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Finding $\int \frac{d x}{x+\sqrt{1-x^{2}}}$.
I have to calculate the following integral:
$$
\int \frac{d x}{x+\sqrt{1-x^{2}}}
$$
An attempt:$$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^{2}}} & \stackrel{x=\sin t}{=} \int \frac{\cos t}{\sin t+\cos t} d t \\
&=\int \frac{\cos t(\cos t-\sin t)}{\cos 2 t} d t
\end{aligned}
$$
I find the solution is
$$\frac{\ln{\left(x + \sqrt{1 - x^{2}} \right)}}{2} + \frac{\sin^{-1}{\left(x \right)}}{2}+C$$
How can I get this without trigonometric substitution?
| $$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^2}} &\stackrel{t=\sin x}{=} \int \frac{\cos t}{\sin t+\cos t} d t \\
&=\frac{1}{2} \int \frac{(\sin t+\cos t)+(\cos t-\sin t)}{\sin t+\cos t} d t \\
&=\frac{1}{2}\left[\int 1 d t+\int \frac{d(\sin t+\cos t)}{\sin t+\cos t}\right] \\
&=\frac{1}{2}[t+\ln |\sin t+\cos t|]+C \\
&=\frac{1}{2}\left[\sin ^{-1} x+\ln \left|x+\sqrt{1-x^2}\right|\right]+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Spivak, Ch. 4 Graphs, Problem 15: Draw the graph of $f(x)=ax^2+bx+c$? I am asking about this problem because the solution manual seems to have a much more limited solution than mine below, and I wonder if they have been lazy or if I have done something incorrect.
Draw the graph of $f(x)=ax^2+bx+c$.Hint: use methods of Problem 1-18.
We would like to draw the graph, but since we are only in chapter 4 of Spivak's Calculus, we have limited methods available to draw conclusions about $f$.
Here is the solution from the solution manual:
$$f(x)=ax^2+bx+c = a\left(x^2+\frac{b}{a}x+\frac{c}{a} \right)$$
$$=a\left[ \left(x+\frac{b}{2a} \right)^2 + \left(\frac{c}{a}-\frac{b^2}{4a} \right) \right]$$
the graph looks like
Here is my solution
Factorize $f(x)$
$$f(x)=a(x^2+\frac{b}{a}x+\frac{c}{a})$$
What do we know about $g(x)=x^2+\frac{b}{a}x+\frac{c}{a}$? Let's factorize it
$$g(x)=x^2+2x\frac{b}{2a}+\left( \frac{b}{2a} \right)^2+c-\left( \frac{b}{2a} \right)^2$$
$$=\left(x+\frac{b}{2a}\right)^2 + c -\left( \frac{b}{2a} \right)^2$$
$$=\left(x+\frac{b}{2a}\right)^2+ \frac{4ac-b^2}{4a^2}\tag{1}$$
Note that $g(x)$ has a minimum value when the term $\left(x+\frac{b}{2a}\right)^2$ is zero, ie when $x=\frac{-b}{2a}$.
Set to zero to find roots
$$g(x)=0 \implies \frac{b^2-4ac}{4a^2}=\left(x+\frac{b}{2a}\right)^2$$
The right side is nonnegative. For there to be a solution (ie for $g(x)$ to have a root), the left side must also be nonnegative.
Case 1: The left side is zero, ie $b^2=4ac$ $\implies$ There is only one root
$$x_r = \frac{-b}{2a}$$
The minimum of $g(x)$ is at $x_r$.
Case 2: $b^2-4ac>0 \implies$ There are two roots
$$x_r=\frac{-b \pm\sqrt{b^2-4ac}}{2a}$$
Case 3: $b^2-4ac<0 \implies$ There are no roots, $g(x)>0\ \forall x$
$$g(x) = \left(x+\frac{b}{2a}\right)^2+ \frac{4ac-b^2}{4a^2}>0,\ \forall x$$
What do we know about $f(x)=a(x^2+\frac{b}{a}x+\frac{c}{a})$ now?
Case 1: $a>0$, $b^2-4ac=0 \implies$ one root at the minimum, $f(x)$ grows without bound as $x \to \pm \infty$.
Case 2: $a<0$, $b^2-4ac=0 \implies$ case 1 multiplied by a negative number. One root at the maximum, $f(x)$ decreases without bound as $x \to \pm \infty$.
Case 3: $a>0$, $b^2-4ac>0 \implies$ similar to case 1, but with two roots
Case 4: $a<0$, $b^2-4ac>0 \implies$ similar to case 2 but with two roots
Case 5: $a>0$, $b^2-4ac<0 \implies$ similar to case 1 but with no roots
Case 6: $a<0$, $b^2-4ac<0 \implies$ similar to case 2 but with no roots
| You're both correct. Spivak's diagram is perfectly fine, as long as you accept the possibility that:
*
*The axes are not necessarily drawn over the lines $x = 0$ and $y = 0$; and
*The axes are not necessarily increasing in the usual directions.
These are both generally considered bad things to do in most cases because of the misleading implications, but they're definitely things you can do.
Notice that your various cases are translations, flips and scalings of the shape given in the answer, which are all effects you can also achieve by redefining the axes appropriately.
Oh, and you're both assuming that $a \neq 0$, otherwise the graph is just a line.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Choosing at least 2 women from 7 men and 4 women In how many different ways can we choose six people, including at least two women, from a group made up of seven men and four women?
Attempt:
As we have to have at least two women in the choices, then $\displaystyle\binom{4}{2}$, leaving a total of $4$ out of $9$ people to be chosen. So $$\binom{9}{4} \cdot \binom{4}{2}=756$$
The answer is $371$. Where is my error?
| If you choose six people and at least two women are selected, then either two women and four men, three women and three men, or four women and two men are selected. The number of ways of selecting $k$ women and $6 - k$ men from four women and seven men is
$$\binom{4}{k}\binom{7}{6 - k}$$
Hence, the number of selections with at least two women is
$$\sum_{k = 2}^{4} \binom{4}{k}\binom{7}{6 - k} = \binom{4}{2}\binom{7}{4} + \binom{4}{3}\binom{7}{3} + \binom{4}{4}\binom{7}{2} = 371$$
You counted each selection with more than two women multiple times.
Your method counts each selection with three women three times, once for each of the $\binom{3}{2}$ ways of designating two of the three selected women as the two women in the group, and each selection with all four women six times, once for each of the $\binom{4}{2}$ ways of designating two of the four women as the four women in the group.
For instance, suppose the women are Anne, Brenda, Charlotte, and Diana and the men are Edward, Frank, George, Henry, Ivan, Jeffrey, and Karl. You count the selection of Anne, Brenda, Charlotte, Edward, Frank, and George three times.
$$\begin{array}{l | l}
\text{designated women} & \text{additional people}\\ \hline
\text{Anne, Brenda} & \text{Charlotte, Edward, Frank, George}\\
\text{Anne, Charlotte} & \text{Brenda, Edward, Frank, George}\\
\text{Brenda, Charlotte} & \text{Anne, Edward, Frank, George}
\end{array}
$$
Similarly, your method counts the selection of Anne, Brenda, Charlotte, Diana, Edward, and Frank six times.
$$\begin{array}{l | l}
\text{designated women} & \text{additional people}\\ \hline
\text{Anne, Brenda} & \text{Charlotte, Diana, Edward, Frank}\\
\text{Anne, Charlotte} & \text{Brenda, Diana, Edward, Frank}\\
\text{Anne, Diana} & \text{Brenda, Charlotte, Edward, Frank}\\
\text{Brenda, Charlotte} & \text{Anne, Diana, Edward, Frank}\\
\text{Brenda, Diana} & \text{Anne, Charlotte, Edward, Frank}\\
\text{Charlotte, Diana} & \text{Anne, Brenda, Edward, Frank}
\end{array}
$$
Notice that
$$\binom{4}{2}\binom{7}{4} + \color{red}{\binom{3}{2}}\binom{4}{3}\binom{7}{3} + \color{red}{\binom{4}{2}}\binom{4}{4}\binom{7}{2} = \color{red}{756}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What´s the value of the area of the triangle below? For reference: The sides of an acute-angled triangle measure
$3\sqrt2$, $\sqrt{26}$ and $\sqrt{20}$.
Calculate the area of the triangle (Answer:$9$)
My progress...
Is there any way other than Heron's formula since the accounts would be laborious or algebraic manipulation for the resolution?
$p=\frac{\sqrt{18}+\sqrt{20}+\sqrt{26}}{2}\implies S_{ABC} =\sqrt{p(p-\sqrt{20})(p-\sqrt{26})(p-\sqrt{18})}$
| Use cosine law,
$c^2=a^2+b^2-2ab\cos{\gamma}$
$(\sqrt{26})^2=(3\sqrt2)^2+(\sqrt{20})^2-2\cdot3\sqrt2\sqrt{20}cos\gamma$
$\implies cos(\gamma) =\frac{1}{\sqrt{10}}\Rightarrow sin(\gamma)=\frac{3}{\sqrt{10}}$
$A=\frac12\cdot 3\sqrt2\cdot \sqrt{20}\cdot sin(\gamma)$
$\implies...A=9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4322288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the largest $t$ such that for all positive $x, y, z$ the following inequality is satisfied Find the largest $t$ such that for all positive $x, y, z$ the following inequality is satisfied:
$(xy+xz+yz) \left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)^2 \geq t$.
If there were such an inequality:
$ t_{1} \leq (xy+xz+yz) \left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)^2 \leq t_{2}$, it is obvious that $t_{1}=0$ is suitable. And if $x=y=z$, then we have:
$(xy+xz+yz) \left(\dfrac{1}{x+y}+\dfrac{1}{x+z}+\dfrac{1}{y+z}\right)^2=3x^2 \cdot \dfrac{9}{4x^2}=\dfrac{27}{4}$. Next question is this value is the maximum or minimum?
If $x=1, y=1, z=0$, then we have $F(1,1,0)=(\dfrac{1}{2}+2)^2=\dfrac{25}{4}< \dfrac{27}{4}$, but if $x=1, y=1, z=n$, then we have $F(1,1,n)=(1+2n)(\dfrac{1}{2}+\dfrac{2}{1+n})^2>(1+2n)(\dfrac{1}{4}) \rightarrow +\infty$, which means, if I'm not mistaken, that this function has no extremum?
What to do with the case when $x,y,z$ are different I have not yet figured out.
Maybe in my case one of the following inequalities could be applied, but I'm stumped:
$1) \ xyz \geq(x+y-z)(x+z-y)(y+z-x)$
$2) \ x^3+y^3+z^3 \geq x^2y+x^2z+y^2x+y^2z+z^2x+z^2y-3xyz$
$3) \ x^3+y^3+z^3 \geq 3xyz$.
Maybe someone can give an idea where to start with my case? Thanks!!
| For $z\rightarrow0^+$ and $x=y=1$ we obtain: $t\leq\frac{25}{4}$.
We'll prove that $\frac{25}{4}$ is valid.
Indeed, we need to prove that $$\sum_{cyc}xy\left(\sum_{cyc}\frac{1}{x+y}\right)^2\geq\frac{25}{4}.$$
Now, let $x+y+z=2u$ and since our inequality is homogeneous, we can assume $xy+xy+xz=1$
and we need to prove that:
$$\sum_{cyc}\frac{1}{x+y}\geq\frac{5}{2}.$$
*
*Let $u\geq1$.
Thus, by AM-GM we obtain:
$$\sum_{cyc}\frac{1}{x+y}=\sum_{cyc}\frac{xy+xz+yz}{x+y}=2u+\sum_{cyc}\frac{xy}{x+y}\geq2u+\sum_{cyc}\frac{xy}{x+y+z}=$$
$$=2u+\frac{1}{2u}=\frac{3}{2}u+\frac{u}{2}+\frac{1}{2u}\geq\frac{3}{2}+2=\frac{5}{2}.$$
2.Let $u\leq1$.
Here, by Schur, we obtain: $$xyz\geq\frac{1}{9}(4(x+y+z)(xy+xz+yz)-(x+y+z)^3)=\frac{8}{9}(u-u^3),$$ which gives
$$\sum_{cyc}\frac{1}{x+y}=\frac{\sum\limits_{cyc}(x+y)(x+z)}{\prod\limits_{cyc}(x+y)}=\frac{\sum\limits_{cyc}(x^2+3xy)}{(x+y+z)(xy+xz+yz)-xyz}=$$
$$=\frac{x^2+y^2+z^2+3}{2u-xyz}=\frac{4u^2+1}{2u-xyz}\geq\frac{4u^2+1}{2u-\frac{8}{9}(u-u^3)}=$$
$$=\frac{9(4u^2+1)}{2(4u^3+5u)}-\frac{5}{2}+\frac{5}{2}=\frac{(1-u)(20u^2-16u+9)}{2(4u^3+5u)}+\frac{5}{2}\geq\frac{5}{2}$$ and we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $\lim\limits_{n→∞}\frac{a^n+b^n}{a^n-b^n}$ with $a ≠ b$
Find $\lim\limits_{n→∞}\dfrac{a^n+b^n}{a^n-b^n}$ with $a ≠ b$.
Notice that I don't have $a>b$ or $b>a$, and also no $b≠0$ or $a≠0$, so I don't know if i even can use my solution.
My solution :
\begin{gather*}
\frac {a^n+b^n}{a^n-b^n} = \frac {a^n}{a^n-b^n} + \frac {b^n}{a^n-b^n}
= \frac{1}{1- \frac{b^n}{a^n}}-\frac {1}{\frac{a^n}{b^n}-1}\\
a>b \implies \lim\limits_{n→∞}\frac{b^n}{a^n}=0,\quad \lim\limits_{n→∞}\frac{a^n}{b^n}=\infty\\
\frac{1}{1- \frac{b^n}{a^n}}+\frac {1}{ \frac{a^n}{b^n}-1} = \frac {1}{1-0} + \frac{1}{\infty-1} = 1+\frac{1}{\infty}= 1+0=1\\
b>a \implies \lim\limits_{n→∞}\frac{b^n}{a^n}=\infty,\quad \lim\limits_{n→∞}\frac{a^n}{b^n}=0\\
\frac {1}{1- \frac{b^n}{a^n}}+\frac {1}{ \frac{a^n}{b^n}-1} = \frac {1}{1-\infty} + \frac {1}{0-1} = \frac{1}{-\infty}-1= 0-1=-1
\end{gather*}
Now I have two convergent solutions for different cases. What should I answer to the question mentioned above?
| Your answer should contain all possible cases, so in general, the answer will not be a single number.
Your solution, by the way, is correct, but it does not cover all cases. If $a>b$, then your proof that the limit is $1$ assumes that $a,b>0$! If $b<a<0$, then the limit $$\lim_{n\to\infty} \frac{a^n}{b^n}$$ is, in fact, $0$, not $\infty$.
Similarly, if $a<b$, then the limit is $-1$ according to your proof, but only if $a,b> 0$.
So, to conclude your examination of the limit, you must check what happens when values are negative.
| {
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Verify if $e^{x^{3}} - \sqrt[4]{1-x^{4}} + x^2 \sin(x)$ is a little-o of $x^{2}$ as $x \to 0^{+}$ using limit I have to verify if $e^{x^{3}} - \sqrt[4]{1-x^{4}} + x \sin(x)$ is a little-o of $x^{2}$ as $x \to 0^{+}$ using limit, i.e. calculating the limit
$$\lim_{x \to 0^{+}} \frac{e^{x^{3}} - \sqrt[4]{1-x^{4}} + x^2 \sin(x)}{x^{2}} $$
and if the result is $0$, then the answer is "yes".
My first question is: is it acceptable that $x \to 0^{+}$? I mean: I always did exercises with limit from left and right. Is the little-o notation defined also with limit only from right side or left side?
My second question is: ho can I calculate the limit?
Thanks
EDIT
My attempt of solution using L'Hopital's rule.
$$\lim_{x \to 0^{+}} \frac{e^{x^{3}} - \sqrt[4]{1-x^{4}} + x^2 \sin(x)}{x^{2}} \\
\lim_{x \to 0^{+}} \frac{e^{x^{3}} - \sqrt[4]{1-x^{4}}}{x^{2}}+\lim_{x \to 0^{+}} \frac{x^{2}\sin(x)}{x^{2}}\\
\lim_{x \to 0^{+}} \frac{\frac{d}{dx}(e^{x^{3}}-\sqrt[4]{1-x^{4}})}{\frac{d}{dx}x^{2}} + \lim_{x \to 0^{+}}\sin(x)\\
\lim_{x \to 0^{+}} \frac{x^{2}\left(\frac{x}{\sqrt[4]{(1-x^{4})^{3}}}+3e^{x^{3}}\right)}{2x} + 0\\
\lim_{x \to 0^{+}}\frac{x}{2}\left(\frac{x}{\sqrt[4]{(1-x^{4})^{3}}}+3e^{x^{3}}\right)\\
\frac{0}{2}\left(\frac{0}{1}+3\right)=
0
$$
Is it OK?
EDIT #2
My attempt of solution following the hint by Claude Leibovici
$$\require{cancel}\lim_{x \to 0^{+}} \frac{e^{x^{3}} - \sqrt[4]{1-x^{4}} + x^2 \sin(x)}{x^{2}} \\
\lim_{x \to 0^{+}} \frac{1+x^{3}-(1-x^{4})+x^{2}x}{x^{2}}\\
\lim_{x \to 0^{+}}\frac{1+x^3-1+x^4+x^3}{x^2}\\
\lim_{x \to 0^{+}}\frac{x^4+2x^3}{x^2}\\
\lim_{x \to 0^{+}}\frac{\cancel{x^{2}}(x^{2}+2x)}{\cancel{x^{2}}}\\
\lim_{x \to 0^{+}} x^2+2x = 0
$$
Is it OK?
| We can simply using asymptotics relations:
$$\lim_{x \to 0} \frac{e^{x^{3}} - \sqrt[4]{1-x^{4}} + x^2 \sin(x)}{x^{2}}\,\,=\,\,\lim_{x \to 0} \frac{\left(e^{x^{3}} - 1\right)+\left(1-\sqrt[4]{1-x^{4}}\right) + x^2 \sin(x)}{x^{2}}\,\,\sim\,\,\lim_{x \to 0^{+}} \frac{x^3+\frac{1}{4}x^4+x^3}{x^2}=\lim_{x\to 0}\frac{o(x^2)}{x^2}=0$$
So, $e^{x^{3}} - \sqrt[4]{1-x^{4}} + x^2 \sin(x) = o(x^2)$ as you wanted to show.
| {
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Finding distribution given bivariate normal $f_{xy}$ Let $X$ and $Y$ be distributed as bivariate normal random variables with pdf
$$f_{X, Y}(x,y)=\frac{1}{2\pi\sqrt{1-\rho^2}}\exp\biggl(\frac{-1}{2(1-\rho^2)}(x^2-2\rho(xy)+y^2)\biggr)$$
Find the distribution of $aX+bY+c$.
Would this be done by finding the marginal distributions of both $X$ and $Y$ and then working with those to find the desired distribution?
I worked through the problem and used the fact that X varies N(0,1) and Y varies N(0,1).
I then considered $$Var(aX+bY+c)=a^2+b^2+2ab*Cov(X,Y)$$
$$=a^2+b^2+2ab\rho\sigma_x\sigma_y$$
$$=a^2+b^2+2ab\rho$$
Is this it? Then I could say $aX+bY+c$ varies $N(c, a^2+b^2+2ab\rho)$? Seems a little off to me.
| That is right. There is a more general way to calculate such affine transformations you probably will see later. We have already established that
$$ \begin{pmatrix} X\\ Y \end{pmatrix} \sim N_2\left(
\begin{pmatrix} 0 \\ 0 \end{pmatrix} ,
\begin{pmatrix} 1& \rho\\ \rho&1 \end{pmatrix}\right) $$
then $aX+bY+c = (a,b)(X,Y)^T + c$ and we find
$$ aX+bY+c \sim N\left( c + (a,b) \begin{pmatrix} 0 \\ 0 \end{pmatrix}, (a,b) \begin{pmatrix} 1& \rho\\ \rho&1 \end{pmatrix}\begin{pmatrix} a \\ b \end{pmatrix} \right) = N(c, a^2+b^2 +2ab\rho). $$
This is from a general calculation of affine transformation $Y = \eta + BX$, where $X\sim N_p(\mu,\Sigma)$ is p-dimensional normal distributed, $B$ is a $k\times p$ matrix and $\eta\in \mathbb{R}^k$. Then
$$Y \sim N_k(\eta+B\mu, B\, \Sigma \,B^T). $$
| {
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Why am I getting the same angle by inputting different values of n in the general solution? Problem:
Solve: $\sqrt{3}\sin x-\cos x=2$, $-2\pi<x<2\pi$
My attempt:
$$\sqrt{3}\sin x-\cos x=2$$
$$\frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x=1$$
$$\sin(\frac{\pi}{3}).\sin x-\cos(\frac{\pi}{3})\cos x=1$$
$$\cos(\frac{\pi}{3})\cos x-\sin(\frac{\pi}{3}).\sin x=-1$$
$$\cos(x+\frac{\pi}{3})=\cos{\pi}$$
$$x+\frac{\pi}{3}=2n\pi\pm\pi, \text{where $n$ belongs to $\mathbb{Z}$}$$
$$x=2n\pi\pm\pi-\frac{\pi}{3}$$
Now, if $n=0$,
$$x=\frac{2\pi}{3},-\frac{4\pi}{3}$$
If $n=1$,
$$x=\frac{8\pi}{3},\frac{2\pi}{3}$$
If n=-1,
$$x=-\frac{4\pi}{3}, -\frac{10\pi}{3}$$
So, in the given interval, $x=\frac{2\pi}{3}, \frac{-4\pi}{3}$ (Ans.)
Question:
When $n=0$, and when $n=1$, I get $\frac{2\pi}{3}$. Similarly, when $n=0$, and when $n=-1$, I get $-\frac{4\pi}{3}$. Why am I getting the same angle twice for different values of $n$?
| $$x=2n\pi\pm\pi-\frac{\pi}{3}\\=-\frac{\pi}{3}+\pi(\color{red}{2n\pm1}).$$
$\color{red}{2n\pm1}:\quad\dots,-7,-7,-5,-5,-3,-3,-1,-1,1,1,3,3,5,5,\ldots.$
When solving trigonometric equations generally and the cosine or secant of the generating solution equals $-1,$ then $\color{red}{2n\pm1}$ occurs, giving rise to repeating/overlapping solutions.
This is inconsequential since the obtained collection of solutions is exhaustive anyway.
| {
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Prove that $(a^6+b^6-1)(a^6+b^6-2)$ is divisible by $252$ if $a$ and $b$ are coprime integers Prove that $(a^6+b^6-1)(a^6+b^6-2)$ is divisible by $252$ if $a$ and $b$ are coprime integers.
I thought about proving that this number is divisible by $2$, $3$ and $7$ but I don't know how should I do that.
| Put $E=(a^6+b^6-1)(a^6+b^6-2)$; the two factors are consecutive so coprime and $252=2^2\cdot3^2\cdot7$.
$1)$ $E\equiv0\pmod7$. By Fermat's Little Theorem it is quite clear, even if $a$ or $b$ is divisible by $7$.
$2)$ $E\equiv0\pmod{3^2}$. We know that $$(X\pm1)^6=X^6\pm6X^5+15X^4\pm20X^3+15X^2\pm6X+1$$ There are two cases to consider:
$► (a,b)=(3m,3n\pm1)\Rightarrow(3m)^6+(3n\pm1)^6-1\equiv\pm6(3n)\equiv0\pmod{3^2}$.
$► (a,b)=(3m\pm1,3n\pm1)\Rightarrow(3m\pm1)^6+(3n\pm1)^6-2\equiv\pm6(3m)\pm6(3n)\equiv0\pmod{3^2}$.
$3) E\equiv0\pmod{2^2}$. There are two cases to consider:
$► (a,b)=(2m,2n+1)\Rightarrow(2m)^6+(2n+1)^6-1\equiv6(2n)\equiv0\pmod4.$
$►(a,b)=(2m+1,2n+1)\Rightarrow(2m+1)^6+(2n+1)^6-2\equiv6(2m+2n)\equiv0\pmod4$.
(In both, $2)$ and $3)$,we have used one of the two factors of $E$ to prove the divisibility).
| {
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"url": "https://math.stackexchange.com/questions/4342628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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} |
Find the area of region triangular BPM For reference: In triangle $ABC$, the external angle bisector $BE$ and the median $AM$ intersect at $P$. Determine the area of triangular region $BPM$; if $AB =3BC$ and $S_{ABC}=40\ \mathrm{m^2}$.
My progress:
$S_{ABM}=S_{AMC}=\frac{40}{2} = 20\\
\frac{S_{ABP}}{S_{APE}}=\frac{BP}{PE}\\
\frac{S_{BPM}}{S_{BCE}}=\frac{BM.BP}{BC.BE}=\frac{BP}{2BE}\\
\frac{S_{ABM}}{S_{ABP}}=\frac{AB.AM}{AB.AP}\implies \frac{20}{S_{ABP}} = \frac{AM}{AP}\\
\frac{S_{ABC}}{S_{BCE}}=\frac{AC}{CE}$
...?
| $S_{\triangle MBP} = \frac 12 \cdot \frac{BC}{2} \cdot BP \sin{(90^\circ - \frac{\angle B}{2})} = \frac{BC \cdot BP}{4} \cos (\frac{\angle B}{2})$
$S_{\triangle ABP} = \frac1 2 \cdot 3 BC \cdot BP \sin{(90^\circ + \frac{\angle B}{2})} = 6 \cdot S_{\triangle MBP}$
$S_{\triangle ABM} = S_{\triangle ABP} - S_{\triangle MBP} = 5 \cdot S_{\triangle MBP} = 20$
$S_{\triangle MBP} = 4$
Or applying external angle bisector theorem in $\triangle ABM$,
$\frac{AB}{BM} = \frac{AP}{MP} \implies AP = 6 \cdot MP$ or $AM = 5 \cdot MP$
$S_{\triangle ABM} = 5 S_{\triangle MBP} = 20$
$ \therefore S_{\triangle MBP} = 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
A composite function problem The question is:
Suppose $$f(x) = x^2+1,$$ $$g(x) = 3-x.$$
Find the values for $x$ for such that $$(g\circ f)(x) = (f \circ g)(x).$$
I tried banging my head for one hour but my answer doesn't match the one given by the book which is $1/\sqrt{2}$ and $-1/\sqrt{2}$.
I think the answer given in the book is wrong because I even tried putting the given answer in $(g\circ f)(x)$ and $(f \circ g)(x)$ and the two don't match up.
My answer: $$ x =\frac{3\pm\sqrt{-7}}{2} $$
| $$f(g(x)) = g(x)^2 + 1 = (3-x)^2 + 1$$
$$g(f(x)) = 3 - f(x) = 3 - x^2 - 1$$
Hence we need to find the points $x$ that satisfy
$$(3-x)^2 + 1 = 2 - x^2$$
We get that:
$$9 -6x+ x^2 + 1 -2 + x^2 = 0 \implies$$
$$2x^2 -6x + 8 = 0 \implies$$
$$x^2 - 3x + 4 = 0 \implies$$
$$...$$
you are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4350747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
limit, quotient of roots How can I find
$$
\lim_{n \to \infty} \frac{\sqrt[3]{n + 2} - \sqrt[3]{n + 1}}{\sqrt{n + 2} - \sqrt{n + 1}} \sqrt[6]{n - 3}?
$$
If I multiply by $\sqrt{n + 2} + \sqrt{n + 1}$ I could get no divisor, but I cannot get any result on this way. On the other hand I could also transform this expression:
$$
\frac{(n + 2)^{1 / 3} \left(1 - {\left(\frac{n + 1}{n + 2}\right)}^{1 / 3}\right)}{(n + 2)^{1 / 2} \left(1 - {\left(\frac{n + 1}{n + 2}\right)}^{1 / 2}\right)} \sqrt[6]{n - 3} = {\left(\frac{n - 3}{n + 2}\right)}^{1 / 6} \frac{1 - {\left(\frac{n + 1}{n + 2}\right)}^{1 / 3}}{1 - {\left(\frac{n + 1}{n + 2}\right)}^{1 / 2}}.
$$
However I also cannot see how continue to see that the limit is $\frac 2 3$. Any help would be appreciated.
| Put $ m=n+1$.
The limit becomes after factorisation and simplification
$$\lim_{m\to\infty}\frac{(1+\frac 1m)^{\frac 13}-1}{(1+\frac 1m)^{\frac 12}-1}(1-\frac 4m)^{\frac 16}$$
Use the result
$$\lim_{N\to \infty}N\Bigl((1+\frac 1N)^\alpha-1\Bigr)=\alpha$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4351218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Show that $\frac{\log_aN-\log_bN}{\log_bN-\log_cN}=\frac{\log_aN}{\log_cN}$ Show that $$\dfrac{\log_aN-\log_bN}{\log_bN-\log_cN}=\dfrac{\log_aN}{\log_cN}$$ where $a,b$ and $c$ are positive and are consecutive terms of а geometric sequence, $a\ne1,b\ne1,c\ne1,N>0,N\ne1$.
$a,b$ and $c$ are consecutive terms of a geometric sequence if and only if $b^2=ac, b=\sqrt{ac}$. Then the LHS is $$\dfrac{\log_aN-\log_\sqrt{ac}N}{\log_\sqrt{ac}N-\log_cN}=\dfrac{\log_aN-2\log_{ac}N}{2\log_{ac}N-\log_cN}$$ This seems useless. What can we do? What is the intuition that will lead to the solution? Thank you!
| From where you left -
$ \displaystyle \frac{\log_aN-2\log_{ac}N}{2\log_{ac}N-\log_cN} = \frac{\log_aN-2(\log_aN)(\log_{ac}a)}{2(\log_cN)(\log_{ac}c)-\log_cN}$
$ \displaystyle = \frac{\log_aN}{\log_cN} \cdot \frac{1 - 2\log_{ac}a}{2\log_{ac}c-1}$
Now note that by dividing by $\log_{ac}c$,
$ \displaystyle \frac{1 - 2\log_{ac}a}{2\log_{ac}c-1} = \frac{\log_{c}{ac} - 2\log_{c}a}{2 - \log_{c}{ac}} = \frac{1 - \log_ca}{1 - \log_ca} = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4353364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
if equation $\log(x^2+2ax)=\log(4x-4a-13)$ has only one solution, then exhaustive set of value of $a$ is?
if equation $\log(x^2+2ax)=\log(4x-4a-13)$ has only one solution, then exhaustive set of value of $a$ is?
Given answer is
$(-13/4,-13/12) \cup [-1]$
My Approach:
for $\log(x^2+2ax)$ to be valid $x^2+2ax>0$
Here we have to cases
case $1$ $a>0$
$x(x+2a)>0$
$\implies$ $x\in(-\infty,-2a)\cup (0,\infty)$
case $2$ $a<0$
$\implies$ $x\in(-\infty,0)\cup (2a,\infty)$
For $\log(4x-4a-13)$ to be valid $4x-4a-13>0$
$\implies$ $x>\frac{4a+13}{4}$
For one solution
$x^2+2ax=4x-4a-13$
and discriminant of above quadratic must be $0$
so i got two value of $a$ those are $a=-1, a=9$
but for $a=9$ I got $x=-7$ which not a valid solution because $\log(x^2+2ax)$ will be invalid.
But how to arrive at other part of solution?
Same question has been asked here but it doesn't solve my query Equation $\log(x^2+2ax)=\log(4x-4a-13)$ has only one solution; then exhaustive set of values of $a$ is
| We want to find $a$ such that there is only one $x$ satisfying
$$x^2+2ax=4x-4a-13\tag1$$
$$x^2+2ax\gt 0\tag2$$
$$4x-4a-13\gt 0\tag3$$
The discriminant of $x^2+(2a-4)x+4a+13=0$ has to be non-negative, so it is necessary that $a\in (-\infty,-1]\cup [9,\infty)$.
*
*$a=-1$ is sufficient since $x=3$ is the only solution.
*$a=9$ is not sufficient since the equation has no solutions.
*For $a\in (-\infty,-1)\cup (9,\infty)$, let $p,q\ (p\lt q)$ be the solutions of $(1)$ where $$p=2-a-\sqrt{(a+1)(a-9)},\quad q=2-a+\sqrt{(a+1)(a-9)}$$ Then, one can see that
$$\begin{align}&\text{$p$ satisfies $(2)$ and $(3)$}
\\\\&\iff 4\bigg(2-a-\sqrt{(a+1)(a-9)}\bigg)-4a-13\gt 0
\\\\&\iff 4\sqrt{(a+1)(a-9)}\lt -8a-5
\\\\&\iff -8a-5\gt 0\quad \text{and}\quad 16(a+1)(a-9)\lt (-8a-5)^2
\\\\&\iff a\lt -\frac{13}{4}\quad \text{or}\quad -\frac{13}{12}\lt a\lt -1\tag4\end{align}$$
and that
$$\begin{align}&\text{$q$ satisfies $(2)$ and $(3)$}
\\\\&\iff 4\bigg(2-a+\sqrt{(a+1)(a-9)}\bigg)-4a-13\gt 0
\\\\&\iff 4\sqrt{(a+1)(a-9)}\gt 8a+5
\\\\&\iff 8a+5\leqslant 0\quad \text{or}\quad \bigg(8a+5\gt 0\quad \text{and}\quad 16(a+1)(a-9)\gt (8a+5)^2\bigg)
\\\\&\iff a\lt -1\tag5\end{align}$$What we want is $a$ such that only one of $(4)(5)$ is satisfied. So, $a\in [-\frac{13}{4},-\frac{13}{12}]$.
Therefore, the answer is
$$\color{red}{a\in\bigg[-\frac{13}{4},-\frac{13}{12}\bigg]\cup [-1]}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4353775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Harmonic series with sign alternates every $n$ terms. Let $A(1)=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots$
Let $A(2)=\frac{1}{1}+\frac{1}{2}-\frac{1}{3}-\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}+\dots$
Let $A(3)=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}-\frac{1}{4}-\frac{1}{5}-\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+\frac{1}{9}-\frac{1}{10}-\frac{1}{11}-\frac{1}{12}+\dots$
In general, for $n \in \mathbb{N}$, let $A(n)$ be the same series as the above with sign alternates every $n$ terms.
I know that $A(1)=\log(2)$, and $A(2)=\frac{1}{4}(\pi+\log(4))$.
How can we evaluate $A(n)$ for $n \in \mathbb{N_{\ge3}}$?
| $A(3) = \lim_{m\to\infty} K_m$ where
$$
K_m = \sum_{k=1}^m\frac{1}{6k-5}
+\sum_{k=1}^m\frac{1}{6k-4}
+\sum_{k=1}^m\frac{1}{6k-3}
-\sum_{k=1}^m\frac{1}{6k-2}
-\sum_{k=1}^m\frac{1}{6k-1}
-\sum_{k=1}^m\frac{1}{6k}
$$
Then use estimates like
$$
\sum_{k=1}^m\frac{1}{6k-a} =
\frac{1}{6}\left(\log m-\psi\left(1-\frac{a}{6}\right)\right)+O(1/m)
$$
with the Gauss digamma theorem for $\psi$ of rational number. Result:
$$
A(3) = \frac{\log 2}{3}+\frac{2\pi}{3\sqrt{3}} .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4355000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove that: $(a^2+2)(b^2+2)(c^2+2)-3(a^2+b^2+c^2)\ge18$ Let $a,b,c$ be real numbers such that $ab+bc+ca=3$. Prove that:$$(a^2+2)(b^2+2)(c^2+2)-3(a^2+b^2+c^2)\ge18$$
I have an imperfect solution:
Let $a+b+c=p;ab+bc+ca=q=3;abc=r$
The problem is:
$$r^2-2(q^2-2pr)+4(p^2-2q)+8-3(p^2-2q)\ge18$$
or:
$$r^2-4pr+p^2+2\ge0 $$
By $p^2\ge3q=9;pr\le\dfrac{q^2}{3}=3$, we have: $r^2-4pr+p^2+2\ge r^2-1$
So we need to prove $$r^2\ge1$$
I have no idea from here, pls help me!
| Some Hints:
$(a^2+2)(b^2+2)(c^2+2)-3(a^2+b^2+c^2)\ge18$
$(a^2+2)(b^2+2)(c^2+2)\ge3(a^2+2+b^2+2+c^2+2)$
Also,
$(a+b+c)^2$= $a^2+b^2+c^2+6$ (as $ab+bc+ca=3$)
$(a+b+c)^2$= $a^2+2+b^2+2+c^2+2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4358917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why isn't $]-1, 1]$ the domain of $f(x) = \arcsin\left(\frac{|x|}{x+1} \right)$? I have to find the domain of this function: $$f(x) = \arcsin\left(\frac{|x|}{x+1} \right)$$
My attempt:
*
*because arcsine is the inverse of sine function, then its domain is the image of sine: $[-1, 1]$.
*the absolute value is defined in $]-\infty, +\infty[$,
*the denominator is defined in $]-1, +\infty[$.
The domain of $f(x)$ is the intersection of these domain. Therefore, it's $]-1, 1]$.
But this is wrong, because the correct domain is $[-1/2, +\infty[$, where does this $-1/2$ come from?
| Lying in bed I thought of probably the clearest answer:
The domain of $\arcsin$ is $[-1,1]$ so the range of $\frac {|x|}{x+1}$ must be restricted to $-1 \le \frac {|x|}{x+1} \le 1$.
That requires that $0\le |\frac {|x|}{x+1}|=\frac {|x|}{|x+1|} \le 1$ and that $|x| \le |x+1|$.
If we consider the four possible positive/negative values of $|x|, |x+1|$ we can have
*
*$x+1< 0$
so $x< -1$ and $|x|=-x$ and $|x+1|= -(x+1)=-x-1$. We need $-x < -x-1$ or $0 < -1$. This never happens. None of $(-\infty, -1)$ is in the domain of $\arcsin\frac{|x|}{x+1}$.
*$-1 \le x < 0$
(Note: $x=-1$ means $\frac {|x|}{x+1}$ will have a forbidden $0$ in the denominator so we can't have $x=-1$ no matter what we conclude below)
If $-1\le x < 0$ then $|x| = -x$ and $|x+1|=x+1$ and we must have $-x \le x+1$ or $-1 \le 2x$ or $x \ge -\frac 12$.
So, of $[-1,0)$ w have $[-\frac 12,0)$ is in the domain of $\arcsin\frac{|x|}{x+1}$.
(We can note that if $x \in (-1,-\frac 12)$ then $|x|\in(\frac 12, 1)$ and $x+1 \in (0,\frac 12)$ so $|x| > x+1> 0$ and $\frac {|x|}{x+1} > 1$.)
(We can further not if $x \in (-\frac 12,0)$ then $|x|\in (0, \frac 12)$ and $x+1\in (\frac 12, 1)$ so $0 < |x| < x+1$ so $0\le \frac {|x|}{x+1}< 1$.)
*$x \ge 0$
so $|x| = x$ and $|x+1| = x+1$ so we need $x \le x+1$ which is always the case.
So of $[0,\infty)$, all of $[0,\infty)$ is in the domain of $\arcsin\frac{|x|}{x+1}$.
Putting 1),2),3) together the domain of $\arcsin\frac{|x|}{x+1}$ is $[-\frac 12,\infty).
=====old answer below====
One thing you can notice is if $-1 < x < -\frac 12$ then $|x|=-x \in (\frac 12, 1)$. So $|x| > \frac 12$.
And $0 < x + 1 < \frac 12$. So $\frac 1{x+1} > 2$.
So $\frac {|x|}{x+1} = |x|\cdot \frac 1{x+1} > \frac 12\cdot 2 = 1$. and is out of the domain for $\arctan$.
.....
But how were you supposed to figure that on your own?
Read on.
=======
Note we need $-1 \le \frac {|x|}{x+ 1} \le 1$.
As the denominator can not be equal to $0$ we must have $x \ne -1$.
If we multiply all terms by $x+1$ we will get $-(x+1) ?? |x| ?? (x+1)$... but we must take into account if $x+1$ is negative of positive.
If $x +1 > 0$ we get $-(x+1) \le |x| \le x+1$. But if $x+1 < 0$ we will flip the inequality sign to get $-(x+1) \ge |x| \ge x+1$.
Okay... let start with $x+1 > 0$ or in other words $x > -1$. Then we have $-(x+1)\le |x| \le x+1$ but as absolute values are never negative that is the same as $0\le |x| \le x+1$ which is the same as saying $-(x+1)=-x-1\le x \le x+ 1$.
As $x \le x+1$ always that condition is always satisfied. But we need $-x-1\le x$ or in other words we need $-1 \le 2x$ or $x \ge -\frac 12$.
So if $x > -1$ the acceptable domain is $[-\frac 12, \infty)$.
But what if $x < -1$?
Then we get $-(x+1) \ge |x| \ge x+1$ but as $x+1$ is negative and absolute values are non-negative we have: $0 \le |x| \le -(x+1)$ or $(x+1) \le x \le -(x+1)$. But note $x+1 \le x$ is never possible. So we can not have $x < -1$.
So the only acceptable domain is $[-\frac 12,\infty)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Height of the hill The angles of elevation of the top of a distant hill in the forest as seen from three consecutive km stones on a straight horizontal road are 30, 45 and 60 degrees. Find the height of the hill.\
My try: Let km stone C,B, A are at distances $x,x+1,x+2$ from base of hill $OP$
Then $\tan 60=\frac{h}{x}$
$\tan45=\frac{h}{x+1}$
$\tan 30=\frac{h}{x+2}$
I solved and got $h=\frac{3+\sqrt3}{2}$ but answer is $\sqrt{\frac{3}{2}}$
| Following up on @Matt's comment, the answer is indeed $\boxed{h = \sqrt{\frac{3}{2}}}$ when you consider a more general scenario.
Setup
(Diagram at the bottom.)
Let $P$ be the peak of the hill that lies $h > 0$ above the origin $O$, which is a horizontal distance $r \geq 0$ from the closest point on the horizontal straight road $R$. $A$, $B$, and $C$ lie on the road distances $d$, $d+1$, and $d+2$ from $R$, respectively, and distances $0 < a < b < c$ from $O$, respectively. The closer a point is to the origin $O$, the bigger the angle to the peak $P$ (i.e. $a < b$ implies $\measuredangle OAP > \measuredangle OBP$); thus the angles from points on the road $A$, $B$, and $C$ to the peak $P$ must be $60° > 45° > 30°$, respectively, as implied by $a < b < c$.
Negative $d$
Note, $d$ can be negative, but it can't be less than or equal to $-\frac{1}{2}$. Otherwise, we'd not have $a < b < c$. For example, if $d = -\frac{1}{2}$, then $a = b$, which implies $\measuredangle OAP = \measuredangle OBP$, which directly contradicts $60° = \measuredangle OAP > \measuredangle OBP = 45°$. Further, if $d < -\frac{1}{2}$, then $a > b$, which implies $\measuredangle OAP < \measuredangle OBP$, which also directly contradicts $60° = \measuredangle OAP > \measuredangle OBP = 45°$.
In summary $d > -\frac{1}{2}$.
Solution
We have
$$
\begin{aligned}
h &= \tan(60°)a = \tan(45°)b = \tan(30°)c\\
&= \sqrt{3} a = b = \frac{1}{\sqrt{3}} c &&[h]
\end{aligned}
$$
Square $[h]$
$$
\begin{aligned}
3 a^2 &= b^2 = \frac{1}{3} c^2\\
&\implies a^2 = \frac{1}{3} b^2 &&[*a \to b]\\
&\implies c^2 = 3b^2 &&[*c \to b]
\end{aligned}
$$
We also have
$$
\begin{aligned}
a^2 &= d^2 + r^2 &&[*a]\\
b^2 &= (d+1)^2 + r^2 &&[*b]\\
c^2 &= (d+2)^2 + r^2 &&[*c]\\
\end{aligned}
$$
These are valid for positive $d$, of course, but, importantly, for $-\frac{1}{2} < d \leq 0$ as well.
We have a system of $6$ linear equations in $6$ unknowns, $h, r, d, a, b, c$. Hurray!
Subtract $[a]$ from $[b]$
$$b^2 - a^2 = 2d + 1 \qquad[*ba] := [b] - [a]$$
Subtract $[b]$ from $[c]$
$$
\begin{aligned}
c^2 - b^2 &= 4d + 4 - (2d + 1) &&[c] - [b]\\
&= 2d + 3 &&[*cb]
\end{aligned}
$$
We have $\operatorname{RHS}[cb] = 2d + 3 = \operatorname{RHS}[ba] + 2$, so
$$
\begin{aligned}
\operatorname{LHS}[cb] &= \operatorname{LHS}[ba] + 2\\
c^2 - b^2 &= b^2 - a^2 + 2\\
c^2 &= 2b^2 - a^2 + 2\\
3b^2 &= 2b^2 - \frac{1}{3} b^2 + 2 &&[* a \to b] \text{ and } [* c \to b]\\
(3 - 2 + \frac{1}{3}) b^2 &= 2\\
\frac{4}{3} b^2 &= 2\\
b^2 &= \frac{3}{2}\\
b &= \sqrt{\frac{3}{2}}\\
&\boxed{h = b = \sqrt{\frac{3}{2}}} &&[h]
\end{aligned}
$$
It's also possible to find the values of the other $5$ variables.
Diagram
| {
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"timestamp": "2023-03-29T00:00:00",
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Is $\sqrt{5-2\sqrt{5}}$ in $\mathbb{Q}(\sqrt{5+2\sqrt{5}},\sqrt{2})$? Is $\sqrt{5-2\sqrt{5}}$ in $\mathbb{Q}(\sqrt{5+2\sqrt{5}},\sqrt{2})$?
I'm told that $\mathbb{Q} \subseteq \mathbb{Q}(\alpha,\sqrt{2})$ is a Galois extension, and so the minimal polynomial of $\alpha$ must split in $\mathbb{Q}(\alpha,\sqrt{2})$.
The minimal polynomial of $\alpha$ over $\mathbb{Q}$ is $x^4-10x^2+5$, whose four roots in $\mathbb{C}$ are:
$\pm \sqrt{5 \pm 2\sqrt{5}}$
And I'm trying to explicitly show that these roots are all contained in $\mathbb{Q}(\sqrt{5+2\sqrt{5}},\sqrt{2})$... I'm pretty sure that they are not in $\mathbb{Q}(\sqrt{5+2\sqrt{5}})$ but I don't know how to make $\sqrt{2}$ useful.
Thanks
| Let $k=\mathbb{Q}(\sqrt{2},\sqrt{5})$
Let $\mu, \nu \in k$ and we ask when is
$\sqrt{\mu}\in k(\sqrt{\nu})$ if this happens,
$$\sqrt{\mu}=a+b\sqrt{\nu}$$ where $a, b\in k$ and so
$$\mu=a^2+b^2\nu+2ab\sqrt{\nu}$$ and assuming $\sqrt{\mu}, \sqrt{\nu}\not\in k$ we have $a=0$ and
thus
$$\mu=b^2\nu$$
So we ask is $\sqrt{5-2\sqrt{5}}\in k(\sqrt{5+2\sqrt{5}})$
This is the same as
$$5-2\sqrt{5}=b^2(5+2\sqrt{5})$$
so $$b^2=\frac{5-2\sqrt{5}}{5+2\sqrt{5}}=9-4\sqrt{5}$$
Thus the problem is reduced to whether $9-4\sqrt{5}$ is a perfect square in $\mathbb{Q}(\sqrt{2},\sqrt{5})$, which in turn can be reduced to
$$9-4\sqrt{5}=(n+m\sqrt{5})^2$$ and this has the solution
$$9-4\sqrt{5}=(2-\sqrt{5})^2$$
So
$$5-2\sqrt{5}=(2-\sqrt{5})^2(5+2\sqrt{5})$$
and
$$\sqrt{5-2\sqrt{5}}=(2-\sqrt{5})\sqrt{5+2\sqrt{5}}$$
And thus in the unlikely event that I have made no mistake the answer is yes, $\sqrt{5-2\sqrt{5}}\in k(\sqrt{5+2\sqrt{5}})$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $f(x) = \sin^{-1}(\frac{2x}{1+x^2})$ Show that $f(x) = 2\tan^{-1}(x)$
Let $$f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) ~~ -\infty<x<\infty.$$
Show that,
(a) $f(x) = 2\tan^{-1}(x)$ for $-1\leq x \leq 1$ and
(b) $f(x) = \pi-2\tan^{-1}(x)$ for $x \geq 1.$
Proof: I started off by equating $$\sin^{-1}\left(\frac{2x}{x^2+1}\right)=2\tan^{-1}(x)$$
(a)
We wish to show that these are equal for $-1\leq x \leq 1$.
For this domain $\displaystyle -1 \leq\frac{2x}{x^2+1} \leq 1 \implies -\frac{\pi}{2} \leq \sin^{-1}\left(\frac{2x}{x^2+1}\right) \leq \frac{\pi}{2}$
$$\frac{2x}{x^2+1} = \sin(2\tan^{-1}(x))$$
The task is now to show $\sin(2\tan^{-1}(x))=\frac{2x}{x^2+1}$
$$2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x))=\frac{2x}{x^2+1}$$
For $-1 \leq x \leq 1 \implies -\frac{\pi}{4}\leq\tan^{-1}(x)\leq\frac{\pi}{4} \implies -\frac{\sqrt{2}}{2}\leq \sin(\tan^{-1}(x)) \leq \frac{\sqrt{2}}{2}$
Also, $\frac{\sqrt{2}}{2}\leq\cos(\tan^{-1}(x)) \leq 1$
$$2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x)) \iff 2(\frac{x}{\sqrt{x^2+1}})(\frac{1}{\sqrt{x^2+1}})= \frac{2x}{x^2+1}$$
Which was to be shown.
(b)
$\displaystyle x \geq 1 \implies 0<\frac{2x}{1+x^2}\leq 1 \implies 0< \sin^{-1}\left(\frac{2x}{x^2+1}\right) \leq \frac{\pi}{2}$
Hence,
We must show that $$\sin^{-1}(\frac{2x}{x^2+1}) = \pi - 2\tan^{-1}(x) \iff \frac{2x}{x^2+1} = \sin(\pi - 2\tan^{-1}(x))$$ for $x\geq 1$
$$\sin(\pi - 2\tan^{-1}(x))=\sin(\pi)\cos(2\tan^{-1}(x)) - \cos(\pi)\sin(2\tan^{-1}(x))=\sin(2\tan^{-1}(x))$$
$$\sin(2\tan^{-1}(x)) = 2\sin(\tan^{-1}(x))\cos(\tan^{-1}(x))=\frac{2x}{x^2+1}$$
Which was to be demonstrated.
Note: This problem didn't flow like I thought it would. I had imagined that during some of the intermediate steps, I would be presented with the option of choosing an $f(x)$ or trig function value that would only be true in one of the intervals. But no such situation presented itself. Did I do something wrong? Did I overlook something?
| For part (a), I would suggest starting as $$\sin(y)=\frac{2x}{1+x}$$ Now, you can use this to build a right triangle with the side of $2x$ and the hypotenuse of $1+x$. Now you can find the other side (How?).
Now let's check the desired part: You need to calculate $tan(y/2)$ (you have $y=f(x)= 2 \tan^{-1} (x)$).
From half angle identities we know that $$\tan(y/2)=\frac{\sin(y/2)}{\cos(y/2)}$$ (Use half angle identities to get these $\sin(\frac{y}{2})$ and $\cos(\frac{y}{2})$ **).
Finally, you will end up $$\tan(\frac{y}{2})=x$$ which means $$f(x)=2 \tan^{-1} (x).$$
**You get those half angles by the right triangle that you got from $$\sin(y)=\frac{2x}{1+x}$$
| {
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"url": "https://math.stackexchange.com/questions/4369535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Suppose X$\sim$ Cauchy(0,1). Then what will be the distribution of $\frac{1-X}{1+X}$? In order to find distribution of $\frac{1-X}{1+X}$ below approach I followed,
Let,
\begin{align}
Y = \frac{1-X}{1+X}
\end{align}
Then, cdf of Y is
\begin{align}
F_{Y}(y) = P(Y \leq y)
\end{align}
\begin{align}
= P\left(\frac{1-X}{1+X} \leq y\right)
\end{align}
\begin{align}
= 1 - P\left(X < \frac{1-y}{1+y}\right)
\end{align}
\begin{align}
= 1 - \int_{-\infty}^{\frac{1-y}{1+y}} f(x) \,dx
\end{align}
\begin{align}
= 1 - \int_{-\infty}^{\frac{1-y}{1+y}} \frac{1}{\pi}\cdot \frac{1}{1+x^2} \,dx
\end{align}
\begin{align}
= 1 - \frac{1}{\pi}\cdot \left[tan^{-1}x\right]_{-\infty}^{\frac{1-y}{1+y}}
\end{align}
\begin{align}
F_{Y}(y) = \frac{1}{\pi}\cdot \left[-\frac{\pi}{2}+tan^{-1}\left({\frac{1-y}{1+y}}\right)\right]
= \frac{1}{2} -\frac{1}{\pi}.tan^{-1}\left({\frac{1-y}{1+y}}\right)
\end{align}
and then
\begin{align}
f_{Y}(y) = \frac{d F_{Y}(y)}{dy} = \frac{1}{\pi}\cdot \frac{1}{1+y^2}
\end{align}
But I have a little confusion here how to find range of Y from X? And CDF of Y is doesn't looks like cdf of a cauchy distribution.
| Let $g(x) = (1-x)/(1+x)$. Then $g$ is self-inverse; i.e., $g^{-1} = g$, but it is not everywhere monotone. Thus, consider $\Pr[g(X) \le y]$ for the case $y \le -1$ versus $y > -1$ separately. In the first case, $g$ is monotone decreasing on $X \in (-\infty, -1)$ and we have $$\Pr[g(X) \le y] = \Pr[g^{-1}(y) \le X < -1] = \Pr[g(y) \le X < -1],$$ with no issues (the inequality reverses because $g$ is order-reversing). You can check this with a numerical example; e.g., for $y = -2$, $$\frac{1-x}{1+x} \le -2 \iff -3 \le x \le -1.$$
However, for $y > -1$, we see that the inequality is compound: $$\begin{align}
\Pr[g(X) \le y] &= \Pr[-\infty < X < -1] + \Pr[g^{-1}(y) \le X < \infty] \\
&= 1 - \Pr[-1 < X \le g(y)]. \end{align}$$
Hence we have $$F_Y(y) = \int_{x=g(y)}^{-1} f_X(x) \, dx$$ when $y \le -1$ and $$F_Y(y) = \int_{x = -\infty}^{-1} f_X(x) \, dx + \int_{x=g(y)}^\infty f_X(x) \, dx = 1 - \int_{x=-1}^{g(y)} f_X(x) \, dx$$ when $y > -1$. I leave the rest of the computation as an exercise.
| {
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Representing the cube of any natural number as a sum of odd numbers I'm expanding my notes on exercises from Donald Knuth's The Art of Computer Programming, and found something rarely mentioned in the Internet, but still useful to prove Nicomachus' Theorem about the sum of cubes.
Knuth phrases this in the following way in exercise 8(a) to chapter 1.2.1:
Prove the following theorem of Nicomachus (A.D. c. 100) by induction: $1^3=1$, $2^3=3+5$, $3^3=7+9+11$, $4^3=13+15+17+19$, etc.
In the answers to exercises, the author gives the following formula: $(n^2-n+1)+(n^2-n+3)+...+(n^2+n-1)=n^3$
My question is, how do I get to that formula from the sample sums given in the problem? It looks kind of odd, especially because the last summand doesn't give me any idea on how it is connected with first ones. Usually one is able to see this clearly, but not here.
I tried to get to that formula by the following set of thoughts:
To get $n^3$ one must sum up $n$ odd numbers, starting from the $(n-1)$th central polygonal number, meaning that a cube of number $n$ is made by summing up odd numbers, starting from $(n-1)$th central polygonal number to $n$th triangular number.
Since odd numbers form arithmetic progression with $a_1=1, d=2$, it's possible to use the following formula of summing up $p$th to $q$th member of this progression:
$$S_{p,q}=\dfrac{a_p+a_q}2\cdot(q-p+1)$$
We can simplify $\dfrac{a_p+a_q}2$, putting $a_p=2p-1$ and $a_q=2q-1$:
$$\dfrac{a_p+a_q}2=\dfrac{(2p-1)+(2q-1)}2=\dfrac{2p+2q-2}2=p+q-1$$
Thus, the formula of summing up $p$th to $q$th member of this progression is:
$$(p+q-1)(q-p+1)=(q+(p-1))(q-(p-1))=q^2-(p-1)^2$$
Substituting in $p=\dfrac{n(n-1)}{2}+1=\dfrac{n^2-n+2}{2}$, and $q=\dfrac{n(n+1)}{2}$, we get the formula:
$$\left(\dfrac{n(n+1)}{2}\right)^2-\left(\dfrac{n(n-1)}{2}+1-1\right)^2=\left(\dfrac{n(n+1)}{2}\right)^2-\left(\dfrac{n(n-1)}{2}\right)^2=\\
\dfrac{(n(n+1))^2-(n(n-1))^2}{4}=\dfrac{(n(n+1)-n(n-1))(n(n+1)+n(n-1))}{4}=\\
\dfrac{n^2(n+1-n+1)(n+1+n-1)}{4}=\dfrac{4n^3}{4}=n^3$$
This... kind of... proves the sum formula for any $n$, really. But it doesn't give out the formula in question, i.e. $(n^2-n+1)+(n^2-n+3)+...+(n^2+n-1)$.
What is the correct way to get this formula? Any hints are greatly appreciated.
| Welcome to MSE!
What we're doing is moving out from $n^2$ in either direction by $2$s. You should have something analogous to gauss's trick for summing $1 \ldots n$ in mind, where we pair up $1$ and $n$, then we pair up $2$ and $n-1$, etc.
So for instance, let's look at $n = 5$. We get the sequence of numbers
$$5^2 - 4 \quad 5^2 - 2 \quad 5^2 \quad 5^2 + 2 \quad 5^2 + 4$$
Of course, if we pair these off from the outside, we see that we get $5$ copies of $5^2$. That is, $5^3$ in total. There's nothing special about $5$ in this example, and every odd number works similarly. We have $\lfloor \frac{n}{2} \rfloor$ entries on either side of $n^2$, moving by $2$s. It's easy to see these are all odd, and that they sum to $n \cdot n^2$ when we pair them off (leaving the one in the middle alone).
For even numbers, we do the obvious thing. For instance, if $n=6$ we get
$$
6^2 - 5 \quad 6^2 - 3 \quad 6^2 - 1 \quad 6^2 + 1 \quad 6^2 + 3 \quad 6^2 + 5
$$
where now we again have $6$ copies of $6^2$, but now everybody has a friend. Since $6^2$ is even, moving to either side gives us odd numbers again.
As for how one might have come up with this formula, if we want to end up with $n^3$ and we want an arithmetic progression of length $n$, we know that the values should average to $n^2$ (do you see why?). Of course, once we have values averaging $n^2$ and we know we want an arithmetic progression, then we're forced to consider the sequence (let's assume $n$ is odd again, so there's a middle)
*
*$n^2$
*$n^2 - k \quad n^2 \quad n^2 + k$
*$n^2 - 2k \quad n^2 - k \quad n^2 \quad n^2 + k \quad n^2 + 2k$
*...
where we continue until we have $n$ numbers total. The case $k=2$ is exactly your sequence of interest.
I'll leave the generalization of $k \neq 2$ with $n$ even to you as a (fun?) exercise!
I hope this helps ^_^
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $a, b, c$ such that $a^3+b^3+c^3-3abc=2017$.
Find all natural numbers $a, b, c$ such that $a\leq b\leq c$ and $a^3+b^3+c^3-3abc=2017$.
My Attempt
$$a^3+b^3+c^3-3abc=2017$$
$$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=2017*1$$
Now, $a+b+c$ can't be equal to $1$ as $a, b, c$ are natural numbers.
So, $$a+b+c=2017$$ $$a^2+b^2+c^2-ab-bc-ca=1$$
How should I proceed after this?
|
These are surfaces over {a,b} expected reals. The calculations are straightforward.
The problems are the second and third solutions are complex. The graph shows only the real parts.
Still calculations are straightforward.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How many 12-letter words are there with no block $5 \times a$, $4 \times b$ and $3 \times c$ We arrange 12-letter words having at our disposal five letters
$a$, four letters $b$ and three letters $c$. How many words are there without any block $5 \times a$, $4 \times b$ and $3 \times c$.
I need to use inclusion - exclusion principle. I counted all possible 12 letter words - $\frac{12!}{5!4!3!}$. Then words with block of letters:
only $a$ blocks - $8\cdot \frac{7!}{4!3!}$
only $b$ blocks - $9\cdot \frac{8!}{5!3!}$
only $c$ blocks - $10\cdot \frac{9!}{5!4!}$.
Now I have to count all words, where block: $a,b$ or $a,c$, or $b,c$ appears together, but I don't now how.
| If we let $N$ denote the total number of distinguishable arrangements of $aaaaabbbbccc$, $A$ denote the set of permutations with five consecutive $a$s, $B$ denote the set of permutations with four consecutive $b$s, and $C$ denote the set of permutations with three consecutive $c$s, then, by the Inclusion-Exclusion Principle, the number of arrangements with no block of five consecutive $a$s or four consecutive $b$s or three consecutive $c$s is
\begin{align*}
|A' \cap B' \cap C'| & = N - |A \cup B \cup C|\\
& = N - (|A| + |B| + |C| - |A \cap B| - |A \cap C| - |B \cap C| + |A \cap B \cap C|)\\
& = N - |A| - |B| - |C| + |A \cap B| + |A \cap C| + |B \cap C| - |A \cap B \cap C|
\end{align*}
You have correctly calculated $N$, $|A|$, $|B|$, and $|C|$.
$|A \cap B|$: We have five objects to arrange: $aaaaa$, $bbbb$, $c$, $c$, $c$. There are $\binom{5}{3}$ ways to select three of the five positions for the three $c$s and $2!$ ways to arrange the remaining two distinct objects in the remaining two positions. Hence, there are
$$\binom{5}{3}2! = \frac{5!}{3!}$$
such arrangements.
Alternatively, arrange the three $c$s in a line. This creates four spaces in which we can place the block $aaaaa$, two between successive $c$s and two at the ends of the row.
$$\square c \square c \square c \square$$
Once we have placed the block $aaaa$, we have four objects in a row, which creates five spaces in which to place the block $bbb$, three between successive objects and two at the ends of the row. Hence,
$$|A \cap B| = 4 \cdot 5$$
Can you finish the argument by counting $|A \cap C|$, $|B \cap C|$, and $|A \cap B \cap C|$?
| {
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Number of car plates having $3$ letters and a $4$-digit number. Car plates have are to be designed by placing $3$ (not necessarily distinct) English letters followed by a $4$-digit number, where the digits of this number can not be simultaneously $0$, that is, the string "$0000$" is excluded. Also, $0$ can be used in the leftmost place of the number.
How many such car plates are there?
For many years, I was thinking that the answer is:
$26 \times 26 \times 26 \times 9999 = 175,742,424$
BUT
Suddenly I have a doubt about that. I think, in that way of calculation, we did not consider the permutation of the letters and numbers.
To clarify, take this as an example:
In how many ways can we arrange the letters if the word "MATHEMATICS"?
The answer is $\frac{11!}{2!2!2!}=6,652,800$ because we have $2$ M's, $2$ A's, and $2$ T's.
So, the word "MATHEMATICS" and the word "MATHEMATICS" are apparently same, except that the $2$ M's exchanged their positions. So we need not over count. Hence we divide by $2!$, and so on.
Going back to the original problem, I have doubt that we are over counting.
For instance, "BHB $1035$" is counted twice in the $175,742,424$ ways because of the "B"
I need your clarification regarding this. Thanks.
| Your original answer is correct.
There are $26$ choices for each letter, giving $26^3$ ways to fill the three positions allocated for letters. Similarly, if we initially ignore the constraint that the final four positions cannot be filled with $0000$, we would have $10$ choices for each digit, giving $10^4$ ways to fill the four positions allocated for digits. Since we cannot use $0000$, one string of digits is eliminated, leaving us with $$26^3(10^4 - 1)$$ possible license plates.
Since you are concerned about repetitions, let's count the license plates a different way.
First, we will count the number of ways of filling the first three positions with letters.
Three distinct letters: There are $26$ ways to fill the first letter, $25$ ways to fill the second letter, and $24$ ways to fill the third letter. Hence, there are $$26 \cdot 25 \cdot 24$$ ways to fill the first three positions with distinct letters.
Exactly two distinct letters: Since we have three positions to fill, one letter must appear twice and another letter must appear once. There are $26$ ways to choose the letter which appears twice, $25$ ways to choose the letter which appears once, and three ways to choose the position of the letter which appears only once. Hence, there are $$26 \cdot 25 \cdot 3$$ ways to fill the first three positions with exactly two distinct letters.
One letter fills all three positions: There are $26$ ways to choose the letter which fills all three positions.
Total: Since these three cases are mutually exclusive and exhaustive, the number of ways we can fill the first three positions with letters is
$$26 \cdot 25 \cdot 24 + 26 \cdot 25 \cdot 3 + 26 = 26^3$$
Next, we will count the number of ways of filling the last four positions with digits.
Four distinct digits: There are $10$ ways to fill the first digit, $9$ ways to fill the second digit, $8$ ways to fill the third digit, and $7$ ways to fill the fourth digit, so there are $$10 \cdot 9 \cdot 8 \cdot 7$$ ways to fill these positions with four distinct digits.
Exactly three distinct digits: Since we have four positions to fill, one digit must appear twice and two other digits must each appear once. There are $10$ ways to select the digit which appears twice, $\binom{4}{2}$ ways to choose two of the four positions for that digit, $\binom{9}{2}$ ways to select two of the remaining nine digits to each appear once, and $2!$ ways to arrange those digits in the remaining two positions. Hence, there are $$\binom{10}{1}\binom{4}{2}\binom{9}{2}2!$$
ways to fill these positions with exactly three distinct digits.
Exactly two distinct digits: Either one digit appears three times and another digit appears once or two digits each appear twice.
One digit appears three times and another digit appears once: There are $10$ ways to choose the digit which appears three times, $9$ ways to choose the digit which appears once, and $4$ ways to choose the position of the digit which appears exactly once. Hence, there are $$10 \cdot 9 \cdot 4$$ such choices.
Two digits each appear twice: There are $\binom{10}{2}$ ways to select the two digits which will each appear twice and $\binom{4}{2}$ ways to select two of the four positions for the smaller of those digits. Hence, there are $$\binom{10}{2}\binom{4}{2}$$ such choices.
One digit appears all four positions: Ignoring the constraint that we cannot use $0000$ gives us $10$ choices for the digit which fills all four positions. Since we have that constraint, we actually have just $9$ choices.
Total: Since these cases are mutually exclusive and exhaustive, the number of admissible ways to fill the last four positions with digits is
$$10 \cdot 9 \cdot 8 \cdot 7 + \binom{10}{1}\binom{4}{2}\binom{9}{2}2! + 10 \cdot 9 \cdot 4 + \binom{10}{2}\binom{4}{2} + 9 = 10^4 - 1$$
Since we can choose letters and digits independently, the number of license plates we can form is indeed $$26^3(10^4 - 1)$$ which demonstrates that your original approach was correct.
| {
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Number of ways to pick $3$ balls from a box with $8$ balls if $3$ of the balls are identical and the other $5$ are all different
A box contains 8 balls, of which 3 are identical and the remaining 5 are different from each other. 3 balls are to be picked out of the box; the order in which they are picked out does not matter. Find the number of different possible selections of 3 balls.
My Solution: I grouped them into cases
Case 1: All three balls are identical in which number of ways $= 1$
Case 2: Two of the balls are identical in which number of ways $= 5$
Case 3: One of the ball is from the $3$ identical ones in which number of ways $= 5 \cdot 4$ since there are $5$ ways to choose $1$ ball from the remaining $5$ balls and another $4$ ways to choose again from the remaining $4$ balls.
Case 4: I do not take from any of the $3$ identical balls so that leaves me with $5$ balls to choose $3$ from in which number of ways $= 5 \cdot 4 \cdot 3 = 60$.
Total number of ways $= 1+5+20+60 = 86$.
However, the answer given is $26$.
| When the order of selection does not matter, you should be thinking in terms of combinations.
In what follows, suppose that the three identical balls are white and that the remaining balls are blue, green, red, orange, and yellow.
Since we wish to determine the number of distinguishable selections of three balls, your first two cases are correct.
Exactly one of the balls is one of the three identical ones: We do not care about which of the identical balls we select. We can select two of the other five balls in $\binom{5}{2}$ ways. Hence, the number of such selections is
$$\binom{5}{2} = \frac{5!}{2!3!} = \frac{5 \cdot 4 \cdot 3!}{2 \cdot 1 \cdot 3!} = 5 \cdot 2 = 10$$
As lulu pointed out in the comments, this is half your answer since first selecting a blue ball and then selecting a green ball results in the same set of balls as first selecting a green ball and then selecting a green ball.
None of the balls is selected from among the three identical ones: There are
$$\binom{5}{3} = \frac{5!}{3!2!} = \frac{5 \cdot 4 \cdot 3!}{3! \cdot 2 \cdot 1} = \frac{5 \cdot 4}{2} = 5 \cdot 2 = 10$$
such selections. You counted each such selection six times, once for each of the $3!$ orders in which you could have picked the same three balls.
With these corrections, we obtain
$$1 + 5 + 10 + 10 = 26$$
distinguishable ways to select three of the eight balls in the box.
| {
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Help on this integral $I=\int_0^1 \frac{x \arctan(x)}{1-x^2}\ln\left(\frac{2}{1+x^2}\right) dx$ $$I=\int_0^1 \frac{x \arctan(x)}{1-x^2}\ln\left(\frac{2}{1+x^2}\right) dx$$
Here is my attempt
$\frac{x}{1-x^2}dx=-\frac{1}{2}d\ln(1-x^2)$, integration by part, we got
$$I=-\frac{1}{2}P-Q$$
Where $P=\int_0^1 \frac{\ln(1-x^2)\ln\left(\frac{1+x^2}{2}\right)}{1+x^2}dx$, and
$Q=\int_0^1 \frac{x\cdot\arctan(x)\cdot\ln(1-x^2)}{1+x^2}dx$
How to proceed then? or should I switch to other ways?
| Here is a more general solution. Use the contour setup by Sangchul Lee, it can be shown that
$$\int_0^1 \frac{x \arctan a x}{1-x^2}\ln\frac{1+a^2}{1+a^2x^2}dx=\frac13 \arctan^3 a
$$
Let $a=1$ to obtain
$$\int_0^1 \frac{x \arctan x}{1-x^2}\ln\frac{2}{1+x^2}dx=\frac13 \arctan^3 (1)=\frac{\pi^3}{192}
$$
| {
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Integrate : $\int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$ Integrate :
$$I = \int \frac{1}{x^{2} \sqrt{2x-x^{2}}} dx$$
My attempt : substitute $\sin t = x-1$, $u = \tan \frac{t}{2}$
$$I = \int \frac{1}{x^{2} \sqrt{1-(x-1)^{2}}} dx = \int \frac{1}{(\sin t + 1)^{2} \cos t} \cos t dt$$
$$= \int \frac{1}{(\sin t + 1)^{2}} dt = \int \frac{2(u^{2}+1)}{(u+1)^{4}} du$$
$$= 2 \int \left(\frac{1}{(u+1)^{2}} - \frac{2}{(u+1)^{3}} + \frac{2}{(u+1)^{4}} \right)$$
$$= - \frac{2}{u+1} + \frac{2}{(u+1)^{2}} - \frac{4}{3(u+1)^{3}} + const.$$
$$= \frac{-2(3u^{2}+3u+2)}{3(u+1)^{3}} + const.$$
where
$$A = \sqrt{2x-x^{2}},~~ u = \sqrt{\frac{1-A}{1+A}}$$
However, Wolfram gives the following answer :
$$ - \frac{\sqrt{2x-x^{2}}~(x+1)}{3x^{2}} + const. $$
which is way more simple than mine. Is there any other way to integrate this? The $u$ in my answer is very complicated, I just couldn't change my answer into Wolfy's.
| If $x>0,$
Let $1-x=\cos2t,dx=2\sin2t\ dt$
so we get $$\int\dfrac{dt}{\sin^4t}dt=\int(\cot^2t+1)\csc^2t\ dt$$
Can you take it home from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4386603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Maximum and Minimum value of an implicit function For the real value of $x$, $f\left( x \right)$ satisfies $f{\left( x \right)^3} - f{\left( x \right)^2} - {x^2}f\left( x \right) + {x^2} = 0$. When the maximum value of $f(x)$ is $1$ and the minimum value of $f(x)$ is $0$, what is the value of $f\left( { - \frac{4}{3}} \right) + f\left( 0 \right) + f\left( {\frac{1}{2}} \right) = \_\_\_\_\_$
My approach is as follow, as it is an implicit function we need to find the roots is $f(x)$.
We end up getting $(f(x)-1)(f(x)+x)(f(x)-x)=0$, so we end up getting three function viz.
$f(x)=1$; $f(x)=-x$ & $f(x)=x$ but how do we proceed further
| If $f(x)^3-f(x)^2-x^2f(x)+x^2=0$ for every $x$, you can input the needeed values $x_0=0,-\frac{4}{3},\frac{1}{2}$ and solve the polynomial for $f(x_0)$. In general, this may give you 3 solutions for each $x_0$ (so 27 possible solutions for the problem), but maybe those 27 are the same.
Edit: The conditions $\max(f) = 1$ and $\min(f) = 0$ make two of the roots for $x_0 = -\frac{4}{3}$ impossible as well as one for $x_0 = \frac{1}{2}$. But still this don't have a unique answer. I found: $$ f(0)+f\left(\frac{1}{2}\right) + f\left(-\frac{4}{3}\right) \in \left\{ \frac{3}{2}, \frac{5}{2}, 3 \right\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4394577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Is it possible to find the $n$-derivative of $\csc(m\pi)?$ I am trying to find the $n$-th derivative of $\csc(m\pi)$, so I took few cases:
for simplicity let $x=\cot(m\pi)$ and $y=\csc(m\pi)$,
$$\frac{d^0}{dm^0}\csc(m\pi)=\pi^0(\color{red}{1}x^0y^1)$$
$$\frac{d^1}{dm^1}\csc(m\pi)=-\pi^1 (\color{red}{1}x^1y^1)$$
$$\frac{d^2}{dm^2}\csc(m\pi)=\pi^2(\color{red}{1}x^2y^1+\color{red}{1}x^0y^3)$$
$$\frac{d^3}{dm^3}\csc(m\pi)=-\pi^3(\color{red}{1}x^3y^1+\color{red}{5}x^1y^3)$$
$$\frac{d^4}{dm^4}\csc(m\pi)=\pi^4(\color{red}{1}x^4y^1+\color{red}{18}x^2y^3+\color{red}{5}x^0y^5)$$
$$\frac{d^5}{dm^5}\csc(m\pi)=-\pi^5(\color{red}{1}x^5y^1+\color{red}{58}x^3y^3+\color{red}{61}x^1y^5)$$
$$\frac{d^6}{dm^6}\csc(m\pi)=\pi^6(\color{red}{1}x^6y^1+\color{red}{179}x^4y^3+\color{red}{479}x^2y^5+\color{red}{61}x^0y^7)$$
and saw that
\begin{align}
\frac{d^n}{dm^n}\csc(m\pi)&=(-\pi)^n\sum_{k=0}^{\lfloor{n/2}\rfloor}\color{red}{a_k} x^{n-2k}
y^{2k+1}\\
&=(-\pi)^n\sum_{k=0}^{\lfloor{n/2}\rfloor}\color{red}{a_k} \cot^{n-2k}(m\pi)\csc^{2k+1}(m\pi)\\
&=(-\pi)^n\csc^{n+1}(m\pi)\sum_{k=0}^{\lfloor{n/2}\rfloor}\color{red}{a_k} \cos^{n-2k}(m\pi)
\end{align}
If we replace $n$ by $2n$ then separate the last term we have
$$\frac{d^{2n}}{dm^{2n}}\csc(m\pi)=\pi^{2n}\csc^{2n+1}(m\pi)\left[a_n+\sum_{k=0}^{n-1}\color{red}{a_k} \cos^{2n-2k}(m\pi)\right]$$
In the cases mentioned above, we notice that when the order of the derivative is $0, 2, 4, 6$, the coefficients of the last terms are $1, 1, 5, 61$ which match the absolute value of the Euler numbers:
$$E_0=1, E_2=-1, E_4=5, E_6=-61$$
and so
$$\frac{d^{2n}}{dm^{2n}}\csc(m\pi)=\pi^{2n}\csc^{2n+1}(m\pi)\left[|E_{2n}|+\sum_{k=0}^{n-1}\color{red}{a_k} \cos^{2n-2k}(m\pi)\right]$$
By the way, if we take the limit to both sides of the last result letting $m$ approach $1/2$, we have
$$\lim_{m\to \frac12}\frac{d^{2n}}{dm^{2n}}\csc(m\pi)=\pi^{2n}(1)\left[|E_{2n}|+0\right]=|E_{2n}|\pi^{2n}.$$
Question: Is it possible to find $\color{red}{a_k}$?
| Thanks to @Domen for his solution, the answer is
$$\frac{d^n}{dm^n}\csc(m\pi)=(-\pi)^n\csc^{n+1}(m\pi)\sum_{k=0}^{\lfloor{n/2}\rfloor}t(n,k) \cos^{n-2k}(m\pi)$$
where
$$t(n,k)=(2k+1)t(n-1,k)+(n-2k+1)t(n-1,k-1)$$
and
$$t(n,0)=1$$
Different form:
The $n$-th derivative of $\sec(x)$ is given by Wolfram:
$$\frac{d^{ n}}{d{x}^{n}}\sec(x)= \sum_{k=0}^{\infty} \frac{|E_{2 k}|}{(2 k-n) !} x^{2 k-n}$$
Proof:
By Taylor series we have
$$\sec(x)=1+\frac1{2!}x^2+\frac{5}{4!}x^4+\frac{61}{6!}x^6+....=\sum_{k=0}^\infty\frac{|E_{2k}|}{(2k)!}x^{2k}$$
Take the $n$-th derivative to both sides
$$ \frac {d^{ n}}{d{x}^{n}}\sec(x)=\sum_{k=0}^\infty\frac{|E_{2k}|}{(2k)!} \frac {d^{ n}}{d{x}^{n}}x^{2k}$$
We have
$$\frac {d}{dx}x^{2k}=2k x^{2k-1}$$
$$\frac {d^{ 2}}{d{x}^{2}}x^{2k}=2k(2k-1) x^{2k-2}$$
$$\frac {d^{ 3}}{d{x}^{3}}x^{2k}=2k(2k-1)(2k-2) x^{2k-3}$$
note that
$$2k(2k-1)(2k-2)*\color{red}{\frac{(2k-3)(2k-4)...}{(2k-3)(2k-4)...}}=\frac{(2k)!}{(2k-3)!}$$
so in general we have
$$\frac {d^{ n}}{d{x}^{n}}x^{2k}=\frac{(2k)!}{(2k-n)!}x^{2k-n}$$
Thus,
$$ \frac {d^{ n}}{d{x}^{n}}\sec(x)= \sum_{k=0}^{\infty} \frac{|E_{2 k}|}{(2 k-n) !} x^{2 k-n}$$
Finally, by using $\sec(x-\pi/2)=\csc(x)$ we have
$$ \boxed{\frac {d^{ n}}{d{x}^{n}}\csc(x)= \sum_{k=0}^{\infty} \frac{|E_{2 k}|}{(2 k-n) !} \left(x-\frac{\pi}{2}\right)^{2 k-n}}$$
Edit. A more rigorous proof: Let $z=x+1/2$,
\begin{gather*}
\lim_{z\to \frac12}\frac{d^{2a}}{d z^{2a}}\csc(z\pi)=\lim_{x\to 0}\frac{d^{2a}}{d x^{2a}}\csc\left(x\pi+\frac{\pi}{2}\right)\\
=\lim_{x\to 0}\frac{d^{2a}}{d x^{2a}}\sec(x\pi)\\
\left\{\text{expand $\sec(x\pi)$ in series}\right\}\\
=\lim_{x\to 0}\frac{d^{2a}}{d x^{2a}}\sum_{k=0}^\infty \underbrace{\frac{|E_{2k}|\pi^{2k}}{(2k)!}}_{f_{2k}}x^{2k}\\
=\lim_{x\to 0}\frac{d^{2a}}{dx^{2a}}\sum_{k=0}^\infty f_{2k}\,x^{2k}\\
=\lim_{x\to 0}\frac{d^{2a}}{dx^{2a}}\left(f_0x^0+f_2x^2+f_4x^4+...\right)\\
=(2a)! f_{2a}\\
=(2a)!\frac{|E_{2a}|\pi^{2a}}{(2a)!}\\
=|E_{2a}|\pi^{2a}.
\end{gather*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4396182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
$A+B+C=2\pi$, prove determinant equals to zero. Given $A$, $B$, $C$ which satisfy $A+B+C=2\pi$, is there an ingenious method to prove that
$$
\det\begin{pmatrix}
1 & 1 & 1 \\
\tan A & \tan B & \tan C \\
\tan 2A & \tan 2B & \tan 2C
\end{pmatrix}=0
$$? By column transformation we have
$$
\det\begin{pmatrix}
1 & 1 & 1 \\
\tan A & \tan B & \tan C \\
\tan 2A & \tan 2B & \tan 2C
\end{pmatrix}=\det\begin{pmatrix}
\tan B-\tan A & \tan C-\tan B \\
\tan 2B-\tan 2A & \tan 2C-\tan 2B
\end{pmatrix}.
$$
Substitute $C=2\pi—A-B$ and the rest computation seems very complicated! Any hint?
| Well here is my working of the problem. I must have made a sign mistake somewhere, however if I have not, then the equation only holds under special cases. Feedback welcome.
Let $a=\tan A, b=\tan B,c=\tan C$ then
$$\tan 2A=\frac{2a}{1-a^2}$$
etc.
So the matrix is
$$X=\begin{vmatrix}
1&1&1\\
a&b&c\\
\frac{2a}{1-a^2}&\frac{2b}{1-b^2}&\frac{2c}{1-c^2}\\
\end{vmatrix}$$
Multiplying the columns the question is transformed into
$$\frac{1}{2}(1-a^2)(1-b^2)(1-c^2)X=\begin{vmatrix}
1-a^2&1-b^2&1-c^2\\
a-a^3&b-b^3&c-c^3\\
a&b&c\\
\end{vmatrix}$$
Which is the same as
$$\frac{1}{2}(1-a^2)(1-b^2)(1-c^2)X=\begin{vmatrix}
a&b&c\\
a^2-1&b^2-1&c^2-1\\
a^3&b^3&c^3\\
\end{vmatrix}$$
Now we calculate,
$$\begin{vmatrix}
a&b&c\\
a^2-1&b^2-1&c^2-1\\
a^3&b^3&c^3\\
\end{vmatrix}=\begin{vmatrix}
a&b&c\\
a^2&b^2&c^2\\
a^3&b^3&c^3\\
\end{vmatrix}-\begin{vmatrix}
a&b&c\\
1&1&1\\
a^3&b^3&c^3\\
\end{vmatrix}$$
$$=abc\begin{vmatrix}
1&1&1\\
a&b&c\\
a^2&b^2&c^2\\
\end{vmatrix}+\begin{vmatrix}
1&1&1\\
a&b&c\\
a^3&b^3&c^3\\
\end{vmatrix}$$
Now by assumption, $a+b+c=abc$
so we have
$$\frac{1}{2}(1-a^2)(1-b^2)(1-c^2)X$$
$$=(a+b+c)\begin{vmatrix}
1&1&1\\
a&b&c\\
a^2&b^2&c^2\\
\end{vmatrix}+\begin{vmatrix}
1&1&1\\
a&b&c\\
a^3&b^3&c^3\\
\end{vmatrix}$$
Note however that the difference of these two matrices is zero,
$$(a+b+c)\begin{vmatrix}
1&1&1\\
a&b&c\\
a^2&b^2&c^2\\
\end{vmatrix}-\begin{vmatrix}
1&1&1\\
a&b&c\\
a^3&b^3&c^3\\
\end{vmatrix}$$
$$=\begin{vmatrix}
1&1&1\\
a&b&c\\
a^2(b+c)&b^2(a+c)&c^2(a+b)\\
\end{vmatrix}=0$$
because
$$bc^2(a+b)-cb^2(a+c)=abc(c-b)$$
$$ca^2(b+c)-ac^2(a+b)=abc(a-c)$$
$$ab^2(a+c)-ba^2(b+c)=abc(b-a)$$
Therefore,
$$\frac{1}{2}(1-a^2)(1-b^2)(1-c^2)X=
2(a+b+c)\begin{vmatrix}
1&1&1\\
a&b&c\\
a^2&b^2&c^2\\\end{vmatrix}$$
So
$$X=\frac{4abc(a-b)(b-c)(c-a)}{(1-a^2)(1-b^2)(1-c^2)}$$
$$=\frac{1}{2}\tan 2A \tan 2B\tan 2C(\tan A-\tan B)(\tan B-\tan C)(\tan C-\tan A)$$
And this $X=0$ happens only in special cases.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find the area of the shaded region. (Apparently this is a Chinese primary school math question!)
Rectangle $ABCD$ has an area of $1$. $AE = ED$. $3BF = AB$. What is the shaded area?
| The simplest way, in my opinion, is to write the equations of lines $BE$, $EC$, $FC$, $FD$. Because scaling doesn't change the ratios of areas, we can assume that we have a square of unit side length, where
$ B = (0, 0)$
$ E = (\frac{1}{2}, 1)$
$ C = (1, 0) $
$ F = (0, \frac{1}{3} )$
$ D = (1,1) $
It is straight forward to derive the following equations
$BE: y = 2 x $
$EC: y = -2 (x - 1) $
$FC: y = -\frac{1}{3} ( x- 1 ) $
$FD: y = \frac{1}{3} + \frac{2}{3} x $
Now, we proceed to find the intersection points
$H: BE \& FC : (\frac{1}{7}, \frac{2}{7} ) $
$G: FD \& BE : (\frac{1}{4}, \frac{1}{2} ) $
$I: EC \& FD : (\frac{5}{8} , \frac{3}{4} ) $
Finally, applying the so-called shoe-lace formula, we get
$[CIGH] = \frac{1}{2} \left( C_x I_y - C_y I_x + I_x G_y - I_y G_x + G_x H_y - G_y H_x + H_x C_y - H_y C_x \right) $
And this is
$[CIGH] = \frac{1}{2} \left( (1)(\frac{3}{4}) - 0 + (\frac{5}{8})(\frac{1}{2}) - (\frac{3}{4})(\frac{1}{4}) + (\frac{1}{4})(\frac{2}{7}) - (\frac{1}{2})(\frac{1}{7}) + 0 - (\frac{2}{7})(1) \right)$
which reduces to
$[CIGH] = \frac{1}{2} \left( \frac{3}{4} + \frac{5}{16} - \frac{3}{16} + \frac{1}{14} - \frac{1}{14} - \frac{2}{7} \right) $
And finally this comes to
$[CIGH] = \frac{1}{2} \left( \frac{ 84 + 35 - 21 - 32 }{(16)(7)} \right) = \boxed{\frac{33}{112}} $
This is assuming we have a square of side $1$. For a rectangle of area $A$, we get
$[CIGH] =\boxed{ \frac{33}{112} A } $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Maximize $\prod x_i$ without Lagrange multipliers
Suppose that $\sum_{i=1}^{6} x_i=0$ and $\sum_{i=1}^{6} x_i^2=6$. We wish to maximize $\prod x_i$.
One can show via Lagrange multipliers that at an extremal point, either $x_i=x_j$ or $x_i x_j=-1$, which leads to the maximum of $1/2$ when 4 of the variables are $1/\sqrt 2$ and the other two are $-\sqrt 2$.
However, given the niceness of the functions involved, it feels like there should be a more elementary approach to showing that $1/2$ is an upper bound.
Question: Is there a simpler alternative approach?
| Let $p$ denote the maximum of $\prod x_i$. Clearly, $p > 0$.
We only need to consider two cases:
Case 1: $x_1, x_2, x_3, x_4 > 0$ and $x_5, x_6 < 0$
Denote $A = x_1 + x_2 + x_3 + x_4 > 0$.
Then $x_5 + x_6 = -A$.
Clearly, we have $x_5^2 + x_6^2 \ge (x_5 + x_6)^2/2 = A^2/2$.
Also, $x_1^2 + x_2^2 + x_3^2 + x_4^2 \ge (x_1 + x_2 + x_3 + x_4)^2/4 = A^2/4$.
Thus, we have $6 = \sum_{i=1}^6 x_i^2 \ge A^2/2 + A^2/4$
which results in $A \le 2\sqrt 2$.
Using AM-GM, we have
$$x_1x_2x_3x_4 \le \frac{(x_1 + x_2 + x_3 + x_4)^4}{4^4} = A^4/4^4.$$
Also, we have
$$x_5x_6 \le \frac{(x_5 + x_6)^2}{4} = A^2/4.$$
Thus, we have
$$\prod_{i=1}^6 x_i
\le \frac{A^4}{4^4}\, \frac{A^2}{4} = \frac{A^6}{4^5}
\le \frac{1}{2}$$
with equality if $x_1 = x_2 = x_3 = x_4 = 1/\sqrt 2$
and $x_5 = x_6 = -\sqrt 2$.
Case 2: $x_1, x_2 > 0$ and $x_3, x_4, x_5, x_6 < 0$
Similar.
Thus, the maximum of $\sum_{i=1}^6 x_i$ is $1/2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How to decompose $\frac{1}{(1 + x)(1 - x)^2}$ into partial fractions Good Day.
I was trying to decompose $$\frac{1}{(1 + x)(1 - x)^2}$$ into partial fractions.
$$\frac{1}{(1 + x)(1 - x)^2} = \frac{A}{1 + x} + \frac{B}{(1 - x)^2}$$
$$1 = A(1 - x)^ 2 + B(1 + x)$$
Substitute $x = 1$, $$B = \frac{1}{2}$$
Substitute $x = -1$, $$A = \frac{1}{4}$$
However, $$\frac{1}{4(1 + x)} + \frac{1}{2 (1 - x)^2}$$ doesn't seem to equal $$\frac{1}{(1 + x)(1 - x)^2}$$
How do we decompose this into partial fractions?
Thanks
| A direct splitting as follows is also possible and sometimes quite quick:
\begin{eqnarray*} \color{blue}{\frac{1}{(1 + x)(1 - x)^2}}
& = & \frac{1+x-x}{(1 + x)(1 - x)^2} \\
& = & \frac{1}{(1 - x)^2} + \frac{1-x -1}{(1 + x)(1 - x)^2} \\
& = & \frac{1}{(1 - x)^2} + \frac{1}{(1 + x)(1 - x)} \color{blue}{- \frac{1}{(1 + x)(1 - x)^2}} \\
\end{eqnarray*}
Hence,
\begin{eqnarray*} \frac{\color{blue}{2}}{(1 + x)(1 - x)^2}
& = & \frac{1}{(1 - x)^2} + \frac 12\frac{1+x+1-x}{(1 + x)(1 - x)}\\
\end{eqnarray*}
So, you get
$$\frac{1}{(1 + x)(1 - x)^2}=\frac 12 \frac{1}{(1 - x)^2} + \frac 14 \frac{1}{1 - x} + \frac 14 \frac{1}{1 + x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Roll a fair 6 faces die. Stop if the sum of the previous rolls is a multiple of 3. What is the expected time to stop. Given there is a regular fair die, roll the die and stop if the sum of all previous rolls is a multiple of 3. What is the expected number of rolls?
Let $X$ denote the number of rolls until the event {the sum of all previous rolls is a multiple of 3} happens.
For $X = 1$, the rolls must be 3 or 6.
For $X = 2$, the rolls must be one of $\{(1,2), (2,1), (1,5), (5,1), (2,4), (4,2), (4,5), (5,4) \}$
For $X = 3$, the rolls must be one of $\{(1,1,1), (1,3,2),(1,3,5),(1,4,1), \ldots\}$
From the above enumeration, the pattern was not clear to me. I suppose there must be some "patterns" to simplify the computation.
| Let $f(x)$ denote the expected number of runs till we get remainder $0$ with $x$ denoting the current remainder mod $3$.
Note that each remainder has $\frac{1}{3}$ chance of occurring.
$$f(1) = \frac{1}{3}(1 + f(2)) + \frac{1}{3}(f(1) + 1) + \frac{1}{3}(1)$$
$$f(2) = \frac{1}{3}(1 + f(1)) + \frac{1}{3}(f(2) + 1) + \frac{1}{3}(1)$$
Note that in case of $f(1)$, if we get remainder $2$ on the roll (numbers $2$ and $5$), we end the process, so we count only $1$ roll.
Similarly, for $f(2)$.
We get $f(1) = f(2)$. Solving, we get $f(1) = f(2) = 3$.
Let us now iterate on the value of the first die roll. (Initial remainder is $0$)
$$ans = \frac{1}{3}(f(2) + 1) + \frac{1}{3}(f(1) + 1) + \frac{1}{3}(1)$$
$$ans = 3$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Polynomial division modulo $7$
We are given polynomials $f(x)=x^5+2x^4+3x^3+4x^2+5x+6,\ g(x)=3x^3+x$ in the polynomial ring $(\mathbb{Z}/7\mathbb{Z})[x]$. I want to find the polynomials $q,r$ for which $f=gq+r$. Note that $deg(r)<deg(q)$
Attempt:
Either $r=0$, or $r\ne 0$ and $\deg(r)<\deg(q)$. Note that $\deg(q)=\deg(f)-\deg(g)=2$. Hence $\deg(r)=1$. As such we can write $q(x)=ax^2+bx+c, \ r(x)=cx+d.$ Performing the multiplication we obtain $$x^5+2x^4+3x^3+4x^2+5x+6=(3x^3+x)(ax^2+bx+c)+dx+e.$$
Expanding and simplifying, this reduces to $$x^5+2x^4+3x^3+4x^2+5x+6=3ax^5+3bx^4+(3c+a)x^3+bx^2+(c+d)x+e$$
We immediately see that $b=4$ as it is the coefficient of $x^2$. Further we have;
$[3a]=1$ and so $[a]=5$
$[3b]=2$ so $[b]=3$
$[3c+a]=3$ so $[c]=4$
$[c+d]=5$ so $[d]=1$, and $[e]=6$
This gives $f(x)=g(x)(5x^2+4x+4)+(x+6)$, but this equality is actually incorrect upon expanding.
Checking $b=4$ in the coefficient of $x^4$ we get $4\times 3\equiv 5\bmod 7$ which isn't what we want. Where have I gone wrong?
| Your coefficient of $24x^2$ in the original problem changed to $4x^2$, but with reduction mod $7$, it should change to $3x^2$. This should also help resolve in your solution that in one place $b=3$ and in another $b=4$.
Edit after revision of problem:
The remainder polynomial is potentially of degree $2$. That is $\deg(r)<\deg(g)=3$. [Not $\deg(r)<\deg(q)=2$.]
| {
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How to show $\int_0^{\infty} \frac{\sin^3 x}{\cosh x\>+\>\cos x}\frac{dx}x=\frac{\pi}{8}$ I would like to evaluate the integral below$$\int_0^{\infty} \frac{\sin^3 x}{\cosh x+\cos x}\frac{dx}x $$
which I found to be $\frac \pi8$ numerically. I was able to evaluate a similarly looking, yet simpler, integral
$$\int_0^{\infty} \frac{\sin x}{\cosh x+\cos x}\frac{dx}x =\frac\pi4$$
by writing the integrand in the Frullani format and apply the theorem. However, it does not work in this case and I am not sure if they are related. Any solution is appreciated.
| It was shown in this answer that if $n$ is a nonnegative integer, then $$\int_{0}^{\infty} \frac{\sin\left((2n+1)x \right)}{\cosh (x) + \cos (x)} \, \, \frac{\mathrm dx}{x} = \frac{\pi}{4}. $$
Therefore, $$\begin{align} \int_{0}^{\infty}\frac{\sin^{3}(x)}{\cosh (x) + \cos (x)} \, \frac{\mathrm dx}{x} &= \frac{1}{4} \int_{0}^{\infty} \frac{3 \sin (x)- \sin(3x)}{\cosh (x) + \cos (x)} \, \frac{\mathrm dx}{x} \\ & = \frac{1}{4} \left(\frac{3 \pi}{4}- \frac{\pi}{4} \right) \\ & = \frac{\pi}{8}. \end{align}$$
In general, if $m$ is a nonnegative integer, then $$ \begin{align} \int_{0}^{\infty}\frac{\sin^{2m+1}(x)}{\cosh (x) + \cos (x)} \, \frac{\mathrm dx}{x} &\overset{(1)}{=} \frac{1}{2^{2m}} \sum_{k=0}^{m} (-1)^{m-k} \binom{2m+1}{k} \int_{0}^{\infty}\frac{\sin \left((2m+1-2k)x \right)}{\cosh(x) + \cos (x)}\frac{\mathrm dx}{x} \\ &= \frac{\pi}{2^{2m+2}}\sum_{k=0}^{m} (-1)^{m-k} \binom{2m+1}{k}. \end{align}$$
$(1)$ https://en.wikipedia.org/wiki/List_of_trigonometric_identities#Power-reduction_formulae
| {
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"url": "https://math.stackexchange.com/questions/4401894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplify $2\sin\frac{\pi}{4}2\cos\frac{\pi}{4}$ I have a question about $2\sin\frac{\pi}{4}2\cos\frac{\pi}{4}$. I ask because when I try to plug the equation in or substitute the double angle equation in for this problem I can never get the answer to come out. The commutative property says $(a\cdot b)\cdot c = a\cdot(b\cdot c)$. Here is where I think I am going wrong
$$2\cdot2\cdot\sin\frac{\pi}{4}\cos\frac{\pi}{4} = 2\cdot2\cdot\frac{\sqrt2}{2}\cdot\frac{\sqrt2}{2}
=2^2\cdot\cos^2\frac{\pi}{4}$$
Can I square the angle of the cos function because the sin and cos both have the same output at $\frac{\pi}{4}$? If I can square the cos function then I can plug it
into the Pythagorean identity, $\sin^2 + \cos^2 = 1$
$$4\sin^2\frac{\pi}{4}+ 4\cos^2\frac{\pi}{4} =1\cdot4$$
I would like to do this because then I could rearrange the terms in the identity to match the double angle cos identity. $$4\cos^2\frac{\pi}{4}=4-4\sin^2\frac{\pi}{4}$$ I think this is where I realize I do not know what is going on again. Here is my attempt to substitute into the cos double angle identity.
$$4\cos\left(2\cdot\frac{\pi}{4}\right) = 4-4\cdot2\sin^2\left(\frac{\pi}{4}\right)$$
When I solve this equation I get $0=0$ and it should be $2=2$. I seemed to have flip flopped the sin and cos functions wrong?
| We wish to simplify the expression
$$2\sin\left(\frac{\pi}{4}\right) \cdot 2\cos\left(\frac{\pi}{4}\right)$$
Let's continue your initial calculation.
\begin{align*}
2\sin\left(\frac{\pi}{4}\right) \cdot 2\cos\left(\frac{\pi}{4}\right) & = 2 \cdot 2\sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right)\\
& = 2 \cdot 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{2}}{2}\\
& = \sqrt{2} \cdot \sqrt{2}\\
& = \sqrt{4}\\
& = 2
\end{align*}
since the factors of $2$ in the numerator and denominator cancel out.
Another way to do the problem is to use the identity $\sin(2x) = 2\sin x\cos x$ with $x = \dfrac{\pi}{4}$.
\begin{align*}
2\sin\left(\frac{\pi}{4}\right) \cdot 2\cos\left(\frac{\pi}{4}\right) & = 2 \cdot 2\sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right)\\
& = 2\sin\left(2 \cdot \frac{\pi}{4}\right)\\
& = 2\sin\left(\frac{\pi}{2}\right)\\
& = 2 \cdot 1\\
& = 2
\end{align*}
You asked whether you could substitute $\cos\left(\dfrac{\pi}{4}\right)$ for $\sin\left(\dfrac{\pi}{4}\right)$ since $\cos\left(\dfrac{\pi}{4}\right) = \sin\left(\dfrac{\pi}{4}\right)$. Yes, you may make that substitution since you can substitute one quantity for another if the two quantities are equal. Doing so yields
\begin{align*}
2\sin\left(\frac{\pi}{4}\right) \cdot 2\cos\left(\frac{\pi}{4}\right) & = 2 \cdot 2\sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right)\\
& = 4\cos^2\left(\frac{\pi}{4}\right)\\
& = 4\left(\frac{\sqrt{2}}{2}\right)^2\\
& = 4\left(\frac{2}{4}\right)\\
& = 2
\end{align*}
as before.
You used the Pythagorean identity $\sin^2x + \cos^2x = 1$ to obtain
$$4\cos^2\left(\frac{\pi}{4}\right) = 4\left[1 - \sin^2\left(\frac{\pi}{4}\right)\right] = 4 - 4\sin^2\left(\frac{\pi}{4}\right)$$
Since
$$\sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}$$
we obtain
\begin{align*}
2\sin\left(\frac{\pi}{4}\right) \cdot 2\cos\left(\frac{\pi}{4}\right) & = 2 \cdot 2\sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right)\\
& = 4\cos^2\left(\frac{\pi}{4}\right)\\
& = 4 - 4\sin^2\left(\frac{\pi}{4}\right)\\
& = 4 - 4\left(\frac{\sqrt{2}}{2}\right)^2\\
& = 4 - 4\left(\frac{2}{4}\right)\\
& = 4 - 2\\
& = 2
\end{align*}
That said, introducing extra steps adds more opportunities to make an error.
Trying to use the double-angle identity $\cos(2x) = 1 - 2\sin^2x$ here is tricky.
\begin{align*}
2\sin\left(\frac{\pi}{4}\right) \cdot 2\cos\left(\frac{\pi}{4}\right) & = 2 \cdot 2\sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right)\\
& = 4\cos^2\left(\frac{\pi}{4}\right)\\
& = 4 - 4\sin^2\left(\frac{\pi}{4}\right)\\
& = 2 + 2 - 4\sin^2\left(\frac{\pi}{4}\right)\\
& = 2 + 2\left[1 - 2\sin^2\left(\frac{\pi}{4}\right)\right]\\
& = 2 + 2\cos\left(2 \cdot \frac{\pi}{4}\right)\\
& = 2 + 2\cos\left(\frac{\pi}{2}\right)\\
& = 2 + 2 \cdot 0\\
& = 2 + 0\\
& = 2
\end{align*}
Again, try to avoid introducing extra steps.
The first two methods are simplest.
| {
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"question_score": "1",
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How to prove $n^2+d$ is not square if $d|2n^2$ ? $n,d$ are positive integers. Obverously,if $n\ge d>0$,then
$$n^2<n^2+d<n^2+2n+1=(n+1)^2$$
So,$n^2+1$ is not square.
Naturelly,we consider $d>n$,but I can't get contradiction.And I do not use $d|2n^2$.
If we assume a positive $r$ s.t. $\sqrt{n^2+d}=r$,then we have $n^2+d = r^2$.From above equation , we can get $d|(n^2+r^2)$ and $d=r^2-n^2$.But I think it just identical transformation.I can't get useful information.
So ,can you give me some hints?Thanks!
| If $n=0$ then take $d$ any square to get a counterexample. So now let’s assume $n \neq 0$ and let $q\neq 0$ be an integer such that $q d = 2 n^2$. Suppose $n^2 + d = m^2$ for some integer $m$. Multiply both sides by $q^2$ to get $$(q^2 + 2q) n^2 = (q m)^2$$ and rewrite this as $$(q+1)^2 - 1 = \left(\frac{q m}n \right)^2.$$ Note that the right hand side is an integer and therefore $qm/n$ must also be an integer. So both $(q+1)^2$ and $(q+1)^2-1$ are squares. This implies $(q+1)^2 = 1$ and therefore $m=0$.
Conclusion: if $d\mid 2n^2$ and $n^2 + d$ is a square then either $n=0$ and $d$ is a square or $n\neq 0$ and $d = -n^2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find the power of the matrix. Let $A = \left( {\begin{array}{*{20}{c}}
0&1&1\\
1&0&1\\
1&1&0
\end{array}} \right)$.
I want to find $A^k,$ where $k \in N$. So far I calculated $A^2, A^3, A^4,...$ but I can not see the general formula for $A^k$. Here are $A^2, A^3, A^4, A^5$.
Not sure if this leads to anything but I found the general formula for $B^k$, where $B = \left( {\begin{array}{*{20}{c}}
1&1&1\\
1&1&1\\
1&1&1
\end{array}} \right)$.
${B^k} = \left( {\begin{array}{*{20}{c}}
{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\\
{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}\\
{{3^{k - 1}}}&{{3^{k - 1}}}&{{3^{k - 1}}}
\end{array}} \right)$
Thanks in advance.
| Given matrix is real symmetric and hence diagonalisable. Hence $A=PDP^{-1}$. Thus $A^n$=$PD^nP^{-1}$ where P is the matrix of eigen vectors. D is diagonal and hence its powers are simply the powers of the eigenvalues which are $2,-1,-1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the all possible values of $x$ such that $\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$ Find possible value of $x$ such that
$$\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) = \tan^{-1}(3)$$
Progress: what I did was to consider a case when $x^2 -1 < 1$ $(xy < 1)$ and $3x>-1$ $(xy > -1)$ and then apply $\tan^{-1}(x) \pm tan^{-1}(y)$ identity and got the range.
But is it correct to say we got all the possible value of $x$ or we need to consider all possibilities of $x^2 - 1 > 1$ , $3x < -1$ etc?
As what my book did was just did the first case and then left it without telling what about the solutions from other cases of $tan^{-1}(x) \pm \tan^{-1}(y)$.
| You do not need to consider cases.
Think first that you are looking for the zeros of function
$$f(x)=\tan^{-1}(x+1) + \tan^{-1}(x) + \tan^{-1}(x-1) -k$$ the first derivative
$$f'(x)=\frac{1}{1+(x+1)^2}+\frac{1}{1+x^2}+\frac{1}{1+(1-x)^2}=\frac {3 x^4+6 x^2+8 } {\left(x^2+1\right) \left(x^2-2 x+2\right) \left(x^2+2 x+2\right) }$$ is always positive; so, whatever is $k$, there is only one root.
$$k=\tan^{-1}(3)\implies f(0)=-\tan^{-1}(3)<0\quad \text{and}\quad f(1)=\frac{\pi }{4}-\cot ^{-1}(7)>0$$ So, the solution is $\in (0,1)$.
If you do not want to make anything complex, expand $f(x)$ as a series around $x=0$
$$f(x)=-\tan ^{-1}(3)+2 x-\frac{x^3}{6}+O\left(x^5\right)$$ Using the expansion to $O\left(x^3\right)$, the simplest estimate is
$$x_0 \sim \frac 12 \tan ^{-1}(3) =0.624523$$ while the exact solution is $5-\sqrt{19}=0.641101$.
If you already now about series reversion, using the expansion to $O\left(x^5\right)$ gives
$$x=\frac{1}{2} \left(f(x)+\tan ^{-1}(3)\right)+\frac{1}{96} \left(f(x)+\tan
^{-1}(3)\right)^3+O\left(\left(f(x)+\tan ^{-1}(3)\right)^5\right)$$ and since we want $f(x)=0$, then the estimate
$$x=\frac{1}{96} \tan ^{-1}(3) \left(48+\tan ^{-1}(3)^2\right)=0.644821$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409707",
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"source": "stackexchange",
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Solving integral $\int \frac{\sqrt{x^2 + x}}{x}dx$ (problem 36 in section $6.25$ in Tom Apostol's calculus) Integrals which involve $\sqrt{(cx + d)^2 - a^2}$ could often be simplified if we do a substitution $cx + d = a \sec t$. If we take a concrete example, $\int \frac{\sqrt{x^2 + x}}{x}dx$, then the substitution would be
$$
x + \frac{1}{2} = \frac{1}{2}\sec t \\
dx = \frac{1}{2} \sec t \tan t dt \\
t = arcsec (2x+1)
$$
If I carry on that substitution, I get to
$$
\frac{1}{2} \int \frac{\sqrt{\tan ^ 2t}}{\sec t - 1} \sec t \tan t dt
$$
As far as I understand, that is
$$
\frac{1}{2} \int \frac{|\tan t|}{\sec t - 1} \sec t \tan t dt
$$
Now, there are two cases, $\tan t \ge 0$ and $\tan t < 0$.
$\tan t \ge 0$ case leads me to the solution also written in the book (and here):
$$
\frac{1}{2} \int \frac{1 + \cos t}{\cos ^ 2 t} dt = \\
\frac{1}{2} \tan t + \frac{1}{2} \log{\frac{1 + \tan \frac{t}{2}}{1 - \tan \frac{t}{2}}} + C = \\
\sqrt{x^2 + x} - \frac{1}{2}\log{|2\sqrt{x^2 + x} + 2x + 1|} + C
$$
But if I try the second case (i.e. $\tan t < 0$), I get to the negation of the previous case, so I wonder where did I go wrong? Should I maybe not even consider the negative case?
Thanks!
| In the first step we try to guess the result basing on our personal experience
and check if the derivative gives back the integrand. If we miss, we can correct our guess and see what happens. In case we are unable to make a good guess, we try to simplify the integrand. In our case
$$x^2+x=\left (x+{1\over 2}\right)^2-{1\over 4}={1\over 4}[(2x+1)^2-1]$$
It seems convenient to perform the substitution $u=2x+1.$ In this way we get
$${1\over 2}\int {\sqrt{u^2-1}\over u-1}\,du$$ The integrated function is defined on two disjoint intervals $u<-1$ and $u>1.$ For $u>1$ it is tempting to make the substitution $u=\cosh t={1\over 2}(e^t+e^{-t}),$ for $t>0,$ having in mind that $\cosh ^2t-\sinh^2 t=1$ (where $\sinh t={1\over 2}[e^t-e^{-t}]).$
Then $du=\sinh t\,dt$ and we end up with the integral
$${1\over 2}\int {\sinh ^2t\over \cosh t-1}\,dt={1\over 2}\int {\cosh^2t-1\over \cosh t-1}\,dt ={1\over 2}\int [\cosh t+1]\,dt={1\over 2}\sinh t +{1\over 2}t$$ Next we have to get back to the original variable $x.$ Solving $$u=\cosh t ={1\over 2}[e^t+e^{-t}],\qquad t>0$$ leads to the quadratic equation for $e^t$
$${1\over 2}\,(e^{t})^2-u\,e^t+{1\over 2}=0$$ from which we get $$e^t=u+\sqrt{u^2-1},\quad \sinh t=\sqrt{u^2-1},\quad t=\log (u+\sqrt{u^2-1}) $$
Remark There are two roots of the equation and their product is equal $1.$ We choose the greater greater root, as $e^t>1$ for $t>0.$
Eventually, as $u=2x+1>1,
$ i.e. $x>0,$ we obtain $$\int {\sqrt{x^2+x}\over x}\,dx = {1\over 2}\log(2x+1+\sqrt{4x^2+4x})+{1\over 2}\sqrt{4x^2+4x}\\ ={1\over 2}\log (2x+1+2\sqrt{x^2+x})+\sqrt{x^2+x}\qquad \qquad (*)$$
It remains to consider the case left behind, when $u<-1,$ i.e. $x<-1.$ But since formula $(*)$ holds for $x>1$ and $2x+1+2\sqrt{x^2+x}<0,$ for $x<-1,$
we get the final result
$$\int {\sqrt{x^2+x}\over x}\,dx ={1\over 2}\log \mid 2x+1+2\sqrt{x^2+x}\mid+\sqrt{x^2+x}\qquad x\notin [-1,0]$$
Remark Mind that if we are looking for a particular antiderivative $F(x),$ satisfying $F(1)=a $ and $F(-2)=b,$ we have to add to the solution above different constants, depending on the cases $x<-1$ or $x>0.$ Therefore I resist from writing $\int f(x)\,dx =F(x)+C,$ when $f(x)$ is defined on several disjoint open intervals. The formula $\int f(x)\,dx =F(x)$ is just equivalent to $F'(x)=f(x).$ For example $\int 2x\,dx =x^2$ and also $\int 2x\,dx =x^2+1,$ but it does not imply $0=1.$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\frac{2^{ab}-1}{2^a-1} $ is composite number let $b>a>1$ be postive integers,show that
$$\dfrac{2^{ab}-1}{2^a-1} $$ is composite number
I try use $$(x^n-1)=(x-1)(x^{n-1}+x^{n-2}+\cdots+x+1)$$
so $x-1|x^n-1$.then let $x=2^a,n=b$ we have
$$\dfrac{2^{ab}-1}{2^a-1}=2^{(b-1)a}+2^{(b-2)a}+\cdots+2^a+1$$
but How to prove $2^{(b-1)a}+2^{(b-2)a}+\cdots+2^a+1$ is composite ？
| We use this $fact*$ that:
if m is a natural number and $a>1$ we have:
$$\big(\frac{a^m-1}{a-1}, a-1\big)= (a-1, m)$$
Proof:
Suppose $\big(\frac{a^m-1}{a-1}, a-1\big)=d$
Using following identity:
$\frac{a^m-1}{a-1}=(x^{m-1}-1)+(x^{m-2}-1)+(x^{m-3}-1)+\cdot\cdot\cdot+(a-1)+m$ $\space\space\space\space\space (1)$
and the fact that ($a^k-1$) is divisible by $(a-1)$ we conclude that m is divisible by d. If numbers $(a-1)$ and $m$ have a common divisor like ($\sigma>d$), considering relation (1) we conclude that ($\frac{a^m-1}{a-1}$) is divisible by ($\sigma$) and numbers ($\frac{a^m-1}{a-1}$) and ($a-1$) have a common divisor $(\sigma>d)$., which results in $(a-1, m)=d$
Now we have :
$A=\frac {2^{ab}-1}{2^a-1}=\big(\frac{(2^a)^b-1}{2^a-1}\big)$ which has a common divisor like d with $(m=b)$ that means $A$ is divisible by d, so it is a composite number.
*
*This fact and its proof is from a number theory book by Sierpinski.
| {
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Let $\left,\left$ be real sequences satisfied as following conditions： $\left<a_i\right>, \left<b_i\right>$ be real sequences with
$$a_1^2+a_2^2+\cdots+a_n^2=1，\\
b_1^2+b_2^2+\cdots+b_n^2=1，\\
a_1b_1+a_2b_2+\cdots+a_nb_n=0.$$
Prove that $(a_1+a_2+\cdots+a_n)^2+(b_1+b_2+\cdots+b_n)^2\leq n$.
My attempt：
I try to prove it by induction，
As $n=2$，$a_1^2+a_2^2=1，b_1^2+b_2^2=1，a_1b_1+a_2b_2=0$
Because $a_1, a_2, b_1, b_2$ cannot be all $0$，W.L.O.G.，Assume $a_1\neq0$，
$b_1=-\frac{a_2b_2}{a_1}\Rightarrow$ $\frac{a_2^2b_2^2}{a_1^2}+b_2^2=1\Rightarrow b_2^2=a_1^2，b_1^2=a_2^2$
\begin{align}
(a_1+a_2)^2+(b_1+b_2)^2 & = a_1^2+a_2^2+b_1^2+b_2^2+2(a_1a_2+b_1b_2)\\
&=2+2(a_1a_2-\frac{a_2b_2^2}{a_1})\\
&=2+2\cdot\frac{a_2}{a_1}(a_1^2-b_2^2)\\
&=2
\end{align}
And the induction hypothesis is that $n=k$，$a_1^2+a_2^2+\cdots+a_k^2=1，b_1^2+b_2^2+\cdots+b_k^2=1，a_1b_1+a_2b_2+\cdots+a_kb_k=0，$
$(a_1+a_2+\cdots+a_k)^2+(b_1+b_2+\cdots+b_k)^2\leq k$ holds；
As $n=k+1$，$a_1^2+a_2^2+\cdots+a_k^2+a_{k+1}^2=1，b_1^2+b_2^2+\cdots+b_k^2+b_{k+1}^2=1，a_1b_1+a_2b_2+\cdots+a_kb_k+a_{k+1}b_{k+1}=0，$
Assume $a_1\neq 0$，
\begin{align}
& b_1=-\frac{a_2b_2+a_3b_3+\cdots a_{k+1}b_{k+1}}{a_1}\\
\Rightarrow & \frac{(a_2b_2+a_3b_3+\cdots+a_{k+1}b_{k+1})^2}{a_1^2}+b_2^2+\cdots+b_{k+1}^2=1\\
\Rightarrow &
(a_2b_2+a_3b_3+\cdots+a_{k+1}b_{k+1})^2+a_1^2b_2^2+\cdots+a_1^2b_{k+1}^2=a_1^2\\
\Rightarrow & ...
\end{align}
I cannot finish the proof. Please help~
| Setup/ rephrasing the question:
Consider the vectors $ a = ( a_1, \ldots, a_n), b = (b_1, \ldots , b_n)$. The conditions state that these are orthogonal unit vectors, $ \angle (a, b) = 90^\circ$.
Consider unit vector $c = ( \frac{1}{\sqrt{n} } , \ldots \frac{1}{\sqrt{n}} )$ and $-c$.
Since $ \angle (a, c) + \angle (a, -c) = 180^\circ$ either $ \angle (a, c) \leq 90^\circ$ or $ \angle (a, -c) \leq 90^\circ$. WLOG, let it hold for $ c$.
The inequality is equivalent to showing that:
$$ (a\cdot c) ^2 + ( b \cdot c)^2 \leq 1 \Leftrightarrow \cos^2 \angle (a, c) + \cos^2 \angle (b, c) \leq 1.$$
Proof: Let $ \alpha = \angle (a, c) $ and $ \beta = \angle (b, c)$.
We have $ 90^\circ = \angle (a, b) \leq \angle (a, c) + \angle (c, b) = \alpha + \beta $ and $ \angle(b, c) \leq \angle (b, a) + \angle (a, c) $, thus$ 90^\circ - \alpha \leq \beta \leq \ \alpha + 90^\circ$.
Recall from above that the choice of $c$ resulted in $ 0^\circ \leq \angle (a, c) \leq 90^\circ$, hence $ \cos^2 \beta \leq \sin^2 \alpha$.
Thus, $ \cos^2 \alpha + \cos^2 \beta \leq \cos^2 \alpha + \sin^2 \alpha = 1$ as desired.
Thus the inequality is true.
Equality holds iff the various "angle at a point inequality" holds, meaning that $a, b, c$ lie on the same plane (and $a, b$ are orthogonal as per the condition). Note that we do not require "$c$ lies in-between $a$ and $b$".
| {
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Showing the AM-GM inequality, $\textit{Problem From The Book}$, 19.17, but using integrals The chapter of this problem is Solving Elementary Inequality Using Integrals. After I typed the problem: I spent several hours trying to solve it, but to no avail, so I am hoping someone here can enlighten me.
For part $a$,I can prove AM-GM inequality by intgeral,But for Weighted AM–GM inequality I can't
My approach maybe
$$\dfrac{A}{G}-1=\sum_{i=1}^{k} r_i \int_{a_i}^{G} (\frac{1}{t} - \frac{1}{G}) dt + \sum_{i=k+1}^{n} r_i \int_{G}^{a_i} (\frac{1}{G} - \frac{1}{t}) dt $$
For part $b$: I can solve it(without use integral),because it must to prove
$$\dfrac{A}{A'}\ge\dfrac{G}{G'}$$ or
$$\dfrac{\sum_{i=1}^{n}r_{i}x_{i}}{\sum_{i=1}^{n}r_{i}(1-x_{i})}\ge \prod_{i=1}^{n}\left(\dfrac{x_{i}}{1-x_{i}}\right)^{r_{i}}$$
so we consider
$$f(x)=\ln{\dfrac{x}{1-x}}=\ln{x}-\ln{(1-x)}(0<x<0.5)$$
and
$$f'(x)=\dfrac{1}{x}+\dfrac{1}{1-x},f''(x)=-\dfrac{1}{x^2}+\dfrac{1}{(1-x)^2}<0(0<x<\frac{1}{2}$$
so By Jenson inequality
$$\sum_{i=1}^{n}r_{i}f(x_{i})\le f(\sum_{i=1}^{n}r_{i}x_{i})$$
so I have prove it part b,
But I can't prove part a
| $\newcommand{\d}{\,\mathrm{d}}$The first step is in evaluating $I$. Let's first note that each $x_i$ and $A$ itself are strictly positive so we only care about evaluating $I$ in the case $x,a\gt0$. A partial fraction decomposition is in order - I leave the derivation at the bottom of the post for readability:
$$\frac{t}{(1+t)(x+at)^2}=\frac{x}{x-a}\cdot\frac{1}{(x+at)^2}+\frac{a}{(x-a)^2}\cdot\frac{1}{x+at}-\frac{1}{(x-a)^2}\cdot\frac{1}{1+t}$$
So:
$$\begin{align}I(x,a)&=\frac{x}{x-a}\int_0^\infty\frac{1}{(x+at)^2}\d t+\frac{1}{(x-a)^2}\int_0^\infty\left(\frac{a}{x+at}-\frac{1}{1+t}\right)\d t\\&\overset{u=at}{=}\frac{x}{x-a}\cdot\frac{1}{a}\int_0^\infty\frac{1}{(x+t)^2}\d t+\frac{1}{(x-a)^2}[\ln(x+at)-\ln(1+t)]_0^\infty\\&\overset{u=t+x}{=}\frac{x}{a(x-a)}\int_x^\infty\frac{1}{t^2}\d t+\frac{1}{(x-a)^2}\left[\lim_{t\to\infty}\ln\left(\frac{x+at}{1+t}\right)-\ln x\right]\\&=\frac{1}{a(x-a)}+\frac{1}{(x-a)^2}[\ln a-\ln x]\end{align}$$
Let's turn our attention to the summand: $$\begin{align}r_i(x_i-A)^2I(x_i,A)&=r_i(x_i-A)^2\left(\frac{1}{A(x_i-A)}+\frac{1}{(x_i-A)^2}\ln A-\frac{1}{(x_i-A)^2}\ln x_i\right)\\&=r_i\left(\frac{x_i}{A}-\frac{A}{A}+\ln A-\ln x_i\right)\\&=-r_i+\frac{1}{A}r_ix_i+\ln A\cdot r_i-r_i\ln x_i\end{align}$$
And let's sum, recalling that $\sum_{i=1}^nr_i=1$:
$$\begin{align}\sum_{i=1}^nr_i(x_i-A)^2I(x_i,A)&=-\sum_{i=1}^nr_i+\frac{1}{A}\sum_{i=1}^nr_ix_i+\ln A\cdot\sum_{i=1}^nr_i-\sum_{i=1}^nr_i\ln x_i\\&=-1+\frac{1}{A}(A)+\ln A-\sum_{i=1}^n\ln(x_i^{r_i})\\&=\ln A-\ln\left(\prod_{i=1}^nx_i^{r_i}\right)\\&=\ln\frac{A}{G}\end{align}$$
How do we conclude? Well, as $r_i,x_i,A\gt0$ always, the integral $I(x_i,A)$ is an integral of purely nonnegative terms and is thus nonnegative; $(x_i-A)^2\ge0$ since the squares of real numbers are nonnegative, and as said $r_i\gt0$ always, so each summand $r_i(x_i-A)^2I(x_i,A)\ge0$ for all $i$, so that: $$\begin{align}\ln\frac{A}{G}&=\sum_{i=1}^nr_i(x_i-A)^2I(x_i,A)\ge0\\\ln\frac{A}{G}&\ge0\\\frac{A}{G}&\ge1\\A&\ge G\quad\blacksquare\end{align}$$
And we are all done!
N.B. On the partial fraction decomposition:
$$\begin{align}\frac{t}{(1+t)(x+at)^2}&=\frac{A}{1+t}+\frac{B}{x+at}+\frac{C}{(x+at)^2}\\\implies t&=A(x+at)^2+B(1+t)(x+at)+C(1+t)\\\implies t&=Ax^2+Bx+C+2Aaxt+Bxt\\&+Bat+Ct+Aa^2t^2+Bat^2\\\implies\tag{1}0&=Ax^2+Bx+C=Ax+B+\frac{C}{x}\\\tag{2}0&=Aa^2+Ba=Aa+B\\\tag{3}1&=2Aax+Bx+Ba+C\end{align}$$
Through comparing coefficients in $1,t,t^2$. Let's use $(1)$ to say that $Ax=-B-\frac{C}{x}$ and use $(2)$ to say that $Aa=-B$, and substitute them into $(3)$: $$\begin{align}\tag{4}1&=2(Aa)x+Bx+Ba+C\overset{(2)}{=}-2Bx+Bx+Ba+C\\&=Ba-Bx+C\\\tag{5}1&=2a(Ax)+Bx+Ba+C\overset{(3)}{=}-2Ba-2a\frac{C}{x}+Bx+Ba+C\\&=Bx-Ba+C\cdot\frac{x-2a}{x}\end{align}$$Now add $(4),(5)$ and note the $Ba-Ba,Bx-Bx$ cancel: $$\tag{6}2=C+C\cdot\frac{x-2a}{x}=C\cdot\frac{2x-2a}{x}\implies C=\frac{x}{x-a}$$Let's put this into $(4)$: $$1=B(a-x)+\frac{x}{x-a}\implies B=\frac{1}{a-x}\cdot\frac{(x-a)-x}{x-a}=\frac{a}{(x-a)^2}$$And finally let's use $(2)$ to say that: $$A=-\frac{B}{a}=-\frac{1}{(x-a)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
An interesting system of simultaneous equations with two unknowns. $ \text{We are going to solve the system (E):}$
$\left\{\begin{aligned} \displaystyle \quad x^{n}+y^{n}&=1 \cdots(1) \\x^{n+1}+y^{n+1}&=1 \cdots(2)\end{aligned}\right., \quad $ where $m,n\in N.$
$\displaystyle \begin{aligned}x&=\frac{x^{n+1}}{x^{n}}\\&=\frac{1-y^{n+1}}{1-y^{n}}\\&=\frac{1-y^{n}+y^{n}-y^{n+1}}{1-y^{n}} \\ \displaystyle &=1+\frac{y^{n}(1-y)}{(1-y)\left(1+y+y^{2}+\cdots+y^{n-1}\right)} \\ \displaystyle &=1+\frac{y^{n}}{1+y+y^{2}+\cdots+y^{n-1}} \cdots \text { (3) } \end{aligned} \tag*{} $
$\textrm{Note that either $n$ or $n+1$ is odd,}\\$
$\displaystyle \quad x^m+y^m=1 \Rightarrow \textrm{ one of x and y must be non-negative for any odd integer m.}$
WLOG, let $y\geq 0. $ $\textrm{ However, if y is positive, then by (3)},$
$$\displaystyle x=1+\frac{y^{2 n}}{1+y+y^{2}+\cdots+y^{2 n-2}}>1\Rightarrow \quad x^n+y^n>1, $$
$\text{which contradicts to one of the equation (1) or (2).}$
$\textrm{Hence }y=0 \textrm{ and } x=1.$
$\textrm{By symmetry, the real solutions are }$$ (1,0)\text{ and }(0,1).$
My question
What can we say for the system $$(F):\left\{\begin{aligned} \displaystyle x^{m}+y^{m}&=1 \cdots(3) \\x^{n}+y^{n}&=1 \cdots(4)\end{aligned}\right.,$$
where $m,n\in N.$
| I believe, there is more simple solution than mine.
My solution is based on evident consequence of system: $(1-x^m)^n=(1-x^n)^m$.
Lemma. If $0\leq x \leq 1$ and $m > n$ then $(1-x^m)^n=(1-x^n)^m$ has only two solutions $x=0$ and $x=1$.
Proof. The fact that $x=0$ and $x=1$ are solutions is easy to check. Let's consider case $0 < x < 1$, then $0 < x^m < x^n <1$, $1 > 1-x^m > 1-x^n > 0$, $1 > (1-x^m)^n > (1-x^m)^m > (1-x^n)^m > 0$. So $(1-x^m)^n \neq (1-x^n)^m$ for $0<x<1$.
Solution. Let's consider system $x^m+y^m=1$, $x^n+y^n=1$, $m,n\in\mathbb{N}$. Without loss of generality $m > n$. Case $m=n$ has infinitely many solutions so we don't consider it.
There are three possible cases: $m$, $n$ are both even, one of them is odd, both $m$ and $n$ are odd.
Case 1. $m$, $n$ are even. Then $|x|^m+|y|^m=1$, $|x|^n+|y|^n=1$. $|y|^m \geq 0 \Rightarrow |x|^m \leq 1 \Rightarrow |x| \leq 1$. $|x|$ should satisfy equation $(1-|x|^m)^n=(1-|x|^n)^m$. Using lemma, we can say that $|x|=0$ or $|x|=1$. Putting this into equations, we can get four solutions $(0;1)$, $(1;0)$, $(0;-1)$, $(-1;0)$.
Case 2. One of $m$ and $n$ is odd. Let's mark this odd number as $k$ and even number as $l$l. $|x|^l+|y|^l=1$, $|x|^l \geq 0 \Rightarrow |y|^l \leq 1 \Rightarrow |y| \leq 1$. $x^k+y^k=1 \Rightarrow y^k=1-x^k \Rightarrow |y|^k =|1-x^k|$. $|y|\leq 1 \Rightarrow |y|^k \leq 1 \Rightarrow |1-x^k| \leq 1 \Rightarrow x^k \geq 0 \Rightarrow x \geq 0$. $|x|^l+|y|^l=1$, $|y|^l\geq 0 \Rightarrow |x|^l\leq 1 \Rightarrow |x|\leq 1 \Rightarrow 0\leq x \leq 1$. $x$ must satisfy equation $(1-x^m)^n=(1-x^n)^m$. Using lemma, we can say that $x=0$ or $x=1$. Putting this into equations, we can get two solutions $(0;1)$ and $(1;0)$.
Case 3. Both $m$ and $n$ are odd. Consider three subcases $x > 1$, $x < 0$ and $0 \geq x \geq 1$.
When $x>1$: $-\left(\frac{y}{x}\right)^{mn}=(1-\left(\frac{1}{x})^m\right)^n=(1-\left(\frac{1}{x}\right)^n)^m$. Then $u=\frac{1}{x}$ must satisfy equation $(1-u^m)^n=(1-u^n)^m$ and $0 < u < 1$. Using lemma, we can say that this subcase has no solutions.
When $x < 0$: $y^m=1-x^m > 1 \Rightarrow y > 1$. Then we can use the same consideration as for subcase $x > 1$.
When $0 \leq x \leq 1$ we can use lemma and get $x=0$ or $x=1$. Putting into equations, we can get two solutions $(0;1)$ and $1;0$.
Final answer: when $m$ and $n$ are both even, there are four solutions $(0;1)$, $(1;0)$, $(0;-1)$, $(-1;0)$, otherwise there are two solutions $(0;1)$ and $1;0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4412285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
why does the golden ratio, Φ=√(81/80)+√(1/80)+1/2? I have been trying to understand the mathematics of the golden ratio starting with the following 345 relationship.
Can anybody help me understand why $ Φ = \sqrt{81/80} + \sqrt{1/80} + 1/2 $ ?
I think it is something to do with these angles but its quite complicated.
I know that $ Φ + 1/Φ = \sqrt{5} $ and $ Φ - 1/Φ = 1 $.
And also that $ (\sqrt{5} + 2) + (\sqrt{5} - 2) = \sqrt{20} $ and $ (\sqrt{5} + 2) - (\sqrt{5} - 2) = 4 $.
Which shows a link to the metallic means (1 and 4). If anybody can provide any help it will be greatly appreciated.
| As suggested by @Blue:
$$\sqrt{\frac{81}{80}} + \sqrt{\frac{1}{80}} + \frac12 \ \ \ = \ \ \ \frac{\sqrt{81}}{\sqrt{80}} + \frac{\sqrt{1}}{\sqrt{80}} + \frac12$$
$$=\frac{9}{\sqrt{80}} + \frac{1}{\sqrt{80}} + \frac12 \ \ \ = \ \ \ \frac{10}{\sqrt{80}} + \frac12$$
$$=\frac{10}{\sqrt{16 \times 5}} + \frac12 \ \ \ = \ \ \ \frac{2 \times \sqrt{5} \times \sqrt{5}}{4 \sqrt{5}} + \frac12=$$
$$=\frac{\sqrt{5}}{2} + \frac12 \ \ \ = \ \ \ \frac{1+\sqrt{5}}{2} \ \ \ = \ \ \ \Phi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4413526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\int_{0}^{2\pi}f(x)\cos(kx)dx \geq 0$ for every $k \geq 1$ given that $f$ is convex. Given $f: [0, 2\pi] \to \mathbb{R}$ convex function, prove that for every $k\geq1$
\begin{align}
\int_{0}^{2\pi}f(x) \cos (kx)dx \geq 0
\end{align}
I am completely stumped. What I have tried to do is return the query for $k=1$, and for that value of $k$ try to write the integral from $0$ to $2\pi$ as a sum of four integrals from $0$ to $\dfrac{\pi}{2}$ and use the the theorem for first derivative monotony. No luck so far.
Any help would be much appreciated.
Edit 1: I saw the link here about a similarly asked topic. However, this process gets the general case as I perceive it and I am really supposed to use the method described above. I will try this and come back with a definitive answer.
Edit 2: I cannot use the $f''(x) \geq 0$ argument due to the simple fact that I have not been formally taught it as part of the class.
Edit 3: Final proof, with thanks to the contributors below.
Let's start by setting $A = \frac{\pi}{2}$ and $B = \frac{3\pi}{2}$. It follows from basic trigonometry that:
\begin{align}
&\cos x \geq 0, \ x \in [A,B] \ \text{and}\\
&\cos x \leq 0, \ x \in [0, A] \cap [B, 2\pi].
\end{align}
And we also set $L$ to be the line segment such that $L(A) = f(A), \ L(B) = f(B)$. We will prove a basic property of said line in regards to the convex function $f$.
*
*I can take for granted that (we proved this in class)
\begin{align}
L(x) = \dfrac{x-A}{B-A}f(B) + \dfrac{B-x}{B-A}f(A)
\end{align}
so for $x \in [A,B]$ there exists $\lambda \in [0,1]$ such that: $x = \lambda A + (1-\lambda) B$. Taking the aforementioned expression and replacing it on $L(x)$ we get (I omit trivial algebra)
\begin{align}
L(x) = (1-\lambda) f(B) + \lambda f(A).
\end{align}
Since $f$ is convex, we can write
\begin{align}
&f(\lambda A + (1-\lambda) B) \leq \lambda f(A) + (1-\lambda) f(B)\\
\implies &f(x) \leq L(x), \ \forall \ x \in [A,B] \ \text{and} \ \lambda \in [0,1].
\end{align} $\blacksquare$
*We assume that there exists $x \in [0,A]: \ f(x) < L(x)$. Then there exists $A \in [x, B] \ \text{and} \ \lambda \in [0,1]: \ A = \lambda x + (1-\lambda) B$. Then
\begin{align}
&L(A) = \dfrac{A-x}{B-x}f(B) + \dfrac{B-A}{B-x}f(x), \ A \in [x,B]\\
\implies &L(A) = (1-\lambda)L(B)+\lambda L(x)\\
\implies &L(A) = (1-\lambda)f(B) + \lambda L(x).
\end{align}
Then assuming that $L(x) > f(x)$ we get
\begin{align}
L(A) > (1-\lambda) f(B) + \lambda f(x) \geq f(A), \ \text{assuming convexity}.
\end{align}
Because $L(A) = f(A)$ the above inequality becomes $L(A) > f(A)$ which is a contradiction.
$\blacksquare$
In the same spirit, for $x \in [B, 2\pi]$ doing the exact same replacements and applying the exact same principles we also get
\begin{align}
&f(\lambda B + (1-\lambda) 2 \pi) \leq \lambda f(B) + (1-\lambda) f(2\pi)\\
\implies &f(x) \leq L(x).
\end{align}
$\blacksquare$
We then set $g(x) = f(x) - L(x)$ and from (1) and (2) above it is safe to assume that $\cos x$ and $g(x)$ will have the same sign in the whole domain, that is:
\begin{align}
&g(x) \geq 0, \ \text{where} \ \cos x \geq 0\\
&g(x) \leq 0, \ \text{where} \ \cos x \geq 0.
\end{align}
*We have
\begin{align}
\int_0^{2\pi} g(x) \cos x dx = \int_0^{\frac{\pi}{2}} g(x) \cos x dx + \int_{\frac{\pi}{2}}^{\frac{3 \pi}{2}} g(x) \cos x dx + \int_{\frac{3\pi}{2}}^{2\pi}g(x) \cos x dx \geq 0,
\end{align}
because
\begin{align}
&\text{in} \left[0, \frac{\pi}{2} \right], \ \cos x \geq 0 \implies g(x) \geq 0\\
&\text{in} \left[\frac{\pi}{2}, \frac{3 \pi}{2} \right], \ \cos x \leq 0 \implies g(x) \leq 0\\
&\text{in} \left[\frac{3 \pi}{2}, 2 \pi \right], \ \cos x \geq 0 \implies g(x) \geq 0.\\
\end{align}
*We have that
\begin{align}
&\int_0^{2 \pi}\cos x dx = 0 \ \text{trivial}\\
&\int_0^{2 \pi}x \cos x dx = \int_0^{2 \pi}x (\sin x)' dx = \left[ x \sin x \right]_0^{2\pi} - \int_0^{2 \pi} \sin x dx = 0.
\end{align}
It then follows that
\begin{align}
\int_0^{2 \pi} f(x) \cos x dx = \int_0^{2 \pi} g(x) \cos x dx \geq 0
\end{align}
which was previously proven. For the case $k=1$, the proof is over.
For $k>1$, we have:
\begin{align}
\int_0^{2\pi}f(x) \cos (kx) dx = \sum_{i=0}^{k-1} \int_{\frac{2\pi i}{k}}^{\frac{2\pi (i+1)}{k}} f(x) \cos (kx) dx.
\end{align}
We perform the change of variable
\begin{align}
x = \dfrac{y+2\pi i}{k}\\
\implies \begin{cases}dx = \dfrac{1}{k}dy\\ x = \dfrac{2\pi i}{k} \to y=0\\ x = \dfrac{2\pi (i+1)}{k} \to y = 2\pi \end{cases}.
\end{align}
So the above sum becomes
\begin{align}
\sum_{i=0}^{k-1} \dfrac{1}{k} \int_{0}^{2\pi}f \left(\dfrac{y+2 \pi i}{k} \right) \cos (y+2\pi i)dy.
\end{align}
Because $f$ is convex in $[0, 2\pi]$ there exists $\lambda \in [0,1]$ such that:
\begin{align}
\theta f\left(\frac{y_1 + 2\pi i}{k}\right)
+ (1 - \theta)f\left(\frac{y_2 + 2\pi i}{k}\right)
&\ge f\left(\theta\frac{y_1 + 2\pi i}{k}
+ (1 - \theta) \frac{y_2 + 2\pi i}{k}\right)\\
&= f\left(\frac{\theta y_1 + (1 - \theta)y_2 + 2\pi i}{k}\right).
\end{align}
Having performed the change of variables:
\begin{align}
x_1 = \dfrac{y_1 + 2\pi i}{k} \ \text{and} \ x_2 = \dfrac{y_2 + 2\pi i}{k}.
\end{align}
So $f \left( \dfrac{y + 2\pi i}{k}\right)$ convex on $[0, 2\pi]$. Using the result from $k=1$ we have
\begin{align}
\int_0^{2\pi} f \left( \dfrac{y + 2\pi i}{k} \right)\cos y dy \geq 0.
\end{align}
$\blacksquare$
| Complement to @ECL's answer:
*
*For $k = 1$:
Let
$$g(x) := f(x) - \frac{f(3\pi/2) - f(\pi/2)}{\pi}x - \frac{\frac{3\pi}{2}f(\pi/2) - \frac{\pi}{2}f(3\pi/2)}{\pi}.$$
We have $g(\pi/2) = g(3\pi/2) = 0$.
Also, $g(x)$ is convex on $[0, 2\pi]$.
For $x \in [\pi/2, 3\pi/2]$, letting $t = \frac{3}{2} - \frac{x}{\pi} \in [0, 1]$, we have $x = \frac{\pi}{2} t
+ \frac{3\pi}{2}(1 - t)$ and
$$g(x) \le t g(\pi/2) + (1 - t)g(3\pi/2) = 0.$$
For $x\in [0, \pi/2]$, using $\pi - x \in [\pi/2, 3\pi/2]$ and $g(x) + g(\pi - x) \ge 2g(\pi/2) = 0$, we have
$$g(x) \ge -g(\pi - x) \ge 0.$$
For $x\in [3\pi/2, 2\pi]$, using $3\pi - x\in [\pi/2, 3\pi/2]$ and $g(x) + g(3\pi - x) \ge 2g(3\pi/2) = 0$, we have
$$g(x) \ge -g(3\pi - x) \ge 0.$$
Using $\int_0^{2\pi} \cos x \mathrm{d} x = 0$ and $\int_0^{2\pi} x \cos x \mathrm{d} x = 0$, we have
\begin{align*}
&\int_0^{2\pi} f(x) \cos x \mathrm{d} x\\
=\,& \int_0^{2\pi} g(x)\cos x \mathrm{d} x\\
=\,& \int_0^{\pi/2} g(x)\cos x \mathrm{d} x
+ \int_{\pi/2}^{3\pi/2} g(x)\cos x \mathrm{d} x
+ \int_{3\pi/2}^{2\pi} g(x)\cos x \mathrm{d} x\\
\ge\,& 0.
\end{align*}
$\phantom{2}$
*For $k > 1$:
We have
\begin{align*}
&\int_0^{2\pi} f(x)\cos k x \mathrm{d} x\\
=\,& \sum_{i=0}^{k-1} \int_{2\pi i/k}^{2\pi(i + 1)/k} f(x)\cos k x \,\mathrm{d} x\\
=\,& \sum_{i=0}^{k-1} \frac{1}{k}\int_0^{2\pi} f\left(\frac{y + 2\pi i}{k}\right)\cos y\, \mathrm{d} y.
\end{align*}
For any $y_1, y_2 \in [0, 2\pi]$ and $\theta \in [0, 1]$, we have
\begin{align*}
\theta f\left(\frac{y_1 + 2\pi i}{k}\right)
+ (1 - \theta)f\left(\frac{y_2 + 2\pi i}{k}\right)
&\ge f\left(\theta\frac{y_1 + 2\pi i}{k}
+ (1 - \theta) \frac{y_2 + 2\pi i}{k}\right)\\
&= f\left(\frac{\theta y_1 + (1 - \theta)y_2 + 2\pi i}{k}\right).
\end{align*}
Thus, $f\left(\frac{y + 2\pi i}{k}\right)$ is convex on $[0, 2\pi]$.
Using the result for $k = 1$, we have
$$\int_0^{2\pi} f\left(\frac{y + 2\pi i}{k}\right)\cos y \,\mathrm{d} y \ge 0.$$
The desired result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 3
} |
How to evaluate $\sum_{k=1}^{\infty}\frac{B\left(k, \frac{1}{2}\right)}{(2k+1)^2}$, where $B(x, y)$ is the Beta function? I am trying to evaluate this sum:
$$\sum_{k=1}^{\infty}\frac{B\left(k, \frac{1}{2}\right)}{(2k+1)^2}$$ where $B(x, y)$ is the Beta function.
Checking with WolframAlpha gives beautiful result: $4-4G$ where $G$ is the Catalan constant.
I tried to use the integral definition of Beta function:
$$B\left(k, \frac{1}{2}\right)=\int_{0}^{1}x^{k-1}(1-x)^{\frac{-1}{2}}dx,$$ so $$\sum_{k=1}^{\infty}\frac{B(k, \frac{1}{2})}{(2k+1)^2}=\sum_{k=1}^{\infty}\frac{1}{(2k+1)^2}\int_{0}^{1}x^{k-1}(1-x)^{\frac{-1}{2}}dx.$$ After changing the order of summation and integration: $$\int_{0}^{1}\frac{1}{\sqrt{1-x}}dx\sum_{k=1}^{\infty}\frac{x^{k-1}}{(2k+1)^2}.$$ And getting stuck, because the latter summation will lead to Lerch transcendent: $$\int_{0}^{1}\frac{\Phi\left(x,2,\frac{3}{2}\right)}{4\sqrt{1-x}}dx.$$ And I don't know how to progress further.
| Let's denote
$$I=\int_{0}^{1}\frac{1}{\sqrt{1-x}}\sum_{k=1}^{\infty}\frac{x^{k-1}}{(2k+1)^2}\,dx$$
Using $\frac{1}{(2k+1)^2}=-\int_0^1t^{2k}\ln t dt$
$$I=-\int_{0}^{1}\frac{dx}{\sqrt{1-x}}\sum_{k=1}^{\infty}x^{k-1}\int_0^1t^{2k}\ln t dt$$
Changing the order of summation and integration and performing summation first hand
$$=-\int_{0}^{1}\frac{dx}{\sqrt{1-x}}\int_0^1\frac{t^2\ln t}{1-xt^2}dt$$
Making the substitutions $x\to 1-x$ and $\,s^2=x$
$$=-\int_{0}^{1}\frac{dx}{\sqrt x}\int_0^1\frac{t^2\ln t}{1-(1-x)t^2}dt=-2\int_0^1ds\int_0^1\frac{t^2\ln t}{1-(1-s^2)t^2}dt$$
Changing the order of integration and integrating over $s$
$$=-2\int_0^1\frac{t^2\ln t}{1-t^2}dt\int_0^1\frac{ds}{1+\frac{t^2}{1-t^2}s^2}=-2\int_0^1\frac{t\ln t}{\sqrt{1-t^2}}\arctan\frac{t}{\sqrt{1-t^2}}\,dt$$
Making the substitution $t=\sin x$ and integrating by part
$$=-2\int_0^\frac{\pi}{2}x\,\ln(\sin x)\,\sin x\,dx=2\cos x\ln(\sin x)\,\Big|_0^\frac{\pi}{2}-2\int_0^\frac{\pi}{2}\Big(\ln(\sin x)+\frac{x\,\cos x}{\sin x}\Big)\cos x\,dx$$
$$I=2-2\int_0^\frac{\pi}{2}\frac{x}{\sin x}(1-\sin^2x)\,dx=4-2\int_0^\frac{\pi}{2}\frac{x}{\sin x}dx$$
To evaluate the last integral we note that $\frac{dx}{\sin x}=d\big(\ln\tan\frac{x}{2}\big)$. Integrating by part
$$J=\int_0^\frac{\pi}{2}\frac{x}{\sin x}dx=x\,\ln\tan\frac{x}{2}\,\Big|_0^\frac{\pi}{2}-\int_0^\frac{\pi}{2}\ln\tan\frac{x}{2}\,dx=-2\int_0^\frac{\pi}{4}\ln\tan x\,dx$$
Making the substitution $x=\arctan t$
$$J=-2\int_0^1\frac{\ln t}{1+t^2}dt=2\,G$$
Taking all together,
$$\boxed{\,\,I=4-2\int_0^\frac{\pi}{2}\frac{x}{\sin x}dx=4-4\,G\,\,}$$
where $G$ denotes Catalan's constant.
| {
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Simplifying $\sqrt{1+\sqrt{5}(6-2\sqrt{5})^{1/4}}$ I have the radical
$$\sqrt{1+\sqrt{5}(6-2\sqrt{5})^{1/4}}$$ for exam preparation (middle school):
I need to simplify it in natural numbers.
My attempt is:
We know the rule:
$(a-b)^2=a^2-2ab+b^2$
Let's $a^2+b^2=6$, then $2ab=2\sqrt5$ and $ab=\sqrt5$, suppose that $b=1$, then $a=\sqrt5$.
$(6-2\sqrt{5})^{1/4}=(1^2-2\cdot1\cdot\sqrt{5}+(\sqrt{5})^2)^{1/4}=(\sqrt{5}-1)^{1/2}$
After first step:
$\sqrt{1+\sqrt{5}\sqrt{\sqrt{5}-1}}$
Question. What is a possible next step?
Or possible error/typo in the task. The answer is 2.
| As noted by Gerry Myerson, the expression is likely supposed to be
$$\sqrt{(1+\sqrt5)\sqrt{6-2\sqrt5}}$$
$$= \sqrt{(\sqrt{5} + 1)(\sqrt{5} - 1)}$$
$$= \sqrt{5 - 1}$$
$$= 2$$
Remark: My intention to answer this question is less to answer the question (which I believe the OP is completely capable of doing themselves) and more to remove this question from the unanswered section.
| {
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Parallelogram with an angle and the midpoint of a side $M$ is the midpoint of the side $AB$ of a parallelogram $ABCD$. Find the perimeter of the parallelogram if $MD=4,MC=6$ and $\measuredangle BAD=60^\circ$.
Let $M$ be the midpoint of $AB\Rightarrow AM=BM=x$ and $AD=BC=y$. We are supposed to find $P_{ABCD}=2(AB+AD)=2(2x+y)=4x+2y$. Two cosine rules for $\triangle AMD$ and $\triangle MBC$ give $$DM^2=AM^2+AD^2-2AM.AD\cos60^\circ\\x^2+y^2-xy=16\\MC^2=MB^2+BC^2-2MB.BC\cos120^\circ\\x^2+y^2+xy=36$$ This seems useless for the perimeter (we can find the area, though).
| There are two possible values of perimeter.
$x^2+y^2-xy=16$, $x^2+y^2+xy=36$ $\Rightarrow x^2+y^2=26$, $xy=10$ $\Rightarrow (x+y)^2=46$, $(x-y)^2=6$ $\Rightarrow$
$x+y=\sqrt{46}$, $x-y=\pm\sqrt{6}$ $\Rightarrow 4x+2y=3(x+y)+(x-y)=3\sqrt{46}\pm\sqrt{6}$.
| {
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Find all integer solutions of $3^a +7 = 2^b$ I want to find all integer solutions of $3^a + 7 = 2^b$
I have found (by brute force) the two solutions
$3^0 + 7 = 2^3$ and
$3^2 + 7 = 2^4$
but I want to see if there are more solutions. I have found that if (b mod 3) = b' then (a mod 6) must be 2b', and that (a mod 4) can't be 1, but that's as far as I get and I have no clue how to make progress.
Any ideas? TIA
| A negative $b$ or a negative $a$ cannot lead to a solution. So $a,b\ge 0$. For $a=0$ we obtain OP's first solution $3^0+7=2^3$. Else $a\ge 1$. Then taken modulo $3$ the powers of two ($2\equiv -1\ [3]$) for $b=0,1,2,3,\dots$ are $1,-1,1,-1,\dots$, so the involved two-power $b$ is even, $b=2B$ for some natural $B$.
Since $b\ge 3$, considering the L.H.S. modulo four, we have for $a=0,1,2,3,\dots$ the values $1+7, -1+7, 1+7, -1+7,\dots$ so the involved three-power is even, $a=2A$ for some natural $A$. We rewrite now the given relation as follows:
$$
7 = 2^a-3^b=2^{2A}-3^{2B}=(2^A-3^B)(2^A+3^B)\ .
$$
This is possible only when $(2^A-3^B)=1$ and $(2^A+3^B)=7$, so $2^A=(7+1)/2=4$, and $3^B=(7-1)/2=3$, leading to the second solution given by the OP.
$\square$
| {
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Need pure geometric solution for proof on 10-20-40-50 angle problem $D$ is a point in $\triangle{ABC}$ so that $\angle{ABD}=10^{\circ}$, $\angle{DBC}=20^{\circ}$, $\angle{BCD}=40^{\circ}$, $\angle{DAC}=50^{\circ}$.
Find $\angle{BAD}$.
This problem is easily done with trigonometric Ceva theorem as:
$$
\begin{aligned}
&\dfrac{\sin x}{\sin50^{\circ}}
\cdot
\dfrac{\sin(60^\circ-x)}{\sin40^{\circ}}
\cdot
\dfrac{\sin20^\circ}{\sin10^\circ}=1
\\
\implies
&
\sin x \cdot \sin(60^\circ-x)
=
\dfrac
{\sin50^\circ\cdot \sin40^\circ\cdot \sin10^\circ}
{\sin20^\circ}
\\
&\qquad
=
\dfrac
{ \cos40^\circ\cdot \sin40^\circ \cdot \sin10^\circ}
{2\sin10^\circ\cdot \cos10^\circ}
=
\dfrac{\sin80^\circ}{4\cos10^\circ}
=\dfrac{1}{4}
\\
\implies
& -\dfrac{\cos60^\circ - \cos(2x-60^\circ)}2 = \dfrac{1}{4}
\\
\implies &
\cos(2x-60^\circ)=1
\\
\implies
&2x-60^\circ=0
\\
\implies
&x=\boxed{30^\circ}\ .
\end{aligned}
$$
Any idea on how to solve this problem in pure geometric approach? Thanks.
| Let's make following construction. $E$ is point of $BC$ such that $BE=BD$, $F$ is point symmetric to $E$ around $CD$. $G$ is point symmetric to $D$ around $AB$. Angular measures which can be restored from problem statement are given in picture.
$\angle DFC=100$°$=2\angle DAC$, then $F$ is center of circumcircle of triangle $ACD$. Then $FA=FD=FC=DE=CE$. Triangle $BDG$ is congruent to triangle $BDE$, then $DG=DE$.
$AF=FD$, $AD=AG$. Then $A$ is common point of perpendicular bisector of $DG$ and circle with center $F$ and radius $FD$.
Let point $H$ is symmetric to $D$ around $FG$, then $FDGH$ is rhombus. Because of $\angle FDG=120$° in this rhombus $DH=GH$ and $H$ is on perpendicular bisector of $DG$ which is $BA$. Then $\angle BHD=30$°, $\angle BHF=90°$ and $FH=FD$. Then perpendicular bisector of $DG$ is tangent to circle with center $F$ and radius $FD$. As tangent has only one common point with circle, then $A=H$ and $\angle BAD=\angle BHD=30$°.
| {
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Showing $\frac{xy+yz+xz}{x+y+z}>5-\sqrt{4(x^2+y^2+z^2)+6}$ for $x,y,z>0$ and $xyz=1$
Let $x,y,z>0$ with $ xyz=1$ then prove that,
$$\frac{xy+yz+xz}{x+y+z}>5-\sqrt{4(x^2+y^2+z^2)+6}$$
Let $$5≤\sqrt{4(x^2+y^2+z^2)+6}\implies x^2+y^2+z^2≥\frac {25-6}{4}=\frac {19}{4}$$ then the inequality is always true.
This also shows that, if $x≥3$ or $x≥\frac {\sqrt {19}}{2}$ then the inequality is true. Since the inequality is cyclic, if $x$ or $y$ or $z≥\frac {\sqrt {19}}{2}$ then the inequality is again true.
I see that if $x,y$ is strictly increasing then $5-\sqrt{4(x^2+y^2+z^2)+6}$ is always negative.
But, I couldn't prove the general case.
| Take $x=y=z=1$ and you will easily see that the right hand side is larger than $0$.
| {
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Closed-form solution of recurrence relation I have the first-order non-homogeneous recurrence (defined for $n \geq 1$):
$$f(n) = f(n-1) \; \frac{n-1}{n} + 1$$
with base case $f(1) = 1$.
Looking at the values of the sequence ‒ $1, 1.5, 2, 2.5$ ‒ one can easily see the closed form is $f(n) = \frac{1}{2}(n+1)$.
But how whould one come to this solution without looking at the values, or if it's not obvious from the values?
If I assume the recurrence is linear, I can
*
*Compute the formula for difference:
$$\Delta_n = \frac{-f(n)}{n+1}+1$$
*Compute difference of differences:
$$\Delta^{(2)}_n = \frac{2f(n-1)-n}{n(n+1)}$$
*Knowing $\Delta^{(2)}_n = 0$ for all $n$, I obtain the closed-form solution from the formula for $\Delta^{(2)}_n$. (I also need to adjust according to the base case, which in this case happens to make no difference).
*I can verify that the closed-form solution is correct (and the recurrence is indeed linear) by substituting the closed-form solution into the original recurrence formula and checking whether the equality holds:
\begin{aligned}
f(n) &= f(n-1) \cdot \frac{n-1}{n} + 1 \\
\frac{1}{2}(n+1) &\stackrel{?}{=} \frac{1}{2}n \cdot \frac{n-1}{n} + 1 \\
\frac{1}{2}(n+1) &= \frac{1}{2}(n+1)
\end{aligned}
So now I know my solution is correct.
But what if I don't want to (cannot) assume the recurrence is linear? The result will obviously be the same, but I don't really care about the result in this particular case, but rather about a more general approach to obtaining closed-form formulas from recurrence relations.
| You just have to use this formula
$$f(n) = f(n-1)\frac{n-1}{n} + 1$$ recursively that is,
\begin{aligned}
f(n) &= f(n-1)\frac{n-1}{n} + 1 \\
&= (f(n-2)\frac{n-2}{n-1} +1)*(\frac{n-1}{n}) + 1 \\
&=
f(n-2)\frac{n-2}{n} + \frac{n-1}{n} + 1
\end{aligned}
If you use it n-1 times and the condition that f(1) = 1 you get
$$f(n) = \frac{1}{n} + \frac{2}{n} + ... + \frac{n-1}{n} + 1 =
\frac{1 + 2 + ... + (n-1)}{n} + 1 = \frac{(n-1)n}{2n} + 1 = \frac{n+1}{2}$$
| {
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Probability of Full House in 5 Cards Given a Pair of Aces What is the probability of a full house in a 5 card hand, given the first 2 cards are aces?
My thought process is that there are 4 other events that could occur: | Method 1: If the first two cards selected from the deck are aces, then of the remaining $50$ cards, two are aces. There are $$\binom{50}{3}$$ ways to select three cards from the $50$ that remain. To obtain a full house, either one of the two remaining aces and two cards from one of the remaining $12$ ranks must be selected or three cards from one of the remaining $12$ ranks must be selected. Thus, the number of favorable cases is
$$\binom{2}{1}\binom{12}{1}\binom{4}{2} + \binom{12}{1}\binom{4}{3}$$
where the first term counts the number of ways of selecting one of the two remaining aces, one of the remaining $12$ ranks, and two cards of that rank and the second term counts the number of ways of selecting one of the remaining $12$ ranks and three of the four cards of that rank. Hence, the probability of obtaining a full house given the first two cards selected are aces is
$$\frac{\dbinom{2}{1}\dbinom{12}{1}\dbinom{4}{2} + \dbinom{12}{1}\dbinom{4}{3}}{\dbinom{50}{3}}$$
Method 2: We correct your attempt.
You did not take into account the rank of the non-aces. Observe that there are $48$ cards which can be selected as the first non-ace. Choosing that card determines the rank of the cards which are non-aces. Hence, there are three choices for the second non-ace and, if there is a third non-ace, there are two choices for that card. The probability of selecting an ace, then two cards of one of the other $12$ ranks in that order is
$$\frac{2}{50} \cdot \frac{48}{49} \cdot \frac{3}{48}$$
Since there are three possible positions for the remaining ace, the probability of obtaining a full house with three aces given that the first two cards are aces is
$$\frac{2}{50} \cdot \frac{48}{49} \cdot \frac{3}{48} + \frac{48}{50} \cdot \frac{2}{49} \cdot \frac{3}{48} + \frac{48}{50} \cdot \frac{2}{49} \cdot \frac{3}{48} = 3 \cdot \frac{2}{50} \cdot \frac{48}{49} \cdot \frac{3}{48}$$
The probability of obtaining a full house containing three cards of another rank given that the first two cards are aces is
$$\frac{48}{50} \cdot \frac{3}{49} \cdot \frac{2}{48}$$
Hence, the probability of a full house given that the first two cards selected are aces is
$$3 \cdot \frac{2}{50} \cdot \frac{48}{49} \cdot \frac{3}{48} + \frac{48}{50} \cdot \frac{3}{49} \cdot \frac{2}{48}$$
| {
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Prove the inequality $9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca)$
Prove the inequality $9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca)$ for $a, b, c \in \mathbb{R_{>0}}$
I tried by first using AM-HM inequality on $a, b, c$ to get the following result.
$\frac{a+b+c}3 \geq \frac 3{\frac 1a+\frac1b+\frac1c}$
$\implies (a+b+c)(\frac1a+\frac1b+\frac1c) \geq 9$
$\implies (a+b+c)(ab+bc+ca) \geq 9abc$
Also I used the inequality
$(a+b)(b+c)(c+a) \geq 8abc$
But then I am not able to proceed further. Can someone please help me.
| Before I start my solution, I want to say that the proof you have just showed was nice try but it's hard to prove the claim. If you want to show that $A>B$, it's meaningless to show that $A>C$ and $B>C$.
Alright, so I'll show my solution.
\begin{align}
& \text{Claim. } 9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca). \\
& \text{pf)} \\
\ \\
\text{Claim} & \Leftrightarrow 9\Bigg(\sum_{sym} a^2b + 2abc \Bigg) \geq 8 \Bigg( \sum_{sym} a^2b + 3abc \Bigg) \\
& \Leftrightarrow \sum_{sym} a^2b \geq 6abc \\
& \Leftrightarrow a^2b+b^2c+c^2a+ab^2+bc^2+ca^2 \geq 6abc \\
& \Leftrightarrow \dfrac {a^2b+b^2c+c^2a+ab^2+bc^2+ca^2} {6} \geq \sqrt[6]{a^2b \cdot b^2c \cdot c^2a \cdot ab^2 \cdot bc^2 \cdot ca^2} \\
& \Rightarrow \text{Proved by AM-GM Inequality.}
\end{align}
| {
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Reference Request: Integral of Gaussian over Unit Sphere I am looking for a reference for integrals of the form
\begin{equation}
\tag{1} \int_{S^{n-1}} \mathcal{N}_{\omega} ( \mu , \Sigma ) d \omega
\end{equation}
where $S^{n-1}$ is the sphere in $\mathbb{R}^n$ and
\begin{equation}
\mathcal{N}_x (\mu , \Sigma) = \frac{e^{- \frac{1}{2} ( x - \mu ) \cdot \Sigma^{-1} (x - \mu) }}{\sqrt{ ( 2 \pi )^n \det{\Sigma} }}
\end{equation}
is the usual multivariate Gaussian distribution evaluated at $x \in \mathbb{R}^n$ with mean $\mu \in \mathbb{R}^n$ and covariance matrix $\Sigma \in \mathbb{R}^{n \times n}$. I am primarily concerned with the $n = 2, 3$ cases, but I find the case for general $n$ to be interesting as well.
There is a substantial simplification that can be made. By the spectral theorem and positive-definite nature of $\Sigma$ we may write $\Sigma = V D^2 V^T$ where $V \in \mathrm{SO}_n (\mathbb{R})$ and $D = \mathrm{diag}(\sigma_1 , \cdots , \sigma_n)$ is a diagonal matrix with positive, real values along the diagonal. Therefore, by spherical symmetry, I am essentially asking for a reference to the integral
\begin{equation}
\tag{2} \int_{S^{n-1}} e^{ - \frac{1}{2} \| D^{-1} (\omega - \tilde{\mu}) \|^2 } d \omega ,
\end{equation}
where $\tilde{\mu} = V^T \mu$.
Question: Does anyone have a reference or suggestion for how to calculate the closed-form expression for (2) and, hence, (1)?
Additional Thoughts:
There are special cases that are simple enough to compute. For example, if $\sigma_1 = \cdots = \sigma_n \equiv \sigma$, then the integral in question becomes
\begin{equation}
\frac{e^{- \frac{1 + |\mu|^2}{2 \sigma^2} }}{(2 \pi)^{n/2} \sigma} \int_{S^{n-1}} e^{\omega \cdot \mu / \sigma^2 } d \omega .
\end{equation}
The previous integral may be evaluated using the formula
\begin{equation}
\int_{S^{n-1}} e^{ \omega \cdot y } d \omega = (2 \pi)^{\frac{n}{2}} | y |^{1 - \frac{2}{n}} I_{\frac{n}{2} - 1} ( |y| ) ,
\end{equation}
where $I_{\nu}$ is the hyperbolic Bessel function of order $\nu$. This formula may be found in, e.g., Loss and Lieb, Analysis, Section 7.11. Therefore,
\begin{equation}
\tag{3} \int_{S^{n-1}} \mathcal{N}_{\omega} ( \mu , \sigma^2 ) d \omega = \frac{e^{- (1 + \mu^2)/(2 \sigma^2) }}{\sigma} \left( \frac{|\mu|}{\sigma^2} \right)^{1 - \frac{2}{n}} I_{\frac{n}{2} - 1} \left( \frac{|\mu|}{\sigma^2} \right) .
\end{equation}
Another special case that I mention in the comments below is $\mu = 0$. As I point out there, the $n = 3$ case of (1) for $\mu = 0$ should be similar to the calculation of the normalization constant of the Kent distribution. For $n = 2$, (2) reduces to
\begin{equation}
\tag{4} \int_0^{2 \pi} e^{- \frac{1}{2} \left( \frac{\cos^2{\theta}}{\sigma_1^2} + \frac{\sin^2{\theta}}{\sigma_2^2} \right) } d \theta = 2 \pi e^{- \frac{1}{4} \left( \frac{1}{\sigma_1^2} + \frac{1}{\sigma_2^2} \right) } I_0 \left( \frac{1}{4} \left( \frac{1}{\sigma_2^2} - \frac{1}{\sigma_1^2} \right) \right) .
\end{equation}
The previous result follows from the cosine and sine reduction formula $\cos^2{x} = (1 + \cos{(2x)})/2$ and likewise for $\sin^2{x}$.
Considering that (3) and (4) have relatively simple expressions, I am hoping this is also the case for (2) with $\tilde{\mu} \neq 0$.
| For the $n = 2$ case, (2) in the OP can be evaluated as follows. Denote $\tilde{\mu} = \mu$. The integral in question may be reduced to
\begin{equation}
\tag{1} e^{- \frac{1}{2} \left( \frac{\mu_1^2}{\sigma_1^2} + \frac{\mu_2^2}{\sigma_2^2} \right) } \int_0^{2 \pi} e^{ - \frac{1}{2} \left( \frac{\cos^2{\theta}}{\sigma_1^2} + \frac{\sin^2{\theta}}{\sigma_2^2} - 2 \frac{\mu_1}{\sigma_1^2} \cos{\theta} - 2 \frac{\mu_2}{\sigma_2^2} \sin{\theta} \right) } d \theta .
\end{equation}
Using the double angle reduction formulas $\cos^2{\theta} = ( 1 + \cos{(2 \theta)} ) / 2$ and $\sin^2{\theta} = ( 1 - \cos{(2 \theta)} ) / 2$, we can simplify (1) to read
\begin{equation}
\tag{2} e^{- \frac{1}{4} \left( \frac{2\mu_1^2 + 1}{\sigma_1^2} + \frac{2\mu_2^2 + 1}{\sigma_2^2} \right) } \int_0^{2 \pi} e^{ - \frac{1}{2} \left(
\left( \frac{1}{2 \sigma_1^2} - \frac{1}{2 \sigma_2^2} \right) \cos{(2 \theta)} - 2 \frac{\mu_1}{\sigma_1^2} \cos{\theta} - 2 \frac{\mu_2}{\sigma_2^2} \sin{\theta} \right) } d \theta .
\end{equation}
By the Jacobi-Anger expansion, we have
\begin{equation}
e^{z \cos{2 \theta}} = I_0 (z) + 2 \sum_{k \geq 1} I_k (z) \cos{(2 k\theta)} ,
\end{equation}
where $z = - \frac{1}{4} \left( \frac{1}{\sigma_1^2} - \frac{1}{\sigma_2^2} \right)$. Plugging this into (2), we have
\begin{equation}
\tag{3} e^{- \frac{1}{4} \left( \frac{2\mu_1^2 + 1}{\sigma_1^2} + \frac{2\mu_2^2 + 1}{\sigma_2^2} \right) } \int_0^{2\pi} \left( I_0 (z) + 2 \sum_{k \geq 1} I_k (z) \cos{(2 k \theta)} \right) e^{\frac{\mu_1}{\sigma_1^2} \cos{\theta} + \frac{\mu_2}{\sigma_2^2} \sin{\theta} } d \theta ,
\end{equation}
where, again, $z = - \frac{1}{4} \left( \frac{1}{\sigma_1^2} - \frac{1}{\sigma_2^2} \right)$. To proceed, note that
\begin{equation}
y_1 \cos{\theta} + y_2 \sin{\theta} = \| y \| \cos{(\theta - \arctan{(y_2/y_1)})} ,
\end{equation}
where $y_1 = \mu_1 / \sigma_1^2$ and $y_2 = \mu_2 / \sigma_2^2$. Plugging this into (3), changing variables, and using the cosine addition formula, we arrive at
\begin{equation}
\int_{S^1} e^{- \frac{1}{2} \| D^{-1} ( \omega - \mu ) \|^2} d \omega = 2 \pi e^{- \frac{1}{4} \left( \frac{2\mu_1^2 + 1}{\sigma_1^2} + \frac{2\mu_2^2 + 1}{\sigma_2^2} \right) } \left( I_0 (z) I_0 (\|y\|) + 2 \sum_{k \geq 1} I_k (z) I_{2k} (\| y \|) \cos{\left(2 k \arctan{\left( \frac{y_2}{y_1} \right)}\right)} \right) , \tag{*}
\end{equation}
where, again, $z = - \frac{1}{4} \left( \frac{1}{\sigma_1^2} - \frac{1}{\sigma_2^2} \right)$, $y_1 = \mu_1 / \sigma_1^2$ and $y_2 = \mu_2 / \sigma_2^2$.
We note how ($\ast$) agrees with some of the special cases outlined above. Indeed, for $\sigma_1 = \sigma_2$, $z = 0$ and all higher order Bessel functions in the sum in ($\ast$) are zero. When $\mu_1 = \mu_2 = 0$, the $\arctan$ isn't defined, so this special case isn't covered.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
solve $\sin(3x)\cos(6x)-\cos(3x)\sin(6x)=-.9$ I made a graph of an equation and also solved the equation algebraically. Even though I can find one answer, I am having a hard time finding all the rest. All of the answers have to fall in the interval $[0, 2\pi)$. Here is my math and graph
$$\sin(3x)\cos(6x)-\cos(3x)\sin(6x)=-.9$$
I used the difference of sin trig identity.
$$\sin(3x-6x)=-.9$$ $$\sin(-3x)=-.9$$ I used the even and odd trig identity $$-\sin(3x)=-.9$$ I did graph this equation because I was lost when looking at the book answers. I labeled the points that may have some relevance. The period $\frac{2\pi}{3}$ seemed important, there are three points in the graph that show it equal to $.9$. I list all the answers at the end of the post and the points labeled on the graph show up in the answer. But, that is as close to the answer I can come.
The book has the answer in $\arcsin$. I solved the equation for that. $$-\arcsin(-.9)=3x$$ I used the even and odd identity for $\sin$. $$\arcsin(.9)=3x$$ Then I divided both sides by $3$ and this gave me the first answer the book was looking for $$\frac13(\arcsin(.9))$$ I realize that since the answer is looking for $3x$ that I have to go around the circle three times. The book says, "Whenever we solve a problem in the form of $\sin(nx) =c$ we must go around the circle $n$ times". This is why I started to graph. I am not sure if the graph has any purpose but I was able to find some points of importance to me. Here are all the answers the book list for the solution. I was wondering if someone could help me in understanding the answer. $$\frac13(\arcsin(.9)), \frac{\pi}{3}-\frac13(\arcsin(.9)), \frac{2\pi}{3}+\frac13(\arcsin(.9)), \pi - \frac13(\arcsin(.9)),$$ $$\frac{4\pi}{3}+\frac13(\arcsin(.9)), \frac{5\pi}{3}-\frac13(\arcsin(.9))$$
| We wish to solve the equation $\sin(3x)\cos(6x) - \sin(6x)\cos(3x) = -0.9$ in the interval $[0, 2\pi)$.
\begin{align*}
\sin(3x)\cos(6x) - \sin(6x)\cos(3x) & = -0.9\\
\sin(3x - 6x) & = -0.9 && \text{since $\sin\alpha\cos\beta - \cos\alpha\sin\beta = \sin(\alpha - \beta)$}\\
\sin(-3x) & = -0.9 && \text{subtract}\\
-\sin(3x) & = -0.9 && \text{since $\sin(-\theta) = -\sin\theta$}\\
\sin(3x) & = 0.9 && \text{multiply both sides of the equation by $-1$}
\end{align*}
A particular solution of this equation is
\begin{align*}
3x & = \arcsin(0.9)\\
\end{align*}
The range of the arcsine function is $[-\pi/2, \pi/2]$. Since $0 < 0.9 < 1$, $0 < \arcsin(0.9) < \pi/2$.
Before we continue, let's consider when $\sin\theta = \sin\varphi$. Clearly, the equation is true if $\theta = \varphi$. Since $\sin\theta$ is the $y$-coordinate of the point where the terminal side of an angle in standard position (vertex at the origin, initial side on the positive $x$-axis) intersects the unit circle, by symmetry, another solution is $\theta = \pi - \varphi$.
Moreover, any angle coterminal with one of these angles satisfies the equation $\sin\theta = \sin\varphi$. Hence, $\sin\theta = \varphi$ if
$$\theta = \varphi + 2k\pi, k \in \mathbb{Z}$$
or
$$\theta = \pi - \varphi + 2m\pi, m \in \mathbb{Z}$$
For the equation above, that means the general solution is
\begin{align*}
3x & = \arcsin(0.9) + 2k\pi, k \in \mathbb{Z} & \pi - 3x & = \arcsin(0.9) + 2m\pi, m \in \mathbb{Z}\\
x & = \frac{\arcsin(0.9)}{3} + \frac{2k\pi}{3}, k \in \mathbb{Z} & 3x - \pi & = -\arcsin(0.9) - 2m\pi, m \in \mathbb{Z}\\
& & 3x & = \pi - \arcsin(0.9) - 2m\pi, m \in \mathbb{Z}\\
& & x & = \frac{\pi}{3} - \frac{\arcsin(0.9)}{3} - \frac{2m\pi}{3}, m \in \mathbb{Z}\\
& & x & = \frac{\pi}{3} - \frac{\arcsin(0.9)}{3} + \frac{2n\pi}{3}, n \in \mathbb{Z}
\end{align*}
where we let $n = -m$ in the preceding step.
We stated above that $0 < \arcsin(0.9) < \dfrac{\pi}{2}$. Hence, $0 < \dfrac{\arcsin(0.9)}{3} < \dfrac{\pi}{6}$. Therefore,
$$\frac{\pi}{3} > \frac{\pi}{3} - \frac{\arcsin(0.9)}{3} > \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6}$$
With that in mind, we select values of $k$ and $n$ such that
\begin{align*}
x & = \frac{\arcsin(0.9)}{3} + \frac{2k\pi}{3}, k \in \mathbb{Z} & x & = \frac{\pi}{3} - \frac{\arcsin(0.9)}{3} + \frac{2n\pi}{3}, n \in \mathbb{Z}
\end{align*}
are in the interval $[0, 2\pi)$. Those values are $k = 0, 1, 2$ and $n = 0, 1, 2$, which correspond to the solutions
\begin{align*}
x & = \begin{cases}
\dfrac{\arcsin(0.9)}{3},\\[2 mm]
\dfrac{\arcsin(0.9)}{3} + \dfrac{2\pi}{3},\\[2 mm]
\dfrac{\arcsin(0.9)}{3} + \dfrac{4\pi}{3}
\end{cases}
& x & = \begin{cases}
\dfrac{\pi}{3} - \dfrac{\arcsin(0.9)}{3},\\[2 mm]
\dfrac{\pi}{3} - \dfrac{\arcsin(0.9)}{3} + \dfrac{2\pi}{3},\\[2 mm]
\dfrac{\pi}{3} - \dfrac{\arcsin(0.9)}{3} + \dfrac{4\pi}{3}
\end{cases}\\
& & x & = \begin{cases}
\dfrac{\pi}{3} - \dfrac{\arcsin(0.9)}{3},\\[2 mm]
\pi - \dfrac{\arcsin(0.9)}{3},\\[2 mm]
\dfrac{5\pi}{3} - \dfrac{\arcsin(0.9)}{3}
\end{cases}
\end{align*}
We get six solutions since there are two solutions within each period and $\sin(3x)$ has three periods within the interval $[0, 2\pi)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4433287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Solution of the recurrence relation $y_n = \frac{1}{2} + \frac{1}{2}y_{n-1}$ $y_0 = 0$ and $y_n = \frac{1}{2} + \frac{1}{2}y_{n-1}$. Solution of this reccurent equation is $y_n = 1 - \frac{1}{2^n}$, accordingly with the software. But I do not understand the minus sign since it would be $y_n = \frac{1}{2} + \frac{1}{2}\left(\frac{1}{2}+\frac{1}{2}y_{n-2}\right)$ and so forth until $y_n = \frac{1}{2} + \frac{1}{2^{n}} +x_{0}$.
Can someone try to explain it to me, I'm having a very tough time trying to understand.
| $$\begin{split}
y_n &= \frac 1 2 + \frac {y_{n-1}} 2\\
&= \frac 1 2 +\frac 1 2\left( \frac 1 2 +\frac {y_{n-2}} 2\right)\\
&= \frac 1 2 + \frac 1 2 \left(\frac 1 2 +\frac 1 2 \left (\frac 1 2 +\frac {y_{n-2}} 2 \right) \right)\\
&= \dots\\
&= \frac 1 2 + \frac 1 4 + \frac 1 8 +\dots + \frac 1 {2^{n}} + \frac{y_0}{2^n}\\
&= \frac 1 2 \cdot\frac {1-2^{-n}}{1-\frac 1 2} \\
&= 1-2^{-n}
\end{split}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4437950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can you please help me whether $N-R$ method or other methods converges? Consider the system of equations
\begin{align}
&f(x,y)=x+\frac{y^4}{2}+\frac{x^{32}}{4}+\frac{y^{128}}{8}=0 \\
&g(x,y)=y+\frac{x^8}{2}+\frac{y^{32}}{4}+\frac{x^{256}}{8}=0.
\end{align}
I want to solve it using Newton-Raphson process or any other methods
Consider the initial guess $(x_0,y_0)=((-2)^{5/31},(-2)^{9/31})$, which is either $( 1.118,-1.223 )$ or
$(x_0,y_0)=(-1.118,1.223)$.
I want to see whether the Newton-Raphson method converges with this
initial guess.
Note that the initial guess $(x_0,y_0)$ is a simultaneous zero of the
following truncated system obtained from the original system
\begin{align}
&x+\frac{y^4}{2}=0\\
&y+\frac{x^8}{2}=0.
\end{align}
By hand calculation, it is laborious. For the jacobian is $$J
=\begin{pmatrix} 1+8x^{31} & 2y^3+16y^{127} \\ 2x^3+32x^{255} &
1+8y^{31} \end{pmatrix} \Rightarrow J((x_0,y_0)) \approx
\begin{pmatrix} 255 & -2.03 \times 10^{12} \\ 7.21 \times 10^{13} &
-4104\end{pmatrix}$$
So the 2nd iteration is given by
\begin{align} \begin{pmatrix} x_1 \\ y_1
\end{pmatrix}&=\begin{pmatrix}1.118 \\ 1.223 \end{pmatrix}
-J((x_0,y_0))^{-1} \begin{pmatrix}f((x_0,y_0)) \\ g((x_0,y_0))
\end{pmatrix}
\end{align}
which seems difficult to calculate because I can not invert the huge matrix.
The zeros of the truncated system should converge to the solutions
of the original system.
Can you please help me whether N-R method or any other numerical methods converges ?
More, specifically, how to show the simultaneous zeroes of the truncated system $
x+\frac{y^4}{2}=0=y+\frac{x^8}{2}$ converges to the simultaneous zeroes of the original system $f(x,y)=0=g(x,y)$ ?
Thanks
| By performing the contour plot for some truncated approximations, we can infer for which the NR algorithm will work.
$$
\cases{
x + \frac{y^9}{3}+\frac{x^{243}}{9}=0 \\
y + \frac{x^{27}}{3}+\frac{y^{243}}{9}=0 \\
}
$$
Here NR works with one solution.
$$
\cases{
x + \frac{y^9}{3}+\frac{x^{243}}{9} + \frac{y^{2187}}{27}=0 \\
y + \frac{x^{27}}{3}+\frac{y^{243}}{9} + \frac{x^{6561}}{27}=0 \\
}
$$
Here NR succeeds with three solutions
$$
\cases{
x + \frac{y^9}{3}+\frac{x^{243}}{9} + \frac{y^{2187}}{27}+\frac{x^{59049}}{81}=0 \\
y + \frac{x^{27}}{3}+\frac{y^{243}}{9} + \frac{x^{6561}}{27}+\frac{y^{59049}}{81}=0 \\
}
$$
Here NR only gives one solution. The two contours only cross one time.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4438130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\frac{1}{|x-y|}=\frac{1}{\pi^3} \int_{R^3}\frac{1}{|x-z|^2}\frac{1}{|y-z|^2}dz$ Prove the following identity for $x,y \in R^3:$ $$\dfrac{1}{|x-y|}=\dfrac{1}{\pi^3} \int_{R^3}\dfrac{1}{|x-z|^2}\dfrac{1}{|y-z|^2}dz$$
I tried multiple ways for example use green function,
put $x=(x_1,x_2,x_3),y=(y_1,y_2,y_3),z=(z_1,z_2,z_3)$ and calculate above formula but I failed
How can I solved this problem? thanks very much
| One of the options, probably, not the most rational one, is to evaluate the integral directly. If we choose this option, we have to define the system of coordinate - to choose the center point and axis direction. We can choose $y$ as a center point, and direct the axis $Z$ along the vector $\vec a=\vec x-\vec y$. Denoting $|\vec a|=a=|\vec x-\vec y|$, our integral takes the form
$$I=\dfrac{1}{\pi^3} \int_{R^3}\dfrac{1}{|x-z|^2}\dfrac{1}{|y-z|^2}dz=\dfrac{1}{\pi^3} \int_{R^3}\dfrac{1}{|s-(x-y)|^2}\dfrac{1}{|s|^2}ds$$
In the polar system of coordinates, and in accordance with our choice of the polar axis $Z$,
$$=\dfrac{1}{\pi^3}\int_0^{2\pi}d\phi\int_0^\infty \frac{s^2ds}{s^2}\int_0^\pi\frac{\sin\theta \,d\theta}{s^2-2sa\cos \theta+a^2}=\frac{1}{a\pi^2}\int_0^\infty \frac{ds}{s}\int_{-1}^1\frac{dx}{\frac{s^2+a^2}{2sa}-x}$$
where we made the substitution $\cos \theta =x$. Integrating with respect to $x$
$$=\frac{1}{a\pi^2}\int_0^\infty \frac{ds}{s}\ln\frac{s^2+2sa+a^2}{s^2-2sa+a^2}=\frac{2}{a\pi^2}\int_0^\infty \frac{ds}{s}\ln\frac{1+s}{|1-s|}$$
Splitting the interval by $[0;1]$ and $[1;\infty)$, and making the change $t=\frac{1}{s}$ in the second integral,
$$I=\frac{4}{a\pi^2}\int_0^1 \frac{ds}{s}\ln\frac{1+s}{1-s}=-\frac{4}{a\pi^2}\int_0^1 \frac{ds}{s}\ln\frac{1-s}{1+s}$$
Making the substitution $t=\frac{1-s}{1+s}$
$$I=-\frac{8}{a\pi^2}\int_0^1\frac{\ln t}{1-t^2}dt=-\frac{8}{a\pi^2}\int_0^1\big(1-t^2-t^4-...\big)\ln t\,dt$$
Integrating by part every term
$$=-\frac{8}{a\pi^2}\Big(-1-\frac{1}{3^2}-\frac{1}{5^2}-...\Big)=\frac{8}{a\pi^2}\Big(1+\frac{1}{2^2}+\frac{1}{3^2}+..\Big)-\frac{8}{a\pi^2}\Big(\frac{1}{2^2}+\frac{1}{4^2}+..\Big)$$
$$=\frac{8}{a\pi^2}\Big(\zeta(2)-\frac{1}{4}\zeta(2)\Big)=\frac{8}{a\pi^2}\frac{3}{4}\frac{\pi^2}{6}=\frac{1}{a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
The solutions of the equation $ \sin{x} + \sin{3x} = \frac{8}{3\sqrt{3}} $ are? I tried this:
$$ \sin{x} + \sin{3x} = \frac{8}{3\sqrt{3}} $$
$$ 2\sin{2x}\cos{x} = \frac{8}{3\sqrt{3}} $$
$$ 4\sin{x}\cos{x}\cos{x} = \frac{8}{3\sqrt{3}} $$
$$ \sin{x}(1-\sin^2{x}) = \frac{2}{3\sqrt{3}} $$
Here, I tried to set $\sin x = t$
$$ t(1-t^2) = \frac{2}{3\sqrt{3}}, $$ but I don't know to resolve this.
| As you obtained,
$ \displaystyle t-t^3 = \frac{2}{3\sqrt{3}}~, \text {where } t = \sin x$
$ \displaystyle t - t^3 = \frac{1}{\sqrt3} - \frac{1}{3 \sqrt3} \implies \left(\frac{1}{\sqrt3}\right)^3 - t^3 = \left(\frac{1}{\sqrt3} - t\right)$
Now using the fact that $a^3 - b^3 = (a-b)(a^2 + b^2 + ab)$
One of the obvious solution is $~t = \dfrac{1}{\sqrt3}$.
If $~t \ne \dfrac{1}{\sqrt3}, $ we have
$ ~\displaystyle t^2 + \frac{1}{3} + \frac{t}{\sqrt3} = 1$
i.e. $~ \displaystyle \left(t + \frac{1}{2 \sqrt3}\right)^2 = \frac{3}{4}~$. That gives $ \displaystyle t = \frac{1}{\sqrt3}, - \frac{2}{\sqrt3}$
As $~- \dfrac{2}{\sqrt3} \lt - 1$, that is not a valid solution.
So the only solution is $\sin x = \dfrac{1}{\sqrt3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Orthogonal projection of an ellipsoïd from N to 2 dimensional space Suppose we have a $N\times N$ symmetric-positive-definite matrix $A$, representing an ellipsoïd in $N$ dimensional space. How to find the matrix $A_{xy}$ corresponding to orthogonal projection of ellispoid on $xy$-plan ?
I have already consulted this pages, they were not that helpful:
How to obtain the equation of the projection/shadow of an ellipsoid into 2D plane?
https://math.stackexchange.com/a/1866994/1054066
edit after some suggestions :
Schur complement method seems to work in 3 dimensions:
here's an exemple:
suppose $A = \begin{pmatrix} 1 & 0.5 & -1 \\ 0.5 & 2 & 0.5 \\ -1 & 0.5 & 3 \end{pmatrix}$
than Schur complement method says that $A_{xy} = \begin{pmatrix} 1 & 0.5 \\ 0.5 & 2\end{pmatrix} - \frac{1}{3} \begin{pmatrix} -1 \\ 0.5\end{pmatrix} \begin{pmatrix} -1 & 0.5\end{pmatrix} = \begin{pmatrix} 0.67 & 0.67 \\ 0.67 & 1.92\end{pmatrix}$
Here's the result :
but I dont see how to make it work in 5 dimansional space. The goal is to project 5D-ellipsoid on xy-plane.
| Suppose M is a symmetric $N \times N$ matrix representing an ellipsoid : $$X^T M X = 1$$ with $X \in \mathbb{R}^N$.
1st method of projection : using a projection matrix P
$$P_{xy} = \begin{pmatrix} 1 & 0 & 0 & ... & 0 \\ 0 & 1 & 0 & ... & 0\end{pmatrix} = \begin{pmatrix} \overrightarrow{i} \\ \overrightarrow{j} \end{pmatrix}$$ where $\overrightarrow{i}$ and $\overrightarrow{j}$ cartesian base vectors for $xy$-plane. For exemple if we want to project an object of 7-dimensional to 3-dimensional space (on any 3 arbitrary axis) than the projection matrix will take the form : $$ P_{7D \to 3D} = \begin{pmatrix} 1&0&0&0&0&0&0 \\ 0&0&0&0&1&0&0 \\ 0&0&1&0&0&0&0 \end{pmatrix} = \begin{pmatrix} \overrightarrow{e_1} \\ \overrightarrow{e_5}\\ \overrightarrow{e_3} \end{pmatrix}$$
The projection of M on $xy$-plane is an ellipse given by $M_{xy}$ : $$ M_{xy} = (P_{xy} M^{-1} P_{xy}^T)^{-1} $$
2nd method : using the Schur complement.
First we divide the matrix M into 4 blocs as follows : $$ M = \begin{pmatrix}
A & B \\ D & C \end{pmatrix}$$ than Schur's complement gives $M_{xy}$ :
$$ M_{xy} = A - BC^{-1}D $$
here's an exemple :
$$ M = \begin{pmatrix} 1 & 0.5 & -1 \\ 0.5 & 2 & 0.5 \\ -1 & 0.5 & 3 \end{pmatrix}$$
1st method: projection matrix
$$P_{xy} = \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$$
$$M_{xy} = (P_{xy} M^{-1} P_{xy}^T)^{-1} = \begin{pmatrix} 0.67 & 0.67 \\ 0.67 & 1.92\end{pmatrix}$$
2nd method : Schur complement
$$M_{xy} = \begin{pmatrix} 1 & 0.5 \\ 0.5 & 2 \end{pmatrix} - \begin{pmatrix} -1 \\ 0.5 \end{pmatrix} \frac{1}{3} \begin{pmatrix} -1 & 0.5 \end{pmatrix} = \begin{pmatrix} 0.67 & 0.67 \\ 0.67 & 1.92\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4442006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Error in computing a limit I was given the following limit to compute,
$$\lim_{k\to\infty}\left[k-\sqrt{k^{2}+1}\right]$$
My approach:
$$= \lim_{k\to\infty}\left[k-\sqrt{k^{2}(1+k^{-2})}\right]$$
$$= \lim_{k\to\infty}[k-k]=0$$
So the following evaluates to $0$. But my book gives that answer is $\frac{-1}{2}$. Where did my process go wrong?
| Another standard approach is to use the Binomial expansion. For large positive $k,$
$$k-\sqrt{k^{2}+1}= k-\sqrt{k^{2}\left(1+k^{-2}\right)} = k-k\left(1+k^{-2}\right)^{1/2}$$
$$ = k - k\left[1+ \left( \frac{1}{2}\right) \frac{1}{k^2} + \frac{\left( \frac{1}{2}\right)\left( - \frac{1}{2}\right)}{2!} \left(\frac{1}{k^2}\right)^2 + \frac{\left( \frac{1}{2}\right)\left( - \frac{1}{2}\right)\left( - \frac{3}{2}\right)}{3!} \left(\frac{1}{k^2}\right)^3 + \ldots \right]\quad (1)$$
With a little bit of manipulation one sees, for example, that
$$ -\frac{1}{k} \left(\frac{1}{1-k^2} \right) = -\frac{1}{k} \sum_{n=0}^{\infty} \left(\frac{1}{k^2}\right)^n < (1) < 0$$
and by letting $k\to\infty$, The Sandwich Theorem gives us that expression $(1)\to 0.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4448007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Prove that $c^3 + a^3 − 2abc = 6b − 11$ Let $a, b, c$ be real numbers such that $(a + b + c)^2 = 3(ab + bc + ca + 1).$ Given,
$$a^3 + b^3 − 2abc = 6c + 2,$$
$$b^3 + c^3 − 2abc = 6a + 9,$$
Prove that $$c^3 + a^3 − 2abc = 6b − 11$$
Note that, subtracting the two equations, we get $$a^3-c^3=6(c-a)+7. $$
Moreover, we have $$a^2+b^2+c^2-ab-bc-ca=3\implies (a-b)^2+(b-c)^2+(c-a)^2=6$$
So, we get $$a^3-c^3=[(a-b)^2+(b-c)^2+(c-a)^2](c-a)+7$$ $$\implies a^3-c^3+2c^3-2abc-11=[(a-b)^2+(b-c)^2+(c-a)^2](c-a)+7+2c^3-2abc-11.$$
And $$3 a^3 - 2 a^2 b - 4 a^2 c + 2 a b^2 + 4 a c^2 - 2 b^2 c + 2 b c^2 - 3 c^3 - 7 = 0 $$
So, it's enough to show that $$[(a-b)^2+(b-c)^2+(c-a)^2](c-a)+7+2c^3-2abc=6b $$
or show that $$2 a^3 - 4 a^2 c + 4 a c^2 + 2 b^3 - 4 b^2 c + 4 b c^2 - 4 c^3=7$$
or show that $$2 (a^3+b^3) - 4 a^2 c + 4 a c^2 - 4 b^2 c + 4 b c^2 - 4 c^3=7$$
I got something ahead, but failed. Any solutions?
| Hint: $2+9-11=0$, also $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$.
| {
"language": "en",
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"answer_count": 3,
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} |
Finding minimum value of $x^2+y^2+xy+x-4y+9$
What is the minimum value of $f(x,y)=x^2+y^2+xy+x-4y+9$ ?
I tried completing squares,
$$x^2+y^2+xy+x-4y+9=\frac12(x^2+2xy+y^2+x^2+2x+1+y^2-8y+16+1)=\frac12[(x+y)^2+(x+1)^2+(y-4)^2+1]$$But not sure how to continue.
| Hint:\begin{align}f(x+y,x-y)&=3x^2-3x+y^2+5y+9\\&=3\left(x-\frac12\right)^2+\left(y+\frac52\right)^2+2\end{align}
| {
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Proving that the following equation does not have integer solutions I want to prove that the following equation has no integer solutions $a,b,c$: $$-a^3 - b^3 - c^3 + ab^2 - ac^2 + bc^2 - 2a^2c + 3abc = 0$$ apart from the naive solution $a=b=c=0$.
The context, in case it helps, is the following: I am dealing with the field $\mathbb{F}_p[x] / (x^3-x-1)$ and this equation appeared when I tried to get a "formula" for the inverse of some element on this field. More precisely, if $g(x) = a + bx +cx^2$ is an element of the field and $g(x)' = c + dx + e^2$ is assumed to be the inverse of $g(x)$ I have tried to compute $c,d$ and $e$ as a function of $a,b$ and $c$. The result being: $$d = \frac{a²+2ac-bc-b²+c²}{-a^3 - b^3 - c^3 + ab^2 - ac^2 + bc^2 - 2a^2c + 3abc}, \\ e = \frac{ba-c²}{-a^3 - b^3 - c^3 + ab^2 - ac^2 + bc^2 - 2a^2c + 3abc}, \\ f = \frac{-b²+ca+c²}{-a^3 - b^3 - c^3 + ab^2 - ac^2 + bc^2 - 2a^2c + 3abc}.$$
However, the expression has a denominator that happens to be the above equation. So, as $\mathbb{F}_p[x] / (x^3-x-1) = \mathbb{F}_{p^3}$ we know that every element on this field has to have a multiplicative inverse (except the $0$) and therefore the above equation cannot be $0$.
How can I be more precise in the proof?
| For a contradiction assume $a, b, c$ is a solution.
First of all, if $a, b, c$ have any common factor, we can factor it out so without loss generality assume $a, b, c$ have no common factor.
Now, reduce the equation $\mod 2$:
\begin{align}
-a^3 - b^3 -c^3 + ab^2 - ac^2 + bc^2 -2a^2c + 3abc &\equiv\\
a + b + c + ab + ac + bc + abc &\equiv \\
(1 + a)(1 + b)(1 + c) - 1 &\equiv 0 \mod 2
\end{align}
But this shows, $a, b, c$ are all even which is impossible because $a, b, c$ have no common factor.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Finding the expected value of $\dfrac{X}{Y}$ Below is a problem I did. I believe I did it correctly and I am hoping that somebody here can either confirm that I did it right or tell me where I went wrong.
Problem:
Let $X$ be a random variable that is uniformly distributed on the interval $[1,2]$. Let $Y$ be a random variable that is uniformly distributed on the interval $[1,2]$. Assuming that $X$ and $Y$ are independent, what is
$E\left( \dfrac{X}{Y}\right) $?
Answer:
The density function of $X$ and $Y$ is the same. It is:
$$ f(x) = \begin{cases}
1 & \text{for } 1 \leq x \leq 2 \\
0 & \text{otherwise}
\end{cases} $$
\begin{align*}
E\left( \dfrac{X}{Y} \right) &= \int_1^2 \int_1^2 \dfrac{x}{y} \,\, dy \, dx \\
E\left( \dfrac{X}{Y} \right) &= \int_1^2 x \ln (y) \Big|_1^2 \,\, dy \, dx \\
E\left( \dfrac{X}{Y} \right) &= \int_1^2 x \ln (2) - x \ln (1) \, dx \\
E\left( \dfrac{X}{Y} \right) &= \int_1^2 \ln (2) \, x \, dx \\
E\left( \dfrac{X}{Y} \right) &= \dfrac{ \ln (2) \, x^2}{2} \Big|_1^2 \\
E\left( \dfrac{X}{Y} \right) &= \dfrac{ 4 \ln (2) }{2} - \dfrac{ \ln (2) }{2} \\
E\left( \dfrac{X}{Y} \right) &= \dfrac{ 3 \ln (2) }{2} \\
E\left( \dfrac{X}{Y} \right) &\doteq 1.0397208 \\
\end{align*}
| Your answer looks perfect to me! I did a quick numerical simulation and it agrees with you.
>> b = 1 + rand(1000000,1);
>> a = 1 + rand(1000000,1);
>> mean(a./b)
ans =
1.0396
| {
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Evaluating $\lim_{x \rightarrow 0} \frac{\sin(\pi x)(1-\cos(\pi x)}{x^2\sin(x)}$ without L'hôpital's rule I need help finding this limit:
$$\lim_{x \rightarrow 0} \frac{\sin(\pi x)(1-\cos(\pi x))}{x^2\sin(x)}$$
I've used L'Hôpital's rule and the solution is $\pi^3/2$. However I'm asked to solved it without using it and I'm stuck since all the trig tricks I can think of are not making the limit any simpler. So what can I do to solve this?
| We use the result that $$\lim_{x \to 0}\frac{\sin x}{x}=1$$
from this result we also have, $$\lim_{x \to 0}\frac{\sin \pi x}{\pi x}=1\text{ and } \lim_{x \to 0} \frac{\sin (\frac{\pi x}{2})^2}{(\frac{\pi x}{2})^2}=1$$
Now,$$\lim_{x \to 0}\frac{\sin \pi x \cdot(1-\cos\pi x)}{x^2 \cdot \sin x}=\lim_{x \to 0}\frac{\sin \pi x \cdot(2\sin (\frac{\pi x}{2})^2)}{x^2 \cdot \sin x}\\=\lim_{x \to 0} \frac{\sin \pi x}{\pi x}\cdot\pi\cdot\frac{x}{\sin x}\cdot\frac{\pi ^2}{4}\cdot2\cdot\frac{\sin (\frac{\pi x}{2})^2)}{(\frac{\pi x}{2})^2}=\frac{\pi ^3}{2}\lim_{x \to 0} \frac{\sin \pi x}{\pi x}\cdot\frac{x}{\sin x}\cdot\frac{\sin (\frac{\pi x}{2})^2}{(\frac{\pi x}{2})^2}=\frac{\pi ^3}{2}$$
| {
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main axis transformation with the conic $Γ:−6x^2−6yx+4x+2y^2−4y+1=0$ The purpose of this exercise is to reduce the conic $Γ:−6x^2−6yx+4x+2y^2−4y+1=0$ to the canonical expression.
What I already have is the Eigenvectors $-7,3$ and the eigenvectors $\begin{pmatrix} 3 \\ 1 \end{pmatrix}, \begin{pmatrix} -1 \\ 3 \end{pmatrix}$. This gives me the rotation matrix $\begin{pmatrix} 3/\sqrt(10) & -1/\sqrt(10) \\ 1/\sqrt(10) & 3/\sqrt(10) \end{pmatrix}$.
After that my next step, determining the mixed term after rotation, I fail. The values before the $x^2,y^2$ are the Eigenvalues and the values before $x,y$ are $$\begin{pmatrix} 4 & -4 \end{pmatrix} \begin{pmatrix} 3/\sqrt(10) & -1/\sqrt(10) \\ 1/\sqrt(10) & 3/\sqrt(10) \end{pmatrix}= \begin{pmatrix} 4 \sqrt{\dfrac{2}{5}} & -8\sqrt{\dfrac{2}{5}} \end{pmatrix}.$$ This gives me $$3y^2-7x^2+4 \sqrt{\dfrac{2}{5}}x-8 \sqrt{\dfrac{2}{5}}y+1.$$
But this is wrong. Why?
| The given conic equation is of the form
$ r^T Q r + r^T b + c = 0 $
with
$Q = \begin{bmatrix} -6 && - 3 \\ - 3 && 2 \end{bmatrix} $
$ b = [ 4, -4]^T $
$ c = 1 $
From $Q$ we determine the rotation angle $\theta $ using the formula
$ \theta = \dfrac{1}{2} \tan^{-1} \left( \dfrac{ 2 Q_{12} }{Q_{11} - Q_{22}} \right) = \dfrac{1}{2} \tan^{-1} \left( \dfrac{3}{4} \right) $
The rotation matrix whose columns are the unit eigenvectors is
$ R = \begin{bmatrix} \cos \theta && -\sin \theta \\ \sin \theta && \cos \theta \end{bmatrix} $
Since $ \tan(2 \theta) = \dfrac{3}{4} $ then $\cos(2 \theta) = \dfrac{4}{5} $
Hence $\cos(\theta) = \sqrt{ \dfrac{1 + \cos(2\theta)}{2} } = \dfrac{3}{\sqrt{10} } $ and $ \sin(\theta) = \dfrac{1}{\sqrt{10}} $, therefore,
$R = \dfrac{1}{\sqrt{10}} \begin{bmatrix} 3 && -1 \\ 1 && 3 \end{bmatrix}$
With this $R$, we can diagonalize $Q$ as follows
$ Q = R D R^T $
The diagonal entries on the diagonal matrix $D$ are given by
$D_{11} = \dfrac{1}{2} \big( Q_{11} + Q_{22} \big) + \dfrac{1}{2} \big( Q_{11} - Q_{22} \big) \cos(2 \theta) + Q_{12} \sin(2 \theta) $
$D_{22} = \dfrac{1}{2} \big( Q_{11} + Q_{22} \big) - \dfrac{1}{2} \big( Q_{11} - Q_{22} \big) \cos(2 \theta) - Q_{12} \sin(2 \theta) $
And these evaluate to
$ D_{11} = \dfrac{-4}{2} + \dfrac{-32}{10} + \dfrac{-9}{5} = - 2 - 3.2 - 1.8 = -7 $
$D_{22} = \dfrac{-4}{2} - \dfrac{-32}{1} - \dfrac{-9}{5} = -2 + 3.2 + 1.8 = 3 $
i.e.
$D = \begin{bmatrix} -7 && 0 \\ 0 && 3 \end{bmatrix}$
Now the equation of the conic is
$ r^T R D R^T r + r^T b + c = 0$
Let $ w = R^T r $ (from which it follows that $r = R w$ ), then
$ w^T D w + w^T R^T b + c = 0 \hspace{40pt}(*)$
The vector $R^T b$ is given by
$R^T b = \dfrac{1}{\sqrt{10}} \begin{bmatrix} 3 && 1 \\ -1 && 3 \end{bmatrix} \begin{bmatrix} 4 \\ - 4 \end{bmatrix} = \dfrac{1}{\sqrt{10}} \begin{bmatrix} 8 \\ -16 \end{bmatrix}$
So, in the transformed coordinate $w = [x, y]$ the equation is
$ - 7 x^2 + 3 y^2 + \dfrac{8}{\sqrt{10}} x - \dfrac{16}{\sqrt{10}} y + 1 = 0$
| {
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Alternate proof that $\cos(3A)+\cos(3B)+\cos(3C)=1-4\sin(\frac{3A}2)\sin(\frac{3B}2)\sin(\frac{3C}2)$ for $A$, $B$, $C$ the angles of a triangle
$A$, $B$, $C$ being the angles of a triangle, we need to prove that:
$$\cos(3A)+\cos(3B)+\cos(3C)=1-4\sin\left(\frac{3A}2\right)\sin\left(\frac{3B}2\right)\sin\left(\frac{3C}2\right)\tag*{}$$
The proof which I got from my prof:
$\begin{align*}& \cos 3A + \cos 3B -\cos 3(A+B) \tag{01}\\ & = 1+ \sin 3A \sin 3B -(1-\cos 3A)(1-\cos 3B)\tag{02}\\ &= 1+4\sin\frac{3A}2\sin\frac{3B}2\left(\cos \frac{3A}2\cos\frac{3B}2-\sin \frac{3A}2\sin\frac{3B}2\right)\tag{03}\end{align*}$
(01) We use the fact that $C=\pi-(A+B) $
(02) $\cos3(A+B)$ is expanded using compound angle formula. 1 is added and subtracted. Following that some factorisation is done.
(03) Half angle formula is applied on $\sin 3A$ and $\sin 3B$. The identity $1-\cos x = 2\sin^2\frac x2$ is aplied on likes of $1-\cos 3A$
Now I believe the rest of the procedure is obvious.
I would like to know if it is possible to get the proof done without a lot of algebra. Does complex definition of sine and cosine help us to get it done nicely?
| It's easier to work backwards. Still very much "a lot of algebra", though.
$$1 - 4 \sin(\frac{3A}{2}) \sin(\frac{3B}{2}) \sin(\frac{3C}{2})$$
$$= 1 - 4(\frac{e^{i(3/2)A} - e^{-i(3/2)A}}{2i})(\frac{e^{i(3/2)B} - e^{-i(3/2)B}}{2i})(\frac{e^{i(3/2)C} - e^{-i(3/2)C}}{2i})$$
$$= 1 - \frac{i}{2} (e^{i(3/2)A} - e^{-i(3/2)A})(e^{i(3/2)B} - e^{-i(3/2)B})(e^{i(3/2)C} - e^{-i(3/2)C})$$
$$= 1 - \frac{i}{2} (e^{i(3/2)(A+B)} - e^{i(3/2)(A-B)} - e^{i(3/2)(-A+B)} + e^{i(3/2)(-A-B)})(e^{i(3/2)C} - e^{-i(3/2)C})$$
$$= 1 - \frac{i}{2} (e^{i(3/2)(A+B+C)} - e^{i(3/2)(A+B-C)} - e^{i(3/2)(A-B+C)} + e^{i(3/2)(A-B-C)} - e^{i(3/2)(-A+B+C)} + e^{i(3/2)(-A+B-C)} + e^{i(3/2)(-A-B+C)} - e^{i(3/2)(-A-B-C)})$$
$$= 1 - \frac{i}{2} (e^{i(3/2)\pi} - e^{i(3/2)(\pi-2C)} - e^{i(3/2)(\pi-2B)} + e^{i(3/2)(-\pi+2A)} - e^{i(3/2)(\pi - 2A)} + e^{i(3/2)(-\pi + 2B)} + e^{i(3/2)(-\pi+2C)} - e^{i(3/2)(-\pi)})$$
$$= 1 - \frac{i}{2} (e^{i(3/2)\pi} - e^{i(3/2)\pi}e^{-i3C} - e^{i(3/2)\pi}e^{-i3B} + e^{-i(3/2)\pi}e^{i3A} - e^{i(3/2)\pi}e^{-i3A} + e^{-i(3/2)\pi}e^{i3B} + e^{-i(3/2)\pi}e^{i3C} - e^{-i(3/2)\pi})$$
$$= 1 - \frac{i}{2} ((-i) - (-i)e^{-i3C} - (-i)e^{-i3B} + ie^{i3A} - (-i)e^{-i3A} + ie^{i3B} + ie^{i3C} - i)$$
$$= 1 - \frac{i}{2} (-i + ie^{-i3C} + ie^{-i3B} + ie^{i3A} + ie^{-i3A} + ie^{i3B} + ie^{i3C} - i)$$
$$= 1 + \frac{1}{2} (-2 + e^{i3A} + e^{-i3A} + e^{-i3B} + e^{i3B} + e^{-i3C} + e^{i3C})$$
$$= 1 + \frac{1}{2} (-2 + 2 \cos(3A) + 2 \cos(3B) + 2\cos(3C))$$
$$= \cos(3A) + \cos(3B) + \cos(3C)$$
| {
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Green's theorem and translated angular form on the path $\gamma:[0,3\pi/2]\to\Bbb R^2,\gamma(t)=(t,\pi\cos t)$-strange result
I would like to compute the following integral $$\int_\gamma-\frac{y}{(x-\pi)^2+y^2}dx+\frac{x-\pi}{(x-\pi)^2+y^2}dy,$$ where $\gamma:\left[0,\frac{3\pi}2\right]\to\Bbb R^2,\gamma(t)=(t,\pi\cos t).$
Here is my answer which I would like to verify.
The differential $1$-form $\omega=-\frac{y}{(x-\pi)^2+y^2}dx+\frac{x-\pi}{(x-\pi)^2+y^2}dy$ is a translated angular form centered at $(\pi,0),$ hence $d\omega=0,$ that is $\omega$ is closed. I would like to apply Green's theorem, so, for that purpose, I considered the following paths:
$$\gamma_2:[0,1]\to\Bbb R^2,\gamma_2(t)=\left(\frac{3\pi}2,t\right),\\\gamma_3:[0,1]\to\Bbb R^2,\gamma_3(t)=\left(\frac{3\pi}2-\frac{3\pi}2t,\pi\right),\\\gamma_4:[0,2\pi]\to\Bbb R^2,\gamma_4(t)=(\pi+\cos t,-\sin t).$$ First, $$\int_{\gamma_2}\omega=\int_0^1\frac{\frac{3\pi}2-\pi}{\left(\frac{3\pi}2-\pi\right)^2+t^2}dt=\arctan\frac{t}{\frac{\pi}2}\Big|_0^1=\arctan\frac2\pi\\\int_{\gamma_3}\omega=\int_0^1\frac{\pi}{(t-\pi)^2+\pi^2}dt=\arctan\frac{t-\pi}\pi\Big|_0^1=\arctan\frac{1-\pi}\pi+\frac\pi4$$ and $$\int_{\gamma_4}\omega=-2\pi$$ because $\gamma_4$ is closed and negatively oriented. Now, let $D$ be the set "inside" the contour $\gamma+\gamma_2+\gamma_3$ and outside the circle $\gamma_4.$ By Green's theorem, $$\begin{aligned}\int_\gamma\omega+\int_{\gamma_2+\gamma_3+\gamma_4}\omega&=\int_{\partial D}\omega=\int_D d\omega=0\\\implies \int_\gamma\omega&=-\int_{\gamma_2+\gamma_3+\gamma_4}\omega=2\pi-\arctan\frac2\pi-\arctan\frac{1-\pi}\pi-\frac\pi4\\&=\frac{7\pi}4-\arctan\left(\tan\left(\arctan\frac2\pi+\arctan\frac{1-\pi}\pi\right)\right)\\&=\frac{7\pi}4-\arctan\left(\frac{\frac2\pi+\frac{1-\pi}\pi}{1-\frac2\pi\cdot\frac{1-\pi}\pi}\right)\\&=\frac{7\pi}4+\arctan\frac{(\pi-3)\pi}{\pi^2+2\pi-2}\\&=\arctan\left(\tan\left(\frac{7\pi}4+\arctan\frac{(\pi-3)\pi}{\pi^2+2\pi-2}\right)\right)\\&=\arctan\left(\frac{-1+\frac{(\pi-3)\pi}{\pi^2+2\pi-2}}{1+\frac{(\pi-3)\pi}{\pi^2+2\pi-2}}\right)\\&=\arctan\frac{2-5\pi}{2\pi^2-\pi-2}\end{aligned}$$
But I'm unsure as I didn't expect such a result. Did I make any mistakes?
| Your $\gamma_2$ ends at $\left(\frac{3\pi}{2},1\right)$, but presumably you want it to end at $(\frac{3\pi}{2},\pi)$ to connect to the start of $\gamma_3$. Correction:
$$ \gamma_2:[0,\pi]\to\mathbb{R}^2,\ \gamma_2(t)=\left(\frac{3\pi}2, t\right) $$
$$ \int_{\gamma_2} \omega = \int_0^\pi \frac{\frac{3\pi}2-\pi}{\left(\frac{3\pi}2-\pi\right)^2+t^2}\, dt = \arctan\frac{t}{\frac{\pi}{2}} \Big|_0^\pi = \arctan 2 $$
(Or you could have $\gamma_2$ keep domain $[0,1]$ and use $y=\pi t$, but then don't forget $dy = \pi\, dt$.)
You didn't properly substitute $x = \frac{3\pi}{2} - \frac{3\pi}{2} t$ when integrating on $\gamma_3$. It's also missing the factor from $dx = -\frac{3\pi}{2} dt$ and/or the minus sign at the beginning of the $dx$ term in the $\omega$ definition. Correction:
$$ \begin{align*} \int_{\gamma_3} \omega &= \int_0^1 \frac{-\pi}{\left(\frac{3\pi}{2} - \frac{3\pi}{2} t - \pi\right)^2 + \pi^2}\left(-\frac{3\pi}{2}\, dt\right)
= \int_0^1 \frac{\frac{3}{2}}{\left(\frac{1}{2} - \frac{3}{2} t\right)^2 + 1}\, dt \\
\int_{\gamma_3} \omega &= \left. \arctan\left(\frac{3}{2} t - \frac{1}{2}\right) \right|_0^1 = \arctan 1 + \arctan \frac{1}{2}
\end{align*} $$
You are correct about the combination of the curves for Green's theorem:
$$ \int_\gamma \omega = - \int_{\gamma_2+\gamma_3+\gamma_4} \omega = 2\pi - \arctan 2 - \arctan 1 - \arctan \frac{1}{2} $$
Note $\arctan 2 + \arctan \frac{1}{2} = \frac{\pi}{2}$ since they form two angles of a right triangle.
$$ \int_\gamma \omega = \frac{5 \pi}{4} $$
And this makes sense since it equals the counter-clockwise change of angle from $(0,\pi)$ to $(\frac{3\pi}{2}, 0)$ with respect to the "center" point $(\pi, 0)$ of the translated angular form $\omega$.
Also, here are different paths I might have used, which are even easier to integrate:
$$\gamma_5: \left[0, \frac{5\pi}{4}\right] \to \mathbb{R}^2,\ \gamma_5(t) = (\pi(1+\sqrt{2} \cos t), \pi \sqrt{2} \sin t)$$
$$\gamma_6: \left[\frac{3\pi}{2}, \pi(1+\sqrt{2})\right] \to \mathbb{R}^2,\ \gamma_6(t) = (t,0)$$
| {
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Solving ${d^2 x \over dt^2}=-\omega^2x +\alpha x^2,$ On solving a Lagrangian, I obtained the Lagrangian equation of motion as
$${d^2 x \over dt^2}=-\omega^2x +\alpha x^2,$$ Where $\omega$ and $\alpha$ are constants and t is the time.
Could anyone please help me to find an analytic solution to this differential equation?
My attempt
Multiplying both sides with $\dot x$, we get
$$\dot x {d\dot x\over dt}=(\alpha x^2-\omega^2 x){dx\over dt}$$
gives
$${1\over2} {d(\dot x)^2\over dt}=(\alpha x^2-\omega^2 x){dx\over dt}$$
gives
$${1\over2} d(\dot x)^2=(\alpha x^2-\omega^2 x)dx$$
gives
$$(\dot x)^2={2\over 3}\alpha x^3-\omega^2x^2$$
Could anyone please assist me to move further
Thanks in advance
| Writing the ode as
\begin{align*}
x^{\prime \prime}&=\left(\alpha x-\omega^{2}\right) x
\end{align*}
Multiplying both sides by $x^{\prime}$ gives
\begin{align*}
x^{\prime} x^{\prime \prime}&=\left(\alpha x-\omega^{2}\right) x x^{\prime}
\end{align*}
Integrating both sides w.r.t. $t$ gives
\begin{align*}
\int{x^{\prime} x^{\prime \prime}\, \mathrm{d}t} &=\int{\left(\alpha x-\omega^{2}\right) x x^{\prime}\, \mathrm{d}t}\\
\int{x^{\prime} x^{\prime \prime}\, \mathrm{d}t} &=\int{\left(\alpha x-\omega^{2}\right) x\, \mathrm{d}x} \tag{1}
\end{align*}
But
$$
\int{x^{\prime} x^{\prime \prime}\, \mathrm{d}t} = \frac{1}{2} \left(x^{\prime}\right)^2
$$
Hence equation (1) becomes
\begin{align*}
\frac{1}{2} \left(x^{\prime}\right)^2 &=\int{\left(\alpha x-\omega^{2}\right) x\, \mathrm{d}x} \tag{2}
\end{align*}
But
$$
\int{\left(\alpha x-\omega^{2}\right) x\, \mathrm{d}x} = \frac{1}{3} \alpha \,x^{3}-\frac{1}{2} \omega^{2} x^{2}
$$
Therefore equation (2) becomes
\begin{align*}
\frac{1}{2} \left(x^{\prime}\right)^2 &=\frac{1}{3} \alpha \,x^{3}-\frac{1}{2} \omega^{2} x^{2} + c_2
\end{align*}
Where $c_2$ is an arbitrary constant of integration.
This is first order ODE which is now solved for $x$.
Solving for $x^{\prime}$ gives
\begin{align*}
x^{\prime}&=\frac{\sqrt{6 \alpha x^{3}-9 \omega^{2} x^{2}+18 c_{2}}}{3}\tag{1} \\
x^{\prime}&=-\frac{\sqrt{6 \alpha x^{3}-9 \omega^{2} x^{2}+18 c_{2}}}{3}\tag{2}
\end{align*}
These are separable. So just need to do integrtion. But there does not
seem to be closed form solution to these elliptical type integrals. At least Maple could not do it. Integrating using Maple gives
$$
x_1(t) = \int_{0}^{x(t)}\frac{3}{\sqrt{6 \textit{_a}^{3} \alpha -9 \textit{_a}^{2} \omega^{2}+18 c_{2}}}d \textit{_a} -t -c_{1} = 0
$$
Similarly to the other ode.
$$
x_2(t) = \int_{0}^{x(t)}-\frac{3}{\sqrt{6 \textit{_a}^{3} \alpha -9 \textit{_a}^{2} \omega^{2}+18 c_{2}}}d \textit{_a} -t -c_{1} = 0
$$
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Determine $a_n :=\frac{2^n+(-3)^n}{(-2)^n+3^n}, \lim \inf a_n,\lim \sup a_n, \lim a_n $. I am to determin $a_n :=\frac{2^n+(-3)^n}{(-2)^n+3^n}, \lim \inf a_n,\lim \sup a_n, \lim a_n.$
I was checking the sequence for the very first elements:
$n=1$: $\;\frac{2^1+(-3)^1}{(-2)^1+3^1}=-1:$
$n=2: \;\frac{2^2+(-3)^2}{(-2)^2+3^2}=1.$
$n=3$: $\;\frac{2^3+(-3)^3}{(-2)^3+3^3}=-1.$
$n=4$: $\;\frac{2^4+(-3)^4}{(-2)^4+3^4}=1.$
$n=5$: $\;\frac{2^5+(-3)^5}{(-2)^5+3^5}=-1.$
So I assume that for odd $n$ values $a_n = -1$ and for even $n$ values $a_n = 1.$
My assumption is that $\lim \inf a_n = -1$,$\lim \sup a_n = 1$ and the $\lim a_n$ doesn't exist, but I'm struggling at this point. I would appreciate your help a lot.
| Note that if $n$ is even, then $(-2)^n = (-1)^n 2^n = 2^n$; similarly, $(-3)^n = (-1)^n 3^n = 3^n$. Hence $$a_n = \frac{2^n + (-3)^n}{(-2)^n + 3^n} = 1, \quad n \text{ even}.$$
And if $n$ is odd, then $(-2)^n = -2^n$, and $(-3)^n = -3^n$, hence $$a_n = \frac{2^n - 3^n}{-2^n + 3^n} = -1, \quad n \text{ odd}.$$
Consequently, for all integers $n$,
$$a_n = \begin{cases}1, & n \text{ even} \\ -1, & n \text{ odd.} \end{cases}$$ We can also write this simply as $a_n = (-1)^n$.
Thus, $$\sup a_n = 1, \quad \inf a_n = -1,$$ and $$\liminf a_n = 1, \quad \liminf a_n = -1.$$ Since $\limsup a_n \ne \liminf a_n$, it follows that $\lim a_n$ does not exist.
How do we justify the supremum and infimum? Well, we established $a_n \le 1$ for all $n$, and $a_2 = 1$ implies that this bound is tight. A similar reasoning applies for the infimum. And since $a_{2m} = 1$ for all positive integers $m$, the limit superior is $1$, and similar reasoning applies for the limit inferior.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given a permutation $\sigma = (13)(254)$, state $\sigma^2$. Given a permutation $\sigma = (13)(254)$, state $\sigma^2$.
$\sigma = (13)(254), \sigma^2=(13)(254)(13)(254) = (13)(13)(254)(254) = (425)
$
Or, in two row format, get:
$$ \sigma = \begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix} $$
Then, $$\sigma^2=\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix}
\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix}
$$
Due to disjoint cycles, they can commute as no affect on map produced.
$$\sigma^2=\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}\begin{pmatrix} 1 & 3 \\ 3 & 1 \end{pmatrix}
\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix}
\begin{pmatrix} 2 & 5 &4 \\ 5 & 4 & 2\end{pmatrix}
$$
My answer is:
$$\sigma^2=\begin{pmatrix} 1 & 3 \\ 1 & 3 \end{pmatrix}
\begin{pmatrix} 2 & 5 &4 \\ 4 & 2 & 5\end{pmatrix}
= e.\begin{pmatrix} 2 & 5 &4 \\ 4 & 2 & 5\end{pmatrix}
= \begin{pmatrix} 2 & 5 &4 \\ 4 & 2 & 5\end{pmatrix}
$$
But, the answer is different: $(245)$?
| Here
$$\begin{align}
\sigma^2&=\begin{pmatrix}1&2&3&4&5\\3&5&1&2&4\end{pmatrix}\cdot\begin{pmatrix}1&2&3&4&5\\3&5&1&2&4\end{pmatrix}\\
&=\begin{pmatrix}1&2&3&4&5\\1&4&3&5&2\end{pmatrix}\\
&=(245).
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Compute $f^{(2020)}(0)$ Problem :
Let
$$f(x)=\frac{x}{(x+1)(1-x^2)}$$
Then find $f^{(2020)}(0)$.
My Attempt :
From partial fraction decomposition,
$$f(x)=\frac{1}{4(1-x)}+\frac{1}{4(x+1)}-\frac{1}{2(x+1)^2}$$
and,
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n,\quad \frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n, \quad\frac{1}{(1+x)^2}=\sum_{n=1}^\infty (-1)^{n+1}nx^{n-1}=\sum_{n=0}^\infty(-1)^n(n+1)x^n$$
From these,
$$f(x)=\sum_{n=0}^\infty \frac{x^n}{4}+\sum_{n=0}^\infty \frac{(-1)^nx^n}{4}+\sum_{n=0}^\infty\frac{(-1)^{n+1}(n+1)x^n}{2} \\ =\sum_{n=0}^\infty\frac{x^n+(-1)^nx^n+(-1)^{n+1}(n+1)x^n}{4} \\ =\sum_{n=0}^\infty\frac{(1+(-1)^n+(-1)^{n+1}(n+1))x^n}{4}$$
$f^{(2020)}(0)=(2020)!\times a_{2020}$ where $f(x)=\sum a_nx^n$
In this case, $a_n = 1+(-1)^n+(-1)^{n+1}(n+1), a_{2020}=-2019$ but there is no same answer.
Where did I make a mistake?
| A bit tricky
$$f(x)=\frac{x}{(x+1)(1-x^2)}$$ Integrate $$g(x)=\int f(x)\,dx=\frac{1}{4} \left(\frac{2}{x+1}-\log (1-x)+\log (x+1)\right)$$
$$g(x)=\frac 12 +\sum_{n=1}^\infty \frac{(-1)^n (2 n-1)+1}{4 n} x^n$$ Differentitate
$$f(x)=g'(x)=\sum_{n=1}^\infty \frac{(-1)^n (2 n-1)+1}{4} x^{n-1}$$
Now, to make $x^{n-1}=x^{2020}$, $n=2021$ and then the coefficient of $x^{2020}$ is $-1010$.
Just multiply by $2020!$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \\
&= (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)
\end{align}$$
now,
$$\begin{align}
a^4 + b^4 &= (a+b)^4 - (4a^3b + 6a^2b^2 + 4ab^3) \\
&= (a+b)^4 - (4ab(a^2 + b^2) + 6(ab)^2) \\
&= (a+b)^4 - (4ab((a + b)^2 - 2ab) + 6(ab)^2) \\
&= 7^4 - (4(2)(7^2 - 2(2)) + 6(2)^2) \\
&= 2017
\end{align}$$
so
$$\begin{align}
&\phantom{=}\; (a+b)^6 - (6ab(a^4 + b^4) + 15a^2b^2 (a^2 + b^2) + 20(ab)^3)\\
&= 7^6 - (6\cdot2\cdot(2017) + 15(2)^2 (7^2 - 2(2)) + 20(2)^3) \\
&= 90585
\end{align}$$
correct?
| From Vieta’s rules, $a+b=7, ab=2$. $$a^3+b^3=(a+b)(a^2+b^2-ab)=(a+b)((a+b)^2-3ab)= 7(49-2\cdot 3)=7\cdot 43=301 $$ Again, $$(a^3+b^3)^2=a^6+b^6+2a^3b^3$$$$301^2=a^6+b^6+2(2)^3$$ so $$a^6+b^6=301^2-16=90585. $$
| {
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"url": "https://math.stackexchange.com/questions/4473560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Is a diagonal matrix with "$12$"s along the diagonal an identity matrix? Is C following an identity matrix?
$$A = \begin{bmatrix}
2& -1 & 12 \\
3 & 6 & -9\\
1& 1& 3
\end{bmatrix}$$
$$B = \begin{bmatrix}
9 & 5 & -21 \\
-6 & -2 & 18\\
-1& -1& 5
\end{bmatrix}$$
$$C = \begin{bmatrix}
12& 0 & 0 \\
0 & 12 & 0\\
0& 0&12
\end{bmatrix}$$
I perform AB and BA and end up with C as the answer. I do know that an identity matrix must consist of only ones and zeros. All the diagonals are 12 and if you pull out 1/12, the identity would remain. Would it alone be considered the identity due the numbers being the same? I’m guessing no but am not sure.
| In this situation, $C$ is a scalar matrix that is, as you say, $12$ times the identity matrix, i.e. $C = 12I$. This is not the same thing as the identity, but it has some special properties in common with the identity matrix. For example, it commutes with any other $3 \times 3$-matrix, i.e. $CX = XC$ for any $X$.
Also, the fact that $AB = BA = C$ allows you write down the inverse of either $A$ or $B$ with ease:
$$
AB = C
\implies
A\, \bigl( \tfrac{1}{12} B \bigr) = \tfrac{1}{12} C = I,
$$
which shows that
$$
A^{-1} = \tfrac{1}{12} B
= \tfrac{1}{12} \begin{bmatrix}
9 & 5 & -21 \\
-6 & -2 & 18 \\
-1 & -1 & 5
\end{bmatrix}
= \begin{bmatrix}
\tfrac{3}{4} & \tfrac{5}{12} & -\tfrac{7}{4} \\
-\tfrac{1}{2} & -\tfrac{1}{6} & \tfrac{3}{2} \\
-\tfrac{1}{12} & -\tfrac{1}{12} & \tfrac{5}{12}
\end{bmatrix}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing $\sum_{cyc}\tan\frac\alpha2\tan\frac\beta2\geq4$ for a cyclic quadrilateral
Let $ABCD$ be a cyclic quadrilateral with sides $a$, $b$, $c$ and $d$. Denote $s$ the semiperimeter and let $\angle{DAB}=\alpha$, $\angle{ABC}=\beta$, $\angle{BCD}=\gamma$ and $\angle{CDA}=\delta$. Then the following inequality holds
$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}+\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}\geq4.\tag{1}$$
Proof. Substituting from the half-angle formula for the tangent (see $(2)$ here) we have that
$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}=\sqrt{\frac{(s-a)(s-d)}{(s-b)(s-c)}}\cdot{\sqrt{\frac{(s-a)(s-b)}{(s-c)(s-d)}}}=\frac{s-a}{s-c}.$$
Similarly,
$$\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}=\frac{s-b}{s-d}\qquad\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}=\frac{s-c}{s-a}\qquad\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}=\frac{s-d}{s-b}$$
Thus, the left-hand side of $(1)$ can be rewritten as follows
$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}+\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}=\frac{s-a}{s-c}+\frac{s-b}{s-d}+\frac{s-c}{s-a}+\frac{s-d}{s-b}.$$
But,
$$\frac{s-a}{s-c}+\frac{s-b}{s-d}+\frac{s-c}{s-a}+\frac{s-d}{s-b}=\frac{a-c}{s-a}+\frac{b-d}{s-b}+\frac{c-a}{s-c}+\frac{d-b}{s-d}+4.$$
Since $\frac{a-c}{s-a}+\frac{c-a}{s-c}=\frac{4(a-c)^2}{(-a+b+c+d)(a+b-c+d)}$, and similarly for $\frac{b-d}{s-b}+\frac{d-b}{s-d}$, then $\frac{a-c}{s-a}+\frac{b-d}{s-b}+\frac{c-a}{s-c}+\frac{d-b}{s-d}$ is positive. Hence,
$$\tan{\frac{\alpha}{2}}\tan{\frac{\beta}{2}}+\tan{\frac{\beta}{2}}\tan{\frac{\gamma}{2}}+\tan{\frac{\gamma}{2}}\tan{\frac{\delta}{2}}+\tan{\frac{\delta}{2}}\tan{\frac{\alpha}{2}}\geq4.$$
Notice equality holds when $ABCD$ is rectangular.
A huge list of inequalities can be seen at Cut-the-knot.org.
Questions: a) Is there a simpler way to prove $(1)$? b) Is $(1)$ known?
| Use AM_GM
$F=\frac{s-a}{s-c}+\frac{s-b}{s-d}+\frac{s-c}{s-a}+\frac{s-d}{s-b}\ge 4 \sqrt[4]{\frac{s-a}{s-c}\frac{s-b}{s-d}\frac{s-c}{s-a}\frac{s-d}{s-b}}=4.$
| {
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Integrating $\int { \arctan{t} \over \sqrt{t} } $ Integrate: $$\int { \arctan{t} \over \sqrt{t} }\,\mathrm dt$$
My attempt:
I first substituted $y = \sqrt{t}$
then $$\int { \arctan{t} \over \sqrt{t} }\,\mathrm dt = 2\int { \arctan{y^2}}\,\mathrm dy$$
then I integrated by parts:
$u = \arctan{y^2}$, $v' = 1$
$$ = 2\left(y\arctan{y^2} - \int { y^2 \over 1+y^4}\,\mathrm dy\right)$$
but I am stuck, I cannot think of a way to solve this other integral.
I believe there is a more straightforward approach than the one I am using.
Thanks in advance.
| Another simplification can be done for the term
$$ \int \frac{y^2}{1+y^4} dy$$
As
Divde by $y^2$
$$\int \frac{dy}{y^2 + \frac{1}{y^2}}$$
Following by
$$ \frac{1}{2}\int \frac{2+\frac{1}{y^2} - \frac{1}{y^2}}{y^2+\frac{1}{y^2}+2 -2}$$
Separating
$$\frac{1}{2}\left [\int \frac{1-\frac{1}{y^2}}{\left(y+\frac{1}{y}\right)^2 - (\sqrt{2})^2}dy + \frac{1+\frac{1}{y^2}}{\left(y-\frac{1}{y}\right)^2 + (\sqrt{2})^2}dy \right]$$
Let
$$I_1 = \frac{1}{2}\int \frac{1-\frac{1}{y^2}}{\left(y+\frac{1}{y}\right)^2 - (\sqrt{2})^2}dy$$
Now
$u =(y+ \frac{1}{y})$
And
$du = 1 - \frac{1}{y^2}$
$$ I_1 = \frac{1}{2}\int \frac{du}{u^2 - (\sqrt{2})^2}$$
Integrating, and
$$I_1 = \frac{1}{4\sqrt{2}} ln\left[\frac{u-\sqrt{2}}{u+ \sqrt{2}}\right] $$
Substituting the value of $u$
$$I_1 = \frac{1}{4\sqrt{2}} ln\left[\frac{y^2-\sqrt{2}y+1}{y^2+\sqrt{2}y+1}\right]. . .(1)$$
Now let
$$I_2 = \frac{1}{2} \int \frac{1+\frac{1}{y^2}}{\left(y-\frac{1}{y}\right)^2 + (\sqrt{2})^2}dy $$
Let,
$v = (y-\frac{1}{y})$
Also
$dv = 1+ \frac{1}{y^2}$
$$I_2 = \frac{1}{2}\int \frac{dv}{v^2 +(\sqrt{2})}$$
Integrating
$$ I_2 = \frac{1}{2} \arctan \left(\frac{v}{\sqrt{2}}\right)$$
Substituting value of $v$
$$I_2 = \frac{1}{2\sqrt{2}} \arctan \left[\frac{y^2 -1}{\sqrt{2}y}\right] . . . (2)$$
From (1) and (2)
$$\int \frac{y^2}{1+y^4} dy = \frac{1}{4\sqrt{2}} ln\left[\frac{y^2-\sqrt{2}y+1}{y^2+\sqrt{2}y+1}\right] + \frac{1}{2\sqrt{2}} \arctan \left[\frac{y^2 -1}{\sqrt{2}y}\right] $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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prove that $\arctan\frac{\cos x-\sin x}{\cos x+\sin x}=\frac{\pi}{4}-x$, where $0I tried to solve this
$$\begin{align} \arctan\frac{\cos x-\sin x}{\cos x+\sin x}&=\arctan\frac{1-\tan x}{1+\tan x}\\&=\arctan\frac{\tan\frac{\pi}{4}-\tan x}{1+\tan\frac{\pi}{4}\tan x}\\&=\arctan\tan\left(\frac{\pi}{4}-x\right)\\&=\frac{\pi}{4}-x \end{align}$$
But suddenly I realised $x$ can be $\frac{\pi}{2}$ where $\cos x=0$ and I can't divide it by $\cos x$.
and also i am wondering that $\frac{\pi}{4}-x$ is only valid fro $\frac{-\pi}{4}<x<\frac{3\pi}{4}$.
what happen when $\pi>x>\frac{3\pi}{4}$.
| Another approach. Let $f(x)=\arctan(\frac{\cos x-\sin x}{\cos x+\sin x})$, with $x\in (-\pi/4,3\pi/4)$.
$$f'(x)=\frac{1}{1+\frac{\cos x-\sin x}{\cos x+\sin x}^2}\frac{(\sin x-\cos x)(\cos x +\sin x)+(\sin x-\cos x)(\cos x-\sin x)}{(\cos x+\sin x)^2}$$
It's easy to simplifie that expression and get that $f'(x)=-1$, it implies that $f(x)=C-x$, where $C$ is some constant. To find it we evaluate the function in a point like for example $\pi/2$, where the function is well defined.
$$C-\frac{\pi}{2}=f(\pi/2)=\arctan(\frac{0-1}{0+1})=\arctan(-1)=-\frac{\pi}{4}$$
so $C=\frac{\pi}{4}$ and $f(x)=\frac{\pi}{4}-x$ in the interval where is well defined
| {
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"url": "https://math.stackexchange.com/questions/4480265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of lattice points with the same parity inside a triangle Given a point $(a,b)$ with positive coordinates, I'd like to count the number of lattice points $(x,y)$ with the same parity (i.e., $x \equiv y \ (mod \ 2)$) inside the triangle $(0,0)(a,0)(a,b)$. How to compute it fast?
Note that without the parity requirement, it can be done using Pick's theorem directly.
| After some trials, I come up with a solution similar to the Euclidean algorithm.
We want to calculate $\sum_{0 \leq x \leq a}\lfloor \frac{\frac{b}{a}x+(x\%2)}{2} \rfloor$ where $\%$ means the standard modulo operator. We separate into two parts: $x$ is odd and $x$ is even. For the first case, we write as $\sum_{0 \leq 2t+1 \leq a}\lfloor \frac{2bt+a+b}{2a} \rfloor$; for the second case, we write as $\sum_{0 \leq 2t \leq a}\lfloor \frac{bt}{a} \rfloor$. Therefore, it is suffice to compute $f(a,b,c,n)=\sum_{0 \leq x \leq n}\lfloor\frac{ax+b}{c} \rfloor$ fast.
If $a \geq c$ or $b \geq c$, we can write $a=q_1 c+r_1$ and $b=q_2 c+r_2$ where $r_1$ and $r_2$ are remainders. Hence, $f(a,b,c,n)=\sum_{0 \leq x \leq n}\lfloor q_1 x + q_2 + \frac{r_{1}x + r_{2}}{c} \rfloor=(\sum_{0 \leq x \leq n}q_1 x + q_2) +\sum_{0 \leq x \leq n}\lfloor \frac{r_1 x + r_2}{c} \rfloor$. The first part can be calculated directly, so we just need to consider the case $a \lt c$ and $b \lt c$.
In this case, it turns out that $f(a,b,c,n)=\sum_{0 \leq x \leq n}\lfloor\frac{ax+b}{c} \rfloor=\sum_{0 \leq x \leq n}\sum_{0 \leq j \leq \lfloor\frac{ax+b}{c} \rfloor-1}1=\sum_{0 \leq j \leq \lfloor\frac{an+b}{c} \rfloor-1}\sum_{0 \leq x \leq n}[j<\lfloor\frac{ax+b}{c} \rfloor]=\sum_{0 \leq j \leq \lfloor\frac{an+b}{c} \rfloor-1}\sum_{0 \leq x \leq n}[cj<ax+b-c+1]=\sum_{0 \leq j \leq \lfloor\frac{an+b}{c} \rfloor-1}(n-\lfloor\frac{cj+c-b-1}{a} \rfloor)=n\lfloor\frac{an+b}{c} \rfloor-\sum_{0 \leq j \leq \lfloor\frac{an+b}{c} \rfloor-1}\lfloor\frac{cj+c-b-1}{a} \rfloor=n\lfloor\frac{an+b}{c} \rfloor-f(c,c-b-1,a,\lfloor\frac{an+b}{c} \rfloor-1)$.
The above runs similar to the Euclidean algorithm, and we solve the problem in $O(log(max(a,b)))$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Closed form of $\sum _{k\ge 1} \frac{(-1)^{\binom{k}{p}}}{k}$, an alternating harmonic series with the signs determined by a binomial coeffcient In a comment to Evaluating $\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$ for $n \in \mathbb{N}$ I proposed to study this alternating harmonic sum
$$s(p) = \sum _{k\ge 1} \frac{(-1)^{\binom{k}{p}}}{k}\tag{1}$$
where the sign of the terms is determined by a binomial coefficient.
This turned out to be interesting, and the results can be formulated in three statements:
Statement 1: the sum is convergent only if $p$ is power of $2$: $p = 2^q, q=0,1,2,...$
Statement 2: the closed expressions of $s$ for the first few values of $q$ are
$$s(1) = -\log (2)\simeq -0.693147$$
$$s(2) = \frac{\pi }{4}-\frac{\log (2)}{2}\simeq 0.438825$$
$$s(4)= \frac{\pi}{8} \left(1+2 \sqrt{2}\right) -\frac{\log (2)}{4}\simeq 1.33013 $$
$$s(8) = \frac{\pi}{16} \left(4 \sqrt{2+\sqrt{2}}+2 \sqrt{2}+1\right) -\frac{\log (2)}{8}\simeq 2.11629$$
The next step was hard, and I did not expect to find a simple expression. But here it is
$$\begin{align}s(16) = \frac{\pi}{16} \left(4 \left(\sqrt{2-\sqrt{2+\sqrt{2}}}+\sqrt{\left(2+\sqrt{2}\right) \left(2-\sqrt{2+\sqrt{2}}\right)}\right)+\sqrt{2}+2 \sqrt{2+\sqrt{2}}+2 \sqrt{2-\sqrt{2+\sqrt{2}}}+2 \sqrt{\sqrt{2+\sqrt{2}}+2}+\frac{1}{2}\right) -\frac{\log (2)}{16}\simeq 2.85436\end{align}$$
Statement 3: The general structure seems to be
$$s(p=2^q) = \frac{\pi}{p}a(p)-\frac{1}{p} \log(2) \tag{2}$$
where $a(p)$ is an algebraic number composed of the number $2$.
Questions
(1) are these results known in the literature?
(2) can you derive/prove the three statements?
(3) can you give the expression $a(p)$ to extend the given range?
| For statement 1:
Write $k = k_N2^N + k_{N-1}2^{N-1} + \cdots + k_12 + k_0$ for the binary expansion of $k$ (and similarly for $p$). We may appeal to Lucas's theorem to show that $\binom{k}{p} \equiv 1 \pmod{2}$ if and only if for every $i$, if $p_i = 1$ then $k_i = 1$.
Now, let $1 < p < 2^q$ which is not a pure power of $2$, and let $b$ denote the number of $1$s appearing in the binary expansion of $p$, noting $b \geq 2$. Then for every integer $N \geq 1$, there are exactly $2^{q-b}$ values of $k$ where $(N-1)2^q < k \leq N2^q$ and $\binom{k}{p} \equiv 1 \pmod{2}$.
It follows that $$\frac{1}{N2^q}\sum_{k=1}^{N2^q}k\frac{(-1)^\binom{k}{p}}{k} = \frac{1}{N2^q}\cdot N(2^q - 2\cdot 2^{q-b}) = 1 - 2^{1-b} \not\to 0 \text{ as } N\to \infty$$
and therefore by Kronecker's lemma,
$$\sum_{k=1}^\infty \frac{(-1)^\binom{k}{p}}{k}$$
doesn't converge.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Trying to solve $|2x-15| = -x^2 - 5x -8$ My first instinct was to take the positive and negative of the right hand side, resulting in
$2x-15 = -x^2 - 5x - 8$, and $2x-15 = x^2 + 5x + 8$, which results in the first giving me two real answers using the quadratic equation, and the second being two imaginary solutions. The problem is that, when graphed, these 2 graphs do not intersect at all, so there should be no real solutions. As for the given complex solutions, neither were considered by wolfram alpha.
Knowing that there are no real solutions, I'm confused as to how I might get complex solutions through means not already attempted and explained above.
| The LHS is always non-negative, but the RHS is negative for real $\,x\,$, so no real solutions exist. Let $\,2x-15=z \in \mathbb C \setminus \mathbb R\,$, then substituting $\,x = (z+15)/2\,$ in the original equation:
$$
z^2 + 40 z + 407 = -4 |z| \tag{1}
$$
Taking complex conjugates and subtracting $\,(1) - \overline{(1)}\,$:
$$
\require{cancel}
\begin{align}
(z^2 + 40 z + \cancel{407}) - (\bar z^2 + 40 \bar z + \cancel{407}) &= \bcancel{-4 |z|} + \bcancel{4 |\bar z|}
\\ \iff\quad\quad (z- \bar z)(z+\bar z + 40) &= 0 \tag{2}
\end{align}
$$
The first factor is non-zero $\,z - \bar z \ne 0\,$ because $\,z \not \in \mathbb R\,$, which leaves:
$$
z + \bar z = -40 \tag{3} \quad\iff\quad \text{Re}(z) = -20
$$
Substituting $\,40 = -(z+\bar z)\,$ from $\,(3)\,$ back in $\,(1)\,$:
$$
\cancel{z^2} - (\cancel{z} + \bar z) z + 407 = -4 |z| \;\;\iff\;\; |z|^2 - 4 |z| - 407 = 0 \tag{4}
$$
The latter is a quadratic in $\,|z|\,$ with the only positive root $\,|z| = 2 + \sqrt{411}\,$. Then:
$$
\text{Im}(z) = \pm \sqrt{|z|^2 - \text{Re}^2(z)} = \pm \sqrt{\left(2 + \sqrt{411}\right)^2 - (-20)^2} = \pm \sqrt{15 + 4 \sqrt{411}} \tag{5}
$$
It follows from $\,(3)\,$ and $\,(5)\,$ that:
$$
z = \text{Re}(z) + i\,\text{Im}(z) = -20 \pm i\sqrt{15 + 4 \sqrt{411}}
\\ \implies\quad\;\; x = \frac{z+15}{2} = \frac{-5 \pm i\sqrt{15 + 4 \sqrt{411}}}{2} \;\;
$$
| {
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Find another method of solving the shaded area. The diagram shows a square of side length $10\;\rm cm$. A quarter circle, of radius $10\;\rm cm$, is drawn from each vertex of the square. Find the exact area of the shaded region.
And This is my answer.
The answer is right, but I am searching for other ways.
Thanks.
|
Consider circle center on D it intesect circl center on C at point G. The area under first circle and axis can be fd as follows:
Equation of corcle :
$x^2+(y-10))^2=10^2$
Or:
$y=10-\sqrt{100-x^2}$
If $y=\sqrt{a^2-x^2}$ then:
$S=\int\sqrt{a^2-x^2}dx=\frac x2\sqrt{a^2-x^2}+\frac {a^2}2 \sin^{-1}\frac xa$
$$S=\int^{5}_0(10-\sqrt{100-x^2})dx$$
WE finally get:
$S_{AGI}=50\big(1-\frac{\sqrt 3}4-\frac{\pi}6\big)$
The area between this circle lines AB and BC which include S is:
$$S_1=10^2-\frac 14 10^2\pi=100(1-\frac {\pi}4)$$
And the area of required region is:
$$S_2=10^2-[4\times 100(1-\frac{\pi}4)-8\times 50(1-\frac{\sqrt3}4-\frac{\pi}6)]=100\big(1-\sqrt 3+\frac {\pi}3\big)$$
| {
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Radicals in a Cyclotomic Field Extension I was looking for the theorems which describe about all the kinds of radicals contained in a cyclotomic extension. With radical I mean the number, say $x$, is not in $\mathbb{Q}$ but some power, say $x^n$, is in $\mathbb{Q}$. For example, $\sqrt{2}$ is contained in $\mathbb{Q}(\zeta_8)$. Definitely there could be other radicals existing.
Now, is there a theorem which describes or classifies a general scenario saying that some/any $n$th root of a rational (we can rather talk about real $n$th root of an integer (because we can always rationalize the denominator)) is contained in a specific cyclotomic extension? In other words, what are all the radicals present in a cyclotomic extension?
Please correct me if I'm stating anything wrong.
| For $a \in \mathbf Z$, $\sqrt{a}$ is contained in some cyclotomic field (specifically, $\mathbf Q(\zeta_{4|a|})$. For nonzero $a \in \mathbf Z$, if $x^n - a$ is irreducible over $\mathbf Q$ and $n > 1$ is not a power of $2$ then $\sqrt[n]{a}$ (that means an $n$th root of $a$, i.e., a root of $x^n - a$) is not contained in a cyclotomic field.
Each $n > 2$ has an odd prime factor or is divisible by $4$.
If $d \mid n$ and $d > 0$ then $\sqrt[n]{a}^{n/d}$ is a $d$th root of $a$ (a root of $x^d - a$), so if an $n$th root of $a$
is contained in a cyclotomic field then so is a $d$th root of $a$ for each positive factor $d$ of $n$. And since $x^n - a$ is irreducible over $\mathbf Q$, so is $x^d - a$ for each positive divisor $d$ of $n$: see Theorem 4.1 here.
Therefore in order to prove no $n$th root of $a$ is contained in a cyclotomic field when $n$ is not a power of $2$ and $x^n - a$ is irreducible over $\mathbf Q$, it suffices to show this when $n = p$ is an odd prime.
We want to show $\sqrt[p]{a}$ is not in a cyclotomic field when $x^p - a$ is irreducible over $\mathbf Q$.
Cyclotomic fields are abelian Galois extensions of $\mathbf Q$, so by Galois theory subfields of cyclotomic fields are also (abelian) Galois extensions of $\mathbf Q$. Therefore if $\sqrt[p]{a}$ is in a cyclotomic field, $\mathbf Q(\sqrt[p]{a})$ would be Galois over $\mathbf Q$, and that implies there are $p$ roots of $x^p - a$ in $\mathbf Q(\sqrt[p]{a})$ by irreducibility of $x^p - a$ over $\mathbf Q$. A ratio of different $p$th roots of $a$ is a nontrivial $p$th root of unity, so $\mathbf Q(\zeta_p)$ is a subfield of $\mathbf Q(\sqrt[p]{a})$. Computing field degrees over $\mathbf Q$, $(p-1) \mid p$ and that is a contradiction. (Another reason $\mathbf Q(\sqrt[p]{a})$ is not Galois over $\mathbf Q$ is that it has a real embedding since $p$ is odd and there is only one real $p$th root of $a$, which contradicts the existence of $p$ roots of $x^p-a$ in $\mathbf Q(\sqrt[p]{a})$.)
What happens if $n$ is a power of $2$ and $n > 2$? Things get tricky here, as they often do when working with $x^n - a$ and $n$ is a power of $2$ (consider the difference between $p$-power cyclotomic extensions of $\mathbf Q$ for prime $p$ when $p = 2$ and $p \not= 2$). I leave it to someone else to sort things out completely, and will just point out one restriction by reducing to the case $n = 4$.
When $n = 2^k$ for $k \geq 2$, irreducibility of $x^n - a$ over $\mathbf Q$ implies irreducibility of $x^4 - a$ over $\mathbf Q$, and
Claim: If $x^4 - a$ is irreducible over $\mathbf Q$ then $\sqrt[4]{a}$ is in a cyclotomic field if and only if $a = -b^2$ for some integer $b$.
Since positive $a$ don't fit the conclusion of the claim, for $a \in \mathbf Z^+$ with $x^n - a$ irreducible over $\mathbf Q$, $\sqrt[n]{a}$ is not in a cyclotomic field if $n > 2$.
Proof of claim:
As in the argument above, if $\sqrt[4]{a}$ is in a cyclotomic field then $\mathbf Q(\sqrt[4]{a})$ is Galois over $\mathbf Q$. The ratio of different 4th roots of $a$ are the 4th roots of unity, so if $\mathbf Q(\sqrt[4]{a})$ is Galois over $\mathbf Q$ then $\mathbf Q(i)$ is inside $\mathbf Q(\sqrt[4]{a})$. Let's look at $\mathbf Q(\sqrt[4]{a})$ as a quadratic extension of $\mathbf Q(i)$.
The number $\sqrt[4]{a}$ has a quadratic minimal polynomial over $\mathbf Q(i)$. What could that minimal polynomial be? It must be a factor of $x^4 - a$ (the minimal polynomial of $\sqrt[4]{a}$ over $\mathbf Q$), so let's just write down all possible monic quadratic factors of $x^4 - a$ with $\sqrt[4]{a}$ as a root and see which ones could have all coefficients in $\mathbf Q(i)$.
By unique factorization in $\mathbf Q(\sqrt[4]{a})[x]$, where $x^4 - a$ splits completely, there are only three monic quadratic factors of $x^4 - a$ with $\sqrt[4]{a}$ as a root:
$$
(x-\sqrt[4]{a})(x+\sqrt[4]{a}) = x^2 - \sqrt{a},
$$
$$
(x-\sqrt[4]{a})(x-i\sqrt[4]{a}) = x^2 - (1+i)\sqrt[4]{a}x + i\sqrt{a},
$$
$$
(x-\sqrt[4]{a})(x+i\sqrt[4]{a}) = x^2 - (1-i)\sqrt[4]{a}x - i\sqrt{a}.
$$
The second and third choices have linear coefficient $-(1\pm i)\sqrt[4]{a}$, and if this is in $\mathbf Q(i)$ then so is $\sqrt[4]{a}$, which is impossible since $\sqrt[4]{a}$ has degree $4$ over $\mathbf Q$. Thus the minimal polynomial of $\sqrt[4]{a}$ over $\mathbf Q(i)$ must be $x^2 - \sqrt{a}$, so
$\sqrt{a} \in \mathbf Q(i)$, which implies $\mathbf Q(\sqrt{a}) = \mathbf Q(i)$.
From $\mathbf Q(\sqrt{a}) = \mathbf Q(i)$, $a = -b^2$ for some integer $b$.
Conversely, suppose $a = -b^2$. Then the roots of $x^4 - a = x^4 + b^2$ are $\pm\sqrt{bi}$ and $\pm\sqrt{-bi}$, which are in cyclotomic fields since $\sqrt{b}$ is in a cyclotomic field and $\sqrt{i}$ is an $8$th root of unity. In fact, $\mathbf Q(\sqrt{bi}) = \mathbf Q(\sqrt{2b},i)$ is a composite of two different quadratic extensions of $\mathbf Q$: $i = (\sqrt{bi})^2/b$ is in $\mathbf Q(\sqrt{bi})$, the number $\alpha = (1-i)\sqrt{bi}$ in $\mathbf Q(\sqrt{bi})$ satisfies $\alpha^2 = -2i(bi) = 2b$, and $x^2 - 2b$ is irreducible over $\mathbf Q$ since otherwise $2b = c^2$ for some integer $c$ and that would reveal $x^4 - a$ to be reducible over $\mathbf Q$: $a = -b^2 = -c^4/4$, so
$$
x^4 - a = x^4 + \frac{c^4}{4} = (x^2 + cx + c^2/2)(x^2 - cx + c^2/2).
$$
From irreducibility of $x^2 - 2b$, $\mathbf Q(\alpha) = \mathbf Q(\sqrt{2b})$ is a quadratic subfield of $\mathbf Q(\sqrt[4]{a})$ and it is not the quadratic subfield $\mathbf Q(i) = \mathbf Q(\sqrt{-1})$ since $-2b$ is not a perfect square: if $-2b$ were a perfect square then (by unique factorization in $\mathbf Z$) $b = -2d^2$, so $a = -b^2 = -4d^4$, but then $x^4 - a = x^4 + 4d^4$ would be reducible over $\mathbf Q$:
$$
x^4 - a = x^4 + 4d^4 = (x^2 + 2dx + 2d^2)(x^2 - 2dx + 2d^2).
$$
Remark. By a similar argument to the case $n = 4$,
if $n = 2^k \geq 4$ and $x^n - a$ is irreducible over
$\mathbf Q$ with $\sqrt[n]{a}$ in a cyclotomic field then $\mathbf Q(\sqrt[n]{a})$ is quadratic over $\mathbf Q(\zeta_{n}) = \mathbf Q(\zeta_{2^k})$ and $\mathbf Q(\sqrt[n/2]{a}) = \mathbf Q(\zeta_n)$. Must $a = -b^{n/2} = -b^{2^{k-1}}$ for some integer $b$?
| {
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Prove that $\sum\limits_{cyc}{} \frac{a^3}{b+c} \geq \sum_\limits{cyc}{} \frac{a^4+b^3c-b^2c^2+bc^3}{(a+b)(a+c)}$ If $ a,b,c>0 $, prove that :$$\sum_{cyc}{} \frac{a^3}{b+c} \geq \sum_{cyc}{} \frac{a^4+b^3c-b^2c^2+bc^3}{(a+b)(a+c)}.$$
my attempt:
After uniting the denominator and dividing by$(a+b)(a+c)(b+c)$
$\Leftrightarrow$$a^5+a^4b+a^4c+a^3bc+b^5+b^4a+b^4c+b^3ac+c^5+c^4b+c^4a+c^3ab\geq a^4b+a^4c+b^4c+b^3c^2-b^3c^2-b^2c^3+b^2c^3+bc^4+b^4a+b^4c+a^2c^3+ac^4-a^3c^2-a^2c^3+a^4c+a^3c^2+c^4a+c^4b+a^4b+a^3b^2-a^3b^2-a^2b^2+ab^4+a^2b^3$$\Leftrightarrow$$a^4(a-c)+b^4(b-c)+c^4(c-a)+a^3(bc-ab)+b^3(ac-ab)+c^3(ab-cb)\geq 0$$\Leftrightarrow$$a^3(a-c)(a-b)+b^3(b-c)(b-a)+c^3(c-a)(c-b)\geq 0$
and this last expression is true(it is schur's inequality).
does my attempt is true?
| Your proof is OK. But it is simpler to manipulate cyclic sum.
It suffices to prove that
$$\sum_{\mathrm{cyc}} [a^3(a + b)(a + c) - (a^4 + b^3c - b^2c^2 + bc^3)(b + c)] \ge 0$$
or
$$\sum_{\mathrm{cyc}} (a^5 + a^3bc - b^4c - bc^4) \ge 0$$
or (because $\sum_{\mathrm{cyc}} b^4c = \sum_{\mathrm{cyc}} a^4 b$
and $\sum_{\mathrm{cyc}} bc^4 = \sum_{\mathrm{cyc}} c a^4$)
$$\sum_{\mathrm{cyc}} (a^5 + a^3bc - a^4b - ca^4) \ge 0$$
or
$$\sum_{\mathrm{cyc}} a^3(a - b)(a - c) \ge 0$$
which is true (Schur).
| {
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$\epsilon-N$ for $\lim\limits_{n \to \infty} \sqrt{n^{2} +3n-3} -n = \frac{3}{2}$ First, I tried to use the triangle inequality only once to find an N:
$$
\left | \sqrt{n^2+3n-3}-n-\frac{3}{2} \right | \leqslant \left | \sqrt{n^2+3n-3}-n \right | + \left | \frac{3}{2} \right | = \epsilon
$$
$$
N=\left \lfloor \frac{(\epsilon -\frac{3}{2})^{2}+3}{6-2\epsilon } \right \rfloor +1
$$
I choose epsilon to be 0.01, and N is 1, which is incorrect.
Then I manipulated the inequality again by using the triangle inequality one more time:
$$
\left | \sqrt{n^2+3n-3}-n-\frac{3}{2} \right | \leqslant \left | \sqrt{n^2+3n-3}-n \right | + \left | \frac{3}{2} \right |
$$
$$
\leqslant \left | \sqrt{n^2+3n-3} \right | +n+ \frac{3}{2} =\epsilon
$$
$$
N=\left \lfloor \frac{(\epsilon -\frac{3}{2})^{2}+3}{2\epsilon } \right \rfloor +1
$$
and this time, when epsilon is 0.01, N is 2624, which is correct
I would like to know why the first approach is wrong and the second one is right, Thank you.
| $$ \lim_{n->\infty}(\sqrt{n^{2}+3n-3} - n) = \lim_{n->\infty}\frac{3n-3}{\sqrt{n^{2}+3n-3} + n} = \lim_{n->\infty}\frac{3-\frac{3}{n}}{\sqrt{1+\frac{3}{n}-\frac{3}{n^2}} + 1} = \frac{3}{2}. $$
Notice that in the last limit as $n$ goes to infinity, the numerator goes to $3$ and the denominator goes to $2$. The second limit is the first one, but the argument of it is multiplied by $$1 = \frac{\sqrt{n^{2}+3n-3} + n}{\sqrt{n^{2}+3n-3} + n}.$$
| {
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What is the number of different sums of the items of a set of consecutive natural numbers? I have been playing around with sets of consecutive natural numbers like say $S=\{1,2,3,...,10\}$ and I have come up with this problem for which however I have not yet found an answer and I don't know if there is one. My problem is the following: Considering a set $S=\{1,2,3,...,n\}$ where $n$ is an arbitrary natural number, can we derive a formula that will indicate the number of different sums that will be produced by summing the items of the set $S$ in all different combinations? For example for the set $S=\{1,2,3,4,5\}$ we get:
$$1+2=3\\
1+2+3=6\\
1+2+3+4=10\\1+3+4+5=13\\1+2+4+5=12\\1+2+3+4+5=15\\
1+2+3+5=11\\1+4+5=10\\1+3+5=9\\1+3+4=8\\1+2+5=8\\1+2+4=7\\1+3=4\\1+4=5\\1+5=6\\
2+3=5\\2+4=6\\2+5=7\\
2+3+4=9\\2+3+5=10\\2+4+5=11\\
2+3+4+5=14\\
3+4=7\\3+5=8\\
3+4+5=12\\
4+5=9\\$$
So we have $26$ different summations and $13$ different sums(I hope I didn't make any mistake)
| Assuming you mean sums of $2$ or more elements, you can get any value between $3$ and $n(n+1)/2.$
If $U\subsetneq \{1,2,3,\ dots,n\},$ we can show either:
*
*$1\not\in U,$ or
*$\exists k<n$ such that $k\in U,$ $k+1\notin U$
In the first case, we can create $U’=U\cup\{1\},$ and the second case we can set $U’=U\setminus \{k\}\cup \{k+1\}.$ In both cases: $$\sum_{j\in U’} j = 1+\sum_{j\in U} j.$$
On the case $n=5,$ we can construct a sequence:
$$\begin{align}3&=1+2\\4&=1+3\\5&=2+3\\6&=1+2+3\\7&=1+2+4\\8&=1+3+4\\9&=2+3+4\\10&=1+2+3+4\\11&=1+2+3+5\\12&=1+2+4+5\\13&=1+3+4+5\\14&=2+3+4+5\\15&=1+2+3+4+5
\end{align}$$
Note this sequence has the values satisfying $n=4$ for the values $3$ to $10,$ the values $3$ to $6$ only use values for $n=3.$
You can see how to extend this for $n=6,$ getting values $16,\dots,21$ via: $$\begin{align} 21-j&=1+2+3+4+5+6-j, j=1,\dots5,\\
21&=1+2+3+4+5+6
\end{align}$$
| {
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Solving the Diophantine system $pqr=a^4$, $p+q+r=b^4$ I am trying to find solutions of the following system of diophantine equations:
$$\left\{\begin{array}{rcl}pqr&=&a^4\\p+q+r&=&b^4\end{array}\right.$$
where $a$, $b$, $p$, $q$ ans $r$ are positive integers such that $\gcd(p,q,r)$ is not divisible by $\theta^4$, $\theta>1$.
I found the following solutions $(p,q,r)$ with a computer program :
$(3\,;6\,;72)$ , $(25\,;60\,;540)$ , $(72\,;576\,;648)$ and $(162\,;448\,;686)$.
The system has infinitely many solutions : take $(p\;q\,;r)=\left(A^4\,;B^4\,;C^4\right)$, where $A^4+B^4+C^4=D^4$ and $A$, $B$, $C$ are coprime (see this article).
But can we prove that there are infinitely many solutions using more elementary ways ?
Thank you for your help !
| Take,
$2p=2n^2+1-w$
$2q=2n^2+1+w$
$2r=16m^2$
Where,
$w^2=4n^4+4n^2-8m^2+1$ ---$(1)$
Eqn (1), is satisfied at, $(m,n,w)=(3,2,3)$
Hence,
$p+q+r=8m^2+2n^2+1$
For, $(m,n,w)=(3,2,3)$
$p+q+r=(3)^4$
$8pqr=(2p)(2q)(2r)$
=$(2n^2+1-w)(2n^2+1+w)(16m^2)$
For, $(m,n,w)=(3,2,3)$, we get:
$pqr=(6)^4$
$(p,q,r)=(3,6,72)$
| {
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A unspotted mistake involving simple Harmonic numbers There is a mistake in the following calculations. However, I can't find it so I'd like to ask for help
\begin{align}
S&=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots+\frac{1}{2n-1}-\frac{1}{2n}\\
&=1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots+\frac{1}{2n-1}-\left(\frac{1}{2}+\frac{1}{4}+\dots+\frac{1}{2n}\right)\\
&=\sum_{k=0}^{n-1}\frac{1}{2k+1}-\frac{1}{2}\sum_{k=0}^{n-1}\frac{1}{k+1}=\sum_{k=0}^{n-1}\int_0^1 x^{2k} dx-\frac{1}{2}\sum_{k=0}^{n-1}\int_0^1 x^k dx\\
&=\int_0^1 \sum_{k=0}^{n-1} x^{2k} dx-\frac{1}{2}\int_0^1 \sum_{k=0}^{n-1} x^k dx=\int_0^1 \frac{1-x^{2n}}{1-x}dx -\frac{1}{2}\int_0^1 \frac{1-x^n}{1-x}dx\\
&=H_{2n}-\frac{H_n}{2}
\end{align}
The result should've been
$$H_{2n}-H_{n}$$
By
\begin{align}
&1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\dots+\frac{1}{2n-1}-\left(\frac{1}{2}+\frac{1}{4}+\dots+\frac{1}{2n}\right)\\
&=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{2n}-\left(1+\frac{1}{2}+\dots+\frac{1}{n}\right)\\
&=H_{2n}-H_{n}
\end{align}
| $1 \over 1-x$ should be $1 \over 1-x^2$ on the fourth line.
$$
\sum_{k=0}^{n-1} x^{2k} = {1-x^{2n}\over 1-x^2}
$$
Disclaimer: I didn't follow through with all the calculations to check that fixing this arrives at the correct answer.
| {
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How can we prove $\sqrt{2+\sqrt{2+....\sqrt{2+\sqrt{2}}}} = 2\cos\left(\frac {\pi }{2^{n+1}}\right)$ without induction I wanted to know the proof without induction / substitution method for the equation
$$\underbrace {\sqrt{2+\sqrt{2+...\sqrt{2+\sqrt{2}}}} }_{\text{n-times}}= 2\cos\left(\frac {\pi }{2^{n+1}}\right)$$
Proof with induction:
\begin{align}
& n=1: \\
& \sqrt{2}=2\cos\left( \dfrac{\pi}{4} \right). \\
\ \\
& \text{Assume that the equation is valid when }n=k. \\
\ \\
&n=k+1; \\
& \underbrace {\sqrt{2+\sqrt{2+...\sqrt{2+\sqrt{2}}}} }_{\text{k+1-times}}=\sqrt{2+\underbrace {\sqrt{2+\sqrt{2+...\sqrt{2+\sqrt{2}}}} }_{\text{k-times}}} \\
& = \sqrt{2+2\cos\left( \dfrac{\pi}{2^{k+1}} \right)} = \sqrt{2}\cdot\sqrt{2\cos^2\left( \dfrac{\pi}{2^{k+2}} \right)} = 2\cos\left(\dfrac{\pi}{2^{k+2}}\right). \blacksquare
\end{align}
| Observe that $2+\sqrt{2} = 2\left(1+\dfrac{1}{\sqrt{2}}\right)=2\left(1+\cos(\frac{\pi}{4})\right)= 2(2\cos^2\left(\frac{\pi}{8}\right))$. Thus when you take the innermost square root you get $2\cos\left(\frac{\pi}{8}\right)$. And repeat this $n$ times you will get to the formula you want to have.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Quadratic formula $x = \frac{- (b +\sqrt{b^2- 4ac})}{ \pm2a}$ In the proof of the quadratic formula
$$x = \frac{- b +\sqrt{b^2- 4ac}}{2a}$$
shouldn't there be $\pm 2a$ instead of $2a$, since both can be the square root of $4a^2$?
| I will make a guess that the proof you were looking at has the following equation in one of its steps:
$$ \left(x+\frac{b}{2a} \right)^2 = \frac{b^2-4ac}{4a^2}. $$
This tells us that $x+\dfrac{b}{2a}$ is one of the square roots
of $\dfrac{b^2-4ac}{4a^2}.$
But it could be either the positive or negative square root.
The square roots of $\dfrac{b^2-4ac}{4a^2}$ are
$\dfrac{\sqrt{b^2 - 4ac}}{2a}$ and $-\dfrac{\sqrt{b^2 - 4ac}}{2a}.$
You can verify this by squaring each one:
\begin{align}
\left(\frac{\sqrt{b^2 - 4ac}}{2a}\right)^2 &= \frac{b^2-4ac}{4a^2}, \\
\left(-\frac{\sqrt{b^2 - 4ac}}{2a}\right)^2 &= \frac{b^2-4ac}{4a^2}. \\
\end{align}
Notice that $-\dfrac{\sqrt{b^2 - 4ac}}{2a}$ is just
$\dfrac{\sqrt{b^2 - 4ac}}{2a}$ with its sign flipped.
That is, one of these is a positive square root and one is a negative square root.
Which one is the positive root and which is the negative root depends on the sign of $a,$ but no matter which sign $a$ has we still have both roots, the positive and the negative.
Therefore we have found that the following statement is true:
$$
\text{$x + \frac{b}{2a}$ is either
$\frac{\sqrt{b^2 - 4ac}}{2a}$ or $-\frac{\sqrt{b^2 - 4ac}}{2a}$}.
$$
Another way to write the same statement is
$$ x + \dfrac{b}{2a} = \pm\dfrac{\sqrt{b^2 - 4ac}}{2a}. $$
Now just subtract $\dfrac{b}{2a}$ from both sides and you have the usual quadratic formula.
At no point in any of this did we ever take a square root of $4a^2.$
Instead, we looked at the expression $\dfrac{b^2-4ac}{4a^2}$ and found its square roots
— both of them — and included both roots in the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Issues with integrating $\int x^2 \sqrt{x^3 +1}$ via integration by parts I want to integrate
$$\int x^2 \sqrt{x^3 +1}~dx$$
I tried it with integration by parts (because we have a product here), but an online calculator did it with integration by substition.
Would this still be correct?
$$\frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int \frac{1}{3} x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} \cdot 3~dx \\
= \frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \int x^3 \frac{1}{2} (x^3+1)^{-\frac{1}{2}} ~dx\\
= \frac{1}{3}x^3 (x^3+1)^\frac{1}{2} - \frac{1}{4} x^4\cdot 2(x^3 +1)^{-\frac{1}{2}} \\
= \frac{1}{3}x^3 \sqrt{x^3+1} - \frac{1}{2} x^4 \frac{1}{\sqrt{x^3+1}}$$
I think this is wrong because when $x=1$ I get a different result than when I insert $x=1$ into
Can someone tell me where I went wrong and why we rather use integration by substition instead of integration by parts here?
| $\int x^2 \sqrt{x^3 +1}~dx = \frac{2}{9}(x^3+1)^{\frac{3}{2}} +c$ by direct integration
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $\sqrt 2 + \sqrt[3]2$ is not rational How do I prove that the following is not rational?
$$x=\sqrt 2 + \sqrt[3]2$$
To prove a simpler case like $\sqrt{2}=a/b$, I can raise both sides to the power of 2 and get $a^2=2b^2$, therefore both $a$ and $b$ must be even numbers which can't be true.
| Let
$$x=\sqrt 2 + \sqrt[3]2$$
then
$$x-\sqrt 2 = \sqrt[3]2$$
$$\left(x-\sqrt 2 \right)^3=2$$
$$x^3-3x^2\left( \sqrt 2 \right)+3x(2)-2\left( \sqrt 2 \right)=2$$
$$\sqrt 2= \frac {x^3+6x-2}{3x^2+2}$$
$x$ cannot be rational because $\frac {x^3+6x-2}{3x^2+2}$ will then be rational and yet $\sqrt 2$ is irrational.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $a^3+5a$ is an integer I've been given the following task. Let
$$a = \sqrt[3]{1+\sqrt{\frac{152}{27}}}-\sqrt[3]{-1+\sqrt{\frac{152}{27}}}$$
Show that $a^3+5a$ is an integer.
I tried calculating it by hand but the small page of my copybook is not large enough for the very long calculations. Is there a trick I could use here instead of calculating it by hand?
I tried factoring $a^3+5a$ into $a(a^2+5)$ to make it more simple to calculate but it still gets a tad bit complicated when multiplying with $a$ again.
Thank you in advance.
| Let $x=\sqrt[3]{1+\sqrt{\frac{152}{27}}}$ and $y=\sqrt[3]{-1+\sqrt{\frac{152}{27}}}$ .
Then
$$\begin{aligned}a^3+5a&=(x-y)^3+5(x-y)\\
&=x^3-3xy(x-y)-y^3+5(x-y)\\
&=x^3-y^3+(5-3xy)(x-y)
\end{aligned}$$
Now,
$\begin{aligned}5-3xy&=5-3\sqrt[3]{\left(1+\sqrt{\frac{152}{27}}\right)\left(-1+\sqrt{\frac{152}{27}}\right)}\\
&=5-3\sqrt[3]{\frac{125}{27}}\\
&=5-3\cdot\frac53\\
&=0
\end{aligned}$
and $x^3-y^3=2$, so $a^3+5a=2$.
| {
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"url": "https://math.stackexchange.com/questions/4505618",
"timestamp": "2023-03-29T00:00:00",
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Interesting integral $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}$ Latest Edit
Inspired by @J.G., I find a formula in general,
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{n}} &=2 \int_{0}^{\pi} \frac{d x}{(3-\cos x)^{n}} \\
&=\left.\frac{2(-1)^{n}}{(n-1) !} \frac{\partial^{n}}{\partial a^{n}}\left(\int_{0}^{\pi} \frac{d x}{a-\cos x} \right)\right|_{a=3} \\
&=\left.\frac{2(-1)^{n} \pi}{(n-1) !} \frac{\partial^{n}}{\partial a^{n}}\left(\frac{1}{\sqrt{a^{2}-1}}\right)\right|_{a=3}
\end{aligned}
$$
Multiplying both the numerator and denominator $\sec^4x$ yields
$\displaystyle I=\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}=\int_{0}^{\frac{\pi}{2}} \frac{\sec ^{4} x}{\left(\sec ^{2} x+\tan ^{2} x\right)^{2}} d x \tag*{} $
Letting $ t=\tan x$ gives
$\displaystyle \begin{aligned}I&= \int_{0}^{\infty} \frac{1+t^{2}}{\left(1+2 t^{2}\right)^{2}} d t\\&=\int_{0}^{\infty} \frac{1+\frac{1}{t^{2}}}{\left(2 t+\frac{1}{t}\right)^{2}} d t\\&=\int_{0}^{\infty} \frac{\frac{3}{4}\left(2+\frac{1}{t^{2}}\right)-\frac{1}{4}\left(2-\frac{1}{t^{2}}\right)}{\left(2 t+\frac{1}{t}\right)^{2}} d t\\&=\frac{3}{4} \int_{0}^{\infty} \frac{d\left(2 t-\frac{1}{t}\right)}{\left(2 t-\frac{1}{t}\right)^{2}+8}-\frac{1}{4} \int_{0}^{\infty} \frac{d\left(2 t+\frac{1}{t}\right)}{\left(2 t+\frac{1}{t}\right)^{2}}\\&=\left[\frac{3}{8 \sqrt{3}} \tan ^{-1}\left(\frac{2 t-\frac{1}{t}}{2 \sqrt{2}}\right)+\frac{1}{4\left(2 t+\frac{1}{t}\right)}\right]_{0}^{\infty}\\&=\frac{3 \pi}{8 \sqrt{2}}\end{aligned}\tag*{} $
Is there any method other than tangent half-angle substitution?
| The obvious alternative is the residue theorem. With $y=2x$ your integral is $$\int_0^\pi\frac{2dy}{(3-\cos y)^2}=\int_0^{2\pi}\frac{dy}{(3-\cos y)^2}\stackrel{z=e^{iy}}{=}\oint_{|z|=1}\frac{-4izdz}{(z^2-6z+1)^2}.$$There is one enclosed pole, the second-order $3-2\sqrt{2}$, so the integral is$$8\pi\lim_{z\to3-2\sqrt{2}}\frac{d}{dz}\frac{z}{(z-3-2\sqrt{2})^2}=8\pi\lim_{z\to3-2\sqrt{2}}\frac{z+3+2\sqrt{2}}{(3+2\sqrt{2}-z)^3}=\frac{3\pi}{8\sqrt{2}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find postive integer $x$ with $x>1$ and $\dfrac{x^6-1}{x-1}$ is perfect square
Find positive integer $x$ with $x>1$ and $\dfrac{x^6-1}{x-1}$ is perfect square.
My try: Let $\dfrac{x^6-1}{x-1}=y^2$, so $(x^4+x^2+1)(x+1)=y^2$.
Let $d= \gcd(x+1,x^4+x^2+1)$.
And
$d \vert x^4+x^2+1=(x^2+x+1)(x^2-x+1)$, because $d\vert x+1$ so $d\vert 2$.
Case 1: $d=1$. If $d=1$ so $(x^4+x^2+1)$,$(x+1)$ is perfect square and $(x^2+\dfrac{1}{2})^2\le x^4+x^2+1\le(x^2+1)^2$ so no satisfactory solution.
Case 2: $d=2$. If $d=2$ so $\dfrac{x^4+x^2+1}{2}$,$\dfrac{x+1}{2}$ is perfect square.
But now I stuck. Please give me a hint. Thank you.
| You've got the right idea, but as Kenta S's comment indicates, it's $d \mid 3$ (since $x \equiv -1 \pmod{x+1}$ means $x^4 + x^2 + 1 \equiv (-1)^4 + (-1)^2 + 1 \equiv 3 \pmod{x + 1}$), not $d \mid 2$, so the other option is $d = 3$. For that case, for some integers $m$ and $n$, we have
$$x^4 + x^2 + 1 = 3m^2, \; \; x + 1 = 3n^2 \; \; \to \; \; x \equiv -1 \pmod{3} \tag{1}\label{eq1A}$$
With your factorization of
$$x^4+x^2+1=(x^2+x+1)(x^2-x+1) \tag{2}\label{eq2A}$$
note that
$$\begin{equation}\begin{aligned}
\gcd(x^2+x+1,x^2-x+1) & = \gcd(x^2+x+1,x^2+x+1-(x^2-x+1)) \\
& = \gcd(x^2+x+1,2x)
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
Since $x^2 + x + 1 = x(x+1) + 1$ is always odd, and $\gcd(x^2+x+1,x) = 1$, this means that $x^2+x+1$ and $x^2-x+1$ are always relatively prime. With \eqref{eq1A} giving that $x \equiv -1 \pmod{3}$, then $x^2 - x + 1 \equiv (-1)^2 - (-1) + 1 \equiv 3 \equiv 0 \pmod{3}$. Thus, from the first part of \eqref{eq1A} and using \eqref{eq2A}, there are integers $s$ and $t$ where $m = st$ and
$$x^2 + x + 1 = s^2, \; \; x^2 - x + 1 = 3t^2 \tag{4}\label{eq4A}$$
However, $x^2 \lt x^2 + x + 1 \lt (x + 1)^2$, so the first part of \eqref{eq4A} is not possible. This means there are no integers $x \gt 1$ where $\frac{x^6 - 1}{x - 1}$ is a perfect square.
| {
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"url": "https://math.stackexchange.com/questions/4510682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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