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Find all the solutions in the complex field of the following equation: Firstly my apologies for anything that should be in LaTex format correctly, I gave it a valiant effort. I have been asked to solve the equation: $(1-z)^6 = (1+z)^6$ A hint given states: do not multiply out!. If the equation was in a more suitable for I would have multiplied by $(1-z)$ and solve it using De Moivre's theorem to find the roots but I am simply unable to get to such a point. I thought about taking the square root of both sides in order to make it simpler.
| Ignoring the given hint, you can solve it by expanding out terms.
Take the square root of both sides (we get two options):
$$(1-z)^6=(1+z)^6\Longleftrightarrow (1-z)^3=\pm(1+z)^3$$
*
*For $(1-z)^3=(1+z)^3$:
$$(1-z)^3=(1+z)^3\Longleftrightarrow$$
$$(1-z)^3=z^3+3z^2+3z+1\Longleftrightarrow$$
$$(1-z)^3-(z^3+3z^2+3z+1)=0\Longleftrightarrow$$
$$-6z-2z^3=0\Longleftrightarrow$$
$$-2z(z^2+3)=0\Longleftrightarrow$$
$$z(z^2+3)=0$$
So, we get two options, $z=0$ or:
$$z^2+3=0\Longleftrightarrow z^2=-3\Longleftrightarrow z=\pm i\sqrt{3}$$
*
*For $(1-z)^3=-(1+z)^3$:
$$(1-z)^3=-(1+z)^3\Longleftrightarrow$$
$$(1-z)^3=-z^3-3z^2-3z-1\Longleftrightarrow$$
$$(1-z)^3+z^3+3z^2+3z+1=0\Longleftrightarrow$$
$$6z^2+2=0\Longleftrightarrow6z^2=-2\Longleftrightarrow z^2=-\frac{1}{3}\Longleftrightarrow z=\pm\frac{i}{\sqrt{3}}$$
So, the solutions we've got are:
*
*$$\color{red}{z=0\space\space\space\vee\space\space\space z=\pm i\sqrt{3}\space\space\space\vee\space\space\space z=\pm\frac{i}{\sqrt{3}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Can you solve $y=\frac{a}{2}x^2\left(y'-\frac{1}{y'}\right)^2+x\left(y'-\frac{1}{y'}\right)+ax^2+c$? I've recently come across this differential equation, but I am having trouble proceeding toward a solution.
$y=\frac{a}{2}x^2\left(y'-\frac{1}{y'}\right)^2+x\left(y'-\frac{1}{y'}\right)+ax^2+c$
where $y'=\frac{dy}{dx}$ and $a$ and $c$ are constants
There appears to be a nice symmetry and I assume some sort of substitution may be in order, but so far, have had no luck.
Originally, it took the form
$y=\frac{a}{2}x^2\left(\frac{(y')^2-1}{y'}\right)^2+x\left(\frac{(y')^2-1}{y'}\right)+ax^2+c$
and I thought that perhaps a trig substitution might help.
Any thoughts?
| I don't know if an implicit solution fits your needs, anyway with this solution you could at least find $x$ as a function of $y$.
Put $y'=\tan z$, $y=-\log k|\cos z|$, $z= \pm \arccos {e^{-y} \over k}$. So: $$
-\log k|\cos z|{\sin^2 z \over \cos^2 z}={a \over 2}{x^2 \over \cos^2 z}+x {\tan z \over \cos^2 z}+ax^2{\sin^2 z \over \cos^2 z}+c {\sin^2 z \over \cos^2 z}
$$
Simplify the denominators and notice that $\sin z=\sqrt{1-{e^{-2y} \over k^2}}$ and $\tan z={\sqrt{1-{e^{-2y} \over k^2}} \over {e^{-y} \over k}}$.
We can write: $$ ax^2 \left( {3 \over 2}-{e^{-2y} \over k^2} \right)+x{\sqrt{1-{e^{-2y} \over k^2}} \over {e^{-y} \over k}}+(c+y)\left(1-{e^{-2y} \over k^2} \right)=0.$$
You can now solve the quadratic to find values for $x$ in function of every value of $y$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the integral of a quotient of rational functions? How do I compute the following integral: $$\int \dfrac{x^4+1}{x^3+x^2}\,dx$$
My attempt:
We can write $$\dfrac{x^4+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$$
It is easy to find that
$A=1$,
$B=2$, and
$C=-1$.
Therefore
$$\frac{x^4+1}{x^3+x^2} = \frac{1}{x^2} + \frac{2}{x} - \frac{1}{x+1}$$
Therefore:
$$\int \frac{x^4+1}{x^3+x^2}\,dx = \int \frac{dx}{x^2} + \int \dfrac{2\,dx}{x} - \int \frac{dx}{x+1} = -\frac{1}{x} +2\log \vert x\vert - \log \vert x+1 \vert + C$$
The problem is I was supposed to find: $$\int \frac{x^4+1}{x^3+x^2}\,dx = \frac{x^2}{2} - x - \frac{1}{2} - \log \vert x \vert + 2 \log \vert x+1 \vert + C$$
Where is my mistake?
| To break a rational polymomial expression into parts, the degree of the numerator must be less than the degree of the denominator.
This is not the case with $\dfrac{x^4+1}{x^3+x^2}$. Using long division, we find
\begin{array}{rcccccccc}
& & x & - & 1 &\\
& & --- & --- & --- & --- & --- & --- & ---\\
x^3 + x^2 & | & x^4 & + & 0x^3 & + & 0x^2 & + & 1 \\
& & --- & --- & --- \\
& & x^4 & + & x^3\\
& & --- & --- & --- & --- & --- \\
& & & & -x^3 & + & 0x^2 \\
& & & & -x^3 & - & x^2 \\
& & & & --- & --- & --- \\
& & & & & & x^2 & + & 1. \\
\end{array}
So $\dfrac{x^4+1}{x^3+x^2} = x - 1 + \dfrac{x^2 + 1}{x^3 + x^2}$.
And you need to solve
$\dfrac{x^2+1}{x^3+x^2} = \dfrac{A}{x^2} + \dfrac{B}{x} + \dfrac{C}{x+1}$
$x^2 + 1 = A(x+1) + Bx(x+1) + Cx^2$
Let $x = -1$ and you get
$C = 2$
Let $C = 2$ and you get
$x^2 + 1 = A(x+1) + Bx(x+1) + 2x^2$
$-x^2 + 1 = A(x+1) + Bx(x+1)$
$1-x = A + Bx$
$A = 1$ and $B = -1$.
So $\dfrac{x^4+1}{x^3+x^2} = x - 1 + \dfrac{1}{x^2} - \dfrac{1}{x} + \dfrac{2}{x+1}$.
etc
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Irreducible representation of $S_3$ How can I show that this representation of $S_3$ is irreducible?
$$\rho\left(e\right)=\left(\begin{array}{cc}
1 & 0\\
0 & 1
\end{array}\right),\,\,\rho\left(a_{1}\right)=\frac{1}{2}\left(\begin{array}{cc}
-1 & -\sqrt{3}\\
\sqrt{3} & -1
\end{array}\right),\,\,\rho\left(a_{2}\right)=\frac{1}{2}\left(\begin{array}{cc}
-1 & \sqrt{3}\\
-\sqrt{3} & -1
\end{array}\right),$$
$$\rho\left(a_{3}\right)=\left(\begin{array}{cc}
-1 & 0\\
0 & 1
\end{array}\right),\,\,\rho\left(a_{4}\right)=\frac{1}{2}\left(\begin{array}{cc}
1 & \sqrt{3}\\
\sqrt{3} & -1
\end{array}\right),\,\,\rho\left(a_{5}\right)=\frac{1}{2}\left(\begin{array}{cc}
1 & -\sqrt{3}\\
-\sqrt{3} & -1
\end{array}\right).$$
| Well, if it were reducible, $\rho(a_3)$ and $\rho(a_4)$ would share an eigenvector. Which is clearly not the case, because the standard basis is the only (up to scalar multiplication and permutation) basis of eigenvectors of $\rho(a_3)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integral with a parameter Hello I try solve a problem on integration
$\int_0^\frac{\pi}{2}\frac{\arctan(y\tan(x))}{\tan(x)}dx$
My stuck is that I use differentiation under integral sign and producing $ I'(y) = \int_0^ \frac{\pi}{2}\frac{dx}{1+(y\tan(x))^2}$ what can I do?
Thanks
The solution is $\frac{\pi}{2}\text{sgn}(y \ln(1+|y|))$
| First, note that the integral given by
$$I(y)=\int_0^{\pi/2}\frac{\arctan(y\tan(x))}{\tan(x)}\,dx$$
is an odd function with $I(y)=-I(-y)$. Therefore, we examine the case for $y\ge 0$.
Next, the integral given by
$$I'(y)=\int_0^{\pi/2} \frac{1}{1+y^2\tan^2(x)}\,dx \tag 1$$
can be evaluated a host of ways.
In doing so, we find that
$$I'(y)=\frac{\pi}{2(y+1)} \tag 2$$
Integrating $(1)$ and using $I(0)=0$ reveals
$$I(y)=\frac{\pi}{2}\log(1+y)$$
If $y<0$, then we use the fact that $I(y)$ is odd to write
$$I(y)=-\frac{\pi}{2}\log(1-y)$$
Putting both cases together yields
$$I(y)=\text{sgn}(y)\frac{\pi}{2}\log(1+|y|)$$
as expected!
EVALUATION OF THE INTEGRAL IN $(1)$
We assume that $y>0$ for the proceeding analysis.
Note that we can write
$$\frac{1}{1+y^2\tan^2(x)}=\frac{1}{1-y^2}-\frac{2y^2}{1-y^2}\frac{1}{(1+y^2)+(1-y^2)\cos(2x)}$$
Then, we have
$$\begin{align}
\int_0^{\pi/2}\frac{1}{1+y^2\tan^2(x)}\,dx&=\frac{\pi}{2(1-y^2)}-\frac{2y^2}{1-y^2}\int_0^{\pi/2}\frac{1}{(1+y^2)+(1-y^2)\cos(2x)}\,dx\\\\
&=\frac{\pi}{2(1-y^2)}-\frac{y^2}{1-y^2}\int_0^{\pi}\frac{1}{(1+y^2)+(1-y^2)\cos(x)}\,dx\\\\
&=\frac{\pi}{2(1-y^2)}-\frac{y^2}{2(1-y^2)}\int_{-\pi}^{\pi}\frac{1}{(1+y^2)+(1-y^2)\cos(x)}\,dx\tag 3
\end{align}$$
The integral in $(3)$ can be evaluated using the classical Tangent Half-Angle Substitution or contour integration. Here, we take the latter approach to write
$$\begin{align}
\int_{-\pi}^{\pi}\frac{1}{(1+y^2)+(1-y^2)\cos(x)}\,dx&=\frac2i\oint_{|z|=1}\frac{1}{(1-y^2)z^2+2(1+y^2)z+(1-y^2)}\,dz\\\\
&=\frac{2}{i(1-y^2)}\oint_{|z|=1}\frac{1}{\left(z-\frac{y+1}{y-1}\right)\left(z-\frac{y-1}{y+1}\right)}\,dz\\\\
&=(2\pi i )\left(\frac{2}{i(1-y^2)}\right)\text{Res}\left(\frac{1}{\left(z-\frac{y+1}{y-1}\right)\left(z-\frac{y-1}{y+1}\right)}, z=\frac{y-1}{y+1}\right)\\\\
&=\frac{\pi}{y}
\end{align}$$
Putting it all together we find that
$$I'(y)=\frac{\pi}{2(1+y)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785767",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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How to calculate $\lim_{x\to 0}\left(\frac{1}{x^2} - \frac{1}{\sin^2 x}\right)^{-1}$? $$f (x) = \frac{1}{x^2} - \frac{1}{\sin^2 x}$$
Find limit of $\dfrac1{f(x)}$ as $x\to0$.
| You don't really need to work with $1/f(x)$; consider
$$
f(x)=
\frac{\sin^2x-x^2}{x^4}\frac{x^2}{\sin^2x}=
\frac{\sin x-x}{x^3}\frac{\sin x+x}{x}\frac{x^2}{\sin^2x}=
\frac{\sin x-x}{x^3}\left(\frac{\sin x}{x}+1\right)
\left(\frac{x}{\sin x}\right)^2
$$
Since the limits of the second and third factors are well known, we just need to consider
$$
\lim_{x\to0}\frac{\sin x-x}{x^3}=
\lim_{x\to0}\frac{\cos x-1}{3x^2}=
\lim_{x\to0}\frac{-\sin x}{6x}=-\frac{1}{6}
$$
Therefore
$$
\lim_{x\to 0}f(x)=-\frac{1}{6}\cdot 2\cdot 1^2=-\frac{1}{3}
$$
and so
$$
\lim_{x\to0}\frac{1}{f(x)}=\frac{1}{-1/3}=-3
$$
The limit above can be computed also with a Taylor expansion:
$$
\lim_{x\to0}\frac{\sin x-x}{x^3}=
\lim_{x\to0}\frac{x-x^3/6+o(x^3)-x}{x^3}=-\frac{1}{6}
$$
Some knowledge of “basic” limits such as this one allows to use l'Hôpital more efficiently.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1786681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Which of the numbers is larger: $7^{94}$ or $9^{91} $? In this problem, I guess b is larger, but not know how to prove it without going to lengthy calculations. It is highly appreciated if anyone can give me a help.
Which number is larger
$$\begin{align}
&\textrm{(a)}\quad 7^{94}
&\quad\textrm{(b)}\quad 9^{91}
\end{align}$$
| I voted for André's answer, but here's another approach, using a different bit of maths.
Note that $7^{94} = 7^3 \times 7^{91}$. $9^{91} = (7 \times \frac{9}{7})^{91}$, where $\alpha = \frac{9}{7} = 1 + \frac{2}{7} > 1$. So
$$
\frac{7^{94}}{9^{91}} = \frac{7^3}{\alpha^{91}}.
$$
What do we make of $\frac{7^3}{\alpha^{91}}$? Well, $7^3 = 49 \times 7 = 343$. Using the binomial theorem, and observing that positive ratios always diminish when the numerators (resp. denominators) are decreased (resp. increased),
\begin{align*}
\alpha^{91} &= \left(1 + \frac{2}{7}\right)^{91} \\
&> 1 + \frac{91}{1!} \times \frac{2}{7} + \frac{91 \times 90}{2!} \times \frac{2^2}{7^2} \\
&\quad= 1 + \frac{182}{7} + \frac{8190 \times 4}{98} \\
&\quad> 1 + 25 + \frac{4 \times 80 \times 100}{100} \\
&\qquad= 1 + 25 + 320 \\
&\qquad= 346 \\
&\qquad> 343.
\end{align*}
So $\alpha^{91} > 7^3$ and thus $9^{91} > 7^{94}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1788290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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"answer_id": 6
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Someone please explain cosine similarity equation to me? I'm trying to understand the cosine similarity in a simple and graphical way, very much similar to this question here but I do not understand how the person got to their answer.
| The vectors are $(x_1 \ x_2 \ x_3 \ x_4 \ x_5 \ x_6 \ x_7 \ x_8)^T=(2 \ 0 \ 1 \ 1 \ 0 \ \ 2 \ 1 \ 1)^T$ and $(y_1 \ y_2 \ y_3 \ y_4 \ y_5 \ y_6 \ y_7 \ y_8)^T=( \ 2 \ 1 \ 1 \ 0 \ 1 \ 1 \ 1 \ 1)^T$
The cosine of angle ($\alpha$) between the two vectors is $cos(\alpha)=\large{\frac{\sum_{i=1}^8 x_i\cdot y_i}{\sqrt{\sum_{i=1}^8 x_i^2}\cdot \sqrt{\sum_{i=1}^8 y_i^2}}}=\frac{2\cdot 2+0 \cdot 1+ 1\cdot 1+ 1\cdot 0 + 0 \cdot 1+ 2\cdot 1+ 1\cdot 1+ 1\cdot 1}{\sqrt{2^2+0^2+1^2+1^2+0^2+2^2+1^2+1^2}\cdot \sqrt{2^2+1^2+1^2+0^2+1^2+1^2+1^2+1^2}}$
$=\frac{4+1+2+1+1}{\sqrt{4+1+1+4+1+1}\cdot \sqrt{4+1+1+1+1+1+1}}=\frac{9}{\sqrt{12}\cdot \sqrt{10}}=0.8216$
Thus the equation is $\cos(\alpha)=0.8216$
$\alpha=arcos(0.8216)=34.76°$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sin^2(10 ^\circ)-\sin^2(20^\circ)-\sin^2(40^\circ)=-\frac{1}{2}$ identity 10 degrees
$$\sin^2(10^\circ)-\sin^2(20^\circ)-\sin^2(40^\circ)=-\frac{1}{2}$$
$$\cos^2(10^\circ)-\cos^2(20^\circ)-\cos^2(40^\circ)=-\frac{1}{2}$$
Why are they both have same answer?
The only time they have same answer is at 45 degrees right?
$\sin(45^\circ)=\cos(45^\circ)=\frac{1}{\sqrt2}$
Can somebody provide me an explanation please?
Also how to prove these two identities
I know all others $15^\circ, 30^\circ, 45^\circ, 60^\circ$, etc, but can't seem to prove these.
| Let $C=\cos^2A-\cos^2B-\cos^2C$
and $S=\sin^2A-\sin^2B-\sin^2C$
$\implies C+S=-1$
and $C-S=\cos2A-\cos2B-\cos2C=\cos2A-2\cos(B+C)\cos(B-C)$ using Prosthaphaeresis Formula
If $C-S=0, C=S=-\dfrac12\ \ \ \ (0)$
Now if $B+C=60^\circ$ or more generally, $360^\circ n\pm60^\circ, \ \ \ \ (1)$
$C-S=\cos2A-\cos(B-C)$ will be $0$ if $B-C=360^\circ m\pm2A\ \ \ \ (2)$
$\implies B=180^\circ n+180^\circ m\pm30^\circ\pm A\ \ \ \ (3)$
and $C=180^\circ n-180^\circ m\pm30^\circ\mp A\ \ \ \ (4)$
Here $B=20^\circ,C=40^\circ\implies B+C=?$
and $A=10^\circ\implies B-C=360^\circ\cdot0+2A$
So, $(0)$ will hold true for $A,B,C$ satisfying $(1),(2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1791258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluation of $\lim\limits_{x\rightarrow 0}\frac1x\left((1+2x+3x^2)^{1/x}-(1+2x-3x^2)^{1/x}\right) $
Evaluation of $$\lim_{x\rightarrow 0}\frac{(1+2x+3x^2)^{\frac{1}{x}}-(1+2x-3x^2)^{\frac{1}{x}}}{x} $$
$\bf{My\; Try::}$ Let $$l=\lim_{x\rightarrow 0}\frac{e^{\frac{\ln(1+2x+3x^2)}{x}}-e^{\frac{\ln(1+2x-3x^2)}{x}}}{x}$$
Using $$\bullet \; \frac{\ln(1+x)}{x}=x-\frac{x^2}{2}+\frac{x^3}{3}-.....\infty$$
But I am not Getting answer.
Now How can I solve after that, Help me
Thanks
| As usual this limit can also be evaluated without the use of L'Hospital's Rule and Taylor's series just by applying standard limits combined with the use of algebra of limits. We have
\begin{align}
L &= \lim_{x \to 0}\frac{(1 + 2x + 3x^{2})^{1/x} - (1 + 2x - 3x^{2})^{1/x}}{x}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + 2x + 3x^{2})}{x}\right) - \exp\left(\dfrac{\log(1 + 2x - 3x^{2})}{x}\right)}{x}\notag\\
&= \lim_{x \to 0}\exp\left(\dfrac{\log(1 + 2x - 3x^{2})}{x}\right)\cdot\dfrac{\exp\left(\dfrac{\log(1 + 2x + 3x^{2}) - \log(1 + 2x - 3x^{2})}{x}\right) - 1}{x}\notag\\
&= \lim_{x \to 0}\exp\left(\dfrac{\log(1 + 2x - 3x^{2})}{2x - 3x^{2}}\cdot(2 - 3x)\right)\cdot\dfrac{\exp\left(\dfrac{1}{x}\log\left(\dfrac{1 + 2x + 3x^{2}}{1 + 2x - 3x^{2}}\right)\right) - 1}{x}\notag\\
&= \exp(1\cdot 2)\lim_{x \to 0}\dfrac{\exp\left(\dfrac{1}{x}\log\left(\dfrac{1 + 2x + 3x^{2}}{1 + 2x - 3x^{2}}\right)\right) - 1}{x}\notag\\
&= \exp(2)\lim_{z \to 0}\frac{\exp(z) - 1}{z}\cdot\lim_{x \to 0}\dfrac{z}{x}\notag\\
&= \exp(2)\lim_{x \to 0}\dfrac{\log\left(1 + \dfrac{6x^{2}}{1 + 2x - 3x^{2}}\right)}{x^{2}}\notag\\
&= \exp(2)\lim_{x \to 0}\dfrac{\log\left(1 + \dfrac{6x^{2}}{1 + 2x - 3x^{2}}\right)}{\dfrac{6x^{2}}{1 + 2x - 3x^{2}}}\cdot\frac{6}{1 + 2x - 3x^{2}}\notag\\
&= e^{2}\cdot 1\cdot 6 = 6e^{2}\notag
\end{align}
We have used the standard limits $$\lim_{x \to 0}\frac{\log(1 + x)}{x} = 1 = \lim_{x \to 0}\frac{e^{x} - 1}{x}$$ and the fact that $$z = \frac{1}{x}\log\left(\frac{1 + 2x + 3x^{2}}{1 + 2x - 3x^{2}}\right) = \frac{1}{x}\log\left(1 + \frac{6x^{2}}{1 + 2x - 3x^{2}}\right)\to 0$$ as $x \to 0$ (which easily follows from the standard limits mentioned above).
| {
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"url": "https://math.stackexchange.com/questions/1792724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show $\frac{2}{\sqrt[3]2}-\frac{1}{2(\sqrt[3]2-1)}+\left(\frac{9}{2\sqrt[3]4}-\frac{9}{4}\right)^{\frac{1}{3}}=\frac{1}{2}$? Prove that:
$$\frac{2}{\sqrt[3]2}-\frac{1}{2(\sqrt[3]2-1)}+\left(\frac{9}{2\sqrt[3]4}-\frac{9}{4}\right)^{\frac{1}{3}}=\frac{1}{2}$$
The LHS is irrational number and RHS is rational number. May be this is almost a $\frac{1}{2}$.
According to the calculator it is a $\frac{1}{2}$
I tried to cube both sides but it looked too messy.
| Let me try.
Note that $$(1+\sqrt[3]{2})^3 = 3(1+\sqrt[3]{2}+\sqrt[3]{4}).$$
Now we have:
$$LHS = \sqrt[3]{4} - \frac{1}{2}(\sqrt[3]{4} + \sqrt[3]{2} + 1) + \left(\frac{9}{4(\sqrt[3]{4}+\sqrt[3]{2}+1) }\right)^{\frac{1}{3}} = \sqrt[3]{4} - \frac{1}{2}(\sqrt[3]{4} + \sqrt[3]{2} + 1) + \left(\frac{27}{4(\sqrt[3]{2}+1)^3 }\right)^{\frac{1}{3}} = \sqrt[3]{4} - \frac{1}{2}(\sqrt[3]{4} + \sqrt[3]{2} + 1) + \frac{3}{\sqrt[3]{4}(\sqrt[3]{2}+1)} = \frac{1}{2}(\sqrt[3]{4} - \sqrt[3]{2} - 1) + \frac{\sqrt[3]{2}(\sqrt[3]{4}-\sqrt[3]{2}+1)}{2} = \frac{1}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796254",
"timestamp": "2023-03-29T00:00:00",
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Prove the fractions aren't integers
Prove that if $p$ and $q$ are distinct primes then $\dfrac{pq-1}{(p-1)(q-1)}$ is never an integer. Is it similarly true that if $p,q,r$ are distinct primes then $\dfrac{pqr-1}{(p-1)(q-1)(r-1)}$ is also never an integer?
I think using a modular arithmetic argument here would help. In other words, we must have for the first fraction $pq-1 \equiv 0 \pmod{(p-1)(q-1)}$. Then I am unsure how to proceed next since we can't really use the Chinese Remainder Theorem.
| I will assume that
$p \le q$,
not just
$p < q$.
Also,
you don't need the assumption
about primality.
I will show that
the only solutions are
$p=q=2$
and
$p=q=3$.
If
$p=2$
then
$\dfrac{pq-1}{(p-1)(q-1)}
=\dfrac{2q-1}{q-1}
=\dfrac{2q-2+1}{q-1}
=2+\dfrac{1}{q-1}
$
which is not an integer
unless $q = 2$.
If
$p=3$
then
$\begin{array}\\
\dfrac{pq-1}{(p-1)(q-1)}
&=\dfrac{3q-1}{2(q-1)}\\
&=\dfrac{3q-3+2}{2q-2}\\
&=\dfrac32+\dfrac{1}{q-1}\\
\end{array}
$
which is not an integer
if $q \ge 4$.
If $q=3$,
then
$\dfrac{pq-1}{(p-1)(q-1)}
=\dfrac{3\cdot 3-1}{2\cdot 2}
=2
$
which is an integer.
If $p > 3$
then
$\begin{array}\\
\dfrac{pq-1}{(p-1)(q-1)}
&=\dfrac{pq-1}{pq-p-q+1}\\
&=\dfrac{pq-p-q+1+p+q-2}{pq-p-q+1}\\
&=1+\dfrac{p+q-2}{pq-p-q+1}\\
\end{array}
$
so we are done if
$p+q-2
<pq-p-q+1
$
or
$0
<pq-2p-2q+3
=(p-2)(q-2)-1
$
and this is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1800384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 5,
"answer_id": 0
} |
Prove $A=B=\pi$ $\gamma=0.57721566...$
$\phi=\frac{1+\sqrt5}{2}$
Let,
$$A=\sum_{n=1}^{\infty}\arctan\left(\frac{(e+e^{-1})(\phi^{\frac{2n-1}{2}}+\phi^{\frac{1-2n}{2}})}{e^2+e^{-2}-1+\phi^{2n-1}+\phi^{1-2n}}\right)$$
$$B=\sum_{n=1}^{\infty}\arctan\left(\frac{(\gamma+\gamma^{-1})(\phi^{\frac{2n-1}{2}}+\phi^{\frac{1-2n}{2}})}{\gamma^2+\gamma^{-2}-1+\phi^{2n-1}+\phi^{1-2n}}\right)$$
(1)
$$A=B=\pi$$
We found (1) accidentally while trying to search for $\pi$ in term of $\arctan(x)$ most idea are from Dr Ron Knott his Fibonacci site.
Can anyone prove (1)
| For any $a \ge 1$, let $f_n(a)$ be the expression
$$f_n(a) \stackrel{def}{=}\frac{(a+a^{-1})(\phi^{n-1/2} + \phi^{1/2-n})}{a^2 + a^{-2} - 1 + \phi^{2n-1} + \phi^{1-2n}}$$
The sums at hand equal to $\sum\limits_{n=1}^\infty \tan^{-1}f_n(e)$ and $\sum\limits_{n=1}^\infty \tan^{-1}f_n(\gamma^{-1})$ respectively.
For any $n \ge 0$, let $p_n = \phi^{n-1/2} - \phi^{1/2-n}$. It is easy to verify
*
*$p_{n+1} - p_{n-1} = (\phi - \phi^{-1})(\phi^{n-1/2} + \phi^{1/2-n}) = \phi^{n-1/2} + \phi^{1/2-n}$
*$p_{n+1} p_{n-1} = (\phi^{n+1/2} - \phi^{-1/2-n})(\phi^{n-3/2} - \phi^{3/2-n})
= \phi^{2n-1} + \phi^{1-2n} - 3$
Let $u = a + a^{-1}$. Notice
$$\frac{\frac{u}{p_{n-1}} - \frac{u}{p_{n+1}}}{
1 + \frac{u}{p_{n-1}} \frac{u}{p_{n+1}}
}
= \frac{u(p_{n+1}-p_{n-1})}{u^2 + p_{n+1}p_{n-1}}
= \frac{(a+a^{-1})(\phi_{n-1/2} + \phi_{1/2-n})}{a^2 + a^{-2} - 1 + \phi^{2n-1} + \phi^{1-2n}} = f_n(a)
$$
There exists integers $N_n$ such that
$$\tan^{-1} f_n(a) = \tan^{-1}
\left(\frac{\frac{u}{p_{n-1}} - \frac{u}{p_{n+1}}}{
1 + \frac{u}{p_{n-1}} \frac{u}{p_{n+1}}
}\right) = \tan^{-1}\frac{u}{p_{n-1}} - \tan^{-1}\frac{u}{p_{n+1}}
+ N_n \pi
$$
When $n > 1$, $p_{n+1} > p_{n-1} > 0$, it is easy to see $N_n = 0$.
When $n = 1$, we have
$$
\begin{cases}
f_1(a), \frac{u}{p_2} > 0\\
\frac{u}{p_0} < 0
\end{cases}
\quad\implies\quad
\begin{cases}
\tan^{-1}f_1(a), \tan^{-1}\frac{u}{p_2} \in (0,\frac{\pi}{2})\\
\tan^{-1}\frac{u}{p_0} \in (-\frac{\pi}{2},0)
\end{cases}
\quad\implies\quad
N_1 = 1
$$
This implies
$$\begin{align}
& \sum_{n=1}^\infty \tan^{-1}f_n(a)\\
= & \sum_{m=1}^\infty \left( \tan^{-1}f_{2m-1}(a) + \tan^{-1}f_{2m}(a)\right)\\
= & \pi + \lim_{N\to\infty}\sum_{m=1}^N\left[\left( \tan^{-1}\frac{u}{p_{2m-2}} - \tan^{-1}\frac{u}{p_{2m}}\right) + \left( \tan^{-1}\frac{u}{p_{2m-1}} - \tan^{-1}\frac{u}{p_{2m+1}}\right)\right]\\
= & \pi + \lim_{N\to\infty}
\left[
\left(\tan^{-1}\frac{u}{p_0} + \tan^{-1}\frac{u}{p_1}\right)
-
\left(\tan^{-1}\frac{u}{p_{2N}} + \tan^{-1}\frac{u}{p_{2N+1}}\right)
\right]\\
= & \pi + \left(\tan^{-1}\frac{u}{p_0} + \tan^{-1}\frac{u}{p_1}\right)
\end{align}
$$
Notice $p_0 = -p_1$, the last bracket vanishes. As a result,
$$\sum_{n=1}^\infty \tan^{-1}f_n(a) = \pi\quad\text{ for all } a \ge 1$$
As pointed out by @Winther, we can relax the condition from $a \ge 1$ to $a > 0$.
This is because for $a \in (0,1)$, $f_n(a) = f_n(a^{-1})$.
We can generalize this a little bit. Let $g_n(u)$ be the expression
$$g_n(u) \stackrel{def}{=} \frac{u(\phi^{n-1/2} + \phi^{1/2-n})}{
u^2 + (\phi^{n-1/2} + \phi^{1/2-n})^2 - 5}$$
We have $f_n(a) = g_n(a + a^{-1})$. If we adopt the convention that $\tan^{-1}(\mathbb{R}) = \left(-\frac{\pi}{2}, \frac{\pi}{2} \right]$,
i.e $\tan^{-1}\infty = \frac{\pi}{2}$ and $\tan^{-1}x < 0$ for $x < 0$, we have
$$\sum_{n=1}^\infty \tan^{-1}g_n(u) =
\begin{cases}
\pi, & u_c \le u\\
0, & 0 < u < u_c
\end{cases}
\quad\text{ where }\quad u_c = \sqrt{3-\sqrt{5}}
$$
For all $u > 0$, the analysis is essentially the same as above. When $u < u_c$, there is one place we need to adjust the argument. Namely, $g_1(u)$ becomes $-ve$
and this forces the corresponding $N_1$ to vanish. At the end, when $u < u_c$, this causes the $\pi$ term disappear from above sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
} |
Solution of $4 \cos x(\cos 2x+\cos 3x)+1=0$ Find the solution of the equation:
$$4 \cos x(\cos 2x+\cos 3x)+1=0$$
Applying trigonometric identity leads to
$$\cos (x) \cos \bigg(\frac{x}{2} \bigg) \cos \bigg(\frac{5x}{2} \bigg)=-\frac{1}{8}$$
But I can't understand what to do from here. Could some suggest how to proceed from here?
| Hint:
$1)$ Note that:
*
*$\cos{2x}=2\cos^2{x}-1$, and
*$\cos{3x}=4\cos^3{x}-3\cos{x}$
$2)$ Using this, setup a cubic equation in $\cos{x}$.
$3)$ Now, solve the quartic equation by making the substitution $t=\cos{x}$.
$4)$ Finally, back-substitute and use $\cos{x}=\cos{\alpha} \Rightarrow x=2n\pi \ \pm \alpha$, where $n \in \mathbb{Z}$
Edit(Showed working of hint)
Using the identities, we get
$16\cos^4x+8\cos^3x-12\cos^2x-4\cos x+1=0$
Set $t=\cos x$.
Our equation is
$16t^4+8t^3-12t^2-4t+1=0 \Rightarrow (2t+1)(8t^3-6t+1)=0$
So, $2t+1=0$ or $8t^3-6t+1=0$
Back-substituting we get,
*
*$2\cos{x}+1=0 \Rightarrow \cos{x}=-1/2=\cos(2\pi/3)$
$\therefore x=2n\pi \pm 2\pi/3$, where $n \in \mathbb{Z}$
*$8\cos^3{x}-6\cos{x}+1=0$
Using the identity $\cos{3x}=4\cos^3{x}-3\cos{x}$, we get $2\cos{3x}+1=0 \Rightarrow \cos{3x}=-1/2=\cos(2\pi/3)$
$ \therefore 3x=2n\pi \pm 2\pi/3 \Rightarrow x=2\pi/3 \pm 2\pi/9$, where $n\in \mathbb{Z}$
Thus, $x=2\pi/3 \pm 2\pi/3$ or $x=2\pi/3 \pm 2\pi/9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1803682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$
Question: If $x, y, z$ are the side lengths of a triangle, prove that $x^2 + y^2 + z^2 < 2(xy + yz + xz)$
My solution: Consider
$$x^2 + y^2 + z^2 < 2(xy + yz + xz)$$
Notice that $x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+xz)$
Hence
$$(x+y+z)^2-2(xy+yz+xz) < 2(xy + yz + xz)$$
$$ (x+y+z)^2 - 4(xy+yz+xz) < 0 $$
As $x,y,z$ are all greater than zero as a side length of a triangle can not be negative. And because $(x+y+z)^2 >0$ for all real $x,y,z$ therefore the whole expression is less than zero. $Q.E.D$
Am I correct? Or could I be more "rigorous" I am a highschool student and getting into proof so any tips would be appreciated as well :)
| $x,y$ and $z$ are sides of a triangle. So, $y+z\gt x,\; x+y\gt z,\; z+x\gt y$.
Now you have to prove that $$x^2+y^2+z^2\lt2\left(\sum xy\right).$$
Just a simple look through the right hand term.
$$2\left(\sum xy\right)=2xy+2yz+2zx\\=x\color{blue}{(y+z)}+y\color{blue}{(z+x)}+z\color{blue}{(x+y)}\gt x\times \color{blue}{x}+y\times \color{blue}{y}+z\times \color{blue}{z}=x^2+y^2+z^2$$
So, now, you derived $$x^2+y^2+z^2\lt2\left(\sum xy\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804220",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Parametric equations for intersection between plane and circle So I was looking at this question Determine Circle of Intersection of Plane and Sphere
but I need to know how to find a parametric equation for intersections such as these. My particular question is to find a parametric equation for the intersection between the plane
$$x+y+z=1$$ and unit sphere centered at the origin.
I started out my question by substituting
$$
z=-x-y+1$$ into
$$x^2+y^2+z^2=1 $$ deriving
$$x^2+y^2+(-x-y+1)^2=1$$
and getting $$2x^2+2y^2+2xy-2x-2y=0$$ but I am unsure how to proceed from here.
I also tried to use the vector equation of the plane
$$r(u, v)=(0,0,1)+(2,1,-3)u+(1,1,-2)v$$ but I am not sure how that would help.
| The plane $ \zeta = \{(x,y,z) : x + y + z = 1\},$ is perpendicular to the vector
$\langle 1,1,1 \rangle$.
Two points on $\zeta$ are
$\mathbf O = \left( \dfrac 13,\ \dfrac 13,\ \dfrac 13 \right)
\text{ and }
\mathbf A = (0,0,1)$.
Since $\overrightarrow{\mathbf{OA}} = \dfrac 13 \langle -1, -1, 2 \rangle$,
the unit vector
$\overrightarrow u = \dfrac {1}{\sqrt 6} \langle -1, -1, 2 \rangle$
is parallel to the plane $\zeta$.
Since $\overrightarrow{\mathbf{OA}} \times
\langle 1,1,1 \rangle = \langle -1, 1, 0 \rangle$, then the unit vector
$\overrightarrow v = \dfrac{1}{\sqrt 2} \langle -1, 1, 0 \rangle$ is parallel to the plane $\zeta$ and
$\overrightarrow u \perp \overrightarrow v$.
So we can parameterize the plane, $\zeta$ as
$$\zeta(s,t)
= \mathbf O + s \overrightarrow u + t \overrightarrow v
= \left(
\dfrac 13 - \dfrac{s}{\sqrt 6} - \dfrac{t}{\sqrt 2},
\dfrac 13 - \dfrac{s}{\sqrt 6} + \dfrac{t}{\sqrt 2},
\dfrac 13 + \dfrac{2s}{\sqrt 6}
\right)$$
As ugly as the following looks
$$\left( \dfrac 13 - \dfrac{s}{\sqrt 6} - \dfrac{t}{\sqrt 2}\right)^2 +
\left( \dfrac 13 - \dfrac{s}{\sqrt 6} + \dfrac{t}{\sqrt 2}\right)^2 +
\left( \dfrac 13 + \dfrac{2s}{\sqrt 6}\right)^2 = 1 $$
It simplifies to $s^2 + t^2 = \dfrac 23$
So we let
$s = \sqrt{\dfrac 23}\cos \theta$ and
$t = \sqrt{\dfrac 23}\sin \theta$ and simplify. We get
$$\zeta(s,t)
= \left(
\dfrac 13 - \dfrac {\cos \theta}{3} - \dfrac{\sin \theta}{\sqrt 3},\;
\dfrac 13 - \dfrac {\cos \theta}{3} + \dfrac{\sin \theta}{\sqrt 3},\;
\dfrac 13 + \dfrac {2\cos \theta}{3}
\right)$$
a more intuitive answer
Let $S$ be the unit sphere $x^2 + y^2 + z^2 = 1$.
Let $P$ be the plane $x + y + z = 1$.
Let $C$ be the circle $C = S \cap P$.
The unit vector $U = \dfrac{1}{\sqrt 3} \langle 1,1,1 \rangle$ is perpendicular to $P$.
The line $x = y = z$
*
*passes through, $(0,0,0)$, the center of $S$
*passes through, $X$, the center of $C$
*is perpendicular to $P$
The distance from $P$ to the center of $S$ is
$ \dfrac{\left| 0 + 0 + 0 - 1 \right|}{\sqrt{1^2 + 1^2 + 1^2}}
= \dfrac{1}{\sqrt 3}$.
The center of $C$ is at
$X = (0,0,0) + \dfrac{1}{\sqrt 3}U
= \left( \dfrac 13,\dfrac 13, \dfrac 13 \right)$.
The radius of $C$ is $r = \sqrt{1 - \dfrac 13} = \sqrt{\dfrac 23}$.
We need to find two points, $A$ and $B$, on $C$ such that
$\overrightarrow{XA} \perp \overrightarrow{XB}$.
*
*$A = (1,0,0)$ is a point on $C$.
*$\left \| \overrightarrow{XA} \right \| =
\left \| \left( \dfrac 23, -\dfrac 13, -\dfrac 13 \right) \right \| =
\sqrt{\dfrac 23}$
*$U \times \sqrt{\dfrac 32} \; \overrightarrow{XA} =
\left( 0, \dfrac{1}{\sqrt 2}, -\dfrac{1}{\sqrt 2} \right)$
is a unit vector, in P, that is perpendicular to $\overrightarrow{XA}$.
*$\overrightarrow{XB} =
\sqrt{\dfrac 23} \left( 0, \dfrac{1}{\sqrt 2}, -\dfrac{1}{\sqrt 2} \right) =
\left( 0, \dfrac{1}{\sqrt 3}, -\dfrac{1}{\sqrt 3} \right)$
\begin{align}
C &=
X
+ \cos \theta \; \overrightarrow{XA}
+ \sin \theta \; \overrightarrow{XB}
\\
C &= \left( \dfrac 13,\dfrac 13, \dfrac 13 \right)
+ \cos \theta \; \left( \dfrac 23, -\dfrac 13, -\dfrac 13 \right)
+ \sin \theta \; \left( 0, \dfrac{\sqrt 3}{3}, -\dfrac{\sqrt 3}{3} \right)
\\
C &= \dfrac 13(1 + 2 \cos \theta, \;
1 - \cos \theta + \sqrt 3 \sin \theta, \;
1 - \cos \theta - \sqrt 3 \sin \theta)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find the value of $a^2-b^2+c^2$ Let $a$, $b$, and $c$ be real numbers such that $a − 7b + 8c = 4$ and $8a + 4b − c = 7$. What is the
value of $a^2-b^2+c^2$ ?
| You can also do it solving the two equations for $b$ and $c$ to be expressed as functions of $a$. This would give $$b=\frac{12}{5}-\frac{13 a}{5}\qquad c=\frac{13}{5}-\frac{12 a}{5}$$ Now, replace in the expression, expand and simplify.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Elasticity of the function Please find the elasticity of the function f(x) in x0:
$$f(x) = \frac{ax^2} {1+ x^2} $$ $$ x0=10$$
Could you please check is done correct ?
$$E_xf(x_0) = \frac{f'(x_0)·x_0} {f(x_0)} $$
$$f'(x)=\frac{2ax·(1+x^2)-2ax^2·2x} {(1+x^2)^2} = \frac{2ax} {(1+x^2)^2} $$
$$E_xf(x) = \frac{\frac{2ax} {(1+x^2)^2}} {\frac{ax^2} {1+ x^2}} = \frac{2} {1+x^2} $$
$$E_xf(10) = \frac{2} {1+10^2} = \frac{2} {101} $$
| As you have mentioned before, the function for elasticity of a function with respect to $x$ is
$$E_x(f(x)) = \frac{f'(x) \cdot x}{f(x)}.$$
We see by the quotient rule that
$$f'(x) = \frac{(2ax)(1+x^2) - (ax^2)(2x)}{(1+x^2)^2}.$$
Thus,
$$E_x(f(x)) = \frac{\frac{(2ax)(1+x^2) - (ax^2)(2x)}{(1+x^2)^2} \cdot x}{\frac{ax^2}{1+x^2}} = \frac{2ax^2(1+x^2) - (ax^3)(2x)}{(ax^2)(1+x^2)}$$
$$=\frac{2(1+x^2) - (x)(2x)}{1+x^2} = \frac{2}{1+x^2}$$
$$\implies E_x(f(10)) = \frac{2}{1+(10)^2} = \frac{2}{101}.$$
Hence, you are correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How should I try to evaluate the integral $\int_a^b \sqrt{1 + \frac{x^2}{r^2 - x^2}} \; dx$ I've tried to evaluate $\displaystyle\int_{-r}^r \sqrt{1 + \frac{x^2}{r^2 - x^2}} \; dx$ on my own, but I have encountered a problem I cannot get around.
The indefinite integral $\sqrt{\frac{r^2}{r^2-x^2}} \sqrt{r^2-x^2} \tan ^{-1}\left(\frac{x}{\sqrt{r^2-x^2}}\right) + C$ is undefined at both limits of integration
(the fractions $\frac{r^2}{r^2-r^2} = \frac{r^2}0 = \text{indetermined}$ and $\frac{({-r})^2}{r^2-r^2} = \frac{r^2}0 = \text{indetermined}$)
so I really don't know what to do. Apparently, the value of this integral should be $\pi r.$ Any help would be appreciated.
| I think you can just find a common denominator and get to $\int \frac{r}{\sqrt{r^2-x^2}}dx=\int \frac{1}{\sqrt{1-(\frac{x}{r})^2}}$. Then you are a u-sub away from finding it as $r\arcsin(\frac{x}{r}) + C $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Joint distribution weird result
We have 3x3 board Each board cell has $0.5$ chance to be white
(there is no dependency between different cell colors)
Let X = number of white rows (a row with only white cells on it)
Let
N = total number of white cells in the whole board
Find $P\{X=0,N=3\}$
Find $P\{X=0 |N=5\}$
What I tried to do is to calculate the $P\{X=0,N=3\}$ , I said that:
$P\{X=0,N=3\}=P(N=3)*P(X=0|N=3)$ is it true?
Because now I got that it equals to $\frac{21}{128}*\frac{1}{168}=\frac{1}{1024}$
Which just seems way too low and also I'm not sure if I would change my answer if the question was to find $P\{X=0|N=3\}$ instead of $P\{X=0,N=3\}$ which seems odd to me
| First note that
$$P(X=0,N=3)=P(X=0\,|\,N=3) P(N=3)$$
We will find $P(X=0 \,|\,N=3)$ then $P(N=3)$ separately, then multiply them together.
Part 1: $P(X=0\,|\,N=3)$
With three white squares, there can be at most one white row. This means that if $N=3$, then $X=0$ or $X=1$ are the only possibilities. Therefore
$$P(X=0\,|\,N=3)=1-P(X=1 \,|\,N=3)$$
The total number of arrangements of the $9$ squares in the grid is $$\binom{9}{3}=\frac{9\cdot 8 \cdot 7}{3 \cdot 2} = 84$$
There are $3$ arrangements such that there is a white row. That means
$$P(X=0\,|\,N=3)=1-\frac{3}{84} = 1-\frac{1}{28} = \frac{27}{28}$$
Part 2: $P(N=3)$
There are $\binom{9}{3}$ arrangements in which there are three white squares. Each of these arrangements has a $(1/2)^3(1/2)^6$ probability of occuring. Therefore
$$P(N=3)=\binom{9}{3}\left(\frac{1}{2}\right)^9=\frac{21\cdot 2^2}{2^9}=\frac{21}{128}$$
Our final answer is therefore
$$P(X=0,N=3)=\left(\frac{27}{28}\right)\left(\frac{21}{128}\right)=\boxed{\frac{81}{512}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluation of $\sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Prove that $\displaystyle \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \frac{\pi}{2}\frac{1}{4^{n}}\binom{2n}{n}$ and also find value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
$\bf{My\; Try::}$ Let $$\displaystyle I_{n} = \int_{0}^{\frac{\pi}{2}}\sin^{2n}xdx = \int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x\cdot \sin^2 xdx = \int_{0}^{\frac{\pi}{2}}\sin^{2n-2}x\cdot (1-\cos^2 x)dx$$
$$I_{n} =I_{n-1}-\int_{0}^{\frac{\pi}{2}}\cos x\cdot \sin^{2n-2}\cdot \cos xdx$$
Now Using Integration by parts, We get $$I_{n} = I_{n-1}-\frac{I_{n}}{2n-1}\Rightarrow I_{n} = \frac{2n-1}{2n}I_{n-1}$$
Now Using Recursively, We get $$I_{n} = \frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}I_{n-2} =\frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}I_{n-3}$$
So we get $$I_{n} = \frac{2n-1}{2n}\cdot \frac{2n-3}{2n-2}\cdot \frac{2n-5}{2n-4}\cdot \frac{2n-7}{2n-6}\cdot \cdot \cdot \cdot \cdot \cdot \cdot\cdot \frac{3}{2}I_{0}$$
and we get $\displaystyle I_{0} = \frac{\pi}{2}$
So we get $$I_{n} = \frac{(2n)!}{4^n\cdot n!\cdot n!}\cdot \frac{\pi}{2}$$
Now I did not understand How can I calculate value of $\displaystyle \sum^{\infty}_{n=0}\frac{1}{16^n}\binom{2n}{n}.$
Help Required, Thanks.
| Maybe it is interesting to see that $$S=\sum_{n\geq0}\dbinom{2n}{n}x^{n}=\sum_{n\geq0}\frac{\left(2n\right)!}{n!^{2}}x^{n}$$ and now note that $$\frac{\left(2n\right)!}{n!}=2^{n}\left(2n-1\right)!!=4^{n}\left(\frac{1}{2}\right)_{n}=\left(-1\right)^{n}4^{n}\left(-\frac{1}{2}\right)_{n}
$$ where $(x)_{n}$ is the Pochhamer' symbol. So we have, by the generalized binomial theorem, that $$S=\sum_{n\geq0}\dbinom{-1/2}{n}\left(-1\right)^{n}4^{n}x^{n}=\frac{1}{\sqrt{1-4x}},\left|x\right|<\frac{1}{4}$$ so if we take $x=\frac{1}{16}$ we have $$\sum_{n\geq0}\dbinom{2n}{n}\frac{1}{16^{n}}=\frac{2}{\sqrt{3}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Extending the ordered sequence of 'three-number means' beyond AM, GM and HM I want to create an ordered sequence of various 'three-number means' with as many different elements in it as possible.
So far I've got $12$ ($8$ unusual ones are highlighted):
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \color{blue}{ \frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}}} \geq $$
$$\geq \color{blue}{\frac{\sqrt{(x+y)^2+(y+z)^2+(z+x)^2}}{2 \sqrt{3}} } \geq \frac{x+y+z}{3} \geq $$
$$ \geq \color{blue}{ \frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2} } \geq \color{blue}{\sqrt{\frac{xy+yz+zx}{3}}} \geq $$
$$\geq \color{blue}{\frac{\sqrt{xy}+\sqrt{yz}+\sqrt{zx}}{3}} \geq \color{blue}{\sqrt{\frac{x\sqrt{yz}+y\sqrt{zx}+z\sqrt{xy}}{3}}} \geq $$
$$ \geq \sqrt[3]{xyz} \geq \color{blue}{ \frac{3\sqrt{xyz}}{\sqrt{x}+\sqrt{y}+\sqrt{z}}} \geq $$
$$ \geq\color{blue}{ 2 \sqrt[3]{\frac{x^2y^2z^2}{(x+y)(y+z)(z+x)}}} \geq \frac{3xyz}{xy+yz+zx}$$
All the inequalities here are proven (note that $x,y,z>0$).
The rules are as follows - only explicit expressions allowed (no AGM for example) and these expressions should contain only addition, multiplication, division, squares, cubes, square and cube roots.
For any new mean we need to find a place between a pair of existing means, so we can keep our sequence precisely ordered.
I partially placed several other means, for example:
$$\frac{\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2}}{3 \sqrt{2}} \geq \\ \geq \color{blue}{ \frac{\sqrt[6]{(x^2+y^2)(y^2+z^2)(z^2+x^2)}}{\sqrt{2}}} \geq \frac{\sqrt[3]{(x+y)(y+z)(z+x)}}{2}$$
$$\sqrt{\frac{xy+yz+zx}{3}} \geq \color{blue}{\frac{xy+yz+zx}{x+y+z}} \geq \sqrt[3]{xyz}$$
But I can't place them exactly among other means so far.
| Let $\mathbf{x}$ denote a vector of positive real numbers (you can take it for your purposes to be 3 such numbers). Let $\mathbf{w}$ denote a vector of positive real numbers of the same length (i.e. 3-dimensional in this case), with $\sum_i w_i=1$; the simplest option is all $w_i$ being equal. Define $M_p:=\left(\sum_i w_i x_i^p\right)^{1/p}$ for $p\neq 0$. We can define $M_0$ by continuity; it's $\prod_i x_i^{w_i}$. Then $M_p$ is a strictly increasing function of $p$. If all $w_i$ are equal, $M_{-1}$ is the HM, $M_0$ is the GM, $M_1$ is the AM etc. Thus $M_p>M_q$ for $p>q$ is called the power means inequality. If you want a greater variety of inequalities, you can use the fact that $M_p\left(\mathbf{x}\right)>M_p\left(\mathbf{y}\right)$ provided each $x_i\geq y_i$ and at least one $x_j>y_j$. In particular, you can replace a vector entry with a power-mean (including geometric) of greater entries to created a complicated "nested" expression. Variations in $\mathbf{w}$ can give you some inequalities too, but only if you assume (for example) that $x_1>x_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1811726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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$\int\frac{\sin x}{\sqrt{1-\sin x}}dx=?$ Calculate this integral $\displaystyle\int\dfrac{\sin x}{\sqrt{1-\sin x}}dx=?$
Effort;
$1-\sin x=t^2\Rightarrow \sin x=1-t^2\Rightarrow \cos x=\sqrt{2t^2-t^4}$
$1-\sin x=t^2\Rightarrow-\cos x dx=2tdt\Rightarrow dx=\frac{2t}{\sqrt{t^4-2t^2}}dt$
$\displaystyle\int\frac{1-t^2}{t}\cdot\frac{2t}{\sqrt{t^4-2t^2}}dt=2\int\frac{1-t^2}{\sqrt{t^4-2t^2}}dt$
$\ = 2\displaystyle\int\frac{1}{\sqrt{t^4-2t^2}}dt-2\displaystyle\int\frac{t}{\sqrt{t^2-2t}}dt$
$\ = 2\displaystyle\int t^{-1}(t^2-2)^{-\frac{1}{2}}dt-2\displaystyle\int t(t^2-2t)^{-\frac{1}{2}}dt$
But after that I don't know how to continue.
| I might be missing something, but I think this is possible to do with a prolonged substitution:
$$
\int \frac{\sin(x)}{\sqrt{1-\sin(x)}} \ dx.
$$
Let $u = 1-\sin(x)$, then $$ du = -\cos(x) dx \implies -\sec(x) du = dx.$$
Then we get $$ \int (1-u) \frac{1}{\sqrt{u}} (-\sec(x) du).$$
We need to find $\sec(x)$ in terms of $u$:
$$
u = 1-\sin(x) \\
\sin(x) = 1-u \\
\sin^2(x) = (1-u)^2 \\
1-\cos^2(x) = 1-2u+u^2 \\
\cos^2(x) = 2u-u^2 \\
\cos(x) = \sqrt{2u-u^2 }\\
\sec(x) = \frac{1}{\sqrt{2u-u^2}}
$$
So the integral becomes
$$
\int \frac{u-1}{\sqrt{u}} \cdot \frac{1}{\sqrt{2u-u^2}} \ du \\
=\int \frac{u-1}{\sqrt{u^2}} \cdot \frac{1}{\sqrt{2-u}} \ du \\
=\int \frac{u-1}{u} \cdot \frac{1}{\sqrt{2-u}} \ du \\
=\int \frac{1}{\sqrt{2-u}} \ du - \int \frac{1}{u\sqrt{2-u}} \ du
$$
The first term is easy to evaluate: $\int \frac{1}{\sqrt{2-u}} \ du = -2\sqrt{2-u}$.
The second term is a bit more involved but with a substitution $v = \sqrt{2-u}$ you'll find that you can use inverse hyperbolic trig to get $\int \frac{1}{u\sqrt{2-u}} \ du = -\sqrt{2} \tanh^{-1}( \frac{\sqrt{2-u}}{\sqrt{2}})$. So altogether, we've got
$$
\int \frac{1}{\sqrt{2-u}} \ du - \int \frac{1}{u\sqrt{2-u}} \ du \\
= -2\sqrt{2-u} - \left( \sqrt{2} \tanh^{-1} \left( \frac{\sqrt{2-u}}{\sqrt{2}}\right) \right) + C \\
= \sqrt{2} \tanh^{-1} \left( \frac{\sqrt{2-u}}{\sqrt{2}} \right) - 2\sqrt{2-u} +C.
$$
And now to replace $u$ with $1-\sin(x)$,
$$\sqrt{2} \tanh^{-1} \left( \frac{\sqrt{2-u}}{\sqrt{2}} \right) - 2\sqrt{2-u} +C \\ =\sqrt{2} \tanh^{-1} \left( \frac{\sqrt{2-(1-\sin(x))}}{\sqrt{2}} \right) - 2\sqrt{2-(1-\sin(x))} +C \\
= \sqrt{2} \tanh^{-1} \left( \frac{\sqrt{\sin(x)+1}}{\sqrt{2}} \right) - 2\sqrt{\sin(x)+1} +C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How is $\lim_{x \to a}\left(\frac{x^n - a^n}{x - a}\right) = n\times a^{n-1}$? In my book this is termed as a theorem and the proof given is as follows :-
$$\begin{align}
\lim_{x \to a}\left(\frac{x^n - a^n}{x - a}\right)
&=\lim_{x \to a}\left(\frac{(x - a)*(x^{x-1} + x^{n-2}*a + x^{n-3}*a^2 + x^{n-4}*a^3 + \cdots + x^1*a^{n-2} + a^{n-1})}{x - a}\right) \\ &=(a^{n-1} + a*a^{n-2} + \cdots + a^{n-1}) \\ &=(a^{n-1} + a^{n-1} + \cdots + a^{n-1}) \\ &=(n*a^{n-1}).
\end{align}$$
Everything made sense to me except
$$x^n - a^n = (x - a)*(x^{x-1} + x^{n-2}*a + x^{n-3}*a^2 + x^{n-4}*a^3 + \cdots + x^1*a^{n-2} + a^{n-1})$$
Somebody please enlighten me on this topic.
| No, this isn't a binomial expansion. This is just long division. To check that it makes sense, we can just expand it out:
\begin{align*}
&(x - a)(x^{n - 1} + ax^{n - 2} + a^2x^{n - 3} + \cdots + a^{n - 2}x + a^{n - 1}) \\
&= x(x^{n - 1} + ax^{n - 2} + a^2x^{n - 3} + \cdots + a^{n - 1}) \\
&~~~~~~~~~~~~~~~~~- a(x^{n - 1} + ax^{n - 2} + \cdots + a^{n - 2}x + a^{n - 1}) \\
&= (x^n \color{red}{+ ax^{n - 1} + a^2x^{n - 2} + \cdots + a^{n - 1}x}) \\
&~~~~~~~+ (\color{red}{-ax^{n - 1} - a^2x^{n - 2} - \cdots - a^{n - 1}x} - a^n) \\
&= x^n - a^n
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving by induction that $\sum\limits_{i=1}^n\frac{1}{n+i}=\sum\limits_{i=1}^n\left(\frac1{2i-1}-\frac1{2i}\right)$ I have a homework problem to prove the following via induction:
$$\sum_{i=1}^n \frac{1}{n+i} = \sum_{i=1}^n \left(\frac{1}{2i-1} - \frac{1}{2i}\right) $$
The base case is true.
So far I've done the following:
$$\sum_{i=1}^{k+1} \frac{1}{(k+1)+i}$$
$$a_{k+1} = \frac{1}{(k+1)+(k+1)} = \frac{1}{2k +2} $$
$$s_k + a_{k+1} = \frac{1}{2k +2} + \sum_{i=1}^k (\frac{1}{2k-1} - \frac{1}{2k}) $$
$$s_{k+1} = \sum_{i=1}^{k+1} (\frac{1}{2(k+1)-1} - \frac{1}{2(k+1)}) $$
$$ \frac{1}{2k +2} + \sum_{i=1}^k (\frac{1}{2k-1} - \frac{1}{2k}) = \sum_{i=1}^{k+1} (\frac{1}{2(k+1)-1} - \frac{1}{2(k+1)}) $$
$$ \frac{1}{2k +2} + \sum_{i=1}^k (\frac{1}{2k-1} - \frac{1}{2k}) = \sum_{i=1}^{k+1} (\frac{1}{2k-1} - \frac{1}{2k+2)}) $$
I'm having a hard time figuring out how to proceed from here. Anyone have any hints or tips for search terms? Or corrections to what I've started above?
| As Thomas Andrews notes, the key part lies in reindexing the sum (so you can apply the inductive hypothesis, IH). More explicitly, the following is the core part of the inductive argument:
\begin{align}
\sum_{i=1}^{k+1}\frac{1}{k+1+i}&=\sum_{i=2}^{k+2}\frac{1}{k+i}\tag{reindex}\\[1em]
&= \sum_{i=1}^k\frac{1}{k+i}+\frac{1}{2k+2}+\frac{1}{2k+1}-\frac{1}{k+1}\tag{rewrite}\\[1em]
&= \sum_{i=1}^k\left(\frac{1}{2i-1}-\frac{1}{2i}\right)+\frac{1}{2k+2}+\frac{1}{2k+1}-\frac{1}{k+1}\tag{IH}\\[1em]
&= \sum_{i=1}^k\left(\frac{1}{2i-1}-\frac{1}{2i}\right)+\left(\frac{1}{2k+1}-\frac{1}{2k+2}\right)\tag{simplify}\\[1em]
&= \sum_{i=1}^{k+1}\left(\frac{1}{2i-1}-\frac{1}{2i}\right).\tag{rewrite}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that: $97|2^{48}-1$ Show that: $97|2^{48}-1$
My work:
$$\begin{align}
2^{96}&\equiv{1}\pmod{97}\\
\implies (2^{48}-1)(2^{48}+1)&=97k\\
\implies (2^{24}-1)(2^{24}+1)(2^{48}+1) &=97k\\
\implies (2^{12}-1)(2^{12}+1)(2^{24}+1)(2^{48}+1)&=97k\\
\implies (2^6-1)(2^6+1)(2^{12}+1)(2^{24}+1)(2^{48}+1) &=97k
\end{align}$$
None of the terms on LHS seem to be divisible by 97!!
Direct calculation shows that: $97\mid 2^{24}+1$ , but how to find it mathematically (of course not using calculator)?
| $$97\equiv 1\pmod8$$
Thus, $2$ is a quadratic residue $\bmod 97$. So, there exists $a$ such that $a^2\equiv 2 \pmod {97}$. Thus, $2^{48}\equiv a^{96}\equiv 1\pmod{97}$, as desired.
This solution will work for any prime number $p$ that is $\pm 1\pmod{8}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n$ Find the rightmost digit of: $1^n+2^n+3^n+4^n+5^n+6^n+7^n+8^n+9^n(n$ arbitrary positive integer)
First of all I checked a few cases for small $n$'s and in all cases the rightmost digit was $5$, so maybe this is the case for all values of $n$.
Then I thought maybe it's better to consider odd and even cases for $n$ but there's no unified rule here, because for example: $8^2\equiv{4}\pmod{10}$ and $8^4\equiv{6}\pmod{10}$. Any ideas??
| Given that the order of any of the components here will divide $4$, since that is the Carmichael function value for $10$, it is only necessary to check the values for $n=\{1,2,3, 4\}$. The results follow a pattern across the range of values (all $\bmod 10$):
$$\begin{array}{c|c}
n & 1^n & 2^n & 3^n & 4^n & 5^n & 6^n & 7^n & 8^n & 9^n & \sum \\ \hline
1 & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 5 \\ \hline
2 & 1 & 4 & 9 & 6 & 5 & 6 & 9 & 4 & 1 & 5 \\ \hline
3 & 1 & 8 & 7 & 4 & 5 & 6 & 3 & 2 & 9 & 5 \\ \hline
4 & 1 & 6 & 1 & 6 & 5 & 6 & 1 & 6 & 1 & 3
\end{array}$$
and you can check that the following line for $n=5$ repeats the values for $n=1$ if required.
Thus the answer is that the last digit is $3$ when $n\equiv 0 \bmod 4$ and is $5$ otherwise.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $-2, -3, -5, -7$ are quadratic non-residues modulo p Prove that if $p$ is prime and $p\equiv 173 \pmod{1680}$, then $-2, -3, -5, -7$ are quadratic non-residues modulo p.
| Since $p\equiv1\pmod4,$ it follows that $\left(\dfrac{-1}{p}\right)=1.$ Then $p\equiv5\pmod8,$ so $\left(\dfrac{2}{p}\right)=-1.$
Also, $\left(\dfrac{-3}{p}\right)=(-1)^{(p-1)/2}(-1)^{(p-1)/2}\left(\dfrac{p}{3}\right)\equiv p\pmod3.$ But $p\equiv-1\pmod3,$ and hence $\left(\dfrac{-3}{p}\right)=-1.$
Now $\left(\dfrac{5}{p}\right)=\left(\dfrac{p}{5}\right)$ while $p\equiv3\pmod5.$ Thus $\left(\dfrac{-5}{p}\right)=\left(\dfrac{3}{5}\right)=\left(\dfrac{2}{3}\right)=-1.$
Finally, $\left(\dfrac{7}{p}\right)=(-1)^{(p-1)/2}\left(\dfrac{p}{7}\right)=\left(\dfrac{p}{7}\right).$ But $p\equiv5\pmod7.$ Therefore $\left(\dfrac{p}{7}\right)=\left(\dfrac{5}{7}\right)=\left(\dfrac{2}{5}\right)=-1.$
This uses a lot of quadratic reciprocity, so tell me if some steps are ambiguous.
Hope this helps.
| {
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"timestamp": "2023-03-29T00:00:00",
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Contour integral for finding $\int_0^\infty \frac{\ln x}{(x+a)^2+b^2} \, dx$ I can't prove the following result:
$$\displaystyle\int_0^\infty \frac{\ln x}{(x+a)^2+b^2} \, dx=\frac{\ln \sqrt{a^2+b^2}}{b}\arctan\frac{b}{a} \text{ for all } a,b \in \mathbb{R}.$$
Well, I consider $\displaystyle\int_C \frac{\operatorname{Ln} z}{(z+a)^2+b^2} \, dz$ where $C$ is the usual contour for this kind of integral with logarithm (actually, I don't know if I'm right). So, assuming there is a simple pole inside the interior $C$, I apply the residue theorem as:
$$\int_{C}\frac{\operatorname{Ln} z}{(z+a)^2+b^2} \, dz\\=
2\pi i\left(\lim_{z\to-a+ib}\dfrac{(z+a-ib) \operatorname{Ln} z}{(z+a-ib)(z+a+ib)}\right)\\=
\frac{\pi}{b} \operatorname{Ln} (-a+ib)=\frac{\pi}{b}(\ln (\sqrt{a^2+b^2})+i\arg (-a+ib))$$
Now, I must consider estimations of integrals on each semicircle, however I have not idea how I can reach it.
Thanks for any hint.
| An alternative, real-analytic solution by symmetry only.
$$ \int_{0}^{+\infty}\frac{\log x}{x^2+2ax+(a^2+b^2)}\stackrel{x\mapsto u\sqrt{a^2+b^2}}{=}\frac{1}{\sqrt{a^2+b^2}}\int_{0}^{+\infty}\frac{\log u+\log\sqrt{a^2+b^2}}{u^2+\frac{2a}{\sqrt{a^2+b^2}}u+1}\,du $$
and since the polynomial $p(u)=u^2+\frac{2a}{\sqrt{a^2+b^2}}u+1$ is quadratic and palindromic, $\int_{0}^{+\infty}\frac{\log u}{p(u)}\,du=0$ by the substitution $u\mapsto \frac{1}{u}$. This leaves us with the elementary integral
$$ \int_{0}^{+\infty}\frac{du}{u^2+2Du+1}=\int_{D}^{+\infty}\frac{du}{u^2+(1-D^2)}=\frac{1}{\sqrt{1-D^2}}\,\left(\frac{\pi}{2}-\arctan\frac{D}{\sqrt{1-D^2}}\right) $$
for any $D\in(-1,1)$. By letting $D=\frac{a}{\sqrt{a^2+b^2}}$ we get
$$\int_{0}^{+\infty}\frac{\log x}{(x+a)^2+b^2}=\frac{\log\sqrt{a^2+b^2}}{|b|}\left[\frac{\pi}{2}-\arctan\left(\frac{a}{|b|}\right)\right] $$
for any $b\neq 0$. If $b=0$ the LHS is convergent only for $a>0$, and in such a case it equals $\frac{\log a}{a}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Given that $\tan 2x+\tan x=0$, show that $\tan x=0$ Given that $\tan 2x+\tan x=0$, show that $\tan x=0$
Using the Trigonometric Addition Formulae,
\begin{align}
\tan 2x & = \frac{2\tan x}{1-\tan ^2 x} \\
\Rightarrow \frac{2\tan x}{1-\tan ^2 x}+\tan x & = 0 \\
\ 2\tan x+\tan x(1-\tan ^2 x) & = 0 \\
2+1-\tan ^2 x & = 0 \\
\tan ^2 x & = 3
\end{align}
This is as far as I can get, and when I look at the Mark Scheme no other Trignometric Identities have been used. Thanks
| Notice that the following conditions are equivalent:
\begin{align*}
\tan x + \tan y &= 0\\
\tan x &= -\tan y\\
\tan x &= \tan(-y)\\
x &= -y + k\pi\\
x+y&= k\pi
\end{align*}
If we use $y=2x$ we get
\begin{align*}
3x &= k\pi\\
x &= k\frac\pi3
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the equations of the tangent planes to the sphere $x^2+y^2+z^2+2x-4y+6z-7=0,$ which intersect in the line $6x-3y-23=0=3z+2.$ Find the equations of the tangent planes to the sphere $x^2+y^2+z^2+2x-4y+6z-7=0,$ which intersect in the line $6x-3y-23=0=3z+2.$
Let the tangent planes be $A_1x+B_1y+C_1z+D_1=0$ and $A_2x+B_2y+C_2z+D_2=0$
As the line $6x-3y-23=0=3z+2$ lies on both the planes,so put $z=\frac{-2}{3},x=\frac{3y+23}{6}$ in the both equations of the planes,
$A_1\frac{3y+23}{6}+B_1y-\frac{2C_1}{3}+D_1=0$ and $A_2\frac{3y+23}{6}+B_2y-\frac{2C_2}{3}+D_2=0$
I am stuck here.Please help.
The answer given in the book is $2x-y+4z-5=0$ and $4x-2y-z=16.$
| Two easy points on the required line are $(0,-\frac{23}{3},-\frac{2}{3}),(\frac{23}{6},0,-\frac{2}{3})$. For these to lie in the plane $Ax+By+Cz+D=0$ we require $23A-4C+6D=0,-46B-4C+6D=0$. Subtracting gives $A=-2B$, and we also have $3D=23B+2C$.
The sphere has equation $(x+1)^2+(y-2)^2+(z+3)^2=21$, so it has centre $(-1,2,-3)$ and radius $\sqrt{21}$. The centre must be a distance $\sqrt{21}$ from the plane, so we have $|-A+2B-3C+D|=\sqrt{21(A^2+B^2+C^2)}$. Substituting the earlier relations for $A,D$ we get $(4B-3C+\frac{23}{3}B+\frac{2}{3}C)^2=21(5B^2+C^2)$, or $7^2(\frac{5}{3}B-\frac{1}{3}C)^2=21(5B^2+C^2)$, and hence $(4B+C)(B-2C)=0$. So $C=\frac{1}{2}B$ or $-4B$.
Thus one plane is $B=2,C=1,A=-4,D=16$ or $4x-2y-z=16$ (note that we can take $B$ to have any non-zero value, it does not affect the equation of the plane). The other plane is $B=1,C=-4,A=-2,D=5$ or $2x-y+4z=5$.
| {
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"url": "https://math.stackexchange.com/questions/1817414",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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proof of $\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\leq\frac{1}{\sqrt{n}}$ for large $n$ I'm trying to conclude this question in which I just need to prove that:
$$0<\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}\leq\frac{1}{\sqrt{n}}$$
and then I'll have that, for large $n$:
$$\begin{align}
\left\lvert\cos\left(\sqrt{x^2+\frac{1}{n}}\right)-\cos(x)\right\rvert &= \left\lvert2\sin\frac{\sqrt{x^2+\frac{1}{n}}-x}{2}\sin\frac{\frac{\frac{1}{n}}{\sqrt{x^2+\frac{1}{n}}+x}}{2}\right\rvert\\
&\leq
2\sin\frac{\sqrt{x^2+\frac{1}{n}}-x}{2}\sin\frac{1}{2\sqrt{n}}\xrightarrow[n\to\infty]{} 0
\end{align}$$
because $\sin\frac{1}{2\sqrt{n}}\to 0$ and $\sin\frac{\sqrt{x^2+\frac{1}{n}}-x}{2}$ is bounded.
I liked the answers given in the question but I really need to solve it this way, I'm trying to investigate the inequation. I know that, the worst case is for $n$ so large that $\frac{1}{n}$ in the LHS of the inequality goes to $0$, and the best case is when $n=1$ and therefore we have:
$$\sqrt{x^2}+x<\sqrt{x^2+\frac{1}{n}}+x<\sqrt{x^2+1}+x\le\sqrt{2}+1$$
*for $0\le x\le 1$
but it doesn't help, I think.
| Cross-multiply, square, subtract ${1\over n}$ from both sides to get
$$
0\le 2x^2+2x\sqrt{x^2+{1\over n}}
$$
Assuming $x\in[0,1]$ the above identity is trivial as each term on the RHS is non-negative. Now backtrack and get your desired identity.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Does this formula ${n^2+3n\over 2}+{2(n+1)(n+2)-1\over 2(n+1)(n+2)}$ generates Pythagorean triples for all n? The idea came from this site
another formula for generating Pythagoras Triples
Let $n\ge1$
$2{11\over 12}, 5{23\over24}, 9{39\over 40},\cdots$ is generated from ${n^2+3n\over 2}+{2(n+1)(n+2)-1\over 2(n+1)(n+2)}.$
$2+{11\over12}={35\over 12}\rightarrow (12,35,37)$
$5+{23\over 24}={143\over24}\rightarrow (24,143,145)$
Where the fractions give the two sides and the hypotenuse is the numerator+2.
| Let $A = n(n+3)$ and $B = (n+1)(n+2)$. The $n$-th element of generated sequense is
$$\frac{A}{2} + \frac{2B - 1}{2B} = \frac{AB + 2B - 1}{2B}$$.
We want to prove that $(AB + 2B - 1)^2 + 4B^2 = (AB + 2B + 1)^2$. Indeed, after moving summand $(AB+2B-1)^2$ to the right side we have
$$4B^2 = 4(AB + 2B)$$
which leads to $B = A + 2$. It's easy to see that $(n+1)(n+2) = n^2+3n+2 = n(n+3)+2$.
So that's true.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For what values of $n$ is $n(n+2)$ a triangle number? A triangle number is a number which can be written in the form $\frac{m(m+1)}{2}$ for some natural number $m$. For what values of $n$ is $n(n+2)$ a triangle number ?
Using a a brute force method I could generate first $24$ such values of $n$ , but beyond that its to slow to generate. Here are the first $24$ values of $n$ such that $n(n+2)$ is a triangle number:
1
3
10
22
63
133
372
780
2173
4551
12670
26530
73851
154633
430440
901272
2508793
5253003
14622322
30616750
85225143
178447501
496728540
1040068260
PS: This is the final step for question swapping counters. I would appreciate hints only.
| You are trying to solve $n(n+2)=\frac{m(m+1)}{2}$. This can be written as $(n+1)^2-1 =\frac{(2m+1)^2-1}{8}$, or:
$$(2m+1)^2-8(n+1)^2 = -7$$
The equation $x^2-8y^2=-7$ is called a Pell-like equation, and, since it has the solution $(x,y)=(1,1)$, it has infinitely many solutions. Specifically, if $(x,y)$ is a solution, $(3x+8y,3y+x)$ is a solution. This gives all solutions starting at $(x,y)=(1,1)$ and $(x,y)=(5,2)$.
All solutions for the Pell-like have $x$ odd, so this will always give you a pair $(m,n)$.
The formulas for the values $n+1$ is:
$$n+1=\frac{1}{2\sqrt{8}}\left((1+\sqrt{8})(3+\sqrt{8})^k -(1-\sqrt{8})(3-\sqrt{8})^k\right)$$
and
$$n+1=\frac{1}{2\sqrt{8}}\left((5+2\sqrt{8})(3+\sqrt{8})^k -(5-2\sqrt{8})(3-\sqrt{8})^k\right)$$
That's a horrible expression, but it indicates you an also find a closed formula for the sum of the first $n$. There is also a linear recursion for the sequence of these values of $n$.
| {
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"url": "https://math.stackexchange.com/questions/1819510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Functional Equation of iterations
Problem: Let $f : \mathbb{Q} \to \mathbb{Q}$ satisfy $$f(f(f(x)))+2f(f(x))+f(x)=4x$$ and $$f^{2009}(x)=x$$ ($f$ iterated
$2009$ times).
Prove that $f(x)=x$.
This is a contest type problem so it is supposed to have an elegant solution.
My approach: I considered $f^{(k)}(x)=a_k$ then we get,
$a_n+2a_{n-1}+a_{n-2}-4a_{n-3}=0$ with $f(a_{n-1})=a_n$ and we need to show that the sequence is constant, $a_i=a_0$.
The characteristic root of that recurrence are not good.
Also I observed that f must be a bijection
Someone please help.
| I'm not sure that my solution is elegant.
So let $g(x) = f(x) - x$. We may rewrite the first equation from task as
$$
f(f^2(x)) - f^2(x) + 3f^2(x)-3f(x) + 4f(x) - 4x = 0,
$$
or using $g(x)$
$$
g(f^2(x)) + 3g(f(x)) + 4g(x) = 0.
$$
Let $a_n = g(f^n(x))$. Then $a_n + 3a_{n-1} + 4a_{n-2} = 0$. We may make this recurrent relation simpler: let $b_n = a_n + \frac{3 - i\sqrt{7}}{3}a_{n-1}$. Then $b_n + \frac{3+i\sqrt{7}}{2}b_{n-1} = a_n + 3a_{n-1} + 4a_{n-1} = 0$ or $$b_n = -\frac{3+i\sqrt{7}}{2}b_{n-1} = \left(-\frac{3+i\sqrt{7}}{2}\right)^{n-1} b_1.$$
Now using relation $f^{2009}(x) = x$ we get that $b_{2010} = b_1$. Thus
$$
b_1 = \left(-\frac{3+i\sqrt{7}}{2}\right)^{2009} b_1 \; \Rightarrow \; b_1 = 0\; \Rightarrow \; b_n = 0.
$$
Thus $a_n + \frac{3 - i\sqrt{7}}{2}a_{n-1} = 0$ or $$a_n = \left(-\frac{3 - i\sqrt{7}}{2}\right)a_{n-1} = \left(-\frac{3 - i\sqrt{7}}{2}\right)^{n}a_0$$
and again, as $f^{2009}(x) = f(x)$ we get $a_{2009} = a_0$ or
$$
a_0 = \left(-\frac{3 - i\sqrt{7}}{2}\right)^{2009}a_0 \;\Rightarrow\; a_0 = 0.
$$
As $a_0 = g(x) = f(x) -x$ we get $$f(x) - x = 0$$ QED.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Use integration by parts $\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$ $$I=\int^{\infty}_{0} \frac{x \cdot \ln x}{(1+x^2)^2}dx$$
Clearly $$-2I=\int^{\infty}_{0} \ln x \cdot \frac{-2x }{(1+x^2)^2} dx$$
My attempt :
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \left(\frac{1}{1+x^2}\right) \cdot \frac{1}{x} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1}{x(1+x^2)} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1+x^2-x^2}{x(1+x^2)} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 - \int^{\infty}_{0} \frac{1}{x}+ \frac{1}{2}\int^{\infty}_{0} \frac{2x}{1+x^2} dx$$
$$-2I=\left[ \ln x \cdot \left(\frac{1}{1+x^2}\right)\right]^\infty_0 -\left[ \ln x -\frac{1}{2}\cdot \ln(1+x^2) \right]^\infty_0 $$
$$-2I=\left[ \frac{\ln x}{1+x^2}\right]^\infty_0 -\left[\ln \left (\frac{x}{\sqrt{1+x^2}} \right) \right]^\infty_0 $$
How can I evaluate the last limits ?
| $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\iff}{\Leftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\, #1 \,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\, #2 \,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$
\begin{align}
\color{#f00}{I} &=
\int_{0}^{\infty}{x\ln\pars{x} \over \pars{1 + x^{2}}^{2}}\,\dd x =
{1 \over 4}\bracks{\overbrace{\int_{0}^{1}{x\ln\pars{x} \over \pars{1 + x}^{2}}\,\dd x}^{\ds{\equiv\ J}}\ +\
\int_{1}^{\infty}{x\ln\pars{x} \over \pars{1 + x}^{2}}\,\dd x}
\end{align}
The second integral in the RHS $\underline{\mbox{is equal to}}$ $\ds{-J}$ after the sub$\ldots$ $\ds{x \to {1 \over x}}$ such that
$\fbox{$\quad\ds{\color{#f00}{I} = J + \pars{-J} = \color{#f00}{0}}\quad$}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1822013",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Angle between 3 points I have three points $(x_1, y_1), (x_c, y_c), (x_3, y_3)$, where I know $(x_1, y_1), (x_c, y_c)$, the angle $\theta$, and $c$ on the dash line in the following figure. How to calculate the point $(x_3, y_3)$?
I think of this form:
$$
\theta = arccos\left(\frac{a\cdot b}{||a||\cdot ||b||}\right)
$$
where
$$
a = (x_1 - x_c, y_1 - y_c)\\
b = (x_1 - x_3, y_1 - y_3)
$$
More information:
| Let's approach the problem via a simplified case, where $(x_c,y_c)$ is at origin, and $(x_1, y_1)$ is on the $x$ axis at $(a, 0)$:
Obviously, we can calculate $a$ from the original coordinates,
$$a = \sqrt{\left(x_1 - x_c\right)^2 + \left(y_1 - y_c\right)^2}$$
We have three unknowns, $x$, $y$, and $b$ (the distance from origin to $(x,y)$), and three equations:
$$\begin{cases}
x = b \cos \theta \\
y = b \sin \theta \\
(x - a)^2 + y^2 = c^2
\end{cases}$$
There is a pair of solutions:
$$\begin{cases}
x = a \left(\cos\theta\right)^2 \pm \cos\theta \sqrt{c^2 - a^2 \left(\sin\theta\right)^2} \\
y = a \sin\theta \cos\theta \pm \sin\theta \sqrt{c^2 - a^2 \left(\sin\theta\right)^2} \\
b = a \cos\theta \pm \sqrt{c^2 - a^2\left(\sin\theta\right)^2}
\end{cases}$$
Pick either the upper or the lower signs for all three, but note that only the triplet for which $b \ge 0$ is actually valid. Indeed, for my illustration above, I've shown the "-" solution; the "+" solution would have $(x,y)$ somewhere near $(a,c)$, making the lower right corner angle somewhat near ninety degrees.
However, now that we do know (possibly two valid values of) $b$, we can go back to looking at the situation in the original coordinates.
Numerical solutions, using atan2():
The simplest way is to use the atan2() function available in most programming languages. ($\operatorname{atan2}(y,x) = \tan(y/x)$, except the former also takes account the quadrant, too.)
With it, in original coordinates,
$$\begin{cases}
b = a \cos\theta \pm \sqrt{c^2 - a^2\left(\sin\theta\right)^2}, & b \ge 0 \\
\theta_0 = \operatorname{atan2}(y_1 - y_c, x_1 - x_c) \\
x_3 = x_c + b \cos \left ( \theta_0 + \theta \right ) \\
y_3 = y_c + b \sin \left ( \theta_0 + \theta \right )
\end{cases}$$
If you want positive $\theta$ to be clockwise, use $(\theta_0 - \theta)$ instead in the formulas for $x_3$ and $y_3$, above.
Symbolic solutions, via coordinate system transformation:
A two-dimensional rotation matrix is defined as
$$\mathbf{R} = \left[\begin{matrix}\cos\varphi & -\sin\varphi \\ \sin\varphi & \cos\varphi\end{matrix}\right] = \left[\begin{matrix}C&-S\\S&C\end{matrix}\right]$$
and rotating a point $\vec{p} = (x, y)$ by matrix $\mathbf{R}$ is
$$\mathbf{R} \vec{p} = \left[\begin{matrix}C&-S\\S&C\end{matrix}\right]\left[\begin{matrix}x\\y\end{matrix}\right]$$
i.e.
$$\begin{cases}
x' = C x - S y \\
y' = S x + C y
\end{cases}$$
In this particular case, we need to rotate our simplified case solutions using a matrix which rotates point $(a,0)$ to $(x_1-x_c, y_1-y_c)$:
$$\begin{cases}
x_1 - x_c = C a \\
y_1 - y_c = S a
\end{cases} \iff \begin{cases}
C = \frac{x_1 - x_c}{a} \\
S = \frac{y_1 - y_c}{a}
\end{cases}$$
Applying the above rotation to our simplified case results, and a translation to move $(x_c, y_c)$ back to its proper place from origin, we get:
$$\begin{cases}
b = a \cos\theta \pm \sqrt{c^2 - a^2\left(\sin\theta\right)^2}, & b \ge 0 \\
x = b \cos\theta \\
y = b \sin\theta \\
C = \frac{x_1 - x_c}{a} \\
S = \frac{y_1 - y_c}{a} \\
x_3 = x_c + C x - S y \\
y_3 = y_c + S x + C y
\end{cases}$$
or equivalently
$$\begin{cases}
b = a \cos\theta \pm \sqrt{c^2 - a^2\left(\sin\theta\right)^2}, & b \ge 0 \\
x_3 = x_c + \frac{b}{a}(x_1 - x_c)\cos\theta - \frac{b}{a}(y_1 - y_c)\sin\theta \\
y_3 = y_c + \frac{b}{a}(y_1 - y_c)\cos\theta + \frac{b}{a}(x_1 - x_c)\sin\theta
\end{cases}$$
or equivalently
$$\begin{cases}
z = \frac{b}{a} = \cos\theta \pm \sqrt{\frac{c^2}{a^2} - \left(\sin\theta\right)^2}, & z \ge 0 \\
x_3 = x_c + z(x_1 - x_c)\cos\theta - z(y_1 - y_c)\sin\theta \\
y_3 = y_c + z(y_1 - y_c)\cos\theta + z(x_1 - x_c)\sin\theta
\end{cases}$$
.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\int x^2 \sqrt{1-x^2}\ dx$ I hope I can find a way to integrate this formula without resorting to numerical techniques.
\begin{equation}
\int x^2 \sqrt{1-x^2}\ dx
\end{equation}
I am not sure if there's actually a closed form for this or not?
I tried integration by parts, but it seems not working! Here's my last resort before numerical solutions.
| Here is a solution with integration by parts and rearranging terms:
First, integrating by parts and adding and subtracting a $1$ in the numerator,
$$
\begin{aligned}
I=\int x^2\sqrt{1-x^2}\,dx & = \frac{x^3}{3}\sqrt{1-x^2}+\int\frac{x^3}{3}\frac{x}{\sqrt{1-x^2}}\,dx\\
&=\frac{x^3}{3}\sqrt{1-x^2}-\int\frac{x^3}{3}\sqrt{1-x^2}\,dx+\int\frac{x^3}{3}\frac{1}{\sqrt{1-x^2}}\,dx.
\end{aligned}
$$
We got $I$ back in the right-hand side, and moving it to the left-hand side, we find
$$
I=\frac{x^3}{4}\sqrt{1-x^2}+\int\frac{1}{4}\frac{x^2}{\sqrt{1-x^2}}\,dx.\tag{1}
$$
We can continue from $(1)$ in two ways. Adding and subtracting a $1$ in the numerator, we get
$$
I=\frac{x^3}{4}\sqrt{1-x^2}-\frac{1}{4}\int\sqrt{1-x^2}\,dx+\frac{1}{4}\int\frac{1}{\sqrt{1-x^2}}\,dx\tag{2}
$$
If we instead integrate by parts, we get
$$
I=\frac{x^3}{4}\sqrt{1-x^2}-\frac{1}{4}x\sqrt{1-x^2}+\frac{1}{4}\int\sqrt{1-x^2}\,dx.\tag{3}
$$
We add $(2)$ and $(3)$, and find that
$$
2I=\frac{x^3}{2}\sqrt{1-x^2}-\frac{1}{4}x\sqrt{1-x^2}+\frac{1}{4}\int\frac{1}{\sqrt{1-x^2}}\,dx,
$$
and hence, finally,
$$
\begin{aligned}
I&=\frac{x^3}{4}\sqrt{1-x^2}-\frac{1}{8}x\sqrt{1-x^2}+\frac{1}{8}\int\frac{1}{\sqrt{1-x^2}}\,dx\\
&=\frac{x^3}{4}\sqrt{1-x^2}-\frac{1}{8}x\sqrt{1-x^2}+\frac{1}{8}\arcsin x+C.
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1825288",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $x$ and $y$ are non-negative integers for which $(xy-7)^2=x^2+y^2$. Find the sum of all possible values of $x$. I am not able to reach to the answer. I have used discriminant as $x$ and $y$ are both integers but it didn't give any hint to reach to answer. I am not able to understand how should I deal with these type of question.
| $$(xy)^2−12xy+49=(x+y)^2\longleftrightarrow(xy−6)^2+13=(x+y)^2.$$ Hence $$13=(x+y+xy−6)(x+y−xy+6).$$ It follows the only possibility (after discarding $−13$ and $−1$) is that $$x+y+xy−6=13,$$ $$x+y−xy+6=1.$$ This implies clearly $x+y=7$ which gives the candidates $(0,7),(1,6),(2,5),(3,4)$ and symmetrics, and it is verified that the only to be accepted are $(0,7)$ and $(4,3)$.
Consequently, taking $(x,y)=(0,7),(7,0),(4,3),(3,4)$, $$S=0+7+4+3=14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 5
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Trigonometric Roots of a Polynomial After wondering on this question, I wondered how would you be able to find the roots of a polynomial, in the form $y=x^3+ax^2+bx+c$ if they are the sums of cosines?
I'm wondering if it can, too, be expressed in the form $b\cos\left(A\right)$ with $A$ larger than $2\pi$.
EDIT: I believe $y=x^3+x^2-4x+1$ to be an example.
| After research, I found a relatively easy method. Given the cubic $y=x^3+ax^2+bx+c$, one can find the roots by finding $p$ and $q$ where $$p=\frac {3b-a^2}{9}\tag{1}$$$$q=\frac {9ab-27c-2a^3}{54}\tag{2}$$
And plugging them into $$\cos(\theta)=\frac {q}{\sqrt{-p^3}}\tag{3}$$So the solutions are given $$\begin{cases}x_1=2\sqrt{-p}\cdot\cos\left(\frac {\theta}{3}\right)-\frac {a}{3}\\x_2=2\sqrt{-p}\cdot\cos\left(\frac {\theta+2\pi}{3}\right)-\frac {a}{3}\\x_3=2\sqrt{-p}\cdot\cos\left(\frac {\theta+4\pi}{3}\right)-\frac {a}{3}\end{cases}\tag{4}$$
With discriminant $\delta=p^3+q^2<0$.
For example: $y=x^3-x^2-9x+1$. We see that $a=-1,b=-9$ and $c=1$. So from $(1)$, $$p=\frac {3(-9)-1^2}{9}=-\frac {28}{9}\tag{5}$$ and from $(2)$, $$q=\frac {9(-1)(-9)-27+2}{54}=\frac {28}{27}\tag{6}$$
So $\cos(\theta)=\frac {1}{2\sqrt{7}}$ and the solutions are $$\begin{cases}x_1=\frac {4\sqrt{7}}{3}\cdot\cos\left(\frac {1}{3}\cdot\arccos\left(\frac {1}{2\sqrt{7}}\right)\right)+\frac {1}{3}=4\cos\left(\frac {2\pi}{7}\right)+1\\x_2=
\frac {4\sqrt{7}}{3}\cdot\cos\left(\frac {1}{3}\left(2\pi+\arccos\left(\frac {1}{2\sqrt{7}}\right)\right)\right)+\frac {1}{3}=4\cos\left(\frac {8\pi}{7}\right)+1\\x_3=\frac {4\sqrt{7}}{3}\cdot\cos\left(\frac {1}{3}\left(4\pi+\arccos\left(\frac {1}{2\sqrt{7}}\right)\right)\right)+\frac {1}{3}=4\cos\left(\frac {4\pi}{7}\right)+1\end{cases}$$
Note: The polynomial has $3$ roots because $\delta$ is less than $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$ Can someone point me in the right direction how to solve this?
$3^{x+2}\cdot4^{-(x+3)}+3^{x+4}\cdot4^{-(x+3)} = \frac{40}{9}$
I guess I have to get to logarithms of the same base. But how? What principle should I use here?
Thx
| Using $a^{-b}=\frac{1}{a^b}$, The equation can be rewritten as
$$\frac{3^{x+2}}{4^{x+3}}+\frac{3^{x+4}}{4^{x+3}}=\frac{40}{9}$$
$$\implies \frac{9}{64}\left(\frac{3}{4}\right)^x+\frac{81}{64}\left(\frac{3}{4}\right)^x=\frac{40}{9}$$
$$\implies \left(\frac{3}{4}\right)^x=\frac{256}{81}$$
$$\implies x=-4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\int_{0}^{\infty}{1\over x^4+x^2+1}dx=\int_{0}^{\infty}{1\over x^8+x^4+1}dx$ Let
$$I=\int_{0}^{\infty}{1\over x^4+x^2+1}dx\tag1$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx\tag2$$
Prove that $I=J={\pi \over 2\sqrt3}$
Sub: $x=\tan{u}\rightarrow dx=\sec^2{u}du$
$x=\infty \rightarrow u={\pi\over 2}$, $x=0\rightarrow u=0$
Rewrite $(1)$ as
$$I=\int_{0}^{\infty}{1\over (1+x^2)^2-x^2}dx$$
then
$$\int_{0}^{\pi/2}{\sec^2{u}\over \sec^4{u}-\tan^2{u}}du\tag3$$
Simplified to
$$I=\int_{0}^{\pi/2}{1\over \sec^2{u}-\sin^2{u}}du\tag4$$
Then to
$$I=2\int_{0}^{\pi/2}{1+\cos{2u}\over (2+\sin{2u})(2-\sin{2u})}du\tag5$$
Any hints on what to do next?
Re-edit (Hint from Marco)
$${1\over x^8+x^4+1}={1\over 2}\left({x^2+1\over x^4+x^2+1}-{x^2-1\over x^4-x^2+1}\right)$$
$$M=\int_{0}^{\infty}{x^2+1\over x^4+x^2+1}dx=\int_{0}^{\infty}{x^2\over x^4+x^2+1}dx+\int_{0}^{\infty}{1\over x^4+x^2+1}dx={\pi\over \sqrt3}$$
$$N=\int_{0}^{\infty}{x^2-1\over x^4-x^2+1}dx=0$$
$$J=\int_{0}^{\infty}{1\over x^8+x^4+1}dx={1\over 2}\left({\pi\over \sqrt3}-0\right)={\pi\over 2\sqrt3}.$$
| Here is an alternative approch using complex analysis. Since both integrands are even one can start with integrating around the singularities in the positive half-plane using the residue theorem which yields
$$
\begin{align*}
\frac{1}{2}\int_{-\infty}^\infty \frac{1}{x^4+x^2+1}\,\mathrm dx
&= \pi\mathrm i\left(\operatorname{Res}_{z=\sqrt[3]{-1}}\left(\frac{1}{z^4+z^2+1}\right) + \operatorname{Res}_{z=(-1)^{2/3}}\left(\frac{1}{z^4+z^2+1}\right)\right)\\
&= \pi\mathrm i\left(\frac{1}{4\sqrt[3]{-1}^3+2\sqrt[3]{-1}} + \frac{1}{4((-1)^{2/3})^3+2(-1)^{2/3}}\right)\\
&= \pi\mathrm i\left(-\frac{\mathrm i}{2\sqrt{3}}\right)\\
&= \frac{\pi}{2\sqrt{3}}
\end{align*}
$$
and similiarly the second integral
$$
\begin{align*}
\frac{1}{2}\int_{-\infty}^\infty \frac{1}{x^8+x^4+1}\,\mathrm dx
&= \pi\mathrm i\left(\operatorname{Res}_{z=\sqrt[6]{-1}}\left(\frac{1}{z^8+z^4+1}\right) + \operatorname{Res}_{z=\sqrt[3]{-1}}\left(\frac{1}{z^8+z^4+1}\right)\right.\\
&\qquad \left.+ \operatorname{Res}_{z=(-1)^{2/3}}\left(\frac{1}{z^8+z^4+1}\right) + \operatorname{Res}_{z=(-1)^{5/6}}\left(\frac{1}{z^8+z^4+1}\right)\right)\\
&= \pi\mathrm i\left(\frac{1}{8(\sqrt[6]{-1})^7+4(\sqrt[6]{-1})^3} + \frac{1}{8(\sqrt[3]{-1})^7+4(\sqrt[3]{-1})^3}\right.\\
&\qquad \left. + \frac{1}{8((-1)^{2/3})^7+4((-1)^{2/3})^3} +\frac{1}{8((-1)^{5/6})^7+4((-1)^{5/6})^3} \right)\\
&= \pi\mathrm i\left(-\frac{\mathrm i}{2\sqrt{3}}\right)\\
&= \frac{\pi}{2\sqrt{3}}
\end{align*}
$$
thus both are equal indeed.
| {
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"url": "https://math.stackexchange.com/questions/1829298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Prove that $\frac{2^a+1}{2^b-1}$ is not an integer
Let $a$ and $b$ be positive integers with $a>b>2$. Prove that $\frac{2^a+1}{2^b-1}$ is not an integer.
This is equivalent to showing there always exists some power of a prime $p$ such that $2^a+1 \not \equiv 0 \pmod{p^a}$ but $2^b-1 \equiv 0 \pmod{p^a}$. How do we prove the statement from this or is there an easier way?
| if $b$ divides $a$ then $2^b - 1$ divides $2^a-1$.
We can use the Euclidean algroithm to find $q,r$ such that $a = qb + r$ with $r < b$.
$2^a+1 = (2^{qb})(2^r) + 2^r - 2^r + 1 = (2^{qb} - 1)(2^r) + 2^r + 1$
$2^b-1$ divides $(2^{qb} - 1)2^r$ leaving a remainder $2^r+1$
If $r<b$ and $b>1, 2^r + 1 < 2^b - 1$ and $2^b - 1$ cannot divide $2^r + 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Surface area of the part of a sphere above a hexagon I want to calculate the surface area of the part of a half-sphere, which lies above a regular 6-gon. (Radius $r=1$)
More formally,
Let $G$ be the region on the $XY$-Plane, bounded by the points $\{P_k=(\cos(\frac{2\pi k}{6}), \sin(\frac{2\pi k}{6}))\}$ for $k=1, ..., 6$. It's just the hexagon whose nodes are on the unit circle.
I want to calculate the area of $S_G = \{(x,y,z)\in \mathbb{R}^3 \ |\ x^2+y^2+z^2 =1, \ z>0, \ and \ (x,y)\in G \}$.
What I did:
Similar to how one computes the sphere's area, I took $F(x,y,z)=x^2+y^2+z^2-1=0$, and eventually got to $Area(S_G)=\int_G{\frac{1}{\sqrt{1-x^2-y^2}}dx dy}$.
Seems kind of nasty since $G$ is not a nice domain to work on with polar coordinates (that would have helped if my domain was "round" in some way, because then it reduces to $\int{\frac{1}{\sqrt{1-s}}ds}$ ).
Any ideas?
| Hint:
By simmetry, the area of the surface of the half sphere which lies above the equilateral triangle $OP_6P_1$ is $\frac16S_G$.
Now, $\triangle OP_6P_1$ is the set bounded by the $x$ axis, and the lines $y=\sqrt{3}x$ and $y=\sqrt{3}(1-x)$. In polar coordinates this region is
$$\left\{(\theta,r)\in\mathbb{R}^2:0\leq\theta\leq\frac{\pi}{3},0\le r\le\frac{\sqrt{3}}{\sqrt{3}\cos\theta+\sin\theta}\right\}$$
Then,
\begin{align*}
\frac{1}{6}S_G&=\int_0^{\frac{\pi}{3}}\int_0^{\frac{\sqrt{3}}{\sqrt{3}\cos \theta+\sin \theta}}\frac{r}{\sqrt{1-r^2}}\,drd\theta\\[3pt]
&=\int_0^{\frac{\pi}3}\left.-\sqrt{1-r^2}\right|_0^{\frac{\sqrt{3}}{\sqrt{3}\cos \theta+\sin \theta}}\,d\theta\\
&=\int_0^{\frac{\pi}{3}}\left(1-\sqrt{1-\frac{3}{4\sin^2\left(\theta+\frac{\pi}{3}\right)}}\right)\,d\theta
\end{align*}
Last integral is hard to evaluate by hand, but Wolfram Mathematica give us $\frac{1}{6} \left(-4+3 \sqrt{3}\right) \pi $, then
$$\color{blue}{S_G=\left(-4+3 \sqrt{3}\right) \pi}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1831085",
"timestamp": "2023-03-29T00:00:00",
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Distribution of throws of die rigged to never produce twice in a row the same result A die is “fixed” so that each time it is rolled the score cannot be the same as the preceding score, all other scores having probablity 1/5. If the first score is 6, what is the probability that the nth score is 6 and what is the probability that the nth score is 1?
| Let $p_n$ be the probability the $n$-th toss is a $6$. Note that $p_1=1$. We obtain a recurrence for $p_n$.
The probability the $n$-th toss is not a $6$ is $1-p_n$. Given that, the probability the $n$-th toss is a $6$ is $\frac{1}{5}$.
Thus
$$p_{n+1}=\frac{1}{5}(1-p_n)=\frac{1}{5}-\frac{1}{5}p_n.$$
The homogeneous recurrence $p_{n+1}=-\frac{1}{5}p_n$ has general solution $A(-1/5)^n$. A particular solution of the non-homogeneous recurrence is $1/6$.
Thus the general solution of our recurrence is $\frac{1}{6}+\frac{A(-1)^n}{5^n}$. Since $p(1)=1$, we have $A=-\frac{25}{6}$ and therefore
$$p_n=\frac{1}{6}\left(1+\frac{(-1)^{n-1}}{5^{n-2}} \right).$$
Now that we know the probability of a $6$, given the first is a $6$, we can easily compute the probability of $1$, since by symmetry $1$ to $5$ are equally likely.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Solve definite integral: $\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$ I need to solve:
$$\int_{-1}^{1}\arctan(\sqrt{x+2})\ dx$$
Here is my steps, first of all consider just the indefinite integral:
$$\int \arctan(\sqrt{x+2})dx = \int \arctan(\sqrt{x+2}) \cdot 1\ dx$$
$$f(x) = \arctan(\sqrt{x+2})$$
$$f'(x) = \frac{1}{1+x+2} \cdot \frac{1}{2\sqrt{x+2}} = \frac{1}{(2x+6)\sqrt{x+2}}$$
$$g'(x) = 1$$
$$g(x) = x$$
So:
$$\bigg[\arctan(\sqrt{x+2}) \cdot x\bigg]_{-1}^{1} - \int_{-1}^{1} \frac{x}{(2x-6)\sqrt{x+2}}\ dx$$
How should I proceed with the new integral?
| You can set $t=\sqrt{x+2}$. For $x=-1$ we have $t=1$, for $x=1$ we have $t=\sqrt{3}$. Moreover $x=t^2-2$, so $dx=2t\,dt$. Therefore the integral is
$$
\int_{1}^{\sqrt{3}}2t\arctan t\,dt=
\underbrace{\Bigl[t^2\arctan t\Bigr]_1^{\sqrt{3}}}_A-
\underbrace{\int_{1}^{\sqrt{3}}\frac{t^2}{1+t^2}\,dt}_B
$$
(by parts). Then
$$
A=3\cdot\frac{\pi}{3}-\frac{\pi}{4}
$$
Let's examine $B$:
$$
B=\int_{1}^{\sqrt{3}}\frac{1+t^2-1}{1+t^2}\,dt=
\Bigl[t-\arctan t\Bigr]_{1}^{\sqrt{3}}=
\left(\sqrt{3}-\frac{\pi}{3}\right)-\left(1-\frac{\pi}{4}\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
} |
I want to show that, ${\Phi\tan{9^\circ}-\phi\tan{27^\circ}\over \sin^2{9^\circ}-\sin^2{27^\circ}}=4$ $\phi$: golden ratio, $\Phi={1\over \phi}$
I want to show that,
$${\Phi\tan{9^\circ}-\phi\tan{27^\circ}\over \sin^2{9^\circ}-\sin^2{27^\circ}}=4$$
Using $\sin^2{x}={1\over 2}(1-\cos{2x})$
$${\Phi\tan{9^o}-\phi\tan{27^o}\over \cos{54^\circ}-\cos{18^\circ}}=2$$
As for numerator
$${\sqrt5(\tan{27^\circ}-\tan{9^\circ})-\tan{27^\circ}-\tan{9^\circ}\over \cos{54^\circ}-\cos{18^\circ}}=4$$
exact trig values
if I was to plug in the corresponding values it would be a lot of work to simply and might not even get to the result.
Any idea would help me to simplify the LHS?
| Noting that
$$ \phi=2\cos(\frac{\pi}{5})=2\sin(\frac{3\pi}{10}), \Phi=2\sin(\frac{\pi}{10})$$
from https://en.wikipedia.org/wiki/Golden_ratio, so one has
\begin{eqnarray}
{\Phi\tan{9^\circ}-\phi\tan{27^\circ}\over \sin^2{9^\circ}-\sin^2{27^\circ}}&=&{2\sin(\frac{\pi}{10})\tan(\frac{\pi}{20})-2\sin(\frac{3\pi}{10})\tan(\frac{3\pi}{20})\over \sin^2(\frac{\pi}{20})-\sin^2(\frac{3\pi}{20})}\\
&=&2{2\sin(\frac{\pi}{20})\cos(\frac{\pi}{20})\tan(\frac{\pi}{20})-2\sin(\frac{3\pi}{20})\cos(\frac{3\pi}{20})\tan(\frac{3\pi}{20})\over \sin^2(\frac{\pi}{20})-\sin^2(\frac{3\pi}{20})}\\
&=&4{\sin^2(\frac{\pi}{20})-\sin^2(\frac{3\pi}{20})\over \sin^2(\frac{\pi}{20})-\sin^2(\frac{3\pi}{20})}\\
&=&4.
\end{eqnarray}
| {
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"url": "https://math.stackexchange.com/questions/1835770",
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"source": "stackexchange",
"question_score": "2",
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I want to show that $\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$ I want to show that
$$\int_{-\infty}^{\infty}{\left(x^2-x+\pi\over x^4-x^2+1\right)^2}dx=\pi+\pi^2+\pi^3$$
Expand $(x^4-x+\pi)^2=x^4-2x^3+2x^2-2x\pi+\pi{x^2}+\pi^2$
Let see (substitution of $y=x^2$)
$$\int_{-\infty}^{\infty}{x\over (x^4-x^2+1)^2}dx={1\over 2}\int_{-\infty}^{\infty}{1\over (y^2-y+1)^2}dy$$
Substituion of $y=x^3$
$$\int_{-\infty}^{\infty}{x^3\over (x^4-x^2+1)^2}dx={1\over 4}\int_{-\infty}^{\infty}{1\over (y^2-y+1)^2}dy$$
As for $\int_{-\infty}^{\infty}{x^2\over (x^4-x^2+1)^2}dx$ and $\int_{-\infty}^{\infty}{x^4\over (x^4-x^2+1)^2}dx$ are difficult to find a suitable substitution. This is the point where I am shrugged with to find a suitable substitution To lead me to a particular standard integral. Need some help, thank.
standard integral of the form
$$\int{1\over (ax^2+bx+c)^2}dx={2ax+b\over (4ac-b^2)(ax^2+bx+c)}+{2a\over 4ac-b^2}\int{1\over ax^2+bx+c}dx$$
And
$$\int{1\over ax^2+bx+c}dx={2\over \sqrt{4ac-b^2}}\tan^{-1}{2ax+b\over \sqrt{4ac-b^2}}$$
| First note that because of
$$
\int_{\mathbb{R}}dx\frac{x^4}{(x^4-x^2+1)^2}=\int_{\mathbb{R}}dx\frac{x^4\overbrace{-x^2+1+(x^2-1)}^{=0}}{(x^4-x^2+1)^2}=\\
\int_{\mathbb{R}}dx\frac{1}{(x^4-x^2+1)}+\int_{\mathbb{R}}dx\frac{x^2-1}{(x^4-x^2+1)^2}=\\
\int_{\mathbb{R}}dx\frac{1}{(x^4-x^2+1)}+\int_{\mathbb{R}}dx\frac{-1}{(x^4-x^2+1)^2}+\color{blue}{\underbrace{\int_{\mathbb{R}}dx\frac{x^2}{(x^4-x^2+1)^2}}_{J}}
$$
we only need to calulate $\color{\blue}{J}$ because everthing else is covered by your standard formulas.
Now let's define
$$
I(a)=\int_{\mathbb{R}}dx\frac{1}{(x^4-a x^2+1)}
$$
from which it follows that
$$
\color{blue}{J}=I'(1)
$$
so it can also derived using ur standard identities ($'$ denotes a derivative w.r.t. $a$)
Edit:
Note that the integrals with numerators containing odd powers of $x$ vanish due to symmetry!
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "13",
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The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + 1/β$ and $β + 1/α$. Quadratic equation question, as specified in the title.
The roots of the equation $x^2 - 6x + 7 = 0$ are $α$ and $β$. Find the equation with roots $α + \frac{1}{β}$ and $β + \frac{1}{α}$.
I gather that $α + β = -\frac{b}{a} = \frac{6}{1} = 6$ and that $αβ = \frac{c}{a} = \frac{7}{1} = 7$. Do I need to convert $α + \frac{1}{β}$ and $β + \frac{1}{α}$ into a format whereby I can sub in the values for adding together or multiplying $α$ and $β$ ? If so, how ?
| $(x-\alpha)(x-\beta) = x^2 - (\alpha + \beta)x + \alpha\beta=0\\
(\alpha + \beta) = 6\\
\alpha\beta = 7$
$(x-\alpha - \frac 1\beta)(x-\beta - \frac 1\alpha) = x^2 - (\alpha + \beta + \frac 1\alpha + \frac 1\beta)x + (\alpha\beta + 2+ \frac {1}{\alpha\beta})=0\\
x^2 - (\alpha + \beta + \frac{(\alpha + \beta)}{\alpha\beta})x + (\alpha\beta + 2+ \frac {1}{\alpha\beta})=0\\
x^2 - (6 + \frac 67)x + (9 + \frac 17)=0$
| {
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"url": "https://math.stackexchange.com/questions/1837156",
"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$
Prove that $\displaystyle \sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$.
I tried using the partial fraction decomposition $a_j = \frac{1}{2j} - \frac{1}{j+1} + \frac{1}{2(j+2)}$, but I don't see how that helps.
| Note that we can write
$$\begin{align}
\frac{1}{m(m+1)(m+2)}&=\frac12\left(\frac{(m+2)-m}{m(m+1)(m+2)}\right)\\\\
&=\frac12\left(\frac{1}{m(m+1)}-\frac{1}{(m+1)(m+2)}\right)
\end{align}$$
So, letting $a_m=\frac{1}{m(m+1)}$, we see that
$$\begin{align}
\frac{1}{m(m+1)(m+2)}&=\frac12\left(a_m-a_{m+1}\right)
\end{align}$$
from which we find
$$\begin{align}
\sum_{m=1}^n \frac{1}{m(m+1)(m+2)}&=\sum_{m=1}^n \frac12\left(a_m-a_{m+1}\right)\\\\
&=\frac12(a_1-a_{n+1})\\\\
&=\frac14-\frac12 \frac{1}{(n+1)(n+2)}
\end{align}$$
as was to be shown!
| {
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Prove by induction $3+3 \cdot 5+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$ My question is:
Prove by induction that $$3+3 \cdot 5+ 3 \cdot 5^2+ \cdots +3 \cdot 5^n = \frac{3(5^{n+1} -1)}{4}$$ whenever $n$ is a nonnegative integer.
I'm stuck at the basis step.
If I started with $1$. I get the right hand side is $18$ which is clearly not even close. It says prove shouldn't it be always true?
| As you want to prove for $n$ nonnegative and integer, your basis is $n=0$
You just have to prove that $\frac{3(5^{0+1} -1)}{4} = 3$ and it is true, because
$\frac{3(5^{0+1} -1)}{4} = \frac{3(5 -1)}{4} = \frac{3.4}{4} = 3$
If you want to dofor $n = 1$:
For $n=1$, you have that:
$3 + 3.5 = 18$ and $\frac{3(5^{1+1} -1)}{4} = \frac{3(5^2 -1)}{4} = 18$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve the limit of this sequence $\lim_{n \to \infty} \left(\frac{1}{3\cdot 8}+\dots+\frac{1}{6(2n-1)(3n+1)} \right)$ $$\lim_{n \to \infty} \left(\frac{1}{3\cdot 8}+\dots+\frac{1}{6(2n-1)(3n+1)} \right)$$
I have tried to split the subset into telescopic series but got no result.
I also have tried to use the squeeze theorem by putting the $a_n$ between $\frac{1}{(2n-1)(2n+1)}$ and $\frac{1}{(4n-1)(4n+1)}$ but it doesn't work.
| $$\frac{1}{6(2n-1)(3n+1)}=\frac{1}{(6n-3)(6n+2)}=\frac{1}{5}\left(\frac{1}{6n-3}-\frac{1}{6n+2}\right)=\frac{1}{5}\int_0^1\left(x^{6n-4}-x^{6n+1}\right)$$
by using partial fractions and noting that $\int_0^1x^a\,dx=\frac{1}{a+1}$. Then:
$$\sum_{n=1}^N\frac{1}{6(2n-1)(3n+1)}=\frac{1}{5}\int_0^1\sum_{n=1}^N\left(x^{6n-4}-x^{6n+1}\right)=\frac{1}{5}\int_0^1(x^2-x^7)\frac{1-x^{6N}}{1-x^6}\,dx$$
and this last integral is equal to:
$$\frac{1}{5}\int_0^1(x^2-x^7)\frac{1}{1-x^6}\,dx-\frac{1}{5}\int_0^1\frac{x^{6N+2}(1-x^5)}{1-x^6}\,dx$$
Upon cancelling fractions, the first of these is equal to $\frac{1}{5}\int_0^1x^2\frac{1+x+x^2+x^3+x^4}{1+x+x^2+x^3+x^4+x^5}\,dx$, and the latter can be shown to be convergent to $0$ by noting that $\frac{1-x^5}{1-x^6}\le1$ on the given interval and $\int_0^1x^{6N+2}\,dx\to0$.
So, the series converges to $\frac{1}{5}\int_0^1x^2\frac{1+x+x^2+x^3+x^4}{1+x+x^2+x^3+x^4+x^5}\,dx$.
This integral can be evaluated exactly, if needed, by expanding the integrand into partial fractions.
| {
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} |
Prove $\frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}$. So I have to prove
$$ \frac{a+b+c}{abc} \leq \frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}.$$
I rearranged it
$$ a^2bc + ab^2c + abc^2 \leq b^2c^2 + a^2c^2 + a^2b^2 .$$
My idea from there is somehow using the AM-GM inequality. Not sure how though. Any ideas?
Thanks
| Note that: $$a^2c^2 + a^2b^2 \ge 2a^2bc \quad\text{ by AM-GM}$$ Now add all the cyclic inequalities and you'll get the wanted inequality.
| {
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"url": "https://math.stackexchange.com/questions/1840148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the area of the triangle having $z_1$, $z_2$ and $z_3$ as vertices in Argand plane? What is the area of the triangle having $z_1$, $z_2$ and $z_3$ as vertices in Argand plane?
Is it
$$\frac{-1}{4i}[z_1(z_2^* - z_3^*)-z_1^*(z_2-z_3)+{z_2(z_3^*)-z_3(z_2^*)}]$$
where $w^*$ denotes the complex conjugate?
| Let $z_j = x_j + iy_j$, $j = 1, 2, 3$. The area of the triangle is given by
\begin{align*}
\frac{1}{2} \begin{vmatrix}
1 & x_1 & y_1 \\
1 & x_2 & y_2 \\
1 & x_3 & y_3
\end{vmatrix}&= \frac{1}{2} \begin{vmatrix}
1 & x_1+iy_1 & y_1 \\
1 & x_2+iy_2 & y_2 \\
1 & x_3+iy_3 & y_3
\end{vmatrix}\\
&= \frac{1}{4i} \begin{vmatrix}
1 & z_1 & z_1-z_1^*\\
1 & z_2 & z_2-z_2^* \\
1 & z_3 & z_3-z_3^*
\end{vmatrix}\\
\end{align*}
Now expand via first column to get the required expression.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
System of Equation
Solve the system of equations
$$-x_1+x_2+x_3=a$$ $$x_1-x_2+x_3=b$$ $$x_1+x_2-x_3=c$$
I have tried writing the augmented matrix of the system of equations above and reducing it into echelon form but that didn't work out. Please help.
| Let's do the Gaussian elimination:
\begin{align}
\left[\begin{array}{ccc|c}
-1 & 1 & 1 & a \\
1 & -1 & 1 & b \\
1 & 1 & -1 & c
\end{array}\right]
&\to
\left[\begin{array}{ccc|c}
1 & -1 & -1 & -a \\
1 & -1 & 1 & b \\
1 & 1 & -1 & c
\end{array}\right]
&& R_1\gets -R_1
\\
&\to
\left[\begin{array}{ccc|c}
1 & -1 & -1 & -a \\
0 & 0 & 2 & b+a \\
0 & 2 & 0 & c+a
\end{array}\right]
&& \begin{aligned}R_2&\gets R_2-R_1\\R_3&\gets R_3-R_1\end{aligned}
\\
&\to
\left[\begin{array}{ccc|c}
1 & -1 & -1 & -a \\
0 & 2 & 0 & c+a \\
0 & 0 & 2 & b+a
\end{array}\right]
&& R_2\leftrightarrow R_3
\\
&\to
\left[\begin{array}{ccc|c}
1 & -1 & -1 & -a \\
0 & 1 & 0 & (c+a)/2 \\
0 & 0 & 1 & (b+a)/2
\end{array}\right]
&& \begin{aligned}R_2&\gets \tfrac{1}{2}R_2\\R_3&\gets \tfrac{1}{2}R_3\end{aligned}
\\
&\to
\left[\begin{array}{ccc|c}
1 & -1 & 0 & (b-a)/2 \\
0 & 1 & 0 & (c+a)/2 \\
0 & 0 & 1 & (b+a)/2
\end{array}\right]
&&R_1\gets R_1+R_3
\\
&\to
\left[\begin{array}{ccc|c}
1 & 0 & 0 & (b+c)/2 \\
0 & 1 & 0 & (c+a)/2 \\
0 & 0 & 1 & (b+a)/2
\end{array}\right]
&&R_1\gets R_1+R_2
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate $\sum_{n=1}^{\infty}a_n$? If
$$a_{n}=1-\frac{1}{2}+\frac{1}{3}-\cdots +\frac{\left ( -1 \right )^{n-1}}{n}-\ln 2$$
then how to eveluate
$$\sum_{n=1}^{\infty}a_n$$
does it converge?
| By using Taylor expansion, we have
\begin{align}
a_{n}=(-1)^{n+1}\left(\frac{1}{n+1}-\frac{1}{n+2}+\frac{1}{n+3}-\cdots\right)
\end{align}
which alternates signs and converges to zero (note that $|a_{n}|<\frac{1}{n+1}$.) Also,
\begin{align}
|a_{n}|&=\left(\frac{1}{n+1}-\frac{1}{n+2}\right)+\left(\frac{1}{n+3}-\frac{1}{n+4}\right)+\cdots \\
&>\left(\frac{1}{n+2}-\frac{1}{n+3}\right)+\left(\frac{1}{n+4}-\frac{1}{n+5}\right)+\cdots=|a_{n+1}|
\end{align}
so the series converges by alternating series test.
But it seems to be difficult to find closed form.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Trinomial Pascal's Triangle I know that there's a trinomial theorem (and a multinomial theorem), but I was wondering if there was a similar structure for trinomials as there is for binomials, like Pascal's triangle.
Thanks in advance!
| Binomial Coefficients
Pascal's Triangle gives the coefficients of the binomial
\begin{align*}
(1+x)^n\qquad\qquad n\geq 0
\end{align*}
we obtain for $n=0,\ldots,4$
\begin{array}{ccccccccccl}
&&&&&1&&&&\qquad&\qquad(1+x)^0=1\\
&&&&1&&1&&&\qquad&\qquad(1+x)^1=1+1x\\
&&&1&&\color{blue}{2}&&\color{blue}{1}&&\qquad&\qquad(1+x)^2=1+\color{blue}{2}x+\color{blue}{1}x^2\\
&&1&&3&&\color{blue}{3}&&1&\qquad&\qquad(1+x)^3=1+3x+\color{blue}{3}x^2+1x^3\\
&1&&4&&6&&4&&1\qquad&\qquad(1+x)^4=1+4x+6x^2+4x^3+1x^4
\end{array}
$$ $$
Let's write the binomial coefficient $\binom{n}{k}$ as $\binom{n}{k,2}$. Using appropriate initial values we have the well-known recurrence relation
\begin{align*}
\binom{n}{k,2}=\binom{n-1}{k-1,2}+\binom{n-1}{k,2}\qquad\qquad n,k\geq 1
\end{align*}
One instance of the recursion is shown in $\rm{\color{blue}{blue}}$ in the triangle above.
$$ $$
A combinatorial interpretation is based upon lattice paths on a $\mathbb{Z}\times\mathbb{Z}$ grid with steps $(1,1),$ and $(1,-1)$. Walks of length $n$ can be represented by the generating function
\begin{align*}
\left(xy+xy^{-1}\right)^n\qquad\text{or}\qquad\left(1+x\right)^n
\end{align*}
Trinomial Coefficients
A corresponding construction with trinomial coefficients is based upon the trinomial
\begin{align*}
(1+x+x^2)^n\qquad\qquad n\geq 0
\end{align*}
we obtain for $n=0,\ldots,4$
\begin{array}{ccccccccccl}
&&&&1&&&&&&(1+x+x^2)^0
=1\\
&&&1&1&1&&&&&(1+x+x^2)^1
=1+1x+1x^2\\
&&1&2&\color{blue}{3}&\color{blue}{2}&\color{blue}{1}&&&&(1+x+x^2)^2
=1+2x+\color{blue}{3}x^2+\color{blue}{2}x^3+\color{blue}{1}x^4\\
&1&3&6&7&\color{blue}{6}&3&1&&&(1+x+x^2)^3
=1+3x+6x^2+7x^3+\color{blue}{6}x^4+3x^5+1x^6\\
1&4&10&16&19&16&10&4&1&&(1+x+x^2)^4
=1+4x+10x^2+16x^3+19x^4\\
&&&&&&&&&&\qquad\qquad\qquad\qquad\qquad+16x^5+10x^6+4x^7+1x^8
\end{array}
Let's write the trinomial coefficient as $\binom{n}{k,3}$. Using appropriate initial values we have the recurrence relation
\begin{align*}
\binom{n}{k,3}=\binom{n-1}{k-1,3}+\binom{n-1}{k,3}+\binom{n-1}{k+1,3}\qquad\qquad n,k\geq 1
\end{align*}
One instance of the recursion is shown in $\rm{\color{blue}{blue}}$ in the triangle above.
$$ $$
A combinatorial interpretation is based upon lattice paths on a $\mathbb{Z}\times\mathbb{Z}$ grid with steps $(1,1),(1,0)$ and $(1,-1)$. Walks of length $n$ can be represented by the generating function
\begin{align*}
\left(xy+x+xy^{-1}\right)^n\qquad\text{or}\qquad\left(1+x+x^2\right)^n
\end{align*}
| {
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"source": "stackexchange",
"question_score": "2",
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How many ways a 9 digit number can be formed using the digits 1 t0 9 without repetition such that it is divisble by $11$. How many ways a 9 digit number can be formed using the digits 1 t0 9 without repetition such that it is divisible by $11$.
My attempt-
A number is divisible by 11 if the alternating sum of its digit is divisible by 11?
The other thing to notice is as it is a 9 digit number formed by digits 1 to 9, exactly once each digit from 1 to 9 will appear in the number.
Basically, the question boils down to how many ways we can arrange 123456789 so that the alternating sum of the digit is divisible by 11.
I am not able to proceed further. Any help would be appreciated.
| Since the raw digit sum of $1..9$ will be $45$, an odd number, we clearly can't have two equal sums for the even-position and odd-position digits. So we need to have one set of digits sum to $17$ and the others to $28$, so that the difference is divisible by $11$.
This can be done either way around; $28$ as the sum of the four even-position digits or the sum of the five odd-position digits.
Making 28 with only four digits is the more restrictive case. The only options are {9,8,7,4} and {9,8,6,5}. With five digits there are more choices: {9,8,7,3,1}, {9,8,6,4,1}, {9,8,6,3,2}, {9,8,5,4,2}, {9,7,6,5,1}, {9,7,6,4,2}, {9,7,5,4,3}, {8,7,6,5,2}, {8,7,6,4,3}.
So we have 11 ways of dividing the digits suitably, and in each case we can permute the five odd-position digits freely and similarly the four even-position digits. So the number of possibilities is:
$$11\cdot 5!\cdot 4! = 11\times 120\times 24 = 31680$$
As a sanity check, this total is reasonably close to $\frac{9!}{11}$, ie. one-eleventh of the total permutations of the digits.
| {
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To prove the given inequality Question:-
If $a,b,c$ are positive real numbers which are in H.P. show that $$\dfrac{a+b}{2a-b} + \dfrac{c+b}{2c-b} \ge 4$$
Attempt at a solution:-
I tried it by AM-GM inequality, but got stuck at a step. My attempt was as follows:-
$$\dfrac{ \dfrac{a+b}{2a-b} + \dfrac{c+b}{2c-b}}{2} \ge \sqrt{\dfrac{a+b}{2a-b} \cdot \dfrac{c+b}{2c-b}}$$
Evaluating the right hand side of the inequality,
$$\left(\dfrac{a+b}{2a-b} \cdot \dfrac{c+b}{2c-b} \right) = \left( \dfrac{b^2+\left(a+c \right)b+ac}{b^2-2b\left(a+c \right)+4ac}\right)$$
Now, as $a,b,c$ are in H.P., hence we get the relation $$b=\dfrac{2ac}{a+c}$$ On substituting this result in the equation, we get
$$\dfrac{10 a^2c^2 + 3ac(a^2+c^2)}{4a^2c^2}=\dfrac{5}{2}+\dfrac{3(a^2+c^2)}{4ac}$$
Now from the above calculations, what I observed is that, that for the proof to be valid, the condition $\dfrac{3(a^2+c^2)}{4ac}=\dfrac{3}{2}$ should be satisfied. The final conclusion to which I arrived was $a=c$. Now although the numbers would still be in H.P. but it doesn't prove the result for the numbers which are in H.P. and are not equal.
Any other approach to the proof is invited too.
| The theorem you are asked to prove is not true in general. The additional condition needed is that $a \geq b$. And then you can do the proof without AM-GM.
Assuming $(a,b,c)$ are in H.P. in the order $$a = \frac{1}{x}, b = \frac{1}{x+d}, c = \frac{1}{x+2d}$$
then the LHS is
$$
\frac{4x^2+8dx+6d^2}{x^2+2dx}
$$
Consider the case of $x^2+2dx > 0$:
$$
\frac{4x^2+8dx+6d^2}{x^2+2dx}-4 = \frac{1}{x^2+2d}\left[4x^2+8dx+6d^2 - 4(x^2+2dx) \right] = \frac{6d^2}{x^2+2d} \geq =0
$$
Now consider the case where $x^2+2dx < 0$ and $d\neq 0$:
$$
\frac{4x^2+8dx+6d^2}{x^2+2dx}-4 = \frac{6d^2}{x^2+2d} < 0
$$
So a good counterexample would be $x=3, d=-1$, giving $a=\frac13, b = \frac12, c=1$.
If you add the condition that $a\geq b$ then you find that forces $x^2 + 2dx \geq 0$ and the statement then is true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Constant such that $\max\left(\frac{5}{5-3c},\frac{5b}{5-3d}\right)\geq k\cdot\frac{2+3b}{5-c-2d}$ What is the greatest constant $k>0$ such that
$$\max\left(\frac{5}{5-3c},\frac{5b}{5-3d}\right)\geq k\cdot\frac{2+3b}{5-c-2d}$$
for all $0\leq b\leq 1$ and $0\leq c\leq d\leq 1$?
The right-hand side looks like a weighted sum of the two terms on the left-hand side, but not quite. If we plug in $b=1$ and $c=d$, then all three terms are equal, so $k\leq 1$.
On the other hand, we have $k\geq 3/5$. Indeed, we will show that $$\frac{5}{5-3c}\geq\frac35\cdot\frac{2+3b}{5-c-2d}.$$
Note that $d\leq 1$ and $2+3b\leq 5$, so it suffices to show $$\frac{1}{5-3c}\geq\frac35\cdot\frac{1}{3-c},$$
or $$5(3-c)\geq 3(5-3c)$$
or $$15-5c\geq 15-9c$$
which is true. But the bound is not tight here, since we must have $b=d=1$ and $c=0$, and we have $\max(5, 5/2)\geq 5/3$. (The term $\frac{5b}{5-3d}$, which we did not use at all, is large.)
Update: By dividing the cases into whether $b\leq 3/5$ (and compare with the first term in the $\max$ if so) or $b\geq 3/5$ (and compare with the second term in the $\max$ if so), we can show that $k\geq 15/19$. Moreover, as Aravind pointed out in the comments, we have $k\leq 15/16$. So the gap is now between $15/19$ and $15/16$.
Update 2: WolframAlpha confirms that $k=15/16$ is the right answer. The question is now how to prove it:
http://www.wolframalpha.com/input/?i=find+minimum+of+max(5%2F(5-3c),(5b)%2F(5-3d))*(5-c-2d)%2F(2%2B3b)+for+0%3C%3Db%3C%3D1+and+0%3C%3Dc%3C%3Dd%3C%3D1
| Note: This is the pretty much the same solution as that of NP-hard.
Let $x=5-3c$, $y=5-3d$, $t=\dfrac{5}{3k}$.
Then we are looking for the minimum $t$ for which:
$min(x,\dfrac{y}{b}) \leq t\dfrac{(x+2y)}{(3b+2)}$.
with $2 \leq y \leq x \leq 5$ and $0 \leq b \leq 1$.
Consider the case $x \leq \dfrac{y}{b}$.
Then sufficient is: $t \geq \dfrac{(3b+2)x}{(x+2y)}$ and since $b \leq \dfrac{y}{x}$, it is sufficient that
$t \geq \dfrac{2x+3y}{x+2y}$ and the max value of the expression on the right is $\dfrac{16}{9}$.
Thus in this case, $t =\dfrac{16}{9}$ suffices which corresponds to $k=\dfrac{15}{16}$.
In the case when $\dfrac{y}{b} \leq x$, it is sufficient to have:
$t \geq \dfrac{(3b+2)}{(x+2y)}y/b=(3+\dfrac{2}{b})\dfrac{y}{x+2y}$ and using $b \geq \dfrac{y}{x}$, we once again get the same
calculation as above.
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $x f \cdot f'' + f \cdot f' - x (f')^2 > 0$ for certain $f(x) = 1 + x - c x^2 + x^3 + x^4$. Consider the polynomial $f(x) = 1 + x - c x^2 + x^3 + x^4$ where $c \ge 0$. Suppose that $|f(z)| < f(|z|)$ for every complex number $z \notin [0, \infty)$. How can we show that $$F(x):= x f(x)\cdot f^{\prime\prime}(x) + f(x) \cdot f^\prime(x) - x (f^\prime(x))^2 > 0$$ for all $0\le x \le 1$?
| The expression that should be positive for $0< x\leq1$ computes to
$$g(x,c):=1+9x^2 +20x^3+9x^4+x^6-cx(4+x+x^3+4x^4)\ .$$
We then can reformulate the question as follows: For which $c\geq0$ do we have
$$p(x):={1+9x^2 +20x^3+9x^4+x^6\over x(4+x+x^3+4x^4)}>c\qquad(0<x\leq1)\ ?$$
This amounts to finding the minimum of $p(x)$ in the given interval. The inherent symmetries of $p$ suggest putting
$$x+{1\over x}=: u\ ,\qquad{\rm resp.}\qquad x={1\over2}\bigl(u-\sqrt{u^2-4}\bigr)\qquad(u\geq2)\ .$$
After some calculations one finds that
$$q(u):=p\bigl(x(u)\bigr)={u^3+6u+20\over 4u^2+u-8}\qquad(u\geq2)\ .$$
A plot of $q$ shows that $q$ is unimodal in the given range with a global minimum between $u=4$ and $u=5$. The minimum is found by solving $q'(u)=0$, which leads to the third degree equation $u^3-12u-34=0$. Fortunately the unique real solution $u_*$ can be expressed as
$$u_*=\tau+2\tau^2\doteq4.4372\ ,$$
with $\tau:=2^{1/3}$. The minimum of $q(u)$ for $u\geq2$, hence the minimum of $p(x)$ for $0<x\leq1$ is then given by
$$q(u_*)={6(2\tau^2+\tau+3)\over 2\tau^2+11\tau+8}\doteq1.78191\ .$$
It follows that the desired inequality holds when $0\leq c<q(u_*)$.
| {
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How to prove by induction that $4^{2n}-3^{2n}-7$ is divisible by 84 for any n starting from 1? How to prove by induction that $4^{2n}-3^{2n}-7$ is divisible by 84 for any n starting from 1 ?
Take n=1 and prove for the base case, assume its true for some n, then the third step went like this: $$4^{2n+2}-3^{2n+2}-7=84k=4^24^{2n}-3^23^{2n}-7=84k=4^2(4^{2n}-3^{2n}-7)+7*3^{2n}+105$$ After using the assumption the expression becomes $$4^284k+63*3^n+105=84k$$ and I'm stuck here, I can't set $63*3^n+105$ to factor with 84.
| for $n=1$ we have $4^{2}-3^{2}-7=0$ and $0$ is divisible by $84$.
Let $4^{2k}-3^{2k}-7$ is divisible by $84$. We show $4^{2k+2}-3^{2k+2}-7$ is divisible by $84$.
$$4^{2k}-3^{2k}-7=84q$$
$$4^{2k+2}-3^{2k}(9+7)-7\times 16=84(16q)$$
$$4^{2k+2}-3^{2k+2}-7=84(16q+1)+7(3+3^{2k})$$
$$4^{2k+2}-3^{2k+2}-7=84(16q+1)+21(1+3^{2k-1})$$
$$4^{2k+2}-3^{2k+2}-7=84(16q+1)+21(4q')$$
$$4^{2k+2}-3^{2k+2}-7=84(16q+1)+84q'$$
Note
$${{3}^{2k-1}}+1\overset{4}{\mathop{\equiv }}\,0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometric identities: $ \frac{1+\cos(a)}{1-\cos(a)} + \frac{1-\cos(a)}{1+\cos(a)} = 2+4\cot^2(a)$ I don't really know how to begin, so if I'm missing some information please let me know what it is and I'll fill you guys in :).
This is the question I can't solve:
$$
\frac{1+\cos(a)}{1-\cos(a)} + \frac{1-\cos(a)}{1+\cos(a)} = 2+4\cot^2(a)
$$
I need to prove their trigonometric identities.
I have the $5$ basic set of rules, I could write them all here but I suppose it's not needed, if it is please let me know since it's not gonna be simple to type.
I have over $40$ questions like these and I just couldn't seem to understand how to prove them equal, my best was $4 \cot^2(a) = 2 + 4 \cot^2(a)$
Thanks for everything!
| Simplify:
$$\frac{1 + \cos a}{1 - \cos a} + \frac{1-\cos a}{1 + \cos a} \equiv \frac{(1+ \cos a)^2 + (1-\cos a)^2}{(1-\cos a)(1+\cos a)}$$
The denominator is the difference of two squares, expand the numerator:
$$\begin{align} \frac{2 + 2\cos^2 a}{1 - \cos^2 a} &\equiv \frac{2(1 + \cos^2 a)}{\sin^2 a} \\ & \equiv \frac{2}{\sin^2 x} + 2\frac{\cos^2 x}{\sin^2 x} \\ & \equiv 2\csc^2 x + 2\cot^2 x \ \\ & \equiv 2(1 + \cot^2 x) + 2\cot^2 x \\ & \equiv 2 + 4\cot^2 x\end{align}$$
Some explanations:
*
*The first line follows from $\frac{a+b}{c} \equiv \frac{a}{c} +
\frac{b}{c}$
*Since $\frac{1}{\sin x} = \csc x \Rightarrow \frac{1}{\sin^2 x}
\equiv \csc^2 x$. We used this in the second line.
*Since $\cot x \equiv \frac{1}{ \tan x} \equiv \frac{\cos x}{\sin x}$
then $\cot^2 x = \frac{\cos^2 x}{\sin^2 x}$ we used this in the
third line.
*Note that since $\sin^2 x + \cos^2 x \equiv 1$ then we can divide
both sides of this identity by $\sin^2 x$ to get $1 + \cot^2 x
\equiv \csc^2 x$, we used this in the penultimate line.
| {
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Find $\sin \theta $ in the equation $8\sin\theta = 4 + \cos\theta$ Find $\sin\theta$ in the following trigonometric equation
$8\sin\theta = 4 + \cos\theta$
My try ->
$8\sin\theta = 4 + \cos\theta$
[Squaring Both the Sides]
=> $64\sin^{2}\theta = 16 + 8\cos\theta + \cos^{2}\theta$
=> $64\sin^{2}\theta - \cos^{2}\theta= 16 + 8\cos\theta $
[Adding on both the sides]
=> $64\sin^{2}\theta + 64\cos^{2}\theta= 16 + 8\cos\theta + 65\cos^{2}\theta$
=> $64 = 16 + 8\cos\theta + 65\cos^{2}\theta$
=> $48 = 8\cos\theta + 65\cos^{2}\theta$
=> $48 = \cos\theta(65\cos\theta + 8)$
I can't figure out what to do next !
| Replacing $cos\theta=\pm\sqrt{1-sin^2\theta}$ in your equation, $8\sin\theta = 4 + \cos\theta$, and considering $u=sin\theta$, we have
$8u=4\pm\sqrt{1-u^2}$ which leads to $64u^2 +16 - 64u=1-u^2\to 65u^2-64u+15=0$.
Now, you only need to solve this equation to find $u$ that is $sin\theta$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Integrate $\int \frac{x^2-2}{(x^2+2)^3}dx$ $$\int \frac{x^2-2}{(x^2+2)^3}dx$$
What I did :
Method $(1) $ Re writing $x^2-2 = (x^2+2)-4 $ and partial fractions.
Method $(2) $ Substituting $x^2 = 2\tan^2 \theta $
Is there any other easy methods ?
Some substitution ?
| A standard method for dealing with a numerator that is a power of positive definite quadratic polynomial is the following.
Observe that for any positive integer $n$ we have
$$
D\frac{x}{(x^2+2)^n}=-\frac{(2n-1)x^2-2}{(x^2+2)^{n+1}}.\qquad(*)
$$
You can write the right hand side as a linear combination of $1/(x^2+2)^n$ and
$1/(x^2+2)^{n+1}$.
Interpreting $(*)$ as an integration formula then gives you a linear equation involving $\int (x^2+2)^{-(n+1)}\,dx$, $\int (x^2+2)^{-n}\,dx$ and $x/(x^2+2)^n$.
So:
*
*Setting $n=2$ in the resulting equation allows you to integrate $1/(x^2+2)^3$ if you know how to integrate $1/(x^2+2)^2$.
*Setting $n=1$ in the resulting equation allows you to integrate $1/(x^2+2)^2$ if you know how to integrate $1/(x^2+2)$.
*But you know how to integrate $1/(x^2+2)$, don't you?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Commutativity and $(a + b)^2$ I read "note that if a and b are commutative, $(a + b)^2 = a^2 + 2ab + b^2$".
Could someone explain how we need commutativity for this to happen?
| Expand it termwise:
$$(a+b)^2 = (a+b) \cdot (a+b) = a \cdot a + \color{blue}{a \cdot b} + \color{red}{b \cdot a} + b\cdot b$$
Commutativity assures us that $a\cdot b = b\cdot a$ so that the expression becomes
$$(a+b)^2 = a^2 + 2ab + b^2$$
Without commutativity, we have to leave our answer as it is in the first expression, we cannot simplify it to the second expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1855169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Another formula for the angle bisector in a triangle I have seen in an old geometry textbook that the formula for the length of the angle bisector at $A$ in $\triangle\mathit{ABC}$ is
\begin{equation*}
m_{a} = \sqrt{bc \left[1 - \left(\frac{a}{b + c}\right)^{2}\right]} ,
\end{equation*}
and I have seen in a much older geometry textbook that the formula for the length of the same angle bisector is
\begin{equation*}
m_{a} = \frac{2}{b + c} \sqrt{bcs(s - a)} .
\end{equation*}
($s$ denotes the semiperimeter of the triangle.)
I did not see such formulas in Euclid's Elements. Was either formula discovered by the ancient Greeks? May someone furnish a demonstration of either of them without using Stewart's Theorem and without using the Inscribed Angle Theorem?
| Here is a proof of the formula:
Let $AD$ be the angle bisector at $A$ (where $D \in BC$).
The area of $\triangle ABC$ is equal to the area of $\triangle ABD$ $+$ the area of $\triangle ACD$:
$$\frac{1}{2} bc \sin\alpha=\frac{1}{2} b m_a \sin(\alpha/2) + \frac{1}{2} c m_a \sin(\alpha/2)$$
$$m_a = \frac{2bc}{b+c}\cos(\alpha/2)$$
We can compute $\cos(\alpha/2)$ in terms of $a,b,c$ by using the cosine rule:
$$ 2\cos^2(\alpha/2) - 1=\cos\alpha=\frac{b^2 + c^2 - a^2}{2bc}$$
$$\cos^2(\alpha/2) = \frac{(b+c)^2 - a^2}{4bc} = \frac{s(s-a)}{bc} $$
Hence, we get: $$m_a = \sqrt{bc\left[ 1 - \left(\frac{a}{b+c}\right)^2\right]} = \frac{2}{b+c}\sqrt{bcs(s-a)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Solve in integers $x,y$ the equation $\left\lfloor\frac{xy-xy^2}{x+y^2} \right\rfloor=a$
Solve in integers the equation
$$\left\lfloor\frac{xy-xy^2}{x+y^2} \right\rfloor=a$$
My work so far:
1) If $a=1$, then $x\in\{-1;-2;-3\}$.
i) $x=-1 \Rightarrow y\ge-3$
ii) $x=-2 \Rightarrow y\ge3$
iii) $x=-3$. No solution.
2) $$a\le\frac{xy-xy^2}{x+y^2}<a+1$$
| HINT.-$x+y^2\ne 0$. Consider two cases $x+y^2\gt 0$ and $x+y^2\lt 0$.
$$\boxed{x+y^2\gt 0}$$
You have $$a(x+y^2)\le xy-xy^2\lt a(x+y^2)+(x+y^2)$$ This gives two quadratic inequalities in $y$
$$\begin{cases}(a+x)y^2-xy+ax\le 0\\(a+x+1)y^2-xy+ax+x\gt 0\end{cases}$$
Now for $x$ fixed, $(\alpha, \beta)$ and $(\gamma,\delta)$ being the corresponding roots supposed reals and $a+x\gt 0$ you have for the values of $y$ the restrictions $$\alpha\le y\le \beta\\y\in]-\infty,\gamma[\space\cup \space]\delta,+\infty[$$ You can see that the roots play a strong role in this and the discusion of possibilities certainly don't stop here. As you say in your comment "Perhaps the problem is not of a "normal" solutions".
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Integrate $\int\sin^2x\cos4x\,dx$ I'm having a difficult time solving this integral.
I tried integrating by parts:
$\int\sin^2x\cos4x\,dx$
$u=\sin^2x$, $dv=\cos4x\,dx$
I used the power reducing formula to evaluate $\sin^2x$
$du = 2\sin x\cos x\,dx$, $v=1/4\sin4x$
$uv - \int\ v\,du$
$\dfrac{1}{4}\sin^2x\sin4x - \dfrac{1}{2}\int\sin x\cos x\sin4x\,dx$
After this step, I tried evaluating the integral by using the $\sin a\sin b$ property.
$\dfrac{1}{4}\sin^2x\sin4x + \dfrac{1}{4}\int\cos x(\cos5x-\cos3x)\,dx$
| Use following trigonometric identities to solve this question,
$\sin^2x=(\frac{1-\cos2x}{2})$
& $\cos A\cos B=\frac{\cos(A+B)+\cos(A-B)}{2}$.
Given integration
$$I=\int \sin^2x\cos 4x\ dx$$
$$put\ \sin^2x=\frac{1-\cos 2x}{2}$$
$$I=\int \Big(\frac{1-\cos 2x}{2}\Big)\cos 4x\ dx$$
$$\Rightarrow I= \frac{1}{2}\int \Big(\cos 4x-\cos 2x \cdot \cos 4x\Big)dx$$
$$\Rightarrow I=\frac{1}{2}\int\Big(\cos 4x-\frac{\cos(2x+4x)+\cos(2x-4x)}{2}\Big)dx$$
$$\Rightarrow I=\frac{1}{2}\int\Big(\cos 4x-\frac{\cos 6x+cos(-2x)}
{2}\Big)dx$$
$$\Rightarrow I=\frac{1}{2}\int\Big(\cos 4x-\frac{\cos 6x+\cos2x}
{2}\Big)dx\ \ \ \ \ \ \ \ \ \ \because \cos(-\theta)=\cos\theta$$
$$\Rightarrow I=\frac{1}{2}\int \cos 4x\ dx-\frac{1}{4}\int \cos 6x\ dx-\frac{1}{4}\int \cos 2x\ dx$$
$$\Rightarrow I=\frac{1}{2}×\frac{\sin 4x}{4}-\frac{1}{4}×\frac{\sin 6x}{6}-\frac{1}{4}×\frac{\sin 2x}{2}+c$$
$$\Rightarrow I=\frac{\sin 4x}{8}-\frac{\sin 6x}{24}-\frac{\sin 2x}{8}+c$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Solving a system of linear congruences
Find all positive integer solutions to \begin{align*}x &\equiv -1 \pmod{n} \\ x&\equiv 1 \pmod{n-1}. \end{align*}
I rewrote the system as $x = nk_1-1$ and $x = (n-1)k_2+1$. Thus, we have $nk_1-1 = (n-1)k_2+1$ and so $n(k_1-k_2) = 2-k_2 \implies n = \frac{2-k_2}{k_1-k_2}$. How do I solve it from here?
| First note that $x_0=2n-1$ is a solution.
Now, if $x$ is a solution, then $x\equiv x_0\pmod n$ and $x\equiv x_0\pmod {n-1}$. Thus, $n|x-x_0$ and $(n-1)|x-x_0$. Also, since $\gcd(n,n-1)=1$, we have $n(n-1)|x-x_0$.
Moreover, for all non-negative integer $k$, we can check that $x=2n-1+kn(n-1)$ is a solution.
Thus, all solutions are $x=2n-1+kn(n-1)$, where $k\in \mathbb{Z}_{\geqslant 0}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find the probability of getting two sixes in $5$ throws of a die.
In an experiment, a fair die is rolled until two sixes are obtained in succession. What is the probability that the experiment will end in the fifth trial?
My work:
The probability of not getting a $6$ in the first roll is $\frac{5}{6}$
Similarly for the second and third throw. Again the probability of getting a $6$ is fourth roll is $\frac{1}{6}$. So the probability of ending the game in the fifth roll is $\frac{5^3}{6^3}\times\frac{1}{6^2}=\frac{125}{6^5}$.
But the answer is not correct. Where is my mistake? Help please.
| Here is simplified solution.
In order experiment to end at 5th trial, the last two rolls must be 6. So we have,
NNN66 $Pr=5^3/6^5$
6NN66 $Pr=5^2/6^5$
N6N66 $Pr=5^2/6^5$
$(N <6)$
Add all probabilities.
PS. The last two roll must be (6,6), so we get $\frac {1}{6^2}$
The third roll can't be 6, so we get $\frac {5}{6}$
And first two rolls can be anything except (6,6), so we get $\frac {35}{6^2}$
Answer is $\frac {175}{6^5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1859138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
Solve an overdetermined system of linear equations I have doubt to solve this system of equations
\begin{cases}
x+y=r_1\\
x+z=c_1\\
x+w=d_1\\
y+z=d_2\\
y+w=c_2\\
z+w=r_2
\end{cases}
Is it an overdetermined system because I see there are more equations than unknowns.
Can we just solve this system in a simple way?
| Your system is described by the augmented matrix
$$
A=
\left[\begin{array}{rrrr|r}
0 & 1 & 1 & 0 & r_{1} \\
0 & 1 & 0 & 1 & c_{1} \\
1 & 1 & 0 & 0 & d_{1} \\
0 & 0 & 1 & 1 & d_{2} \\
1 & 0 & 1 & 0 & c_{2} \\
1 & 0 & 0 & 1 & r_{2}
\end{array}\right]
$$
Row-reducing the system gives
$$
\DeclareMathOperator{rref}{rref}\rref A=
\left[\begin{array}{rrrr|r}
1 & 0 & 0 & 0 & -\frac{1}{2} \, c_{1} + d_{1} + \frac{1}{2} \, d_{2} - \frac{1}{2} \, r_{1} \\
0 & 1 & 0 & 0 & \frac{1}{2} \, c_{1} - \frac{1}{2} \, d_{2} + \frac{1}{2} \, r_{1} \\
0 & 0 & 1 & 0 & -\frac{1}{2} \, c_{1} + \frac{1}{2} \, d_{2} + \frac{1}{2} \, r_{1} \\
0 & 0 & 0 & 1 & \frac{1}{2} \, c_{1} + \frac{1}{2} \, d_{2} - \frac{1}{2} \, r_{1} \\
0 & 0 & 0 & 0 & c_{1} + c_{2} - d_{1} - d_{2} \\
0 & 0 & 0 & 0 & -d_{1} - d_{2} + r_{1} + r_{2}
\end{array}\right]
$$
This implies that your system is solvable if and only if
\begin{align*}
c_1+c_2-d_1-d_2 &= 0 \\
-d_1-d_2+r_1+r_2 &= 0
\end{align*}
If these conditions are satisfied, then your system is solved by
\begin{align*}
w &=-\frac{1}{2} \, c_{1} + d_{1} + \frac{1}{2} \, d_{2} - \frac{1}{2} \, r_{1} \\
x &= \frac{1}{2} \, c_{1} - \frac{1}{2} \, d_{2} + \frac{1}{2} \, r_{1}\\
y &= -\frac{1}{2} \, c_{1} + \frac{1}{2} \, d_{2} + \frac{1}{2} \, r_{1}\\
z &=\frac{1}{2} \, c_{1} + \frac{1}{2} \, d_{2} - \frac{1}{2} \, r_{1}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Problem calculating the sine of a matrix Given the matrix $A=\begin{pmatrix}-\frac{3\pi}{4} & \frac{\pi}{2}\\\frac{\pi}{2}&0\end{pmatrix}$, I want to calculate the sine $\sin(A)$.
I do so by diagonalizing A and plugging it in the power series of the sine:
\begin{align}
\sin (A) = \sum_{k=0}^\infty \frac{(-1)^k}{(2k+1)!} A^{2k+1}.
\end{align}
The diagonalization leads to:
\begin{align}
A = \frac{1}{5}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\begin{pmatrix}-\pi & 0\\0&\frac{\pi}{4}\end{pmatrix}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\end{align}
and thus:
\begin{align}
A^n = \frac{1}{5}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\begin{pmatrix}-\pi & 0\\0&\frac{\pi}{4}\end{pmatrix}^n
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}.
\end{align}
Hence:
\begin{align}
\sin (A) &= \begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\begin{pmatrix}\sin(-\pi) & 0\\0&\sin(\frac{\pi}{4})\end{pmatrix}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}\\
&= \begin{pmatrix}-2 & 1\\1&2\end{pmatrix}
\begin{pmatrix}0 & 0\\0&\frac{1}{\sqrt{2}})\end{pmatrix}
\begin{pmatrix}-2 & 1\\1&2\end{pmatrix}\\
&= \frac{1}{5}\begin{pmatrix}\frac{1}{\sqrt{2}} & \sqrt{2}\\\sqrt{2}&2\sqrt{2}\end{pmatrix},
\end{align}
which differs from "Wolfram Alpha's result"
\begin{align}
\sin(A) &= \begin{pmatrix}-\frac{1}{\sqrt{2}} & 1\\ 1 & 0 \end{pmatrix} .
\end{align}
How can this happen?
| $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\half}{{1 \over 2}}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\iff}{\Longleftrightarrow}
\newcommand{\imp}{\Longrightarrow}
\newcommand{\Li}[1]{\,\mathrm{Li}_{#1}}
\newcommand{\ol}[1]{\overline{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\ul}[1]{\underline{#1}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\,\mathsf{A} \equiv
\pars{\begin{array}{rc}
\ds{-\,{3\pi \over 4}} & \ds{\pi \over 2}
\\
\ds{\pi \over 2} & \ds{0}
\end{array}}\,,\qquad\sin\pars{\mathsf{A}} =\, ?}$
$$
\mbox{Note that}\quad\,\mathsf{A} =
-\,{3\pi \over 8}\,\sigma_{0} + {\pi \over 2}\,\sigma_{x} -
{3\pi \over 8}\,\sigma_{z} =
-\,{3\pi \over 8}\,\sigma_{0} + \vec{b}\cdot\vec{\sigma}\,,\qquad
\left\lbrace\begin{array}{rcr}
\ds{b_{x}} & \ds{=} & \ds{\pi \over 2}
\\[1mm]
\ds{b_{y}} & \ds{=} & \ds{0}
\\[1mm]
\ds{b_{z}} & \ds{=} & \ds{-\,{3\pi \over 8}}
\end{array}\right.
$$
where $\ds{\sigma_{0}}$ is the $\ds{2 \times 2}$ identity matrix. $\ds{\braces{\sigma_{i},\ i = x,y,z}}$ are the $\ds{2 \times 2}$ Pauli Matrices which satisfies
$$
\sigma_{i}^{2} = \sigma_{0}\,,\qquad
\left\lbrace\begin{array}{rcccl}
\ds{\sigma_{x}\sigma_{y}} & \ds{=} & \ds{-\sigma_{y}\sigma_{x}} & \ds{=} & \ds{\ic\sigma_{z}}
\\
\ds{\sigma_{y}\sigma_{z}} & \ds{=} & \ds{-\sigma_{z}\sigma_{y}} & \ds{=} & \ds{\ic\sigma_{x}}
\\\ds{\sigma_{z}\sigma_{x}} & \ds{=} & \ds{-\sigma_{x}\sigma_{z}} & \ds{=} & \ds{\ic\sigma_{y}}
\end{array}\right.
$$
$\ds{\expo{\mu\vec{b}\cdot\vec{\sigma}}}$ satisfies
$\ds{\pars{\partiald[2]{}{\mu} - \vec{b}\cdot\vec{b}}\expo{\mu\vec{b}\cdot\vec{\sigma}} = 0}$ with
$\ds{\left.\expo{\mu\vec{b}\cdot\vec{\sigma}}\right\vert_{\ \mu\ =\ 0} = \sigma_{0}}$ and
$\ds{\left.\partiald{\expo{\mu\vec{b}\cdot\vec{\sigma}}}{\mu}
\right\vert_{\ \mu\ =\ 0} = \vec{b}\cdot\vec{\sigma}}$ such that
$\ds{\pars{~\mbox{note that}\ \vec{b}\cdot\vec{b} = \pars{5\pi \over 8}^{2}~}}$
\begin{align}
\expo{\mu\vec{b}\cdot\vec{\sigma}} & =
\cosh\pars{{5\pi \over 8}\,\mu}\sigma_{0} +
{8 \over 5\pi}\,\sinh\pars{{5\pi \over 8}\,\mu}\vec{b}\cdot\vec{\sigma}
\\[8mm] \mbox{and}\
\expo{\mu\,\mathsf{A}} & =
\expo{-3\pi\mu/8}\,\cosh\pars{{5\pi \over 8}\,\mu}\sigma_{0} +
{8 \over 5\pi}\,\expo{-3\pi\mu/8}\,
\sinh\pars{{5\pi \over 8}\,\mu}\vec{b}\cdot\vec{\sigma}
\\[4mm] & =
\half\,\exp\pars{{\pi \over 4}\mu}
\bracks{\sigma_{0} + {8 \over 5\pi}\,\vec{b}\cdot\vec{\sigma}} +
\half\,\exp\pars{-\pi\mu}
\bracks{\sigma_{0} - {8 \over 5\pi}\,\vec{b}\cdot\vec{\sigma}}
\end{align}
\begin{align}
A^{n} & = n!\bracks{\mu^{n}}\expo{\mu\,\mathsf{A}} =
\half\pars{\pi \over 4}^{n}\bracks{%
\sigma_{0} + {8 \over 5\pi}\,\vec{b}\cdot\vec{\sigma}} +
\half\pars{-\pi}^{n}\bracks{%
\sigma_{0} - {8 \over 5\pi}\,\vec{b}\cdot\vec{\sigma}}
\end{align}
\begin{align}
\color{#f00}{\sin\pars{A}} & = \sum_{n = 0}^{\infty}{\pars{-1}^{n} \over \pars{2n + 1}!}\,
A^{2n + 1} =
\half\sin\pars{\pi \over 4}\bracks{%
\sigma_{0} + {8 \over 5\pi}\,\vec{b}\cdot\vec{\sigma}} +
\half\sin\pars{-\pi}\bracks{%
\sigma_{0} - {8 \over 5\pi}\,\vec{b}\cdot\vec{\sigma}}
\\[4mm] & =
\half\,{\root{2} \over 2}\bracks{\sigma_{0} + {8 \over 5\pi}\,\pars{A + {3\pi \over 8}\,\sigma_{0}}} =
{\root{2} \over 4}\ \underbrace{%
\bracks{{8 \over 5}\,\sigma_{0} + {8 \over 5\pi}\,A}}
_{\ds{{2 \over 5}
\pars{\begin{array}{cc}\ds{1} & \ds{2}\\ \ds{2} & \ds{4}\end{array}}}}
\\[3mm] & =
\color{#f00}{{1 \over 5}
\pars{\begin{array}{cc}\ds{1 \over \root{2}} & \ds{\root{2}}
\\[2mm]
\ds{\root{2}} & \ds{2\root{2}}
\end{array}}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860675",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Difficult Functions Evaluation Problem I have a question about finding the value of a certain function that I cannot wrap my head around.
The question is: Given a function $f(x)$ satisfying $$f(x) + 2f\left(\frac{1}{1-x}\right) = x,$$ Then find $f(2).$
So far, I have tried plugging 2 into the original equation to yield $f(2) + 2f(-1) = 2.$
Next, I plugged $\displaystyle \frac{1}{2}$ into the original equation to yield $\displaystyle f\left(\frac{1}{2}\right) + 2f(2) = \frac{1}{2}$.
However, I do not know how to solve this system of equations for $f(2).$
Please let me know of any hints you may have. Many thanks.
| Given $$f(x)+2f\left(\frac{1}{1-x}\right) = x...................(1)$$
Replace $\displaystyle x\rightarrow \frac{1}{1-x}\;,$ We get
$$f\left(\frac{1}{1-x}\right)+2f\left(\frac{x-1}{x}\right) = \frac{1}{1-x}.......(2)$$
Again replace $\displaystyle x\rightarrow \frac{1}{1-x}\;,$ We get
$$f\left(\frac{1}{1-x}\right)+2f(x) = \frac{x-1}{x}..............(3)$$
Now from $(1)$ and $(2)$
$$f(x)-4f\left(\frac{x-1}{x}\right)=x-\frac{2}{1-x}..........(4)$$
Multiply $(3)$ by $4\;,$ We get
$$4f\left(\frac{1}{1-x}\right)+8f(x) = \frac{4(x-1)}{x}..........(5)$$
Add $(4)$ and $(5)$
$$9f(x) = x-\frac{2}{1-x}+\frac{4(x-1)}{x}$$
So $$9f(2)=2+2+2=6\Rightarrow f(2) = \frac{2}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1860910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve recurrence relation $a(n) = -a(n - 2) + \cos({n} \cdot {\frac{\pi}{2}})$ Given recurrence equation
$$a(n) = -a(n - 2) + \cos({n} \cdot {\frac{\pi}{2}})$$
find the closed form solution.
Here is my attempt.
First solve the homogeneous equation:
$$a^{(0)}(n) = -a^{(0)}(n - 2)$$
My solution is:
$$a^{(0)}(n) = k_1 \cos({n} \cdot {\frac{\pi}{2}}) + k_2 \sin({n} \cdot {\frac{\pi}{2}})$$
Now the main concern is how should the particular solution look like.
I understand that if we have a recurrence equation $b(n) = c_1 b(n - 1) + c_2 b(n - 2) + 2^n$ and $f(n) = 2^n$ resonates with the solutions to $b^{(0)}(n) = c_1 b^{(0)}(n - 1) + c_2 b^{(0)}(n - 2)$, we just multiply it by $n$, so we look for $b(n) = k_3 n 2^n$.
If I repeat blindly the same multiplication, I get
$$a(n) = k_3 n \cos({n} \cdot {\frac{\pi}{2}}) + k_4 n \sin({n} \cdot {\frac{\pi}{2}})$$
Is it the correct way to look for particular solution?
If I proceed this way, I get the following general solution:
$$a(n) = k_1 \cos({n} \cdot {\frac{\pi}{2}}) + k_2 \sin({n} \cdot {\frac{\pi}{2}}) + \frac {n}{2} \cos({n} \cdot {\frac{\pi}{2}})$$
Is it the correct solution?
| The $\cos$ seems to add a lot of complexity. Why not consider instead separately the even and odd subsequences, which do not "interact" and — rewriting the recurrence relation — satisfy:
$$\begin{align}
a_{2n+2} &= -a_{2n} + \cos (n+1)\pi = -a_{2n} - (-1)^n\\
a_{2n+1} &= -a_{2n-1}
\end{align}$$
for all $n$. Solving this gives directly
$$\begin{align}
a_{2n} &= (-1)^n a_0 + (-1)^n n\\
a_{2n+1} &= (-1)^{n}a_1
\end{align}$$
(If the first one is not straightforward, you can compute the first few terms to get an idea, then show it by induction.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1862261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
If $x+y=10^{200}$ then prove that 50 divides $x$ Let $x$ be a positive integer and $y$ is another integer obtained after rearranging the digits of $x$. If $x+y=10^{200}$ then prove that $x$ is divisible by 50.
My attempt
Since $y$ is the digit rearrangement of $x$ so $x$ $\cong$ $y$ $\bmod{9}$ from there we get $x$ $\cong$ $5$ $\bmod{9}$ and $y$ $\cong$ $5$ $\bmod{9}$. Also possible last digits of $x$ and $y$ are
$(0,0), (1,9),(2,8),(3,7) , (4,6),(5,5),(6,4),(7,3),(8,2),(9,1)$.
For last digits $(0,0), (2,8), (4,6),(6,4),(8,2)$ divisibility by $2$ is ensured but divisibility by $25$ and general case is eluding me. Can someone help? Thanks in advance.
| We can assume $x$ and $y$ are non-zero. So, with suitable initial $0$-padding, each has $200$ digits.
If $x$ ends in $00$ we are finished. Suppose now that $x$ ends in $0$ but not $00$. Then the next to last digits of $x$ and $y$ are $10$'s complements of each other, and non-zero. Each of $x$ and $y$ has $198$ digits that are $9$'s complements of each other. These come in pairs, since the digits are permuted. So the next to last digit of $x$ and $y$ are equal, and therefore each is $5$.
And we cannot have last digit non-zero, for in the rest of $x$ and $y$ the digits come in $9$'s complement pairs, and $199$ is odd.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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} |
Prove the following trigonometric result If $\theta_1,\theta_2(0\leq\theta_1,\theta_2<2\pi)$ are two solutions of $\sin(\theta+\phi)=\frac{1}{2}\sin(2\phi)$, prove that
$$\frac{\sin(\theta_1)+ \sin(\theta_2) }{ \cos(\theta_1)+ \cos(\theta_2)} =\cot\phi$$
I have tried with the following process:
Since
$\theta_1,\theta_2(0\leq\theta_1,\theta_2<2\pi)$ are two solutions of $\sin(\theta+\phi)=\frac{1}{2}\sin(2\phi)$, we have
$\sin(\theta_i+\phi)=\frac{1}{2}\sin(2\phi)$, I=1,2.
This gives
$\sin\theta_i\cos\phi+ \cos\theta_i \sin\phi= \frac{1}{2}\sin(2\phi)$, I=1,2.
I don't know what will be the next process.
| $$\sin(\theta+\phi)=\frac{1}{2}\sin(2\phi)$$
$$\sin\theta \cos \phi + \sin\phi \cos \theta = \sin \phi \cos\phi$$
$$\sin\theta \cot \phi + \cos \theta = \cos\phi$$
$$\sin\theta \cot \phi - \cos \phi = -\cos\theta$$
Squaring both sides, we get
$$\sin^2\theta \cot^2 \phi + \cos^2 \phi - 2\sin\theta \cot \phi \cos \phi= \cos^2\theta=1-\sin^2\theta \tag1$$
$$\sin^2\theta (1+\cot^2 \phi) + \cos^2 \phi - 2\sin\theta \cot \phi \cos \phi -1=0$$
$$\sin^2\theta \csc^2 \phi - 2\sin\theta \cot \phi \cos \phi - \sin^2 \phi =0$$
$$\sin^2\theta - 2\sin\theta \sin \phi \cos^2 \phi - \sin^4 \phi =0$$
Hence $\sin \theta_1+\sin \theta_2=2 \sin \phi \cos^2 \phi$
and accordingly, if instead of converting the $\cos$ to $\sin$ in $(1)$, you convert the $\sin$ to $\cos$ , you will get $\cos \theta_1+\cos \theta_2=2 \sin^2 \phi \cos \phi$
The answer will follow. Hope this is clear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For what value of $(a+b)$ will all roots of $f(x)=x^4-8x^3+ax^2+bx+16$ be positive? I was thinking of using Descartes' rule of signs, from which I find there are at most 2 positive roots and 2 negative roots of the given equation.
Also, $f(\infty)>0$ and $f(0)>0$ imply that either there are no real roots or 2 real roots in $(0,\infty)$. Similar is the case for $(-\infty,0)$.
If I could factor $f'(x)=4x^3-24x^2+2ax+b$ and find its roots, then the real and distinct roots of $f(x)$ will be separated by those of $f'(x)$. How do I solve the problem?
| Let us consider first the case when $b=-32$.
$$f_1(x)=x^4-8x^3+ax^2-32x+16=0\iff a=\frac{-x^4+8x^3+32x-16}{x^2}$$
Now, let
$$g(x):=\frac{-x^4+8x^3+32x-16}{x^2}$$
Then,
$$g'(x)=\frac{-2 (x-2)^3 (x+2)}{x^3}$$
So, $g(x)$ is increasing for $x\lt -2$ or $0\lt x\lt 2$, and is decreasing for $-2\lt x\lt 0$ or $x\gt 2$ with $\lim_{x\to\pm\infty}g(x)=-\infty,\lim_{x\to 0^{\pm}}g(x)=-\infty$.
Now considering the graph of $y=g(x)$ gives that
$$\begin{align}&\text{all roots of $f_1(x)$ are positive}\\&\implies \text{all $x$ coordinates of the intersection points of $y=a$ with $y=g(x)$ are positive}\\&\implies a=24\ (=g(2))\\&\implies a+b=-8\end{align}$$
By the way, for $a=28, b=-36$ where $a+b=-8$,
$$f_2(x)=x^4-8x^3+28x^2-36x+16$$
$$f'_2(x)=4x^3-24x^2+56x-36=4(x-1)\left(\left(x-\frac 52\right)^2+\frac{11}{4}\right)$$
so $f_2(x)$ is decreasing for $x\lt 1$ and is increasing for $x\gt 1$ with $f_2(1)=1\gt 0$ from which $f_2(x)\gt 0$ follows. Hence, $f_2(x)$ has no real roots.
Therefore, there are no $a+b$ such that all roots of $f(x)$ are positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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On real part of the complex number $(1+i)z^2$
Find the set of points belonging to the coordinate plane $xy$, for which the real part of the complex number $(1+i)z^2$ is positive.
My solution:-
Lets start with letting $z=r\cdot e^{i\theta}$. Then the expression $(1+i)z^2$ becomes $$\large\sqrt2\cdot|z|^2\cdot e^{{i}\left(2\theta+\dfrac{\pi}{4}\right)}$$
Now, as $\sqrt2\cdot|z|^2\gt0$, so $\Re{((1+i)z^2)}\gt 0 \implies\cos{\left(2\theta+\dfrac{\pi}{4}\right)}\gt 0$. So, we get
$$-\dfrac{\pi}{2}\lt\left(2\theta+\dfrac{\pi}{4}\right)\lt\dfrac{\pi}{2}
\implies-\dfrac{3\pi}{4}\lt2\theta\lt\dfrac{\pi}{4}
\implies-\dfrac{3\pi}{8}\lt\theta\lt\dfrac{\pi}{8}$$
Now, lets find the equation of the lines which would help us show these inequalities in the coordinate plane.
The inequality can be represented by
$$\begin{equation}
y\lt \tan{\dfrac{\pi}{8}}x\implies y\lt(\sqrt2-1)x \tag{1}
\end{equation}$$
$$\begin{equation}
y\gt \tan{(-\dfrac{3\pi}{8})}x \implies y\gt-(\sqrt2+1)x \tag{2}
\end{equation}$$
So, the inequality can be represented in the coordinate plane as in the following portion of the graph with the cross-hatched part.
My deal with the question:-
The book I am solving gives the answer as the (cross-hatched part + un-hatched part), so what am I missing in my solution. And, as always more elegant solutions are welcome.
|
what am I missing in my solution
After having $\cos\left(2\theta+\frac{\pi}{4}\right)\gt 0$, you have
$$-\dfrac{\pi}{2}\lt\left(2\theta+\dfrac{\pi}{4}\right)\lt\dfrac{\pi}{2}$$
which is incorrect.
To make it easy to understand why this is incorrect, let $\alpha:=2\theta+\frac{\pi}{4}$.
Then, we want to solve
$$\cos\alpha\gt 0\quad\text{and}\quad -\pi\le\theta\lt \pi,$$
i.e.
$$\cos\alpha\gt 0\quad\text{and}\quad -\frac{7}{4}\pi\le\alpha\lt \frac{9}{4}\pi$$
which sould be easier to solve, to have
$$-\frac{7}{4}\pi\le \alpha\lt -\frac{3}{2}\pi\quad\text{or}\quad -\frac{\pi}{2}\lt\alpha\lt \frac{\pi}{2}\quad\text{or}\quad \frac 32\pi\lt\alpha\lt\frac{9}{4}\pi,$$
i.e.
$$-\pi\le \theta\lt -\frac78\pi\quad\text{or}\quad -\frac 38\pi\lt \theta\lt\frac{\pi}{8}\quad\text{or}\quad \frac{5}{8}\pi\lt \theta\lt\pi$$
as always more elegant solutions are welcome
(not sure if this is more elegant, but) another solution :
Let $z=x+iy$ where $x,y\in\mathbb R$. Then,
$$\Re((1+i)z^2)=\Re((1+i)(x+iy)^2)=x^2-y^2-2xy\tag1$$
When we solve $y^2+2xy-x^2=0$ for $y$, we get
$$y=-x\pm\sqrt{x^2+x^2}=-x\pm\sqrt 2\ x=(\pm\sqrt 2-1)x$$
so from $(1)$,
$$\begin{align}&\Re((1+i)z^2)\gt 0\\&\iff ((\sqrt 2-1)x-y)((\sqrt 2+1)x+y)\gt 0\\&\iff -(\sqrt 2+1)x\lt y\lt (\sqrt 2-1)x\quad\text{or}\quad (\sqrt 2-1)x\lt y\lt -(\sqrt 2+1)x\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find Area Enclosed by Curve I want to find the area enclosed by the plane curve $x^{2/3}+y^{2/3}=1$. My attempt was to set $x=\cos^3t, \ y=\sin^3t$ so:$$x^{2/3}+y^{2/3}=\cos^2t+\sin^2t=1$$
Then the area is $$2A=\oint_Cxdy-ydx=3\oint_C\cos^3ty'dy+\sin^3tx'dx=3\int_0^{2\pi}\cos^2t\cdot \sin^2tdt=\frac{3\pi}{4}\implies A=\frac{3\pi}{8}$$
However, when I did a level curve plot I got the following figure:
so does "area enclosed by figure" even make sense? For the graph above, my calculator gives me $A=\frac{3\pi}{32}$.
| Notice that we have symmetry about the $y$ and $x$ axes. Hence we can use the formula
$$y=(1-x^{2/3})^{3/2}$$
Which gives the top half, integrate from $x=0$ to $1$, and then multiply by $4$, exploiting the symmetry. Hence we have
$$A=4\int_0^1(1-x^{2/3})^{3/2}dx$$
Letting $x=\cos(u)^{6/2}=\cos^3(u)$, $dx=-3\cos^2(u)\sin(u)du$ we have
$$\begin{align}
A &=4\cdot 3\int_0^{\pi/2}\sin^4(u)\cos^2(u)du \\
&= 4\cdot\frac{3\pi}{32}
\end{align}$$
Hence we have the area $3\pi/32$ like you got, but we must account for the other pieces.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1868613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Show that $\log(1+y) \approx y- \frac {y^2}2 + \cdots$ without Taylor Series
For small $y$, prove that $\log(1+y)\approx y -\frac {y^2}2 + \cdots $
I have no idea to solve it.
| One approach directly based on a definition of $\log x$ is as follows. We use the following definition of $\log x$ $$\log x = \lim_{n \to \infty}n(x^{1/n} - 1)$$ Replacing $x$ by $(1 + x)$ and for $|x| < 1$ using binomial theorem we get
\begin{align}
\log (1 + x) &= \lim_{n \to \infty}n((1 + x)^{1/n} - 1)\notag\\
&= \lim_{n \to \infty}n\left(\frac{x}{1!}\frac{1}{n} - \frac{x^{2}}{2!}\frac{1}{n}\left(1 - \frac{1}{n}\right) + \frac{x^{3}}{3!}\frac{1}{n}\left(1 - \frac{1}{n}\right)\left(2 - \frac{1}{n}\right) + \cdots\right) \notag\\
&= \lim_{n \to \infty}\left(x - \frac{x^{2}}{2!}\left(1 - \frac{1}{n}\right) + \frac{x^{3}}{3!}\left(1 - \frac{1}{n}\right)\left(2 - \frac{1}{n}\right) + \cdots\right)\notag\\
&= x - \frac{x^{2}}{2} + \frac{x^{3}}{3} - \cdots\notag
\end{align}
The last step where I took limit of an infinite series term by term requires some justification which might not be suitable at your stage of learning (as guessed from the content of question and comments to it). Moreover since you need only an approximation it is better to just take the terms till $x^{2}$ in the binomial expansion and then take limit.
Another approach is available here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1869449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $\sum_{r=0}^{\left\lfloor\frac{n-1}3\right\rfloor}\binom{n}{3r+1}$
Show that $$\binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\ldots=\dfrac{1}{3}\left[ 2^{n-2} + 2\cos{\dfrac{(n-2)\pi}{3}}\right]$$
My solution:-
$$(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+\binom{n}{3}x^3+\ldots=\sum_{r=0}^{n}{\binom{n}{r}x^r} \\ \therefore x^2(1+x)^n=\binom{n}{0}x^2+\binom{n}{1}x^3+\binom{n}{2}x^4+\binom{n}{3}x^5+\ldots=\sum_{r=0}^{n}{\binom{n}{r}x^{r+2}}$$
In the above Binomial Expansion on substituting $x=1,\omega,\omega^2$, $\omega$ being a complex cube root of unity, we get the following three equations
$$\tag{1}(1)^2(1+1)^n=\sum_{r=0}^{n}{\binom{n}{r}}=2^n$$
$$(\omega)^2(1+\omega)^n=\sum_{r=0}^{n}{\binom{n}{r}\omega^{r+2}}=(-1)^n(\omega)^{2n+2} \tag{2}$$
$$(\omega)^4(1+\omega^2)^n=(\omega)(1+\omega^2)^n=\sum_{r=0}^{n}{\binom{n}{r}\omega^{2r+4}}=(-1)^n(\omega)^{n+1} \tag{3}$$
On adding $(1),(2) \text{ and }(3)$, we get
$$\dfrac{1}{3}\left(2^n+(-1)^n(\omega^{n+1}+\omega^{2n+2})\right)=\sum_{r=0}^{\left\lfloor\frac{n-1}3\right\rfloor}\binom{n}{3r+1}$$
Now, as $\omega=-e^{i(\pi/3)}$ ($\omega$ being the cube root of unity)
$$\begin{aligned}
\therefore (\omega^{n+1} +\omega^{2n+2})
&= \left(\left(-e^{i(\pi/3)}\right)^{n+1}+\left(-e^{-i(\pi/3)}\right)^{n+1}\right) \\
&=(-1)^{n+1}\left(e^{i(\pi(n+1)/3)}+e^{-i(\pi(n+1)/3)}\right) \\
&=(-1)^n\left(2\cos{\left(\dfrac{\pi(n+1)}{3}\right)}\right)
\end{aligned}$$
Now, substituting the value of $(\omega^{n+1}+\omega^{2n+2})$ back into $(4)$, we get
$$2^n+(-1)^n(\omega^{n+1}+\omega^{2n+2})=2^n-2\cos{\left(\dfrac{\pi(n+1)}{3}\right)}=\sum_{r=0}^{n}{\binom{n}{3r+1}}$$
$$\therefore \sum_{r=0}^{\left\lfloor\frac{n-1}3\right\rfloor}\binom{n}{3r+1}=\boxed{\dfrac{1}{3}\left(2^n-2\cos{\left(\dfrac{\pi(n+1)}{3}\right)}\right)}$$
So, where did I go wrong, or is it that the book has provided the wrong answer.
| Let $\alpha=e^{2\pi i/3}=\frac{-1+i\sqrt3}2$, that is $\alpha^3=1$, then if $k\not\equiv0\pmod3$, $\alpha^k+1+\alpha^{-k}=\frac{\alpha^{3k}-1}{\alpha^k\left(\alpha^k-1\right)}=0$. If $k\equiv0\pmod3$, then $\alpha^k+1+\alpha^{-k}=1+1+1=3$. That is,
$$
\frac{\alpha^{k-1}+1+\alpha^{1-k}}3=\left\{\begin{array}{}
1&\text{if }k\equiv1\pmod3\\
0&\text{if }k\not\equiv1\pmod3
\end{array}\right.
$$
Furthermore,
$$
1+\alpha=\frac{1+i\sqrt3}2=e^{\pi i/3}
$$
Therefore,
$$
\begin{align}
\sum_{k=0}^{\left\lfloor\frac{n-1}3\right\rfloor}\binom{n}{3k+1}
&=\sum_{k=0}^n\binom{n}{k}\frac{\alpha^{k-1}+1+\alpha^{1-k}}{3}\\
&=\frac1{3\alpha}(1+\alpha)^n+\frac13\cdot2^n+\frac\alpha3\left(1+\frac1\alpha\right)^n\\
&=\frac1{3\alpha}e^{\pi in/3}+\frac13\cdot2^n+\frac\alpha3e^{-\pi in/3}\\
&=\frac13e^{\pi i(n-2)/3}+\frac13\cdot2^n+\frac13e^{-\pi i(n-2)/3}\\
&=\frac13\left(2^n+2\cos\left(\pi\frac{n-2}3\right)\right)\\
&=\frac13\left(2^n-2\cos\left(\pi\frac{n+1}3\right)\right)
\end{align}
$$
As far as the periodic part goes, neither is wrong: $\cos\left(\pi\frac{n-2}3\right)=-\cos\left(\pi\frac{n+1}3\right)$.
However,
$$
\sum_{k=0}^{\left\lfloor\frac{n}3\right\rfloor}\binom{n}{3k}
+\sum_{k=0}^{\left\lfloor\frac{n-1}3\right\rfloor}\binom{n}{3k+1}
+\sum_{k=0}^{\left\lfloor\frac{n-2}3\right\rfloor}\binom{n}{3k+2}
=2^n
$$
and since each of the sums above are approximately equal, the non-periodic part of the sum should be $\frac13\cdot2^n$.
| {
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"url": "https://math.stackexchange.com/questions/1869622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Length of side of biggest square inscribed in a triangle I have seen that the length of each side of the biggest square that can be inscribed in a right triangle is half the harmonic mean of the legs of the triangle. I have not seen a rigorous explanation for it, though. I would appreciate such an explanation. (This is intriguing - it is an optimization problem that does not require Calculus to explain.)
| Let we assume that the legs $CB$ and $CA$ of our right triangle $ACB$ have lengths $a$ and $b$.
First case: we consider the largest inscribed with a vertex at $C$. Assuming that its side length is $l$,
by triangle similarities we have
$$ c = \sqrt{a^2+b^2} = l\cdot\left(\frac{c}{b}+\frac{c}{a}\right), $$
from which $l=\frac{1}{2}HM(a,b)$.
Second case: we consider the largest inscribed square with a side on $AB$. By triangle similarities we have:
$$ c = l+l\cdot\left(\frac{a}{b}+\frac{b}{a}\right)=l\cdot\frac{ab+c^2}{ab} $$
from which $l=\frac{abc}{ab+c^2}=\frac{ab\sqrt{a^2+b^2}}{a^2+ab+b^2}$. I leave to you to check if the inequality
$$ \frac{ab}{a+b}\geq \frac{ab\sqrt{a^2+b^2}}{a^2+ab+b^2} $$
holds or not. Hint:
$$ (a^2+ab+b^2)^2-(a+b)^2(a^2+b^2) = a^2b^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum to infinity of trignometry inverse: $\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)$ If we have to find the value of the following (1)
$$
\sum_{r=1}^\infty\arctan \left(\frac{4}{r^2+3} \right)
$$
I know that
$$
\arctan \left(\frac{4}{r^2+3} \right)=\arctan \left(\frac{r+1}2 \right)-\arctan \left(\frac{r-1}2 \right)
$$
I tried it lot and got a result but then stuck! (2)
(1) http://i.stack.imgur.com/26hA4.jpg
(2) http://i.stack.imgur.com/g2vBb.jpg
| $$\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=1}^N\arctan\left(\frac{r-1}2\right)=\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=-1}^{N-2}\arctan\left(\frac{r+1}2\right)$$
$$=\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\sum_{r=1}^N\arctan\left(\frac{r+1}2\right)-\arctan\left(0\right)-\arctan\left(\frac{1}2\right)+\arctan\left(\frac{N}2\right)+\arctan\left(\frac{N+1}2\right)$$
Then let $N\rightarrow \infty$...
You then get $$\boxed {\pi -\arctan\left(\frac12\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1870535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Show that $3^n+4^n+\dots+(n+2)^n=(n+3)^n$ has no answers for $n\ge 6$. Considering
$$3^n+4^n+\dots+(n+2)^n=(n+3)^n$$
Clearly $n=2$ and $n=3$ are solutions of this equation and this equality does not hold for $n=4$ and $n=5$.
How can I show this equation has no solutions for $n>5$.
Thanks.
| A direct proof for $n\ge7$:
One can very easily show by induction $\sum_{k=1}^{n-1} k^n<n^n$. This implies \begin{align} \sum_{k=3}^{n+2}\left(\frac{k}{n}\right)^n &=1+\left(1+\frac1n\right)^n+\left(1+\frac2n\right)^n+\sum_{k=3}^{n-1} \left(\frac{k}{n}\right)^n \\ &< 2+e+e^2<\left(\frac{10}{7}\right)^7\le\left(1+\frac3n\right)^n.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Solve $\cos^n x + \sin^n x =1 $ the solutions of this equation as a function of the value of $n$??
\begin{align}
\cos^n x + \sin^n x =1
\end{align}
I already found the solution if n is odd,
| If $n$ is odd, we can write it as:$n = 2k+1, k \ge 1$, $1 = |\sin^{2k+1}x + \cos^{2k+1}x| \le |\sin x|\cdot \sin^{2k}x + |\cos x|\cdot \cos^{2k}x \le \sin^{2k}x+\cos^{2k}x \le \sin^2 x + \cos^2x = 1$. thus $|\sin x| = 1, 0$ , and you can find $x$ from this. If $n = 1$, then $\sin x + \cos x = 1\implies (\sin x+\cos x)^2 = 1 \implies \sin (2x) = 0 \implies 2x = m\pi \implies x = \dfrac{m\pi}{2}, m \in \mathbb{Z}$. If $n$ is even, then $n \ge 2 \implies 1 = \sin^2x + \cos^2 x \ge \sin^n x+\cos^n x = 1\implies \cos^2 x = 1, 0 \implies \cos x = 0, \pm 1 \implies x = m\pi, \pm\dfrac{\pi}{2} + 2m\pi, m \in \mathbb{Z}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1874442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Integration by means of partial fraction decomposition I'm trying to solve this indefinite integral by means of partial fraction decomposition:
$\int\dfrac{x+1}{\left(x^2+4x+5\right)^2}\ dx$.
The denominator has complex (but not real) roots because $\Delta<0$; so, according with my calculus book, i try to decompose the integrand function in this form:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{Ax+B}{\left(x^2+4x+5\right)}+\dfrac{Cx+D}{\left(x^2+4x+5\right)^2}$.
I get:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{\left(Ax+B\right)\left(x^2+4x+5\right)+Cx+D}{\left(x^2+4x+5\right)^2}$.
Multiplying the right term:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{Ax^3+4Ax^2+5Ax+Bx^2+4Bx+5B+Cx+D}{\left(x^2+4x+5\right)^2}$.
Now i collect the terms with the same pwer of $x$:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{Ax^3+\left(4A+B\right)x^2+\left(5A+4B+C\right)x+D+ 5B}{\left(x^2+4x+5\right)^2}$.
Now i equate the two numerators:
$x+1=Ax^3+\left(4A+B\right)x^2+\left(5A+4B+C\right)x+D$
and equate term by term: i get:
$A=0$, $B=0$, $C=1$, $D=1$.
With these values i get a correct identity:
$\dfrac{x+1}{\left(x^2+4x+5\right)^2}=
\dfrac{x+1}{\left(x^2+4x+5\right)^2}$
but this is unuseful in order to solve the integral.
Where is my mistake ?
| Note that we can write the partial fraction expansion of $\frac{1}{(x+4x+5)^2}$ as
$$\frac{1}{(x+4x+5)^2}=\frac{1}{(x-r)^2(x-r^*)^2}=\frac{A}{x-r}+\frac{B}{(x-r)^2}+\frac{C}{x-r^*}+\frac{D}{(x-r^*)^2}$$
where $r=-2+i$ and $r^*=-2- i$.
Multiplying by $(x-r)^2$ and letting $x\to r$ reveals that $B=\frac{1}{(r-r^*)^2}=-\frac14$.
Multiplying by $(x-r)^2$, taking a derivative with respect to $x$, and letting $x\to r$ reveals that $A=-\frac{2}{(r-r^*)^3}=-\frac{i}{4}$.
Multiplying by $(x-r^*)^2$ and letting $x\to r^*$ reveals that $D=\frac{1}{(r^*-r)^2}=-\frac14$.
Multiplying by $(x-r^*)^2$, taking a derivative with respect to $x$, and letting $x\to r^*$ reveals that $C=\frac{2}{(r-r^*)^3}=\frac{i}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If the roots of the equation are integers then find the value of $k$. The question says roots of $x^2-kx+36=0$ are integers then the number of values for $k$ are?
I know roots are integral if discriminant is a perfect square of an integer, but using this I get infinite values for $k$. Which values should I reject?
| HINT:
Let $k^2-4\cdot1\cdot36=y^2\iff(k+y)(k-y)=144$
As $k+y\pm(k-y)$ are even, $k-y,k+y$ have the same parity, hence both must be even
$$\dfrac{k+y}2\cdot\dfrac{k-y}2=\dfrac{144}4=?$$
What are the positive divisors of $36?$
Also as $k+y\ge k-y,36=\dfrac{k+y}2\cdot\dfrac{k-y}2\le\dfrac{(k+y)^2}4$
$\implies k+y\ge\sqrt{4\cdot36}$ as $k+y>0$
$\implies\dfrac{k+y}2\ge6$ and must divide $36$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875694",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value $\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$ The value of $$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{5 - 2{\sqrt6}}}{1+ \sqrt{6}}\right)$$is equal to
*
*$\frac{\pi}{6}$
*$\frac{\pi}{4}$
*$\frac{\pi}{3}$
*$\frac{\pi}{12} $
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) - \tan^{-1}\left(\frac{\sqrt{3} -\sqrt{2}}{1+ \sqrt{6}}\right)$$
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) -\tan^{-1}{\sqrt3} + \tan^{-1} {\sqrt2} $$
$$\implies\frac{\pi}{2} -\frac{\pi}{3}=\frac{\pi}{6}$$
Another possibility is
$$\tan^{-1}\left(\frac{1}{\sqrt2}\right) +\tan^{-1}{\sqrt3} - \tan^{-1} {\sqrt2} $$
How to solve this ?
| One should be careful when using identities related to the arctangent, because $\arctan\tan t=t$ only holds for $-\pi/2<t<\pi/2$. On the other hand, $\tan\arctan s=s$ holds without any condition.
Set
$$
\alpha=\arctan\frac{1}{\sqrt{2}},
\qquad
\beta=\arctan\frac{\sqrt{5-2\sqrt{6}}}{1+ \sqrt{6}}=
\arctan\frac{\sqrt{3}-\sqrt{2}}{1+ \sqrt{6}}
$$
Note that
$$
1\cdot(1+\sqrt{6})>\sqrt{2}\,(\sqrt{3}-\sqrt{2})
$$
so $\alpha>\beta$ and $0<\alpha-\beta<\pi/2$. Then
$$
\tan(\alpha-\beta)=
\frac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}=
\frac{\dfrac{1}{\sqrt{2}}-\dfrac{\sqrt{3}-\sqrt{2}}{1+ \sqrt{6}}}
{1+\dfrac{1}{\sqrt{2}}\dfrac{\sqrt{3}-\sqrt{2}}{1+ \sqrt{6}}}=
\frac{1}{\sqrt{3}}
$$
Since $0<\alpha-\beta<\pi/2$, we can conclude
$$
\alpha-\beta=\arctan\frac{1}{\sqrt{3}}=\frac{\pi}{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1875861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Find the least squares solution for rank deficient system
Find the least squares solution to the system
$$x - y = 4$$
$$x - y = 6$$
Normally if I knew what the matrix $A$ was and what $b$ was I could just do $(A^TA)^{-1} A^Tb$, but in this case I'm not sure how to set up my matrices. How can I find the least square solution to the system?
| Problem statement: underdetermined system
Start with the linear system
$$
\begin{align}
\mathbf{A} x &= b \\
%
\left[
\begin{array}{cc}
1 & -1 \\
1 & -1 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
x \\
y
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
4 \\
6
\end{array}
\right]
%
\end{align}
$$
The system has matrix rank $\rho = 1$; therefore, if a solution exists, it will not be unique.
Provided $b\notin \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)}$, we are guaranteed a least squares solution
$$
x_{LS} = \left\{ x\in\mathbb{C}^{2} \colon \lVert \mathbf{A} x_{LS} - b \rVert_{2}^{2} \text{ is minimized} \right\}
\tag{1}
$$
Subspace resolution
By inspection, we see that the row space is resolved as
$$
\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)} \oplus
\color{red}{\mathcal{N} \left( \mathbf{A} \right)}
=
\color{blue}{\left[
\begin{array}{r}
1 \\
-1
\end{array}
\right]} \oplus
\color{red}{\left[
\begin{array}{c}
1 \\
1
\end{array}
\right]}
$$
The column space is resolved as
$$
\color{blue}{\mathcal{R} \left( \mathbf{A} \right)} \oplus
\color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)}
=
\color{blue}{\left[
\begin{array}{c}
1 \\
1
\end{array}
\right]} \oplus
\color{red}{\left[
\begin{array}{r}
-1 \\
1
\end{array}
\right]}
$$
The coloring indicates vectors in the $\color{blue}{range}$ space and the $\color{red}{null}$ space.
Finding the least squares solution
Since there is only one vector in $\color{blue}{\mathcal{R} \left( \mathbf{A}^{*} \right)}$, the solution vector will have the form
$$
\color{blue}{x_{LS}} = \alpha
\color{blue}{\left[
\begin{array}{r}
1 \\
-1
\end{array}
\right]}
$$
The goal is to find the constant $\alpha$ to minimize (1):
$$
\color{red}{r}^{2} = \color{red}{r} \cdot \color{red}{r} =
\lVert
\color{blue}{\mathbf{A} x_{LS}} - b
\rVert_{2}^{2}
=
8 \alpha ^2-40 \alpha +52
$$
The minimum of the polynomial is at
$$
\alpha = \frac{5}{2}
$$
Least squares solution
The set of least squares minimizers in (1) is then the affine set given by
$$
x_{LS} = \frac{5}{2}
\color{blue}{\left[
\begin{array}{r}
1 \\
-1
\end{array}
\right]}
+
\xi
\color{red}{\left[
\begin{array}{r}
1 \\
1
\end{array}
\right]}, \qquad \xi\in\mathbb{C}
$$
The plot below shows how the total error $\lVert \mathbf{A} x_{LS} - b \rVert_{2}^{2}$ varies with the fit parameters. The blue dot is the particular solution, the dashed line homogeneous solution as well as the $0$ contour - the exact solution.
Addendum: Existence of the Least Squares Solution
To address the insightful question of @RodrigodeAzevedo, consider the linear system:
$$
\begin{align}
\mathbf{A} x &= b \\
%
\left[
\begin{array}{cc}
1 & 0 \\
0 & 0 \\
\end{array}
\right]
%
\left[
\begin{array}{c}
x \\
y
\end{array}
\right]
%
&=
%
\left[
\begin{array}{c}
0 \\
1
\end{array}
\right]
%
\end{align}
$$
The data vector $b$ is entirely in the null space of $\mathbf{A}^{*}$:
$b\in \color{red}{\mathcal{N} \left( \mathbf{A}^{*} \right)}$
As pointed out, the system matrix has the singular value decomposition. One instance is:
$$\mathbf{A} = \mathbf{U}\, \Sigma\, \mathbf{V}^{*} = \mathbf{I}_{2}
\left[
\begin{array}{cc}
1 & 0 \\
0 & 0 \\
\end{array}
\right]
\mathbf{I}_{2}$$
and the concomitant pseudoinverse,
$$\mathbf{A}^{\dagger} = \mathbf{V}\, \Sigma^{\dagger} \mathbf{U}^{*} =
\mathbf{I}_{2}
\left[
\begin{array}{cc}
1 & 0 \\
0 & 0 \\
\end{array}
\right]
\mathbf{I}_{2} = \mathbf{A}$$
Following least squares canon, the particular solution to the least squares problem is computed as
$$
\color{blue}{x_{LS}} = \mathbf{A}^{\dagger} b =
\color{red}{\left[
\begin{array}{c}
0 \\
0 \\
\end{array}
\right]}
\qquad \Rightarrow\Leftarrow
$$
The color collision (null space [red] = range space [blue]) indicates a problem. There is no component of a particular solution vector in a range space!
Mathematicians habitually exclude the $0$ vector a solution to linear problems.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find $\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $ Find $$\int \frac{x^2}{(x \cos x - \sin x)(x \sin x + \cos x)}dx $$
Any hints please?
Could'nt think of any approach till now...
| $$ \begin{aligned}
\int\frac{x^{2}}{(x\cos x - \sin x)(x\sin x + \cos x)}\,\mathrm{d}x&=\int\left(\frac{x\sin x + \cos x}{x\cos x - \sin x}\right)^{-1}\frac{x^{2}}{(x\cos x - \sin x)^{2}}\,\mathrm{d}x\\&=\int\left(\frac{x\sin x + \cos x}{x\cos x - \sin x}\right)^{-1}\,\mathrm{d}\!\left(\frac{x\sin x + \cos x}{x\cos x - \sin x}\right)\\
&=\log\left|\frac{x\sin x + \cos x}{x\cos x - \sin x}\right|+C.
\end{aligned} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1878650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
What is the $\lim_\limits{x \to 1+}\left(\frac{3x}{x-1}-\frac{1}{2 \ln(x)}\right)$?
What is the limit of $$\lim_\limits{x \to 1+} \left(\frac{3x}{x-1}-\frac{1}{2 \ln(x)} \right)$$
I attemped the problem using L^Hopital's Rule.
My Work
$$\lim_\limits{x \to 1+} \left(\frac{3x}{x-1}-\frac{1}{2 \ln(x)} \right)$$
$$\lim_\limits{x \to 1+} \left(-\frac{3}{(x-1)^2}+\frac{1}{2 \ln^2(x)\cdot x} \right)$$
$$\frac{-3+1}{0}=\frac{-2}{0}$$
The answer is suppose to be $\infty$. I know what I did is probably not right.
| It is easier to put $x = 1 + h$ and then $h \to 0^{+}$. We have then
\begin{align}
L &= \lim_{x \to 1^{+}}\left(\frac{3x}{x - 1} - \frac{1}{2\log x}\right)\notag\\
&= \lim_{h \to 0^{+}}\left(\frac{3 + 3h}{h} - \frac{1}{2\log (1 + h)}\right)\notag\\
&= \lim_{h \to 0^{+}}\frac{(3 + 3h)2\log(1 + h) - h}{2h\log(1 + h)}\notag\\
&= \frac{1}{2}\lim_{h \to 0^{+}}\dfrac{6\log(1 + h) + 6h\log(1 + h) - h}{h^{2}\cdot\dfrac{\log(1 + h)}{h}}\notag\\
&= \frac{1}{2}\lim_{h \to 0^{+}}\dfrac{6\log(1 + h) - 6h + 6h\log(1 + h) + 5h }{h^{2}}\notag\\
&= \frac{1}{2}\lim_{h \to 0^{+}}\left(6\cdot\frac{\log(1 + h) - h}{h^{2}} + 6\cdot\frac{\log(1 + h)}{h} + \frac{5}{h}\right)\notag\\
\end{align}
Using Taylor expansion $$\log(1 + h) = h - \frac{h^{2}}{2} + o(h^{2})$$ we can see that the first term in parentheses tends to $6(-1/2) = -3$ and second term clearly tends to $6$. But the last term tends to $\infty$ so that the whole limit tends to $\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Minimum of $ f(\alpha) = \left(1+\frac{1}{\sin^{n}\alpha}\right)\cdot \left(1+\frac{1}{\cos^{n}\alpha}\right)$
Minimum value of $\displaystyle f(\alpha) = \left(1+\frac{1}{\sin^{n}\alpha}\right)\cdot \left(1+\frac{1}{\cos^{n}\alpha}\right)\;,$ Where $n\in \mathbb{N}$ and $\displaystyle \alpha \in \left(0,\frac{\pi}{2}\right)$
I have solved It using Derivative Test, But that nethod is very Lengthy,
Can we solve it Using Inequality
If yes ,Then plz explain me how can we solve it, Thanks
| Thanks friends got it
Let $\displaystyle f(x) = \ln\left(1+\frac{1}{t}\right)\;, t>0$ Using Jesan Inequality function $f(t)$ is convex function.
So $$\ln\left(1+\frac{1}{x}\right)+\ln\left(1+\frac{1}{y}\right)\geq 2\ln\left(1+\frac{2}{x+y}\right)$$
Where $x,y\in (0,1)$
So $$\ln\left(1+\frac{1}{x}\right)+\ln\left(1+\frac{1}{y}\right)\geq \ln\left(1+\frac{2}{x+y}\right)^2\geq \ln \left(1+\frac{1}{\sqrt{xy}}\right)^2$$
So $$\left(1+\frac{1}{x}\right)\cdot \left(1+\frac{1}{y}\right)\geq \left(1+\frac{1}{\sqrt{xy}}\right)^2$$ and equality hold when $x=y$
So $$\left(1+\frac{1}{\sin^ n\alpha}\right)\cdot \left(1+\frac{1}{\cos^n \alpha}\right)\geq \left(1+\frac{2^{\frac{n}{2}}}{\sqrt{\sin 2 \alpha}}\right)^2\geq \left(1+2^{\frac{n}{2}}\right)^2$$
and equality hold when $\displaystyle \sin^n \alpha = \cos^n \alpha\Rightarrow \alpha = \frac{\pi}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1879895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve $\int_{0}^{1}\frac{1}{1+x^6} dx$ Let $$x^3 = \tan y\ \ \text{ so that }\ x^2 = \tan^{2/3}y$$
$$3x^2dx = \sec^2(y)dy$$
$$\int_{0}^{1}\frac{1}{1+x^6}dx = \int_{1}^{\pi/4}\frac{1}{1+\tan^2y}\cdot \frac{\sec^2y}{3\tan^{2/3}y}dy = \frac{1}{3}\int_{1}^{\pi/4} \cot^{2/3}y\ dy$$
How should I proceed after this?
EDITED: Corrected the final integral and the limit from $45$ to $\pi/4$
| It is easier to solve this integral using partiel fraction decomposition
$\frac{1}{1+x^6} =\frac{1}{3(1+x^2)}+\frac{2-x^2}{3(x^4-x^2+1)}$
and solving the 3 integrals seperatly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1882650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
How to find the singular value decomposition of $A^TA$ & $(A^TA)^{-1}$ I want to find the singular value decomposition of $A^TA$ & $(A^TA)^{-1}$.
The singular value decomposition of $A$ is $$A=U \Sigma V^T$$
Basically, I want to find the singular values of $A^TA$ & $(A^TA)^{-1}$
$$A^TA=V{\Sigma}^2 V^T$$.
Does this mean that the singular values of $A^TA$ is equal to square of the singular values of $A$?
How to find the $(A^TA)^{-1}$?
| Finding the singular value decomposition
Start with a matrix with $m$ rows, $n$ columns, and rank $\rho$:
$$
\mathbf{A} = \mathbb{C}^{m \times n}_{\rho}.
$$
Since the question concerns the normal equations, let's fix $\rho = m$ and $m\ge n$. The matrix $\mathbf{A}$ is tall and has full column rank.
The singular value decomposition is
$$
\begin{align}
\mathbf{A} &= \mathbf{U}\, \mathbf{\Sigma} \, \mathbf{V}^{*} \\
&=
\left[ \begin{array}{cc}
\color{blue}{\mathbf{U}_{\mathcal{R}\left( \mathbf{A} \right)}} &
\color{red}{\mathbf{U}_{\mathcal{N}\left( \mathbf{A}^{*} \right)}}
\end{array} \right]
%
\left[ \begin{array}{c}
\mathbf{S} \\ \mathbf{0}
\end{array} \right]
%
\color{blue}{\mathbf{V}_{\mathcal{R}\left( \mathbf{A}^{*} \right)}}^{*}
%
\end{align}
$$
The coloring distinguishes $\color{blue}{range}$ spaces from $\color{red}{null}$ spaces.
The diagonal matrix of singular values, $\mathbf{S}\in\mathbb{R}^{\rho\times\rho}$ is
$$
\mathbf{S}_{k,k} = \sigma_{k}, \quad k=1,\rho.
$$
Manipulating the singular value decomposition
The Moore-Penrose pseudoinverse is
$$
\begin{align}
\mathbf{A}^{\dagger}
&= \mathbf{V}\, \Sigma^{\dagger} \mathbf{U}^{*} \\
%
&=
%
\color{blue}{\mathbf{V}_{\mathcal{R}\left( \mathbf{A}^{*} \right)}}
%
\left[ \begin{array}{cc}
\mathbf{S}^{-1} & \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{l}
\color{blue}{\mathbf{U}_{\mathcal{R} \left( \mathbf{A} \right)}}^{*} \\
\color{red}{\mathbf{U}_{\mathcal{N} \left( \mathbf{A}^{*} \right)}}^{*}
\end{array} \right].
%
\end{align}
$$
The Hermitian conjugate is
$$
\begin{align}
\mathbf{A}^{*}
&= \mathbf{V}\, \Sigma^{\mathrm{T}} \mathbf{U}^{*} \\
%
&=
%
\color{blue}{\mathbf{V}_{\mathcal{R}\left( \mathbf{A}^{*} \right)}}
%
\left[ \begin{array}{cc}
\mathbf{S} & \mathbf{0}
\end{array} \right]
%
\left[ \begin{array}{l}
\color{blue}{\mathbf{U}_{\mathcal{R} \left( \mathbf{A} \right)}}^{*} \\
\color{red}{\mathbf{U}_{\mathcal{N} \left( \mathbf{A}^{*} \right)}}^{*}
\end{array} \right].
%
\end{align}
$$
Resolving the product matrix
The product matrix has a simple expression:
$$
\begin{align}
\mathbf{A}^{*} \mathbf{A} &=
\left( \mathbf{V} \, \Sigma^{\mathrm{T}} \mathbf{U}^{*} \right)
\left( \mathbf{U} \, \Sigma \mathbf{V}^{*} \right) \\
%
&=
%
\color{blue}{\mathbf{V}_{\mathcal{R}\left( \mathbf{A}^{*} \right)}}
\, \mathbf{S}^{2} \,
\color{blue}{\mathbf{V}_{\mathcal{R}\left( \mathbf{A}^{*} \right)}}^{*}.
%
\end{align}
$$
The pseudoinverse of the product matrix is then
$$
\begin{align}
\left( \mathbf{A}^{*} \mathbf{A} \right)^{-1}
&=
%
\left(
\color{blue}{\mathbf{V}_{\mathcal{R}\left( \mathbf{A}^{*} \right)}}
\, \mathbf{S}^{-2} \,
\color{blue}{\mathbf{V}_{\mathcal{R}\left( \mathbf{A}^{*} \right)}}^{*}
\right)^{-1} \\[3pt]
%
&=
%
\left(
\color{blue}{\mathbf{V}_{\mathcal{R}\left( \mathbf{A}^{*} \right)}}^{*}
\right)^{-1}
\left( \mathbf{S}^{2} \right)^{-1}
\left(
\color{blue}{\mathbf{V}_{\mathcal{R}\left( \mathbf{A}^{*} \right)}}
\right)^{-1} \\[3pt]
%
&=
%
\color{blue}{\mathbf{V}_{\mathcal{R}\left( \mathbf{A}^{*} \right)}}
\, \mathbf{S}^{-2} \,
\color{blue}{\mathbf{V}_{\mathcal{R}\left( \mathbf{A}^{*} \right)}}^{*}.
%
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Semisimple, connected Lie groups generated by unipotent elements. Let $G$ be a linear, semisimple Lie group with no compact factors.
The unipotent elements of $G$ are those that have only eigenvalue 1.
I've seen it asserted that $G$ is generated by its unipotent elements: see Exercise #2 $\S 4.5$ in Dave Witte Morris' book on Arithmetic Groups.
The hint in the book is that you consider the simple factors.
But even considering $SL(2, \mathbb{R})$, it is unclear to me why it is generated by its unipotent elements.
I am aware that any matrix in $G$ can be written as a product of commuting hyperbolic, elliptic, and unipotent element, but I am unsure of how you might generally express the hyperbolic and elliptic elements as unipotent elements.
| I think I did the case $SL_2(\mathbb{R}):$
Let
$$A=\begin{pmatrix}
a&b\\c&d
\end{pmatrix} \in SL_2(\mathbb{R}).$$
Notice that
$$A\cdot\begin{pmatrix}
0&1\\-1&0
\end{pmatrix} = \begin{pmatrix}
-b&a\\-d&c
\end{pmatrix}, $$
and
$$\begin{pmatrix}
0&1\\-1&0
\end{pmatrix}\cdot A = \begin{pmatrix}
c&d\\-a&-b
\end{pmatrix},$$
and also
$$\begin{pmatrix}
0&1\\-1&0
\end{pmatrix} = \begin{pmatrix}
1&0\\-1&1
\end{pmatrix}\begin{pmatrix}
1&1\\0&1
\end{pmatrix}\begin{pmatrix}
1&0\\-1&1
\end{pmatrix}.$$
So, using these facts, you'll be able to reduce your problem to show that
$$ A = \begin{pmatrix}
a&0\\0&1/a
\end{pmatrix}$$
is generated by unipotent elements.
But
$$ \begin{pmatrix}
a&0\\0&1/a
\end{pmatrix}
=
\begin{pmatrix}
1&0\\1/a-1&1
\end{pmatrix}
\begin{pmatrix}
1&1\\0&1
\end{pmatrix}
\begin{pmatrix}
1&0\\a-1&1
\end{pmatrix}
\begin{pmatrix}
1&-1/a\\0&1
\end{pmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
A horrid-looking integral $\int_{0}^{5} \frac{\pi(1+\frac{1}{2+\sqrt{x}} )}{\sqrt{10}\sqrt{\sqrt{x}+x}} $
$$
\mathbf{\mbox{Evaluate:}}\qquad
\int_{0}^{5} \frac{\pi(1+\frac{1}{2\sqrt{x}} )}{\sqrt{10}\sqrt{\sqrt{x}+x}}
\,\,\mathrm{d}x
$$
This is a very ugly integral, but appears to have a very simple closed form of: $$\Gamma(\frac15)\Gamma(\frac45)$$ Mathematica can evaluate this integral, but WolframAlpha doesn't even give a correct numerical answer. I have tried many techniques on this integral but have not been able to crack it at all.
Any help on this integral would be greatly appreciated. Thank you!
| Let $u = \sqrt{x} + x$. Then we have
$$ \frac\pi{\sqrt{10}} \int_0^5 \frac{1 + \frac{1}{2\sqrt x}}{\sqrt{\sqrt x + x}} \, dx = \frac\pi{\sqrt{10}} \int_0^{\sqrt{5}+5} \frac{1}{\sqrt{u}} \, du$$
If you want to be pedantic (and who doesn't?!) then we need to note that this is an improper integral because the integrand is not defined at the lower limit of integration. Therefore:
\begin{align}
\frac\pi{\sqrt{10}} \int_0^{\sqrt{5}+5} \frac{1}{\sqrt{u}} \, du
&= \lim_{B\to0^+}\frac\pi{\sqrt{10}} \int_B^{\sqrt{5}+5} \frac{1}{\sqrt{u}}\, du \\[0.3cm]
&= \frac{2\pi}{\sqrt{10}} \lim_{B \to 0^+} \sqrt{u}\bigg|_B^{\sqrt5 + 5} \\[0.3cm]
&= \frac{2\pi}{\sqrt{10}} \lim_{B\to0^+} \left(\sqrt{\sqrt5 + 5} - \sqrt B\right)\\[0.3cm]
&= \frac{2\pi}{\sqrt{10}} \left(\sqrt{\sqrt5 + 5} - 0\right)\\[0.3cm]
&= \frac{2\pi\sqrt{\sqrt5 + 5}}{\sqrt{10}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1883532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Does $2764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} = 0$? In the process of my numerical computations I have found a very special identity:
*
*$\;\;1264483 + 1707789 \,\sqrt[3]{7} - 1238313\,\sqrt[3]{7^2} = 9.313225746154785 \times 10^{-10}$
*$
-1500493 - 2026256\,\sqrt[3]{7} + 1469290\,\sqrt[3]{7^2}
= 9.313225746154785 \times 10^{-10}$
Therefore if we subtract these two equations one should find the difference is zero. On my computer I found:
$$
2764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} \stackrel{?}{=} \left\{
\begin{array}{cl} 0 & \text{by hand} \\
-1.862645149230957 \times 10^{-9} & \text{by computer} \end{array}\right.
$$
Subtracting these two numbers - which might be the same - we have gotten twice the number!
| Assume
$$ \tag 12764976 + 3734045\,\sqrt[3]{7} -2707603\,\sqrt[3]{7^2} = 0$$
Multiply by $\sqrt[3]7$ to obtain
$$ \tag 22764976\,\sqrt[3]7 + 3734045\,\sqrt[3]{7^2} -18953221 = 0.$$
The system of linear equations
$$\begin{align}
3734045 x-2707603 y&=-12764976\\
22764976 x+3734045 y&=18953221
\end{align} $$
has rational solutions, i.e., we are forced to conclude that
$$\sqrt[3]7=\frac{-12764976\cdot 3734045+2707603\cdot 18953221}{3734045\cdot 3734045+2707603\cdot 22764976} =\frac{3652803231343}{75581609374553}$$
is rational, which is absurd! (Not to mention that this rational number is $\approx 0.05$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1884231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
How to prove that $2 \min(|a−b|,|b−c|,|c−a|)$ ≤ $R$ How to prove that $2 \min(|a−b|,|b−c|,|c−a|)$ ≤ $R$
I know that min is for $\sqrt{7}-1,\sqrt{7},\sqrt{7}+1$, but how to prove this?
| We may assume without loss of generality that the side lengths are given by $2x-y,2x,2x+y$ with $y\leq x$ in order to fulfill the triangle inequality. With such assumptions, the area of $ABC$, by Heron's formula, is given by:
$$ \Delta = \frac{\sqrt{3}}{2}\sqrt{x^2(2x-2y)(2x+2y)} $$
and the circumradius is given by:
$$ R = \frac{abc}{4R} = \frac{1}{2\sqrt{3}}\sqrt{\frac{(2x-y)^2(2x+y)^2}{(2x-2y)(2x+2y)}}=\frac{1}{4\sqrt{3}}\sqrt{\frac{(4x^2-y^2)^2}{x^2-y^2}} $$
so the problem boils down to studying a quadratic form in $x^2$ and $y^2$, pretty easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1887079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the value of the constants If $\displaystyle
\frac{\left( \frac{2x^2}{3a} \right)^{n-1}}
{\left( \frac{3x}{a} \right)^{n+1}} =
\left( \frac{x}{4} \right)^3$, determine the values of the constants $a$ and $n$
I could find the value of $a$, i.e, $\displaystyle \frac{\sqrt{x^6 \times 3^{2n}}}{2^n x^n 2^5 }$ and substituted the same to find the value of $n$, to no avail
The right values for $a$ is $\pm(3^6 2^{-11/2})$, and $n$ is equal to $6$
| \begin{align*}
\frac{\left( \frac{2x^2}{3a} \right)^{n-1}}
{\left( \frac{3x}{a} \right)^{n+1}} &=
\left( \frac{x}{4} \right)^3 \\
\frac{2^{n-1}}{3^{2n}a^{2n}} x^{n-3} &=
\frac{x^3}{2^6}
\end{align*}
Comparing $x$ terms,
$$n-3=3 \implies n=6$$
Comparing coefficient of $x^3$,
$$\frac{2^5 a^2}{3^{12}}=\frac{1}{2^6} \implies a^{2}=\frac{3^{12}}{2^{11}} \implies a=\pm \frac{3^6}{2^{11/2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1891464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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