv 0 \pmod{10}$
If $6x \equiv 1 \pmod{10}$ then $5 \cdot (6x) \equiv 5 \pmod{10}$, so there are no solutions.
| {
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"url": "https://math.stackexchange.com/questions/1794078",
"timestamp": "2023-03-29T00:00:00",
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For a,b in a group G, $|a| = 2$, $b ≠ e$, and $aba = b^2$. What is $|b|$? If it was $aba^{-1}$ = $b^2$ I could use the theorem that if $aba^{-1}$ = $b^n$ then $a^kba^{-k}$ = $b^{n^k}$ so that $b$ = $ebe$ = $a^2ba^{-2}$ = $b^4$ so that $|b|$ = 3. Is there a similar method here?
| Since $a^2=e$, we have $a=a^{-1}$. From the equation $aba=b^2$
\begin{align*}
ba &= ab^2 \text{(composing a on the left)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)\\
ab &= b^2a \text{(composing a on the right)}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)\\
ab^2&=b^2ab \text{(composing b on the right of (2))}\,\,\,\,\,\,(3)\\
\text{Thus,} ba&=b^2ab\\
b&=b^2aba\\
b&=b^4\text{(}aba=b^2\text{)}\\
b^3&=e
\end{align*}
Thus $|b|$ divides $3$. But since $b \neq e$, we have $|b|=3$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{ab}{4-d}+ \frac{bc}{4-a}+\frac{cd}{4-b}+\frac{da}{4-c} \leqslant \frac43$ for $a^2 + b^2 + c^2 + d^2 = 4$
Let $a,b,c,d \geqslant 0$ and $a^2+b^2+c^2+d^2=4$. Prove that
$$\frac{ab}{4-d}+ \frac{bc}{4-a}+\frac{cd}{4-b}+\frac{da}{4-c} \leqslant \frac43.$$
I try some reverse AM-GM techniques but fail. I don't think rearrangement inequality works because of cyclic nature of this inequality. I try to homogenize this inequality to remove the constrain, but the square root in the denominator kills me.
| Note: in this solution, $\sum$ denotes the cyclic sum, so that $\sum f(a, b, c, d)=f(a, b, c, d)+f(b, c, d, a)+f(c, d, a, b)+f(d, a, b, c)$.
Lemma: $\sum c^4+2\sum abc^2+6\sum ab\le 36$.
Proof: Expanding the inequality $\sum[2(a-b)^2+(ab-1)^2]\ge 0$ gives:
$$6\sum ab\le 20+\sum a^2b^2$$.
Expanding $\sum (ab-ac)^2\ge 0$ gives:
$$2\sum abc^2\le \sum a^2b^2+2(a^2c^2+b^2d^2)$$
Summing these, we have:
\begin{align*}
\sum c^4+2\sum abc^2+6\sum ab &\le 20+\sum c^4+2\sum a^2b^2+2(a^2c^2+b^2d^2)
\\ &= 20+(a^2+b^2+c^2+d^2)^2
\\ &= 36
\end{align*}
so the lemma is true as required.
Now we may use the lemma, along with Cauchy Schwarz of the form $\sum \frac{x_i^2}{y_i}\ge \frac{(\sum x_i)^2}{\sum y_i}$:
\begin{align*}
\sum\frac{ab}{4-d}&= \sum\frac{2ab}{7-d^2+(d-1)^2}
\\ &\le \sum\frac{2ab}{7-d^2}
\\ &= \sum\frac{2ab}{3+2ab+c^2+(a-b)^2}
\\ &\le \sum\frac{2ab}{3+2ab+c^2}
\\ &=4-\sum \frac{c^2+3}{c^2+2ab+3}
\\ &=4-\sum \frac{(c^2+3)^2}{(c^4+2abc^2+6ab)+6c^2+9}
\\ &\le 4-\frac{(\sum c^2+12)^2}{\sum(c^4+2abc^2+6ab)+6\sum c^2+36}
\\ &\le 4-\frac{16^2}{36+60}
\\ &=4-\frac{8}{3}
\\ &=\frac{4}{3}
\end{align*}
so the inequality is proved. Equality holds at $a=b=c=d=1$.
| {
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Find the condition such that the roots of the polynomial are in AP
$f(x)=x^3+3px^2+3qx+r$ has roots in AP.Find the relation between $p,q$ and $r$.
[Answer:$-2p^2-3pq+r=0$]
My attempt:-
Taking $d$ as the common difference of the roots in AP we have $f(x)=(x-(a-d))(x-a)(x-(a+d))=x^3-3x^2a+(3a^2-d^2)x-a(a^2-d^2)$.
Comparing this with,the given equation,we get,
$(p=-a),q=(\frac{3a^2-d^2}{3}),r=-a(a^2-d^2)$.
But,after this I have no idea how to eliminate $a$ and $d$ from the equation and find a relation between $p,r$ and $r$.
Thanks for any help!!
| Let $x_1=a, x_2=a+d, x_3=a+2d$.
Use Vieta's Formulas
Then $$\begin{cases}a+a+d+a+2d=-3p
\\
a(a+d)+a(a+2d)+(a+d)(a+2d)=3q
\\
a(a+d)(a+2d)=-r
\end{cases}$$
$$\begin{cases}a+d=-p
\\
3a^2+6ad+2d^2=3q
\\
a(a+d)(a+2d)=-r
\end{cases}$$
Then $-2p^2-3pq+r=-2(a+d)^2+(a+d)(3a^2+6ad+2d^2)-a(a+d)(a+2d)=0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Most natural way to prove $\sum_{n=1}^{\infty}\frac{1}{n+2}$ diverges I don't know how my teacher wants me to prove that
$$\sum_{n=1}^{\infty}\frac{1}{n+2}$$
diverges. All I know is that I have to use the $a_n>b_n$ criteria and prove that $b_n$ diverges.
I tried this:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} =
\frac{1}{1+2}+\frac{1}{2+2}+\frac{1}{3+2} + \cdots + \frac{1}{n+2} = \left(\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{n}\right)-\left(\frac{1}{1}+\frac{1}{2}\right) = \left(\sum_{n=1}^{\infty}\frac{1}{n}\right)-\frac{3}{2}$$
but I can't get a relationship between $a_n$ and $b_n$ of these series.
If I go to the root of the comparsion criterion, I know that:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} =
\frac{1}{1+2}+\frac{1}{2+2}+\frac{1}{3+2} + \cdots + \frac{1}{n+2} = \left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\cdots+\frac{1}{n}> \frac{2}{4}+\frac{4}{8}+\cdots+g_n$$ for some $g_n$ that I'm lazy to calculate.
Therefore, by the properties of limits, since the righthand sum $p_n$ diverges, the lefthand sum $s_n$ diverges too, because $s_n>p_n$. But I don't know if my teacher would accept that, I think she would only accept that I use the argument for the $a_n$ of the sum, not for the partial sum.
I also thought about rewriting:
$$\sum_{n=1}^{\infty} \frac{1}{n+2} = \sum_{n=3}^{\infty} \frac{1}{n}$$
but I don't see how it helps because the indexes are different.
All I need is an argument that will work with the $a_n$ of the $\sum_{n=1}^{\infty}a_n$, in relation with $b_n$ from $\sum_{n=1}^{\infty} b_n$
| You can do in this way:
If $\sum_{n=1}^{\infty} \frac{1}{n+2} $ converges, say $\sum_{n=1}^{\infty} \frac{1}{n+2} = S$.
$\sum_{n=1}^{\infty} \frac{1}{n}=1+\frac12+\sum_{n=1}^{\infty} \frac{1}{n+2}=\frac 32+S$,
Which will contradics with $\sum_{n=1}^{\infty} \frac{1}{n}$ diverges.
| {
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"url": "https://math.stackexchange.com/questions/1801588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
If $9 \mid 2^b-2^a$, then $7\mid2^b-2^a$
Prove that if $9 \mid 2^b-2^a$, then $7\mid2^b-2^a$.
I am not sure how to prove this statement, but it seems that from $9 \mid 2^b-2^a$ we have $b-a = 6n$. Then what should I do from here to prove the statement?
| If $9\mid(2^b-2^a),$ then, taking $2$ out, we may suppose that $a=0.$ So $9\mid 2^b-1.$
But we see the powers of $2$ are $2,4,-1,-2,-4,1.$ Thus the order of $2$ modulo $9$ is $6.$ Hence $6\mid b.$
Also $2^6\equiv1\pmod7$ by Fermat's little theorem. Therefore $2^b-1\equiv1-1\equiv0\pmod7.$
Hope this helps.
| {
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"url": "https://math.stackexchange.com/questions/1801667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Permutations conjugated
Show that the permutations:
$\alpha=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
2 & 5 & 3 & 6 & 1 & 4 \\
\end{pmatrix}
$
and
$\beta=
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 3 & 4 & 2 & 1 & 6 \\
\end{pmatrix}
$
Are conjugated in $S_6$ and you hrite all permutations $\gamma\in S_6$ such that $\gamma\alpha\gamma^{-1}=\beta$
I know that $\alpha=(1,2,5)(4,6)(3)$ and $\beta=(2,3,4)(1,5)(6)$ also if $\sigma_1=\bigl( \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 1 & 5 & 4 & 6 \end{smallmatrix} \bigr)$ then $\sigma_1(1,2,5)\sigma_1^{-1}=(2,3,4)$; if $\sigma_2=\bigl( \begin{smallmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 2 & 3 & 4 & 1 & 6 & 5 \end{smallmatrix} \bigr)$ then $\sigma_2(4,6)\sigma_2^{-1}=(1,5)$ and if $\sigma_3=(3,6)$ then $\sigma_3(3)\sigma_3^{-1}=(6)$.
But I don't know how to build $\gamma$ such that $\gamma\alpha\gamma^{-1}=\beta$. Can you help me, please?
| Let $\gamma$ be the function that takes a number in the cycle of $\beta$ to the number in $\alpha$ in the corresponding position. That is, send $2 \rightarrow 1$, and $3 \rightarrow 2$, and $4 \rightarrow 5$ et cetera.
Intuitively you can think of this as as you temporarily relabelling the numbers in $\beta$ (applying $\gamma$), applying the permutation $\alpha$ and then "undoing" this relabelling at the end ( applying $\gamma ^{-1} $).
Of course, the choice of $\gamma$ isn't unique. We could have sent $2 \rightarrow 2$, $3 \rightarrow 5$ and $4 \rightarrow 1$, for example.
This method should work in general for finding $\gamma$ for any permutations of any order and type.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can you prove that $1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$ without using induction? Using mathematical induction, I have proved that
$$1+ 5+ 9 + \cdots +(4n-3) = 2n^{2} - n$$
for every integer $n > 0$.
I would like to know if there is another way of proving this result without using PMI. Is there any geometric solution to prove this problem? Also, are there examples of problems where only PMI is our only option?
Here is the way I have solved this using PMI.
Base Case: since $1 = 2 · 1^2 − 1$, the formula holds for $n = 1$.
Assuming that the
formula holds for some integer $k ≥ 1$, that is,
$$1 + 5 + 9 + \dots + (4k − 3) = 2k^2 − k$$
I show that
$$1 + 5 + 9 + \dots + [4(k + 1) − 3] = 2(k + 1)^2 − (k + 1).$$
Now if I use hypothesis I observe.
$$
\begin{align}
1 + 5 + 9 + \dots + [4(k + 1) − 3]
& = [1 + 5 + 9 + \dots + (4k − 3)] + 4(k + 1) −3 \\
& = (2k^2 − k) + (4k + 1) \\
& = 2k^2 + 3k + 1 \\
& = 2(k + 1)^2 − (k + 1)
\end{align}
$$
$\diamond$
| The most straightforward way is of course
$$\frac n2(a+\ell))=\frac n2 \big(1+(4n-3)\big)=\color{red}{2n^2-n}\qquad\blacksquare$$
Another method:
The well-known result of the sum of the first $n$ integers:
$$1+2+3+4+\cdots+n=\frac{n(n+1)}2$$
Subtract $1$ from each term ($n$ in total):
$$0+1+2+3+\cdots+(n-1)=\frac{n(n-1)}2$$
Multiply by $4$:
$$0+4+8+12+\cdots+4(n-1)=2n(n-1)$$
Add $1$ to each term ($n$ in total):
$$1+5+9+13+\cdots+(4n-3)=\color{red}{2n^2-n}\quad\blacksquare$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "53",
"answer_count": 14,
"answer_id": 3
} |
numbers from $1$ to $2046$ We have randomly taken $21$ integers from $1$ to $2046$.
Show that we can take $a$, $b$ and $c$ from the previous $21$ integers in a way such that the following inequality holds
\begin{equation}
bc<2a^2<4bc \end{equation}
My Attempt:
I have found $3$ triplets of such integers, but I do not think the above inequalities hold for every triplet.
Any hints ?
| Note that for any three positive integers $a,b,c$, if $2^k\leqslant b<a<c<2^{k+1}$ for some $k\in\mathbb{Z}_{>0}$, then
$$bc<ac<a\cdot 2^{k+1}=2a\cdot 2^k\leqslant 2a\cdot a=2a^2,$$
and
$$bc>ba\geqslant 2^ka=\frac{1}{2}\cdot 2^{k+1}a>\frac{1}{2}\cdot a\cdot a=\frac{1}{2}a^2.$$
Thus, $2^k\leqslant b<a<c<2^{k+1}$ for some $k\in\mathbb{Z}_{>0}\Rightarrow bc<2a^2<4bc.$
And $b=1,a=2,c=3$ also satisfies the requirement.
Note that we can divide the integers from $1$ to $2046$ into ten parts:
\begin{align*}
& A_1=\{1,2,3<2^2\}\\
& A_2=\{2^2=4,5,6,7<2^3\}\\
& \dots \\
& A_{10}=\{2^{10}=1024, 1025,\dots, 2046<2^{11}\}
\end{align*}
If we choose $21=10\times 2+1$ integers in $[1,2046]$,we can always find three of them $x_1<x_2<x_3$ falls in the same part $A_i$. Then we can let $b=x_1,a=x_2,c=x_3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating $\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)}$ I would like to evaluate the following integral:
$$
\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)}
$$
I tried various methods but without success.
| Substitution $x \rightarrow -x$ gives $$\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)}=\int_{-1}^{1} \frac{dx}{(e^{-x}+1)(x^2+1)}=\int_{-1}^{1} \frac{e^xdx}{(e^x+1)(x^2+1)}$$
Therefore
$$
\int_{-1}^{1} \frac{dx}{(x^2+1)}=\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)} + \int_{-1}^{1} \frac{e^xdx}{(e^x+1)(x^2+1)}=2\int_{-1}^{1} \frac{dx}{(e^x+1)(x^2+1)}
$$
Should be easy now.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding $\int_{0}^{1}{ \frac{ e^x}{ \cosh(x)} \,dx}$ using substitution $t = \cosh(x)$ $$\int_{0}^{1}{ \frac{ e^x}{ \cosh(x)} \,dx}$$
Hello,
I have to substitute with $$t = \cosh (x)$$
I just don´t know what to do with the e-function.
Thanks.
| Note that $\cosh(x) = \frac{e^x + e^{-x}}{2}$, so $$\frac{e^{x}}{\cosh(x)} = \frac{2 e^{x}}{e^{x}+e^{-x}} = \frac{2}{1+e^{-2x}}$$
Now subsitute $u=e^{-x}$ (which gives $du = -e^{-x}dx = -u dx$) to find: $$\int \frac{2}{1+e^{-2x}} dx = \int \frac{-2}{u(1+u^2)} du$$
The rest is straightforward.
The remainder of the work can be completed using partial fraction decomposition:
$$\frac{1}{u(1+u^2)} = \frac{A}{u} + \frac{Bu+C}{1+u^2}$$
Which gives:
$$1=A(1+u^2) + (Bu+C)u$$
Setting $u=0$ tells us $A=1$. Then $$1=1+u^2 + Bu^2 + Cu$$ $$0=(1+B)u^2 + Cu$$ and we see that $B=-1$ and $C=0$.
Therefore $$\frac{1}{u(1+u^2)} = \frac{1}{u} - \frac{u}{1+u^2}.$$
Now we have $$\int \frac{-2}{u(1+u^2)} du = -2 \int \frac{1}{u} -\frac{u}{1+u^2} du$$
$$= -2 \left( \ln(u) - \int \frac{u}{1+u^2} du \right)$$
Finish by using another substitution $w=1+u^2$. This gives $dw = 2u du$ and $$= -2 \left( \ln(u) - \frac{1}{2}\int \frac{1}{w} dw \right) = -2 \left( \ln(u) - \frac{\ln(1+u^2)}{2}\right)$$
Finally, replace $u$ by $e^{-x}$ to find:
$$\int \frac{2}{1+e^{-2x}} dx = 2x + \ln(1+e^{-2x}) +C.$$
Then plug in the bounds.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\int_{0}^{\infty}{x^2\over (e^{x}-x^2)^2}dx=\sum_{n=1}^{\infty}{n(2n)!\over (n+1)^{2n+1}}$ $$I=\int_{0}^{\infty}{x^2\over (e^{x}-x^2)^2}dx=\sum_{n=1}^{\infty}{n(2n)!\over (n+1)^{2n+1}}$$
Applying partial fractions
$x^2=A(e^x-x^2)+B$
$$I=\int_{0}^{\infty}{e^x\over (e^x-x^2)^2}-{1\over e^x-x^2}dx$$
Let
$$J=\int_{0}^{\infty}{1\over e^x-x^2}dx$$
Applying partial fractions
$1=A(e^x-x)+B(e^x+x)$
$$J={1\over 2}\int_{0}^{\infty}{e^{-x}\over e^x+x}+{e^{-x}\over e^x-x}dx={1\over 2}(A+B)$$ respectively.
$$\sum_{n=1}^{\infty}x^{n-1}e^{-nx}={1\over e^x-x}$$
Let us concentrate only on B for the moment
$${1\over 2}B={1\over 2}\sum_{n=1}^{\infty}\int_{0}^{\infty}x^{n-1}e^{-x(n+1)}dx={1\over 2}\sum_{n=2}^{\infty}\int_{0}^{\infty}x^{n-2}e^{-xn}dx$$
We can apply Laplace transform
$$B=\sum_{n=2}^{\infty}{(n-2)!\over n^{n-1}}$$
The method I am doing right now seem lengthy, so I don't think this is right approach to tackle this problem. Can someone show me another way to tackle this problem?
| \begin{align}
I = \int_0^\infty {\frac{{x^2 }}{{\left( {e^x + x^2 } \right)^2 }}dx} = \int_0^\infty {\frac{{x^2 e^{ - 2x} }}{{\left( {1 + x^2 e^{ - x} } \right)^2 }}dx}
\end{align}
Using the geometric series we have $$ \frac{1}{{1 - x}} = \sum\limits_{n = 0}^\infty {x^n } \Rightarrow \frac{1}{{\left( {1 - x} \right)^2 }} = \sum\limits_{n = 1}^\infty {nx^{n - 1} } $$
Substituting
\begin{align}
I = \int_0^\infty {x^2 e^{ - 2x} \cdot \sum\limits_{n = 1}^\infty {(-1)^{n-1}n\left( {x^2 e^{ - x} } \right)^{n - 1} } dx} \\
= \int_0^\infty {\sum\limits_{n = 1}^\infty {(-1)^{n-1}nx^{2n} e^{ - (n+1)x} } dx} \\
= \sum\limits_{n = 1}^\infty {(-1)^{n-1}n\int_0^\infty {x^{2n} e^{ - (n+1)x} dx} } \,\, \\
\end{align}
which is the gamma function
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the equation $x^2+y^2+z^2=x^2y^2$ has no integer solution,except $x=y=z=0$ Show that the equation $x^2+y^2+z^2=x^2y^2$ has no integer solution,except $x=y=z=0.$
Let one of the $x,y,z$ be even number.Let $x=2p$
$x^2+y^2+z^2=x^2y^2$
This gives $y^2+z^2$ is also even,which means either $y,z$ are both even or $y,z$ are both odd.
So either $x=2p,y=2q,z=2r$ is the solution or $x=2p,y=2q+1,z=2r+1$ is the solution.I put $x=2p,y=2q+1,z=2r+1$ in the equation $x^2+y^2+z^2=x^2y^2$ and see that it does not satisfy both sides,so it is not the solution.
When I put $x=2p,y=2q,z=2r$,i get $p^2+q^2+r^2=4p^2q^2$
I am confused here how to prove that $p^2+q^2+r^2=4p^2q^2$ does not hold true.
Also i cannot solve when $x$ is odd which is the second case.
Is my method correct?Please help me complete the solution.
| This is a Pythagorean quadruple problem.there is a good way to prove that the only solution is (0,0,0)
assuming the equation is $x^2+y^2+z^2=t^2$ ,∀ t=xy
all integer positive solution given by
$x=(l^2+m^2-n^2)/n$ , $y=2l$, $z=2m$ ,and $t=(l^2+m^2+n^2)/n$
Then $2l(l^2+m^2-n^2)=l^2+m^2+n^2$
one can find that $(l^2+m^2-n^2)(2l-1)=0$
suppose $2l-1=0$ refused
then $l^2+m^2-n^2=0 ⇒x=0$ ,and $xy=2n$
but $xy=0 ⇒ n=0 ⇒ l^2+m^2=0 ⇒ y=0 ,z=0$
therefore the only solution is (0,0,0)
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Asymptotic expansion of ratio function I want to expand the following function:
$$
f(x)=\frac{1}{(1-e^{-x})}
$$
$f(x)$ can be rewritten as
$$
f(x) \sim \frac{1}{x-x^2/2 + x^3/2/3}
$$
But I want to express big-oh notation such that
$$
f(x) = \frac{1}{x} + .... +O(x^2)
$$
up to $x^2$ order.
How to do it?
| $$\frac{1}{1-e^{-x}}=\frac{1}{x-x^2/2+x^3/6+O(x^4)} \\
=\frac{1}{x} \frac{1}{1-x/2+x^2/6+O(x^3)} \\
=\frac{1}{x} (1+x/2-x^2/6+x^2/4+O(x^3))$$
where in the last step we used the identity $\frac{1}{1-y}=\sum_{m=0}^\infty y^m$ whenever $|y|<1$. Note that in this case $y=(-x/2+x^2/6+O(x^3))$ and so the $x^2$ term has a contribution from two factors of $x/2$ as well as a contribution from one factor of $x^2/6$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the hypotenuse ( Trigonometric Identities!) If the sides of a right-angled triangle are $\cos 2a + \cos 2b + 2\cos(a+b)$ and $\sin 2a + \sin 2b + 2\sin(a+b)$, find the hypotenuse-
I can simplify this but I end up having a lot of terms in the end :/
| HINT:
Use Prosthaphaeresis Formulas
$$\cos2A+\cos2B+2\cos(A+B)=2\cos(A+B)[1+\cos(A-B)]$$
$$=4\cos(A+B)\cdot\sin^2\dfrac{A-B}2$$
Similarly, $$\sin2A+\sin2B+2\sin(A+B)=\cdots=4\sin(A+B)\cdot\sin^2\dfrac{A-B}2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a,b$ or $c$ is divisible by $7$.
Show that if $a^3+b^3+c^3\equiv{0}\pmod7$ then at least one of $a$, $b$ or $c$ is divisible by $7$.
Here it seems Fermat's theorem has no use. We could consider many different cases of remainders of $a,b,c$ modulo $7$ but that's tedious.
Any ideas for a much shorter solution?
| By Fermat's theorem $x^6\equiv1\pmod7$ holds for all $x$ coprime to $7$, so $(x^3)^2\equiv1\pmod7$ hence
$$(x^3)^2-1=(x^3-1)(x^3+1)\equiv0\pmod7$$
and hence $x^3\equiv\pm1\pmod{7}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving Chinese Remainder Theorem Algebraically I am doing a practice problem for my final which asks:
Solve the following Chinese Remainder Theorem:
$$
x \equiv 2 \pmod{3}, \\
x \equiv 3 \pmod{5}, \\
x \equiv 5 \pmod{7}, \\
x \equiv 7 \pmod{11} \\
x \equiv 11 \pmod{13}
$$
From the first I can conclude that $x = 3k + 2$ for some $k \in \mathbb{Z}$.
Now I can apply that to the second equation which gives $ 3k+2 \equiv 3 \pmod{5}.$
Then I get lost here. Do I subtract $2$ and solve $ 3k \equiv 1 \pmod{5}$?
I don't have a solid understanding of solving the Chinese Remainder Theorem algebraically in general.
| I shall solve the system of modulo equations by Chinese Remainder Theorem. I firstly solve the following system:
$$
\left\{\begin{array}{ll}
5 \times 7 \times 11 \times 13 A \equiv 2 & (\bmod 3) \\
3 \times 7 \times 11 \times 13 B \equiv 3 & (\bmod 5) \\
3 \times 5 \times 11 \times 13 C \equiv 5 & (\bmod 7) \\
3 \times 5 \times 7 \times 13 D \equiv 7 & (\bmod 11) \\
3 \times 5 \times 7 \times 11 E \equiv 11 & (\bmod 13)
\end{array}\right.
\iff \left\{\begin{aligned}
A & \equiv 2 \quad (\bmod 3) \\
3 B & \equiv 3 \quad (\bmod 5) \\
-4 C & \equiv 5 \quad (\bmod 7) \\
D & \equiv 7 \quad (\bmod 11) \\
11E & \equiv 11 \quad(\bmod 13)
\end{aligned}\right.
$$
$$
\iff\left\{\begin{array}{l}
A \equiv 2 \quad (\bmod 3) \\
B \equiv 1 \quad (\bmod 5) \\
C \equiv 4 \quad (\bmod 7) \\
D \equiv 7 \quad (\bmod 11) \\
E \equiv 1 \quad (\bmod 13)
\end{array}\right.
$$
By the Chinese Remainder Theorem, the general solution is
$$
\begin{aligned}
x \equiv & 5005 \times 2+3003 \times 1+2145 \times 4 +1365 \times 7+1155 \times 1 \quad(\bmod 15015) \\
\equiv & 2273 \quad(\bmod 15015)
\end{aligned}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Express the function $f(x)= (1-\sin x)/(1+\sin x)$ as the sum of an even and odd function. Express the function $f(x)= (1-\sin x)/(1+\sin x)$ as the sum of an even and odd function.
| $f(x)= \frac{1-\sin x}{1+\sin x} = \frac{(1-\sin x)^2}{(1+\sin x)\times (1-\sin x)} = \frac{1+\sin^2 x-2sinx}{\cos^2 x} = \frac{1}{\cos^2 x}+\tan^2 x -2 \tan x \sec x = sec^2 x+\tan^2 x -2 \tan x \sec x$
Let the even function be $sec^2 x+\tan^2 x $
Let the odd function be $-2 \tan x \sec x $
Because the parity of $\sec^2 x$ and $\tan^2 x$ and $\tan x$ and $\sec x $, it is easy to verify the parity of the above functions.
| {
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"timestamp": "2023-03-29T00:00:00",
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how many answers do this equation have$3^{2x}-34(15^{x-1})+5^{2x}=0$ How many answers do this equation have?
$3^{2x}-34(15^{x-1})+5^{2x}=0$
My Attempt:$3^{2x}+5^{2x}=34(15^{x-1})$.Now what to do?
| Substitute $a=3^x$ and $b=5^x$ to get
$a^2-\frac{34}{15}ab+b^2=0$
Multiply both sides by $15$ to get,
$15a^2-34ab+15b^2=0 \Rightarrow 15a^2-25ab-9ab+15b^2=0 \Rightarrow 5a(3a-5b)-3b(5a-3b)=0 \Rightarrow (3a-5b)(5a-3b)=0$
Now, let $\frac{a}{b}=t$
Dividing both sides by $b^2=5^{2x} \neq 0$, we get
$t=\frac{3}{5}$ or $t=\frac{5}{3}$
Back-substituting, we get $t=\frac{3^x}{5^x}=(\frac{3}{5})^x$
So we are just left to check two cases
*
*$(\frac{3}{5})^x=\frac{3}{5}$. Dividing both sides by $\frac{3}{5}$ we get, $(\frac{3}{5})^{x-1}=1 \Rightarrow x-1=0 \Rightarrow x=1$, as $a^x=1 \Rightarrow x=0$, if $a \neq 0$
*$(\frac{3}{5})^x=\frac{5}{3}=(\frac{3}{5})^{-1} $. Multipying both sides by $\frac{3}{5}$ we get, $(\frac{3}{5})^{x+1}=1 \Rightarrow x+1=0 \Rightarrow x=-1$, as $a^x=1 \Rightarrow x=0$, if $a \neq 0$
Thus, $x= \pm 1$ and there are two solutions.
| {
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"url": "https://math.stackexchange.com/questions/1814368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$9 \mid a^2 +b^2+ab$. Show that $3$ divides both $a$ and $b$. $a$ and $b$ are integers.
$a^2 +b^2+ab$ is a multiple of $9$.
I have to prove that $3$ divides both $a$ and $b$.
Converse is very easy. Put $a=3k$ and $b=3l$ and that's it.
I was trying factorisation but didn't get anything from it.
| Assuming $a,b$ are distinct,
$9 \mid a^2 +b^2+ab$
$\Rightarrow 9\mid \{(a-b)^2+3ab\} \,\,\,\,\,\,\,\,\, - (1)$
$\Rightarrow 3\mid \{(a-b)^2+3ab\}$
$\Rightarrow 3\mid(a-b)^2$
$\Rightarrow 3\mid(a-b)$
$\Rightarrow 9\mid(a-b)^2 \,\,\,\,\,\,\,\,\, - (2)$
From $(1)$ and $(2)$, we can conclude that
$9\mid3ab \Rightarrow 3\mid ab$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\varepsilon - \delta$ proof that $f(x) = x^2 - 2$ is continuous - question concerning the initial choice of $\delta$ I just realized I did non really internalized $\varepsilon - \delta$ proofs. Here there is an attempt, with some general questions I have.
Proposition: $f(x) := x^2 - 2$ is continuous.
Skratch Work: We assume that $|x - c| < \delta$. Thus, we have
\begin{align}
| x^2 - 2 - c^2 + 2 | & = | x^2 - c^2 | \\
& = | x-c| \cdot |x+c|.
\end{align}
Assume that $\delta = \frac{c}{2}$. Then we have
$$ |x - c | < \frac{c}{2} \Longleftrightarrow \frac{c}{2} < x < \frac{3c}{2} .$$
By adding $c$ we obtain
$$\frac{3c}{2} < x +c < \frac{5c}{2} .$$
Hence, $x + c < \frac{5c}{2}$, which implies that
$$ | x - c | \cdot | x+ c| < \frac{5c}{2} | x - c| < \varepsilon,$$
if $\delta = \frac{2\varepsilon}{5c}$.
Proof: Let $c$ be arbitrary, and assume that $| x - c | < \delta$. Let $\delta := \min \{ \frac{c}{2} , \frac{2\varepsilon}{5c} \}$. Hence,
\begin{align}
| x^2 - 2 - c^2 + 2 | & = | x^2 - c^2 | \\
& = | x-c| \cdot |x+c| < \frac{5c}{2} | x - c| < \frac{5c}{2} \delta = \varepsilon.
\end{align}
Being $c$ arbitrary, this establishes the continuity of $f(x)$. $\square$
Questions:
*
*Is this proof correct?
*I chose the initial value of $\delta$ arbitrarily, but in such a way that depends on the initial choice of $c$ (in doing so I did not follow a proof of this proposition I found, that starts by taking $\delta = 1$). I did this by following a well-known proof of the continuity of $f(x) = \frac{1}{x}$.
Is this sound?
| Your proof didn't cover the case $c < 0$. I would do it as follows:Let $c \in \mathbb{R}, \epsilon > 0$ be arbitrary, choose $\delta = \min \{1,\frac{\epsilon}{1+2|c|}\}$. Thus if $|x-c| < \delta $, then
$$|x-c||x+c| \leq |x-c|(|x-c| + 2|c|)< |x-c|(1+2|c|)< \dfrac{\epsilon}{1+2|c|}\cdot (1+2|c|) = \epsilon .$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} \leq \frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)$ for $x,y,z \geq 0$ This inequality is wrong - see the accepted answer (it appears there is no general inequality for these two expressions).
On the left we have harmonic mean of pairwise geometric means, which obeys:
$$\sqrt[3]{xyz} \geq \color{blue}{ \frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}} \geq 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}} \geq \frac{3xyz}{xy+yz+zx}$$
On the right we have arithmetic means of pairwise harmonic means, obeying the same inequality (the first inequality here may not be true as well, waiting for proof in another question):
$$\sqrt[3]{xyz} \geq \color{blue}{ \frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)} \geq 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}} \geq \frac{3xyz}{xy+yz+zx}$$
According to Wolfram Alpha the following is true:
$$\frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}} \leq \frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)$$
But I have not been able to prove it so far.
It may help to transform the RHS in the following way:
$$\frac{2}{3} \left(\frac{xy}{x+y}+\frac{yz}{y+z}+\frac{zx}{z+x} \right)=\frac{2}{3} \frac{xyz(x+y+z)+(xy+yz+zx)^2}{(x+y+z)(xy+yz+zx)-xyz}$$
| Your inequality is wrong! For $y=z=1$ and $x\rightarrow+\infty$ we obtain:
$3\leq\frac{2}{3}\left(1+\frac{1}{2}+1\right)$, which is wrong.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to express $2x^3-x^2-3x+2$ as a linear combination of Legendre polynomials I have used the formula
\begin{align}p_0(x)&=1\\
p_1(x)&=x\\
p_2(x)&=\frac12(3x^2−1)\\
p_3(x)&=\frac12(5x^3−3x)
\end{align}
$$2x^3-x^2-3x+2=Ap_3(x)+Bp_2(x)+Cp_1(x)+Dp_0(x)$$
EDIT-
\begin{align}&=\frac A2(5x^3−3x) + \frac B2(3x^2−1) + Cx +D \end{align}
but don't know how to continue after this.
| Continuing from your steps, we compare the coefficients of each term on the L.H.S and R.H.S.
$$2x^3-x^2-3x+2=\dfrac A2(5x^3−3x) + \dfrac B2(3x^2−1) + Cx +D$$
So, $$ 2x^3-x^2-3x+2=\dfrac{5}{2}Ax^3+\dfrac{3}{2}Bx^2+(C-\dfrac{3}{2}A)x+(D-\dfrac{B}{2})$$
*
*For the $x^3$ term, we have
$\dfrac{5}{2}A=2 \Rightarrow A=\dfrac{4}{5}$
*For the $x^2$ term, we have
$\dfrac{3}{2}B=-1 \Rightarrow B=\dfrac{-2}{3}$
*For the $x$ term, we have
$C-\dfrac{3}{2}A=-3 \Rightarrow C-\dfrac{3}{2}\cdot\dfrac{4}{5}=-3 \Rightarrow C-\dfrac{6}{5}=-3 \Rightarrow C=\dfrac{6}{5}-3\Rightarrow C=\dfrac{-9}{5}$
*For the constant term, we have
$D-\dfrac{1}{2}B=2 \Rightarrow D-\dfrac{1}{2} \cdot \dfrac{(-2)}{3}=2 \Rightarrow D=2-\dfrac{1}{3} \Rightarrow D=\dfrac{5}{3}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sum of combinatorics sequence $\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1}$ I need to find sum like $$\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1},\qquad \text{ for even } n$$
Example:
Find the sum of $$\binom{20}{1} + \binom{20}{3} +\cdots+ \binom{20}{19}=\ ?$$
| Hint: Use the fact that ${n \choose k} = {n-1 \choose k-1} + {n-1 \choose k}$. Since $n$ is even, let $n = 2p$ for some $p$. Then we have:
$$\sum_{r\ \text{ odd}}{n \choose r} = \sum_{i=1}^p{2p \choose 2i-1} = \sum_{i=1}^p{2p-1 \choose 2i-2}+{2p-1 \choose 2i-1} = \sum_{i=0}^{2p-1}{2p-1 \choose i}$$
This may be easier to see with a specific example:
$${20 \choose 1} + {20 \choose 3} + \ldots + {20 \choose 19} = \left[{19 \choose 0} + {19 \choose 1}\right] + \left[{19 \choose 2} + {19 \choose 3}\right] + \left[{19 \choose 18} + {19 \choose 19}\right]$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all real numbers x, y, z, u and v in $\sqrt{x}+\sqrt{y}+2\sqrt{z-2}+\sqrt{u}+\sqrt{v}=x+y+z+u+v$ And thanks in advance for your answers. Sorry if the text is badly formatted, I'm new here. Anyway, here is the question:
Find all real numbers x, y, z, u and v in $\sqrt{x}+\sqrt{y}+2\sqrt{z-2}+\sqrt{u}+\sqrt{v}=x+y+z+u+v$
Could I use the method of completing the square (I'm familiar with that) or is there some better way?
| Yes, you can solve your equation by completing the squares.
$$\sqrt{x}+\sqrt{y}+2\sqrt{z-2}+\sqrt{u}+\sqrt{v}=x+y+z+u+v\tag{*1}$$
For each variable, subtract its appearance in LHS from corresponding term in RHS, you have:
$$
\begin{array}{rl}
x - \sqrt{x} &= (\sqrt{x}-\frac12)^2 - \frac14\\
y - \sqrt{y} &= (\sqrt{y}-\frac12)^2 - \frac14\\
z - 2\sqrt{z-2} &= (\sqrt{z-2}-1)^2 + 1\\
u - \sqrt{u} &= (\sqrt{u}-\frac12)^2 - \frac14\\
v - \sqrt{v} &= (\sqrt{v}-\frac12)^2 - \frac14\\
\end{array}
$$
Now sum over both sides and compare result with $(*1)$, you find:
$$\verb/RHS/(*1) - \verb/LHS/(*1) =
(\sqrt{x}-\frac12)^2 +
(\sqrt{y}-\frac12)^2 +
(\sqrt{z-2}-1)^2 +
(\sqrt{u}-\frac12)^2 +
(\sqrt{v}-\frac12)^2$$
This leads to
$$\begin{cases}
\sqrt{x} = \sqrt{y} = \sqrt{u} = \sqrt{v} = \frac12,\\
\sqrt{z-2} = 1
\end{cases}
\quad\implies\quad (x,y,z,u,v) = \left(\frac14,\frac14,3,\frac14,\frac14\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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New Maths 9-1 GCSE for 2017 Sample Question My teacher gave me some practice questions for my end of year exam which will be like the new GCSE and this question is very tackling to me. Could any with clear working solve the question and show me how they did it:
A(-2,1), B(6,5), and C(4,k) are the vertices of a right angled triangle ABC.
Angle ABC is the right angle.
Find an equation of the line that passes through A and C.
Give your answer in the form ay+bx=c where a, b and c are integers.
(Total for Question = 5 marks)
| Since $\angle ABC$ is the right angle, according to Pythagorean theorem, we have $AB^2+BC^2=AC^2$
$AB^2=(6-(-2))^2+(5-1)^2=80$
$BC^2=(k-5)^2+(4-6)^2=(k-5)^2+4$
$AC^2=(k-1)^2+(4-(-2))^2=(k-1)^2+36$
So we have $80+(k-5)^2+4=(k-1)^2+36 \Rightarrow k=9$
The line passing through A and C has the form $y=\alpha x+\beta$
Calculate the slope $\alpha$:
$\alpha=\frac{k-1}{4-(-2)}=\frac{9-1}{4-(-2)}=\frac{4}{3}$
Plug in $A(-2,1)$ into the equation $y=\frac{4}{3}x+\beta$:
$1=\frac{4}{3}(-2)+\beta \Rightarrow \beta =\frac{11}{3}$
So, $y=\frac{4}{3}x+\frac{11}{3}$. We can rewrite this as $y-\frac{4}{3}x=\frac{11}{3}$, and multiply it by 3 to get $3y-4x=11$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to express $\binom{a+b}{n}$ as a sum of regular coefficients I am trying to prove that $(1 + f(x))^a(1 + f(x))^b = (1 + f(x))^{a+b}$ in the world of formal power series. At a certain point in the prove I get
\begin{align*}
(1 + f(x))^a(1+f(x))^b& = \ldots \\
& = \lim\limits_{N \to \infty} \left(\left[\sum_{n=0}^{N}\binom{a}{n} f(x)^n\right] \cdot \left[\sum_{n=0}^{N}\binom{b}{n} f(x)^n\right]\right) \\
& = \lim\limits_{N \to \infty} \left(\left[\binom{a}{0} \binom{b}{0} \right] f(x)^0 + \left[\binom{a}{1} \binom{b}{0} + \binom{a}{0} \binom{b}{1}\right] f(x)^1 + \ldots \right) \\
& = \lim\limits_{N \to \infty} \sum_{n=0}^{N} \binom{a+b}{n} f(x)^n
\end{align*}
I tried $\binom{a}{0} \binom{b}{0} = \frac{1}{0!} \cdot \frac{1}{0!} = \frac{(a+b)^{\underline{0}}}{0!}$, $\binom{a}{1} \binom{b}{0} + \binom{a}{0} \binom{b}{1} = \frac{a}{1!} \cdot \frac{1}{0!} + \frac{1}{0!} \frac{b}{1!} = \frac{(a+b)^{\underline{1}}}{1!}$, $\ldots$, but I don't seem to be able to generalize and prove this.
Could anybody get me in the right direction of how to prove that $\left[\sum_{n=0}^{N}\binom{a}{n} f(x)^n\right] \cdot \left[\sum_{n=0}^{N}\binom{b}{n} f(x)^n\right] = \sum_{n=0}^{N} \binom{a+b}{n} f(x)^n$?
Or am I approaching this problem the wrong way?
| The development chain goes like this:
$$
\eqalign{
& \left( {1 + f(x)} \right)^a \left( {1 + f(x)} \right)^b = \sum\limits_k {\left( \matrix{
a \cr
k \cr} \right)f(x)^k } \sum\limits_j {\left( \matrix{
b \cr
j \cr} \right)f(x)^j } \cr
& = \sum\limits_j {\sum\limits_k {\left( \matrix{
a \cr
k \cr} \right)\left( \matrix{
b \cr
j \cr} \right)f(x)^k } } f(x)^j = \cr
& = \sum\limits_n {\left( {\sum\limits_k {\left( \matrix{
a \cr
k \cr} \right)\left( \matrix{
b \cr
n - k \cr} \right)} } \right)\;} f(x)^n = \cr}
$$
here the Chu-Vandermonde identity comes into play,(https://en.wikipedia.org/wiki/Vandermonde's_identity) giving:
$$
\eqalign{
& = \sum\limits_n {\left( \matrix{
a + b \cr
n \cr} \right)\;} f(x)^n = \cr
& \left( {1 + f(x)} \right)^{a + b} \cr}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show $\frac{\pi^2}{8}=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}$ Knowing $$\frac{\pi^2}{sin^2(\pi z)}=\sum_{n=-\infty}^{\infty}\frac{1}{(z-n)^2}$$ how can I prove $$\frac{\pi^2}{8}=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}\qquad?$$
| We know that,
$\sum_{n=0}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$
We also know,
$\sum_{n=0}^{\infty}\frac{1}{n^2}=\sum_{n=0}^{\infty}\frac{1}{(2n)^2}+\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}$.
Now,$\sum_{n=0}^{\infty}\frac{1}{(2n)^2}=\sum_{n=0}^{\infty}\frac{1}{4n^2}=\frac{1}{4}\sum_{n=0}^{\infty}\frac{1}{n^2}=\frac{1}{4}.\frac{\pi^2}{6}=\frac{\pi^2}{24}$.
Thus,
$\frac{\pi^2}{6}=\frac{\pi^2}{24}+\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}$
So you get, $\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Partial fraction decomposition of $\pi\cdot \tan(\pi z)$
Evaluate the partial fraction decomposition of $\pi \tan(\pi z)$
$$2\pi \tan(\pi z)=\cot\left(\frac{\pi}{2}-\pi z\right)-\cot\left(\frac{\pi}{2}+\pi z\right)$$ $$=\frac{2}{1-2z}+\sum_{k=1}^\infty \frac{1-2z}{(\frac{1}{2}-z)^2-k^2}-\frac{2}{1+2z}-\sum_{k=1}^\infty \frac{1+2z}{(\frac{1}{2}-z)^2-k^2}$$
$$=\frac{8z}{1-4z^2}+\sum_{k=1}^\infty \frac{4-8z}{1-2z+4z^2-4k^2}-\sum_{k=1}^\infty \frac{4+8z}{1+2z+4z^2-4k^2}$$
I know that the result should be $$\pi \tan(\pi z)=\sum_{k=1}^\infty \frac{8z}{(2k+1)^2-4z^2}$$ but I dont know how to get there.
Any suggestions?
| Hint. If you know that, for $0<z<1/2$,
$$
\pi \cot (\pi z)=\sum_{k=1}^\infty \left(\frac{1}{z-k}+\frac{1}{z+k-1}\right)
$$ then by putting $z \to 1/2-z$ you get
$$
\begin{align}
\pi \cot (\pi (1/2-z))&=\pi \tan(\pi z)
\\\\&=\sum_{k=1}^\infty \left(\frac{1}{1/2-z-k}+\frac{1}{1/2-z+k-1}\right)
\\\\&=\sum_{k=1}^\infty\frac{8 z}{(2k-1)^2-4 z^2}
\end{align}
$$ as wanted.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to prove this inequality? $a^{2}+b^{2}+c^{2}\leq 3$ How to prove this inequality?
If $a^{2}+b^{2}+c^{2}\leq 3$ and $a,b,c\in \Bbb R^+$, then
$$\left( a+b+c\right) \left( a+b+c-abc\right)\geq 2\left( a^{2}b+b^{2}c+c^{2}a\right) $$
I tried AM>GM but I couldn't get result
| Alternative solution:
Recall the known inequality $a^2b+b^2c+c^2a \le \frac{4}{27}(a+b+c)^3 - abc$ for $a, b, c \ge 0$.
See: https://artofproblemsolving.com/community/c6h478168p2677448,
https://artofproblemsolving.com/community/c4h1705936p10979968,
https://artofproblemsolving.com/community/c6h479531p2685016
It suffices to prove that
$(a+b+c)(a+b+c-abc) \ge \frac{8}{27}(a+b+c)^3 - 2abc$
or
$$(a+b+c)^2 + 2abc \ge \frac{8}{27}(a+b+c)^3 + (a+b+c)abc.$$
By using the method of @Aravind, since $a+b+c \le 3\sqrt{\frac{a^2+b^2+c^2}{3}} \le 3$, it suffices to prove that
$$(a+b+c)^2 + 2abc \ge \frac{8}{27}(a+b+c)^2 \cdot 3 + 3abc$$
or
$$(a+b+c)^2 \ge 9abc.$$
Since $\frac{a+b+c}{3} \ge \sqrt[3]{abc}$ and $a+b+c\le 3$, we have $abc\le 1$ and $(a+b+c)^2 \ge 9(abc)^{2/3} \ge 9abc$.
We are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers
Prove $a^2+b^2+c^2 \ge a+b+c$ if $abc=1$, and $a$, $b$, $c$ are positive real numbers
It is in the exercises of the AM-GM inequality chapter of a book, and that is why I believe it will be solved by that. Can anyone give me a proof using that or otherwise, too?
| We can also proceed as follows:
\begin{align*}
a^2+b^2+c^2 &\ge \frac{(a+b+c)^2}{3}\\
&= (a+b+c)\,\frac{a+b+c}{3}\\
&\ge (a+b+c) \sqrt[3]{abc}\\
&=a+b+c.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of the sequence $a_{n+1}=\frac{1}{2} (a_n+\sqrt{\frac{a_n^2+b_n^2}{2}})$ - can't recognize the pattern Consider the sequence:
$$a_0=x,~~~b_0=y$$
$$a_{n+1}=\frac{1}{2} \left(a_n+\sqrt{\frac{a_n^2+b_n^2}{2}} \right),~b_{n+1}=\frac{1}{2} \left(b_n+\sqrt{\frac{a_n^2+b_n^2}{2}}\right)$$
$$\lim_{n \to \infty} a_n=\lim_{n \to \infty} b_n=l(x,y)$$
I can't pin down the pattern for the limit. Numerically, I've got the following result:
$$\frac{1}{l(x,y)}=\arctan (f(x,y))$$
What is the explicit expression for $f(x,y)$?
Numerical examples:
$$\begin{array}( x & y & \frac{1}{l(x,y)} \\ 1 & 2 & \arctan \left( \frac{3}{4} \right) \\ 1 & 3 & \arctan \left( \frac{1}{2} \right) \\ 2 & 3 & \arctan \left( \frac{5}{12} \right) \\ 3 & 5 & \arctan \left( \frac{1}{4} \right) \\ 1 & 5 & \arctan \left( \frac{\sqrt{13}-3}{2} \right) \\ 4 & 5 & \arctan \left( \frac{9}{40} \right) \\ 3 & 4 & \arctan \left( \frac{7}{24} \right) \\ 3 & 7 & \arctan \left( \frac{\sqrt{29}-5}{2} \right) \end{array}$$
Edit
To summarize @IvanNeretin's comment (and adding my own numerical results), so far we have:
$$l(x,0)=\frac{2x}{\pi}$$
$$f(x,x+1)=\frac{2x+1}{2x(x+1)}$$
$$f(x,x+2)=\frac{1}{x+1}$$
$f(x,x+3)$ - see my answer below
$$f(x,x+4)=\frac{\sqrt{x^2+4x+8}-(x+2)}{2}$$
Here $x \in \mathbb{R}^{+}$, $x \neq 0$
Also by symmetry we have of course:
$$f(x,y)=f(y,x)$$
At this point it's obvious that no ordinary function works here.
My hope is to connect $f(x,y)$ to some special function or an integral.
| Geometrically:
$(a_{n+1}, b_{n+1})= \frac 12 (a_n,b_n) + \frac 12 \sqrt{a_n^2+b_n^2} (\cos \pi/4, \sin \pi/4)$
So we draw a line between $(a_n,b_n)$ and connect it to a point equidistant from the origin on the line $x = y$, and find the mid-point.
$r_n = \sqrt{a_n^2 + b_n^2}\\
\phi_n = \tan^{-1}(\frac{b_n}{a_n})$
$\phi_{n+1} = \frac{\frac{\pi}{4} + \phi_n}{2}\\
r_{n+1} = r_n\cos(\frac{\frac{\pi}{4} - \phi_n}{2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
} |
How do I solve for a matrix in this linear algebra expression? In the problem I am working on I have an expression that looks like
$$B^T\operatorname{diag}(Bx) y$$
where $B$ is a known $m \times n$ matrix, $x$ is an unknown $n \times 1$ vector of variables, and $y$ is a known $m \times 1$ vector, all of real numbers.
The expression above can always be rewritten in the form $Px$ by expanding and collecting like terms, where $P$ is an $n \times n$ matrix. I would like to know how to calculate $P$ using only matrix and vector products.
As an example, take $B = \left( \begin{array}{ccc}
-1 & 1 & 0 \\
-1 & 0 & 1\end{array} \right), y = \left( \begin{array}{c}
2 \\
3\end{array} \right) $.
Then
\begin{align}
B^T \operatorname{diag}(Bx) y & = \left( \begin{array}{cc}
-1 & -1 \\
1 & 0 \\
0 & 1\end{array} \right) \left( \begin{array}{ccc}
-x_1 + x_2 & 0\\
0 & -x_1 + x_3\end{array} \right) \left( \begin{array}{c}
2 \\
3\end{array} \right) \\[10pt]
& = \left( \begin{array}{c}
-5x_1 -2x_2 -3x_3 \\
-2x_1 + 2x_2 \\
-3x_1 + 3x_3\end{array} \right) = \left( \begin{array}{cc}
-5 & -2 & -3 \\
-2 & 2 & 0 \\
-3 & 0 & 3\end{array} \right) \left( \begin{array}{c}
x_1 \\
x_2 \\
x_3\end{array} \right)
\end{align}
Therefore $P = \left( \begin{array}{cc}
-5 & -2 & -3 \\
-2 & 2 & 0 \\
-3 & 0 & 3\end{array} \right)$
But I do not know how to calculate $P$ as an expression of the other known quantities, $P = F(B,y)$. Thanks.
| Note that $x \mapsto B^T \operatorname{diag}(Bx)y$ is a linear transformation on $x$. To find the matrix of this transformation (that is, the matrix $P$ for which $Px = B^T \operatorname{diag}(Bx)y$), we can see what happens to the standard basis vectors in order to get the columns of $P$.
Long story short: let $B_k$ denote the $k$th column of $B$. Let $P_k = B^T \operatorname{diag}(B_k)y$. We then have
$$
P = [P_1 \quad P_2 \quad \cdots \quad P_n]
$$
If we use $\circ$ to denote the Hadamard product, then we could also say
$$
P_k = B^T(B_k \circ y) = B^T(y \circ B_k) = B^T \operatorname{diag}(y) B_k
$$
From which one may deduce that
$$
P = B^T \operatorname{diag}(y) B
$$
(as inspired by the other answer)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Recursively counting divisors of a number I want to make a recursive function f that counts all (not only prime) different divisors of a given natural number:
$f(n): = |{a ∈ ℕ | ∃ b ∈ ℕ : a . b = n }| $ ; with $ f(0)=0 $
for example $ f(3) = 2$, $f(6) = 4$, $f(16) = 5 $ etc.
Theoretically, how could I do that?
Thanks.
| You need to use the Möbius function $\mu(n)$:
Recurrence:
Let $z=1$
$$T(n, k) = \text{ if } k = 1 \text{ then } 1 - \sum_{i=2}^{i=n} \mu(i) \frac{T(n, i)}{i^{(s - 1)}} \text{ else }
\text{ if } \text{mod(n, k)} = 0 \text{ then } T(n/k, 1) \text{ else } 0 \text{ else } 0.$$
Then:
$$\lim\limits_{s \rightarrow z} T(n,1) = \text{number of divisors of n}$$
The first few entries of the matrix $T$ then starts:
$$T=\left(
\begin{array}{cccccccccccc}
1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \dots \\
2 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\
2 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\
3 & 2 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \\
2 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 & \\
4 & 2 & 2 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & \\
2 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & \\
4 & 3 & 0 & 2 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & \\
3 & 0 & 2 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & \\
4 & 2 & 0 & 0 & 2 & 0 & 0 & 0 & 0 & 1 & 0 & \\
2 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & \\
\vdots & & & & & & & & & & & \ddots
\end{array}
\right)$$
which has the number of divisors tau A000005 recursively defined in the first column, starting:
$$\text{A000005} = 1, 2, 2, 3, 2, 4, 2, 4, 3, 4, 2, ...$$
As a Mathematica 8 program this recurrence is:
(*recurrence*)
Clear[t, s, n, k, z, nn];
z = 1;
nn = 12; t[n_, k_] :=
t[n, k] =
If[k == 1, 1 - Sum[MoebiusMu[i] t[n, i]/i^(s - 1), {i, 2, n}],
If[Mod[n, k] == 0, t[n/k, 1], 0], 0]; A =
Table[Table[Limit[t[n, k], s -> z], {k, 1, nn}], {n, 1, nn}];
MatrixForm[A]
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How do you simplify this square root of sum: $\sqrt{7+4\sqrt3}$? I came around this expression when solving a problem.
$$\sqrt{7+4\sqrt{3}}$$
WolframAlpha says it equals $2+\sqrt{3}$. We can confirm it like this
$$\left(2+\sqrt{3}\right)^2 \;=\; 4+4\sqrt{3} + 3 \;=\; 7 + 4\sqrt{3}.$$
However, the only way I can think of how to simplify that expression in hand is guessing. Is there a better way of calculating square root of a sum like that one?
| $$\sqrt { 7+4\sqrt { 3 } } =\sqrt { 7+2\cdot 2\sqrt { 3 } } =\sqrt { { \left( \sqrt { 3 } \right) }^{ 2 }+{ 2 }^{ 2 }+4\sqrt { 3 } } =\sqrt { { \left( \sqrt { 3 } +2 \right) }^{ 2 } } =\sqrt { 3 } +2\\ \\ $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 9,
"answer_id": 0
} |
Find all positive integer roots of $\,5xy=19x+96y$
Find all positive integer roots of : $5xy=19x+96y$
I tried using decomposition technique but no success...,it seems suitable factorization of this equation is IMPOSSIBLE!!
Handy calculations show that it has as many as 6 answers.
| We have $19x=(5x-96)y$, so $y=0\bmod19$ or $x=4\bmod19$.
If $y=19n$, then we have $\frac{96}{x}=5-\frac{1}{n}$. So $x\le24$, and $x(5n-1)=96n\ (*)$. Checking $n=1,2,3,4,5$ we find that $n=1$ works giving solution $(24,19)$. $n=2,3,4$ do not work. $n=5$ gives the solution $(20,95)$. If $n>5$ then since $5n-1,n$ are coprime we must have $5n-1|96$ and hence $5n-1=48,96$, neither of which work.
If $x=4\bmod19$, then $(5n-4)y=19n+4$. If $n=1$ we have $y=23$, giving the solution $(23,23)$. If $n\ge2$, then we have $5=\frac{96}{x}+\frac{19}{y}=\frac{1}{y}\left(\frac{96}{5n-4}+19\right)\le\frac{35}{y}$, so $y\le7$. If $y=7$, we get the solution $(42,7)$. If $y=6$ we get $11n=28$ which fails. If $y=5$ we get the solution $(80,5)$. If $y=4$ we get the solution $(384,4)$. If $y\le3$ we get $(19-5y)n=-4-4y$ which fails.
So in total we have the six solutions $(20,95),(23,23),(24,19),(42,7),(80,5),(384,4)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 3
} |
Evaluate $\int \frac{1}{3+4\tan x} \, dx$ I'm am trying to evaluate $$\int \frac{1}{3+4\tan x} \, dx.$$
I tried to use the universal trig substitution, that is
$$t = \tan{x \over 2}, \quad \tan x = {2t \over 1-t^2}, \quad dx = {2 \over t^2 + 1}dt$$
and after manipulation I got the integral to be
$$\int \frac{2(1 - t^2)}{-3t^4 + 8t^3 + 8t +3} \, dt.$$
Is there a more clever approach to this problem or do I have to continue and face partial fractions?
| For sure, as shown in comments and answers, the tangent half-angle substitution is not the easiest way for this problem. However, computing $$I=\int \frac{2(1 - t^2)}{-3t^4 + 8t^3 + 8t +3} \, dt$$ is doable noticing that the roots of the denominator are $-\frac 13$, $i$,$-i$ and $3$. Using it, by partial fraction decomposition, we can arrive to $$ \frac{2(1 - t^2)}{-3t^4 + 8t^3 + 8t +3}=-\frac{2 (4 t-3)}{25 \left(t^2+1\right)}+\frac{4}{25 (t-3)}+\frac{12}{25 (3 t+1)}$$ and get $$I=\frac{6}{25} \tan ^{-1}(t)-\frac{4}{25} \log \left(t^2+1\right)+\frac{4}{25} \log (t-3)+\frac{4}{25} \log (75 t+25)$$ and simplifying $$I=\frac{2}{25} \left(3 \tan
^{-1}(t)+2\log \left(\frac{25 (t-3) (3 t+1)}{t^2+1}\right)\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Upper bound for $\gcd(a,b)$ if $\frac{a+1}{b}+\frac{b+1}{a}\in\Bbb{N}$
Suppose that $a,b$ are two positive integers so that $\frac{a+1}{b}+\frac{b+1}{a}$ is also a positive integer.Find the best upper bound for $\gcd(a,b)$.
My work:
$\frac{a+1}{b}+\frac{b+1}{a}=\frac{a(a+1)+b(b+1)}{ab} \in \Bbb{N} \implies ab|a^2+b^2+a+b , 2ab \implies ab|(a+b)(a+b+1)\implies ab|a+b\ or\ ab|a+b+1$
Now may I infer that $a=b=1\ or\ a=b=2$? If yes how may we set an upper bound for $\gcd(a,b)$??!!
| For the equation. $$\frac{m+1}{n}+\frac{n+1}{m}=a$$
If are solutions of the equation Pell. $p^2-(a^2-4)s^2=1$
Then the formula is:
$$n=2(p-(a+2)s)s$$
$$m=-2(p+(a+2)s)s$$
And another solution:
$$n=\frac{2p(p+(a-2)s)}{a-2}$$
$$m=\frac{2p(p-(a-2)s)}{a-2}$$
If are solutions of the equation Pell. $p^2-(a^2-4)s^2=4$
Then the formula is:
$$n=\frac{p-(a-2)s+2}{2(a-2)}$$
$$m=\frac{p+(a-2)s+2}{2(a-2)}$$
You can write a more General formula. http://www.artofproblemsolving.com/community/c3046h1046841___
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I solve this System of Equations? How do I begin to solve this system? $$x^2=y+a$$$$y^2=z+a$$$$z^2=x+a$$
Do I take the square roots of $x,y$ and $z$? If so, we get $$x=\pm\sqrt{a+y}$$$$z=\pm\sqrt{a\pm\sqrt{a+y}}$$$$y=\pm\sqrt{a\pm\sqrt{a\pm\sqrt{a+y}}}$$
What do I do now? I'm confused as to what to do.
| Proceeding from Sonnhard's solution, besides $z = (1 \pm \sqrt{1+4a})/2$ from the factor $-z^2 + z + a$, we get the two cubics
$$z^3 + \frac{1\pm r}{2} z^2 + \frac{\pm r - 2a - 1}{2} z + \frac{a \mp ra -2}{2}$$
where $r = \sqrt{4a-7}$. If you're interested in real solutions, you need $
a \ge -1/4$ for the first pair of solutions, and $a \ge 7/4$ for the cubics.
The solutions for $z = (1 \pm \sqrt{1+4a})/2$ have the same values for $x$ and $y$. The solutions for the cubics have $x, y, z$ the three roots of the cubic, in decreasing order (or cyclically permuted).
Here's a plot: the black curve is $ z = (1 \pm \sqrt{1+4a})/2$, the red and blue curves are the roots of the two cubics.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Distance from a set to a point There is this exercise I cannot understand well. It asks me for the distance between this set in $\mathbb{R}^3$
$$U = \{(x, y, z)\ |\ ax + y - 2z = 0, z = 0 \}$$
and the point $(0, b, 1)$.
Also it tells me to say "what is that".
Any hint or help?
| This is a least-squares problem. Note that $U$ contains the origin and is parametrized by
$$\left\{ \gamma \begin{bmatrix} 1\\ -a\\ 0\end{bmatrix} : \gamma \in \mathbb R \right\}$$
The projection matrix that projects onto $U$ is
$$P := \begin{bmatrix} 1\\ -a\\ 0\end{bmatrix} \left(\begin{bmatrix} 1\\ -a\\ 0\end{bmatrix}^T \begin{bmatrix} 1\\ -a\\ 0\end{bmatrix}\right)^{-1} \begin{bmatrix} 1\\ -a\\ 0\end{bmatrix}^T = \frac{1}{1+a^2} \begin{bmatrix} 1 & -a & 0\\ -a & a^2 & 0\\ 0 & 0 & 0\end{bmatrix}$$
and the projection matrix that projects onto the orthogonal complement of $U$ is
$$I_3 - P = \frac{1}{1+a^2} \begin{bmatrix} a^2 & a & 0\\ a & 1 & 0\\ 0 & 0 & 1+a^2\end{bmatrix}$$
The Euclidean distance between $U$ and $y := \begin{bmatrix} 0\\ b\\ 1\end{bmatrix}$ is given by
$$\| (I_3 - P) y \|_2 = \left\|\frac{1}{1+a^2} \begin{bmatrix} a^2 & a & 0\\ a & 1 & 0\\ 0 & 0 & 1+a^2\end{bmatrix} \begin{bmatrix} 0\\ b\\ 1\end{bmatrix}\right\|_2 = \frac{1}{1+a^2} \left\| \begin{bmatrix} ab\\ b\\ 1+a^2\end{bmatrix} \right\|_2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the smallest positive integer which can be written as the sum of the squares of two positive integers in two different ways
Find the smallest positive integer which can be written as the sum of the squares of two positive integers in two different ways.
I took extremely long to solve this
I got
$50= 7^2 + 1^2 $
$50= 5^2 + 5 ^2 $
I did it with the trial and error method.
I'm just curious if this came out during my exams, is there a quicker way to find the answer?
| I do wonder why it took so long... I added all pairs from the first ten squares on a piece of scrap paper in a few minutes. Maybe that's a long time.
\begin{array}{c|c|c}
&& +1 & +4 & +9 & +16 & +25 & +36 & +49 & +64 & +81 & +100 \\ \hline
1 && 2 \\ \hline
4 && 5 & 8\\ \hline
9 && 10 & 13 & 18 \\ \hline
16 && 17 & 20 & 25 & 32 \\ \hline
25 && 26 & 29 & 34 & 41 & \color{red}{50} \\ \hline
36 && 37 & 40 & 45 & 52 & 61 & 72 \\ \hline
49 && \color{red}{50} & 53 & 58 & \color{blue}{65} & 74 & \color{magenta}{85} & 98 \\ \hline
64 && \color{blue}{65} & 68 & 73 & 80 & 89 & 100 & 113 & 128 \\ \hline
81 && 82 & \color{magenta}{85} & 90 & 97 & 106 & 117 & 130 & 145 & 162 \\ \hline
100 && 101 & 104 & 109 & 116 & 125 & 136 & 149 & 164 & 181 & 200 \\ \hline
\end{array}
This also gives the first case when all the squares in question are distinct, at $65 = 8^2+1^2 = 7^2+4^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
How to find specific variables that cause vectors to be linearly independent / dependent The vectors
$v= \begin{bmatrix}
5\\
2\\
7\\
\end{bmatrix}, u = \begin{bmatrix}
4\\
4\\
13+k\\
\end{bmatrix}, \text{and } w = \begin{bmatrix}
-4\\
-2\\
-6\\
\end{bmatrix} $
are linearly independent if and only if $k \neq$ what?
How exactly can I go about solving this?
| As Fly by Night mentioned, these vectors will be linearly dependent if and only if $$\det \begin{pmatrix}
5 & 4 & -4\\
2 & 4 & -2\\
7 & 13+k & -6
\end{pmatrix} = 0.$$
\begin{align*}
\det \begin{pmatrix}
5 & 4 & -4\\
2 & 4 & -2\\
7 & 13+k & -6
\end{pmatrix} &= 5[(4)(-6) - (13+k)(-2)] - 4[(2)(-6) - (7)(-2)] + (-4)[(2)(13+k)-(7)(4)] \\
&=2k+10
\end{align*}
Therefore the vectors will be linearly dependent if and only if $$ 2k+10 = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A common term for $a_n=\begin{cases} 2a_{n-1} & \text{if } n\ \text{ is even, }\\ 2a_{n-1}+1 & \text{if } n\ \text{ is odd. } \end{cases}$ When I was answering a question here, I found a sequence as a recursive one as given below.
$a_1=1$, and for $n>1$,
$$a_n=\begin{cases}
2a_{n-1} & \text{if } n\ \text{ is even, }\\
2a_{n-1}+1 & \text{if } n\ \text{ is odd. }
\end{cases}$$
I need to find a common term for this sequence. For example, for the sequence $a_1=2$ and $a_n=2a_{n-1}$, for $n>1$, the common term is $a_n=2^n$.
I appreciate any answer or hint in advance.
| Note that in binary we have
$$\begin{align*}
a_1&=1\\
a_2&=10\\
a_3&=101\\
a_4&=1010\\
a_5&=10101\;,
\end{align*}$$
displaying a pattern easily shown by induction to be real. Now note that the binary expansion of $\frac23$ is $\frac23=0.\overline{10}_{\text{two}}$, so that
$$\begin{align*}
2\cdot\frac23&=1.\overline{01}_{\text{two}}\\
2^2\cdot\frac23&=10.\overline{10}_{\text{two}}\\
2^3\cdot\frac23&=101.\overline{01}_{\text{two}}\\
2^4\cdot\frac23&=1010.\overline{10}_{\text{two}}\\
2^5\cdot\frac23&=10101.\overline{01}_{\text{two}}\;,
\end{align*}$$
and therefore
$$a_n=\left\lfloor 2^n\cdot\frac23\right\rfloor=\left\lfloor\frac{2^{n+1}}3\right\rfloor\;.$$
If you really want to get rid of the floor function, observe that $2^{n+1}\equiv 1\pmod3$ when $n$ is odd, and $2^{n+1}\equiv2\pmod3$ when $n$ is even, so
$$\left\lfloor\frac{2^{n+1}}3\right\rfloor=\begin{cases}
\dfrac{2^{n+1}-1}3,&\text{if }n\text{ is odd}\\
\dfrac{2^{n+1}-2}3,&\text{if }n\text{ is even}\;.
\end{cases}$$
Now
$$\frac12\big(1+(-1)^n\big)=\begin{cases}
0,&\text{if }n\text{ is odd}\\
1,&\text{if }n\text{ is even}\;,
\end{cases}$$
so
$$\begin{align*}
\left\lfloor\frac{2^{n+1}}3\right\rfloor&=\frac13\left(2^{n+1}-1-\frac12\big(1+(-1)^n\big)\right)\\
&=\frac13\left(2^{n+1}-\frac12\left(3+(-1)^n\right)\right)\\
&=\frac16\left(2^{n+2}-3-(-1)^n\right)\;.
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\ln(1+x) \approx A+Bx+Cx^2$, differentiate twice both sides and show that $\ln(1+x) \approx x-\frac{1}{2}x^2$
Question: $\ln(1+x) \approx A+Bx+Cx^2$, for $-1<x\leq1$, where $A,B,C$ are constants.
Differentiate twice both sides of the approximation above and hence show that $$ \ln(1+x) \approx x-\frac{1}{2}x^2$$
Using logs, estimate the smallest value of the positive integer $n$ for which $$ \left(1+\frac{1}{2n} \right)^{n+3} < \left(1+\frac{1}{n} \right)^{n-1}$$
My working:
For the first part:
$$\ln(1+x) \approx A+Bx+Cx^2$$
$$ \Leftrightarrow \frac{1}{1+x} \approx B+2Cx$$
$$ \Leftrightarrow \frac{-1}{(1+x)^2} \approx 2C $$
But now I do not know what to do with this....
| \begin{align}
\ln(1+x) & = A + Bx +Cx^2. & & \text{When } x = 0 \text{ then } \ln 1 = A. \\
& & & \text{So } A=\ln 1 = 0. \\[10pt]
\frac 1 {1+x} & = B + 2Cx. & & \text{When } x=0 \text{ then } \frac 1 {1+0} = B. \\
& & & \text{So } B = 1. \\[10pt]
\frac {-1}{(x+1)^2} & = 2C. & & \text{When } x=0 \text{ then } \frac{-1}{(1+0)^2} = 2C. \\
& & & \text{So } C= \frac{-1}2.
\end{align}
Thus
$$
\ln(1+x) = x - \frac{x^2} 2 + \text{higher-degree terms}.
$$
After finding the derivatives on both sides you need to set $x=0$ on both sides, at each step finding one coefficient.
The higher-degree terms are microscopic compared to the initial terms if $x$ is very close to $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 3
} |
Find all pairs such that $x^2 y + x + y$ is divisible by $xy^2 + y + 7$
Find all pairs $(x, y)$ of positive integers such that $x^2 y + x + y$ is divisible by $xy^2 + y + 7$. If there are too many to write, write a generic form.
I was thinking of rewriting the divisibility in a simpler way such as $xy^2 + y + 7\mid x^2 y + x + y \implies xy^2 + y + 7\mid x^2y-xy^2+x-y$ but I am not sure if that helps. Is there an easier way?
| Hint:
$$\begin{array}{ccc}xy^2+y+7&|&x^2y+x+y\\
xy^2+y+7&|&x^2y^2+xy+y^2\\
xy^2+y+7&|&y^2-7x
\end{array}$$
Compare $y^2-7x$ with $xy^2+y+7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove: if $|x-1|<\frac{1}{10}$ so $\frac{|x^2-1|}{|x+3|}<\frac{1}{13}$
Prove: $$|x-1|<\frac{1}{10} \rightarrow \frac{|x^2-1|}{|x+3|}<\frac{1}{13}$$
$$|x-1|<\frac{1}{10}$$
$$ -\frac{1}{10}<x-1<\frac{1}{10}$$
$$ \frac{19}{10}<x+1<\frac{21}{10}$$
$$|x+1|<\frac{19}{10}$$
Adding 4 to both sides of $$ -\frac{1}{10}<x-1<\frac{1}{10}$$ gives:
$$\frac{39}{10}<x+3<\frac{41}{10}$$
$$|x+3|<\frac{39}{10}$$
Plugging those results in $$\frac{|x-1|*|x+1|}{|x+3|}<\frac{1}{13}$$
We get: $$\frac{\frac{1}{10}*\frac{19}{10}}{\frac{39}{10}}<\frac{1}{13}$$
$$\frac{19}{390}<\frac{1}{13}$$ Which is true, is this proof is valid as I took the smallest intervals, like $|x+3|<\frac{39}{10}$ and not $|x+3|<\frac{41}{10}$?
| The gradient of $f(x)=\dfrac{x^2-1}{x+3}$ near $x=1$ is $$f'(x)=1-\dfrac{8}{(x+3)^2},$$which is positive in the range $0.9<x<1.1$; so, in this range, $|f(x)|$ ranges from $0$ (at $x=1$) to $\max\{|f(0.9)|,|f(1.1)|\}=0.0512...<1/13.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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} |
Value of $\left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
If $a+b+c=0,$ Then value of $\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right)$
$\bf{My\; Try::}$ Given $$\displaystyle \left(\frac{a}{b-c}+\frac{b}{c-a}+\frac{c}{a-b}\right)\cdot \left(\frac{b-c}{a}+\frac{c-a}{b}+\frac{a-b}{c}\right) = 1+1+1+\frac{b-c}{a}\left(\frac{b}{c-a}+\frac{c}{a-b}\right)+\frac{c-a}{b}\left(\frac{a}{b-c}+\frac{c}{a-b}\right)+\frac{a-b}{c}\left(\frac{a}{b-c}+\frac{b}{c-a}\right)$$
Now How can I Solve after that, Is there is any less complex method plz explain here , Thanks
| I think your method is efficient and pretty neat.
Continuing from where you left.
Observe that
$$\dfrac{b-c}{a}\left(\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)=\dfrac{b-c}{a}\left(\dfrac{b(a-b)+c(c-a)}{(a-b)(c-a)}\right)=\dfrac{(b-c)^2 \cdot (a-b-c)}{a(a-b)(c-a)}$$
Now, using $a+b+c=0$ we get,
$$\dfrac{b-c}{a}\left(\dfrac{b}{c-a}+\dfrac{c}{a-b}\right)=\dfrac{2(b-c)^3}{(a-b)(b-c)(c-a)}$$
Similarly, other terms are dealt with.
Now, substitute $x=b-c,y=c-a$ and $z=a-b$
And then use $x^3+y^3+z^3=3xyz$ as $x+y+z=0$
Finally, we get the answer as $\boxed{9}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Is the least-squares solution unique? I am looking for a line closest to $(-5, -2)$, $(-2, 0)$, $(-1, 0)$, $(2, 3)$, $(5, 4)$ using the least square solution. So I set the line as $$ax+by+c=0$$ let $a=1$ (where $a$ is not $0$ obviously) and got
$$\begin{pmatrix}
-2 & 1 \\
0 & 1 \\
0 & 1 \\
3 & 1 \\
4 & 1\\
\end{pmatrix}
\begin{pmatrix}
b \\
c \\
\end{pmatrix}=
\begin{pmatrix}
5 \\
2 \\
1 \\
-2 \\
-5\\
\end{pmatrix}
$$
Then I solved it by multiplying $A^T$on the both sides.
But the $(b, c)$ I got here was different from that of the usual solution using $$y=ax+b$$ Although the two lines are almost identical (which implies I am not that wrong), they are still different. What's the matter?
| By inspection, no the least squares is not unique. The nullspace $\mathcal{N}\left( \mathbf{A}^{*} \right)$ is not trivial. The column space has dimension $m=5$, and we only have 2 linearly independent vectors.
To continue the problem, a guess was made about the notation and true problem. Start with the $m=5$ data points $\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$ and the trial function.
$$
y(x) = a_{0} + a_{1} x
$$
The linear system is
$$
\begin{align}
\mathbf{A} x &= y \\
%
\left[
\begin{array}{rr}
1 & -5 \\
1 & -2 \\
1 & -1 \\
1 & 2 \\
1 & 5 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
a_{0} \\ a_{1}
\end{array}
\right]
%
&=
%
\left[
\begin{array}{r}
-2 \\
0 \\
0 \\
3 \\
4 \\
\end{array}
\right]
%
\end{align}
$$
The general least squares problem is defined as
$$
x_{LS} = \left\{ x\in\mathbb{C}^{n} \colon \lVert \mathbf{A} x_{LS} - b \rVert_{2}^{2} \text{ is minimized} \right\}
$$
which has the general solution
$$
x_{LS} = \mathbf{A}^{\dagger} b + \left( \mathbf{I}_{n} - \mathbf{A}^{\dagger}\mathbf{A} \right) z, \quad z\in\mathbb{C}^{n}
$$
which is in general an affine space shown with the dashed red line below.
Pose the normal equations
$$
\begin{align}
\mathbf{A}^{*} \mathbf{A} x &= \mathbf{A}^{*} b \\
\left[
\begin{array}{cc}
\mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x \\
x \cdot \mathbf{1} & x \cdot x
\end{array}
\right]
%
\left[
\begin{array}{c}
a_{0} \\
a_{1}
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
\mathbf{1} \cdot y \\
x \cdot y
\end{array}
\right] \\[3pt]
%
\left[
\begin{array}{rr}
5 & -1 \\
-1 & 59
\end{array}
\right]
%
\left[
\begin{array}{c}
a_{0} \\
a_{1}
\end{array}
\right]
%
&=
%
\left[
\begin{array}{r}
1 \\
-59
\end{array}
\right].
%
%
\end{align}
%
$$
The solution is
$$
\begin{align}
x_{LS}
%
&= \left( \mathbf{A}^{*}\mathbf{A} \right)^{-1} \mathbf{A}^{*} y \\[3pt]
%
&=
%
\frac{1}{294}
\left[
\begin{array}{rr}
59 & 1 \\
1 & 5 \\
\end{array}
\right]
%
\left[
\begin{array}{r}
1 \\
-59
\end{array}
\right] \\[3pt]
%
&=
%
\frac{1}{294}
\left[
\begin{array}{r}
331 \\
185
\end{array}
\right]
%
\end{align}
$$
As shown in Difference between orthogonal projection and least squares solution, the normal equations solution in the pseudoinverse solution. but we need the rest of the minimizers on the dashed line.
Knowing that
$$
\mathcal{N}\left( \mathbf{A}^{*} \right) = \text{span }
\left\{ \
%
\left[
\begin{array}{r}
7 \\
-10 \\
0 \\
0 \\
3
\end{array}
\right], \
%
\left[
\begin{array}{r}
4 \\
-7 \\
0 \\
3 \\
0
\end{array}
\right],\
%
\left[
\begin{array}{r}
1 \\
-4 \\
3 \\
0 \\
0
\end{array}
\right]\
\right\}
%
$$
the full least squares solution is
$$
x_{LS} =
%
\frac{1}{294}
\left[
\begin{array}{r}
331 \\
185
\end{array}
\right]
%
+ \alpha
%
\left[
\begin{array}{r}
7 \\
-10 \\
0 \\
0 \\
3
\end{array}
\right]
%
+ \beta
%
\left[
\begin{array}{r}
4 \\
-7 \\
0 \\
3 \\
0
\end{array}
\right]
%
+ \gamma
%
\left[
\begin{array}{r}
1 \\
-4 \\
3 \\
0 \\
0
\end{array}
\right]
$$
where $\alpha$, $\beta$, and $\gamma$ are arbitrary complex constants.
For more insight: Is the unique least norm solution to Ax=b the orthogonal projection of b onto R(A)?
| {
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"url": "https://math.stackexchange.com/questions/1839221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Rearrange the series $ \sum_{n=1}^\infty (-1)^{n+1}\frac{1}{n}$ to converge to $1$. I have studied the Riemann's theorem about rearrangement of conditionally convergent series. Also I have seen other rearrangements of the given series on this site that converge to different sums $\ln2,\;\frac{3}{2}\ln2 $, etc. But Iam not able to visualize the rearrangement that converges to $1$. Please help. Thank You.
| The positive terms are $$ 1, \frac 1 3, \frac 1 5, \frac 1 7, \frac 1 9, \frac 1 {11}, \frac 1 {13}, \ldots $$ Their sum diverges to $+\infty$. The negative terms are $-1$ multiplied by $$ \frac 1 2, \frac 1 4, \frac 1 6, \frac 1 8, \frac 1 {10}, \frac 1 {12}, \ldots $$ Their sum diverges to $-\infty$.
$1 + \dfrac 1 3$ exceeds $1$. Now add enough negative terms to that to get a sum less than $1$:
$$
1 + \frac 1 3 - \frac 1 2 = \frac 5 6 < 1.
$$
Then add enough positive terms after that to make the sum more than $1$:
$$
1 + \frac 1 3 - \frac 1 2 + \frac 1 5 = \frac{31}{30} >1.
$$
Then add enough negative terms after that to make the sum less than $1$:
$$
1 + \frac 13 - \frac12 + \frac 1 5 - \frac 1 4 = \frac{47}{60} <1.
$$
Then add enough positive terms after that to make the sum more than $1$:
$$
1 + \frac 13 - \frac12 + \frac 1 5 - \frac 1 4 \underbrace{{} + \frac 1 7 + \frac 1 9} = \frac{1307}{1260} > 1.
$$
This last time we needed two terms. Is there a pattern to the number of terms we have to add at each step? Maybe not. But we know that it will always be possible to make the sum more than $1$ or less than $1$ as the case may be, because the series of positive terms and the series of negative terms both diverge to infinity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1840623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove that given a binary operator $*$, $a * b = a^2 - ab + b^2$ is associative? No matter how hard I try I cannot seem to prove that, given the binary operator $*$, the operation $a * b = a^2 - ab + b^2$ is associative.
This is what I have tried:
$(a * b) * c = a * (b * c) [Associativity];
$(a^2 - ab + b^2) * c = a * (b^2 - bc + c^2)$;
I have tried using $(a - b)^2$: $(a - b)^2 * c = a * (b - c)^2$;
Like other associativity problems, I could prove that but reaching an equality after expanding both sides, but that doesn't seems to be the case here. What am I doing wrong? Is this even an associative operation?
| $$1 * (2 * 3)=1 * (2^2-2*3+3^2)=1 * 7=1^2-1*7+7^2=43$$
$$(1 * 2) * 3=(1^2-1*2+2^2) * 3=3 * 3=3^2-3*3+3^2=9$$
Therefore, $*$ is not associative. Always remember to guess and check values of $a,b,c$ to see if an operation is associative before trying to prove it!
Now, to show you why this is not associative, let's expand this out. In the following equations, the quantities are not actually equal, but we are going to show this by expanding each side out separately.
As you said, we have:
$$(a^2-ab+b^2) * c=a * (b^2-bc+c^2)$$
Substitute into the formula:
$$(a^2-ab+b^2)^2-c(a^2-ab+b^2)+c^2=a^2-a(b^2-bc+c^2)+(b^2-bc+c^2)^2$$
Distribute the $-c$ on the left side and the $-a$ on the right sode:
$$(a^2-ab+b^2)^2-ca^2+abc-cb^2+c^2=a^2-ab^2+abc-ac^2+(b^2-bc+c^2)^2$$
Use the $(u+v+w)^2=u^2+v^2+w^2+2uv+2uw+2vw$ identity to square the trionmials:
$$a^4+a^2b^2+b^4-2a^3b+2a^2b^2-2ab^3-ca^2+abc-cb^2+c^2 \\ =a^2-ab^2+abc-ac^2+b^4+b^2c^2+c^4-2b^3c+2b^2c^2-2bc^3$$
Add like terms and rearrange terms:
$$abc+a^4-2a^3b+3a^2b^2-2ab^3+b^4-ca^2-cb^2+c^2 \\ =abc+b^4-2b^3c+3b^2c^2-2bc^3+c^4-ab^2-ac^2+a^2$$
Hopefully, you see now that wherever there is $a$ on the left side, there is $b$ on the right side, wherever there is $b$ on the left side, there is $c$ on the right side, and wherever there is $a$ on the right side. Therefore, the right side is the left side with those substitutions, or $(b * c) * a$. However, originally, the right side was $a* (b * c)$, so:
$$a * (b * c)=(b * c) * a$$
However, this just shows commutativity, not associativity, so this operation isn't associative since the left and right side in the above equations aren't the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
When is $\sqrt{x/y^2}$ equal to $\sqrt{x}/y$? The solution to the quadratics is given by
$r = -\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$, which is shortened to
$r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$, but I'm wondering how if this is justified, given that $4a^2$ can be negative if $a \in \mathbb{C}$, and $\sqrt{\dfrac{x}{y}} \neq \dfrac{\sqrt{x}}{\sqrt{y}}$ if $x$ and $y$ are negative, but given that we have $a^2$, is this justified?
Is $\sqrt{x/y^2} = \sqrt{x}/y$? Is $r = -\dfrac{b}{2a}\pm\dfrac{\sqrt{b^2-4ac}}{2a}$ always true?
| $$\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}$$
Is "the solution" to $ax^2 + bx + c$, provided that $a,b,c$ are real numbers and $a \ne 0$.
$4a^2 \ge 0$ for all real numbers $a$. Since we are excluding $a = 0$, then $4a^2 > 0$. We can then say
$$\sqrt{4a^2} =
\begin{cases}
2a, & \text{if $a \ge 0$} \\
-2a, & \text{if $a < 0$}
\end{cases}$$
The real problem has to do with $b^2 - 4ac$ since it can be positive, negative, or $0$. But, in all three cases, we will have
$$\sqrt{\dfrac{b^2-4ac}{4a^2}}= \pm \dfrac{\sqrt{b^2-4ac}}{2a}$$
so we end up with
\begin{align}
\dfrac{b}{2a}\pm\sqrt{\dfrac{b^2-4ac}{4a^2}}
&= \dfrac{b}{2a}\pm \left( \pm \dfrac{\sqrt{b^2-4ac}}{2a} \right) \\
&= \dfrac{b}{2a}\pm \dfrac{\sqrt{b^2-4ac}}{2a} \\
\end{align}
By the way, for any real number $x$, $\sqrt{x^2} = |x|$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1842693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
The double ${-6}$, from ${|3x + 7| = 11}$ Solve for $x$ such that $|3x + 7| = 11$.
Answer. "${x = \frac {4}{3}, -6, -6}$". First rewrite the absolute value equation as two separate linear equations. In the first equation, assume that the ${3 + 7}$ is positive and set it equal to 11. In the second equation, also equal to 11, assume that the ${(3x + 7)}$ is negative. For that one, negate (multiply by -1) the whole binomial, and then solve the equation.
${3x + 7 = 11}$
${3x = 4}$
${x = \frac {4}{3}}$
$-(3x + 7) = 11$
${-3x - 7 = 11}$
$-3x = 18$
${x = -6}$
Where does a double -6 apply.
| ${3x + 7 = 11}$
$-(3x + 7) = 11$
$3x + 7 = -11$
$3x = -18$
$x = -6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1845060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Induction Sum of all odd Numbers Show that
$\sum_{k=1}^{n}(2k-1)=n^2$
Beginning: n=1
$\sum_{k=1}^{1}(2k-1)=(2*1-1)=1=1^2$
Let $\sum_{k=1}^{n}(2k-1)=n^2$ be true, then for n=p+1
$\sum_{k=1}^{p+1}(2k-1)=(p+1)^2$ has to be true too.
$\sum_{k=1}^{p+1}(2k-1)=1+3+...+(2(p+1)-3)+(2(p+1)-1)$
$=1+3+...+(2p-1)+(2p+1)$
$=1+3+...+(2p-3)+(2p-1)+(2p+1)$
(Using our assumption that $1+3+...+(2p-3)+(2p-1)=p^2$)
$=p^2+2p+1$
$=(p+1)^2$ which is exactly what we wanted to show.
Is this ok?
| Hint:
$1$ $ +3+5+\cdots+2k-1=x$
$2k-1+2k-3++\cdots+3+1=x$
$2k+2k+\ldots+2k=2x$
| {
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"url": "https://math.stackexchange.com/questions/1846093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Compute $\int_{0}^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta$ I''m stuck in a exercise in complex analysis concerning integration of rational trigonometric functions. Here it goes:
We want to evaluate $\int_{0}^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta$.
Here's my work:
Let $z=e^{i\theta}$ so that $d\theta=dz/iz$ and $\cos \theta = \frac12 (z+z^{-1})$. We have
$$\begin{align}
I&=\int_{0}^{2\pi}\frac{1}{(2+\cos\theta)^2}\,d\theta \quad (1)\\\\
&=\oint_C \frac{1}{(2+\frac{z+z^{-1}}{2})^2}\frac{dz}{iz} \quad (2)\\\\
&=\frac1i\oint_C \frac{4}{(4+z+z^{-1})^2}\frac{dz}{z} \quad (3) \\\\
&=\frac4i\oint_C \frac{1}{(z^2+4z+1)^2}\,dz \quad (4) \\\\
&=\frac4i\oint_C \frac{1}{(z-(-2+\sqrt3)^2(z-(-2-\sqrt3)^2}\,dz \quad (5)\\\\
&=\frac4i\oint_Cf(z)\,dz \quad (6)
\end{align}$$
where $C$ is the unit circle in the complex $z$-plane.
The function $f(z)$ has singularities at $(-2\pm\sqrt3)$ but only $(-2+\sqrt3)\in int(C)$. Therefore,
$$\oint_Cf(z)\,dz=2\pi i (Res(f, (-2+\sqrt3))) \quad (*)$$
Since $(-2+\sqrt3)$ is a pole of order $2$, after a quick calculation I get
$$Res(f, (-2+\sqrt3)=-\frac{\sqrt3}{3}$$
and so for $(*)$ I get
$$\oint_Cf(z)\,dz=2\pi i (-\frac{\sqrt3}{3})=-\frac{2\pi\sqrt3i}{3}.$$
Hence, by $(6)$, I find
$$\frac4i\oint_Cf(z)\,dz=\frac4i \left(\frac{-2\pi\sqrt3i}{3}\right)=-\frac{8\pi\sqrt3}{3}.$$
My solution however is not correct. The given integral evaluates to $\frac{4\pi}{3\sqrt3}$ (verified with Wolfram). I suspect there must be a flaw in one of my steps from $(1)$ to $(6)$ but I can't find it.
| Edit
$$I(a)=\int_{0}^{2\pi}\frac{1}{a+\cos x}dx=\int_{-\pi}^{\pi}\frac{1}{a-\cos x}dx=2\int_{0}^{\pi}\frac{1}{a-\cos x}dx$$
We know $\cos x=\large \frac{1-\tan^2\frac x 2}{{1-\tan^2\frac x 2}}$,thus
$$I(a)=2\int_{0}^{\pi}\frac{{1+\tan^2\frac x 2}}{(a-1)+(a+1)\tan^2\frac{x}{2}}dx=\frac{2}{(a+1)}\int_{0}^{\pi}\frac{1+\tan^2\frac x 2}{\frac{a-1}{a+1}+\tan^2\frac{x}{2}}dx$$
set $u=\tan\frac{x}{2}$, we have
$$I(a)=\frac{4}{a+1}\int_{0}^{+\infty}\frac{1}{\frac{a-1}{a+1}+u^2}du=\frac{4}{a+1}\sqrt{\frac{a+1}{a-1}}\tan^{-1}\left(\frac{a+1}{a-1}u\right)\large|_{0}^{\infty}=\frac{2\pi}{\sqrt{a^2-1}}$$
$$I(a)=\int_{0}^{2\pi}\frac{1}{a+\cos x}dx=\frac{2\pi}{\sqrt{a^2-1}}$$
$$I'(a)=-\int_{0}^{2\pi}\frac{1}{(a+\cos x)^2}dx=-\frac{2\pi a}{(a^2-1)^\frac32}$$
In the other words
$$\int_{0}^{2\pi}\frac{1}{(a+\cos x)^2}dx=\frac{2\pi
a}{(a^2-1)^\frac32}$$
now set $a=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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On the proof that $\sum\limits_{k=0}^{n-1}\frac {a^k}{(1+a^k x) (1+ a^{k+1}x)}=\frac 1 {1-a} \left( \frac 1 {1+x} -\frac {a^n}{ 1+a^n x }\right)$ Question:-
Find the sum to $n$ terms of the following series
$$\frac{1}{(1+x) (1+ax)} + \frac{a}{(1+ax) (1+a^2 x)} + \frac{a^2}{(1+a^2 x) (1+a^3 x)} + \cdots$$
My solution:-
First of all I found out the general term of the series
$$t_n=\frac {a^n}{(1+a^n x) (1+ a^{n+1}x)} $$
Now to find out the partial fraction, I multiplied and divided the $t_n$ by $(a-1) x$. The partial fraction of the $t_n$ looks like this
$$t_n =\frac {a^n (a-1) x }{(( a-1) x) [ (1+a^n x) (1+ a^{n+1} x)] } =\frac{( 1+ a^{n+1} x ) -( 1+ a^n x )}{((a-1)x) [(1+ a^n x) (1+ a^{n+1}x)]} =\frac 1 {(a-1) x} \left(\frac{1}{1+ a^n x} -\frac {1}{1+a^{n+1} x}\right) $$
As, $n \ge 0$,son on summing till all the $n$ terms the diagonal terms start to cancel out and we are left with
$$ S_n=\sum_{n=0}^{n-1} t_n = \sum_{n=0}^{n-1}{ \frac 1 {(a-1) x } \left( \frac 1 { 1+ a^n x } -\frac 1 { 1+ a^{n+1} x } \right) } = \left( \frac 1 {(a-1) x } \right) \left( \frac 1 {1+x} -\frac 1 { 1+a^n x} \right) $$
The answer given in the book:-
$$\left( \frac 1 {1-a} \right) \left( \frac 1 {1+x} -\frac {a^n}{ 1+a^n x }\right)$$
Now, where did the $x$ in the $\dfrac{1}{(a-1)x}$ go and from where did the $a^{n}$ come in $\displaystyle \dfrac { { a }^{ n } }{ 1+{ a }^{ n }x }$
| Since the general term is $$t_j=\frac{a^j}{(1+a^jx)(1+a^{n+1}x)} \ ,\qquad j=0,1,2,\dots,$$
we can decompose it into partial fractions as follows (so as to using finite telescoping series). Suppose $$t_j=\frac{a^j}{(1+a^jx)(1+a^{n+1}x)}=\frac{A}{1+a^jx}+\frac{B}{1+a^{n+1}x},$$
then $$A=\frac{a^j}{1-a^{j+1}\cdot\frac{1}{a^j}}=\frac{a^j}{1-a}, \quad B=\frac{a^j}{1-a^j\cdot \frac{1}{a^{j+1}}}=\frac{a^{j+1}}{a-1},$$
and so
\begin{gather*}
t_j=\frac{1}{1-a}\left(\frac{a^j}{1+a^jx}-\frac{a^{j+1}}{1+a^{j+1}x}\right).
\end{gather*}
Therefore,
\begin{align*}
&\sum_{j=0}^{n-1}t_j=\sum_{j=0}^{n-1} \frac{1}{1-a}\left(\frac{a^j}{1+a^jx}-\frac{a^{j+1}}{1+a^{j+1}x}\right)\\
=&\frac{1}{1-a}\left(\sum_{j=0}^{n-1}\frac{a^j}{1+a^jx}-\sum_{j=0}^{n-1}\frac{a^{j+1}}{1+a^{j+1}x}\right)\\
=&\frac{1}{1-a}\left(\sum_{j=0}^{n-1}\frac{a^j}{1+a^jx}-\sum_{j=1}^{n}\frac{a^{j}}{1+a^{j}x}\right)\\
=&\frac{1}{1-a}\left(\frac{1}{1+x}+\sum_{j=1}^{n-1}\frac{a^j}{1+a^jx}-\sum_{j=1}^{n-1}\frac{a^j}{1+a^jx}-\frac{a^{n}}{1+a^nx}\right)\\
=&\frac{1}{1-a}\left(\frac{1}{1+x}-\frac{a^{n}}{1+a^nx}\right).
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Eliminate $\theta$
Eliminate $\theta$ in
$$\sin \theta + \mbox{cosec} \, \theta = m$$
$$\sec \theta - \cos \theta = n$$
My approach-
I multiplied the first equation by $\sin \theta$ and the second equation by $\cos \theta$ but it doesn't give me the desired answer..
| We have
$$\frac{1}{\cos (\theta)} - \cos (\theta) = n \qquad \qquad \qquad \qquad \sin (\theta) + \frac{1}{\sin(\theta)} = m$$
Let $x := \cos (\theta)$ and $y := \sin(\theta)$. Hence, $x^2 + y^2 = 1$. The two equations above can be rewritten as
$$\frac{1}{x} - x = n \qquad \qquad \qquad \qquad y + \frac{1}{y} = m$$
or, as follows
$$x^2 = 1 - n x \qquad \qquad \qquad \qquad y^2 = m y - 1$$
Since $x^2 + y^2 = 1$, we obtain the equation $m y - nx = 1$. Thus, we have the intersection of the unit circle and a line
$$x^2 + y^2 = 1\qquad \qquad \qquad \qquad m y - nx = 1$$
If $m, n$ are such that the intersection is not empty, then from $x, y$ we obtain the angle $\theta$. Note that the cardinality of the intersection is $0$, $1$, or $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1847605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
} |
Show that $\int_0^1 \left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm dx=\alpha \ln\alpha$
Show that the improper integral $\int_0^1 \left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm dx=\alpha \ln\alpha$, for $\alpha\in(0,1)$.
This is an integral of Riemann. My work:
*
*The set of discontinuities of the integral is
$$D=\left\{\frac1k:k\in\Bbb N\right\}\cup\left\{\frac{\alpha}{k}:k\in\Bbb N\right\}$$
*And when we have that $x>\alpha$ the integral can be simplified to
$$\int_0^1 \left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm dx=\int_0^\alpha \left(\left\lfloor\frac{\alpha}{x}\right\rfloor-\alpha\left\lfloor\frac{1}{x}\right\rfloor\right)\mathrm dx-\alpha\int_{\alpha}^1 \left\lfloor\frac{1}{x}\right\rfloor\mathrm dx$$
I dont know how to continue from here, it is not clear how to handle the partition $D$ to simplify the integral. What I did here is just see what is the value of $\int_{\alpha}^1 \left\lfloor\frac{1}{x}\right\rfloor\mathrm dx$ to see if I get some clue.
If there is no weird mistake somewhere:
$$\int_{\alpha}^1 \left\lfloor\frac{1}{x}\right\rfloor\mathrm dx=\int_\alpha^{\frac1{\left\lfloor 1/\alpha\right\rfloor}}\frac{\mathbf 1_{\Bbb N}(1/\alpha)\mathrm dx}{\lfloor 1/\alpha\rfloor}+\sum_{k=1}^{\lfloor1/\alpha\rfloor}\int_{\frac1{k+1}}^{\frac1k}\frac{\mathrm dx}{k}=\\=\mathbf 1_{\Bbb N}(1/\alpha)\frac{1-\alpha\lfloor 1/\alpha\rfloor}{\lfloor 1/\alpha\rfloor^2}+\sum_{k=1}^{\lfloor1/\alpha\rfloor}\frac1{k^2(k+1)}$$
what is not useful at all. So I get stuck with this problem, can you help me to show this identity (not going deeper than a Riemann integral background)? Thank you in advance.
| The improper integral can be evaluated as the limit of an integral over $[1/n,1]$ as $n \to \infty.$
Making the change of variables $u = 1/x,$ we get
$$\alpha\int_{1/n}^1 \left\lfloor\frac{1}{x}\right\rfloor dx = \alpha\int_{1}^{n } \frac{\left\lfloor u\right\rfloor}{u^2} \, du \\ = \alpha \sum_{k=1}^{n-1}\int_{k}^{k+1 } \frac{\left\lfloor u\right\rfloor}{u^2} \, du \\ = \alpha \sum_{k=1}^{n-1}\int_{k}^{k+1 } \frac{k}{u^2} \, du \\ = \alpha \sum_{k=1}^{n-1} k\left(\frac{1}{k} - \frac{1}{k+1} \right) \\ = \alpha \sum_{k=2}^{n} \frac{1}{k},\tag{1}$$
and, making the change of variables $u = \alpha/x,$ we get
$$\int_{1/n}^1 \left\lfloor\frac{\alpha}{x}\right\rfloor dx= \alpha \int_{\alpha}^{n \alpha} \frac{\left\lfloor u\right\rfloor}{u^2} \, du \\ = \alpha \int_{\alpha}^{n} \frac{\left\lfloor u\right\rfloor}{u^2} \, du - \alpha \int_{n \alpha}^{n} \frac{\left\lfloor u\right\rfloor}{u^2} \, du \\ = \alpha \int_{1}^{n} \frac{\left\lfloor u\right\rfloor}{u^2} \, du - \alpha \int_{n \alpha}^{n } \frac{\left\lfloor u\right\rfloor}{u^2} \, du\tag{2}$$
Subtracting (1) from (2) we eliminate the divergent harmonic sum and obtain
$$\int_{1/n}^1 \left(\left\lfloor\frac{\alpha}{x}\right\rfloor - \alpha \left\lfloor\frac{1}{x}\right\rfloor \right) dx = - \alpha \int_{n \alpha}^{n} \frac{\left\lfloor u\right\rfloor}{u^2} \, du \\ = - \alpha \int_{1}^{n} \frac{\left\lfloor u\right\rfloor}{u^2} \, du + \alpha \int_{1}^{\lfloor n \alpha \rfloor} \frac{\left\lfloor u\right\rfloor}{u^2} \, du + \alpha \int_{\lfloor n \alpha \rfloor}^{n \alpha} \frac{\left\lfloor u\right\rfloor}{u^2} \, du \\ = -\alpha \sum_{k=2}^{n} \frac{1}{k} + \alpha \sum_{k=2}^{\lfloor n \alpha \rfloor}\frac{1}{k} + \alpha \frac{n \alpha - \lfloor n \alpha \rfloor}{n \alpha} \\ = -\alpha \left(\sum_{k=1}^{n} \frac{1}{k} - \log n \right) + \alpha \left( \sum_{k=1}^{\lfloor n \alpha \rfloor}\frac{1}{k} - \log\lfloor n \alpha \rfloor \right) - \alpha \log n + \alpha \log \lfloor n \alpha \rfloor + \alpha \frac{n \alpha - \lfloor n \alpha \rfloor}{n \alpha}$$
From here, it should be relatively straightforward to take the limit as $ n \to \infty$ to obtain $\alpha \log \alpha$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1848321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 2,
"answer_id": 1
} |
Is difference of two consecutive sums of consecutive integers (of the same length) always square? I am an amateur who has been pondering the following question. If there is a name for this or more information about anyone who has postulated this before, I would be interested about reading up on it. Thanks.
If $x$ is the sum of $y$ integers, and $z$ is the sum of the next $y$ integers, then is it always true that $z$ minus $x$ equals $y$ squared? Perhaps only starting at one.
For example:
$y = 3$
$x = (1+2+3) = 6$
$z = (4+5+6) = 15$
$z - x = 9 = y^2$
Brad
| If you start at 1, then yes.
The sum of the first $n$ integers is $\frac{n(n+1)}{2}$.
Note that the next $n$ integers are simply $n^2 + \frac{n(n+1)}{2}$. That is, you can remove $n$ from each of the subsequent n integers, and just wind up with the sum of the first $n$ integers again.
For example:
$y = 5$
$x = 1 + 2 + 3 + 4 + 5 = 15$
$z = 6 + 7 + 8 + 9 + 10 = (1+5) + (2+5) + (3+5) + (4+5) + (5+5)$
$ = (1 + 2 + 3 + 4 + 5) + (5 + 5 + 5 + 5 + 5) = 15 + 5\times 5 $
So $z-x = (15 + 5\times 5) - 15 = 5\times 5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1851508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 9,
"answer_id": 4
} |
Area of a square whose one part is in circle A square has two of its vertices on a circle and the other two on a tangent to the circle. If the diameter of the circle is $10$ cm, then what is the area of the square is?
My solution:
I figured out this diagram
What to do after this ?
| Observe the diagram below:
Let $x =$ the length of one side of the square.
Since the diameter of the circle is $10 \ \text{cm}$, we know that $OA=OC=OD=5 \ \text{cm}$, since they are all radii of circle $O$. Thus, $OB = x-5$. We also know that $AB = \frac{x}{2}$. Using the Pythagorean theorem:
$$
\begin{align*}
(AB)^2+(OB)^2&=(OA)^2 \\
\left(\frac{x}{2}\right)^2 + (x-5)^2&=5^2 \\
\frac{1}{4}x^2 + x^2-10x+25&=25 \\
\frac{5}{4}x^2-10x&=0 \\
x\left( \frac{5}{4}x-10 \right) &= 0 \\
x=0 \text{ or } x&=8
\end{align*}
$$
Thus one side of the square is $8 \ \text{cm}$, and the area of the square is $\color{blue}{64 \ \text{cm}^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to find this function, and what method to use? The function is $f(x-\frac{1}{x})= x^3-\frac{1}{x^3}$ and they are asking us to find out what $f(-x)$ is?
| As per Parth Kohli's comment we can write $$f\left(x - \frac{1}{x}\right) = \left(x - \frac{1}{x}\right)^3 + 3\left(x - \frac{1}{x}\right).$$
That is, $f(x) = x^3 + 3x$ so $f(-x) = (-x)^3 + 3(-x) = -(x^3 + 3x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
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.