Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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What will be the minimum value of $\frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ}$? What will be the minimum value of
$$\frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ}$$ if
$$p+q+r+s=5$$ where $p, q, r, s$ are positive reals?
I tried applying AM-GM inequality but it didn't help.
| $(\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ) \left ( \frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ} \right ) \geq {(p+q+r+s)}^2=5^2$
From Cauchy–Schwarz
Hence
$$\frac{p^2}{\tan9^\circ} + \frac{q^2}{\tan27^\circ} + \frac{r^2}{\tan63^\circ} + \frac{s^2}{\tan81^\circ} \, \, \, \,\geq \frac{ 5^2 }{\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ }$$
Then take $$p = \frac{ 5 \tan9^\circ}{\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ}$$
$$q = \frac{ 5 \tan27^\circ}{\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ}$$
$$r = \frac{ 5 \tan63^\circ}{\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ}$$
$$s = \frac{ 5 \tan81^\circ}{\tan9^\circ+\tan27^\circ+\tan63^\circ+\tan81^\circ}$$
And sum up.
Note that those $ \tan $ are positive
| {
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"url": "https://math.stackexchange.com/questions/180727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to see $\sin x + \cos x$ $$\sin x + \cos x = \sqrt{2} \sin(x + \pi/4)$$
Is there an easy way to visualize this identity or to convert the left-hand side to the right-hand side?
In general, can $p \sin x + q \cos x$ for some integers $p$ and $q$ be easily expressed as $\sin($something$)$ or $\cos($something$)$? If so, how can that be done?
| Given $p \sin x + q \cos x$, divide the expression by $\sqrt{p^2+q^2}$ to get $a \sin x + b \cos x$ for $a^2 + b^2 = 1$. Now name $a = \cos \alpha$ and $b = \sin \alpha$. Notice that what you got is the expansion of $\sin(x+\alpha)$. In the given case, $a = b = \frac{\sqrt{2}}{2}$ so $\alpha = \frac{\pi}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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Calculating $x$ from $x^3 = 2x^2 - x$, getting wrong result I am working with a text book problem, trying to find the solutions to $x^3 = 2x^2 - x$, but I keep ending up with a different result than the book gives. I'm thinking the problem lie in getting the wrong square root towards the end, seeing as the text book gives different results for $x_1$ and $x_2$, but I can't find any mistakes.
$$x^3 = 2x^2 - x$$
$$\iff x^2 = 2x - 1$$
$$\iff x^2 - 2x + 1 = 0$$
$$\iff (x - \frac{2}{2})^2 = (\frac{2}{2})^2 - 1$$
$$\iff (x - 1)^2 = 0$$
$$\iff x - 1 = \pm \sqrt{0}$$
$$\iff x = 1 + \pm \sqrt{0}$$
$$\iff x = 1 + \pm 0$$
$$x_1 = 1 + 0 = 1$$
$$x_2 = 1 - 0 = 1$$
| You can not just divide by $x$, since $x$ could be $0$.
You have $x^3=2x^2-x$, which is equivalent to:
$$0=x^3-2x^2+x=x(x^2-2x+1)=x(x-1)^2$$
Hence $x=0$ or $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/183955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluate $\sum_{i=1}^n \frac{i}{n^2}$ My book asks to simplify this problem:
$$\sum_{i=1}^n\frac{i}{n^2}$$
This equals:
$$\sum_{i=1}^n\frac{1}{n^2}i$$
Now, shouldn't
$$\sum_{i=1}^n\frac{1}{n^2}=\frac{1}{n}\text{ ?}$$
Because you are summing $1/n^2$, $n$ times, so that $n / n^2 = 1/n$.
So it should be
$$\frac{1}{n}\sum_1^n i = \frac{1}{n} \cdot \frac{n(n+1)}{2} = \mathbf{\frac{n+1}{2}}$$
But the answer in my book is $\mathbf{(n+1)/2n}$.
| It is well known that $\sum_{i = 1}^n i = \frac{n(n + 1)}{2}$. Therefore by factoring out $n^2$.
$\sum_{i = 1}^n \frac{i}{n^2} = \frac{1}{n^2}\sum_{i = 1}^n i = \frac{1}{n^2}\frac{n(n + 1)}{2} = \frac{n + 1}{2n} = \frac{1}{2} + \frac{1}{2n}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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limit of a sequence $(-1)^n$ instead $1$ I should calculate the limit of a sequence. A friend told me, that the solution is $1$. But I got $(-1)^n$.
The exercise is: $\lim\limits_{n \to \infty} \frac{1}{n^2} + (-1)^n \cdot \frac{n^2}{n^2+1}$
I did following:
$$\begin{align*}
&=\frac{n^2 ((-1)^n n^2 + 1 + \frac{1}{n^2})}{n^2(n^2+1)}\\
&=\frac{(-1)^n n^2 + 1 + \frac{1}{n^2}}{(n^2+1)}\\
&=\frac{n^2(\frac{(-1)^n n^2}{n^2} + \frac{1}{n^2} + \frac{1}{n^4})}{n^2(1 + \frac{1}{n^2})}\\
&=\frac{(-1)^n + 0 +0}{1}\\
&=\lim\limits_{n \to \infty} (-1)^n
\end{align*}$$
What did I wrong?
Edit
Well, some answers confused me. Here the complete exercise.
I should check if the sequence is convergent for ${n \to \infty}$ and determine the limit if it exist. Also for a sequence which is $\infty$ or $-\infty$.
My friend got $1$ as limit. I got $(-1)^n$. I would say, that this sequence has no limit, just limit points $1$ and $-1$.
| The limit of a sequence indexed by $n$ cannot contain $n$, as a limit is a number, not a sequence. So your answer $(-1)^n$ as outcome of the limit cannot be correct. However by writing
$$
\frac1{n^2} + (-1)^n \frac{n^2}{n^2+1}=\frac1{n^2} + (-1)^n -(-1)^n\frac1{n^2+1}
$$
you can see that you sequence is the sum of three sequences $\frac1{n^2}$, $(-1)^n$, and $-(-1)^n\frac1{n^2+1}$, of which the first and the last are convergent (to $0$), and the middle one is divergent, which implies that the sequence diverges (it has no limit).
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Partial fraction: is my final answer and computation correct? I need some help at this exercise (partial fraction): $$\int{\frac{4x + 1}{x^3(x+2)}} dx$$
First of all I calculated the roots of the denominator.
$$
x_{1, 2, 3} = 0 \\
x_4 = -2
$$
After that I assigned the roots to the partial fraction.
$$
\frac{A_1}{x}, \frac{A_2}{x^2}, \frac{A_3}{x^3}, \frac{B}{(x+2)} \\
\frac{4x + 1}{x^3(x+2)} = \frac{4x + 1}{x^3(x+2)} = \frac{A_1}{x} + \frac{A_2}{x^2} + \frac{A_3}{x^3} + \frac{B}{(x+2)} \\
4x + 1 = A_1x^2(x+2) + A_2x(x+2) + A_3(x+2) + Bx^3 \\
4x + 1 = A_1(x^3+2x^2) + A_2(x^2+2x) + A_3(x+2) + Bx^3 \\
$$
Then I insert the root values.
For $-2$, I get $B = \frac{7}{8}$. For $0$ I get $A_3 = \frac{1}{2}$.
After that I insert two additional numbers, I choose $1$ and $-1$.
I transformend the equation to $A_1$ and $A_2$ and get $A_1 = \frac{23}{20}$ and $A_2 = \frac{197}{120}$.
$$
\frac{4x + 1}{x^3(x+2)} = \frac{\frac{23}{20}}{x} + \frac{\frac{197}{120}}{x^2} + \frac{\frac{1}{2}}{x^3} + \frac{\frac{7}{8}}{(x+2)}
$$
Could anyone tell me if it's correct? I belive it is wrong :), where is my misstake?
| Reduce the RHS to a common denominator
$$\begin{eqnarray*}
\frac{4x+1}{x^{3}(x+2)} &=&\frac{A_{1}}{x}+\frac{A_{2}}{x^{2}}+\frac{A_{3}}{
x^{3}}+\frac{B}{x+2} \\
&=&\frac{A_{1}x^{2}\left( x+2\right) +A_{2}x\left( x+2\right) +A_{3}\left(
x+2\right) +Bx^{3}}{x^{3}(x+2)} \\
&=&\frac{\left( A_{1}+B\right) x^{3}+\left( 2A_{1}+A_{2}\right) x^{2}+\left(
2A_{2}+A_{3}\right) x+2A_{3}}{x^{3}(x+2)}.
\end{eqnarray*}$$
You can proceed by equating coefficients of
$$\begin{equation*}
4x+1=\left( A_{1}+B\right) x^{3}+\left( 2A_{1}+A_{2}\right) x^{2}+\left(
2A_{2}+A_{3}\right) x+2A_{3}.
\end{equation*}$$
So$^1$
$$\begin{eqnarray*}
2A_{3} &=&1\Leftrightarrow A_{3}=\frac{1}{2} \\
2A_{2}+\frac{1}{2} &=&4\Leftrightarrow A_{2}=\frac{7}{4} \\
2A_{1}+\frac{7}{4} &=&0\Leftrightarrow A_{1}=-\frac{7}{8} \\
-\frac{7}{8}+B &=&0\Leftrightarrow B=\frac{7}{8},
\end{eqnarray*}$$
and
$$\begin{equation*}
\frac{4x+1}{x^{3}(x+2)}=-\frac{7}{8}\frac{1}{x}+\frac{7}{4}\frac{1}{x^{2}}+
\frac{1}{2}\frac{1}{x^{3}}+\frac{7}{8}\frac{1}{x+2}.
\end{equation*}$$
--
$^1$ It corresponds to solve the system
$$\left\{
\begin{array}{c}
A_{1}+B=0 \\
2A_{1}+A_{2}=0 \\
2A_{2}+A_{3}=4 \\
2A_{3}=1
\end{array}
\right. $$
ADDED. Your method works just as well, but you have made some mistake in the computation.
Then I insert the root values. For $-2$, I get $B = \frac{7}{8}$. For
$0$ I get $A_3 = \frac{1}{2}$. After that I insert two additional
numbers, I choose $1$ and $-1$.
Using
$$\begin{equation*}
4x+1=A_{1}x^{2}\left( x+2\right) +A_{2}x\left( x+2\right) +A_{3}\left(
x+2\right) +Bx^{3}
\end{equation*}$$
you get for $x=1$
$$\begin{equation*}
5=3A_{1}+3A_{2}+\frac{3}{2}+\frac{7}{8},
\end{equation*}$$
and for $x=-1$
$$\begin{equation*}
-3=A_{1}-A_{2}+\frac{1}{2}-\frac{7}{8},
\end{equation*}$$
whose solutions are $A_{2}=\frac{7}{4},A_{1}=-\frac{7}{8}$ as above and not $A_1 = \frac{23}{20}$, $A_2 = \frac{197}{120}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\int_0^\infty \sin x^2\, dx$ with real methods? I have seen the Fresnel integral
$$\int_0^\infty \sin x^2\, dx = \sqrt{\frac{\pi}{8}}$$
evaluated by contour integration and other complex analysis methods, and I have found these methods to be the standard way to evaluate this integral. I was wondering, however, does anyone know a real analysis method to evaluate this integral?
| Let $$F(t)=\int_0^{\infty} \sin(tx^2)\ dx$$ Taking the Laplace Transform of $F(t)$: $$\mathscr{L}\{F(t)\}=\int_0^{\infty}F(t)e^{-st}\ dt$$ By being slightly less than rigorous this can be rewritten as $$\int_0^{\infty}\mathscr{L}\{\sin (tx^2)\}\ dx=\int_0^{\infty}\frac{x^2}{s^2+x^4}\ dx$$ This can be put into partial fractions as $$\frac{x^2}{s^2+x^4}\to\frac{Ax+B}{(s+ix^2)}+\frac{Cx+D}{(s-ix^2)}$$ $$(Ax+B)(s-ix^2)+(Cx+D)(s+ix^2)=x^2$$ $$A=C=0,\ B=\frac{i}{2},\ D=-\frac{i}{2}$$ $$\begin{equation}\begin{aligned}
\int_0^{\infty}\frac{x^2}{s^2+x^4}\ dx&=\int_0^{\infty}\frac{i}{2(s+ix^2)}\ dx\ -\int_0^{\infty}\frac{i}{2(s-ix^2)}\ dx \\
&=\frac{1}{2}\int_0^{\infty}\frac{1}{x^2-is}\ dx\ +\frac{1}{2}\int_0^{\infty}\frac{1}{x^2+is}\ dx \\
&=\lim_{c\to\infty}\frac{1}{2}\bigg[\frac{1}{\sqrt{is}}\tan ^{-1} \bigg(\frac{x}{\sqrt{is}}\bigg)-\frac{1}{\sqrt{is}}\tanh ^{-1}\bigg(\frac{x}{\sqrt{is}}\bigg)\bigg]_0^c \\
&=\frac{\pi}{4}\bigg(\frac{1}{\sqrt{is}}+\frac{1}{\sqrt{-is}}\bigg)=\frac{\pi}{4}\bigg(\frac{\sqrt{2}}{2\sqrt{s}}(1-i)+\frac{\sqrt{2}}{2\sqrt{s}}(1+i)\bigg)=\frac{\pi\sqrt{2}}{4\sqrt{s}}
\end{aligned}\end{equation}$$
Now taking the inverse Laplace Transform:
$$\mathscr{L}^{-1}\bigg\{\frac{\pi\sqrt{2}}{4\sqrt{s}}\bigg\}=\frac{\pi\sqrt{2}}{4}\mathscr{L}^{-1}\bigg\{\frac{1}{\sqrt{s}}\bigg\}=\frac{\pi\sqrt{2}}{4\sqrt{\pi t}}$$
By noting that $$F(1)=\int_0^{\infty}\sin (x^2)\ dx$$
$$\int_0^{\infty}\sin (x^2)\ dx=\frac{\pi\sqrt{2}}{4\sqrt{\pi\times 1}}=\sqrt{\frac{\pi}{8}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/187729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "75",
"answer_count": 8,
"answer_id": 0
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Showing that $ \left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3 $ I would like to show that:
$$ \left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3 $$
Using AM-GM inequality:
$$ \left( \frac{a+2b}{a+2c}\right)^3+\left( \frac{b+2c}{b+2a}\right)^3+\left( \frac{c+2a}{c+2b}\right)^3 \geq3\frac{(a+2b)(b+2c)(c+2a)}{(a+2c)(b+2a)(c+2b)} $$
It suffices to show that:
$$ \frac{(a+2b)(b+2c)(c+2a)}{(a+2c)(b+2a)(c+2b)} \geq1 $$
$$ \Longleftrightarrow ab^2+bc^2+ca^2-(a^2b+b^2c+c^2a)\geq0$$
$$ \Longleftrightarrow x+y+z\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}$$
$$ xyz=1$$
($x=\frac{a}{b}$, $y=\frac{b}{c}$, $z=\frac{c}{a}$)
How can I prove this last inequality?
Is there any simpler proof?
| I will prove that this is true for $a,b,c>0$. If they're allowed to be non-positive, there are counterexamples; e.g., $(a,b,c)=(1,0,-1)$.
Since this expression is homogeneous in $a,b,c$, we may assume without loss of generality that $a+b+c=1$. Then setting $x=c-b$, $y=a-c$, $z=b-a$ yields:
$$
\left(\frac{a+2b}{a+2c}\right)^3+\left(\frac{b+2c}{b+2a}\right)^3+\left(\frac{c+2a}{c+2b}\right)^3=\left(\frac{1-x}{1+x}\right)^3+\left(\frac{1-y}{1+y}\right)^3+\left(\frac{1-z}{1+z}\right)^3 \, .
$$
But if $a,b,c$ are positive and $a+b+c=1$, then $0<a,b,c<1$, from which it follows that $-1<x,y,z<1$. As the function $f(t)=\left(\frac{1-t}{1+t}\right)^3$ is concave upward on $(-1,1)$, we have:
$$
\left(\frac{1-x}{1+x}\right)^3+\left(\frac{1-y}{1+y}\right)^3+\left(\frac{1-z}{1+z}\right)^3=f(x)+f(y)+f(z) \geq 3f\left(\frac{x+y+z}{3}\right)=3f(0)=3 \, .
$$
This completes the proof.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Solve $x^2+x^2t^2+2xt^2+t^2=1$ for $x$. Solving
$$x^2+x^2t^2+2xt^2+t^2=1$$
for $x$ yields
$$x=\frac{1-t^2}{t^2+1}.$$
I had to use software because I could not manipulate it algebraically, and sadly the program did not tell me how it reached that solution.
What is the method?
| Collect the coefficients in $x$ to get a polynomial in $x$:
$$ (t^2+1) x^2 +(2t^2)x+(t^2-1) = 0 $$
Now solve using quadratic formula:
$$ \frac{-2t^2 \pm \sqrt{(2t^2)^2 - 4(t^2+1)(t^2-1)}}{2(t^2+1)} = \\
\frac{-2t^2 \pm 2}{2(t^2+1)} = \frac{-t^2 \pm 1}{t^2 + 1}$$
So either $x = -1$ or $x = \frac{-t^2 + 1}{t^2 + 1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/189283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Linear Algebra: Change of Basis Let $A[a,b,c]$ and $B[d,e,f]$ be two non-standard bases. I have to find the $3\times3$ matrix that will convert a vector defined in terms of $A$ to $B$.
My solution is:
Let's assume a standard basis $O[e_1,e_2,e_3]$.
Now, the rotation matrix $R(A\to O)$ to change basis from $A$ to $O$ is:
$
\begin{bmatrix}
ax & bx & cx \\
ay & by & cy \\
az & bz & cz
\end{bmatrix}
$
and $R(O\to B)$ is
$
\begin{bmatrix}
ax & ay & az \\
bx & by & bz \\
cx & cy & cz
\end{bmatrix}.
$
So, $R(A \to B)=R(A \to O)\cdot R(O\to B)$.
Is this the correct solution?
Thanks a lot.
| Let $\alpha=\{a,b,c\}$ and $\beta=\{d,e,f\}$ be bases for $\mathbb{R}^3$ (thinking of elements of $\mathbb{R}^3$ as column vectors). Then the matrix $A=[a|b|c]$ transitions from $\alpha$-coordinates to standard-coordinates. Likewise the basis $B=[d|e|f]$ transitions from $\beta$-coordinates to the standard-coordinates. So $B^{-1}$ goes from standard-coordinates to $\beta$-coordinates. Thus $B^{-1}A$ goes from $\alpha$-coordinates (to standard and then standard) to $\beta$-coordinates and is the matrix you're looking for.
Example: Let $\alpha=\left\{\begin{bmatrix} 1\\0\\0 \end{bmatrix},\begin{bmatrix} 1\\1\\0 \end{bmatrix},\begin{bmatrix} 1\\1\\1 \end{bmatrix}\right\}$ and $\beta=\left\{\begin{bmatrix} 1\\-1\\0 \end{bmatrix},\begin{bmatrix} 0\\1\\-1 \end{bmatrix},\begin{bmatrix} 1\\0\\1 \end{bmatrix}\right\}$.
Then ${\bf v}=\begin{bmatrix} 1\\2\\3 \end{bmatrix} = (-1)\begin{bmatrix} 1\\0\\0 \end{bmatrix}+(-1)\begin{bmatrix} 1\\1\\0 \end{bmatrix}+3\begin{bmatrix} 1\\1\\1 \end{bmatrix}$. So ${\bf v}$ in $\alpha$ coordinates is $[{\bf v}]_\alpha = \begin{bmatrix} -1\\-1\\3\end{bmatrix}$.
Let $A=\begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}$. Notice that $A[{\bf v}]_\alpha = \begin{bmatrix} 1 & 1 & 1 \\ 0 & 1 & 1 \\ 0 & 0 & 1 \end{bmatrix}\begin{bmatrix} -1\\-1\\3\end{bmatrix} = \begin{bmatrix} 1\\2\\3\end{bmatrix}$.
Let $B=\begin{bmatrix} 1 & 0 & 1 \\ -1 & 1 & 0 \\ 0 & -1 & 1\end{bmatrix}$. Then $B^{-1} = \frac{1}{2}\begin{bmatrix} 1 & -1 & -1 \\ 1 & 1 & -1 \\ 1 & 1 & 1 \end{bmatrix}$ and so $B^{-1}A = \begin{bmatrix} \frac{1}{2} & 0 & -\frac{1}{2} \\ \frac{1}{2} & 1 & \frac{1}{2} \\ \frac{1}{2} & 1 & \frac{3}{2} \end{bmatrix}$ goes from $\alpha$ to $\beta$-coordinates.
An alternate way to compute this change of basis matrix is to create the augmented matrix: $[B|A] = [d|e|f|a|b|c]$ and row-reduce. This will compute $[I|(B^{-1}A)]$ where $B^{-1}A$ is the desired matrix.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Little help with some algebra Note: This isn't homework, I'm skipping ahead of class. Please answer all these equations, I'm deathly stuck on them.
Use the substitution method only please. (Find $x$ and $y$.)
\begin{align}
ax\left({\frac {1}{a-b}-\frac {1}{a+b}}\right)+by\left({\frac {1}{b-a}-\frac {1}{b+a}}\right)=2
\end{align}
$a$ is not equal to $b$ or $-b$
Note - It's $ax$ as the first word, I am worried the latex might mess up there.
Equation 2:
\begin{align}
6x+5y=7x+3y+1=2\left({x+6y-1}\right)
\end{align}
Equation 3:
\begin{align}
\sqrt{2}x+\sqrt{3}y=0
\end{align}
\begin{align}
\sqrt{3}x-\sqrt{8}y=0
\end{align}
Thank you for the help!
Note: Please provide the full solution, the more hints i get the more confused i get...
| First Equation:
$$ax\left({\frac {1}{a-b}-\frac {1}{a+b}}\right)+by\left({\frac {1}{b-a}-\frac {1}{b+a}}\right)=2 $$
$$ax\left({\frac {a+b-a+b}{a^2-b^2}}\right)+by\left({\frac {b+a-b+a}{b^2-a^2}}\right)=2$$
$$2abx-2aby=2(a^2-b^2)$$
$$x-y=\frac{a}{b}-\frac{b}{a}$$
From this the obvious solution is $x=a/b \text{ and } y=b/a$. But there will be infinite solutions $x=y+(a/b)-(b/a)$. Just put values for $y$ and get corresponding values for $x$.
Second Equation:
$$6x+5y=7x+3y+1=2(x+6y-1)$$
Break it into $2$ equations
$$6x+5y=7x+3y+1 \text{ and } 6x+5y=2(x+6y-1)$$
$$-x+2y=1 \text{ and } 4x-7y=-2$$
The value of $x$ from the first equation is $x=2y-1$. Put it in second equation.
$$4(2y-1)-7y=-2$$
$$y=2$$
Hence $x=2y-1=3$.
Third equation:
$$\sqrt2x+\sqrt3y=0 $$
From this $x=-\sqrt{3/2}y$. Put this in the second equation.
$$\sqrt3x-\sqrt8y=0$$
$$y(3/\sqrt2 - \sqrt8)=0$$
$$y=0$$
Hence $x=0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 0
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Limit (without series expansion and l'Hôpital's rule) $$\lim_{x \to \infty}\ln{\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}}}\cdot \left(\ln{\frac{x+1}{x-1}}\right)^{-2}=\frac{1}{8}$$
Any suggestion to find this limit without series expansion and l'Hôpital's rule? Thanks and regards.
Note:
WolframAlpha confirms that the result is $\frac{1}{8}$.
| I guess the only way to avoid using series expansion or l'Hôpital's rule is to reduce the limit to known limits. Let $y=1/x$. Then
$$
\lim_{x \to \infty}\ln{\frac{x+\sqrt{x^2+1}}{x+\sqrt{x^2-1}}}\cdot \left(\ln{\frac{x+1}{x-1}}\right)^{-2} = \lim_{y \to 0} \frac{\ln \frac{1+\sqrt{1+y^2}}{1+\sqrt{1-y^2}}}{\left(\ln \frac{1+y}{1-y}\right)^2} = \lim_{y \to 0} \frac{\frac{1}{y^2}\ln\left(\frac{1+\sqrt{1+y^2}}{2}\right) - \frac{1}{y^2}\ln\left(\frac{1+\sqrt{1-y^2}}{2}\right)}{\left(\frac{1}{y} \ln(1+y) - \frac{1}{y} \ln(1-y)\right)^2}
$$
Now, using:
$$
\lim_{y \to 0} \frac{1}{y} \ln(1\pm y) = \pm 1, \qquad \lim_{y \to 0} \frac{1}{y^2} \ln\left(\frac{1+\sqrt{1 \pm y^2}}{2}\right) = \pm \frac{1}{4}
$$
we readily arrive at the result. Limits above can be reduced to one of the classic limits involving the exponential function, namely:
$$
\lim_{t \to 0} \frac{\mathrm{e}^t-1}{t} = 1
$$
Indeed:
$$
\lim_{y \to 0} \frac{1}{y} \ln(1\pm y) \stackrel{y = \pm \left(\exp(t)-1\right)}{=} \pm \left( \lim_{t \to 0} \frac{t}{\mathrm{e}^t-1}\right) = \pm 1
$$
Using using the substitution $t = \ln \left(\frac{1+\sqrt{1+y^2}}{2}\right)$ we have
$$
\lim_{y \to 0} \frac{1}{y^2} \ln \left(\frac{1+\sqrt{1+y^2}}{2}\right) = \lim_{t\to 0} \frac{t}{4 \left(\mathrm{e}^{2t} - \mathrm{e}^t\right)} = \frac{1}{4} \cdot \lim_{t \to 0} \mathrm{e}^{-t} \cdot \lim_{t\to 0} \frac{t}{\left(\mathrm{e}^{t} - 1\right)} = \frac{1}{4}
$$
and similarly for the case with $\sqrt{1-y^2}$.
| {
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"url": "https://math.stackexchange.com/questions/191379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Inequality. $ab^2+bc^2+ca^2 \geq a+b+c.$ Using rearrangement inequalities prove the following inequality:
Let $a,b,c$ be positive real numbers satisfying $abc=1$. Prove that
$$ab^2+bc^2+ca^2 \geq a+b+c.$$
Thanks :)
| Just by Hölder inequality;
$(ab^2+bc^2+ca^2)(\frac1a+\frac1b+\frac1c)\ge(a+b+c)^2\tag1$
If $(\frac1a+\frac1b+\frac1c)\le(a+b+c)$ we're done. Otherwise;
$(ab^2+bc^2+ca^2)(a+b+c)\ge(ab+bc+ca)^2=(\frac1a+\frac1b+\frac1c)^2\tag2$
since $(a+b+c)<(\frac1a+\frac1b+\frac1c)$ we're done again.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the minimum values of x, y, and z If:
\begin{aligned}
\ {n} & = \ 2x ^2 \\
\ {n} & = \ 3y^3 \\
\ {n} & = \ 5z^5 \\
\end{aligned}
Find the minimum value for $x, y,$ and $z$.
Note: $n$ is the minimum value you can get by solving the set of equations.
| I take it that $n$ is a positive integer, as are $x$, $y$, and $z$. Then $n$ must be divisible by the primes $2$, $3$, and $5$. To minimize $n$, we allow no other primes.
Let $n=2^a3^b5^c$. Since $n=2x^2$, $a$ must be odd, and $b$ and $c$ even. This is because in the prime power factorization of a perfect square, each prime must occur to an even power.
Since $n=3y^3$, $a$ must be divisible by $3$, and $b-1$ must be divisible by $3$, and $c$ must be divisible by $3$. This is because in the prime power factorization of a perfect cube, each prime must occur to a power divisible by $3$.
Since $n=5z^5$, $a$ must be divisible by $5$, and $b$ must be divisible by $5$, and $c-1$ must be divisible by $5$.
Now we look for the smallest non-negative integers $a$, $b$, and $c$ that satisfy our conditions.
Start with $a$. We must have $a$ odd, $a$ divisible by $3$ and by $5$. It is clear that the cheapest $a$ is $15$.
We want $b$ to be even, and a multiple of $5$, and we want $b-1$ to be divisible by $3$. So $b$ is a multiple of $10$. Easily, $b=10$ satisfies the additional condition $b-1$ is divisible by $3$.
Do something similar to find the cheapest $c$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factoring the expression $x^6 + 64$ Alright, so apparently I've factored this out wrong...
$x^6 + 64 =$
$x^6 + 2^6$
Then I continued, using $a^3 + b^3 = (a + b)(a^2 - ab + b^2)$ to get ...
$$(x^2)^3 + 64 =
(x^2)^3 + 4^3 =
(x^2 + 4)(x^4 - 4x^2 + 16)$$
How is this incorrect?
| A different approach is to first factorize it over complex numbers:
$$x^6+2^6=\prod_{k=0}^5(x-2\cdot e^{\pi i(1/6+k/3)}) $$
And then pair conjugate roots, yielding:
$$\prod_{k=0}^2(x-2\cdot e^{\pi i(1/6+k/3)})(x-2\cdot e^{-\pi i(1/6+k/3)})=\prod_{k=0}^2(x^2-4\cos(\pi(1/6+k/3))+4)$$
Which equals
$$(x^2-2\sqrt 3x+4)(x^2+4)(x^2+2\sqrt 3x+4)$$
This method will work for any polynomial over $\bf R$ you can factor over $\bf C$ into linear factors (such a factorization always exists, but sometimes it might be impossible to express it in the way you would like).
The reason it will work is that for any real polynomial, it's non-real complex roots occur always occur in conjugate pairs, and for a nonreal complex $z$ we have that $(x-z)(x-\overline z)=x-2\Re(z)+\lvert z\rvert^2$ and the latter is indecomposable over $\bf R$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality. $\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3$
Let $a,b,c$ be positive numbers . Prove the following inequality:
$$\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}} \leq 3.$$
What I tried:
I used Cauchy-Schwarz in the following form
$\sqrt{Ax}+\sqrt{By}+\sqrt{Cz} \leq \sqrt{(a+b+c)(x+y+z)}$ for:
$$A=11a, \quad{} B=11b, \quad{} C=11c$$
and
$$x=\frac{1}{5a+6b}, \quad{} y=\frac{1}{5b+6c}, \quad{} z=\frac{1}{5c+6a}$$ but still nothing. Thanks for your help :)
I tried something else:
$$\large\frac{\sqrt{\frac{11a}{5a+6b}}+\sqrt{\frac{11b}{5b+6c}}+\sqrt{\frac{11c}{5c+6a}}}{3} \leq \sqrt{\frac{\frac{11a}{5a+6b}+\frac{11b}{5b+6c}+\frac{11c}{5c+6a}}{3}}$$ and what we have to prove become:
$$\large\sqrt{\frac{\frac{11a}{5a+6b}+\frac{11b}{5b+6c}+\frac{11c}{5c+6a}}{3}} \leq 1 \Leftrightarrow \sqrt{\frac{11a}{5a+6b}+\frac{11b}{5b+6c}+\frac{11c}{5c+6a}} \leq \sqrt{3}$$
Another attempt
$$\large\sqrt{\frac{1}{xy}} \leq \frac{\frac{1}{x}+\frac{1}{y}}{2}=\frac{x+y}{2xy}$$
and $y=1$ and $\displaystyle x=\frac{5a+6b}{11a}$. So:
$$\large\sqrt{\frac{11a}{5a+6b}} \leq \frac{\frac{5a+6b}{11a}+1}{2 \cdot \frac{5a+6b}{11a}}=\frac{8a+3b}{5a+6b}.$$ Now we have to prove:
$$\sum_{cyc}{\frac{8a+3b}{5a+6b}} \leq 3.$$
But still nothing .
| It's a bit ugly (actually more than a bit), but it works.
By AM-GM we get
$$
abc = \sqrt[3]{ab^2\cdot bc^2 \cdot ca^2} \leq \frac 1 3 (ab^2 + bc^2 + ca^2)
$$
and therefore
$$
9(5a + 6b)(5b + 6c)(5c + 6a) - 11^3(a + b + c)(ab + bc + ca) =\\ 289(ab^2 + bc^2 + ca^2) + 19(a^2b + b^2c + c^2a) - 924abc =\\ 289(ab^2 + bc^2 + ca^2 - 3abc) + 19(a^2b + b^2c + c^2a - 3abc) \geq 0
$$
From the previous inequality and applying Cauchy-Schwarz we arrive to
$$
\sum_{cyc}\sqrt{\frac {11(5c + 6a)} {(5a + 6b)(5b + 6c)(5c + 6a)} \cdot a(5b + 6c)} \leq\\
\sqrt{\frac {11^2 (a + b+ c)} {(5a + 6b)(5b + 6c)(5c + 6a)} \sum_{cyc} a(5b + 6c) } =\\
\sqrt{\frac {11^3 (a + b+ c) (ab + bc + ca)} {(5a + 6b)(5b + 6c)(5c + 6a)}} \leq \sqrt 9 = 3
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Expressing in the form $A \sin(x + c)$
Express in the form $A\sin(x+c)$
a) $\sin x+\sqrt3\cos x$; b) $\sin x-\cos x$
sol: a) $A=\sqrt{1+3}=2$, $\tan c=\frac{\sqrt 3}1$, $c=\frac\pi3$. So $\sin x+\sqrt3\cos x=2\sin(x+\frac\pi3)$
b) $\sqrt 2\sin(x-\frac\pi4)$
Can someone please explain the method used in the provided solution above? (I'm not familiar with this way of solving whatsoever.)
Thanks in advance =]
| Solution for item b of the question (this is the item on Problem Set 1 on MITOCW 18.01SC:
$$A\sin(x+c) = A\sin x \cos c + A \sin c \cos x$$
We can see that if $A \cos c = 1$ in the first term and $A \sin c = -1$ in the second term then we will end up with $\sin x - \cos x$, so we will have shown that $f(x)$ can be written in the form $A \sin(x+c)$.
We have two equations in two unknowns (A and c) and so we can solve for these variables. Square both sides of each equation:
$$A^2 \cos^2 c = 1$$
$$A^2 \sin^2 c = 1$$
Sum the two equations
$$A^2(\cos^2 x + \sin^2 x) = 2$$
Use $\cos^2 x + \sin^2 x = 1$
$$A^2 = 2 \Rightarrow A = \pm \sqrt{2}$$
Solution 1: $A = \sqrt{2}$, $c = -\frac{\pi}{4}$
$$\cos c = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} \Rightarrow c = \frac{\pi}{4} \text{ or } c = \frac{7 \pi}{4}$$
$$\sin c = -\frac{1}{\sqrt{2}}= -\frac{\sqrt{2}}{2} \Rightarrow c = \frac{5\pi}{4} \text{ or } c = \frac{7 \pi}{4}$$
Therefore the value of $c$ that satisfies both equations is $\frac{7\pi}{4}$, which is the same as $-\frac{\pi}{4}$
Solution 2: $A = -\sqrt{2}$, $c = \frac{3\pi}{4}$
$$\cos c = -\frac{1}{\sqrt{2}}= -\frac{\sqrt{2}}{2} \Rightarrow c = \frac{3\pi}{4} \text{ or } c = \frac{5 \pi}{4}$$
$$\sin c = \frac{1}{\sqrt{2}}= \frac{\sqrt{2}}{2} \Rightarrow c = \frac{\pi}{4} \text{ or } c = \frac{3 \pi}{4}$$
Therefore the value of $c$ that satisfies both equations is $\frac{3\pi}{4}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $ 1 + 2 + 4 + 8 + \dots = -1$
Possible Duplicate:
Infinity = -1 paradox
I was told by a friend that $1 + 2 + 4 + 8 + \dots$ equaled negative one. When I asked for an explanation, he said:
Do I have to?
Okay so, Let $x = 1+2+4+8+\dots$
$2x-x=x$
$2(1+2+4+8+\dots) - (1+2+4+8+\dots) = (1+2+4+8+\dots)$
Therefore, $(2+4+8+16+\dots) + (-1-2-4-8+\dots) = (1+2+4+8+\dots)$. Now $-2$ and $2$, $-4$ and $4$, $-8$ and $8$ and so on, cancel out, and the only thing left is $-1$.
Therefore, $1+2+4+8+\dots = -1$.
I feel that this conclusion is not right, but I cannot express it. Can anyone tell if this proof is wrong, and if it is, how it is wrong?
| The infinite series $\displaystyle 1 + 2 + 4 + 8 + \ldots$ diverges. However, the sum
$f(z) = 1 + z + z^2 + z^3 + \ldots$, which converges to $1/(1-z)$ for $|z| < 1$, has an
analytic continuation to the complex plane with the point $1$ removed, and indeed $f(2) = -1$.
So in that sense you could regard $-1$ as the value of the divergent series.
For more on this, see http://en.wikipedia.org/wiki/1_%2B_2_%2B_4_%2B_8_%2B_%C2%B7_%C2%B7_%C2%B7 and references there.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\cos(2\pi/n)+\cos(4\pi/n)+\cdots+\cos(2(k-1)\pi/n)=-1$ How do you prove that
$$\cos(2\pi/n)+\cos(4\pi/n)+\cdots+\cos(2(n-1)\pi/n)=-1,$$ where $n \geq 2$?
| A proof using trigonometric identity proceeds by multiplying the identity with $2 \sin\left(\frac{\pi}{n}\right)$ and using $2 \cos(\alpha) \sin(\beta) = \sin(\alpha+\beta) - \sin(\alpha-\beta)$:
$$ \begin{eqnarray}
\sum_{m=1}^{n-1} 2 \sin\left(\frac{\pi}{n} \right) \cos\left(\frac{2 \pi}{n} m\right) &=& \sum_{m=1}^{n-1} \left(\underbrace{\sin\left( \frac{\pi}{n} (2m+1) \right)}_{f_{m+1}} - \underbrace{\sin\left( \frac{\pi}{n} (2m-1) \right)}_{f_m} \right) = f_n-f_1 \\
&=& \sin\left( \frac{\pi}{n} \left(2n-1\right) \right) - \sin\left( \frac{\pi}{n} \right) \\
&=& \sin\left( 2\pi - \frac{\pi}{n} \right) - \sin\left( \frac{\pi}{n} \right) = -2 \sin\left(\frac{\pi}{n}\right)
\end{eqnarray}
$$
We thus proved:
$$
2 \sin\left(\frac{\pi}{n} \right) \left\{ \cos\left(\frac{2\pi}{n}\right) + \cos\left(\frac{4\pi}{n}\right) + \cdots + \cos\left(\frac{2\pi}{n} (n-1)\right) \right\} = -2 \sin\left(\frac{\pi}{n} \right)
$$
Choosing $n>1$, $\sin\left(\frac{\pi}{n} \right) \not=0$, establishing the identity.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Limit of quotients with square roots: $\lim_{x\to2} \frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ You can't use L'Hospital's rule :S
$$\lim_{x\to2} {\sqrt{6-x}-2\over\sqrt{3-x}-1}$$
I've tried to multiply by conjugates but ended up with a so complex equation, please help, anyone? :S
| The key is to multiply by the conjugates of both radicals, in order to eliminate what makes the problem annoying: the two radicals whose limits both go to zero. And then you see what cancels out.
$$\begin{align*}
\lim_{x \to 2} \;\frac{\sqrt{6-x} - 2}{\sqrt{3-x} - 1}
\;&=\;
\lim_{x \to 2} \;\frac{\bigl((6-x) - 4\bigr)\bigl(\sqrt{3-x} + 1\bigr)}{\bigl( \sqrt{6-x} + 2 \bigr)\bigl((3-x) - 1\bigr)}
&\qquad\qquad\tag{multiply by conjugates}
\\[2ex]&=\;
\lim_{x \to 2} \;\frac{\bigl(2-x\bigr)\bigl(\sqrt{3-x} + 1\bigr)}{\bigl( \sqrt{6-x} + 2 \bigr)\bigl(2 - x\bigr)}
&\qquad\qquad\tag{simplify}
\\[2ex]&=\;
\lim_{x \to 2} \;\frac{\sqrt{3-x} + 1}{\sqrt{6-x} + 2}
&\qquad\qquad\tag{simplify more}
\\[2ex]&=\;
\frac{\lim_{x \to 2} \bigl(\sqrt{3-x} + 1\bigr)}{\lim_{x \to 2} \bigl(\sqrt{6-x} + 2\bigr)}
\tag{as both limits exist $\ldots$}
\\[2ex]&=\;
\frac{1 + 1}{2 + 2} \;=\; \frac{1}{2}.
\tag{$\ldots$ by substitution}
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating Average Case Complexity I am trying to find the average case complexity of a sequential search. I know that the value is calculated as follows:
Probability of the last element is $\frac{1}{2}$
Probability of the next to last is $\frac{1}{4}$
Probability of any other elements is $\frac{1}{4n-8}$
I can assume the I have a list of n $\geq$ 3. How Would I go about solving this?
| Let $X \in \{1,2,\dots,n\}$ be the position of the element you're searching for. This random variable has the following probability mass function
$$p_X (k) = \left\{\begin{array}{cl} \frac{1}{4 n - 8} & \textrm{if} \quad{} k \in \{1,2,\dots,n-2\}\\ \frac{1}{4} & \textrm{if} \quad{} k = n-1\\ \frac{1}{2} & \textrm{if} \quad{} k = n\end{array}\right.$$
where $n \geq 3$ is the size of the list / array over which you're searching. Note that
$$\displaystyle\sum_{k=1}^{n-2} p_X (k) = \frac{n-2}{4 n - 8} = \frac{1}{4}$$
and therefore
$$\displaystyle\sum_{k=1}^{n} p_X (k) = \frac{1}{4} + \frac{1}{4} + \frac{1}{2} = 1$$
as it should be. The average case complexity is given by the expected value of $X$, which is
$$\mathbb{E} (X) = \displaystyle\sum_{k=1}^{n} k \, p_X (k) = \left[\frac{1}{4 n - 8}\displaystyle\sum_{k=1}^{n-2} k \right] + \frac{n-1}{4} + \frac{n}{2} = \frac{7 n - 3}{8}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to show that that $\sum_{n=1}^{\infty}\left( \frac{1}{3n-1} + \frac{1}{3n-2}- \frac{2}{3n}\right)= \ln\left(3\right)$? $$
\mbox{How to show that that}\qquad
\sum_{n = 1}^{\infty}\left({1 \over 3n - 1} + {1 \over 3n - 2} - {2 \over 3n}\right)
=
\ln\left(3\right)\ {\large ?}
$$
$$
\mbox{or}\quad
1 + \frac{1}{2} -\frac{2}{3} + \frac{1}{4} + \frac{1}{5} - \frac{2}{6} +\frac{1}{7} + \frac{1}{8} - \frac{2}{9} \cdots
=
\ln\left(3\right)
$$
| Let $H_n=\sum_{k=1}^n \frac1k$ denote the $n$-th harmonic number.
It is well known that $H_n \sim \ln n + \gamma$, where $\gamma$ is Euler-Mascheroni constant. (To be more precise, $\lim\limits_{n\to\infty} (H_n-\ln n) = \gamma$, which is what we actually need below.)
You are interested in
$$\sum_{k=1}^n \left(\frac1{3k-2}+\frac1{3k-1}+\frac1{3k}-\frac3{3k}\right) = H_{3n}-H_n \sim \ln{3n}-\ln n = \ln 3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 2
} |
finding the sum of the absolute values for the roots How to find the sum of the absolute values for the roots of this equation:
$$x^4-4x^3-4x^2+16x-8=0$$
| Notice that
$$\begin{align}x^4-4x^3-4x^2+16x-8 &= (x-1)^4 - 10(x-1)^2 + 1 \\
&= ((x-1)^2-5)^2-24 \end{align}$$
so you can actually calculate the roots explicitly and sum their absolute values.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a,b,c>0$ and $a+b+c=3$. Find the minimum of $P$. Let $a,b,c>0$ and $a+b+c=3$. Prove that:
$$P=\frac{{a}^{2}}{\sqrt{{b}^{3}+1}}+ \frac { { b }^{ 2 } }{ \sqrt{ { c }^{ 3 }+1 } }+\frac{{c}^{2}}{\sqrt{{a}^{3}+1}}\geq \frac{3\sqrt{2}}{2}$$
| By AM-GM, C-S and Rearrangement we obtain:
$$\sum_{cyc}\frac{a^2}{\sqrt{b^3+1}}=\sum_{cyc}\frac{2\sqrt2a^2}{2\sqrt{2(b^2-b+1)(b+1)}}\geq\sum_{cyc}\frac{2\sqrt2a^2}{2(b^2-b+1)+(b+1)}=$$
$$=\sum_{cyc}\frac{2\sqrt2a^2}{2b^2-b+3}=\sum_{cyc}\frac{2\sqrt2a^4}{2a^2b^2-a^2b+3a^2}\geq\frac{2\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^2b^2-a^2b+3a^2)}=$$
$$=\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(6a^2b^2-3a^2b+9a^2)}=\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(6a^2b^2-(a+b+c)a^2b+(a+b+c)^2a^2)}=$$
$$=\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(6a^2b^2-a^3b-a^2b^2-a^2bc+a^4+2a^2b^2+2a^3b+2a^3c+2a^2bc)}=$$
$$=\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+a^3b+2a^3c+7a^2b^2+a^2bc)}\geq\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^4+a^3b+a^3c+7a^2b^2+a^2bc)}.$$
Thus, it remains to prove that
$$\frac{6\sqrt2(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(2a^4+a^3b+a^3c+7a^2b^2+a^2bc)}\geq\frac{3\sqrt2}{2}$$
or
$$4(a^2+b^2+c^2)^2\geq\sum\limits_{cyc}(2a^4+a^3b+a^3c+7a^2b^2+a^2bc)$$ or
$$\sum_{cyc}(2a^4-a^3b-a^3c+a^2b^2-a^2bc)\geq0$$ or
$$\sum_{cyc}(a-b)^2(a^2+ab+b^2)+\frac{1}{2}\sum_{cyc}c^2(a-b)^2\geq0.$$
Done!
| {
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Trigonometric proof If $4 \tan(\alpha - \beta) = 3 \tan \alpha $, then prove that
$$\tan \beta = \frac{\sin(2 \alpha)}{7 + \cos(2 \alpha)}$$
This is not homework and I've tried everything so I would just like a straight answer thank you in advance.
| Expanding out the left side of the first equation,
$$ \frac{4 \tan \alpha - 4 \tan \beta}{1 + \tan \alpha \tan \beta} = 3 \tan \alpha$$
Thus
$$ \tan \beta = \frac{\tan \alpha}{4 + 3 \tan^2 \alpha}$$
Writing $\tan \alpha = \dfrac{\sin \alpha}{\cos \alpha}$, this becomes
$$ \tan \beta = \frac{\sin \alpha \cos \alpha}{4 \cos^2 \alpha + 3 \sin^2 \alpha}$$
Now use $\sin (2\alpha) = 2 \sin \alpha \cos \alpha$, $\cos^2 \alpha = \dfrac{1+\cos (2\alpha) }{2}$ and $\sin^2 \alpha = \dfrac{1-\cos (2\alpha) }{2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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which of the following is true, number theory problem
$a$ is correct as $n=2m+1 \text{(say)} \Rightarrow n^2=4m^2+4m+1\equiv1(\mod 8),\ m\ge1$
$b$ is true ingeneral, say $n=5, m=3$, for $c$ I dont know how to prove, please help
| (a)$(2a+1)^2-1=4a^2+4a+1=8\frac{a(a+1)}2$
(b)$(2x+1)^2+(2y+1)^2=4(x^2+y^2+x+y)+2\equiv2\pmod 4$
But $(2z)^2\equiv 0\pmod 4,(2z\pm1)^2\equiv1\pmod 4$
(c)$$\frac{n^5}5+\frac{n^3}5+\frac{7n}{15}=\frac{n^5-n}5+\frac{n^3-n}3+\frac{7n}{15}+\frac n 5+\frac n 3$$
$$=\frac{n^5-n}5+\frac{n^3-n}3+n$$
Now using Femat's little theorem , prime $p\mid (n^p-n),$ so $5\mid (n^5-n)$ and
$5\mid (n^3-3)$
| {
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prove $\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$ by mathematical induction How to prove
$$\frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}<\frac{25}{36}$$
by Mathematical induction，n$\ge $1
| EDIT: This approach is unfortunately wrong, as pointed out in the comments.
Let $S(n) = \frac{1}{ n+1}+\frac{1}{ n+2}+\cdots+\frac{1}{2n}.$
(1) Base step. Let $n=1$. Then $n+1 = 2n$, so the series only has one term. $S(1) = \frac{1}{2} < \frac{25}{36}$, hence the statement is true for $n=1$.
(2) Inductive step. Suppose the induction hypothesis is true. Then
$S(n+1) = \frac{1}{ (n+1)+1}+\frac{1}{ (n+1)+2}+\cdots+\frac{1}{2(n+1)}$
$= \frac{1}{ n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}$
$= - \frac{1}{n+1} + \left(\frac{1}{n+1} + \frac{1}{ n+2}+\frac{1}{n+3}+\cdots+\frac{1}{2n}\right) + \frac{1}{2n+1} + \frac{1}{2n+2}$
$= - \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} + S(n)$
$< - \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} + \frac{25}{36}$
where the inequality comes from the induction hypothesis.
Now $S(n+1) < \frac{25}{36}$, as we want to show, requires
$- \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} < 0.$
As $- \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+2} < - \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+1}$, this is certainly true if
$- \frac{1}{n+1} + \frac{1}{2n+1} + \frac{1}{2n+1} < 0$, or $\frac{1}{n+1} < \frac{2}{2n+1}$. The last inequality is trivial to show.
| {
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When is $\left\lfloor \frac {7^n}{2^n} \right\rfloor \bmod {2^n} \ne 0\;$? Is
$$\left\lfloor \frac {7^n}{2^n} \right\rfloor \bmod{2^n} \ne 0\;$$
always true when $n \ge 3$.
Baker's theorem on transcendental numbers that provide bounds for diophantine equations may be useful, but I will leave that to the experts.
| The following may be a starting point:
Write $7^n \mod 4^n$:
$$ 7^n =a\cdot 4^n + b \quad b<4^n $$
$$ \left\lfloor \frac{7^n}{2^n} \right\rfloor \equiv \left\lfloor \frac b{2^n} \right\rfloor \pmod{2^n}$$
and this is $0$ iff $b<2^n$.
Well, $7^n=(2^3-1)^n=\displaystyle\sum_{0\le i\le n} \binom{n}{i} 2^{3i} \cdot (-1)^{n-i} $, and those with indices $3i\ge 2n$ vanish, thus
$$7^n \equiv (-1)^n\sum_{ i < 2n/3 } \binom ni (-8)^{i} = ... \pmod{4^n}$$
| {
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Finding the limit $ \lim_{x\to 0} \frac{1-\cos x \cos(2x)}{x^2}$ I cannot find the following limit:
$$ \lim_{x\to 0} \frac{1-\cos x \cos(2x)}{x^2} \, . $$
Please, help me.
| $$\cos(x) \cos(2x) = \left(1 - \dfrac{x^2}{2!} + \dfrac{x^4}{4!} - \cdots \right)\left(1 - \dfrac{(2x)^2}{2!} + \dfrac{(2x)^4}{4!} - \cdots \right)\\ = 1 - \dfrac{x^2 + 4x^2}{2!} + \mathcal{O}(x^4)$$
Hence, $$1 - \cos(x) \cos(2x) = \dfrac52 x^2 + \mathcal{O}(x^4)$$
Can you now find the limit?
EDIT
Alternately, you could also proceed as below.
$$\cos(x) \cos(2x) = \cos(x) \left(2 \cos^2(x)-1 \right) = 2 \cos^3(x) - \cos(x)$$
Hence, $$1- \cos(x) \cos(2x) = 1 + \cos(x) - 2\cos^2(x) = (1-\cos(x))(1+2\cos(x) + 2\cos^2(x))\\ = 2 \sin^2(x/2)(1+2\cos(x) + 2\cos^2(x))$$
Hence, $$f(x) = \dfrac{1- \cos(x) \cos(2x)}{x^2} = \dfrac{2 \sin^2(x/2)(1+2\cos(x) + 2\cos^2(x))}{x^2}$$
$$\lim_{x \to 0} f(x) = \lim_{x \to 0}\dfrac{2 \sin^2(x/2)}{x^2} \times \lim_{x \to 0}(1+2\cos(x) + 2\cos^2(x)) = 2 \times \dfrac14 \times (1+2+2) = \dfrac52$$
| {
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} |
Coefficent of an equation I am suppose to find the coefficent of x^2 in this equation
after doing the calculation I ended up with this
but that is the wrong answer what is that I am missing?
| It is actually a bit less confusing (you don't have to expand the brackets with three terms) if you write
$$
f(f(x))=(x^2+bx+c)^2+b(x^2+bx+c)+c=(x^2+bx+c)(x^2+bx+c+(b+c))
=x^4+bx^3+(c+b)x^2+bx^3+b^2 x^2+b(c+b)x+cx^2+bcx+c(b+c)+c
$$
The it is clear tha the coefficient at $x^2$ is $c+b+b^2+c=2 \cdot 15 -23+23^2$
| {
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$\frac{a+b}{b+c} + \frac{c+d}{d+a} ≤ 4(\frac{a+c}{b+d}) ; a , b , c , d ∈ [1 , 2]$ Is it true that
if all the real numbers $a , b , c , d$ are from the closed interval $[1 , 2]$ then we always have the inequality
$$
\frac{a+b}{b+c} + \frac{c+d}{d+a} ≤ 4\Big(\frac{a+c}{b+d}\Big)
$$
| Since $a, c \geq 1$, we have
$$
a - \frac {a + b} {b + c} = \frac{ab + ac - a - b} {b + c} = \frac {b(a - 1) + a(c - 1)} {b + c} \geq 0
$$
A similar argument leads to
$$
c - \frac{c + d} {d + a} \geq 0
$$
Finally $b + d\leq 4$ implies
$$
\frac {a + b} {b + c} + \frac{c + d} {d + a} \leq a + c \leq \frac 4 {b + d} (a + c)
$$
| {
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Where do I go wrong when solving this integral involving a quadratic? Simple question. I have an arithmetic mistake somewhere in here but can't find it.
$\int \frac{3x-1}{x^2+10x+28}dx$
$\frac{1}{3}\int \frac{3x-1}{\frac{1}{3}(x+5)^2+1}dx$
Set:
$u=x+5$
$x=u-5$
$dx = du$
$\frac{1}{3}\int \frac{3u-16}{\frac{u^2}{3}+1}du$
$\frac{1}{3}\int \frac{3u}{\frac{u^2}{3}+1}du - \frac{1}{3}\int\frac{16}{\frac{u^2}{3}+1}du$
Set:
$u = \sqrt3 tan\theta$
$du = \sqrt3 sec^2\theta d\theta$
$\theta = arctan(\frac{u}{\sqrt3})$
$3\int tan\theta d\theta - \frac{1}{3}\int 16\sqrt3d\theta$
$3 ln(sec(arctan(\frac{u}{\sqrt3}))) - \frac{16}{\sqrt3}(arctan(\frac{u}{\sqrt3}))+c $
$3 ln(\frac{\sqrt{u^2+3}}{\sqrt3})-\frac{16}{\sqrt3}arctan(\frac{u}{\sqrt3})+c$
$3 ln(\frac{\sqrt{(x+5)^2+3}}{\sqrt3})-\frac{16}{\sqrt3}arctan(\frac{x+5}{\sqrt3})+c$
$\frac{3}{2}ln(\frac{(x+5)^2}{3})-\frac{16}{\sqrt3}arctan(\frac{x+5}{\sqrt3})+c$
I think it's correct but there shouldn't be a $3$ in the denominator in the first term. But I can't see where I went wrong.
| Note that $\ln(g(x)/c)=ln(g(x))-ln(c)$ for any function $g(x)$. Since you're finding an antiderivative, any extra added/subtracted constants are irrelevant.
| {
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Sides of triangle and an altitude Let $a$, $b$, $c$ be the lengths of the sides of a triangle. Let $h$ be the altitude drawn on the side of length $a$ Then is $a^2 + 4h^2 - (b+c)^2$ always negative ?
| I also have been able to come up with a proof that the expression indeed is negative ;
$-(b-c)^2 ≤ 0 $
$\implies a^2 - (b-c)^2 ≤ a^2 $
$\implies (a+b-c)(a-b+c) ≤ a^2 $
$\implies (a+b+c)(b+c-a)(a+b-c)(a-b+c) ≤ a^2 ( a+b+c)(b+c-a) $
{since $a+b+c >0$ and for a triangle $b+c > a$ , the multiplication in the above line doesn't change sign of inequality}
$\implies 16 A^2 ≤ a^2 \{ (b+c)^2 - a^2 \} $ [ A is the area of the triangle , by Heron's formula ]
$\implies 16 ( \frac 12 ah )^2 ≤ a^2 \{(b+c)^2 - a^2\} $ [ since h is altitude on a , A= ah/2 ]
$\implies 4 (ah)^2 ≤ a^2 \{ (b+c)^2 - a^2 \} $
$\implies 4 h^2 ≤ (b+c)^2 - a^2 $
| {
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Find the greatest powers of $2$ dividing $10!$, $20!$, $30!$, $40!$ I'm trying to find the greatest powers of $2$ dividing $10!$, $20!$, $30!$, $40!$, as part of a basic number systems course.
I'm rather lost with this question. For $10!$ I tried writing the terms out and just extracting powers of $2$ manually, getting $2^8$ as the highest powers of $2$, with $10! = (2^8)(14175)$ as the result.
I'm fairly confident that the answer is correct (although I'm not sure, so confirmation of that would be great!), but this method is rather crude for larger numbers, so I suspect that it isn't the right way to do it.
If anyone can point me in the right direction I would be very grateful.
| For any prime $p$ and integer $n$, let
$$
n=d_0+d_1p+d_2p^2+d_3p^3+\dots+d_kp^k\tag{1}
$$
where $0\le d_j\lt p$. $(1)$ is the base-$p$ representation of $n$.
The number of multiples of $p$ not greater than $n$ would be
$$
\left\lfloor\frac np\right\rfloor=d_1p^0+d_2p^1+d_3p^2+d_4p^3+\dots+d_kp^{k-1}\tag{2}
$$
The number of multiples of $p^2$ not greater than $n$ would be
$$
\left\lfloor\frac n{p^2}\right\rfloor=\hphantom{d_1p^0+}d_2p^0+d_3p^1+d_4p^2+\dots+d_kp^{k-2}\tag{3}
$$
The number of multiples of $p^3$ not greater than $n$ would be
$$
\left\lfloor\frac n{p^3}\right\rfloor=\hphantom{d_1p^0+d_2p^1+}d_3p^0+d_4p^1+\dots+d_kp^{k-3}\tag{4}
$$
and so forth.
$(2)$ only counts each multiple of $p^2$ once. To count each multiple of $p^2$ twice, we need to add $(3)$. This only counts each multiple of $p^3$ twice. To count each multiple of $p^3$ three times, we need to add $(4)$, and so on.
After adding up $(2)$, $(3)$, $(4)$, and so on, the coefficient of $d_j$ is
$$
p^{j-1}+p^{j-2}+p^{j-2}+\dots+p^0=\frac{p^j-1}{p-1}\tag{5}
$$
Thus, the sum of $(2)$, $(3)$, $(4)$, and so on is
$$
d_0\frac{p^0-1}{p-1}+d_1\frac{p^1-1}{p-1}+d_2\frac{p^2-1}{p-1}+\dots+d_k\frac{p^k-1}{p-1}=\frac{n-\sum d_j}{p-1}\tag{6}
$$
Therefore, the number of factors of $p$ in $n!$ is
$$
\frac{n-\sum d_j}{p-1}\tag{7}
$$
where $\sum d_j$ is the sum of the base-$p$ digits of $n$.
Examples
For $p=2$, $10=1010_{\text{two}}$, so $\sum d_j=2$. There are $\frac{10-2}{2-1}=8$ factors of $2$ in $10!$
For $p=2$, $20=10100_{\text{two}}$, so $\sum d_j=2$. There are $\frac{20-2}{2-1}=18$ factors of $2$ in $20!$
For $p=2$, $30=11110_{\text{two}}$, so $\sum d_j=4$. There are $\frac{30-4}{2-1}=26$ factors of $2$ in $30!$
For $p=2$, $100=1100100_{\text{two}}$, so $\sum d_j=3$. There are $\frac{100-3}{2-1}=97$ factors of $2$ in $100!$
| {
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$a+b=c \times d$ and $a\times b = c + d$ There is a 'nice' relationship between the integers (1,5) and (2,3) as
$$1+5=2 \times 3;$$
$$1\times 5 = 2 + 3.$$
So I tried to find all positive integers pairs $(a, b)$ and $(c, d)$ such that $$a+b=c \times d;$$
$$a\times b = c + d.$$
To find this, $a, b, c, d$ must satisfy $$(a+1)(b+1)=(c+1)(d+1).$$ However, this condition is only necessary but not sufficient.
Any idea?
| Define $r$, $s$ by $(x-a)(x-b)=x^2-rx+s$. Then $r=a+b=cd$ and $s=ab=c+d$, so $(x-c)(x-d)=x^2-sx+r$. Thus, the polynomials $x^2-rx+s$ and $x^2-sx+r$ both have only integer roots. Thus, $r^2-4s$ and $s^2-4r$ are both squares: $r^2-4s=u^2,s^2-4r=v^2$.
EDIT: Continuing, from $r^2-4s=u^2$ we get $4s=r^2-u^2\ge r^2-(r-2)^2=4r-4$, so $s\ge r-1$. By symmetry, $r\ge s-1$, so we have $r=s$ or $r=s\pm1$. If $r=s$ then $r^2-4r$ is a square, but $r^2-4r+4$ is a square, so we have two squares differing by $4$, which gives $r^2-4=0$, $r=0$ or $r=4$. These lead to the solutions $a=b=c=d=0$ and $a=b=c=d=2$.
$r=s+1$ implies $s^2-4s-4=(s-2)^2-8$ is a square. The only pair of squares differing by $8$ is $3^2-1^2$, and this leads to the $(1,5)$, $(2,3)$ solution (and its permutations) that we already have.
| {
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Integrating a partial fraction How do I compute the integral
$$\int_0^{\pi/2} \cfrac 1 {12\cos x + 9 \sin x} dx?$$
Please can you help.
| If all else fails, the Weierstrass substitution will do it.
However, let's try something that begins with what Marvin suggested in his answer. First notice that $12^2+9^2=15^2$, so $\left(\frac{12}{15}\right)^2+\left(\frac{9}{15}\right)^2=1$. Hence we can find $\theta$ such that $\cos\theta=12/15$ and $\sin\theta=9/15$. Then
$$
12\cos x + 9\sin x = 15\left(\frac{12}{15}\cos x+\frac{9}{15}\sin x\right) = 15(\cos\theta\cos x + \sin\theta\sin x) = 15\cos(x-\theta).
$$
Therefore
$$
\int\frac{dx}{12\cos x+9\sin x} = \int \frac{dx}{15\cos(x-\theta)}= \int \frac{\cos(x-\theta)\,dx}{15(\cos(x-\theta))^2} = \frac{1}{15}\int \frac{\cos(x-\theta)\,dx}{1- (\sin(x-\theta))^2}
$$
$$
= \frac{1}{15}\int \frac{du}{1-u^2} = \frac{1}{15}\int \frac{A}{1-u} + \frac{B}{1+u} \, du.
$$
Then use partial fractions.
| {
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Contour integration of $\int_{-\infty}^{\infty} \frac{1-b+x^2}{(1-b+x^2)^2 + 4bx^2}dx = \pi$ Given $0< b <1$, derive the equality:
$\displaystyle \int_{-\infty}^{\infty} \frac{1-b+x^2}{(1-b+x^2)^2 + 4bx^2}dx = \pi$
by integrating the function $(1+z^2)^{-1}$ around the rectangle with vertices $R, -R, R+i\sqrt{b}, -R+i\sqrt{b}$ for $R>0$ and taking $R \to \infty$. What happens if $b>1$?
Could you give me some help with the 4 integrals that I should evaluate? Also, if there is any other trick I would appreciate it if could give me some help.
| Note
$$ (1-b+x^2)^2+4bx^2=(1-b+x^2+2\sqrt{b}xi)(1-b+x^2-2\sqrt{b}xi)=[(x+\sqrt bi)^2+1][(x+\sqrt bi)^2+1] $$
and hence
$$ \frac{1-b+x^2}{(1-b+x^2)^2+4bx^2}=\frac12\left(\frac{1}{(x+\sqrt bi)^2+1}+\frac{1}{(x+\sqrt bi)^2+1}\right). $$
It is easy to use contour integral to check
$$ \int_{-\infty}^\infty \frac{1}{(x+\sqrt bi)^2+1}dx=\int_{-\infty}^\infty \frac{1}{(x-\sqrt bi)^2+1}dx=\int_{-\infty}^\infty\frac{1}{x^2+1}dx=\pi.$$
Thus
$$ \int_{-\infty}^\infty \frac{1-b+x^2}{(1-b+x^2)^2+4bx^2}dx=\pi.$$
| {
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Partial Fractions continued... Hi asked the following question yesterday: Obtaining the sum of a series
Given the answers to that question by wj32, I am now trying to solve the following problem:
Consider the series
$$\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$$
Use partial fractions to write the general term
$$u_n=\frac{n}{n^4+n^2+1}$$
as a difference of two simpler terms
My attempt at a Solution:
The partial fractions are
$$\begin{align}
\frac{n}{n^4+n^2+1}&=\frac{n+n-n}{n^4+n^2+1}\\
&=\frac{n+n}{n^4+n^2+1}-\frac{n}{n^4+n^2+1}\\
&=\frac{2}{n^3+n+\frac{1}{n}}-\frac{2}{2(n^3+n+\frac{1}{n})}
\end{align}$$
Then
$$\begin{align}
S_n&=\left[\sum_{n=1}^k\frac{2}{n^3+n+\frac{1}{n}}-\sum_{n=1}^k\frac{2}{2(n^3+n+\frac{1}{n})}\right]\\
&=\left[\left(\frac{2}{3}+\frac{4}{21}+\frac{6}{91}+...+\frac{2}{n^3+n+\frac{1}{n}}\right)-\left(\frac{1}{3}+\frac{2}{21}+\frac{3}{91}+...+\frac{2}{2(n^3+n+\frac{1}{n})}\right)\right]
\end{align}$$
Here the terms in the left do not cancel the terms in the right?
I'm guessing I need simpler/different partial fractions up at the top?
| First note that we have a nice factorization of $n^4 + n^2 + 1$.
$$n^4 + n^2 + 1 =(n^2+n+1)(n^2-n+1)$$
Hence, $n = \dfrac{(n^2+n+1) - (n^2-n+1)}2$. This gives us $$u_n = \dfrac{(n^2+n+1) - (n^2-n+1)}{2(n^2+n+1)(n^2-n+1)} = \dfrac12 \left(\dfrac1{n^2-n+1} - \dfrac1{n^2+n+1}\right) = \dfrac12 \left(\dfrac1{(n-1)n+1} - \dfrac1{n(n+1)+1}\right)$$
Now telescopic summation should do the job for you
Hence,
\begin{align}
S_N = \sum_{n=1}^{N} u_n & = \dfrac12 \sum_{n=1}^N\left(\dfrac1{(n-1)n+1} - \dfrac1{n(n+1)+1}\right)
\end{align}
\begin{align}
2S_N & = \sum_{n=1}^N\left(\dfrac1{(n-1)n+1} - \dfrac1{n(n+1)+1}\right)\\
& = \left(1 - \dfrac13 \right) + \left(\dfrac13 - \dfrac17 \right) + \left(\dfrac17 - \dfrac1{13} \right) + \cdots\\
& + \left(\dfrac1{(n-2)(n-1)+1} - \dfrac1{(n-1)n+1}\right) + \left(\dfrac1{(n-1)n+1} - \dfrac1{n(n+1)+1}\right)\\
& = 1 - \dfrac1{n(n+1)+1}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\log_9 71$ or $\log_8 61$ I am trying to know which one is bigger :$$\log_9 71$$ or $$\log_8 61$$ how can i know without using a calculator ?
| If we are allowed to use calculus, we can get a somewhat formal estimate of both numbers: $$\log_9 (71) = \log_9 (81) + \log_9(71/81) = 2 + \log_9 \left(1 - \frac{10}{81}\right) \approx 2 - \frac{1}{2 \ln 3}\left(\frac{10}{81}\right), \\ \log_8 (61) = \log_8 (64) + \log_8 (61/64) = 2 + \log_8 \left(1 - \frac{3}{64}\right) \approx 2 - \frac{1}{3 \ln 2}\left(\frac{3}{64}\right).$$
Using rough estimates like $\ln 2 \approx 0.7$, $\ln 3 \approx 1.1$, $\frac{10}{81} \approx \frac{1}{8} = 0.125$ and $\frac{3}{64} \approx \frac{3}{60} = 0.05$ we get
$$\log_9 (71) \approx 2 - \frac{1}{2.1}(0.125) \approx 2 - 0.06 \approx 1.94, \qquad (\text{exact: } \log_9 (71) = 1.940\ldots) \\ \log_8(61) \approx 2 - \frac{1}{2.2}(0.05) \approx 2 - 0.025 \approx 1.975. \qquad (\text{exact: } \log_9(61) = 1.976\ldots)$$
So $\log_9 71 < \log_8 61$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Squaring inequality $a>b+c$ Given $a,b,c > 0$ and $a > b + c$ , is it true that $a^2 > b^2 + c^2$?
Im tried proving it. I followed the below steps and not sure whether im right or wrong?
Squaring on both sides
$\implies a^2 > (b+c)^2$
$\implies a^2 > b^2 + c^2 + 2*b*c$
since $b,c > 0 $, we have $2*b*c > 0$, then
$\implies a^2 > b^2 + c^2$ Q.E.D
| Another way of proving is ; since $a$ , $b$ , $c$ $>$ $0$ then $a$ $>$ $b$ $+$ $c$ implies $a$ $>$ $b$ and $a$ $>$ $c$ ; so $ab$ $>$ $b^2$ and $ca$ $>$ $c^2$
, and $a$ $>$ $b$ $+$ $c$ $\implies$ $a^2$ $>$ $ab$ $+$ $ca$ $>$ $b^2$ $+$ $c^2$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to represent fraction $\frac{1}{a+b\cdot \sqrt[3]{2} + c\cdot (\sqrt[3]{2})^2}$ as $a_1 + b_1 \cdot \sqrt[3]{2} + c_1 \cdot (\sqrt[3]{2})^2$? What do I have to multiply both numerator and denominator with to get the representation is asked? $a, b, c, a_1, b_1, c_1$ are rational numbers.
| I think, the easiest if you solve this:
$$(a_1+b_1\cdot \sqrt[3]2 + c_1\sqrt[3]{2^2})(a+b\cdot \sqrt[3]2 + c\sqrt[3]{2^2}) =1 $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Factorise $f(x)=x^5-x$ into a product of irreducibles in $\mathbb Z_{5}[x]$ So plugging in $1$ gives $f(1) = 0$ which means $1$ is a root and $f$ has a factor $(x-1)$ which is $\equiv (x+4)$ in $\mathbb Z_{5}[x]$ ?
I then divide $f(x)$ by $(x+4)$ using polynomial long division and I get
$x^4-4x^3+16x^2-64x+255$ with remainder $-1020$
And that's looking kinda strange.. not sure what I'm doing at this point
I notice $f(0)$ also $= 0$ so maybe if I divide $f(x)$ by $x$ instead I would get $(x^4-1)$ but then I'm still stuck and don't know what else to do
| $f(x) = x (x-1)(x^3+x^2 + x+1)=x(x-1)(x+1)(x^2+1)$
$(x+a)(x+b)=x^2 + (a+b)x + ab = x^2 +1$ in ${\bf Z}_5[x]$
So $a+b =0(5)$ and $ab=1(5)$ Hence $a=2$ and $b=3$
$ f(x) =x(x-1)(x+1)(x+2)(x+3)$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim\limits_{x\to\infty}\frac{(x+7)^2\sqrt{x+2}}{7x^2\sqrt{x}-2x\sqrt{x}}$
Determine
$$\lim\limits_{x\to\infty}\frac{(x+7)^2\sqrt{x+2}}{7x^2\sqrt{x}-2x\sqrt{x}}.$$
Multiplying and dividing by $\sqrt{x}$ yields
$$\frac{(x+7)^2\sqrt{x^2+2x}}{7x^3-2x^2}$$
where I would like to approximate the squareroot for sufficiently large $x$ with
$$\frac{(x+7)^2\sqrt{x^2+2x+1}}{7x^3-2x^2}=\frac{(x+7)^2(x+1)}{7x^3-2x^2}=\frac{x^3(49/x^3+63/x^2+15/x+1)}{x^3(7-2/x)}\longrightarrow 1/7.$$
Can anyone confirm that my approximation is valid and does anyone know how to solve this in a more "usual" way like I did?
| Hint: A standard thing to do is to divide top and bottom by $x^2\sqrt{x}$.
The new top is
$$\left(1+\frac{7}{x}\right)^2 \sqrt{1+\frac{2}{x}},$$
and its behaviour for large $x$ is clear.
The new bottom is $7$ plus something tiny.
Remark: In the solution given in the OP, $\sqrt{x^2+2x}$ was replaced by $x+1$. True, this is fine, the change is indeed small. But if we are doing things formally, the replacement leaves a gap in the argument.
| {
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Prove $1/x^n$ is differentiable Consider the function $g(x)=\frac{1}{x^n}$ where $n \in\Bbb N$, prove that g is differentiable.
I tried to use the definition,
Let $c \in\Bbb R$, then:
$$\frac{g(x)-g(c)}{x-c}=\frac{\frac{1}{x^n}-\frac{1}{c^n}}{x-c}=\frac{c^n-x^n}{(x-c)(x^n \cdot c^n)}=\frac{(c-x)^n}{(x-c)(x^n \cdot c^n)}=-\frac{(c-x)^{n-1}}{(x^n \cdot c^n)}$$
Hence
$$\lim_{x \to c} -\frac{(c-x)^{n-1}}{(x^n \cdot c^n)}=-\lim_{x \to c} \frac{(c-x)^{n-1}}{(x^n \cdot c^n)}=-\lim_{x \to c} \frac{(0)^{n-1}}{(c^n \cdot c^n)}=-\lim_{x \to c} \frac{0}{c^{2n}}=0$$
But I dont think it's okay since I know the derivative of $\frac{1}{x^n} \neq 0$
So how should I do this then?
Thanks in advance,
| You should evidently assume $x\neq 0$.
Then, for $|h|<\delta$ so that $x+h\neq 0$
$$\begin{align}
\mathop {\lim }\limits_{h \to 0} \frac{{\dfrac{1}{{{{\left( {x + h} \right)}^n}}} - \dfrac{1}{{{x^n}}}}}{h} &= \mathop {\lim }\limits_{h \to 0} \frac{{\dfrac{{{x^n} - {{\left( {x + h} \right)}^n}}}{{{x^n}{{\left( {x + h} \right)}^n}}}}}{h} \cr
\\&= \mathop {\lim }\limits_{h \to 0} \frac{1}{{{x^n}{{\left( {x + h} \right)}^n}}}\frac{{{x^n} - {{\left( {x + h} \right)}^n}}}{h} \cr
\\&= - \mathop {\lim }\limits_{h \to 0} \frac{1}{{{x^n}{{\left( {x + h} \right)}^n}}}\mathop {\lim }\limits_{h \to 0} \frac{{{{\left( {x + h} \right)}^n} - {x^n}}}{h} \cr
\\&= - \frac{1}{{{x^{2n}}}}\left( {\frac{d}{{dx}}{x^n}} \right) \cr
\\&= - \frac{1}{{{x^{2n}}}}n{x^{n - 1}} \cr
\\&= - n{x^{ - n - 1}} \end{align} $$
In fact, in a neighborhood where $g(x)\neq 0$, if $g(x)$ is differentiable, then so is $g(x)^{-1}$ and $$\frac d {dx}g(x)^{-1}=-\frac{g'(x)}{g(x)^2}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to find the other vertices of an equilateral triangle given one vertex and centroid If I know the coordinates of the center and one vertex of an equilateral triangle, how do I find the coordinates of the other vertices?
I'm thinking I need to find (x,y) such that the distance to the known vertex is the square root of 3 times the distance to the center, so just need to solve those two distance equations simultaneously, but I keep coming up with too many unknowns.
| Suppose we have centroid $M = (x_0,\ y_0)$ and vertex $A=(x_1,\ y_1)$.
First let us center the triangle at the origin with shifted vertex $$A' = (x_1',\ y_1') = (x_1 - x_0,\ y_1 - y_0)$$
The other vertices will be reached from this one by a rotation about the origin $120^\circ$ clockwise and counter-clockwise. The counter-clockwise rotation matrix is
$$R_{120^\circ} = \begin{pmatrix}\cos120^\circ & -\sin120^\circ \\ \sin120^\circ & \cos120^\circ\end{pmatrix} = \begin{pmatrix}-\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}$$
with the clockwise rotation matrix as
$$R_{-120^\circ} = \begin{pmatrix}\cos120^\circ & \sin120^\circ \\ -\sin120^\circ & \cos120^\circ\end{pmatrix} = \begin{pmatrix}-\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}$$
Your vertices are then
$$B' = \begin{pmatrix}-\frac{1}{2} & -\frac{\sqrt{3}}{2} \\ \frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}\begin{pmatrix}x_1' \\ y_1'\end{pmatrix}=\begin{pmatrix}-\frac{1}{2}x_1' - \frac{\sqrt{3}}{2}y_1' \\ \frac{\sqrt{3}}{2}x_1' - \frac{1}{2}y_1'\end{pmatrix}$$
$$C' = \begin{pmatrix}-\frac{1}{2} & \frac{\sqrt{3}}{2} \\ -\frac{\sqrt{3}}{2} & -\frac{1}{2}\end{pmatrix}\begin{pmatrix}x_1' \\ y_1'\end{pmatrix}=\begin{pmatrix}-\frac{1}{2}x_1' + \frac{\sqrt{3}}{2}y_1' \\ -\frac{\sqrt{3}}{2}x_1' - \frac{1}{2}y_1'\end{pmatrix}$$
Adding $M$ to each coordinate shifts back the triangle to the original spot.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\cot^2 (2x) + \cos^2 (2x) + \sin^2 (2x) = \csc^2 (2x)$ I'm having massive issues proving this identity:
$$\cot^2 (2x) + \cos^2 (2x) + \sin^2 (2x) = \csc^2 (2x)$$
How is this proven?
| Recall the following identities
$$\cos^2{\alpha} + \sin^2{\alpha} = 1$$
$$\cot^2{\beta} + 1 = \csc^2 \beta$$
Hence, $$\cot^2 (2x) + \cos^2 (2x) + \sin^2 (2x) = \underbrace{\cot^2 (2x) + \underbrace{\left(\cos^2 (2x) + \sin^2 (2x) \right)}_{=1}}_{\cot^2(2x)+1 = \csc^2(2x)}$$
| {
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"url": "https://math.stackexchange.com/questions/242403",
"timestamp": "2023-03-29T00:00:00",
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Solving for unknown When solving for unknowns why does the symbols sometimes change from $a + $ to $a -$ or vice versa or stays the same?
For example (using solutions from text):
$2x+6 = x+16\quad$ and the solution is $2x-x = 16-6\quad \Rightarrow \quad x=10$.
Or
$8x-12 = -3x+21\quad $Solution: $8x+3x = 21+12\quad \Rightarrow \quad
11x = 33\quad \Rightarrow \quad
x=3$
|
For example: $2x+6 = x+16$, and the solution is $2x-x = 16-6\Rightarrow x=10$.
Here, let's start by adding $-x$ to both sides of the equation $2x +6 = x+16$: $$-x + 2x+6 = -x + x + 16$$ $$\leftrightarrow x+6 = 16.$$
Next: Adding $-6\,$ to each side of the equation gives us: $$x + 6 + (-6) = 16 + -6$$ $$\leftrightarrow x + 0 = 16 - 6 $$ $$\leftrightarrow x = 10$$
Solving for $x$ given $8x-12 = -3x+21$:
Add $3x$ to each side of the equation, AND add $12$ to both sides of the equation.
This gives: $$3x + 8x -12 + 12 = 3x + -3x + 21 + 12$$ $$\leftrightarrow 11x + 0 = 0 + 21 + 12$$ $$\leftrightarrow 11x =33$$ $$\leftrightarrow x = 3.$$
Note that in the last step, we divided both sides of the equation by $11$ (which is the same as multiplying both sides of the equation by $\frac{1}{11}$).
The objective when solving for one unknown, say $x$, is to get all multiples of $x$ on the left hand side, and all constants on the right hand side. Then simplify. As long as you are adding (or subtracting) the same value from each side of the equation, and as long as you multiply (or divide) both sides of an equation by the same value, the equality remains unchanged.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why ${ \sum\limits_{n=1}^{\infty} \frac{1}{n} }$ is divergent , but ${ \sum\limits_{n=1}^{\infty} \frac{1}{n^2} }$ is convergent? I don't understand why ${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n} }$ is divergent, but ${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n^2} }$ is convergent and its limit is equal to ${ \displaystyle\frac{\pi^2}{6} }$. In both cases, ${n^{th}}$ term tends to zero, so what makes these series different?
${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + }$ ...
${ \displaystyle \sum\limits_{n=1}^{\infty} \frac{1}{n^2} =1 + \frac{1}{4} + \frac{1}{9} + \frac{1}{16} + \frac{1}{25} + \frac{1}{36} + }$ ...
| We can see that the harmonic series, $\displaystyle\sum_{k=1}^\infty\frac1k$, diverges using the classical observation:
$$
\frac11+\frac12+\underbrace{\frac13+\frac14}_{\mbox{$2$ terms}}+\underbrace{\frac15+\frac16+\frac17+\frac18}_{\mbox{$4$ terms}}+\dots+\underbrace{\frac1{2^n+1}+\dots+\frac1{2^{n+1}}}_{\mbox{$2^n$ terms}}+\dots
$$
where each grouping of terms totals at least $\frac12$.
The series $\displaystyle\sum_{k=1}^\infty\frac1{k^2}$ converges to a value $\le2$ by comparison:
$$
\frac1{1^2}+\underbrace{\frac1{2^2}}_{\Large\lt\frac1{1\cdot2}}+\underbrace{\frac1{3^2}}_{\Large\lt\frac1{2\cdot3}}+\underbrace{\frac1{4^2}}_{\Large\lt\frac1{3\cdot4}}+\dots+\underbrace{\frac1{n^2}}_{\Large\lt\frac1{(n-1)n}}+\dots
$$
and
$$
\begin{align}
&\frac1{1\cdot2}+\frac1{2\cdot3}+\frac1{3\cdot4}+\dots+\frac1{(n-1)n}+\dots\\
&=\left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\left(\frac13-\frac14\right)+\dots+\left(\frac1{(n-1)}-\frac1n\right)+\dots\\
&=1
\end{align}
$$
This last series is called a "telescoping sum" since the last part of each term is cancelled by the first part of the next term, leaving only the first part of the first term and the last part of the last term. Since the last part of the last term vanishes, this series converges to the first part of the first term.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if $ |a_n-a_{n-1}| < \frac{1}{2^{n+1} }$ and $a_0=\frac12$, then $\{a_n\}$ converges to $0I try to solve this question but I don't know how.
given $ a_0 = \frac12 $ and for each $n\geq 1$:
$$ |a_n-a_{n-1}| < \frac{1}{2^{n+1}} $$
show that $\{a_n\}$ converges and the limit is $a$ such that $0<a<1$
Update (Edited):
I showed by cauchy that $ |a_m-a_n| < |a_m-a_{m-1}+a_{m+1}-...+a_{n+1}-a_n| < \frac{1}{2^{m+1}} + \frac{1}{2^{m}}+...+\frac{1}{2^{n+2}}$
by the sum of Geometric series, $q=2, a_1=\frac{1}{2^{m+1}}$ then $s_n=\frac{1}{2^{m+1}}[\frac{2^{m-n}-1}{2-1}]$, so $$\frac{1}{2^{m+1}} + \frac{1}{2^{m}}+...+\frac{1}{2^{n+2}} = \frac{1}{2^{m+1}}[\frac{2^{m-n}-1}{2-1}] = \frac{1}{2^{n+1}}-\frac{1}{2^{m+1}}\leq\frac{1}{2^{n+1}}$$
now, it converges!
Can someone help me please to show that the limit is $a$ with $0<a<1$?
Thank you!
| You can also prove this generalization the same way:
if $|a_{n+1}-a_n| < c_n$
where $c_n$ is decreasing
and $\sum_{n=1}^{\infty} c_n$ converges,
then $\lim_{n \to \infty} a_n$ exists
and is less than
$a_1+\sum_{n=1}^{\infty} c_n$.
| {
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Finding the sum of a series $\sum_{n=4}^{\infty}\frac{1}{n^{2}-1}$ I was thinking about this problem:
Given a series $\sum_{n=4}^{\infty}\frac{1}{n^{2}-1}$ ,how can i show that its sum is a/an rational/irrational number,given that the series converges?
Could someone point me in the right direction? Thanks everyone in advance.
| $1/(n-1)-1/(n+1) = 2/(n^2-1)
$, so
$$\sum_{n=4}^{\infty} \frac{1}{n^2-1}
= (1/2)\sum_{n=4}^{\infty} \left(\frac{1}{n-1} - \frac{1}{n+1}\right)
= \frac{1}{3} + \frac{1}{4}
$$
since all the later terms are cancelled out.
Whoops - as pointed out by Limitless, this should be
$\frac{1}{2}\left(\frac{1}{3} + \frac{1}{4}\right)$.
Note that this also allows you to get an explicit expression for
$\sum_{n=a}^b \frac{1}{n^2-1}$
for any integers $a$ and $b$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$(a - 1)x^2+3(a + 1)x+4(a - 1) = 0$ has real solutions iff $7a^2 - 50a + 7\leq 0 $ How can we show that $(a - 1)x^2+3(a + 1)x+4(a - 1) = 0$ has real solutions if and only if $7a^2 - 50a + 7\leq 0$?
I know these are quadratics and can solve them, but I'm not entirely sure what the question is asking of me and how to lay out the logic.
| From the quadratic formula you know that the solutions of
$$(a-1)x^2+3(a+1)x+4(a-1)=0$$
are given by
$$x=\frac{-3(a+1)\pm\sqrt{9(a+1)^2-16(a-1)^2}}{2(a-1)}\;.$$
These will be real if and only if
$$9(a+1)^2-16(a-1)^2\ge 0\;.$$
Expanding the lefthand side, we see that this inequality reduces to
$$-7a^2+50a-7\ge 0\;.$$
Now just multiply the inequality by $-1$.
| {
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Taylor Polynomial of $\arctan$ of given Degree and Error
Replace the following function by its taylor polynomial of the given
grade, and approximate the error in the given interval:
$$f(x) = \arctan(x) \textrm{ by } T_3(f,x,0) \textrm{ in } |x| \le\frac{1}{10}$$
My solution and thoughts
We only need the first three derivatives and the fourth one for the error
$$
f'(x) = \frac{1}{1+x^2} \\
f''(x) = \frac{2x}{(x^2+1)^2} \\
f'''(x) = \frac{6x^2-2}{(x^2+1)^3} \\
f^{(iv)}(x) = \frac{24x(x^2-1)}{(x^2+1)^4}
$$
And by definition we know that
$$
T_3(f,x,0) = \sum\limits_{k=0}^3 \frac{f^{(k)}(0)}{k!}x^k
$$
we get
$$
T_3(f,x,0) = x - \frac{x^3}{3}
$$
We know that
$$R_3(x) = \frac{f^{(iv)}(c)}{4!}x^4, |x| \le \frac{1}{10}$$
by plugging in the maximal value of $x = c = \frac{1}{10}$ we get for any $c$:
$$
\left|f^{(iv)}(c)\right| = \left|\frac{24c(c^2-1)}{(c^2+1)^4}\right| \le \\
\le \frac{24\cdot 99 \cdot 100^4}{1000\cdot 101^4}
$$
$$
\Rightarrow \left|R_3(x)\right| \le \frac{24\cdot 99 \cdot 100^4}{1000\cdot 101^4 \cdot 4!}\cdot\left|x^4\right| \Leftrightarrow \\
\Leftrightarrow \left|R_3(x)\right| \le \frac{10}{101^4}
$$
Is this solution correct? How can I make it more formally right? (I know I lack some mathematical formalism).
| This should be considered a minor addition to the answer by Mike Spivey.
Note that the fourth derivative is $0$ at $0$. So the Taylor polynomial that we get by expanding until the third derivative is exactly the same as the Taylor polynomial we get by expanding until the fourth derivative!
Thus you can use the formula for the remainder that involves the fifth derivative. There is not a lot of gain for the extra work, but there is some. The fifth derivative, I think, is $\dfrac{24(5x^4-10x^2+1)}{(x^2+1)^5}$.
The analysis leads to an upper bound on the error which is about one-fifth of the error upper bound we get by using the Lagrange estimate without noting that stopping at $3$ gives the same polynomial as stopping at $4$.
Remark: You may have worked too hard in doing the arithmetic. Let's do exactly as you did, using the fourth derivative instead of the better fifth.
At a certain stage, you reached
$$
\left|f^{(iv)}(c)\right| = \left|\frac{24c(c^2-1)}{(c^2+1)^4}\right|
$$
Note that $c^2+1\gt 1$ and $|c^2-1|\lt 1$. So our fourth derivative has absolute value $\lt 24\cdot \dfrac{1}{10}$. We have harly given away anything in making this simple estimate. Now divide by $24$, multiply by $\left(\dfrac{1}{10}\right)^4$. We find that the absolute value of the error is $\lt 10^{-5}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/252360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
$\lim\limits_{x\rightarrow+\infty}\left[(x^3+x^2+1)^{1/3}- (x^2+x+1)^{0.5}\right] = -1/6$? Every method I use seem to get me to something to the extent of $0/0$, stuff I can't work with. Wolfram Alpha claims the answer to this is $-1/6$ but they offer no step by step solution.
Would appreciate any tips and help.
| Since $$\alpha-\beta=\frac{\alpha^6-\beta^6}{\alpha^5+...+\beta^5}$$
we have that
\begin{gather}\lim\limits_{x\rightarrow+\infty}({x^3+x^2+1})^{1/3}-{(x^2+x+1)}^{1/2}=\lim\limits_{x\rightarrow+\infty}x[{(1+1/x+1/{x^2})}^{1/3}-{(1+1/x+1/{x^2})^{1/2}}]=
\lim\limits_{x\rightarrow+\infty}x\frac{(1+1/x+1/{x^2})^2-(1+1/x+1/{x^2})^3}{(1+1/x+1/{x^2})^{5/3}+...+{(1+1/x+1/{x^2})}^{5/2}}
\end{gather}
The denominator goes to $6$ and the numerator:
\begin{gather}\lim\limits_{x\rightarrow+\infty}x[(1+1/x+1/{x^2})^2-(1+1/x+1/{x^2})^3]=\\
\lim\limits_{x\rightarrow+\infty}x(1+1/x+1/{x^2})^2[-1/x-1/{x^2}]=\\
\lim\limits_{x\rightarrow+\infty}-(1+1/x+1/{x^2})^2[1+1/x]=-1
\end{gather}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/254496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
5 linear equations in 5 unknowns I need an example of 5 linearly independent equations with 5 variables. How can I write such a equation set. As an example:
0 0 0 0 1 | -4
0 0 1 -1 0 | 3
1 0 0 0 1 | 2
0 7 -8 0 0 | -14
2 0 0 0 0 | 2
But this is wrong. I need to build a correct one.
| Begin with a 5x5 random matrix, with positive diagonal. For example,
$$
A=\begin{pmatrix}
1&0&-3&2&4\\
5&6&7&8&-9\\
1&1&1&1&1\\
0&0&0&1&0\\
2&-3&2&-3&4
\end{pmatrix}
$$
Then compute the absolute row sums (i.e. sum up the absolute values of the entries in each row):
\begin{align}
1+0+3+2+4=10\\
5+6+7+8+9=35\\
1+1+1+1+1=5\\
0+0+0+1+0=1\\
2+3+2+3+4=14.
\end{align}
Add these row sums to the diagonal of $A$:
$$
A=\begin{pmatrix}
1+10&0&-3&2&4\\
5&6+35&7&8&-9\\
1&1&1+5&1&1\\
0&0&0&1+1&0\\
2&-3&2&-3&4+14
\end{pmatrix}.
$$
The rows of $A$ are then linearly independent. If you want to make $A$ looks more random, further scramble the rows or columns of $A$. Then form a random but nice looking solution vector $x$ (so that you may control the look of the solution). Compute $b=Ax$ and let the coefficients of $b$ be the numbers on the RHS of the equations. The resulting system will always have a unique solution, which is your nice $x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/255750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
What is the greatest value... The sum of the first three terms of a geometric progression is 21, and the sum of their squares is 1281. What is the greatest value can be the sum of the cubes?
| Let $a,ar,ar^2$ be the three terms
So, $a+ar+ar^2=21$
So, $(a)^2+(ar)^2+(ar^2)^2=1281\implies a(r^2+r+1)a(r^2-r+1)=1281$
On division, $a-ar+ar^2=\frac{1281}{21}=61$
So, $ar=-20,a+ar^2=41$
$a^3+(ar)^3+(ar^2)^3=a^3+(ar^2)^3+(-20)^3$
$=(a+ar^2)^3-3a\cdot ar^2(a+ar^2)+(-20)^3$
$=(41)^3-3(-20)^2(41)+(-20)^3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/256119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Scale modified Bessel functions to then unscale later So I have some variables $\,x_{1},\, x_{2},\, \nu\, =\, 12.654,\, 13.487,\, 0\,$ and the following function:
$\dfrac{(x_{1}\cdot(-BesselK(\nu,x_{1}\cdot125))\cdot BesselI(\nu,x_{2}\cdot125))-(x_{2}\cdot BesselK(\nu+1,x_{1}\cdot 125)\cdot BesselI(\nu+1,x_{2}\cdot 125))}{\dfrac{-1}{125\cdot x_{1}}}$
What's my problem? Well even though the final output is a reasonable number, the output of the Bessel functions are either super-huge or super-small and the calculation requires quad-precision (which is very sluggish). I need to scale down the value going into the Bessel function but I need to be able to get the end result to be the same as if I used quad-precision.
Thank you.
| If $125 x_1$ and $125 x_2$ are fairly large compared to $\nu^2$, you might want to use the asymptotic forms of these Bessel functions:
$$ \eqalign{K_\nu(x) &\sim \sqrt{\frac{\pi}{2x}} e^{-x} \left(1+{\frac {4\,{\nu}^{2}-1^2}{1!\; 8x}}+{\frac {(4 \nu^2-1^2)(4 \nu^2-3^2)}{2!\; (8x)^{2}}}\right.\cr &\left.+{\frac {(4 \nu^2 - 1^2)(4 \nu^2 - 3^2)(4 \nu^2 - 5^2)}{3! \;(8x)^{3}}}+{\frac {(4 \nu^2 - 1^2)(4 \nu^2 - 3^2)(4 \nu^2 - 5^2)(4 \nu^2 - 7^2)}{4! \;(8x)^{4}}}+\ldots
\right)\cr
I_\nu(x) &\sim \frac{e^x}{\sqrt{2\pi x}}\left(1-{\frac {4\,{\nu}^{2}-1^2}{1!\; 8x}}+{\frac {(4 \nu^2-1^2)(4 \nu^2-3^2)}{2!\; (8x)^{2}}}\right.\cr &\left.-{\frac {(4 \nu^2 - 1^2)(4 \nu^2 - 3^2)(4 \nu^2 - 5^2)}{3! \;(8x)^{3}}}+{\frac {(4 \nu^2 - 1^2)(4 \nu^2 - 3^2)(4 \nu^2 - 5^2)(4 \nu^2 - 7^2)}{4! \;(8x)^{4}}}-\ldots
\right)\cr}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Bound of a function $f_n=\frac{x^2}{x^2+(1-nx)^2}$ Let $\displaystyle f_n=\frac{x^2}{x^2+(1-nx)^2}$ where ($0\le x\le 1,\; n=1,2,3,...$)
Then $|f_n(x)| \le M$.
Find this $M$.
The answer is $1$.
Without any restriction of n, how can we find that bound?
| Let $\frac{x^2}{x^2+(1-nx)^2}=y$
or $x^2(y+yn^2-1)-2nyx+y=0$
As $x$ is real, $(2ny)^2\ge 4y(y+yn^2-1)$
$\implies y^2-y\le 0\implies y(y-1)\le 0\implies 0\le y\le 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/257626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving square root of a square is the same as absolute value Lets say I have a function defined as $f(x) = \sqrt {x^2}$. Common knowledge of square roots tells you to simplify to $f(x) = x$ (we'll call that $g(x)$) which may be the same problem, but it isn't the same equation. For example, say I put $-1$ into them:
$\begin{align}
f(x) &= \sqrt {x^2} \\
f(-1) &= \sqrt {(-1)^2} \\
f(-1) &= \sqrt {1} \\
f(-1) &= 1
\end{align}$
$\begin{align}
g(x) &= x \\
g(-1) &= -1
\end{align}$
thereby, we conclude that $f(x)$ and $g(x)$ do not produce the same results even though they are mathematically the same. This is also shown when we try to graph the functions:
$y = \sqrt {x^2}$:
$y = x$:
$y = \mid x \mid$:
From this, we can see that given $f(x) = \sqrt {x^2}$, when simplified is not the same as $f(x) = x$. So, is there any way to prove that $y = \sqrt {x^2}$ is not the same as $y = x$ for negative values, but is infact the same as $y = \mid x \mid$?
| It is the definition of square root of a number. The square root is defined in the sense that $s(x^2) = \sqrt{ x^2 } = |x|$ for all real $x$. Thus, the domain is the real numbers and the codomain is the non-negative real numbers. The reason it is defined this way is to make sure that $s$ is a function. Assume for a minute that $s(x^2) = x$, then:
$$\sqrt{(-5)^2} = -5, \qquad \sqrt{5^2} = 5$$
But we know that $\sqrt{(-5)^2} = \sqrt{25} = \sqrt{5^2}$. Thus we see that $s(25) = -5, 5$. And thus $s$ is not a function. To keep it as a function, we have to 'sacrifice' and say that $s(x^2) \neq x$. Rather, $s(x^2) = |x|$. This will be consistent with the definition of a function.
Since it is a definition, it cannot be proven. The problem is that many think that $\sqrt {x^2} = x$ because we study positive numbers before we study negative numbers, which is understandable, because I used to make that mistake all the time.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Finding the sum of all solutions $2x + 3y = n$ has exactly $2011$ non-negative integral solutions. Determine the SUM of the possible values of $n$.
| The integer solutions to $2x+3y=1$ are given by
$$\left\{\begin{align*}
&x=-1+3k\\
&y=1-2k
\end{align*}\right.$$
for $k\in\Bbb Z$, so the integer solutions to $2x+3y=n$ are given by
$$\left\{\begin{align*}
&x=-n+3k\\
&y=n-2k
\end{align*}\right.$$
for $k\in\Bbb Z$. Clearly $n>0$, so $x\ge 0$ iff $3k\ge n$ iff $k\ge\frac{n}3$, and $y\ge 0$ iff $2k\le n$ iff $k\le\frac{n}2$. Thus, for a given $n>0$ you get one solution to $2x+3y=n$ in non-negative integers for each integer $k$ satisfying $$\frac{n}3\le k\le\frac{n}2\;.\tag{1}$$
$(1)$ is equivalent to $$\left\lceil\frac{n}3\right\rceil\le k\le\left\lfloor\frac{n}2\right\rfloor\;,$$
so there are $\left\lfloor\frac{n}2\right\rfloor-\left\lceil\frac{n}3\right\rceil+1$ solutions. Thus, you’re looking for the sum of all positive integers $n$ such that
$$\left\lfloor\frac{n}2\right\rfloor-\left\lceil\frac{n}3\right\rceil+1=2011$$ or, equivalently,
$$f(n)\triangleq\left\lfloor\frac{n}2\right\rfloor-\left\lceil\frac{n}3\right\rceil=2010\;.$$
Clearly $$f(n)\approx\frac{n}2-\frac{n}3=\frac{n}6\;,$$ so the desired values of $n$ will be near $6\cdot2010=12060$. One of these values is $12060$, since $f(12060)=6030-4020=2010$. Now note that
$$\begin{align*}
f(k+6)&=\left\lfloor\frac{k+6}2\right\rfloor-\left\lceil\frac{k+6}3\right\rceil\\
&=\left(\left\lfloor\frac{k}2\right\rfloor+3\right)-\left(\left\lceil\frac{k}3\right\rceil+2\right)\\
&=\left\lfloor\frac{k}2\right\rfloor-\left\lceil\frac{k}3\right\rceil+1\\
&=f(k)+1
\end{align*}$$
for all $k$, so each arithmetic progression with constant difference $6$ contains exactly one term $n$ such that $f(n)=2010$, and there are therefore exactly $6$ such values of $n$. From here you should have no trouble finding the $6$ values of $n$ such that $f(n)=2010$ and their sum.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to make a matrix a magic square? Suppose I have a matrix
$$\begin{pmatrix}
* & 3 & 6\\
5 & * & 5\\
4 & 7 & *
\end{pmatrix}$$
How can I find the three numbers on the main diagonal such that the sum of the numbers on every row and every column is equal (i.e., it's a magic square). Thank you.
| Let's say we put in the numbers $a$, $b$, and $c$:
$$\begin{pmatrix}
a & 3 & 6\\
5 & b & 5\\
4 & 7 & c
\end{pmatrix}$$
The sums of the rows are $a+9$, $b+10$, and $c+11$. The sums of the columns are the same (indeed, if that were not the case, it'd be impossible to make it into a magic square). Thus, we want the numbers $a$, $b$, and $c$ to satisfy
$$a+9=b+10=c+11.$$
Such triples of numbers are precisely those of the form $a=x$, $b=x-1$, and $c=x-2$ for some number $x$. But we also want the diagonals to add up to the same value; thus, we want
$$a+b+c=4+b+6$$
$$x+(x-1)+(x-2)=4+(x-1)+6$$
$$3x-3=x+9$$
$$x=6$$
Thus, the unique entries we can put in to make the matrix a magic square are
$$\begin{pmatrix}
\fbox{6} & 3 & 6\\
5 & \fbox{5} & 5\\
4 & 7 & \fbox{4}
\end{pmatrix}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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upper and lower bounds of a complex expression How do I prove that $$\sqrt{\frac{7}{2}}\leq |1+z|+|1-z+z^2|\leq 3\sqrt{\frac{7}{6}}$$ for all complex numbers $|z|=1$?
I don't really know how to grapple with it.
P.. I am extremely sorry, the condition should actually be $|z|=1$ and it was previously incorrectly stated as $|z|\leq 1$.
| Let $z=\cos 2t+i\sin 2t,$
$|1+z|=|1+\cos 2t+i\sin 2t|=|2\cos^2t+i2\sin t\cos t|=|2\cos t||\cos t+i\sin t|=|2\cos t|$
$|1-z+z^2|=|1-(\cos 2t+i\sin 2t)+(\cos 2t+i\sin 2t)^2$
$=|1-(\cos 2t+i\sin 2t)+(\cos 4t+i\sin 4t)|$
$=|2\cos^22t-\cos 2t+i(2\sin2t\cos2t-\sin2t)|$
$=|(2\cos2t-1)(\cos 2t+i\sin2t)|=|(2\cos2t-1)||(\cos 2t+i\sin2t)|$
$=|(2\cos2t-1)|=|4\cos^2t-3|$
So, $|1+z|+|1-z+z^2|=2|\cos t|+|4\cos^2t-3|$
Now $|\cos t|= \cos t$ if $\cos t\ge 0$ else $-\cos t$
and $|4\cos^2t-3|=-(4\cos^2t-3)$ if $\cos^2t\le\frac34$ ie., if $-\frac{\sqrt3}2\le \cos t \le \frac{\sqrt3}2,$ else $=4\cos^2t-3$
(i)If $\cos t\le-\frac{\sqrt3}2, |1+z|+|1-z+z^2|=-2\cos t+4\cos^2t-3$
(ii)If $-\frac{\sqrt3}2< \cos t <0, |1+z|+|1-z+z^2|=-2\cos t-(4\cos^2t-3)$
(iii)If $0\le \cos t \le \frac{\sqrt3}2, |1+z|+|1-z+z^2|=2\cos t-(4\cos^2t-3)$
(iv)If $\cos t > \frac{\sqrt3}2 , |1+z|+|1-z+z^2|=2\cos t+(4\cos^2t-3)$
In (i), $\cos^2t\ge\frac34, -\cos t\ge \frac{\sqrt3}2\implies |1+z|+|1-z+z^2|\ge 4\cdot\frac34+2\cdot\frac{\sqrt3}2-3={\sqrt3}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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asymptotics of the expected number of edges of a random acyclic digraph with indegree and outdegree at most one A recent discussion, which may be found here, examined the problem of counting the number of acyclic digraphs on $n$ labelled nodes and having $k$ edges and indegree and outdegree at most one. It was established that the bivariate mixed generating function of this class $\mathcal{G}$ of graphs on $n$ nodes and with $k$ edges is
$$ G(z, u) = \exp \left(\frac{z}{1-uz} \right).$$
This immediately implies that the expected number of edges of a random graph from $\mathcal{G}$ is
$$\epsilon_n = [z^n]\left. \frac{d}{du} G(z,u) \right|_{u=1}.$$
Evaluating this quantity we obtain
$$ \left. \exp \left(\frac{z}{1-uz} \right) z (-1)\frac{1}{(1-uz)^2} (-z)\right|_{u=1}
= \left. \exp \left(\frac{z}{1-uz} \right) \frac{z^2}{(1-uz)^2} \right|_{u=1}
= \exp \left(\frac{z}{1-z}\right) \left(\frac{z}{1-z}\right)^2$$
Continuing with this calculation we find
$$ \epsilon_n = [z^n] \sum_{m\ge 0} \frac{1}{m!} \left(\frac{z}{1-z}\right)^{m+2} =
\sum_{m=0}^{n-2} \frac{1}{m!} [z^n] \left(\frac{z}{1-z}\right)^{m+2} =
\sum_{m=0}^{n-2} \frac{1}{m!} [z^{n-m-2}] \left(\frac{1}{1-z}\right)^{m+2} =
\sum_{m=0}^{n-2} \frac{1}{m!} \binom{n-m-2+m+1}{m+1} =
\sum_{m=0}^{n-2} \frac{1}{m!} \binom{n-1}{m+1}.$$
This closed form is actually quite nice, but it does not answer the question that is the most obvious one for this problem, namely Is there an asymptotic expansion of $\epsilon_n$ and if yes, what is the first term?
| We can expand this calculation to the variance in addition to the expected value. To do so we compute $$E[X(X-1)] = \frac{ [z^n] \left. \left(\frac{d}{du}\right)^2 G(z, u) \right|_{u=1}}{[z^n] G(z, 1)}$$ where $X$ is the RV representing the number of edges.
Now $$\left. \left(\frac{d}{du}\right)^2 G(z, u) \right|_{u=1} =
\left.\left( \frac{d}{du}\right)\exp\left(\frac{z}{1-uz}\right) \left(\frac{z}{1-uz}\right)^2 \right|_{u=1} =
\left.\exp\left(\frac{z}{1-uz}\right) \left(\frac{z}{1-uz}\right)^4 +
2 \exp\left(\frac{z}{1-uz}\right) \left(\frac{z}{1-uz}\right)^3 \right|_{u=1} =
\exp\left(\frac{z}{1-z}\right) \left(\frac{z}{1-z}\right)^4 +
2 \exp\left(\frac{z}{1-z}\right) \left(\frac{z}{1-z}\right)^3$$
Now expand the two terms in turn.
Left term,
$$[z^n] \exp\left(\frac{z}{1-z}\right) \left(\frac{z}{1-z}\right)^4 =
[z^n] \sum_{m\ge 0} \frac{1}{m!} \left( \frac{z}{1-z}\right)^{m+4} =
[z^n] \sum_{m=0}^{n-4} \frac{1}{m!} \left( \frac{z}{1-z}\right)^{m+4} $$ or $$
\sum_{m=0}^{n-4} \frac{1}{m!} [z^{n-m-4}] \left( \frac{1}{1-z}\right)^{m+4} =
\sum_{m=0}^{n-4} \frac{1}{m!} \binom{n-m-4+m+3}{m+3} =
\sum_{m=0}^{n-4} \frac{1}{m!} \binom{n-1}{m+3}.$$
Right term,
$$2 [z^n] \exp\left(\frac{z}{1-z}\right) \left(\frac{z}{1-z}\right)^3 =
2 [z^n] \sum_{m\ge 0} \frac{1}{m!} \left( \frac{z}{1-z}\right)^{m+3} =
2 [z^n] \sum_{m=0}^{n-3} \frac{1}{m!} \left( \frac{z}{1-z}\right)^{m+3} $$ or $$
2 \sum_{m=0}^{n-3} \frac{1}{m!} [z^{n-m-3}] \left( \frac{1}{1-z}\right)^{m+3} =
2 \sum_{m=0}^{n-3} \frac{1}{m!} \binom{n-m-3+m+2}{m+2} =
2 \sum_{m=0}^{n-3} \frac{1}{m!} \binom{n-1}{m+2}.$$
It follows that $$E[X(X-1)] =
\frac{\sum_{m=0}^{n-4} \frac{1}{m!} \binom{n-1}{m+3} + 2 \sum_{m=0}^{n-3} \frac{1}{m!} \binom{n-1}{m+2}}{\sum_{m=1}^n \frac{1}{m!} \binom{n-1}{m-1}}.$$
The question now becomes, what is the asymptotic expansion of this quotient of sums? That would put the variance within reach, given that $E[X^2] = E[X(X-1)] + E[X].$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I find maximum and and minimum values of $ f(x,y)=xye^{-(x+y)}$? How can I find maximum and and minimum values of $ f(x,y)=xye^{-(x+y)}$ in the region $(x-1)^2+(y-1)^2 \leq4$ ?
| First look for critical points of $f$ in the interior $(x-1)^2 + (y-1)^2 < 4$. Do this by comparing the partial derivatives of $f$ to zero. Next search for critical values on the boundary $(x-1)^2 + (y-1)^2 = 4$ using Lagrange multipliers. Compute $f$ in these points and, by putting everything together, you will find the maximum and minimum of $f$ in the region.
EDIT: Computing the derivatives:
$$f_x(x,y) = x(1-y)e^{-(x+y)} \\\\
f_y(x,y) = y(1-x)e^{-(x+y)}$$
Comparing both to zero we get:
$$x(1-y) = 0 \\\\
y(1-x) = 0$$
Simultaneous solutions are $(0,0)$ and $(1,1)$, which are both within the interior (but as we will see these points don't matter anyway). Computing $f$ at these points gives $0$ and $e^{-2} \approx 0.135$ correspondingly.
Now to the boundary. It is given by $g(x,y) = 4$ where $g(x,y) = (x-1)^2 + (y-1)^2$ and the partial derivatives:
$$g_x(x,y) = 2(x-1) \\\\ g_y(x,y) = 2(y-1)$$
Using the Lagrange multipliers method, we have to solve
$$g(x,y) = 4 \\\\ f_x(x,y) - \lambda g_x(x,y) = 0 \\\\ f_y(x,y) - \lambda g_y(x,y) = 0$$
to find the critical points on the boundary. Substituting:
$$(x-1)^2 + (y-1)^2 = 4 \\\\
y(1-x)e^{-(x+y)} - \lambda 2(x-1) = 0\\\\
x(1-y)e^{-(x+y)} - \lambda 2(y-1) = 0$$
Assuming $x,y \neq 1$ we can divide the last two equations by $(x-1)$ and $(y-1)$ correspondingly, so they become:
$$ ye^{-(x+y)} = - 2\lambda \\\\
xe^{-(x+y)} = - 2\lambda $$
which implies $x = y$. Substituting into $g(x,y) = 4$, we have:
$$2(x-1)^2 = 4 \quad \implies \quad x = 1 \pm \sqrt{2}$$
Corresponding values of $f$ are
$$f(1 - \sqrt{2}, 1 - \sqrt{2}) = (3 - 2 \sqrt{2}) e^{-2 + 2\sqrt{2}} \approx 0.393 \\\\
f(1 + \sqrt{2}, 1 + \sqrt{2}) = (3 + 2 \sqrt{2}) e^{-2 - 2\sqrt{2}} \approx 0.0466$$
For the case where $x = 1$ or $y = 1$ we plug them directly into $g(x,y) = 4$ to get
$$x = 1 \quad \implies \quad (y-1)^2 = 4 \quad \implies \quad y = -1, 3 \\\\
y = 1 \quad \implies \quad (x-1)^2 = 4 \quad \implies \quad x = -1, 3$$
Corresponding values of $f$ are
$$f(-1,1) = f(1,-1) = -e^0 = -1 \\\\
f(3,1) = f(1,3) = 3e^{-4} \approx 0.0549$$
We conclude that the minimum value of $-1$ is attained at $(-1,1)$ and $(1,-1)$, and the maximum value of $(3 - 2 \sqrt{2}) e^{-2 + 2\sqrt{2}}$ is attained at $(1 - \sqrt{2}, 1 - \sqrt{2})$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/265105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
How to find the square root of a permutation Observe that
$$\begin{pmatrix}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 1 & 5 & 3
\end{pmatrix}
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 \\
2 & 4 & 1 & 5 & 3
\end{pmatrix} =
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 \\
4 & 5 & 2 & 3 & 1
\end{pmatrix},$$
so $\begin{pmatrix}
1 & 2 & 3 & 4 & 5 \\
4 & 5 & 2 & 3 & 1
\end{pmatrix}$ is the square of a permutation. More generally, how do I check if a permutation is square and how can I find its root?
| Any permutation can be written as a product of disjoint cycles $\sigma = c_1\cdot\ldots\cdot c_n$. Because disjoint cycles commute, $\sigma^2 = c_1^2\cdot\ldots\cdot c_n^2$.
So a square permutation is one that consists of a product of disjoint square cycles. Now when is a cycle square?
If $c=(i_1 i_2\ldots i_k)$, then $c^2$ takes $i_1$ to $i_3$, $i_2$ to $i_4$, and so on. In other words:
If $k$ is odd, $c^2$ is another $k$-cycle, which means that every $k$-cycle is a square.
If $k$ is even, then $c^2=(i_1i_3\ldots i_{k-1})(i_2i_4\ldots i_{k})$.
Therefore: a permutation is a square if and only if the number of cycles of any even length in its disjoint cycle decomposition is even, and the algorithm you can use to find its root is as following:
*
*Find the roots of all odd cycles, $\sqrt{c} = (i_1 i_{(k+3)/2} i_2 i_{(k+5)/2} \ldots i_k i_{(k+1)/2})$.
*Take pairs $c=(i_1\ldots i_k),d=(j_1\ldots j_k)$ of the even cycles of the same length and construct $\sqrt{c,d} = (i_1j_1\ldots i_kj_k)$.
Your permutation is an example of an odd cycle with a root that matches what would follow this method.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/266569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
"answer_count": 2,
"answer_id": 1
} |
Integral of $1/\sqrt{x^2 - a^2}$ where $a > 0$, why does $a$ has to be greater than $0$? I know how to solve the integral (set $x = a\sec(\theta)$ then $dx = a\sec(\theta)\tan(\theta)\,d\theta$ where $0 < \theta < \pi/2$ in order to have a one-to-one function), anyway, the problem specifies that $a > 0$ but I don't see how this changes anything or affects the substitution because $x = a\sec(\theta)$ still remains a one-to-one function, correct? I may be wrong but by seeing the graph of asec(theta) it appears that the sign of a does not change the fact that sec(theta) is one to one. Does it affect something else?
Then we could also use the substitution of $x = a\cosh(t)$ then $dx = a\sinh(t)\,dt$ but the problem then specifies that $x > 0$. Why these restrictions?
| If $x=a\sec \theta, dx=a\sec \theta\tan\theta \,d\theta $
$\sqrt{x^2-a^2}=\sqrt{a^2(\sec^2\theta-1)}=|a\tan \theta|=|a||\tan \theta|$
As $0< \theta< \frac\pi 2, |\tan \theta|=\tan \theta$
So, $\sqrt{x^2-a^2}=|a|\tan \theta$ if $0< \theta< \frac\pi 2$
$$\int \frac{dx}{\sqrt{x^2-a^2}}=\frac{a\sec\theta\tan\theta d\theta}{|a|\tan \theta}$$
=sign$(a)\int\sec\theta d\theta=$sign$(a)\ln|\sec\theta+\tan\theta|+C$
$=\ln|\frac xa+\frac{\sqrt{x^2-a^2}}{|a|}|+C$
If $\operatorname{sign}(a)>0, \int \frac{dx}{\sqrt{x^2-a^2}}=\ln|\frac xa+\frac{\sqrt{x^2-a^2}}a|+C=\ln|x+\sqrt{x^2-a^2}|+C-\ln a$
If $\operatorname{sign}(a)<0, \int \frac{dx}{\sqrt{x^2-a^2}}=-\ln|\frac xa-\frac{\sqrt{x^2-a^2}}a|+C=-\ln|x-\sqrt{x^2-a^2}|+C+\ln a$
But, $\ln|x+\sqrt{x^2-a^2}|+\ln|x-\sqrt{x^2-a^2}|=\ln|x^2-(x^2-a^2)|=\ln |a^2|$ which is constant.
So, $\operatorname{sign}(a)<0, \int \frac{dx}{\sqrt{x^2-a^2}}=\ln|x+\sqrt{x^2-a^2}|-\ln |a^2|+C=\ln|x+\sqrt{x^2-a^2}|+C'$ where $C'=C-\ln |a^2|$ is also a constant.
So, the value of the integration in the above two cases differ only by constant,hence the sign of $a$ does not matter.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/268042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Demonstrate: $ \sqrt{2}=\lim_{n\rightarrow\infty}{\sum_{i=0}^n{\frac{\left(-1\right)^i\left(-\frac{1}{2}\right)_i}{i!}}} $ Demonstrate that $\sqrt2$ can be expressed as:
$$ \sqrt{2}=\lim_{n\rightarrow\infty}{\sum_{i=0}^n{\frac{\left(-1\right)^i\left(-\frac{1}{2}\right)_i}{i!}}} $$ Where $\left(z\right)_i$ is the Pochhammer symbol
$\left(z\right)_i=z(z+1)(z+2)...(z+i-1); (z)_0=1$
This is a nice problem, just wanted to share it.
| If we Taylor expand $f(x)=\sqrt{1+x}$, around $x=0$, we get
$$
\sqrt{1+x}=\sum_{k=0}^\infty \frac{(-1)^k\big(-\frac{1}{2}\big)_k}{k!}x^k=\sum_{k=0}^\infty
a_kx^k, \tag{1}
$$
and this series converges for $|x|<1$, as
$$
|a_k|=\frac{1}{k!}\cdot\frac{1}{2}\left(\frac{1}{2}\frac{3}{2}\cdots\frac{k-1-\frac{1}{2}}{2}\right)=\frac{1}{2k}\prod_{j=1}^{k-1}\frac{j-1/2}{j}<1.
$$
Also
$$
\left|\frac{a_{k+1}}{a_{k}}\right|=\frac{k}{k+1}\cdot
\frac{k-\frac{1}{2}}{k}=\frac{k-\frac{1}{2}}{k+1}=1-\frac{3}{2(k+1)},
$$
and hence
$$
\lim_{k\to\infty}k\left(\left|\frac{a_{k+1}}{a_{k}}\right|-1\right)=-\frac{3}{2.}
$$
which, due to Raabe's test, converges absolutely, even for $|x|=1$, and therefore, the powerseries $(1)$ defines a continuous function for $x\in [-1,1]$. Thus, $(1)$ holds even for $x=1$, i.e.,
$$
\sqrt{2}=\sqrt{1+1}=\sum_{k=0}^\infty \frac{(-1)^k\big(-\frac{1}{2}\big)_k}{k!}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/268957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Can the integral $\int \frac{6x-1}{3x^3-4} \, dx$ solved with a simple solution? I got the integral
$$\int \frac{6x-1}{3x^3-4} \, dx$$
as an exercise for exam preparation. I tried to get closer to the result with third binomial formula, but could not get anywhere. So I tried to find the solution with wolframalpha, but this one seems overly complex. Is there a simple solution for this integral?
| $$\int \frac{6x-1}{3x^3-4} \, dx$$
$$\frac{1}{4}\int \frac{6x-1}{\frac{3}{4}x^3-1} \, dx$$
$\frac{3}{4}x^3=u^3$
$\sqrt[3]{\frac{3}{4}}x=u$
$\sqrt[3]{\frac{4}{3}}u=x$
$\sqrt[3]{\frac{4}{3}}du=dx$
$$\frac{\sqrt[3]{\frac{4}{3}}}{4}\int \frac{6\sqrt[3]{\frac{4}{3}}u-1}{u^3-1} \, du=\frac{\sqrt[3]{\frac{4}{3}}}{4}\int \frac{A}{u-1} +\frac{B}{u-\epsilon}+\frac{C}{u-\epsilon^2}\, du$$
where $\epsilon=e^{\frac{2\pi i}{3} }=-\frac{1}{2}+i\frac{\sqrt3}{2}$
$\epsilon^2=e^{\frac{4\pi i}{3} }=-\frac{1}{2}-i\frac{\sqrt3}{2}$
$$\frac{6\sqrt[3]{\frac{4}{3}}u-1}{u^3-1}=\frac{A}{u-1} +\frac{B}{u-\epsilon}+\frac{C}{u-\epsilon^2}$$
$f(u)=6\sqrt[3]{\frac{4}{3}}u-1$
$A=f(1)=6\sqrt[3]{\frac{4}{3}}-1$
$B=f(\epsilon)=6\epsilon\sqrt[3]{\frac{4}{3}}-1$
$C=f(\epsilon^2)=6\epsilon^2\sqrt[3]{\frac{4}{3}}-1$
$$\frac{\sqrt[3]{\frac{4}{3}}}{4}\int \frac{A}{u-1} +\frac{B}{u-\epsilon}+\frac{C}{u-\epsilon^2}\, du=$$
$$\frac{A\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-1} \, du+\frac{B\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-\epsilon}\, du+\frac{C\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-\epsilon^2}\, du=$$
$$\frac{A\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-1} \, du+\frac{B\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-\epsilon}\, du+\frac{C\sqrt[3]{\frac{4}{3}}}{4}\int \frac{1}{u-\epsilon^2}\, du=$$
$$\frac{A\sqrt[3]{\frac{4}{3}}}{4}\ln {(u-1)} +\frac{B\sqrt[3]{\frac{4}{3}}}{4}\ln {(u-\epsilon)}+\frac{C\sqrt[3]{\frac{4}{3}}}{4}\ln {(u-\epsilon^2)}+c=$$
$$\frac{A\sqrt[3]{\frac{4}{3}}}{4}\ln {(\sqrt[3]{\frac{3}{4}}x-1)} +\frac{B\sqrt[3]{\frac{4}{3}}}{4}\ln {(\sqrt[3]{\frac{3}{4}}x-\epsilon)}+\frac{C\sqrt[3]{\frac{4}{3}}}{4}\ln {(\sqrt[3]{\frac{3}{4}}x-\epsilon^2)}+c=$$
After that you will need to do some calculations and also need to do transform of the complex value of $\ln$ couple to $arctan$
Use the formula for that transform $\arctan x = \frac{1}{2}i\left(\ln\left(1-i\,x\right)-\ln\left(1+i\,x\right)\right) $
Ref :http://en.wikipedia.org/wiki/Inverse_trigonometric_functions
The most easiest way for me is that way. Maybe someone else can offer quicker way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/269113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Help in manipulating Integrals I try to express : $\displaystyle 1+2\sum _{ k=1 }^n \cos(2k\theta ) $
as : $\dfrac { \sin\left( \theta +2\theta n \right) }{ \sin\left( \theta \right) } $
I tried to use the exponential function :
$$I=\sum _{ k=1 }^n \cos(2k\theta ) ;\quad J=\sum _{ k=1 }^n \sin(2k\theta) $$
$I+i\cdot J=\displaystyle\sum _{ k=1 }^n e^{ 2ki\theta }$ and then get only the real part, but without success.
I also tried multiplying by $\dfrac {\sin(p)}{\sin(p)}$ (with $p$ an unknown variable I would determine later), but still, no success.
If someone could help, or identify the problem (I know it's a pretty famous one)
Thank you
| There are many equivalent ways to proceed. Below are a couple of equivalent methods.
First method:
Let $$S = \sum_{k=1}^n \cos(2k \theta)$$
We then have
$$S \sin(\theta) = \sum_{k=1}^n \sin(\theta) \cos(2k \theta)$$
Now recall that
$$\sin(A) \cos(B) = \dfrac{\sin(A+B) - \sin(B-A)}2$$
Hence, we get that
$$S \sin(\theta) = \sum_{k=1}^n \dfrac{\left(\sin((2k+1) \theta) - \sin((2k-1) \theta) \right)}2$$
Now you can see a telescopic cancellation
\begin{align}
2S \sin(\theta) & = \sin((2n+1) \theta) - \sin((2n-1) \theta)\\
& {}+ \sin((2n-1) \theta) - \sin((2n-3) \theta)\\
& {}+ \sin((2n-3) \theta) - \sin((2n-5) \theta) + \cdots\\
& {}+ \sin(3 \theta) - \sin(\theta)
\end{align}
Hence, we get that
$$2S \sin(\theta) = \sin((2n+1) \theta) - \sin(\theta) = 2 \sin(n \theta) \cos((n+1) \theta)$$
Hence, we get that
$$S = \dfrac{\sin(n \theta) \cos((n+1) \theta)}{\sin(\theta)}$$
Second method:
Another way is to look at the real part of $e^{2i k \theta}$ and sum the geometric progression.
$$S = \text{Re} \left( \sum_{k=1}^n e^{2ik \theta} \right)$$
$$\sum_{k=1}^n e^{2ik \theta} = e^{2i \theta} \left(\dfrac{e^{2in \theta} - 1}{e^{2i \theta} - 1}\right) = e^{2i \theta} \left(\dfrac{1-e^{2in \theta}}{1-e^{2i \theta}}\right)$$
$$1-e^{2i \theta} = 1 - \cos(2 \theta) - i \sin(2 \theta) = 2\sin^2(\theta) - 2i \sin(\theta) \cos(\theta) = -2i \sin(\theta) e^{i \theta}$$
$$1-e^{2i n\theta} = -2i \sin(n\theta) e^{i n\theta}$$
$$\sum_{k=1}^n e^{2ik \theta} = e^{2i \theta} \times \dfrac{-2i \sin(n\theta) e^{i n\theta}}{-2i \sin(\theta) e^{i \theta}} = \dfrac{\sin(n \theta)}{\sin(\theta)} e^{i(n+1) \theta}$$
Hence, $$\text{Re} \left( \sum_{k=1}^n e^{2ik \theta} \right) = \dfrac{\sin(n \theta) \cos((n+1) \theta)}{\sin(\theta)}$$
Third method:
You can also use induction directly to prove $$\sum_{k=1}^n \cos(2k \theta) = \dfrac{\sin(n \theta) \cos((n+1) \theta)}{\sin(\theta)}$$
For $n=1$, we have $$\cos(2 \theta) = \dfrac{\sin(\theta)\cos(2 \theta)}{\sin(\theta)}$$
Assume it is true for $n=m$ i.e. $$\sum_{k=1}^m \cos(2k \theta) = \dfrac{\sin(m \theta) \cos((m+1) \theta)}{\sin(\theta)}$$
Now at the induction step, we have that
\begin{align}
\sum_{k=1}^{m+1} \cos(2k \theta) & = \sum_{k=1}^{m} \cos(2k \theta) + \cos(2(m+1) \theta)\\
& = \underbrace{\dfrac{\sin(m \theta) \cos((m+1) \theta)}{\sin(\theta)}}_{\text{From induction hypothesis}} + \cos(2(m+1) \theta)\\
& = \dfrac{\sin(m \theta) \cos((m+1) \theta) + \cos(2(m+1) \theta) \sin(\theta)}{\sin(\theta)}
\end{align}
$$\sin(\theta) \cos(2(m+1) \theta) = \dfrac{\sin((2m+3)\theta) - \sin((2m+1)\theta)}2$$
$$\sin(m \theta) \cos((m+1) \theta) = \dfrac{\sin((2m+1)\theta) - \sin(\theta)}2$$
Hence,$$\sin(\theta) \cos(2(m+1) \theta) + \sin(m\theta) \cos((m+1) \theta) = \dfrac{\sin((2m+3)\theta) - \sin(\theta)}2 = \sin((m+1)\theta) \cos((m+2) \theta)$$
$$\sum_{k=1}^{m+1} \cos(2k \theta) = \dfrac{\sin((m+1)\theta) \cos((m+2) \theta)}{\sin(\theta)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/271694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
For $n∈ N$, determine the real part of $(1 + i\sqrt{3})^{n}$ For $n∈ N$, determine the real part of $(1 + i\sqrt{3})^{n}$.
I just can't find the regularity within it. Thanks.
| Use De Moivre's formula
\begin{align}
\left[ 1+i\cdot \sqrt{3} \right]^n =
&
2^n \left[ \frac{1}{2}+i\cdot \frac{\sqrt{3}}{2} \right]^n
&
\\
=
&
2^n\cdot \left[\cos\left(\frac{\pi}{3}\right)+i\cdot \sin\left(\frac{\pi}{3}\right) \right]^n
&
\\
=
&
2^n\cdot \left[\cos\left(n\cdot\frac{\pi}{3}\right)+i\cdot \sin\left(n\cdot\frac{\pi}{3}\right) \right]
&
\\
\end{align}
Let's $n= 3\cdot [2\cdot k]+r$. We have $r\in\{0,1,2,3,4,5\}$ and
\begin{align}
\left[ 1+i\cdot \sqrt{3} \right]^n =
&
2^n\cdot \left[\cos\left(r\cdot\frac{\pi}{3}\right)+i\cdot \sin\left(r\cdot\frac{\pi}{3}\right) \right]
\\
\end{align}
Then we have,
$$
Re\Big(\left[ 1+i\cdot \sqrt{3} \right]^{3(2k)+r}\Big)=
\begin{cases}
2^n\cdot \left(+\frac{1}{1} \right) & \mbox{ if } r=0 \\
2^n\cdot \left(+\frac{1}{2} \right) & \mbox{ if } r=1,5 \\
2^n\cdot \left(-\frac{1}{2} \right) & \mbox{ if } r=2,4 \\
2^n\cdot \left(-\frac{1}{1} \right) & \mbox{ if } r=3 \\
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/273099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Solve this inequality $\prod_{i=1}^{50} \frac {2i-1}{2i} < \frac {1}{10}$ Prove that
$ \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \frac{7}{8}\cdot \frac{9}{10}\cdot \frac{11}{12}\cdot \frac{13}{14}...\cdot \frac{91}{92}\cdot \frac{93}{94}\cdot \frac{95}{96}\frac{97}{98}\cdot \frac{99}{100} <\frac{1}{10}$
| Multiply the numerators of the RHS and LHS by $49!2^{49}$. This converts the numerator on the LHS to $99!$. We then have $$\frac{99!}{2\cdot4\cdot8...\cdot100}<\frac{49!2^{49}}{10}$$ Next multiply the denominators on the RHS and LHS by $$\frac{100!}{50!2^{50}}$$ The formula to convert the above multiplications of even integers to factorials is $$\frac{(2n)!}{n!2^n}$$ where $n$ is the number of terms in the multiplication sequence. See OEIS A001147 for formula and sequence of multipliers to convert the sequential multiplications of even integers to factorials. This converts the denominator on the LHS to $100!$ to give $$\frac{99!}{100!}<\frac{2^{99}50!49!}{100!10}$$ The final result is $$\frac{1}{100}<\frac{19807040628566084398385987584}{1576427258524440520856445269625}$$ or $$.01< 0.012564513...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/274753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
calculate $\lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos2x}{\cos x-\sin x}$ $$\lim_{x\rightarrow\frac{\pi}{4}}\frac{\cos(2x)}{\cos(x)-\sin(x)}=\lim_{x\rightarrow\frac{\pi}{4}}\frac{2\cos^{2}(x)-1}{\cos(x)-\sqrt{1-\cos^{2}(x)}}$$
$$t=\cos(x)$$
$$\lim_{x\rightarrow\frac{\pi}{4}}\frac{2t^{2}-1}{t-\sqrt{1-t^{2}}}=\lim_{x\rightarrow\frac{\pi}{4}}\frac{(2t^{2}-1)(t+\sqrt{t^{2}-1})}{2t^{2}-1}$$
$$\lim_{x\rightarrow\frac{\pi}{4}}(2t^{2}-1)(t-\sqrt{t^{2}-1})=3(\sqrt{2}-1)$$
Please help me to find an error. Correct answer is $\sqrt{2}$. Thanks
| writing $\cos 2x = \cos^2 x - \sin^2x $ and expanding $a^2-b^2$ and cancelling $\cos x - \sin x$ you get $\cos x + \sin x$. Putting value $x \to \frac{\pi}{4}$ we get $ \frac{1}{\sqrt 2} + \frac{1}{\sqrt 2} = \sqrt 2$.
EDIT::
Looks like you've been writing $\sqrt{t^2-1}$ for $\sqrt{1 -t^2}$ and also there is substitution so you must be putting $ t \to \frac{1}{\sqrt 2}$
$\underset{x\rightarrow\frac{\Pi}{4}}{\lim}\frac{2t^{2}-1}{t-\sqrt{1-t^{2}}}=\underset{x\rightarrow\frac{\Pi}{4}}{\lim}\frac{(2t^{2}-1)(t+\sqrt{t^{2}-1})}{2t^{2}-1}$
After this you should get
$\lim_{t \to \frac{1}{\sqrt 2}} t + \sqrt{1-t^2}$
this is essentially same as multiplying by $\cos x + \sin x $ and cancelling the $\cos 2x$ up there, so substitution quite unnecessary though.
| {
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"url": "https://math.stackexchange.com/questions/275215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality. $\frac{1}{\sqrt{x^2+yz+3}}+\frac{1}{\sqrt{y^2+zx+3}}+\frac{1}{\sqrt{z^2+xy+3}} \geq 1$ Prove that :
$$\frac{1}{\sqrt{x^2+yz+3}}+\frac{1}{\sqrt{y^2+zx+3}}+\frac{1}{\sqrt{z^2+xy+3}} \geq 1$$ if $x^2+y^2+z^2 \leq9$.
I try to apply Cauchy-Buniakowski and I obtaine the followin:
$$\sum_{x,y,z}{\frac{1}{\sqrt{x^2+yz+3}}}\cdot \sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}\geq 9$$
So I have to prove that : $$\displaystyle\frac{9}{\sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}} \geq 1$$ if $x^2+y^2+z^2 \leq9$.
Another trying :
$$\left(\sum_{x,y,z}{\sqrt{x^2+yz+3}}\right) \leq \sqrt{\left(\sum{x^2+yz+3}\right)(1+1+1)} $$ so we have to prove that:
$$\frac{9}{\sqrt{\left(\sum{x^2+yz+3}\right)(1+1+1)}} \geq 1$$ hence:
$$3(x^2+y^2+z^2+xy+yz+zx+9) \leq 81$$ or
$$(x^2+y^2+z^2+xy+yz+zx+9) \leq 27$$ or
$$x^2+y^2+z^2+xy+yz+zx \leq 18$$
$$x^2+y^2+z^2+xy+yz+zx \leq 2\left(x^2+y^2+z^2\right) \leq 2 \cdot 9 =18.$$
Yes, it is ok :)
thanks :)
| I try to apply Cauchy-Buniakowski and I obtaine the followin:
$$\sum_{x,y,z}{\frac{1}{\sqrt{x^2+yz+3}}}\cdot \sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}\geq 9$$
So I have to prove that : $$\displaystyle\frac{9}{\sum_{x,y,z}{\left(\sqrt{x^2+yz+3}\right)}} \geq 1$$ if $x^2+y^2+z^2 \leq9$.
$$\left(\sum_{x,y,z}{\sqrt{x^2+yz+3}}\right) \leq \sqrt{\left(\sum{x^2+yz+3}\right)(1+1+1)} $$ so we have to prove that:
$$\frac{9}{\sqrt{\left(\sum{x^2+yz+3}\right)(1+1+1)}} \geq 1$$ hence:
$$3(x^2+y^2+z^2+xy+yz+zx+9) \leq 81$$ or
$$(x^2+y^2+z^2+xy+yz+zx+9) \leq 27$$ or
$$x^2+y^2+z^2+xy+yz+zx \leq 18$$
$$x^2+y^2+z^2+xy+yz+zx \leq 2\left(x^2+y^2+z^2\right) \leq 2 \cdot 9 =18.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/275326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding all solutions of the equation $(x^2 + y^2 + z^2 − 1)^2 + (x + y + z − 3)^2 = 0$. Find all solutions of the equation
$(x^2 + y^2 + z^2 − 1)^2 + (x + y + z − 3)^2 = 0$.
not getting any clue.can somebody help me please.thanks.
| I assume we are looking for real solutions. Then our equation is equivalent to the system $x^2+y^2+z^2=1$, $x+y+z=3$.
Note that because $x^2+y^2+z^2=1$, the absolute values of $x$, $y$, and $z$ are $\le 1$, and they are not all $1$.
Thus $|x+y+z|\le |x|+|y|+|z| \lt 3$.
| {
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How do we solve the system of equations?18-1-2013 Giải hệ phương trình:
(Editor's note: This translates to "Solve a system of equations:")
$$ \begin{cases}\sqrt[4]{x}\left(\dfrac{1}{4}+\dfrac{2\sqrt{x}+\sqrt{y}}{x+y}\right)=2 \\[8pt] \sqrt[4]{y}\left(\dfrac{1}{4}-\dfrac{2\sqrt{x}+\sqrt{y}}{x+y}\right) =1\end{cases} $$
| Divide the first equation by $\sqrt[4]{x}$ and the second by $\sqrt[4]{y}$. Then adding and subtracting the two resulting equations gives us the new pair of simultaneous equations:
$$
\frac{2}{\sqrt[4]{x}}+\frac{1}{\sqrt[4]{y}}=\frac{1}{2} \\
\frac{2}{\sqrt[4]{x}}-\frac{1}{\sqrt[4]{y}}=2\frac{2\sqrt{x}+\sqrt{y}}{x+y} \, .
$$
Multiplying these two equations together, we have:
$$
\frac{4}{\sqrt{x}}-\frac{1}{\sqrt{y}}=\frac{2\sqrt{x}+\sqrt{y}}{x+y} \, .
$$
Clearing denominators yields:
$$
(4\sqrt{y}-\sqrt{x})(x+y)=2x\sqrt{y}+y\sqrt{x}
$$
which after some algebra can be reduced to
$$
(x+2y)(2\sqrt{y}-\sqrt{x})=0 \, .
$$
So either $x=-2y$ or $\sqrt{x}=2\sqrt{y}$. If $x$ and $y$ are positive and real the first is clearly impossible and the second is equivalent to $x=4y$. (If they're not positive and real, we have to worry more about branch cuts than I really want to.) Now, if $x=4y$ we have
$$
\frac{\sqrt{2}}{\sqrt[4]{y}}+\frac{1}{\sqrt[4]{y}}=\frac{1}{2} \, ,
$$
from the top equation in this answer. So $\sqrt[4]{y}=2(1+\sqrt{2})$, yielding the solution
$$
x=64(1+\sqrt{2})^4 \\
y=16(1+\sqrt{2})^4 \, ,
$$
which upon expanding is precisely what Robert Israel got from Maple in the other answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Two different solutions to integral Given the very simple integral
\begin{equation}
\int -\frac{1}{2x} dx
\end{equation}
The obvious solution is
\begin{equation}
\int -\frac{1}{2x} dx = -\frac{1}{2} \int \frac{1}{x} dx = -\frac{1}{2} \ln{|x|} + C
\end{equation}
However, by the following integration rule
\begin{equation}
\int \frac{1}{ax + b} dx = \frac{1}{a} \ln{|ax + b|} + C
\end{equation}
the following solution is obtained
\begin{equation}
\int -\frac{1}{2x} dx = -\frac{1}{2}\ln{|-2x|} + C
\end{equation}
Why are these solutions different? Which is correct?
The second solution can be simplified
\begin{equation}
-\frac{1}{2}\ln{|-2x|} + C = -\frac{1}{2}\ln{|-2|} -\frac{1}{2}\ln{|x|} + C= -\ln{\frac{1}{\sqrt{2}}} - \frac{1}{2}\ln{|x|} + C
\end{equation}
but they still differ.
| They are both correct: they differ only in terms of their constants of integration.
$$\begin{align}
-\frac{1}{2}\ln|-2x| + \color{red}{C} &= -\frac{1}{2}(\ln 2 + \ln |x|) + \color{red}{ C} \\&= -\frac{1}{2}\ln |x| + \color{red}{(C - \frac{1}{2} \ln 2)}\\ &= -\frac{1}{2}\ln |x| + \color{red}{K}
\end{align}
$$
$C \neq K\;$ but $\;C, K\;$ are constants nonetheless!
TIP: One can always check two apparently different solutions to an integral by differentiating each of them; if the respective derivatives are equal to the original integrand, then you can conclude that the two apparently different solutions are, in fact, solutions that differ only in their constants of integration.
EDIT: You were almost there (in obtaining the adjusted constant of integration), but : $$-\frac{1}{2}\ln{|-2|} + C\; \ne \;-\ln{\frac{1}{\sqrt{2}}} + C$$
Rather, since $\;|-2| \;= \;2,\;$ we have
$$-\frac{1}{2}\ln{|-2|}+C \; =\; -\frac{1}{2}\ln 2 + C\;=\; +\ln{\frac{1}{\sqrt{2}}} + C = K$$
The important thing to note is that all of $K$ is a constant term.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the limit of $\frac{e^{6x}-2e^{3x} + 1}{x^2}$, as $x \rightarrow 0$? I am to calculate $\frac{e^{6x}-2e^{3x} + 1}{x^2}$ when $x$ goes towards $0$.
I find that
$$\frac{e^{6x}-2e^{3x} + 1}{x^2} = \frac{(e^{3x}-1)^2}{x^2} = \left(\frac{e^{3x}-1}{x}\right)^2$$
$$\left(\frac{e^{3x}-1}{x}\right)^2 \rightarrow 1^2$$
but according to the answer in the book I am incorrect. It agrees with me halfway through, but ends with
$$\frac{(e^{3x}-1)^2}{x^2} = 9\left(\frac{e^{3x}-1}{3x}\right)^2 \rightarrow 9 \times 1^2$$
While this is correct mathematically, why would it be $3$ and $9$ instead of for example $4$ and $16$ or, as in my case, $1$ and $1$? I don't see the relevance of adding the $3$ and $3^2$.
| Another approach to write $$\frac{e^{3x}-1}{x}=\frac{e^{3x}-e^{3\cdot 0}}{x}=\frac{f(x)-f(0)}{x}$$
Where $f(x)=e^{3x}$ then$$\lim_{x\to 0} \frac{f(x)-f(0)}{x} = f'(0)=3e^{0}=3$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A trigonometric integral inequality $$\displaystyle\frac{4\sin 1}{\pi }<\int_{0}^{1}{\frac{\cos x}{\sqrt{1-{{x}^{2}}}}}\text{d}x\le \frac{\pi }{2}\ln \left( \sec 1+\tan 1 \right)$$
I've got no ideas for this one.
| Numerically, the integral evaluates to $\approx 1.201969715$. Your RHS is $1.926096588$ which seems like a fairly bad estimate. Here's one way to do better:
Partial integration gives
\begin{align}
\int_0^1 \frac{\cos x}{\sqrt{1-x^2}}\,dx &= \left[ \arcsin x \cos x \right]_0^1 + \int_0^1 \sin x \arcsin x \,dx \\
&\le \frac{\pi}{2} \cos 1 + \sin 1 \int_0^1 \arcsin x\,dx \\
&= \frac{\pi}2 \cos 1 + \sin 1\left(\frac{\pi}2 -1\right) \approx 1.329013425
\end{align}
(we use that $\sin x$ is increasing on $[0,1]$).
For the other direction, your LHS is $\approx 1.071394134$. The same partial integration trick and the inequality $\arcsin x \ge x$ gives
\begin{align}
\int_0^1 \frac{\cos x}{\sqrt{1-x^2}}\,dx &= \left[ \arcsin x \cos x \right]_0^1 + \int_0^1 \sin x \arcsin x \,dx \\
&\ge \frac{\pi}{2} \cos 1 + \int_0^1 x \sin x\,dx \\
&= \frac{\pi}2 \cos 1 + \sin 1 - \cos 1 \approx 1.149873556
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $a$ and $b$ in the given cubic polynomial Find $a$ and $b$ such that $x+1$ and $x+2$ are factors of the polynomials $x^3+ax^2-bx+10$.
Here I am not sure that how can I obtain the value of $a$ and $b$, I tried to multiply $x+1$ and $x+2$ to obtain a quadratic equation by which I divided the obtained quadratic polynomial with the above given cubic polynomial, But it didn't worked. How can I overcome my answer.
| Let $x+c$ be the 3rd/last factor as the expression is of degree three.
So, $$(x+c)(x+1)(x+2)=x^3+ax^2-bx+10$$
But, $$(x+c)(x+1)(x+2)=(x+c)(x^2+3x+2)=x^3+x^2(c+3)+x(3c+2)+2c$$
So, $$x^3+x^2(c+3)+x(3c+2)+2c=x^3+ax^2-bx+10$$
Comparing the coefficients of the different powers of $x,$
$2c=10\implies c=5$
$a=c+3=5+3=8$
$-b=3c+2=3\cdot5+2=17,b=-17$
| {
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Product of sums of square is a sum of squares. Given $a,b,c,d \in \mathbb{Z}$, there is $x,y$ such that
$$(a^2 + b^2)(c^2 + d^2) = x^2 + y^2$$
One can show this by considering the complex number $a + bi$ and $c+ di$, using complex properties to deduce that $x = ac - bd, y = ad + bc$ is a solution. However, given that either $a^2, b^2$ are distinct or $c^2,d^2$ are distinct, then how can one can find nonzero solutions $x,y$?
Sorry, I am assuming $a,b,c,d$ are nonzero.
| If $a=b=0$ or $c=d=0$, obviously we cannot do it.
If one of (say) $a$ or $b$ is $0$, and one of $c$ or $d$ is $0$, there will be situations in which we cannot do it, for instance when $a=c=3$, $b=d=0$.
If one of $a$ or $b$ is $0$, but the other is not, and neither $c$ nor $d=0$, there is no difficulty.
Now examine the case all of $a$, $b$, $c$, $d$ are non-zero. Without loss of generality we may assume they are all positive.
Use the formulas
$$(a^2+b^2)(c^2+d^2)=(ac+bd)^2+(ad-bc)^2=(ad+bc)^2+(ac-bd)^2.$$
We need only check that at least one of $ad-bc$ and $ac-bd$ is non-zero.
Suppose to the contrary that $ad=bc$ and $ac=bd$. Then (i) $a^2cd=b^2cd$ and (ii) $c^2ab=d^2ab$. Since none of $a$, $b$, $c$, $d$ is $0$, by cancellation we get $a^2=b^2$ and $c^2=d^2$.
| {
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Is a monotone differentiable function continuously differentiable? If $f:\mathbb{R}^+\to\mathbb{R}$ is monotonic and differentiable, does it follow that $f$ is continuously differentiable?
(This question arose from discussion here: problem on continuous and differentiable function)
| or every $x\in (-1,1).$ Define,
$$ f(x) =100000+ 100x+ x^2\sin\frac{1}{x} \qquad\text{with $f(0)=100000$}$$
which is differentiable, and
$$ f'(x) = 100+ 2x\sin\frac{1}{x}-\cos\frac{1}{x} \qquad\text{with $f'(0)=100$}$$
$f'$ is not continuous a $x=0$. Also $f'\ge 0$ indeed,
$$ f'(x)=100+2x\sin\frac{1}{x}-\cos\frac{1}{x} \ge 100+ 2x\sin\frac{1}{x}-1$$
Since $|\sin a| \le |a|$ and $-1\le-\cos a\le 1$ then $|2x\sin\frac{1}{x}|\le 2$ i.e $$2x\sin\frac{1}{x}\ge -2 $$
therefore
$$ f'(x)=100+2x\sin\frac{1}{x}-\cos\frac{1}{x} \ge 100-2-1= 97>0$$ for every $x\in \mathbb R$. so $f $ is increasing and $f'$ is not continuous at $x=0$.
Also $|x^2\sin\frac{1}{x}|\le 1$ for all $x\in (-1,1)$ so that
$$f(x) =100000+ 100x+ x^2\sin\frac{1}{x} > 100000-100 -1 >0$$ for every $x\in (-1,1).$
| {
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Asymptotic expansion of $ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $ I'm trying to compute the asymptotic expansion of
$$ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $$
Here is what I've done:
Change of variable $$ t= \tan x $$
$$ I_n = \int_0^1 \frac{t^n \mathrm dt}{1+t^2} = \int_0^1 \frac{(1-t)^n \mathrm dt}{t^2-2t+2} $$
Change of variable
$$ t=\frac{x}{n}$$
$$ I_n = \frac{1}{2n}\int_0^n \frac{\left(1-\frac{x}{n}\right)^n \mathrm dx}{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)}$$
Taylor expansions:
$$ \left(1-\frac{x}{n}\right)^n = e^{-x} \left(1-\frac{x^2}{2n}+\frac{3x^4-8x^3}{24n^2}+\mathcal{O} \left(\frac{1}{n^3} \right) \right)$$
$$ \frac{1}{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)} = 1+\frac{x}{n}+\frac{x^2}{2n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) $$
$$ \frac{\left(1-\frac{x}{n}\right)^n }{1-\left(\frac{x}{n}-\frac{x^2}{2n^2} \right)}=e^{-x} \left( 1+\frac{x}{n}+\frac{x^2}{2n^2}-\frac{x^2}{2n}-\frac{x^3}{2n^2}+\frac{3x^4-8x^3}{24n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) \right)$$
So
$$ I_n = \frac{1}{2n} \int_0^n e^{-x} \left( 1+\frac{x}{n}+\frac{x^2}{2n^2}-\frac{x^2}{2n}-\frac{x^3}{2n^2}+\frac{3x^4-8x^3}{24n^2} + \mathcal{O} \left(\frac{1}{n^3} \right) \right) \mathrm dx $$
$$ I_n = \frac{1}{2n} \left(1+\frac{1}{n}+\frac{1}{2n^2}\times 2-\frac{1}{2n}\times 2 - \frac{1}{2n^2} \times 6 + \frac{1}{8n^2} \times 24 - \frac{1}{3n^2} \times 6+ \mathcal{O} \left(\frac{1}{n^3} \right) \right) $$
$$ I_n = \frac{1}{2n}-\frac{1}{2n^3}+\mathcal{O} \left(\frac{1}{n^4} \right)$$
For example Wolfram gives:
$$ 1-1000^2+ 2\times1000^3\int_0^{\pi/4} \tan(x)^{1000} \mathrm dx \approx 4.9\times 10^{-6}$$
I'm quite sure of my work, I would just like to know if everything is correct!
| You get that
$$I_n = \int_0^1 \dfrac{t^n}{1+t^2} dt = \int_0^1 t^n \left(\sum_{k=0}^{\infty} (-t^2)^k\right) dt = \sum_{k=0}^{\infty}(-1)^k \int_0^1 t^{n+2k} dt$$
Now $$\int_0^1 t^{n+2k} dt = \dfrac1{n+1+2k}$$
Hence,
$$I_n = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{n+1+2k}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Calculus integration problem: $\int \sin^5 (x) \cos^2 (x)\,dx$ What's the integration of $$\int \sin^5 (x) \cos^2 (x)\,dx?$$
| Using trig identities, you can show that:
$$\sin ^5(x) \cos ^2(x)=\frac{5 \sin (x)}{64}+\frac{1}{64} \sin (3
x)-\frac{3}{64} \sin (5 x)+\frac{1}{64} \sin (7 x)$$
To do this, first use the "Power-reduction formulas" to reduce to get:
$$\sin^5(x)=\frac{10 \sin x - 5 \sin 3 x+ \sin 5 x}{16}$$
$$\cos^2(x)=\frac{1 + \cos (2 x)}{2}$$
And then use:
$$\cos (2 x) \sin (nx) = {{\sin((n+2)x) - \sin((n-2)x)} \over 2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to go by this, to solve for X $$\begin{bmatrix}3 & -2\\-1 & -3\end{bmatrix}X + \begin{bmatrix}1 & 5\\0 & 7\end{bmatrix} = \begin{bmatrix}-5 & 8\\-1 & 3\end{bmatrix}X$$
Not sure how I am supposed to go at this ? Do I move the + to the other side first ? Can someone let me know how to tackle this, thanks.
| Taking the $x$ to LHS, we get $$\begin{pmatrix}3+5&-2-8\\-1+1&-3-3\end{pmatrix}X=-\begin{pmatrix}1&5\\0&7\end{pmatrix}$$ $$\Rightarrow \begin{pmatrix}8&-10\\0&-6\end{pmatrix}X=-\begin{pmatrix}1&5\\0&7\end{pmatrix}$$
Now note that $\begin{pmatrix}8&-10\\0&-6\end{pmatrix}$ is invertible and let A be its inverse. (To find the inverse you may look here). So premultiplying with $A$, we get $$X=-A\begin{pmatrix}1&5\\0&7\end{pmatrix}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I integrate $\int x\sqrt{A^2 - x^2} \mathrm dx$ $$ \int x\sqrt{A^2 - x^2} \mathrm dx $$
I never dealt with 2 variables on integration, what should I do here?
| Alternative way is using the trigonometric substitution $x=A\sin{t}.$
Then $$dx=A\cos{t}\;dt,\\
{A^2 - x^2}=A^2-A^2{\sin^2{t}}=A^2{\cos^2{t}},$$ therefore
$$\int x\sqrt{A^2 - x^2}\; dx = \int {A\sin{t}\,A{\cos{t}}}A\cos{t}\;dt =\\
= -A^3 \int {\cos^2{t}\;d(\cos{t})}=-\frac{A^3}{3}\cos^3{t}+C= -\frac{A^3}{3} \left(1-\sin^2{t} \right)^{\tfrac{3}{2}}+C =\\
=-\frac{A^3}{3}\left({1-\frac{x^2}{A^2}}\right)^{\tfrac{3}{2}}+C=-\frac{1}{3}\left({A^2-{x^2}}\right)^{\tfrac{3}{2}}+C.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $3^{2^{k}} - 1$ is divisible by $2^{k+2}$? Prove that $3^{2^{k}} - 1$ is divisible by $2^{k+2}$ for $k \ge 1$?
Suppose $p,q,r$ are positive integers satisfying $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}<1$, prove that $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}\le \dfrac{41}{42} $ ?
| $$(1)$$
$$3^{2^{k}}-1=(3^2-1)(3^2+1)(3^4+1)\cdots (3^{2^{k-1}}+1)=2^3 \cdot\displaystyle\prod_{m=1}^{k-1} (3^{2^m}+1)$$
Since each $3^{2^m}+1$ is even, $2$ divides each of these $k-1$ terms.
It follows that $2^3\cdot 2^{k-1}=2^{k+2}$ divides $\displaystyle 2^3\cdot \prod_{m=1}^{k-1} (3^{2^m}+1)=3^{2^k}-1$
$$(2)$$
We wish to maximise $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r},$ without loss of generality assume $p\geq q\geq r$
Since we are dealing with three unit fractions summing to $<1$ we want $r\leq 3$. When $r=3$ the optimum solution is clearly $q=3,r=4$ which agrees with the desired inequality.
Now let $r=2$ and we have $\dfrac{1}{p}+\dfrac{1}{q}<\dfrac{1}{2}$
Since we are now dealing with two unit fractions summing to $<1/2$ we want $q\leq 4$. When $q=4$ the optimum solution is $p=5$ which again, agrees with the inequality.
Now let $q=3$ and we have $\dfrac{1}{p}<\dfrac{1}{6}$
Here the maximum is when $p=7,$ and this last case in fact gives the greatest of all values we investigated, hence:
$$\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}\leq \dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{7}=\dfrac{41}{42}$$
| {
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Find a basis for a solution set of a linear system I'm trying hard with this exercise, but it is breaking my back.
Find a basis for the solution set of the given homogeneous linear system
$3x_1+x_2+x_3=0$
$6x_1+2x_2+2x_3=0$
$-9x_1-3x_2-3x_3=0$
I do what I know I need to do. First I get the solution set of the system by reducing like this:
$\begin{pmatrix}
3 & 1 & 1 \\
6 & 2 & 2 \\
-9 & -3 & -3 \end{pmatrix} \leadsto \begin{pmatrix}
3 & 1 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \end{pmatrix} \leadsto\begin{pmatrix}
1 & 1/3 & 1/3 \\
0 & 0 & 0 \\
0 & 0 & 0 \end{pmatrix}$
So I know $\vec x = \begin{bmatrix}
x_1\\
x_2\\
x_3\end{bmatrix} = \begin{bmatrix}
1-\frac{1}{3}r-\frac{1}{3}s\\
r\\
s\end{bmatrix}$
That being the general solution.
Now, giving the values for $r$ and $s$ according to the standard vectors $i$, $j$
$\vec x = \begin{bmatrix}
x_1\\
x_2\\
x_3\end{bmatrix} = \begin{bmatrix}
1-\frac{1}{3}r-\frac{1}{3}s\\
r\\
s\end{bmatrix} = r \begin{bmatrix}
\frac{2}{3}\\
1\\
0\end{bmatrix} + s\begin{bmatrix}
\frac{2}{3}\\
0\\
1\end{bmatrix}$
From my results, the basis will be:
$ ( \begin{bmatrix}
\frac{2}{3}\\
1\\
0\end{bmatrix}, \begin{bmatrix}
\frac{2}{3}\\
0\\
1\end{bmatrix})$
But instead, the book answer (I'm self-studying )is:
$ ( \begin{bmatrix}
-1\\
3\\
0\end{bmatrix}, \begin{bmatrix}
-1\\
0\\
3\end{bmatrix})$
Any idea on what I'm doing wrong? Thank you :)
| Check your expression for $\vec x$. I think it should be $\vec x=\begin{bmatrix} x_1 \\x_2\\x_3\end{bmatrix}=\begin{bmatrix} -\frac{1}{3}r-\frac{1}{3}s \\r \\s \end{bmatrix}=r\begin{bmatrix} -\frac{1}{3} \\1 \\0 \end{bmatrix}+s\begin{bmatrix} -\frac{1}{3} \\0 \\1 \end{bmatrix}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/296600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Prove $p^n\ne a^3+b^3$ where $p\in\mathbb{P}\ge5$, $n,a,b\in\mathbb{N}$
Given prime $p\geq5$, prove that $p^n$ can't be represented as sum of two positive cubes for any $n\in\mathbb{N}$. What about $p=2$ and $p=3$?
| Let $p\ge 5$, and suppose that there is an $n\ge 1$ such that $p^n$ can be represented as $a^3+b^3$, where $a$ and $b$ are positive. Then there is a smallest such $n$.
From the factorization $a^3+b^3=(a+b)(a^2-ab+b^2)$, we conclude that $a+b$ and $a^2-ab+b^2$ are each powers of $p$. Neither can be the $0$-th power, since for positive $a$ and $b$, we have $a^2-ab+b^2=1$ only if $a=b=1$.
But $3ab=(a+b)^2-(a^2-ab+b^2)$. So $p$ divides $3ab$, and since $p\gt 3$, we conclude that $p$ divides one of $a$ or $b$. But since $p$ divides $a+b$, it follows that $p$ divides both $a$ and $b$. Let $a=pc$, $b=pd$. Then $c^3+d^3=p^{n-3}$.
This contradicts the minimality of $n$ unless $n-3=0$. But $n-3=0$ is impossible, since $c$ and $d$ are positive.
The cases $p=2$ and $p=3$ are different, since $1^3+1^3=2$ and $1^3+2^3=3^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/298303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find max and min of $2\sqrt{x+y} - \sqrt x - \sqrt y$ Find $max$ and $min$ of $M=2\sqrt{x+y} - \sqrt{x} - \sqrt {y}$ with $x + y = constant$
I tried to call $c = \sqrt{x+y} \Rightarrow M= 2c - (\sqrt x+\sqrt {y})\ge (2-\sqrt{2})c$ ?
| Since $x+y$ is a constant, to maximize $2\sqrt{x+y}-\sqrt{x}-\sqrt{y}$, we need to minimize what we take away from $2\sqrt{x+y}$, so we need to minimize $\sqrt{x}+\sqrt{y}$.
Similarly, to minimize $2\sqrt{x+y}-\sqrt{x}-\sqrt{y}$, we need to maximize $\sqrt{x}+\sqrt{y}$.
So let us first minimize $\sqrt{x}+\sqrt{y}$. This is equivalent to minimizing its square, which is $x+y+2\sqrt{x}\sqrt{y}$. But $x+y$ is fixed. What is the minimum possible value of $2\sqrt{x}\sqrt{y}$?
Next we maximize $\sqrt{x}+\sqrt{y}$. Equivalently, we maximize its square $x+y+2\sqrt{x}\sqrt{y}$. So we want to maximize $4xy$.
Note that $(x+y)^2=(x-y)^2 +4xy$. So $4xy=(x+y)^2-(x-y)^2$. Given that $x+y$ is fixed, what does this say about the maximum value of $4xy$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How many solutions to prime = $2 b^2 c^2 + 2 c^2 a^2 + 2 a^2 b^2 - a^4 - b^4 - c^4$ Let $a,b,c$ be integers, no sign restriction.
Let $p$ be a given prime.
How to find the number of solutions to $p = 2 b^2 c^2 + 2 c^2 a^2 + 2 a^2 b^2 - a^4 - b^4 - c^4$ ?
Note, from Heron's formula,
$$ 2 b^2 c^2 + 2 c^2 a^2 + 2 a^2 b^2 - a^4 - b^4 - c^4 = (a + b + c) ( b + c - a) ( c + a - b) (a + b - c). $$
Another question is ; let $w$ be a positive integer. Let $f(w)$ be the number of primes of type prime = $2 b^2 c^2 + 2 c^2 a^2 + 2 a^2 b^2 - a^4 - b^4 - c^4$ below $w$. How does the function $f(w)$ behave ? How fast does it grow ? Are those primes of type $A$ $mod$ $B$ for some integers $A$ and $B$ ?
How to deal with this ?
Can this be solved without computing the class number ?
| It seems the only prime representable this way is $p=3$, which has eight such representations by using $a,b,c=\pm 1$ (independent sign choices). [Note that $-3$, which is prime, has no such representations.] I ran a check of all choices of $a,b,c$ between $-100$ and $100$ just to make sure of this.
It wouldn't be hard to prove this, since three of the four factors must be $\pm 1$ and the fourth $\pm p$ for some prime $p$. By looking at all 32 ways of setting three of the four factors to $\pm 1$, and solving each such system, one gets (at most) 32 cases to look at, and for each one computes the value of the fourth factor.
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that $\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3 \cdot 4} +\ldots + \frac{1}{n(n+1)} = \frac{n}{n+1}$ How would we go about proving that $$\frac{1}{1\cdot 2} + \frac{1}{2\cdot 3} + \frac{1}{3 \cdot 4} +\ldots +\frac{1}{n(n+1)} = \frac{n}{n+1}$$
| Hint: $\frac{1}{n}-\frac{1}{n+1}=\frac{1}{n(n+1)}$
| {
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Construct a convergent series of positive terms with $\displaystyle\limsup_{n\to\infty} {a_{n+1}\over{a_n}}=\infty$ Construct a convergent series of positive terms with $\displaystyle\limsup_{n\to\infty} {a_{n+1}\over{a_n}}=\infty$
My thoughts:
By the theorem: Suppose $a_n\ge0$ for all $n$, and let $l=\displaystyle\limsup{\sqrt[n]{a_n}}$. If $l<1$, then $\displaystyle\sum_{n=1}^\infty a_n$ converges; and if $l>1$, then $\displaystyle\sum_{n=1}^\infty a_n$ diverges.
Thus, we know $\displaystyle\sum_{n=1}^\infty {a_{n+1}\over{a_n}}$ diverges
I guess it might works, $(x_n)_{n=1}^{+\infty}$, $x_{n+1} = x_n^2 + x_n$ for all $n\ge1$..
| Try the sequence $( \frac{1}{2})^3 + ( \frac{1}{2})^2+ ...+ ( \frac{1}{k})^3+ (\frac{1}{k})^2+ ...$.
Since $\frac{a_{2n+2}}{a_{2n+1}}=\frac{(\frac{1}{n+1})^2}{(\frac{1}{n+1})^3} = n+1$, we have $\limsup_n \frac{a_{n+1}}{a_n} = \infty$. The sequence is convergent since both $\sum \frac{1}{k^2}$ and $\sum \frac{1}{k^3}$ are convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/300846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Two improper log integrals Evaluate
$$\int_0^{\frac{\pi}{2}}\ln ^2(\tan x)\text{d}x$$
$$\int_0^{\frac{\pi}{2}}\ln ^2(\sin x)\text{d}x$$
| Per @julien's comment: Let $u=\tan{x}$ and transform the integral into
$$\int_0^{\infty} du \frac{\log^2{u}}{1+u^2} $$
This can be evaluated via the residue theorem. Consider the integral
$$\oint_C dz \frac{\log^3{z}}{1+z^2}$$
where $C$ is a keyhole contour about the positive real axis. The integral about the contour arcs that go to infinity and zero, respectively, go to zero in those limits. We are left with
$$\begin{align}\oint_C dz \frac{\log^3{z}}{1+z^2} &= \int_0^{\infty} du \frac{\log^3{u}}{1+u^2} - \int_0^{\infty} du \frac{(\log{u}+ i 2 \pi)^3}{1+u^2}\\ &= -i 6 \pi \int_0^{\infty} du \frac{\log^2{u}}{1+u^2} + 12 \pi^2 \int_0^{\infty} du \frac{\log{u}}{1+u^2} + i 8 \pi^3 \int_0^{\infty} du \frac{1}{1+u^2} \end{align}$$
The value of this integral is equal to $i 2 \pi$ times the sum of the residues of the poles of the integrands, which are $e^{i \pi/2}$ and $e^{i 3\pi/2}$. The residues at these poles are
$$\mathrm{Res}_{z=e^{i \pi/2}} = \frac{-i \pi^3/8}{2 i}$$
$$\mathrm{Res}_{z=e^{i 3\pi/2}} = \frac{-i 27\pi^3/8}{-2 i}$$
We may then write
$$i \left [- 6 \pi \int_0^{\infty} du \frac{\log^2{u}}{1+u^2} + 8 \pi^3 \int_0^{\infty} du \frac{1}{1+u^2} \right ] + 12 \pi^2 \int_0^{\infty} du \frac{\log{u}}{1+u^2} = i \frac{13 \pi^4}{4}$$
Now use the fact that
$$\int_0^{\infty} du \frac{1}{1+u^2} = \frac{\pi}{2}$$
and equate real and imaginary parts to get
$$\int_0^{\infty} du \frac{\log^2{u}}{1+u^2} = \frac{\pi^3}{8}$$
$$\int_0^{\infty} du \frac{\log{u}}{1+u^2} = 0$$
Therefore, the value of the stated integral is
$$\int_0^{\pi/2} dx \: \log^2{(\tan{x})} = \frac{\pi^3}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/303919",
"timestamp": "2023-03-29T00:00:00",
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Using Parseval' s theorem to evaluate a sum.. On the function $f(x)$ = $x^3$ on $(-1,1)$ find Fourier coefficients for this function and then use Parseval's Theorem to evaluate:
$$\sum_{n=1}^{\infty}\frac{1}{n^6}$$
Current work:
I have used Mathematica to find the sum, which results in $\frac{\pi^6}{945}$
Since $a_0$ and $a_n = 0$ as it is an odd function, my problem is calculating $b_n$.
I am using Mathematica and resulting in:
$$\int_a^b x^3 \sin{(n x \pi)} \,dx= \frac17$$ in which $a$ is $-1$ and $b$ is $1$, the right hand side is calculated from the average value formula, by squaring Integral$[f(x)dx]$.
This is the problem, as my result for the $b_n$ coefficients, is resulting in a nasty rational:
$$-2(-6+n^2\pi^2)cos(n\pi) / n^3\pi^3$$
where I can't seem to factor out a $\frac{1}{n^6}$ term to evaluate the sum using the theorem.
| When you integrate (using integration by parts or a table)
\begin{equation*}
b_n=\int_{-1}^{1}x^3\sin (n \pi x) \, dx = 2\int_{0}^{1}x^3 \sin (n \pi x) \, dx
\end{equation*}
you get
\begin{equation*}
-\frac{2(-1)^n}{n\pi}\left( 1-\frac{6}{n^2\pi^2} \right).
\end{equation*}
Squaring this, we obtain
\begin{equation*}
\frac{4}{n^2\pi^2}-\frac{48}{n^4\pi^4}+\frac{144}{n^6\pi^6}.
\end{equation*}
If we use the fact that
\begin{equation*}
\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}, \quad \sum_{n=1}^{\infty}\frac{1}{n^4}=\frac{\pi^4}{90}
\end{equation*}
together with Parseval's identity
\begin{equation*}
\frac{1}{1}\int_{-1}^{1}|x^3|^2 \, dx = \sum_{n=1}^{\infty}|b_n|^2\quad (\text{since } a_n=0)
\end{equation*}
the result follows.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplification of $\sqrt{(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta}$ Simplify the expression $$\sqrt{(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta}$$
I have done this way : $(1-\cos\alpha \cos\beta)^2 = 1-2\cos\alpha \cos\beta +\cos^2\alpha \cos^2\beta$
Please guide further....
| $$\begin{align*}(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta&=1+\cos^2{\alpha}\cos^2{\beta}-2\cos{\alpha}\cos{\beta}-\sin^2\alpha \sin^2\beta\\
&=1+(1-\sin^2{\alpha)\cos^2{\beta}}-2\cos{\alpha}\cos{\beta}-\sin^2\alpha \sin^2\beta\\
&=1+\cos^2{\beta}-\sin^2{\alpha\cos^2{\beta}}-2\cos{\alpha}\cos{\beta}-\sin^2\alpha \sin^2\beta\\
&=1+\cos^2{\beta}-\sin^2{\alpha}-2\cos{\alpha}\cos{\beta}\\
&=\cos^2{\beta}+\cos^2{\alpha}-2\cos{\alpha}\cos{\beta}\\
&=(\cos {\alpha}-\cos {\beta})^2\end{align*}
$$
So the answer will be $|\cos {\alpha}-\cos {\beta}|$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Verifying Trigonometric Identities: $2\cos^2x-1 = \frac{1-\tan^2x}{1+\tan^2x}$ Verify that $$ 2\cos^2x-1 = \frac{1-\tan^2x}{1+\tan^2x}$$
| Putting $z=e^{ix}$ we have
$$ \cos x = \frac{1}{2} \left( z + \frac{1}{z} \right) =
\frac{1}{2z} (z^2+1)$$
and
$$ \tan x = \frac{1}{i} \frac{z - \frac{1}{z}}{ z + \frac{1}{z} } =
\frac{1}{i} \frac{z^2-1}{z^2+1}.$$
Now $$2\cos^2 x - 1 = \frac{1}{2z^2} (z^2+1)^2 -1 $$
and $$\frac{1-\tan^2 x}{1 + \tan^2 x} =
\frac{1+\left(\frac{z^2-1}{z^2+1}\right)^2}{1-\left(\frac{z^2-1}{z^2+1}\right)^2} =
\frac{(z^2+1)^2+(z^2-1)^2}{(z^2+1)^2-(z^2-1)^2}=
\frac{2z^4 + 2}{4z^2} = \frac{z^4+1}{2z^2} \\
= \frac{z^4+2z^2+1}{2z^2} -1 = \frac{1}{2z^2} (z^2+1)^2 -1.$$
I decided to post this computation because it is algorithmic. It should be easy to see that verifying trigonometric identities is equivalent to factoring rational polynomials in one variable and that these proofs can be produced automatically by a computer.
| {
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"url": "https://math.stackexchange.com/questions/308136",
"timestamp": "2023-03-29T00:00:00",
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The number of real roots of $(x+3)^4 + (x+5)^4 = 16$ I was solving some problems and I came across this question:
Q: The number of real roots of $(x+3)^4 + (x+5)^4 = 16$ is
(a) 0 (b) 2
(c) 4 (d) none of these
Solution: put $y = x + (3+5)/2 = x+4$
the equation becomes
=> $(y-1)^4 + (y+1)^4 = 16$ ---- (i)
=> $2{y^4 + 6(y)^2 + 1 } = 16$ --------(ii)
My question is how was (i) converted to (ii)? I just couldn't get it. Please help?
| $$(y-1)^4+(y+1)^4$$
$$=y^4-\binom 41y^3+\binom 42y^2-\binom 43y+1+(y^4+\binom 41y^3+\binom 42y^2+\binom 43y+1)$$
$$=2(y^4+6y^2+1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/309895",
"timestamp": "2023-03-29T00:00:00",
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Trig substitution integral I am trying to find $$\int{ \frac {5x + 1}{x^2 + 4} dx}$$
The best approach would be to split up the fraction. According to Wolfram Alpha, the answer is $\frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\displaystyle\arctan\left(\frac x2\right)$ which seems OK, but when I try the trig substitution: $x = 2\tan\theta$, I get an answer that is slightly different but not equivalent, and I've looked at this over and over and I couldn't quite figure out what I did wrong.
$$x = 2\tan\theta$$
$$dx = 2\sec^2\theta d\theta$$
$$\int{ \frac {5x + 1}{x^2 + 4}dx} = \frac{1}{4}\int{\frac{10\tan\theta + 1}{\sec^2\theta} 2\sec^2\theta \,d\theta}$$
$$ = \frac{1}{2}\int{10 \tan\theta + 1}\space d\theta$$
$$ = 5 \ln|\sec\theta| + \frac{\theta}{2} + C$$
We know $\theta = \displaystyle\arctan\left(\frac x2\right)$ and since $\tan\theta = \displaystyle\frac{x}{2}$, we can draw a triangle to see that $\sec\theta = \displaystyle\frac{\sqrt{x^2 + 4}}{2}$.
$$5 \ln|\sec\theta| + \frac{\theta}{2} = 5\ln\left({\frac{\sqrt{x^2 + 4}}{2}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right) $$
$$= \frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right)$$
But $$\frac{5}{2}\ln\left({\frac{x^2 + 4}{4}}\right) + \frac{1}{2}\arctan\left({\frac x2}\right) \neq \frac{5}{2}\ln\left(x^2 + 4\right) + \frac{1}{2}\arctan\left(\frac x2\right)$$
There seems to be a small difference between the answer provided by Alpha and the trig substitution method, but I cannot see where I made the mistake.
| The first approach is easier in fact:
$$
I= \int \frac{(5x +1)dx}{x^2+4}=I_1 + I_2
$$
where
$$
I_1 = \int\frac{5x dx}{x^2+4}\\
I_2 = \int \frac{dx}{x^2+4}
$$
hence,
$$
I_1 = \frac{5}{2} \int \frac{2xdx}{x^2+4} = \frac{5}{2} \int \frac{d(x^2+4)} {x^2+4}=\frac{5}{2} \log( x^2 +4) +C_1
$$
In fact you can see that the numerator is a derivative of the denominator.
$$
I_2 = \int \frac{dx}{x^2+ 2^2} = \frac{\arctan(\frac{x}{a})}{a} +C_2
$$
This approach is more efficient as it does not require you to make any substitutions/
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Tangent line to a hyperbola Can anyone explain and prove the following statement?
Any tangent line to a hyperbola touches the hyperbola halfway between the points of intersection of the tangent and the asymptotes.
Thank you very much for the help!
| Derivating implicity the equation of the canonical hyperbola:
$$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\Longrightarrow \frac{2x}{a^2}dx-\frac{2y}{b^2}dy=0\Longrightarrow \frac{dy}{dx}=\frac{b^2x}{a^2y}$$
Thus, at any point $\,(x_1,y_1)\,$ on the hyperbola, the tangent line is
$$\text{I}\;\;\;y-y_1=\frac{b^2x_1}{a^2y_1}(x-x_1)\Longleftrightarrow y=\frac{b^2x_1}{a^2y_1}x-\frac{b^2}{y_1}$$
The hyperbola's asymptotes are the lines
$$\text{II}\;\;\;\;y=\pm\frac{b}{a}x$$
The intersection points between I and II are
$$\frac{b^2x_1}{a^2y_1}x-\frac{b^2}{y_1}=\pm\frac{b}{a}x\Longrightarrow\left(\frac{b^2x_1}{a^2y_1}\mp\frac{b}{a}\right)x=\frac{b^2}{y_1}\Longrightarrow$$
$$x=\frac{a^2b}{bx_1\mp ay_1}\Longrightarrow y=\pm\frac{ab^2}{bx_1\mp ay_1}$$
so the points are
$$A=\left(\frac{a^2b}{bx_1-ay_1}\;,\;\frac{ab^2}{bx_1-ay_1}\right)\;\;,\;\;B=\left(\frac{a^2b}{bx_1+ay_1}\;,\;-\frac{ab^2}{bx_1+ay_1}\right)$$
The midpoint of $\,AB\,$ is
$$\frac{1}{2}\left(\left[\frac{a^2b}{bx_1-ay_1}+\frac{a^2b}{bx_1+ay_1}\right]\;,\;\left[\frac{ab^2}{bx_1-ay_1}-\frac{ab^2}{bx_1+ay_1}\right]\right)=$$
$$\frac{1}{2}(2x_1\,,\,2y_1)=(x_1,y_1)\;\;\;\;\;\;\;\;\;\;\;\;\square$$
Note: Be sure you can prove all the steps above, and remember: $\,(x_1,y_1)\,$ is a point on the hyperbola!
| {
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A two variable Inequality Show that $(a^2+b^2)(a^4+b^4) \geq (a^3+b^3)^2$
Can we go like this : $(a^2+b^2)=(a+b)^2-2ab \& (a^4+b^4) = (a^2+b^2)^2-2a^2b^2$ and right hand side of the inequality : $[(a+b)^3-3ab(a+b)]^2$
| Multiplying out and cancelling leaves $$a^2b^4+a^4b^2\ge2a^3b^3$$ which let's write as $${a^2b^4+a^4b^2\over2}\ge a^3b^3$$ which is true by the inequality of the arithmetic and geometric means.
| {
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Solve $x^2+x+7 \equiv 0\pmod{81}$ Solve $x^2+x+7\equiv 0 \pmod{81}$
My work:
Prime factorization $81 = 9^2 = 3^4$
Test the value $x\equiv0,1,2$ for $x^2+x+7\equiv0\mod{3}$
we have $x\equiv1\mod{3}$ works.
Now life this to $\mod{3^2} = \mod{9}$
Let $x=1+3k$ for some integer$k$
$(1+3k)^2 + (1+3k) +7 \equiv0\mod{9}$
$1+6k+0+1+3k+7\equiv0\mod{9}$
$9+9k\equiv0\mod9$
$1+k\equiv0\mod9$
$k\equiv-1\mod9$
$k=-1+9m$ for some integer m
Lift again to $\mod81$
$(-1+9m)^2 + (-1+9m)+7\equiv0\mod81$
$-9m+7\equiv0\mod81$
I cant continue... how can I make it work???
thanks!!
| You can write the equation as $(x +41)^{2} \equiv -27$ (mod 81). But this means there is no solution, as $x+41$ would need to be divisible by $9.$ I'll put in more explanation, in view of comments and questions, including correcting a sign error pointed out in the comments.
We have $x^{2} + x + 7 \equiv x^{2}+ 82x + 7 $ (mod 81). This is (mod 81), equal to $(x+41)^{2}+ 7 -1600,$ and also $1600 \equiv -20$ (mod 81). Hence we need
$(x+41)^{2} \equiv -27$ (mod 81), which is impossible in integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/313259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.