Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Number of binary strings with $n$ ones and $m$ zeros $f(n,m)$ is the number of binary strings with up to $n$ ones and up to $m$ zeros.
Prove that the number of possible strings is: $${n+m+2 \choose n+1} -1$$
I got to the point that:
$$\sum_{a=0}^n \sum_{b=0}^m {a+b \choose a}$$
And I also understand that there are $(n+1)$ options for the amount of ones and $(m+1)$ options for the amount of zeros.
| We can find the sum by generating functions
\begin{align*}
\sum_{b=0}^{m} \binom{a+b}{b} &= [x^m] \frac{1}{(1-x)}\frac{1}{(1-x)^{a+1}} \\
\implies \sum_{a=0}^{n} \sum_{b=0}^{m} \binom{a+b}{b} &= [x^m]\sum_{a=0}^{n}\frac{1}{(1-x)^{a+2}} \\
&= [x^m]\left(\frac{1}{x\left(1-x\right)^{n+2}}-\frac{1}{x\left(1-x\right)}\right) \\
&= \binom{m+n+2}{m+1}-1
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
If this is a telescoping series then how does it collapse? $\frac{3r+1}{r(r-1)(r+1)}$ Express $$\frac{3r+1}{r(r-1)(r+1)}$$ in partial fractions. Hence, or otherwise, show
$$\sum_{r=2}^n\frac{3r+1}{r(r-1)(r+1)}=\frac52-\frac2n-\frac{1}{n+1}$$
So, I have obtained the partial fractions $$\frac{3r+1}{r(r-1)(r+1)}=\frac{2}{r-1}-\frac{1}{r}-\frac{1}{r+1}$$
then for $r=(2,3,4,5...(n-1),n)$
$$
\begin{align}
& {}+2-\frac12-\frac13 \\[8pt]
& {}+1-\frac13-\frac14 \\[8pt]
& {}+\frac23-\frac14-\frac15 \\[8pt]
& {}+\frac12-\frac15-\frac16 \\[8pt]
& {}+\cdots \\[8pt]
& {}+\frac{2}{n-2}-\frac{1}{n-1}-\frac{1}{n} \\[8pt]
& {}+\frac{2}{n-1}-\frac{1}{n}-\frac{1}{n+1}
\end{align}
$$
If this is a telescopic series how does it collapse?
| Here is a breakdown into a single telescoping series.
$$
\begin{align}
\frac{3r+1}{r(r-1)(r+1)}
&=\frac2{r(r-1)}+\frac1{(r+1)r}\\
&=\left(\frac2{r-1}-\frac2r\right)+\left(\frac1r-\frac1{r+1}\right)\\
&=\underbrace{\left(\frac2{r-1}+\frac1r\right)}_{\large a_r}-\underbrace{\left(\frac2r+\frac1{r+1}\right)}_{\large a_{r+1}}
\end{align}
$$
Thus,
$$
\begin{align}
\sum_{r=2}^n\frac{3r+1}{r(r-1)(r+1)}
&=\sum_{r=2}^n(a_r-a_{r+1})\\
&=a_2-a_{n+1}\\[3pt]
&=\frac52-\left(\frac2n+\frac1{n+1}\right)
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/475256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Uniform convergence of series $\sum_{n=1}^{\infty}\frac{x}{n(1+nx^2)}$
Discuss the uniform convergence of the series $$\sum_{n=1}^{\infty}\frac{x}{n(1+nx^2)}$$ for real values of $x$.
Let $f(x)$ denote the given series. Then $f(0)=0+0+\ldots = 0$, and $f(-x)=-f(x)$ for all real $x$. Also, $$f(1) = \sum_{n=1}^{\infty}\frac{1}{n(1+n)} = \sum_{n=1}^{\infty}\left(\frac{1}{n}-\frac{1}{n+1}\right) = 1.$$ For any $x$ we can write $$\frac{x}{n(1+nx^2)} = \frac{x}{n}-\frac{x^3}{1+nx^2},$$ so $$f(x) = \sum_{n=1}^{\infty}\left(\frac{x}{n}-\frac{x^3}{1+nx^2}\right).$$ How can we check which values of $x$ this sum converges?
| Hint: $$n(1+nx^2) \ge 2n \sqrt{n}|x|.$$
| {
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Is there any solution to the following system of equations? Is there any solution to the following system of diophantine equations?
$$
\left\{\begin{array}{l}
2.a^2 = b^2+c^2+d^2 \\
a^2 = e^2+f^2+g^2 , & \mbox{with }((a,b,c,d,e,f,g)>2)\in N\mbox{ and differents among them}
\end{array}
\right.
$$
| set of equations:
$\left\{\begin{aligned}&X^2+Y^2+Z^2=2(R^2+W^2)\\&F^2+G^2+T^2=R^2+W^2\end{aligned}\right.$
has solutions
$X=4(t^2+q^2-a^2-v^2)a^2+(v^2+t^2+q^2-z^2)^2$
$Y=4(v^2-a^2-t^2-q^2)a^2+(v^2+t^2+q^2-z^2)^2$
$Z=4az(v^2+t^2+q^2-2a^2-z^2)$
$F=4(a^2+z^2-2t^2-2q^2)a^2+(v^2+t^2+q^2-z^2)^2$
$G=4aq(v^2+t^2+q^2-2a^2-z^2)$
$T=4at(v^2+t^2+q^2-2a^2-z^2)$
$R=4(z^2+a^2-2v^2)a^2+(v^2+t^2+q^2-z^2)^2$
$W=4av(v^2+t^2+q^2-2a^2-z^2)$
$a,v,t,q,z$ - integers and sets us.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Perform the indicated operations and simplify $${2x^2+5x-3\over16x^2+26x+9}\cdot{10x^2+13x+4\over5x^2+11x-12}\cdot{35x^2-38x+8\over x^2+x-2}$$
Please help me. I'm having a hard time with 3 terms.
| HINT: Factor!
$${2x^2+5x-3\over16x^2+26x+9}\cdot{10x^2+13x+4\over5x^2+11x-12}\cdot{35x^2-38x+8\over x^2+x-2} $$ $$= \dfrac{(2x -1)\color{red}{\bf (x+3)}}{(8x+9)\color{blue}{\bf (2x + 1)}}\cdot \dfrac{(5x + 4)\color{blue}{\bf (2x + 1)}}{\color{green}{\bf (5x-4)}\color{red}{\bf (x+3)}} \cdot \dfrac{(7x - 2)\color{green}{\bf (5x - 4)}}{(x+2)(x-1)}$$
Cancel common factors in numerator and denominator.
| {
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Don't know how to find all the roots So i got this problem :
Find all the roots of $r^{3}=(-1)$
i can only think to use :
$\sqrt[n]{z} =\sqrt[n]{r}\left[\cos \left(\dfrac{\theta + 2\pi{k}}{n}\right) + i \sin\left(\dfrac{\theta + 2\pi k}{n}\right)\right] $
i know that $\theta = \tan^{-1}(\dfrac{b}{a}) $as in from $ z=a + bi$
$r=\sqrt[]{a^2 + b^2}$ , $k=0,1,2,3,4,.. n-1$
so anyone could explain this to me ?
lets say that $r^{3}=(-1) == z^{3}=(-1)$ a since i dont like it and it messes me up using the formula
so far :
$ z^{3} = (-1) $ so $z = \sqrt[3]{(-1 + 0 \cdot i)}$
$ r= \sqrt[]{(-1)^{2} + 0^{2}}=1$
$\tan^{-1}(\dfrac{b}{a})=\tan^{-1}(\dfrac{0}{-1})=\tan^{-1}(0)=0$ <- not sure about his part
k=0;
$ z^{3}=\sqrt[3]{1}(\cos(\dfrac{\pi + 2\pi \cdot 0}{3})+i \cdot \sin(\dfrac{\pi + 2\pi \cdot 0}{3})$
$ =\cos(\dfrac{\pi}{3}) +i\cdot \sin(\dfrac{\pi}{3})= \dfrac{\sqrt[]{3}}{2} +\dfrac{i}{2}$
k=1;
$ \sqrt[3]{1}(\cos(\dfrac{\pi+2\pi \cdot 1}{3})+i \cdot \sin(\dfrac{\pi+2\pi \cdot 1}{3})$
$=\cos(\pi) + i \cdot \sin(\pi)=(-1) + 0 = -1$
k=2;
$\sqrt[3]{1}(\cos(\dfrac{\pi + 2\pi \cdot 2}{3})+i \cdot \sin(\dfrac{\pi + 2\pi \cdot 2}{3})$
$=\cos(\dfrac{5\pi}{3})+i \cdot \sin(\dfrac{5\pi}{3})=-\dfrac{\sqrt[]{2}}{2}-i\cdot\dfrac{\sqrt[]{2}}{2}$
so is this correctet or im missing something here?
Roots $(\dfrac{\sqrt[]{3}}{2} +\dfrac{i}{2};-1;-\dfrac{\sqrt[]{2}}{2}-i\cdot\dfrac{\sqrt[]{2}}{2})$
| As an alternative to finding "roots of unity" using the "$\operatorname{cis} \theta$" approach, you can easily solve the equation $$r^3 = -1 \iff r^3 + 1 = 0 \iff (r+1)(r^2 - r + 1)$$
That gives a real root $$r_1 = -1,$$
and two complex roots, which can be solved by applying the quadratic formula $$\left(\dfrac {-b \pm \sqrt {b^2 - 4ac}}{2}\right)\;\;\text{to}\;\;r^2 -r + 1 = 0\quad\text{to find}\;\;r_2, r_3$$
$$r_2, r_3 = \frac 12 \pm \frac{\sqrt {-3}}{2} = \frac 12 \pm \frac{\sqrt 3}2\;i$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For what $n$ can $\pm 1\pm 2\pm 3 ... \pm (n-1) \pm n = n+1$? More explicitly,
for what values of $n$
can the signs be chosen
in the equation
$$\pm 1\pm 2\pm 3 ... \pm (n-1) \pm n = n+1$$
so the equation is true?
For all $n$,
if the equation cannot be satisfied,
prove it;
if the equation can be satisfied,
exhibit a solution.
Note:
I have a solution,
and will present it in two days
if no one else submits a similar one.
| As others have stated,
we want a partition of
$\{1, 2, ..., n+1\}$
into two sets with equal sums.
The sum is
$\frac{(n+1)(n+2)}{2}$.
If $n=4k$,
the sum is
$(4k+1)(2k+1)$
which is odd
and therefore impossible.
If $n = 4k+1$,
the sum is
$(2k+1)(4k+3)$
which is also odd,
and therefore impossible.
If $n = 4k+2$,
the sum is
$(4k+3)(2k+2)$,
so it is not ruled out,
and each sum must be
$(4k+3)(k+1)$.
if $n = 4k+3$,
the sum is
$(2k+2)(4k+5)$
which is also not ruled out,
and each sum must be
$(k+1)(4k+5)$.
Here are my solutions
for the not impossible cases.
For the $n=4k+2$ case,
the sum must be
$(4k+3)(k+1)
=(4k+4-1)(k+1)
=4(k+1)^2-(k+1)
=(2k+2)^2-(k+1)
$.
The square there suggests,
to me,
the formula for
the sum of consecutive odd numbers
$1+3+...+(2m-1)=m^2$,
so $1+3+...+(4k+3) = (2k+2)^2$.
If $k+1$ is odd,
remove it from the sum
so it is
$(2k+2)^2-(k+1)$.
If $k+1$ is even,
both $1$ and $k$ are odd,
so remove them from the sum.
In either case, we have the desired partition.
For the $n=4k+3$ case,
the sum must be
$(4k+5)(k+1)
=(4k+4+1)(k+1)
=4(k+1)^2+(k+1)
=(2k+2)^2+(k+1)
$.
Again,
$1+3+...+(4k+3) = (2k+2)^2$.
If $k+1$ is even,
add it to the sum
so it is
$(2k+2)^2+(k+1)$.
If $k+1$ is odd,
$k+2$ is even,
so remove $1$
and add $k+2$ to the sum.
In either case, we have the desired partition.
I do not know how many partitions
can be found.
| {
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"timestamp": "2023-03-29T00:00:00",
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Derivative of $\frac{d}{dx}\ln (x+\sqrt{x^2+1})$ Show that $\frac{d}{dx}\ln (x+\sqrt{x^2+1}) = \frac{1}{\sqrt{x^2+1}}$
So I've done this so far:
$$\frac{d}{dx}\ln (x+\sqrt{x^2+1}) = \frac{1+x(x^2+1)^{-0.5}}{x+(x^2+1)^{0.5}}$$
I have tried a few combinations but cannot attain the desired result. Could someone offer just a hint to get me going?
Thanks
| Your differentiation is correct, and you applied the chain rule correctly. But as you note, your result can be simplified.
Let's rewrite the fraction, expressing the term $x(x^2 + 1)^{-0.5}$ in the numerator as $\frac{x}{\sqrt {x^2 + 1}}$, then simply multiply the numerator and denominator by $\sqrt{x^2 + 1}$:
$$\begin{align}\frac{1+x(x^2+1)^{-0.5}}{x+(x^2+1)^{0.5}} & = \frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}\\ \\
& = \frac{\sqrt{x^2 +1} + x}{{(x+\sqrt{x^2+1})\sqrt{x^2+1}}}\\ \\
& =\frac{\color{blue}{\bf x+\sqrt{x^2+1}}}{(\color{blue}{\bf x+\sqrt{x^2+1}})\sqrt{x^2+1}} \\ \\
& =\frac1{\sqrt{x^2+1}}\end{align}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$az+b\overline{z}+c=0$ represents a line
When does $az+b\overline{z}+c=0$ represent a line?
All of $a,b,c$ are complex numbers.
I know that a line in the complex plane is usually represented by $z=a+bt$, where the parameter $t$ runs through all real values.
So suppose we're given the equation $az+b\overline{z}+c=0$. If $z$ represents a line $d+et$, we must have $a(d+et)+b(\overline{d}+\overline{e}t)+c=0$. Since the equation holds for all real values $t$, it must be that $ae+b\overline{e}=0$, and as a result $ad+b\overline{d}+c=0$. How can we get to a condition involving only $a,b,c$ from here?
| Your approach is a good idea, but let's instead just write $z = x + iy$. Then
$az + b\overline{z} + c = 0$ becomes
\begin{align*}
a(x + iy) + b(x - iy) + c &= 0 \\
\iff (a + b)x + (ai - bi)y + c &= 0
\end{align*}
At this point we have what looks a lot like a basic line in $\mathbb{R}^2$ of the form $Ax + By + C = 0$, with one problem: $A,B,$ and $C$ are complex. We want this line to be an actual line in $\mathbb{R}^2$, which means we want to be able to reduce it to $A'x + B'y + C' = 0$, with $A', B', C'$ real. In other words, we need to be able to multiply $Ax + By + C$ by a constant nonzero (complex) number to make all the coefficients real.
Excluding the case $C = 0$ ($c = 0$), such a number exists if and only if $A$ and $B$ are both real multiples of $C$. So we have the necessary and sufficient conditions
\begin{align*}
\frac{A}{C} &= \frac{a + b}{c} \in \mathbb{R} \\
\frac{B}{C} &= \frac{ai - bi}{c} \in \mathbb{R}
\end{align*}
That is, $az + b\overline{z} + c, \; c \ne 0$ is a line if and only if $\frac{a + b}{c},\frac{ai - bi}{c} \in \mathbb{R}$.
You should be able to deal with the case $c = 0$ in a similar fashion.
| {
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Reduced row-echelon form of a matrix with variables I have been staring at this for an hour. How would you reduce such a matrix?
\begin{bmatrix}
p & 0 & a \\
b & 0 & 0 \\
q & c & r
\end{bmatrix}
$abc\neq0$
| Because $a,b,c \ne 0$ then
$$
\left( {\begin{array}{*{20}{c}}
p & 0 & a \\
b & 0 & 0 \\
q & c & r \\
\end{array}} \right) \to \left( {\begin{array}{*{20}{c}}
b & 0 & 0 \\
q & c & r \\
p & 0 & a \\
\end{array}} \right)\mathop \to \limits_{ - \frac{p}{b}{\rho _1} + {\rho _3}}^{ - \frac{q}{b}{\rho _1} + {\rho _2}} \left( {\begin{array}{*{20}{c}}
b & 0 & 0 \\
0 & c & r \\
0 & 0 & a \\
\end{array}} \right)\mathop \to \limits^{ - \frac{r}{a}{\rho _3} + {\rho _2}} \left( {\begin{array}{*{20}{c}}
b & 0 & 0 \\
0 & c & 0 \\
0 & 0 & a \\
\end{array}} \right)\mathop \to \limits^{\frac{1}{b}{\rho _1},\frac{1}{c}{\rho _2},\frac{1}{a}{\rho _3}} \left( {\begin{array}{*{20}{c}}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}} \right).
$$
| {
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Find the number $b$ such that the line $y = b$ divides the region bounded by the curves $y = x^2$ and $y = 4$ into two regions with equal area. Find the number $b$ such that the line $y = b$ divides the region bounded by the curves $y = x^2$ and $y = 4$ into two regions with equal area.
I could find the area of this region but i have no clue how to split it parallel to the $x$ axis perfectly... this is too abstract for me...
I found the whole area and got $\dfrac{32}{3}$... I tried plugging that back into the definite integral but that definitely wasn't right...
| The intersection of $y=4$ and $y=x^2$ is at $(-2, 4)$ and $(2, 4)$ so the whole area between the two curves is (by the symmetry of $y=x^2$)
$$
\int_{-2}^2 4-x^2dx=2\int_0^24-x^2dx=\frac{32}{3}
$$
as you noted. Now if the upper boundary is $y=b$ the intersection will be at $(-\sqrt{b}, b)$ and $(\sqrt{b}, b)$ so you'll have the area
$$\begin{align}
\int_{-\sqrt{b}}^{\sqrt{b}} b-x^2dx&=2\int_0^{\sqrt{b}}b-x^2dx\\
&= 2(bx-\frac{1}{3}x^3)\left|_0^\sqrt{b}\right.\\
&=2(b^{3/2}-\frac{1}{3}b^{3/2})=\frac{4}{3}b^{3/2}
\end{align}$$
and you want to find $b$ so that
$$
\frac{4}{3}b^{3/2}=\frac{16}{3}
$$
which I'll bet you can do.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the minimum value of this product Is it true that $f(n) = \prod _{ i=1 }^{ n }{ (1-\frac { 1 }{ { 2 }^{ i } } ) } \ge \frac{1}{4} \quad \forall n$? I came up with this expression while trying to find an alternative way to solve a physics problem. The official solution has a minimum value of $\frac{1}{4}$, but I couldn't prove that my solution also has a minimum of $\frac{1}{4}$.
| For $0 < x \leqslant \frac12$, we have the estimate
$$\lvert \log (1-x)\rvert = \sum_{n=1}^\infty \frac{x^n}{n} < x + \frac{x^2}{2}\sum_{k=0}^\infty x^k = x + \frac{x^2}{2(1-x)} \leqslant x + x^2,$$
so we find
$$\log f(n) = \sum_{i=1}^n \log \left(1 - \frac{1}{2^i}\right) > - \sum_{i=1}^n \left(\frac{1}{2^i} + \frac{1}{2^{2i}}\right) > - \left(1 + \frac13\right),$$
whence
$$f(n) > e^{-4/3} \approx 0.26359713811572677 > \frac14.$$
A less crude estimate of the logarithm yields larger bounds.
| {
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Help in factoring polynomials Please help in factoring:
*
*$x^3 - 13x + 12$
*$x^5 - 3x^3 - 4x$
*$x^3 - 6x^2 + 5x + 12$
Thank you in advance.
| Hint:
*
*$1^3 - 13\cdot1 + 12=0$
*$x\cdot\left((\pm2)^4 - 3\cdot(\pm2)^2 - 4\right)=0$
*$(-1)^3 - 6(-1)^2 + 5(-1) + 12=0$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Simple Uniform Continuity Show that:
$\cfrac1{x^2+1}$ is uniformly continuous on $\mathbb R$!
I'm having difficulty finding an inequality between $1+x^2+c^2+x^2c^2$ and $x+c$.
| Hint: Let $f$ be the relevant function, and $c$ a small number. Then
\begin{align} |f(x + c) - f(x)| &= \left|\frac{1}{(x + c)^2 + 1} - \frac{1}{x^2 + 1}\right| \\
&= \left|\frac{x^2 + 1 - ((x + c)^2 + 1)}{((x + c)^2 + 1)(x^2 + 1)}\right| \\
&= \frac{|2cx + c^2|}{x^2 + 1} \\
&= |c| \frac{|2x + c|}{x^2 + 1} \\
&\le |c| \frac{|2x|}{x^2 + 1} + \frac{|c|^2 }{x^2 + 1} \\
&\le |c| \frac{|2x|}{x^2 + 1} + |c|
\end{align}
provided $|c| \le 1$. Now show that the quantity $\frac{|2x|}{x^2 + 1}$ is bounded by some fixed constant $\alpha$, and complete the proof using $\delta = \epsilon /\alpha$.
| {
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Prove that $1^3 + 2^3 + 3^3 +\cdots+ n^3 = \frac14n^4 + \frac12n^3 + \frac14n^2$ I have to prove that this is true using mathematical induction.
I have this:
for every $n \in \mathbb N$: $1^3 + 2^3 + 3^3 + ... + n^3 = \frac 14n^4 + \frac 12n^3 + \frac 14n^2$
for $n = 1: 1^3 = 1/4 + 1/2 + 1/4$, hence $P(1)$ is true.
Let $N \in \mathbb N$ be given and assume that $P(N)$ is true, that is $$1^3 + 2^3 + 3^3 + ... + N^3 = \frac 14N^4 + \frac 12N^3 + \frac 14N^2$$
For n = $N$ + 1:
And now what? I just couldn't solve it.
| Let $P(N)$ is true, that is $1^3+\ldots+n^3=\frac{1}{4}n^4+\frac{1}{2}n^3+\frac{1}{4}n^2$
Then for $P(N+1)$, $1^3+\ldots+n^3+(n+1)^3=\frac{1}{4}n^4+\frac{1}{2}n^3+\frac{1}{4}n^2+n^3+3n^2+3n+1$
$=\frac{1}{4}n^4+n^3+\frac{3}{2}n^2+n+\frac{1}{4}+\frac{1}{2}n^3+\frac{3}{2}n^2+\frac{3}{2}n+\frac{1}{2}+\frac{1}{4}n^2+\frac{1}{2}n+\frac{1}{4}$
$=\frac{1}{4}(n+1)^4+\frac{1}{2}(n+1)^3+\frac{1}{4}(n+1)^2$
That is $P(N+1)$ is true.
| {
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"question_score": "4",
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Limit of the sequence For all natural numbers $n$
$$\begin{align}
a_n &= 2^{1/2}\cdot 2^{1/4}\cdot 2^{1/8}\cdot \dotsc\cdot 2^{1/{2^n}}\\
b_n &= \cos(x/2)\cdot\cos(x/4)\cdot\dotsc \cdot\cos(x/2n)
\end{align}$$
Find limit of $a_n$ and $b_n$ when $n$ approach to infinity.
I know that $\lim a_n$ is $2$, but I don't know how to show it.
| For the first one, note that
$$2^{1/2} \cdot 2^{1/4} \cdot 2^{1/8} \cdots 2^{1/2^n} = 2 ^{1/2+1/4+\cdots1/2^n} = 2^{1-1/2^n}$$
For the second one, I assume you have
$$b_n = \cos(x/2)\cos(x/4) \cdots\cos(x/2^n)$$
If $x = (2n+1) \pi \cdot 2^{k-1}$, where $k \in \mathbb{Z}^+$, then $b_k$ onwards is zero and the limit is zero. If not, note that
\begin{align}
b_n & = \cos(x/2)\cos(x/4) \cdots\cos(x/2^n) = \dfrac{\cos(x/2)\cos(x/4) \cdots\cos(x/2^n) \cdot \sin(x/2^n)}{\sin(x/2^n)}\\
& = \dfrac{\cos(x/2)\cos(x/4) \cdots\cos(x/2^{n-1}) \sin(x/2^{n-1})}{2\sin(x/2^n)}\\
& = \dfrac{\sin(x/2^{n-1})}{2 \sin(x/2^n)} b_{n-1} = \dfrac{\sin(x/2^{n-1})}{2 \sin(x/2^n)} \dfrac{\sin(x/2^{n-2})}{2 \sin(x/2^{n-1})} b_{n-2} = \dfrac{\sin(x/2^{n-2})}{2^2 \sin(x/2^n)} b_{n-2}\\
& = \dfrac{\sin(x)}{2^n \sin(x/2^n)}
\end{align}
| {
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Generating Functions in Discrete Mathematics in Computer Science Hey Guys can anyone help me with the following question in Discrete Structures in Mathematics, relating generating functions
Find a closed form for the generating function for the sequence $\{a_n\}$, where
$$a_n=\binom{n+4}n$$
for $n=0,1,2,\dots\;$.
| So the generating function is, by definition
$$g(x):=\sum_{n \geq 0} \binom{n+4}{n} x^n = \sum_{n \geq 0} \binom{n+4}{4} x^n$$
using the binomial identity $\binom{x}{y}=\binom{x}{x-y}$ for natural numbers $0 \leq y \leq x$
We can rewrite this as $$g(x)=\sum_{n \geq 0} \frac{1}{4!}(n+1)(n+2)(n+3)(n+4) x^n.$$ Since $\frac{d}{dx}x^{n+1}=(n+1)x^n$, we find $$g(x)=\sum_{n \geq 0} \frac{1}{4!}(n+2)(n+3)(n+4) \frac{d}{dx} x^{n+1}$$ and if we keep doing this, we get $$g(x)=\sum_{n \geq 0} \frac{1}{4!}\frac{d^4}{dx^4} x^{n+4}.$$
We can rearrange this to obtain $$g(x)=\frac{1}{4!} \frac{d^4}{dx^4} \sum_{n \geq 0} x^{n+4}.$$
We know the generating function for $(1,1,1,\ldots,)$ is $$\frac{1}{(1-x)}=\sum_{n \geq 0} x^n,$$ which we substitute into the above to give $$g(x) = \frac{1}{4!} \frac{d^4}{dx^4} \frac{x^4}{(1-x)}=\frac{1}{4!}\frac{24}{(1-x)^5}=\frac{1}{(1-x)^5}.$$
(We can check this using Wolfram|Alpha: "Taylor series for 1/(1-x)^5".)
| {
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How find this nice limit It is well kown this following
$$\lim_{x\to+\infty}\left(\dfrac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)^x=\sqrt{ab}(a,b>0)$$
and also kown this general
$$\lim_{x\to+\infty}\left(\dfrac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+a^{\frac{1}{x}}_{n}}{n}\right)^x=\sqrt[n]{a_{1}a_{2}\cdots a_{n}},(a_{i}>0,i=1,2,\cdots,n)$$
and some hours ago,I found this nice and Hard limit
$$\lim_{x\to+\infty}x\left[\left(\dfrac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)^x-\sqrt{ab}\right]=\dfrac{\sqrt{ab}}{8}\left(\ln{a}-\ln{b}\right)^2$$
my proof
$$a^{\frac{1}{x}}+b^{\frac{1}{x}}=e^{\frac{1}{x}\ln{a}}+e^{\frac{1}{x}\ln{b}}=1+\dfrac{1}{x}\ln{a}+\dfrac{1}{2x^2}\ln^2{a}+1+\dfrac{1}{x}\ln{b}+\dfrac{1}{2x^2}\ln^2{b}+o(1/x^2)$$
then
\begin{align*}
\left(\dfrac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)^x&=\left[1+\dfrac{1}{2x}\ln{ab}+\dfrac{1}{4x^2}\left(\ln^2{a}+\ln^2{b}\right)+o(\frac{1}{x^2})\right]^x\\
&\approx e^{x\ln{\left(1+\dfrac{1}{2x}\ln{ab}+\dfrac{1}{4x^2}\left(\ln^2{a}+\ln^2{b}\right)\right)}}\\
&\approx e^{x\left(\dfrac{1}{2x}\ln{ab}+\dfrac{1}{4x^2}\left(\ln^2{a}+\ln^2{b}\right)\right)}\\
&=\cdots\cdots\\
&\approx \sqrt{ab}\left(1+\dfrac{1}{8x}\left(2\ln^2{a}+2\ln^2{b}-\ln^2{(ab)}\right)\right)
\end{align*}
so
$$\lim_{x\to+\infty}x\left[\left(\dfrac{a^{\frac{1}{x}}+b^{\frac{1}{x}}}{2}\right)^x-\sqrt{ab}\right]=\dfrac{\sqrt{ab}}{8}\left(\ln{a}-\ln{b}\right)^2$$
My question:
$$\lim_{x\to+\infty}x\left[\left(\dfrac{a_{1}^{\frac{1}{x}}+a_{2}^{\frac{1}{x}}+\cdots+a^{\frac{1}{x}}_{n}}{n}\right)^x-\sqrt{a_{1}a_{2}\cdots a_{n}}\right]=?$$
| We can compute
$$\lim_{x\to +\infty} x\left[\left(\frac{a_1^{1/x} + a_2^{1/x} + \dotsb + a_n^{1/x}}{n}\right)^x - \sqrt[n]{a_1a_2\dotsb a_n}\right]$$
in the same way. For brevity, write $\lambda_\nu = \log a_\nu$, then we have
$$\begin{align}
\frac1n\sum_{\nu=1}^n a_\nu^{1/x} &= \frac1n\sum_{\nu=1}^n \exp\left(\frac1x\lambda_\nu\right)\\
&= \frac1n\sum_{\nu=1}^n \left(1 + \frac{\lambda_\nu}{x} + \frac{\lambda_\nu^2}{2x^2} + O(x^{-3})\right)\\
&= 1 + \frac{1}{nx}\sum_{\nu=1}^n\lambda_\nu + \frac{1}{2nx^2}\sum_{\nu=1}^n\lambda_\nu^2 + O(x^{-3}).
\end{align}$$
Write $\mu := \frac1n\sum\limits_{\nu=1}^n\lambda_\nu$ and $s = \frac1n\sum\limits_{\nu=1}^n \lambda_\nu^2$, so we have
$$A := \frac1n\sum_{\nu=1}^n a_\nu^{1/x} = 1 + \frac{\mu}{x} + \frac{s}{2x^2} + O(x^{-3}).$$
Thus
$$\begin{align}
\log A &= \frac{\mu}{x} + \frac{1}{2x^2}\left(s - \mu^2\right) + O(x^{-3}),\\
x\log A &= \mu + \frac{1}{2x} (s-\mu^2) + O(x^{-2}),\\
A^x - \sqrt[n]{a_1a_2\dotsb a_n} &= \exp \left(x\log A\right) - e^\mu\\
&= e^\mu\left(\exp (x\log A - \mu) - 1\right)\\
&= e^\mu\left(\exp \left(\frac{s-\mu^2}{2x} + O(x^{-2}) \right)-1\right)\\
&= e^\mu\left(\frac{s-\mu^2}{2x} + O(x^{-2})\right)\\
x\left(A^x - e^\mu\right) &= e^\mu\left( \frac{s-\mu^2}{2} + O(x^{-1})\right),
\end{align}$$
whence
$$\begin{align}
\lim_{x\to +\infty}&\; x\left[\left(\frac{a_1^{1/x} + a_2^{1/x} + \dotsb + a_n^{1/x}}{n}\right)^x - \sqrt[n]{a_1a_2\dotsb a_n}\right]\\
&= e^\mu\frac{s-\mu^2}{2}\\
&= \sqrt[n]{a_1a_2\dotsb a_n}\frac{n\sum\limits_{\nu=1}^n \log^2 a_\nu - \left(\sum\limits_{\nu=1}^n \log a_\nu\right)^2}{2n^2}.
\end{align}$$
| {
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"question_score": "8",
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Proving inequality $(a+\frac{1}{a})^2 + (b+\frac{1}{b})^2 \geq \frac{25}{2}$ for $a+b=1$ If $a, b$ are positive real numbers and $a+b = 1$, prove that :
$$\left(a+\frac{1}{a}\right)^2 + \left(b+\frac{1}{b}\right)^2 \geq \frac{25}{2}$$
I can see that the value $\frac{25}2$ is attained for $a=b=\frac12$. But I do not know how to show that this is the minimal possible value.
Thank you.
| First Method.
$a^{2}+\dfrac{1}{a^{2}}\geq -15a+\dfrac{47}{4}$ $~$ $\Longleftrightarrow$ $~$ $(2a-1)^{2}(a^{2}+16a+4)\geq 0$ : evident
$\therefore$ $\left(a+\dfrac{1}{a}\right)^{2}+\left(b+\dfrac{1}{b}\right)^{2}=4+a^{2}+\dfrac{1}{a^{2}}+b^{2}+\dfrac{1}{b^{2}}\geq 4-15a+\dfrac{47}{4}-15b+\dfrac{47}{4}=\dfrac{25}{2}$
Second Method.
$\left(a+\dfrac{1}{a}\right)^{2}+\left(b+\dfrac{1}{b}\right)^{2}=4+\underline{a^{2}+b^{2}}+\underline{\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}}\geq 4+\underline{\dfrac{1}{2}}+\underline{8}=\dfrac{25}{2}$
*
*$(1^{2}+1^{2})(a^{2}+b^{2})\geq (a+b)^{2}$ : Cauchy-Schwarz
*$(a+b)(a+b)\left(\dfrac{1}{a^{2}}+\dfrac{1}{b^{2}}\right)\geq (1+1)^{3}$ : Holder
| {
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Help verify $\lim_{x\to\infty} \frac{20x^2+6x+4}{7x-6-5x^2}$ Help verify this answer. Determine the limit $$\lim_{x\to\infty} \frac{20x^2+6x+4}{7x-6-5x^2}$$
I say the answer is $-4$ because what I did was took the highest $x$'s on both sides and divided them. So $\frac{20x^2}{-5x^2}$ and because the $x$'s cancel out so I can't use $\infty$ in place of $x$, I'm left with $-4$.
If anyone can verify that I got the right answer, I would appreciate it.
| You're absolutely on the right track. Note that for non-zero $x,$ we can rewrite $$\frac{20x^2+6x+4}{7x-6-5x^2}=\cfrac{\frac{20x^2+6x+4}{x^2}}{\frac{7x-6-5x^2}{x^2}}=\cfrac{20+\frac6x+\frac4{x^2}}{\frac7x-\frac5{x^2}-5}.$$ The limit of the numerator on the far right as $x\to\infty$ is $20$ and the corresponding limit of the far-right denominator is $-5.$ Hence, since $x$ is eventually non-zero as we let $x$ grow without bound, we have $$\lim_{x\to\infty}\frac{20x^2+6x+4}{7x-6-5x^2}=\frac{20}{-5}=-4,$$ as you said.
| {
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"source": "stackexchange",
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Math inequality proof If $a, b$ are positive real numbers and $a + b = 1$, prove that
$$
\left(a +\frac{1}{a}\right)^2 + \left(b +\frac{1}{b}\right)^2 \geq \frac{25}{2}
$$
Thank you.
| Let $\displaystyle \phi(x) = \left(x + \frac{1}{x}\right)^2$. Since
$$\frac{d^2}{dx^2} \phi(x) = 2 + \frac{6}{x^4} > 0,$$
$\phi(x)$ is a strictly convex function for $x > 0$.
By Jensen's inequality, we have
$$\left(a + \frac{1}{a}\right)^2 + \left(b + \frac{1}{b}\right)^2 = \phi(a)+\phi(b) \ge 2 \phi(\frac{a+b}{2}) = 2 \left(\frac12 + 2\right)^2 = \frac{25}{2}$$
| {
"language": "en",
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"source": "stackexchange",
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Can anyone solve this integral function? I want to know the what is integral function of $\displaystyle y=\frac{\exp(a+bx+cx^2)}{1+\exp(a+bx+cx^2)}$.
Thanks so much. I can remember how to integrate.
| Assume $c\neq0$ for the key case:
$\int\dfrac{e^{a+bx+cx^2}}{1+e^{a+bx+cx^2}}dx$
$=\int\dfrac{1}{e^{-a-bx-cx^2}+1}dx$
$=\int\dfrac{1}{e^{-c\left(x^2+\frac{bx}{c}+\frac{a}{c}\right)}+1}dx$
$=\int\dfrac{1}{e^{-c\left(x^2+\frac{bx}{c}+\frac{b^2}{4c^2}+\frac{a}{c}-\frac{b^2}{4c^2}\right)}+1}dx$
$=\int\dfrac{1}{e^{-c\left(\left(x+\frac{b}{2c}\right)^2+\frac{4ac-b^2}{4c^2}\right)}+1}dx$
$=\int\dfrac{1}{e^\frac{b^2-4ac}{4c}e^{-c\left(x+\frac{b}{2c}\right)^2}+1}dx$
$=\int\dfrac{1}{e^\frac{b^2-4ac}{4c}e^{-cu^2}+1}du$ $\left(\text{Let}~u=x+\dfrac{b}{2c}\right)$
$=\int\dfrac{1}{2\sqrt{v}\left(e^\frac{b^2-4ac}{4c}e^{-cv}+1\right)}dv$ $(\text{Let}~v=u^2)$
Which may relates to the incomplete polylogarithm function
| {
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How prove this trigonometric identity Show that
$$\sum_{k=0}^{n-1}\dfrac{\tanh{\left(x\dfrac{1}{n\sin^2{\left(\dfrac{2k+1}{4n}\pi\right)}}\right)}}{1+\dfrac{\tanh^2{x}}{\tan^2{\left(\dfrac{2k+1}{4n}\pi\right)}}}=\tanh{(2nx)}$$
Thank you ,This problem I take some hours,and at last I don't prove it
and This problem is from
This book have some same problem.all is not true? if not true,and we how find it or edit it somewhere?
Thank you achille hui,he told me this following maybe is true,Now How prove it?
$$\sum_{k=0}^{n-1}\dfrac{\dfrac{\tanh{x}}{n\sin^2{\left(\dfrac{2k+1}{4n}\pi\right)}}}{1+\dfrac{\tanh^2{x}}{\tan^2{\left(\dfrac{2k+1}{4n}\pi\right)}}}=\tanh{(2nx)}$$
| Notice
$$\cosh(2nx) = T_{2n}(\cosh x)\quad\text{ and }\quad \cos(2nx) = T_{2n}(\cos x)$$
where $T_{2n}(z)$ is a
Chebyshev polynomial of the first kind. Using the $2^{nd}$ relation above, it is clear the roots of $T_{2n}(x)$ has the
form:
$$\pm\cos(\frac{2k+1}{4n}\pi),\quad\text{ for } k = 0,\ldots, n-1$$
From this, we arrive following expansion of $\cosh(2n x)$:
$$\cosh(2n x ) = A \prod_{k=0}^{n-1}\left(\cosh^2 x - \cos^2(\frac{2k+1}{4n}\pi)\right)$$
for some constant $A$ we don't care.
Taking logarithm, differentiate w.r.t $x$ and divide by $2n$ for both sides, we find:
$$\begin{align}
\tanh(2n x)
= & \frac{1}{2n} \sum_{k=0}^{n-1}\frac{ 2\sinh x\cosh x}{\cosh^2 x - \cos^2(\frac{2k+1}{4n}\pi)}\\
= & \frac{1}{n}\sum_{k=0}^{n-1}\frac{ \sinh x\cosh x}{\sinh^2 x + \sin^2(\frac{2k+1}{4n}\pi)}\\
= & \frac{1}{n}\sum_{k=0}^{n-1}\frac{ \tanh x}{\tanh^2 x + \sin^2(\frac{2k+1}{4n}\pi)(1 - \tanh^2 x)}\\
\\
= & {\Large \sum_{k=0}^{n-1}\frac{\frac{\tanh x}{n \sin^2(\frac{2k+1}{4n}\pi)}
}{1 + (\frac{1}{\sin^2(\frac{2k+1}{4n}\pi)} - 1 ) \tanh^2 x} }\\
\\
= & {\Large \sum_{k=0}^{n-1}\frac{\frac{\tanh x}{n \sin^2(\frac{2k+1}{4n}\pi)}
}{1 + \frac{\tanh^2 x}{\tan^2(\frac{2k+1}{4n}\pi)}} }\\
\end{align}$$
| {
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Show in between steps in this Riemann zeta function equivalence/reduciton In the answer chosen by the OP in this question I had trouble understanding the steps taken to get the equivalences/reduce the zeta function into another one. Can somebody show me the steps to go from one step to the next in this:
$$\begin{align}
\zeta(z)&=\frac 1{1^z}+\frac 1{2^z}+\frac 1{3^z}+\frac 1{4^z}+\frac 1{5^z}+\cdots\\
&=\frac 1{1^z}-\frac 1{2^z}+\frac 1{3^z}-\frac 1{4^z}+\frac 1{5^z}+\cdots+\frac 2{2^z}+\frac 2{4^z}+\cdots\\
&=\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n^z}+\frac 2{2^z}\left(1+\frac 1{2^z}+\frac 1{3^z}\cdots\right)\\
\zeta(z)&=\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n^z}+2^{1-z}\zeta(z)\\
\end{align}$$
By step, I mean to go from one equals sign to the next.
| I have added three intermediate states.
\begin{align}
\zeta(z)&=\frac 1{1^z}+\frac 1{2^z}+\frac 1{3^z}+\frac 1{4^z}+\frac 1{5^z}+\frac 1{6^z}+\frac 1{7^z}+\cdots\qquad\qquad\qquad(1)\\
&=\frac 1{1^z}+\left(-\frac 1{2^z}+\frac 2{2^z}\right)+\frac 1{3^z}+\left(-\frac 1{4^z}+\frac 2{4^z}\right)+\frac 1{5^z}+\left(-\frac 1{6^z}+\frac 2{6^z}\right)+\frac 1{7^z}+\cdots\\
&=\left(\frac 1{1^z}-\frac 1{2^z}+\frac 1{3^z}-\frac 1{4^z}+\frac 1{5^z}-\frac 1{6^z}+\frac 1{7^z}+\cdots\right)+\frac 2{2^z}+\frac 2{4^z}+\frac 2{6^z}+\cdots\\
&=\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n^z}+\frac 2{2^z\,1^z}+\frac 2{2^z\,2^z}+\frac 2{2^z\,3^z}+\cdots\\
&=\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n^z}+\frac 2{2^z}\left(\frac 1{1^z}+\frac 1{2^z}+\frac 1{3^z}+\cdots\right)\\
\zeta(z)&=\sum_{n=1}^\infty \frac {(-1)^{n-1}}{n^z}+2^{1-z}\;\zeta(z)\qquad\text{from}\ (1)\ \,\text{and since}\ \,2^{1-z}=\frac {2^1}{2^z}\\
\end{align}
If this is not cleared please let me know (I'll try a more classical $\sum$ formulation...)
| {
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Monotone and Bounded Sequences: Proof Say $s_1=1$ and $s_{n+1}=\frac{1}{5}(s_n+7)$ for $n\geq 1$. Prove that the sequence is monotone and bounded, then find the limit.
| We have
$$
s_{n+1}-s_n=\frac15(s_n-s_{n-1}) \quad \forall n \ge 2,
$$
and so we have
$$\tag{*}
s_{n+1}-s_n=\frac{s_2-s_1}{5^{n-1}}=\frac{3}{5^n} \quad \forall n \ge 1.
$$
Thus $s_n<s_{n+1}$ for every $n \ge 1$, i.e. $(s_n)$ is increasing.
Thanks to (*) we have for every $n \ge 1$:
\begin{eqnarray}
s_n&=&1+\sum_{i=1}^{n-1}(s_{i+1}-s_i)=1+\sum_{i=1}^{n-1}\frac{3}{5^i}=-2+3\sum_{i=0}^{n-1}\frac{1}{5^i}=-2+3\cdot\frac{1-\frac{1}{5^n}}{1-\frac15}\\
&=&-2+\frac{15}{4}\left(1-\frac{1}{5^n}\right)=\frac74-\frac34\cdot\frac{1}{5^{n-1}}<\frac74.
\end{eqnarray}
We deduce that $(s_n)$ is a convergent sequence and
$$
\lim_ns_n=\lim_n\left(\frac74-\frac34\cdot\frac{1}{5^{n-1}}\right)=\frac74.
$$
| {
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Contest math integer doublet equation Can anyone help me with this?
Find all ordered pairs $(x, y)$ of positive integers $x$, $y$ such that $$x^2 + 4y^2 = (2xy − 7)^2$$
| $$x^2 + 4y^2 - 4xy +4xy = (2xy-7)^2$$
$$(x+2y)^2 = (2xy-7)^2 + 4xy$$
$$ = 4x^2y^2 - 28xy +4xy +49$$
$$= 4x^2y^2 - 24xy +49$$
$$= (4x^2y^2 - 24xy +36) +13$$
$$(x+2y)^2 - (2xy -6)^2 +13$$
$$(x+2y +2xy -6) (x+2y - 2xy +6) = 13$$
The product of these two terms can only be 13 and 1
either
$(x+2y +2xy -6) = 13$ and $(x+2y - 2xy +6) = 1$
or
$(x+2y+ 2xy - 6) = 1$ and $(x+2y - 2xy + 6) = 13$
Case i)
$x+2y+2xy - 6 = 13$
$x+2y - 2xy + 6) = 1$
Adding these two
We get
$2x+4y = 14$
Subtracting
We get
$4xy -12 = 12$ This gives $xy = 6$
Case ii)
$x+2y+2xy - 6 = 1$
$x+2y - 2xy + 6 = 13$
Adding these two
We get
$2x+4y = 14$
Subtracting
We get
$4xy -12 = -12$ This gives $xy = 0$
Case ii) is not possible because x and y postive integers
Case i) will lead to possible combinations of $x= 6$ or $y =1$, or $x=3$ and $y = 2$ or $x=2$ and $y = 3$ or $x= 1$ and $y = 6$ as xy are positive integers
Substituting back you will see that the ordered pair(x,y) is (3,2) is the only solution
Thanks
Satish
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Olympiad inequality: is this reasoning sound? I am trying to show that for $a,b,c>0,\;abc=1:$
$$\underbrace{\frac{1}{b(a+b)}+\frac{1}{c(b+c)}+\frac{1}{a(c+a)}}_{X}\geq \frac{3}{2}$$
This problem is from the Zhautykov Olympiad of 2008.
Attempt:
If $X\geq \frac{3}{2}$ is satisfied for all $a,b,c$ satisfying the initial conditions then by letting $a\mapsto b$ and $b\mapsto a$, the following inequality must also be satisfied:
$$\underbrace{\frac{1}{a(a+b)}+\frac{1}{b(b+c)}+\frac{1}{c(c+a)}}_{Y}\geq\frac{3}{2}$$
The reasoning can be reversed, thus $X\geq\frac{3}{2}\iff Y\geq\frac{3}{2}$. Now, in any case:
$$X+Y=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\geq 3\sqrt[3]{\frac{1}{(abc)^2}}=3$$
By AM-GM. Hence either $X\geq \frac{3}{2}$ or $Y\geq \frac{3}{2}$ is true, and therefore they are both true.
Is this logic valid? It seems too easy.
| No, your reasoning is wrong. The inequality
$$X+Y \geq 3$$ is ok. But you did not proof that if $ a,b,c \gt 0 \land abc=1 \land X(a,b,c) \ge \frac{3}{2}$ for a special triple $(a,b,c)$ then $Y(a,b,c) \ge \frac{3}{2}$ for the same triple (and vice versa) but you only proved that if $ a,b,c \gt 0 \land abc=1 \land X(a,b,c) \ge \frac{3}{2}$ for all triples $(a,b,c)$ then $Y(a,b,c) \ge \frac{3}{2}$ (and vice versa) which is only a change in notation. So you assume $X(a,b,c) \ge \frac{3}{2}$ to prove $X(a,b,c) \ge \frac{3}{2}$
| {
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Prove that: $\left | 2x-y-4 \right |\geq 4\sqrt{2}+4$ Let $x,y\in \mathbb{R}$ know that $4x^2-9y^2=36$
Prove that:
$$\left | 2x-y-4 \right |\geq 4\sqrt{2}+4$$
| Let $u=2x-3y$ and $v = 2x+3y$. Then $uv = 4x^2-9y^2 = 36$, so
$$2x-y-4 = {2u+v \over 3} - 4 = {2 \over 3}u + 12u^{-1} - 4$$
${\partial \over \partial u}$ of this is ${2 \over 3}-12u^{-2}$, which is zero at $u=\pm3\sqrt{2}$. Plugging these values into ${2 \over 3}u + 12u^{-1} - 4$ yields the local extrema $\pm4\sqrt{2}-4$. It follows that $\left|2x-y-4\right| \geq 4\sqrt{2}-4$. $\square$
| {
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Factor $(x+y)^7-(x^7+y^7)$ I encountered the following problem while preparing for upcoming math contests.
Factor $(x+y)^7-(x^7+y^7)$.
I got zero for $(x+y)^7-(x^7+y^7)$, however, the solutions said it's
$$7xy(x+y)(x^2+xy+y^2)(x^2+xy+y^2)$$
Can someone explain how this is possible?
| Users have approached the problem using the Binomial Theorem, namely,
$$(a+b)^n=\sum_{k=0}^n\binom nka^{n-k}b^k,\tag1$$
such that the formula for a binomial coefficient is expressed thus: $$\dbinom nk\stackrel{\small\text{def}}{=}\frac{n!}{k!(n-k)!}.\tag{$n!=1\times 2\times 3\times\cdots\times n$}$$ It follows, then, that by substituting $(a,b,n)=(x,y,7)$ respectively, the answer is as follows: $$\begin{align}(x+y)^7&=x^7+7x^6y+21x^5y^2+35x^4y^3+35x^3y^4+21x^2y^5+7xy^6+y^7 \\ &= x^7+y^7+7(x^6y+xy^6)+21(x^5y^2+x^2y^5)+35(x^4y^3+x^3y^4) \\ &=x^7+y^7+7\big(x^6y+xy^6+3(x^5y^2+x^2y^5)+5(x^4y^3+x^3y^4)\big) \\ &=x^7+y^7+7xy\big(x^5+y^5+3xy(x^4+y^4)+5xy(x^3+y^3)\big),\end{align}$$ to which we can then factor $-$ but there is a better way!
Of course now, $(x+y)^7-(x^7+y^7)$ is a multiple of $x+y$. Let $x^7+y^7=x^7-(-y)^7$, then you can use the following formula from my answer to this post in order to factor the equation:
$$a^n-b^n=(a-b)\sum_{k=1}^na^{n-k}b^{k-1}.\tag2$$
It follows, then, that $$(x+y)^7-(x^7+y^7)=(x+y)\bigg((x+y)^6-\sum_{k=1}^7x^{n-k}y^{k-1}\bigg).$$ The following step requires you to use the binomial theorem for the algebraic expression $(x+y)^6$, which is much easier than calculating for $(x+y)^7$ by hand.
| {
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Sum of all $4$ Digit no. using the digit $1,2,2,3,4,5,5$, The Sum of all $4$ Digit no. using the digit $1,2,2,3,4,5,5$, when repetition of digit is not allowed
My Try::
Here we are selecting $4$ Digit from a set $\{1,2,2,3,4,5,5\}$ in which we select $2$ or $5$ up to twice, each
So we will form $2$ cases::
case (I) : When $4$ Digit selected no. contain one duplicate element (Like $\{1,2,2,3,4,5\}$ or $\{1,2,3,4,5,5\}$)
case (II) : When $4$ Digit selected no. contain Two duplicate element (Like $\{2,2,3,4,5,5\}$ or $\{1,2,2,4,5,5\}$ or $\{1,2,2,3,5,5\}$)
Now I Did not understand how can i proceed after that
plz help me , Thanks
| First, we approach it for the case where there are no repeated digits. We have to use $\{1, 2, 3, 4, 5\}$. There are $5*4*3*2 = 120$ such integers. The average value of each digit is $\frac{1+2+3+4+5}{5}=3$. So this yields a sum of $120*3*1111$.
Now, we approach it for the case where there are repeated digits.
Consider the case where we use 2 2's, and 2 other distinct digits. There are $4*3*{4 \choose 2} = 72$ such numbers. The average value of each digit is $\frac{1}{2} \times 2 + \frac{1}{2} \times \frac{1+3+4+5}{4} = 2.625$. Hence, the sum is $72 * 2.625 * 1111$.
Cosnider the case where we use 2 5's, and 2 other distinct digits. As before, there are 72 such numbers. The average value of each digit is $\frac{1}{2} \times 5 + \frac{1}{2} \times \frac{1+2+3+4}{4} = 3.75$. Hence the sum is $72 * 3.75 * 1111$.
Finally, add the cases where we use 2 2's and 2 5's. This is similar to the previous case and left to you.
| {
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A closed form for $\int_0^\infty e^{-a\,x} \operatorname{erfi}(\sqrt{x})^3\ dx$ Let $\operatorname{erfi}(x)$ be the imaginary error function
$$\operatorname{erfi}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{z^2}dz.$$
Consider the following parameterized integral
$$I(a)=\int_0^\infty e^{-a\,x} \operatorname{erfi}(\sqrt{x})^3\ dx.$$
I found some conjectured special values of $I(a)$ that are correct up to at least several hundreds of digits of precision:
$$I(3)\stackrel{?}{=}\frac{1}{\sqrt{2}},\ \ I(4)\stackrel{?}{=}\frac{1}{4\sqrt{3}}.$$
*
*Are these conjectured values correct?
*Is there a general closed-form formula for $I(a)$? Or, at least, are there any other closed-form special values of $I(a)$ for simple (e.g. integer or rational) values of $a$?
| Let $I = [0,1]$ and notice
$$\text{erfi}(t) = \frac{2}{\sqrt{\pi}} \int_0^t e^{z^2} dz \quad\implies\quad \text{erfi}(\sqrt{t}) = \frac{2}{\sqrt{\pi}}\sqrt{t} \int_I e^{tz^2} dz$$
The integral $\mathscr{I}$ we want can be rewritten as:
$$\begin{align}
\mathscr{I}
= & \int_0^\infty e^{-at} \sqrt{t}^3 \left(\int_I e^{tz^2}dz\right)^3 dt\\
= & \frac{8}{\sqrt{\pi}^3}
\int_{I^3} dx dy dz\left[\int_0^\infty \sqrt{t}^3e^{-(a-x^2-y^2-z^2)t}) dt\right]\\
= & \frac{8}{\sqrt{\pi}^3} \Gamma(\frac52)
\int_{I^3} \frac{dx dy dz}{\sqrt{a - x^2 - y^2 - z^2}^5}\\
= & \frac{8}{\pi} \frac{\partial^2}{\partial a^2}
\int_{I^3} \frac{dx dy dz}{\sqrt{a - x^2 - y^2 - z^2}}
\end{align}$$
Since the maximum value of $x^2 + y^2 + z^2$ on $I^3$ is $3$, above rewrite is valid
when $a > 3$.
To compute the integral, we will split the cube $I^3$ into 6 simplices according to the
sorting order of $x, y, z$. On any one of these simplices, say the one for
$1 \ge x \ge y \ge z \ge 0$, introduce parameters $(\rho,\lambda,\mu) \in I^3$ such that $(x, y, z) = (\rho,\rho\lambda,\rho\lambda\mu)$, we have:
$$\begin{align}
\mathscr{I}
= & \frac{48}{\pi} \frac{\partial^2}{\partial a^2} \int_{I^3}
\frac{\rho^2 \lambda d\rho d\lambda d\mu}{\sqrt{a - \rho^2 - \lambda^2 \rho^2 ( 1 + \mu^2})}\\
= & \frac{48}{\pi} \frac{\partial^2}{\partial a^2} \int_{I^2}
\frac{\rho^2 d\rho d\mu}{\rho^2(1+\mu^2)} \left[ - \sqrt{a - \rho^2 - \lambda^2 \rho^2 ( 1 + \mu^2}) \right]_{\lambda=0}^1\\
= & \frac{48}{\pi} \frac{\partial^2}{\partial a^2} \int_{I^2}
\frac{\rho^2 d\rho d\mu}{\rho^2(1+\mu^2)} \left[ \sqrt{a - \rho^2 } - \sqrt{a - \rho^2 ( 2 + \mu^2}) \right]\\
= & \frac{24}{\pi} \frac{\partial}{\partial a} \int_{I^2} \frac{d\rho d\mu}{1+\mu^2}
\left[
\frac{1}{\sqrt{a - \rho^2 }} -
\frac{\frac{1}{\sqrt{2+\mu^2}}
}{\sqrt{\frac{a}{2+\mu^2} - \rho^2})}
\right]\\
= & \frac{24}{\pi} \frac{\partial}{\partial a} \int_{I} \frac{d\mu}{1+\mu^2}
\left[
\arcsin(\frac{1}{\sqrt{a}}) -
\frac{1}{\sqrt{2+\mu^2}} \arcsin(\sqrt{\frac{2+\mu^2}{a}})
\right]\\
= & \frac{24}{\pi} \int_{I} \frac{d\mu}{1+\mu^2}
\left[
\frac{-\frac{1}{2\sqrt{a}^3}}{\sqrt{1 - \frac{1}{a}}} -
\frac{-\frac{1}{2\sqrt{a}^3}}{\sqrt{1 - \frac{2+\mu^2}{a}}}
\right]\\
= & \frac{12}{\pi a } \int_{I} \frac{d\mu}{1+\mu^2}
\left[ \frac{1}{\sqrt{a-2-\mu^2}} - \frac{1}{\sqrt{a-1}} \right]\\
\stackrel{\color{blue}{[1]}}{=} & \frac{12}{\pi a}
\left[
\frac{1}{\sqrt{a-1}}\arctan \frac{\sqrt{a-1} \mu}{\sqrt{a - 2 -\mu^2}}
- \frac{1}{\sqrt{a-1}} \arctan \mu
\right]_{\mu=0}^1
\\
= & \frac{12}{\pi a\sqrt{a-1}} \left[ \arctan\left(\sqrt{\frac{a-1}{a-3}}\right) - \frac{\pi}{4}
\right]\\
= & \frac{12}{\pi a\sqrt{a-1}}
\arctan\left(
\frac{\sqrt{\frac{a-1}{a-3}}-1}{\sqrt{\frac{a-1}{a-3}}+1}
\right)\\
\stackrel{\color{blue}{[2]}}{=} &
\frac{6}{\pi a\sqrt{a-1}} \arctan\frac{1}{\sqrt{(a-1)(a-3)}}
\end{align}$$
Notes
*
*$\color{blue}{[1]}$ I am lazy, I get the leftmost integral in RHS from Wolfram alpha instead of deriving it myself.
*$\color{blue}{[2]}$ Let $u = \sqrt{\frac{a-1}{a-3}}$, we have
$$\begin{align}
2\arctan\frac{u-1}{u+1}
= & \arctan\frac{2\frac{u-1}{u+1}}{1-\left(\frac{u-1}{u+1}\right)^2}
= \arctan\frac{u^2 - 1}{2u}\\
= & \arctan\frac{\frac{a-1}{a-3}-1}{2\sqrt{\frac{a-1}{a-3}}}
= \arctan\frac{1}{\sqrt{(a-1)(a-3)}}
\end{align}$$
| {
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Derivative of trigonometric functions
Find the derivative of $$\dfrac{5\sin x}{1-\cos x}.$$
I tried to do this myself by applying the product rule and got $$\dfrac{5\cos x}{1-\cos x}-\dfrac{5\sin x}{(1-\cos x)^2}.$$ I checked this with Wolfram Alpha but it was wrong. According to Wolfram Alpha, the derivative is $$\dfrac{5}{\cos(x) -1}.$$ Am i missing some simplification here?
| This is how I would simplify it:
if $$f(x)=g(x)h(x)$$
then $$f'(x) = h'(x)g(x)+g'(x)h(x)$$
so$$\dfrac{d}{dx}(\dfrac{5\sin x}{1-\cos x})$$
$$={\dfrac{d}{dx}(5\sin x)}(1-\cos x)^{-1}$$
$$=({\dfrac{d}{dx}(5\sin x)})*(1-\cos x)^{-1}+{\dfrac{d}{dx}(1-\cos x)^{-1}}*(5\sin x)$$
$$= (5\cos x)(1-\cos x)^{-1}+(-1)(1-\cos x)^{-2}(\sin x)(5\sin x)$$
$$=\dfrac{5\cos x}{1-\cos x}-\dfrac{5\sin ^2x}{(1-\cos x)^2}$$
Now we can multiply $\dfrac{5\cos x}{1-\cos x}$ by $\dfrac{1-\cos x}{1-\cos x}$
$$\dfrac{5\cos x}{1-\cos x} = \dfrac{5\cos x}{1-\cos x}*\dfrac{1-\cos x}{1-\cos x}$$
$$=\dfrac{5\cos x-5\cos ^2 x}{(1-\cos x)^2}$$
Now we can add the two fractions together
$$\dfrac{5\cos x}{1-\cos x}-\dfrac{5\sin ^2x}{(1-\cos x)^2}=\dfrac{5\cos x-5\cos ^2 x}{(1-\cos x)^2}-\dfrac{5\sin ^2x}{(1-\cos x)^2}$$
$$ = \dfrac{5\cos x-5\cos ^2 x -5\sin ^2x}{(1-\cos x)^2}$$
| {
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Is this operation legal? Is this operation allowed?
Going from this:
$\left ( \frac{x^{2}+6}{x^{2}-4} \right )^{2}= \left ( \frac{5x}{4-x^{2}} \right )^{2}$
To this: $\left ( \frac{\left (x^{2}+6 \right )\left ( 4-x^{2} \right )}{\left (x^{2}-4 \right )5x} \right )^{2}= 1$
| As others have noted, it is valid so long as $x\ne0,$ since otherwise you're dividing by $0$. Let me offer another approach, using the difference of squares formula $a^2-b^2=(a-b)(a+b)$, together with the fact that $(-c)^2=c^2.$ The following, then, are equivalent:
$$\left(\frac{x^2+6}{x^2-4}\right)^2=\left(\frac{5x}{4-x^2}\right)^2$$
$$\left(\frac{x^2+6}{x^2-4}\right)^2=\left(\frac{5x}{x^2-4}\right)^2$$
$$\left(\frac{x^2+6}{x^2-4}\right)^2-\left(\frac{5x}{x^2-4}\right)^2=0$$
$$\left(\frac{x^2+6}{x^2-4}-\frac{5x}{x^2-4}\right)\left(\frac{x^2+6}{x^2-4}+\frac{5x}{x^2-4}\right)=0$$
$$\frac{x^2-5x+6}{x^2-4}\cdot\frac{x^2+5x+6}{x^2-4}=0.$$
Now, note that none of the above equations make sense when $x=\pm 2,$ since $x^2-4=(x-2)(x+2).$ So, we must assume that $x\neq 2$ and $x\neq-2$. At that point, the last equation is readily equivalent to $$(x^2-5x+6)(x^2+5x+6)=0,$$ which is true if and only if $x^2-5x+6=0$ or $x^2+5x+6=0$. These two quadratic equations give the solutions $x=\pm 2,\pm3,$ and so since we had to rule out $x=\pm 2,$ we get $x=\pm 3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove the identity $3\sin^4x-2\sin^6x=1-3\cos^4x+2\cos^6x$? I'm trying to prove a trigonometric identity but I can't. I've been trying a lot but I can't prove it. The identity says like this:
$$3\sin^4x-2\sin^6x=1-3\cos^4x+2\cos^6x$$
The identity would be easy if $1-\cos^4x=\sin^4x$ and $1-\cos^6x=\sin^6x$ but we know that $\sin^4x+\cos^4x$ isn't equal with $1$ and $\sin^6x+\cos^6x$ isn't equal with $1$.
Can anybody help me?!
Thank you!
| HINT:
As $\sin^2x+\cos^2x=1,$
$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-2\sin^2x\cos^2x \ \ \ \ (1)$
$\sin^6x+\cos^6x=(\sin^2x+\cos^2x)^3-3\sin^2x\cos^2x(\sin^2x+\cos^2x)$
$\implies \sin^6x+\cos^6x=1-3\sin^2x\cos^2x \ \ \ \ (2)$
Now equate the values of $\sin^2x\cos^2x$ from $(1),(2)$
| {
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How does one prove that $6$ divides ${(3n)!}/{(n!)^3}$? How does one prove that $6$ divides the multinomial coefficient $\displaystyle \frac{(3n)!}{\;(n!)^3}$?
| $\displaystyle
\frac{(3n)!}{\;(n!)^3} = 3n\frac{(3n-1)\cdots(2n)}{n!} \, \frac{(2n-1)\cdots n}{n!} \, \frac{(n-1)\cdots 1}{n!} = 3 {3n-1 \choose n} {2n-1 \choose n}
$
$\displaystyle
\frac{(3n)!}{\;(n!)^3} = \frac{(3n)\cdots(2n+1)}{n!} \, 2n \, \frac{(2n-1)\cdots n}{n!} \, \frac{(n-1)\cdots 1}{n!} = 2 {3n \choose n} {2n-1 \choose n}
$
So
$\displaystyle
\frac{(3n)!}{\;(n!)^3}$ is a multiple of $2$ and of $3$, hence a multiple of $6$.
Here is a simpler solution:
$\displaystyle
\frac{(3n)!}{\;(n!)^3} = {3n \choose n} {2n \choose n} = \frac{3n}{n}{3n-1 \choose n-1} \frac{2n}{n} {2n-1 \choose n-1} = 6 {3n-1 \choose n-1} {2n-1 \choose n-1}
$
This solution can be generalized to prove that $\displaystyle
\frac{(kn)!}{\;(n!)^k}$ is a multiple of $k!$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove this inequality $\frac{2}{(a+b)(4-ab)}+\frac{2}{(b+c)(4-bc)}+\frac{2}{(a+c)(4-ac)}\ge 1$ let $a,b,c>0$,and such
$a+b+c=3$,
show that
$$\dfrac{2}{(a+b)(4-ab)}+\dfrac{2}{(b+c)(4-bc)}+\dfrac{2}{(a+c)(4-ac)}\ge 1$$
I think this inequality use this
$$ab\le\dfrac{(a+b)^2}{4}$$
| It is possible to prove a slightly weaker bound by cutting down the dimension of the problem.
For any $K\in[0,3]$, define:
$$f_K(x)=\frac{2}{(3-x)(4-x(K-x))},\qquad g_K(x)=f_{K}(x)+f_{K}(K-x).$$
By differentiating with respect to $x$, we have that the minima of $g_K(x)$ over $[0,3]$ are located in $\frac{K}{2}\pm\frac{1}{2}\sqrt{1+(3-K)^2}$, so
$$g_K(x) \geq g_K\left(\frac{K}{2}\pm\frac{1}{2}\sqrt{1+(3-K)^2}\right)=\frac{8(6-K)}{(13-3K)^2}$$
holds over $[0,K]$. By taking $x=b,K=b+c$, $a=3-K$ follows and we have:
$$\sum_{cyc}\frac{2}{(3-a)(4-bc)}=\frac{2}{K(4-x(K-x))}+g_K(x).\tag{1}$$
The first term in the RHS is a non-negative and convex function $h_K(x)$ on $[0,K]$ whose graphics is symmetric with respect to $x=K/2$. Since $h_K(K/2)>h_K(0)$, $h_K(x)\geq h_K(0)=\frac{1}{2K}$ follows. This gives:
$$\sum_{cyc}\frac{2}{(3-a)(4-bc)}\geq\frac{1}{2K}+\frac{8(6-K)}{(13-3K)^2}=j(K).\tag{2}$$
$j(K)$ is a convex function over $[0,3]$, whose minimum is attained in $x=1.4638\ldots$. This gives:
$$\sum_{cyc}\frac{1}{(3-a)(4-bc)}\geq 0.831262\ldots > \frac{4}{5}.\tag{3}$$
We can improve this to:
$$\sum_{cyc}\frac{1}{(3-a)(4-bc)}\geq h_K\left(\frac{K}{2}-\frac{1}{2}\sqrt{1+(3-K)^2}\right)+\frac{8(6-K)}{(13-3K)^2}=\frac{4(13+9K-2K^2)}{K(13-3K)^2},\tag{4}$$
where the RHS is increasing on $[2,3]$. Since we can assume $K=(b+c)\geq 2$ without loss of generality, we have:
$$\sum_{cyc}\frac{1}{(3-a)(4-bc)}\geq\frac{46}{49}.\tag{5}$$
| {
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Prove that if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ So, if $\gcd(a,b)=1$, then $\gcd(a^2,b^2)=1$ means $1=ax+by$, and want to show $a^2x+b^2y=1$.
By squaring $1=ax+by$ both sides, I get, $1=(ax)^2+2(ax)(by)+(by)^2$, but this doesn't help my proof.
| As $1 = ax + by$, then $ab = a^2bx + ab^2y$.
Therefore,
$1 = (ax + by)^2 = a^2x + b^2y + 2abxy$
$= a^2x + b^2y + 2(a^2bx + ab^2y)xy$
$= a^2(x + 2bx^2y) + b^2(y + 2axy^2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/512924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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if $f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$ If $\displaystyle f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$
$\underline{\bf{My\;\; Try}}::$ Given $\displaystyle f(x) = x-\frac{1}{x} = \frac{x^2-1}{x}.$ Now Replace $\displaystyle x\rightarrow \frac{1}{x}\;,$ We Get.
$\displaystyle f(f(x)) = x-\frac{1}{x}-\frac{x}{x^2-1} = \frac{x^2-1}{x}-\frac{x}{x^2-1} = \frac{(x^2-1)^2-x^2}{x.(x^2-1)} = \frac{x^4-3x^2+1}{x.(x^2-1)}$
again Replace $\displaystyle x\rightarrow \frac{1}{x}\;,$ We Get
$\displaystyle f(f(f(x))) = \frac{\left(\frac{x^2-1}{x}\right)^4-3\left(\frac{x^2-1}{x}\right)^2+1}{\left(\frac{x^2-1}{x}\right).\left\{\left(\frac{x^2-1}{x}\right)^2-1\right\}}$
Now I did not understand how can I solve This $8^{th}$ Degree equation,
So Help Required,
Thanks
| You want to solve $f(f(f(x)))=1$ for $x$. You can solve it by solving three equations in turn: Set $a:=f(f(x))$, then find $a$ via $f(a)=1$. Then set $b:=f(x)$ and find $b$ via $f(b)=a$. Lastly, find $x$ via $f(x)=b$. Notice also that because $f(-\frac{1}{x_0})=f(x_0)$, for each solution $x_0$, the number $-\frac{1}{x_0}$ is a solution too.
There are $8$ solutions and they all lie in the interval $(-1.9,2.5)$. You can e.g. take Wolfram alpha and plot your ugly rational function and see how often it takes the value $1$. Solving the equation is a one-liner in a program like Mathematica. One solution is $x=-0.40065\dots$, convince yourself by plugging in $-0.4$ and you find $f(f(f(-0.4)))=1.00797\dots$.
| {
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"url": "https://math.stackexchange.com/questions/518683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Doing Some Asymptotic Integration: Problem with a Taylor Series I evaluated this integral (with some help from you guys):
$$\int_0^{\arcsin\left(\frac{a}{b}\right)} \exp\left(-\beta\left(b\cos(x)-\sqrt{a^2 - b^2 \sin^2(x)}\right)^2 \right) \, dx$$
and got an answer in terms of an F1 Appell hypergeometric series (with some sums), so I tried using a Taylor expansion to simplify things...
I only need the integral to be valid for small $b\cos(x)-\sqrt{a^2-b^2 \sin^2(x)}$ so I Taylor expand:
$$
\int_{0}^{\arcsin\left(\frac{a}{b}\right)}\left(1-\beta\left(b\cos(x)-\sqrt{a^{2}-b^{2}\sin(x)}\right)^2\cdots\right)dx$$
$$
=\arcsin\left(\frac{a}{b}\right)-\beta\left[\left(a^{2}x+b\cos(x)\sin(x)-b\sin(x)\left(\sqrt{a^{2}-b^{2}\sin^{2}(x)}\right)-a^{2}\arctan\left(\frac{b\sin(x)}{\sqrt{a^{2}-b^{2}\sin^{2}(x)}}\right)\right)\right]_{0}^{\arcsin\left(\frac{a}{b}\right)}$$
according to Mathematica. Notice the $a/0$ in the $\arctan$ term at the upper limit, which makes the integral undefined at this point so the approach doesn't work. I know this integral has a definite value (from the hypergeometric version), so what's gone wrong with this approach?
| Following the approach like Difficult Gaussian Integral Involving Two Trig Functions in the Exponent: Any Help?, you will get the two key types of integral: $\int_0^{\sin^{-1}\frac{a}{b}}\cos^{2n-2m}x\left(a^2-b^2+b^2\cos^2x\right)^m~dx$ and $\int_0^{\sin^{-1}\frac{a}{b}}\cos^{2n-2m+1}x\left(a^2-b^2\sin^2x\right)^{m-\frac{1}{2}}~dx$ , where $m$ and $n$ are any non-negative integers
For $\int_0^{\sin^{-1}\frac{a}{b}}\cos^{2n-2m}x\left(a^2-b^2+b^2\cos^2x\right)^m~dx$ , where $m$ and $n$ are any non-negative integers, it can expand to the polynomial of $\cos x$ , so the indefinte integral will have $x$ term and polynomial of $\sin x$ and $\cos x$ , substitution of $0$ and $\sin^{-1}\dfrac{a}{b}$ should be no problem.
For $\int_0^{\sin^{-1}\frac{a}{b}}\cos^{2n-2m+1}x\left(a^2-b^2\sin^2x\right)^{m-\frac{1}{2}}~dx$ ,
$\int\cos^{2n-2m+1}x\left(a^2-b^2\sin^2x\right)^{m-\frac{1}{2}}~dx$
$=\int\cos^{2n-2m}x\left(a^2-b^2\sin^2x\right)^{m-\frac{1}{2}}~d(\sin x)$
$=\int(1-\sin^2x)^{n-m}\left(a^2-b^2\sin^2x\right)^{m-\frac{1}{2}}~d(\sin x)$
According to http://en.wikipedia.org/wiki/List_of_integrals_of_irrational_functions, the indefinte integral will have polynomial of $\sin x$ and $\sqrt{a^2-b^2\sin^2x}$ and $\sin^{-1}\dfrac{b\sin x}{a}$ term, substitution of $0$ and $\sin^{-1}\dfrac{a}{b}$ should be no problem.
How you get the $\tan^{-1}\dfrac{b\sin(x)}{\sqrt{a^2-b^2\sin^2x}}$ term? It should be not contained!
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{n\rightarrow \infty}\left(\sqrt{n^2+n+1}-\big\lfloor \sqrt{n^2+n+1} \big\rfloor \right)$ I need to find the limit
$$
\displaystyle \lim_{n\rightarrow \infty}\left(\sqrt{n^2+n+1}-\big\lfloor \sqrt{n^2+n+1} \big\rfloor \right),$$ where $n\in \mathbb{N}$.
My attempt. As $\displaystyle \lim_{n\rightarrow \infty} (n^2+n+1)\approx n^2$, then $\displaystyle \lim_{n\rightarrow \infty}\sqrt{n^2+n+1}\approx \displaystyle \lim_{n\rightarrow \infty}\sqrt{n^2} = n$.
So $$\displaystyle \lim_{n\rightarrow \infty}\left(\sqrt{n^2+n+1}-n\right) = \displaystyle \lim_{n\rightarrow \infty}\frac{\left(\sqrt{n^2+n+1}-n\right).\left(\sqrt{n^2+n+1}+n\right)}{\left(\sqrt{n^2+n+1}+n\right)}.$$
So $$\displaystyle \lim_{n\rightarrow \infty}\frac{n\cdot\left(1+\frac{1}{n}\right)}{n \left(\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}+1\right)} = \frac{1}{2}.$$
My Question is , Is my Process is Right OR Not ,OR Is there is any error .
If Not Then How can I Solve it
Help Required
Thanks
| Note that
$$
n^2<n^2+n+1<n^2+2n+1=(n+1)^2,
$$
and hence
$$
n<\sqrt{n^2+n+1}<n+1,
$$
and therefore $\lfloor\sqrt{n^2+n+1}\rfloor=n$. Therefore
\begin{align}
\sqrt{n^2+n+1}-\lfloor\sqrt{n^2+n+1}\rfloor&=\sqrt{n^2+n+1}-n
=\frac{\big(\sqrt{n^2+n+1}-n\big)\big(\sqrt{n^2+n+1}+n\big)}{\big(\sqrt{n^2+n+1}+n\big)}\\ &=
\frac{n+1}{\sqrt{n^2+n+1}+n}=\frac{1+\frac{1}{n}}{\sqrt{1+\frac{1}{n}+\frac{1}{n^2}}+1}\to\frac{1}{2},
\end{align}
as $n\to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/521230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Radical expression for Cosine formulas Is there nice radical expression for $$\cos\left(\frac{\pi}{2^k+1}\right)?$$
Example: $\cos\left(\dfrac{\pi}{5}\right)=\dfrac{\sqrt{5}+1}{4}$.
Please provide some concrete examples.
Also please provide a general procedure.
I would like to handle $$\cos\left(\frac{a\pi}{2^k+1}\right)$$ as well for $a\in\{0,1,2,\dots,k-1\}$.
| For the form $p=2^k+1$, you can find an expression in real (not complex) radicals for,
$$\cos\Big(\frac{n\pi}{p}\Big)$$
and $n$ a positive integer only when $p$ is a Fermat prime, hence $p=3,5,17,257,65537$. For higher $p$ there can be several ways to express it.
I. For $p=17$:
$$\cos\frac{2\pi}{17}=\frac{1}{4}\Bigl(\frac{1}{x}+\sqrt{x}\,(17+4\sqrt{17})^{1/4}\Bigr),\quad x =\frac{1}{4}\Bigl(1-\sqrt{17}+\sqrt{\bigl(1-\sqrt{17}\bigr)^2+16}\Bigr)$$
or, from this post,
$$\cos\frac{2\pi}{17}=\frac{-1+\sqrt{17}+\sqrt{34-2\sqrt{17}}+2\sqrt{17+3\sqrt{17}-\sqrt{34-2\sqrt{17}}-2\sqrt{34+2\sqrt{17}}}}{16}$$
II. For $p=257$:
See this post.
| {
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"timestamp": "2023-03-29T00:00:00",
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integrate $ \frac {7x^2 + 2x − 7}{ x^3 − x}$ dx I keep messing up with the integration part of this I think. Before that I have to factor out the $x$ on the bottom, and then set up the $A$ and $B$ right?
Evaluate the integral. (Remember to use $\ln|u|$ where appropriate. Use $C$ for the constant of integration.)
$$\int\frac{7x^2+2x-7}{x^3-x}\,dx$$
| Hint : Apply http://en.wikipedia.org/wiki/Partial_fraction_decomposition#Procedure
$$ \frac{7x^2+2x-7}{x^3-x} = \frac{7x^2+2x-7}{x(x^2-1)} = \frac{7x^2+2x-7}{x(x-1)(x+1)} $$
$$ \frac{7x^2+2x-7}{x(x-1)(x+1)} = \frac{A}{x} + \frac{B}{(x-1)} + \frac{C}{(x+1)} $$
You can obtain A,B,C by setting the value of x are 1,-1,0 respectively and compare the coefficient of different power of x i.e 2,1 and 0.
$$ \int (\frac{A}{x} + \frac{B}{(x-1)} + \frac{C}{(x+1)}) dx = \int \frac{A}{x} + \int \frac{B}{(x-1)} + \int \frac{C}{(x+1)}$$
Now, apply this, to get your final answer
$$ \int \frac{1}{x+k} = ln|x+k| + C$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometry : Simplify and find the value of $\tan\theta(1-\sec\frac{\theta}{2})(1-\sec\theta)(1-\sec2\theta)\dots(1-\sec2^{n-1}\theta)$ at n =1,2,3 Problem :
Simplify and find the value of $\tan\theta(1-\sec\frac{\theta}{2})(1-\sec\theta)(1-\sec2\theta)\dots(1-\sec2^{n-1}\theta)$ at n =1,2,3
My approach :
$\tan\theta(1-\sec\frac{\theta}{2})(1-\sec\theta)(1-\sec2\theta).....(1-\sec2^{n-1}\theta)$ ....(i)
$\tan\theta = \frac{2\tan\frac{\theta}{2}}{1-\tan^2\frac{\theta}{2}}$
=$\frac{2\sin\frac{\theta}{2} \cos\frac{\theta}{2}}{\cos^2\frac{\theta}{2}- \sin^2\frac{\theta}{2}}$
Putting this value in (i) we get :
$$ = \frac{2\sin\frac{\theta}{2} \cos\frac{\theta}{2}}{\cos^2\frac{\theta}{2}- \sin^2\frac{\theta}{2}}(1-\sec\frac{\theta}{2})(1-\sec\theta)(1-\sec2\theta).....(1-\sec2^{n-1}\theta)$$
$$ =\frac{2\sin\frac{\theta}{2}}{\cos^2\frac{\theta}{2}- \sin^2\frac{\theta}{2}}(\cos\frac{\theta}{2} - \sin\frac{\theta}{2} )(1-\sec\theta)(1-\sec2\theta).....(1-\sec2^{n-1}\theta)$$
$$ =\frac{2\sin\frac{\theta}{2}}{\cos\frac{\theta}{2} + \sin \frac{\theta}{2}}(1-\sec\theta)(1-\sec2\theta).....(1-\sec2^{n-1}\theta)$$
Please suggest whether this is the right approach of solving this or we can use some other method.. thanks..
| Use the following identity:
$1-\sec(2^m\theta)=-(1-\cos(2^m\theta))(\sec(2^m\theta))
=-\frac{2\cos(2^{m-1}\theta)}{cos(2^m\theta)}$
The given equation becomes:
$\tan(\theta)\frac{(-2)^{n+1}cos(\frac{1}{4}\theta)}{cos(2^{n-1}\theta)}$
If $n=1$, $4\cos(\theta/4)\tan(\theta)\sec(\theta)$
If $n=2$, $-8 \cos(\theta/4) \tan(\theta) \sec(2 \theta)$
If $n=3$, $16 \cos(\theta/4) \tan(\theta) \sec(4 \theta)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Can't solve following limit: $\lim_{x \to \infty}x \left( \sqrt[3]{5+8x^3} - 2x\right)$ Need to solve following problem:
$$\lim_{x \to \infty}x \left( \sqrt[3]{5+8x^3} - 2x\right) $$ I've tried to do something like this: $$\lim_{x \to \infty} x\left(\sqrt[3]{5+8x^3} - 2x\right) =\lim_{x \to \infty}x\left( \sqrt[3]{ \left( \frac {5}{x^3}+8 \right)x^3 } - 2x\right) = \lim_{x \to \infty} x\left( \left ( \sqrt[3] { \frac{5}{x^3}+8} \right)x - 2x\right) $$
It seems to be the right way, but I can't do my next step.
| Hint:
$$\sqrt[3]{5+8x^3} - 2x = \sqrt[3]{5+8x^3} - \sqrt[3]{8x^3} = \frac{\left(5+8x^3\right)-8x^3}{\left(\sqrt[3]{5+8x^3}\right)^2 + \sqrt[3]{5+8x^3} \cdot \sqrt[3]{8x^3} + \left(\sqrt[3]{8x^3}\right)^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/525028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Triangle inequality- complex I am trying to prove the triangle inequality purely algebraically.
Let $z=x+iy$, $w=u+iv$.
Then,
$|z+w|^2$=$|(x+u)+i(y+v)|^2$=$(x+u)^2+(y+v)^2$=$x^2+2xu+u^2+y^2+2yv+v^2$
I tried the other way:
$(|z|+|w|)^2$=$(\sqrt{x^2+y^2}+\sqrt{u^2+v^2})^2$=$x^2+y^2+u^2+v^2+2 \sqrt{x^2+y^2} \sqrt{u^2+v^2}$
I'm trying to do this without using the complex conjugate or anything.
Any help is appreciated.
| Hint: Use Cauchy-Schwarz inequality:
$$
\sqrt{x^2+y^2} \sqrt{u^2+v^2}\geq xu+yv
$$
To prove it directly:
$$
\sqrt{x^2+y^2} \sqrt{u^2+v^2}\geq xu+yv\implies\\
x^2u^2+x^2v^2+y^2u^2+y^2v^2\geq x^2u^2+2xvyu+y^2v^2\implies\\
x^2v^2+y^2u^2 - 2xvyu\geq 0 \implies\\
(xv-yu)^2 \geq 0 \\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/527043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Inequality $\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$+$\frac{1}{d}$+$\frac{9}{a+b+c+d}\geq 25/4$
Prove that $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{9}{a+b+c+d}\geq \frac{25}{4}$$
given that $a, b, c, d > 0$ and $abcd = 1$
I reach to a point that says $\frac{1}{a}$+$\frac{1}{b}$+$\frac{1}{c}$+$\frac{1}{d}$+$\frac{9}{a+b+c+d}\geq \frac{25}{a+b+c+d}$
| $$\frac{1}{a^4} + \frac{1}{b^4} + \frac{1}{c^4} + \frac{1}{d^4} + \frac{9}{a^4 + b^4 + c^4 + d^4} $$
$$\geq\frac{8}{9}\left(\frac{1}{a^2b^2}+\frac{1}{a^2c^2}+\frac{1}{a^2d^2}+\frac{1}{b^2c^2}+\frac{1}{b^2d^2}+\frac{1}{c^2d^2}\right)+ \frac{11}{3\left(a^4 + b^4 + c^4 + d^4\right)}$$
$$\geq\frac{25}{4abcd},$$
where the last inequality follows from
\begin{align*}
&36a^2b^2c^2d^2(a^4 + b^4 + c^4 + d^4)\left[\dfrac{11}{3t(a^4 + b^4 + c^4 + d^4)}-\frac{25}{4abcd}+\frac{8}{9}\sum_{sym}\frac{1}{a^2b^2}\right]\\
&=\sum_{sym}\dfrac{(a-b)^2}{12}\left((a-b)^2[192(a^2+b^2)(d^2+c^2)+ab(331(c^2-186cd+331d^2)+80cd(c+d)(a+b)
+80c^2d^2]+3cd(c+d)^2(18ab+119c^2-194cd+119d^2)\right)\ge 0
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/527285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to solve $\lim_{x\to 16} \frac{4-\sqrt{x}}{16x-x^2}$ Solve the following question :
\begin{eqnarray}
\\\lim_{x\to 16} \frac{4-\sqrt{x}}{16x-x^2}\\
\end{eqnarray}
The answer should be $\frac{1}{128}$.
I try that:
\begin{eqnarray}
\\\lim_{x\to 16} \frac{4-\sqrt{x}}{16x-x^2} &=& \lim_{x\to 16} \frac{(4-\sqrt{x})(4+\sqrt{x})}{(4\sqrt{x}-x)(4\sqrt{x}+x)(4+\sqrt{x})}\\
\\ &=& \lim_{x\to 16} \frac{16-x}{(4\sqrt{x}-x)(4\sqrt{x}+x)(4+\sqrt{x})}\\
\end{eqnarray}
What can I do?
Thank you for your attention.
| Putting $\sqrt x=h,$
We have $$\lim_{h\to4}\frac{4-h}{16h^2-h^4}=\lim_{h\to4}\frac{4-h}{h^2(16-h^2)}=\lim_{h\to4}\frac{4-h}{h^2(4-h)(4+h)}=\lim_{h\to4}\frac1{h^2(4+h)}=\cdots$$ Cancelling out $4-h$ as $4-h\ne0$ as $h\to4$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What's wrong with my answer? $\int \frac{3x^2-2}{x^2-4x-12} \mathrm dx$ Here is the original problem: $\int \frac{3x^2-2}{x^2-4x-12}\ \mathrm dx$
After doing polynomial division and factoring the denominator I got this:
$$\int 3 + \frac{12x+36}{(x-6)(x+2)}\ \mathrm dx$$
Then using partial fraction decomposition I got the following:
$$\int 3\ \mathrm dx+ \frac{27}{2}\int \frac{\ \mathrm dx}{x-6} -\frac{3}{2}\int \frac{\ \mathrm dx}{x+2}$$
For the final answer I got this:
$$3x+\frac{27}{2}ln|x-6|-\frac{3}{2}ln|x+2|+C$$
But it says my answer is incorrect. Can you all spot my error or should I provide more details?
| You did the calculation wrong.
$$\frac{3x^2-2}{x^2-4x-12} = \frac{3(x^2-4x-12)-2+12x+36}{x^2-4x-12} = 3+\frac{12x+34}{x^2-4x-12}$$
Now, try to evaluate your integral
| {
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Fix $a \in \mathbb{Z} \setminus \{0\}$. Find all integers $n$ such that $\frac{n^3+a}{n^2+a}\in \mathbb{Z}? $ May I know what is the correct approach to tackle the a/m problem? Since $n^3+a$ and $n^2+a$ have no common divisors, in order for $\dfrac{n^3+a}{n^2+a}\in \mathbb{Z},$ we must have $ n^3 +a = n^2 + a \implies n = 0 \lor 1 ?$
| One of the ways is to find the common divisors of $n^3+a,n^2+a$
Let integer $d$ divides both
$\implies d$ divides $n(n^2+a)-(n^3+a)=an-a$
Again, $d$ will divide $a(n^2+a)-n(an-a)=an+a^2$
Again, $d$ will divide $na+a^2-(an-a)=a^2+a$
One of the necessary condition for $\displaystyle \frac{n^3+a}{n^2+a}$ be a positive integer is
that $n^2+a$ must divide $a^2+a$
| {
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A triangle has side lengths 4,6,8. A tangent is drawn to incircle parallel to side 4 cutting ..... Problem :
A triangle has side lengths $4,6,8$. A tangent is drawn to incircle parallel to side $4$ cutting other two sides at M and N, than length of MN is
(a) $\frac{10}{9}$
(b) $\frac{20}{ 9}$
(c) $\frac{5}{3}$
(d) $\frac{4}{3}$
I have no clue how to proceed such problem however, please help on this I will be greatful to you thanks...
| I was waiting for a more elegant solution to come up, but has not, so I will go ahead. Let the triangle be $ABC$, such that $M$ is on $AC$, and $N$ is on $AB$. Let $AM = x$, $AN = y$, and $MN = z$, and $\angle BAC = \theta$. Since, triangles $ABC$ and $ANM$ are similar due to AA criterion, we have the following relations: $y = \frac{4x}{3}$ and $z= \frac{2x}{3}$. $(1)$
Now, the area of the triangle $AMN$ can be given by $\frac{1}{2}xy\cdot \sin\theta$. Since $BNMC$ is a tangential quadrilateral, its area is given by the formula $rs_1$, where $r$ is the inradius and $s$ is the semiperimeter. Summing both of them gives us the area of triangle $ABC$, which can be given by the formula, $rs_2$, as well, but note that their semiperimeters would be different though the inradius would be the same. We can find that $s_2=9$, which gives us the equation:
$$rs_1 + \frac{1}{2}xy\cdot \sin\theta = rs_2 = 9r$$
So, our first task is to eliminate $\sin\theta$. In order to that, we have to eliminate $s_1$ as well:
$$s_1=\frac{4 + (6 -x)+z+(8-y)}{2} = 9 - \frac{x+y}{2}+\frac{z}{2} =9 - \frac{x+y}{2}+\frac{x}{3}$$
This gives:
$$ \sin\theta = \frac{2r}{xy}(\frac{x+y}{2} - \frac{x}{3})$$
Notice that the area of $ABC$ can be given by $\frac{1}{2}\cdot6\cdot8\cdot \sin\theta =24\cdot \sin\theta$. So,
$$24\cdot \sin\theta = \frac{48r}{xy}(\frac{x+y}{2} - \frac{x}{3}) =9r $$
Cancelling the $r$ [and a bit of manipulation] and substituting for $y$ from $(1)$ gives us:
$$\frac{1}{2}(x+\frac{5x}{3})= \frac{x^2}{4}$$
Solving the [actually linear] equation gives us , $x = \frac{10}{3}$. Now we have reached our final destination, to get the final answer, substitute back from (1), which gives:
$$MN = z = \frac{2x}{3} = \frac{2\cdot\frac{10}{3}}{3}= \frac{20}{9}$$
BTW: Although the length of explanation suggests it took 10 - 20 minutes to solve the question, it took me only 2 - 3. Its surely a long method, but fast and effective nonetheless. When trying to solve questions like this, and not looking to come up with especially elegant solutions, equating [and summing if required] some property of the geometric figures from different aspects will always help, and area is always a good try.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\gcd(c^a + 1, c^b + 1)$ for even $a$ and $b$? Following on this question, what is the Greatest Common Denominator of $c^a + 1$ and $c^b + 1$, where $a, b, c \in N$. I know that for odd a and b, we have $\gcd(c^a + 1, c^b + 1) = c^{\gcd(a, b)} + 1$ Thanks, Aleks Vlasev
I also found that for a odd and b even (or vice versa), the result is 1 or 0 depending on odd or even c.
The last case remains: For both $a$ and $b$ even: How can $gcd(c^a +1, c^b+1)$ be simplified such that it can be computed more quickly?
I tried to adapt this answer to my case but got nowhere*. Or maybe I should be using Fermat's little theorem?
Edit:
I did use the last linked answer to simplify this. Let a > b, $(c^b + 1, c^a + 1) = (c^b + 1, -c^{a-b} + 1)$.
| Let $2^r$ be the highest power of 2 dividing $a$ and $2^s$ the highest power of 2 dividing $b$. Then if $r\ne s$, $\gcd(c^a+1,c^b+1)=2$ if $c$ is odd and 1 if $c$ is even. On the other hand, if $r=s$, $\gcd(c^a+1, c^b+1)=c^{\gcd(a,b)}+1$.
Proof: Suppose $n>2$ is a factor of $c^a+1$ and $c^b+1$. Then let $k$ be the smallest non-negative integer such that $n$ is a factor of $c^k+1$. Then by considering the powers of $c$ ($\bmod n$), it is clear that $n$ is a factor of $c^t+1$ if and only if $t=(2r+1)k$ for some $r$.
Thus $a$ and $b$ must be multiples of $k$ with $\frac ak$ and $\frac bk$ both odd. This means the same powers of $2$ are factors of $a$ and $b$, and $k$ is a factor of $\gcd(a,b)$ with $\frac{\gcd(a,b)}k$ odd, so $n$ is a factor of $c^{\gcd(a,b)}+1$.
QED
| {
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Find all entire functions that satisfy $f(2z) = (1-2z)f(z)$ This is for homework, and I could use a little help. The question asks
Find all entire functions that satisfy $f(2z) = (1-2z)f(z)$.
Here is what I have done so far. Since $f$ is entire, I wrote
$$ f(z) = \sum_{n=0}^{\infty} a_n z^n = a_0 + a_1z + a_2z^2 + a_3z^3 + a_4z^4 + \dotsb $$
for some $z \in \mathbb{C}$. Then
$$ f(2z) = a_0 + 2a_1z + 4a_2z^2 + 8a_3z^3 + 16a_4z^4 + \dotsb $$
and
$$ (1-2z)f(z) = a_0 + (a_1-2a_0)z + (a_2-2a_1)z^2+(a_3-2a_2)z^3 + (a_4-2a_3)z^4 + \dotsb. $$
Comparing coefficients, I find that
\begin{align*}
a_0 &= a_0 \\
a_1 &= -2a_0 \\
a_2 &= \frac{2^2}{3}a_0 \\
a_3 &= -\frac{2^3}{7 \cdot 3}a_0 \\
a_4 &= \frac{2^4}{15 \cdot 7 \cdot 3}a_0 \\
a_5 &= \frac{2^5}{31 \cdot 15 \cdot 7 \cdot 3} a_0 \\
&\vdots
\end{align*}
Now $f$ looks like
$$ f(z) = a_0 \left( 1 - 2z + \frac{2^2}{3}z^2 - \frac{2^3}{7 \cdot 3}z^3 + \frac{2^4}{15 \cdot 7 \cdot 3}z^4 - \frac{2^5}{31 \cdot 15 \cdot 7 \cdot 3}z^5 + \dotsb \right). $$
Does the series in parenthesis represent any elementary function? Besides the denominators, it looks like the Taylor expansion of $e^{-2z}$.
| Observations:
1) $f(1)=f(2*0.5)=(1-2*0.5)f(0.5)=0$. Hence $f(2)=(1-2*1)f(1)=0$, and so, by induction, you have $f(2^k)=0$ for all $k\geq 0$.
2) If there is another zero, say $z_0\neq 2^k$, then $z_k=z_0/2^k$ will also be zeroes of $f(z)$, and hence $f(z)=0$. So:
3) If $f(z)$ is not identically $0$, then $f(z)$ vanishes at precisely $2^k, k\geq0$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $x_{n+1}=\frac{2}{9}(x_n^3+3)$ converges Let $x_1=1/2$ and $x_{n+1}=\frac{2}{9}(x_n^3+3)$ for $n\geq 1$. We want to prove that the sequence $(x_n)$ converges to real number $r\in (0,1)$ satisfying the equation $2r^3-9r+6=0$.
First part
For the sequence to converge, it must be bounded. A sequence $X:=(x_n)$ of real numbers is said to be bounded if there exists a real number $M>0$ such that $|x_n| \leq M$ for all $n\in \mathbb{N}$.
First, we show that $(x_n)$ converges.
This is where I struggle
*
*I tried showing by induction that $(x_n)$ < 1 for all $n$** but it didn't work
*I also tried the ratio test...
Second part
Claim The real number $r$ satisfies the equation $2r^3-9+6=0$
Proof - Since $x_n$ and $x_{n+1}$ convergence to the same real number $r$, we can substitute $r$ in the equation for $x_{n+1}$ when $n$ gets sufficiently large :
$$
\begin{aligned}
x_{n+1}&=\frac{2}{9}(x_n^3+3)\\
r&=\frac{2}{9}(r^3+3))\\
r&=\frac{2}{9}r^3+\frac{6}{9}\\
0&=\frac{2}{9}r^3-r+\frac{6}{9}\\
0&=2r^3-9r+6 \quad \text{multiply by 9}
\end{aligned}
$$
| Induction works fine. $x_n<1$ implies that $x_n^3<1$. Hence $$x_{n+1}<\frac{2}{9}(1+3)=\frac{8}{9}<1$$
In fact the same proof shows $\frac{8}{9}$ as an upper bound.
| {
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Problem based on Algebraic identities Algebraic identities:
$$(a+b)^2 = a^2 + b^2 + 2ab$$
$$(a-b)^2 = a^2 + b^2 - 2ab$$
Other identities can also come to solve this question?
If $x + 1/x = 5$ and $x^2 + 1/x^3 = 8$, then what would be the value of $x^3 + 1/x^2$?
Possible answers:
a- 215
b- 125
c- 256
d- 525
| let $$a=x^2+\dfrac{1}{x^3},b=x^3+\dfrac{1}{x^2}$$
then $$a+b=x^2+\dfrac{1}{x^2}+x^3+\dfrac{1}{x^3}$$
and
$$x^2+\dfrac{1}{x^2}=(x+\dfrac{1}{x})^2-2=23$$
$$x^3+\dfrac{1}{x^3}=(x+\dfrac{1}{x})^3-3(x+\dfrac{1}{x})=125-15=110$$
so
$$a+b=133$$
then
$$b=133-8=125$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Smallest such $n \in \mathbb{N}$ that $2^{n} \equiv 1 \pmod{5\cdot 7\cdot 9\cdot 11\cdot 13}$ Can anybody give me a hint about how to find smallest such $n \in \mathbb{N}$ that $2^{n} \equiv 1 \pmod{5\cdot 7\cdot 9\cdot 11\cdot 13}$?
I thought that I will find it piece by piece with help from my friend Fermat's Little Theorem, so :
*
*$2^{n_{1}} \equiv 1 \mod 5$, so $n_{1}=4$.
*$2^{n_{2}} \equiv 1 \mod 7$, so $n_{2}=6$.
*$2^{n_{3}} \equiv 1 \mod 3$, so $n_{3}=2$.
*$2^{n_{4}} \equiv 1 \mod 11$, so $n_{4}=10$.
*$2^{n_{5}} \equiv 1 \mod 13$, so $n_{5}=12$.
So, since I know that $x \equiv y \mod mz \Leftrightarrow x \equiv y \mod m$ when $z \neq 0$, so my lucky was to take the lowest common multiple of $4,6,2,10,12$, which is $60$ and lo and behold it fits the bill. But is it the smallest such $n$? If so, how to explain it?
| You are looking for the smallest $n_1,..,n_5$, FLT only gives you some $n$. But you should know that the order (the smallest such $n_i$) divides any other $n$.
So in each line, the smallest one must be a divisor of the one given by FLT.
*
*You know $2^{n_{1}} \equiv 1 \mod 5$ and $n_1$ is a divisor of $4$. Since $1,2$ don't work $n_1=4$.
*You know $2^{n_{2}} \equiv 1 \mod 7$ and $n_1$ is a divisor of $6$. We test $1,2,3,6$ and the smallest one which works is....
Continue.
At the end, the smallest number which works must be divisible by all of them, thus the lcm is the right choice.
| {
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Prove that $m=(x+y^2, y+x^2+2xy^2+y^4)$ is a maximal ideal of $\mathbb{C}[x,y]$.
Prove that $m=(x+y^2, y+x^2+2xy^2+y^4)$ is a maximal ideal of $\mathbb{C}[x,y]$.
I can show that the ideal $(x,y)$ of $\mathbb{C}[x,y]$ contains $m$ and $(x,y)$ is a maximal ideal. Therefore to show that $m$ is itself a maximal ideal I only need to show that $m=(x,y)$. I tried to prove it by looking for two polynomials $f,g$ such that $x=f(x,y)(x+y^2)+g(x,y)(y+x^2+2xy^2+y^4)$ but wasn't able to find them. And I'm not sure whether this is the best way of proving this. I would appreciate any help. Thank you.
| Indeed, $x=\color{red}{f(x,y)}(x+y^2)+\color{green}{g(x,y)}(y+x^2+2xy^2+y^4)$ with, for example, $\color{red}{f(x,y)}=1-(x+y^2)^3$ and $\color{green}{g(x,y)}=(x+y^2)^2-y$.
This can be found by noting that $a=x+y^2$ and $b=y+x^2+2xy^2+y^4$ are such that $b=y+a^2$ hence $x=a-y^2=a-(b-a^2)^2$. Thus, $x=\color{red}{(1-a^3)}a+\color{green}{(2a^2-b)}b$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Median for continuous distribution
Consider a continuous random variable X with probability density function given by $f(x)=cx$ for $1 \le x \le 5$, zero otherwise. Find the median.
First I calculate the CDF: $F(x)=cx^2/2$ for $1 \le x \le 5$, zero otherwise.
Now we have to solve for constant c by using the definition of PDF, namely:
$$\int\limits_{-\infty}^{\infty}f(x)dx=1 \implies \frac{c}{2}x^2\Big{|}_1^5=1 \implies c=\frac{1}{12} $$
Then to calculate the median, we set the CDF = 0.5:
$$\frac{1}{2}=\frac{1}{12}\cdot \frac{1}{2} \cdot x^2 \implies x=\sqrt{12}$$
But the book solution is $\sqrt{13}$. Can someone tell me what I am doing wrong?
Thank you.
| For median $m$:
$$
\int\limits_{-\infty}^{m} f(x) dx = \int\limits_{m}^{+\infty} f(x) dx\\
\int\limits_{1}^{m} f(x) dx = \int\limits_{m}^{5} f(x) dx\\
c \frac{x^2}{2}\Biggr|_{1}^{m} = c \frac{x^2}{2}\Biggr|_{m}^{5}\\
x^2\Biggr|_{1}^{m} = x^2\Biggr|_{m}^{5}\\
m = \sqrt{\frac{1^2 + 5^2}{2}}=\sqrt{13}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding the limit of $\frac{e^{x}+x-\cos(2x)}{x^2}$ How would one find the limit for the following problem.
$x\rightarrow\infty$
$\frac{e^{x}+x-\cos(2x)}{x^2}$
I did the hospital rule.
$\frac{e^x+1+2\sin(2x)}{2x}$
But now I am stuck I did this but I feel it diverges.
$e^x+1+2\sin(2x)*\frac{1}{2x}$
| Since $\cos(2x) \leq 1,\,\forall x$, and $e^x = 1+x+x^2/2+x^3/6+\cdots \geq 1+x+x^2/2+x^3/6$, we obtain, $$\frac{e^{x}+x-\cos(2x)}{x^2} \geq \frac{2x+x^2/2+x^3/6}{x^2} \geq \frac{x^3/6}{x^2} \geq \frac{x}{6},\,\forall x > 0.$$ The lower bound diverges, hence the original series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/545827",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Find the Minimum value of this expression Let $x,y,z>0$ such that: $x+y+z=1$. Find the minimum value of this expression.
$P=(\frac{x+1}{x})^3.\frac{z^2}{z^2+1}+(\frac{y+1}{y})^3.\frac{x^2}{x^2+1}+(\frac{z+1}{z})^3.\frac{y^2}{y^2+1}$
| Hint :
use Lagrange multipliers
| {
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$xy=22$ and $yz=26$: What is $x+y+z $ equal to? Given the following: $$xy=22,\qquad yz=26,$$ where $x,y,z\in\mathbb{N}$. Which of the following is a possible value of $ x + y + z $?
$ \textbf {(A) } 22 \qquad \textbf {(B) } 24 \qquad \textbf {(C) } 26 \qquad \textbf {(D) } 48 $
| Shortcut. Since both given products are even, we can just assume that $y$ is even, so let it be $y=2$. Then, $x=11$ and $z=13$. Then, $$x+y+z=11+2+13=\boxed{\textbf{(C) } 26}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this limit $\lim\limits_{x\to 0^{+}}\frac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$
Find the limit
$$\lim_{x\to 0^{+}}\dfrac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$$
My attempt: Since
$$\sin{x}=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+o(x^7)$$
$$\tan{x}=x+\dfrac{1}{3}x^3+\dfrac{2}{15}x^5+\dfrac{1}{63}x^7+o(x^3)$$
So
$$\sin{(\tan{x})}=\tan{x}-\dfrac{1}{3!}(\tan{x})^3+\dfrac{1}{5!}(\tan{x})^5-\dfrac{1}{7!}(\tan{x})^7+o(x^7)$$
Though this method might solve, I think this problem has nicer methods. Thanks.
| The function is of an indeterminate form, so use l'Hopital's Rule $7$ times.
Note that
.
At $x=0$, this is equal to $-168$. Taking the derivative of the denominator $7$ times, we get $7!=5040$. Thus, the answer is $$ - \dfrac {168}{5040} = \boxed {- \dfrac {1}{30}}. $$
| {
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"timestamp": "2023-03-29T00:00:00",
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How prove this inequality $a^2+b^2+c^2+8(ab+bc+ac)+3-10(a+b+c)\ge 0$ let $a,b,c\ge 0$,and such $abc=1$,show that
$$a^2+b^2+c^2+8(ab+bc+ac)+3-10(a+b+c)\ge 0$$
My solution: Without loss of generality,assume that
$a=\max{(a,b,c)}$, since $abc=1$,we have
$a\ge 1$,
we will show that
$$f(a,b,c)\ge f(a,t,t)\ge 0, t=\sqrt{bc},0<t\le 1$$
since $$f(a,b,c)-f(a,t,t)=(\sqrt{b}-\sqrt{c})^2[(\sqrt{b}+\sqrt{c})^2+8a-10]$$
then equivalent to
$$(\sqrt{b}+\sqrt{c})^2+8a\ge 10$$
which is true because
$$(\sqrt{b}+\sqrt{c})^2+8a\ge 4\sqrt{bc}+8a=4(a+\sqrt{bc})+4a\ge 8\sqrt{a\sqrt{bc}}+4a=8\sqrt[4]{a}+4a\ge 12$$
Now,since $a=\dfrac{1}{t^2}$,we have
$$f(a,t,t)=f(\dfrac{1}{t^2},t,t)=\dfrac{(10t^4-7t^2+2t+1)(t-1)^2}{t^4}$$
which is clearly nonnegative because
$$10t^4-7t^2+2t+1=(3t^2-1)^2+t(1-t)+t^4+t>0$$
Have other nice methods? Thank you
| Let $f(x) = x^2+\dfrac8x + 1 - 10x + 16 \log x$ for $x > 0$.
Then the given inequality is $f(a)+f(b)+f(c) \ge 0$, and it is sufficient to show $f(x) \ge 0$.
We note $f'(x) = \dfrac{2(x-2)^2 (x-1)}{x^2}$.
Thus for $x < 1, f'(x)<0$ and for $x > 1, f'(x) \ge 0$. Hence $\forall x >0, \; f(x) \ge f(1) = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sin 2\theta +\sin \theta =1$ I tried Wolframalpha to solve this equation. The solution is $\theta\approx 0.355$. Since once a wise guy at MSE told me not to trust machines, I would like to know what methods can be used to solve this equation. I would appreciate for a brief explanation.
By the way $0\leq\theta\leq\pi/4$.
My second question is for any $x>0$ and $x\in\mathbb{R}$, Does $$x\sin 2\theta +\sin \theta =1$$ have a real valued solution for $0\leq\theta\leq\pi/4$?
| $$\begin{align}
x \sin 2\theta + \sin\theta = 1 \quad &\implies \quad 2 x \sin\theta \cos\theta = 1 - \sin\theta \\
&\implies \quad 4 x^2 \sin^2\theta \left(1-\sin^2\theta\right)=\left(1-\sin\theta\right)^2 \\
&\implies \quad \left( 1 - s \right) \left( 4 x^2 s^3 + 4 x^2 s^2 + s - 1 \right) = 0 \\
\end{align}$$
where $s := \sin\theta$. The first factor's root, $s = 1$, corresponds to $\theta = \pi/2 + 2 n \pi$, which is outside the specified domain of interest. Roots via the second factor
$$p(s) := 4 x^2 s^3 + 4 x^2 s^2 + s - 1 = 0$$
can be found via the messy cubic formula. Notice that, by the Descartes Rule of Signs, the polynomial $p$ has exactly one positive root (for any real $x$). Notice also that
$$p(\sin 0) = p(0) = -1 \qquad \text{and} \qquad p\left(\sin\frac{\pi}{4}\right) = p\left(\frac{\sqrt{2}}{2}\right) = x^2 \left( 2 + \sqrt{2} \right) - \frac{1}{2}\left( 2 - \sqrt{2} \right)$$
The polynomial's single positive root lies between $\sin 0$ and $\sin\frac{\pi}{4}$ (inclusive) when, and only when, the signs at the endpoints disagree or the value at the larger endpoint vanishes; thus, that latter value must be non-negative,
$$x^2 \left( 2 + \sqrt{2} \right) \geq \frac{1}{2} \left( 2 - \sqrt{2} \right)$$
so that
$$x^2 \geq \frac{2 - \sqrt{2}}{2\left( 2 + \sqrt{2} \right)}=\frac{2-\sqrt{2}}{2\left(2 + \sqrt{2}\right)}\frac{2-\sqrt{2}}{2-\sqrt{2}} = \frac{\left( 2 - \sqrt{2}\right)^2}{4}$$
whereupon, the condition
$$|x| \geq \frac{2 - \sqrt{2}}{2}$$
is necessary and sufficient for the equation $x \sin 2\theta + \sin\theta = 1$ to have a solution in the interval $\left[0, \frac{\pi}{4}\right]$.
| {
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Why is normalization of inequalities possible? I have seen, in many proofs for inequalities, the author does something called normalization. I believe this is only possible for homogeneous inequalities. I saw this in a proof of Nesbitt's inequality:
$$\frac{a}{b+c} + \frac{b}{a +c} + \frac{c}{a+b} \geq \frac32$$
The above can be transformed as:
$$\frac{a+b+c}{b+c} + \frac{a+b+c}{a+c} + \frac{a+b+c}{a+b} - 3$$
$$=(a+b+c)\left(\frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b}\right) - 3$$
$$=\frac{1}{2}[(a+b) + (b+c) + (c+a)]\left(\frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b}\right) - 3$$
Now, the author says, since the inequality is homogeneous, we may normalize $a+b+c = 1$
From the $AM-HM$ inequality:
$$\frac{\frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b}}{3} \geq \frac{3}{2(a+b+c)}$$
$$\implies \frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b} \geq \frac92$$
Since $a+b+c = 1$, we may multiply the LHS with $\frac12(a+b+b+c+c+a)$, without changing its value
$$\implies \frac{1}{2}[(a+b) + (b+c) + (c+a)]\left(\frac{1}{b+c} + \frac{1}{a+c} + \frac{1}{a+b}\right) \geq \frac92$$
Subtracting $3$ from both sides gives us the result.
I don't understand the normalization procedure at all. I can't see why it is possible to do this. It also seems a bit too good to be true.
| Say $a + b+ c = k$. Let $a = k\alpha$, $b = k\beta$, $c = k\gamma$. Express the inequality in terms of $\alpha, \beta, \gamma$. What happens to $k$?
| {
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Sum of real roots of the equation $x^2 + 5|x| + 6 = 0$? Sum of real roots of the equation $x^2 + 5|x| +6 = 0$
| Hint
$|x|=x\forall x\ge0$
$|x|=-x\forall x<0$
Case 1 $x\ge 0$
$x^2+5x+6=(x+2)(x+3)=0\Rightarrow x=-2,-3$ but we already assumed $x\ge 0$ so $(\Leftrightarrow)$
Case 2 $x<0$
then the equation becomes according to the definition of $|x|$
$x^2-5x+6=(x-2)(x-3)=0\Rightarrow x=2,3$ again $(\Leftrightarrow)$
$(\Leftrightarrow)$ is the sign of contradiction
| {
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What is $\lim\limits_{i\to 0} \frac{2^n}{\frac{(n+1)\sin((n+1)\theta)}{\sin\theta } - \frac{(n-1)\sin((n-1)\theta)}{\sin\theta }} $? What is $$\lim\limits_{i\to 0} \dfrac{2^n}{\frac{(n+1)\sin((n+1)\theta)}{\sin\theta } - \frac{(n-1)\sin((n-1)\theta)}{\sin\theta }} $$ where $$\theta=\frac{i\pi}{n} $$ The second page of this document says, that it's $$\frac{2^{n-2}}{n}$$ but I'm at a complete loss as to how to come to this conclusion.
| It's not too hard to show that
$$ \lim_{x \to 0} \frac{\sin{mx}}{nx} = \frac{m}{n}$$
Thus
$$\lim\limits_{i\to 0} \dfrac{2^n}{\frac{(n+1)\sin((n+1)\theta)}{\sin\theta } - \frac{(n-1)\sin((n-1)\theta)}{\sin\theta }} = \frac{2^n}{(n+1)^2-(n-1)^2} = \frac{2^n}{4n} = \frac{2^{n-2}}{n}$$
| {
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Third-degree cosine inequality for obtuse triangle Suppose $\triangle ABC$ is an obtuse triangle with side lengths $a=BC, b=CA, c=AB$. I want to show that $$a^3\cos A+b^3\cos B+c^3\cos C<abc.$$
My idea is to use the cosine rule. I have $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, etc. Plugging into the inequality I get
$$a^4b^2+a^4c^2-a^6+b^4a^2+b^4c^2-b^6+c^4a^2+c^4b^2-c^6<2a^2b^2c^2.$$
How can I show this?
| here is another way follow op's idea:
for easy, $x=a^2,y=b^2,z=c^2,$ WOLG, let $C$ is obtuse triangle, then $ x+y<z$, we want to prove :
$x^2y+x^2z-x^3+y^2x+y^2z-y^3+z^2x+z^2y-z^3<2xyz$
that is to prove: when $z>x+y$
$f(z)=-(x-y)^2(x+y)+(x-y)^2z+(x+y)z^2-z^3 <0$
now we prove $f(z)$ is mono decreasing function:
$f'(z)=(x-y)^2+2(x+y)z-3z^2$
$f''(z)=2(x+y)-6z<0 \implies f'(z)<f'(x+y)=-4xy<0 \implies f(z) <f(x+y)=0$
| {
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Integral $\int_0^{\pi/2}\arctan^2\left(\frac{6\sin x}{3+\cos 2x}\right)\mathrm dx$ Is it possible to evaluate this integral in a closed form?
$$I=\int_0^{\pi/2}\arctan^2\left(\frac{6\sin x}{3+\cos 2x}\right)\mathrm dx$$
| I will refer to the following result from my previous answer:
\begin{align*}
I(r, s)
&= \int_{0}^{\frac{\pi}{2}} \arctan (r \sin\theta) \arctan (s \sin\theta) \, d\theta \\
&= \pi \chi_{2} \left( \frac{\sqrt{1+r^{2}} - 1}{r} \times \frac{\sqrt{1+s^{2}} - 1}{s} \right),
\end{align*}
where $\chi_{2}$ is the Legendre chi function. Using the addition formula for the arctangent, it follows that
$$ \arctan\left(\frac{6\sin x}{3 + \cos 2x} \right) = \arctan \left( \frac{\frac{3}{2}\sin x}{1 - \frac{1}{2}\sin^{2} x} \right) = \arctan (\sin x) + \arctan ( \tfrac{1}{2}\sin x). $$ So it follows that
$$ \int_{0}^{\frac{\pi}{2}} \arctan^{2}\left(\frac{6\sin x}{3+\cos 2x}\right) \, dx = I(1,1) + 2I(1,\tfrac{1}{2}) + I(\tfrac{1}{2},\tfrac{1}{2}), $$
which reduces to a combination of Legendre chi functions
$$ \pi \left\{ \chi_{2}(3 - 2\sqrt{2}) + \chi_{2}(9 - 4\sqrt{5}) + 2\chi_{2}\left( (\sqrt{2} - 1)(\sqrt{5} - 2) \right) \right\}. $$
| {
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How find this minimum $\sum_{i=1}^{n}a^2_{i}-2\sum_{i=1}^{n}a_{i}a_{i+1},a_{n+1}=a_{1}$ let $a_{1},a_{2},\cdots,a_{n}\ge 0$,and such $a_{1}+a_{2}+\cdots+a_{n}=1$.
Find this follow minimum
$$I=a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-\cdots-2a_{n-1}a_{n}-2a_{n}a_{1}$$
My try:since
$$a^2_{1}+a^2_{2}+\cdots+a^2_{n}\ge a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}+a_{n}a_{1}$$
this is true because
$$\Longleftrightarrow \dfrac{1}{2}[(a_{1}-a_{2})^2+(a_{2}-a_{3})^2+\cdots+(a_{n-1}-a_{n})^2+(a_{n}-a_{1})^2]\ge0$$
so
\begin{align*}
&a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-\cdots-2a_{n-1}a_{n}-2a_{n}a_{1}\\
&\ge a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2(a^2_{1}+a^2_{2}+\cdots+a^2_{n})\\
&=-(a^2_{1}+a^2_{2}+\cdots+a^2_{n})
\end{align*}
if we use Cauchy-Schwarz inequality,we have
$$(a^2_{1}+a^2_{2}+\cdots+a^2_{n})(1+1+\cdots+1)\ge (a_{1}+a_{2}+\cdots+a_{n})^2=1$$
But this is not usefull.
and Then I can't, yesterday I have ask this problem：How find this inequality minimum $\sum_{i=1}^{n}a^2_{i}-2\sum_{i=1}^{n-1}a_{i}a_{i+1}$,
and the Ewan Delanoy use nice method to solve it.I don't know this problem have this nice methods to solve it too(maybe can use previous question methods?but I can't use it.) and this problem have someone have research? if no,I think this is nice problem.
and Now I guess This problem when $a_{i}=\dfrac{1}{n}$ then $I$ is minimum?and the minimum is
$$I_{min}=-\dfrac{1}{n}?$$
Thank you
| UPDATE 11/14/2013 : Part of the first proposed answer was wrong. The method
proposed only works for $n<7$, and I currently have no complete solution for $n\geq 7$. All that is corrected in the updated version below.
Let us put
$$
Q_n(a_1,a_2,\ldots ,a_n)=
a^2_{1}+a^2_{2}+\cdots+a^2_{n}-2a_{1}a_{2}-2a_{2}a_{3}-\cdots-2a_{n-2}a_{n-1}-2a_{n-1}a_{n}-2a_na_1
$$
and
$$
T_n(a_1,a_2,a_3,\ldots,a_n)=
Q_n(a_1,a_2,\ldots,a_n)+\frac{(a_1+a_2+a_3+\ldots +a_n)^2}{n}
$$
For $3\leq n\leq 6$, the minimum is $-\frac{1}{n}$ (attained when all coordinates
are equal to $\frac{1}{n}$), because of
$$
\begin{array}{lcl}
T_6(a_1,a_2,\ldots,a_6)&=&
\frac{1}{42}\bigg(-5a_1+a_2+a_3+a_4-5a_5+7a_6\bigg)^2 \\
& & +\frac{1}{28}\bigg(-3a_1+2a_2+2a_3-5a_4+4a_5\bigg)^2
+\frac{1}{4}\bigg(-a_1+2a_2-2a_3+a_2\bigg)^2 \\
T_5(a_1,a_2,\ldots,a_5) &=&
\frac{1}{30}\bigg(-4a_1+a_2+a_3-4a_4+6a_5\bigg)^2
+\frac{1}{6}\bigg(-a_1+a_2-2a_3+2a_4\bigg)^2 \\
& &
+\frac{1}{2}\bigg(-a_2+a_3\bigg)^2+\frac{1}{2}\bigg(-a_1+a_2\bigg)^2 \\
& & \\
T_4(a_1,a_2,a_3,a_4) &=&
\frac{1}{20}\bigg(-3a_1+a_2-3a_3+5a_4\bigg)^2
+\frac{1}{20}\bigg(-a_1-3a_2+4a_3\bigg)^2 \\
& &
+\frac{3}{4}\bigg(-a_1+a_2\bigg)^2 \\
& & \\
T_3(a_1,a_2,a_3) &=&
\frac{1}{3}\bigg(-a_1+a_2+2a_3\bigg)^2
+\bigg(-a_1+a_2\bigg)^2 \\
\end{array}
$$
Unfortunately, this method does not work any more for $n \geq 7$. Indeed, in that
case the minimum is $\leq -\frac{1}{6}$ (because
$Q_n(0,\ldots,0,\frac{1}{6},\frac{1}{3},\frac{1}{3},\frac{1}{6})=
-\frac{1}{6}$) ; on the other hand, the polynomial
$$
R_n(a_1,a_2,\ldots ,a_n)=Q_n(a_1,a_2,\ldots,a_n)+\frac{(a_1+a_2+a_3+\ldots +a_n)^2}{6}
$$
is not nonnegative on ${\mathbb R}^n$ any more. For example, for $n=7$ we have
$$
R_7(4, 1, 0, -1, 0, 3, 5)=(-2) < 0
$$
| {
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"source": "stackexchange",
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Finding a limit results in division by 0. $$
\lim_{x\to0}\dfrac{\sqrt{1+x}-1}{x^2}.
$$
Tried multiplying by $\sqrt{1+x}+1$ but got $1/(x(\sqrt{1+x}+1))$ where substitution results in $1/0$ which is illegal (or maybe $1/0$ is Infinity?). Second approach is to divide by $x^2$ which leads to the $\sqrt{1/x^2 +1)}- 1/x^2$. How to solve it?
| $$\lim_{x\to 0 }\frac{\sqrt{x+1}-1}{x^2}=\lim_{x\to 0 }\frac{\sqrt{x+1}-1}{x^2}\cdot 1=\lim_{x\to 0 }\frac{\sqrt{x+1}-1}{x^2}\frac{\sqrt{x+1}+1}{\sqrt{x+1}+1}$$
$$=\lim_{x\to 0 }\frac{x+1-1}{x^2(\sqrt{x+1}+1)}=\lim_{x\to 0 }\frac{x}{x^2(\sqrt{x+1}+1)}$$
$$=\lim_{x\to 0 }\frac{1}{x(\sqrt{x+1}+1)}$$
If $x\rightarrow 0^- \Rightarrow \frac{1}{x(\sqrt{x+1}+1)}\rightarrow -\infty$, and
$x\rightarrow 0^+ \Rightarrow \frac{1}{x(\sqrt{x+1}+1)}\rightarrow +\infty$
| {
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Triangle numbers problem
Find the smallest three-digit triangular number that can be
represented both as the sum of three different triangular
numbers, and as the sum of two different triangular numbers.
So the answer is $120$, but I found it by trial and error. What is the clever 'mathematical' way of answering this question?
| I went through it. With $T(x) = (x^2 + x)/2,$ we start with
$$ T(u) = T(v)+T(w) = T(x) + T(y) + T(z). $$
Multiply by $8.$
Adding $1$ to all three positions and considering the first equality we get
$$ (2u+1)^2 + 1 = (2v+1)^2 + (2w+1)^2. $$
This is always solvable, except that you want all three letters positive and $v \neq w.$ The short version is that we must have $ (2u+1)^2 + 1$ divisible by two distinct primes $\equiv 1 \pmod 4.$ For example, if $2u+1 = 13,$ we have $170,$ which can indeed be written as $11^2 + 7^2.$
Second thing to check: instead of $1,$ add $2$ to all positions and equate the first position and the third. We have
$$ (2u+1)^2 + 2 = (2x+1)^2 + (2y+1)^2 + (2z+1)^2. $$
Once again this is always possible, except you want $x,y,z$ positive and distinct. So we strike out as
$$ 171 = 11^2 + 5^2 + 5^2 = 9^2 + 9^2 + 3^2. $$
When $2u+1 = 21,$ we do get $442 = 17 \cdot 13 \cdot 2,$ and indeed
$$ 21^2 + 1 = 442 = 9^2 + 19^2. $$ Then
$$ 443 = 15^2 + 13^2 + 7^2. $$
So that works. And the first occasion is actually
$$ \color{magenta}{55 = 10+45 = 6+21+28}. $$
I see, they want at least $100.$
As $u$ gets larger, the ability to solve this with the full prohibitions should become the norm rather than the exception.
| {
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Find the area enclosed by $\sqrt{(x-2)^2+(y-3)^2} + 2\sqrt{(x-3)^2+(y-1)^2} = 4$ Question: What is the area of the interior of the simple closed curve described by the equation
$\sqrt{(x-2)^2+(y-3)^2} + 2\sqrt{(x-3)^2+(y-1)^2} = 4$?
Comments: I came up with this specific problem myself in response to my earlier question, which I don't think was well posed, or at least it was not clear what I was after: to see how to find the area of the interior of a Jordan curve that is described by an implicit function. To see how this area looks like, I uploaded a picture from WolframAlpha:
As you can see, it is quite egg-like. We can generalize; the object could be called a weighted ellipse [edit: usually called a Cartesian oval] with an equation of the form
$\sqrt{(x-x_1)^2+(y-y_1)^2} + a \cdot \sqrt{(x-x_2)^2+(y-y_2)^2} = k$
where $(x_1,y_1)$ and $(x_2,y_2)$ are the Cartesian coordinates of the focal points of this weighted ellipse and $a$ is a weight (it equals $1$ in the case of an ordinary ellipse). As a bonus question, I would like to see how to find the area of this object.
| from geometry point, the equation $\sqrt{(x-2)^2+(y-3)^2} + 2\sqrt{(x-3)^2+(y-1)^2} = 4$ is equal to:
$\sqrt{x^2+y^2} + 2\sqrt{(x-\sqrt{5})^2+y^2} = 4$ or $\sqrt{(x-\sqrt{5})^2+y^2} + 2\sqrt{x^2+y^2} = 4$ which is symmetry to $x$ axis
so for first one:
$y = \dfrac{ \sqrt{-9 x^2+24 \sqrt{5} x-16 \sqrt{6 \sqrt{5} x+1}+20}}{3}$ for the half area.
but the integration is difficult. I have no idea except the numeric method.
edit: the area is:
$2\int_{x_1}^{x_2}ydx=3.2,x_1=2 (\sqrt{5}-2),x_2=\dfrac{2 (\sqrt{5}+2)}{3}$
and for general a,b,if it is not rational,it is complex. if it is rational, then we can simplify in same idea.
| {
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Find all matrices that satisfy $\mathrm B \mathrm A = \mathrm I_2$
Given the matrix $$A=\begin{pmatrix}1&8\\3&5\\2&2\\ \end{pmatrix}$$ find all $2 \times 3$ matrices in $B \in M_{2 \times 3}(\mathbb R)$ with $BA=I_2$.
Here's what I did:
$$\begin{pmatrix}a&b&c\\d&e&f\\ \end{pmatrix} \begin{pmatrix}1&8\\3&5\\2&2\\ \end{pmatrix} = \begin{pmatrix}1&0\\0&1\\ \end{pmatrix}$$
and then multiplying things out:
$$\begin{pmatrix} {a+3b+2c}&{8a+5b+2c}\\{d+3e+2f}&{8d+5e+2f}\\ \end{pmatrix} = \begin{pmatrix}1&0\\0&1\\ \end{pmatrix}$$
So would I just set
$$a+3b+2c=1$$
$$8a+5b+2c=0$$
$$d+3e+2f=0$$
$$8d+5e+2f=1$$
But then this gives $4$ equations in $6$ unknowns, so wouldn't there be infinitely many solutions? Did I do this correctly? Matrices aren't my strong point...
Thanks.
| The variables of the first two equations do not occur in the last two equations and vice versa. You therefore can independently solve two systems of two equations with three unknowns each. This gives you two solutions with one free variable in each of them:
$$
B = \pmatrix{ a & -\frac{1}{2}(7a+1) & \frac{1}{4}(19a+5)\cr
d & \frac{1}{2}(1-7d) & \frac{1}{4}(19d-3)\cr}
$$
| {
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How to Prove Below Inverse Sin How to prove below equation .
$$ \sin^{-1}\frac{3}{5}+\sin^{-1}\frac{8}{17} = \sin^{-1}\frac{77}{85}.$$
I am not able to prove the above equation how can we prove it.
As we know
$ \sin^{-1} = y$ if $\sin y = x$ where $-1\leq x\leq 1, -\pi/2 \leq y \leq \pi/2.$
| Let $\displaystyle\sin^{-1}\frac35= A$
$\displaystyle\implies (i)\sin A=\frac35$ and $(ii)-\frac\pi2< A<\frac\pi2 $ as the Principal value of sine inverse lies in that range $\displaystyle\implies\cos A>0 \implies \cos A=+\sqrt{1-\left(\frac35\right)^2}=+\frac45$
Similarly for $\displaystyle\sin^{-1}\frac5{13}= B$
Use $\sin(A+B)=\sin A\cos B+\cos A\sin B$
From this, $$\sin^{-1}x+\sin^{-1}y=$$
$$ \begin{cases} \sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2}) &\mbox{if } x^2+y^2\le1\text{ or if } x^2+y^2>1\text{ and } xy<0 \\ \pi-\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2}) & \mbox{if } x^2+y^2>1\text{ and }x,y>0 \\ -\pi-\sin^{-1}(x\sqrt{1-y^2}+y\sqrt{1-x^2}) & \mbox{if } x^2+y^2>1\text{ and }x,y<0 \end{cases} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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recurrence relation for squares of fibonacci numbers I have a problem finding a proof that the squares of the Fibonacci numbers satisfy the recurrence relation $a_{n+3} - 2*a_{n+2} - 2*a_{n+1} + a_n = 0$ and solving this recurrence relation.
Some help would be great!
| Let $(u_n)$ denote the sequence of Fibonacci numbers. By definition,
$$
u_{n+2} = u_{n+1} + u_n\quad\text{and}\quad u_{n+3} = u_{n+2} + u_{n+1} = 2u_{n+1} + u_n,
$$
so that using $(x+y)^2 = x^2 + y^2 + 2xy$,
$$
a_{n+2} = a_{n+1} + a_n + 2u_n u_{n+1},\quad\text{and}\quad
a_{n+3} = 4a_{n+1} + a_n + 4u_nu_{n+1}.
$$
Finally, for all $n$ one has:
$$
a_{n+3} - 2a_{n+2} - 2a_{n+1} + a_n = (4-2-2)a_{n+1} + (1-2+1)a_n + (4-4)u_nu_{n+1}.
$$
| {
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Find $x$ and $y$ in $2^{x-y} + 1 = 2^x,$ where $x,y$ are integers I have no idea what to do now.
Is there any way to find the integers $x$ and $y$ by factoring?
Thank you.
| As an alternative (using factoring), note that $2^{x-y}=\frac{2^x}{2^y},$ so multiplying both sides by the non-zero number $2^y$ yields the equivalent equation $$2^x+2^y=2^x2^y,$$ which in turn yields $$0=2^x2^y-2^x-2^y\\0=2^x(2^y-1)-2^y\\0=2^x(2^y-1)-2^y+1-1\\0=2^x(2^y-1)-(2^y-1)-1\\0=(2^x-1)(2^y-1)-1\\1=(2^x-1)(2^y-1)$$
Note that we cannot have $x=0$ or $y=0$. (Why?) Also $x$ and $y$ must have the same sign, for if not, then $(2^x-1)(2^y-1)$ is negative. If $x,y$ are negative, then $(2^x-1)(2^y-1)$ is positive, but less than $1,$ so we must have $x,y$ positive.
Now, for any positive integers $x,y,$ we have $x,y\ge 1$ (since $x,y$ integers), so $2^x,2^y\ge 2,$ so $2^x-1,2^y-1\ge 1,$ and so $(2^x-1)(2^y-1)\ge 1.$ Equality holds in the last case precisely when it holds in the first case--that is, precisely when $x,y=1.$
| {
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How to solve the equations $\sqrt{x-3}+\sqrt{y-3}=\sqrt{y-12}+\sqrt{z-12}=\sqrt{z-27}+\sqrt{x-27}=12$
Let $x,y,z\in R$,
and
$$\begin{cases}
\sqrt{x-3}+\sqrt{y-3}=12\\
\sqrt{y-12}+\sqrt{z-12}=12\\
\sqrt{z-27}+\sqrt{x-27}=12
\end{cases}$$
Find the $x,y,z$.
My try: I want use The geometry to solve it. ( Norbert have solved it)
and I think this problem have algebra methods.Thank you
we only find this $x,y,z$,
then we have
$$OP^2=ON^2+NP^2-ON\cdot NP=39$$
and
$$\dfrac{OP}{\sin{A}}=x\Longrightarrow \sqrt{x}=2\sqrt{13}$$
and use the same methods we easy to find $y,z$
I think of seeing algebra methods? maybe use if $x>y$,then use
$$\sqrt{x-3}+\sqrt{y-3}=\sqrt{y-12}+\sqrt{z-12}<\sqrt{x-12}+\sqrt{z-12}?$$
then I can't,
Thank you
| Here's another algebraic solution, similar to Norbert's but taking a somewhat different tack.
For all the square roots to be real, we must have $x,z\ge27$ and $y\ge12$, so let's let
$$\begin{align}
x&=27+u^2\\
y&=12+v^2\\
z&=27+w^2\\
\end{align}$$
with $u,v,w\ge0$. Then the equations become
$$\begin{align}
\sqrt{u^2+24}+\sqrt{v^2+9}&=12\\
v+\sqrt{w^2+15}&=12\\
u+w&=12\\
\end{align}$$
Thus
$$w=12-u \quad\text{and}\quad v=12-\sqrt{(12-u)^2+15}$$
so the equation to be solved is
$$\sqrt{u^2+24}+\sqrt{\left(12-\sqrt{(12-u)^2+15}\right)^2+9}=12$$
This "simplifies" first to
$$\left(12-\sqrt{u^2-24u+159}\right)^2+9=\left(12-\sqrt{u^2+24}\right)^2$$
which boils down to
$$\sqrt{u^2-24u+159}=\sqrt{u^2+24}-(u-6)$$
This squares and simplifies to
$$2(u-6)\sqrt{u^2+24}=u^2+12u-99$$
and this squares to produce a quartic that factors as
$$3(u-5)(u^3-19u^2+3u+423)=0$$
We thus have the solution $u=5$, which leads to $x=52$, $y=28$ and $z=76$, as Norbert found. The cubic factor looks unpleasant, but we don't have to worry about any of its roots that are less than $0$ or greater than $12$, since those do not satisfy the non-negativity conditions on $u$ and $w$. Now if we go back to the original equation and let
$$f(u)=\sqrt{u^2+24}+\sqrt{\left(12-\sqrt{(12-u)^2+15}\right)^2+9}-12$$
we have
$$f'(u)={u\over\sqrt{u^2+24}}+{12-\sqrt{(12-u)^2+15}\over\sqrt{\left(12-\sqrt{(12-u)^2+15}\right)^2+9}}\cdot{(12-u)\over\sqrt{(12-u)^2+15}}$$
which is clearly positive for $1\le u\le 12$, so we don't have to worry about any roots of the cubic in that range either. Finally, it's easy to see that the cubic decreases from $423$ to $408$ in the interval $0\lt u\lt1$. Thus $u=5$ is the only solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Using the Maclaurin series of $\arctan(x)$ to evaluate $\int_{0}^{\frac{\pi}{2}} \arctan( a \sin x) \, \mathrm dx$ I want to use the Maclaurin series of $\arctan (x)$ to show that
\begin{align}\int_{0}^{\pi/ 2} \arctan (a \sin x) \, \mathrm dx &= 2 \sum_{k=0}^{\infty} \frac{\left(\frac{\sqrt{1 + a^{2}}- 1}{a}\right)^{2k+1}}{(2k+1)^{2}} \\ &= \operatorname{Li}_{2} \left(\frac{\sqrt{1+a^{2}}-1}{a} \right) - \operatorname{Li}_{2} \left(\frac{1-\sqrt{1+a^{2}}}{a} \right).
\end{align}
I guess we should first impose the restriction $ |a| \le 1$.
Then we have
$$ \begin{align} \int_{0}^{\pi/ 2} \arctan (a \sin x) \ dx &= \int_{0}^{\pi /2} \sum_{k=0}^{\infty} (-1)^{k} \, \frac{(a \sin x)^{2k+1}}{2k+1} \, \mathrm dx \\ &= \sum_{k=0}^{\infty} (-1)^{k} \, \frac{a^{2k+1}}{2k+1} \int_{0}^{\pi /2} \sin^{2k+1} (x) \, \mathrm dx \\ &= \sum_{k=0}^{\infty} (-1)^{k} \frac{a^{2k+1}}{(2k+1)^{2}} \frac{2^{2k}}{\binom{2k}{k}}. \end{align}$$
But how do we proceed from here?
| Sangchul Lee showed that the problem reduces to showing that $$\sum_{n=0}^{m} (-1)^{n} \, \frac{(m+n)!}{(m-n)!} \left( \frac{n!2^{2n}}{(2n+1)!} \right)^{2} = \frac{1}{(2m+1)^{2}}. \tag{1} $$
The following is a way to prove $(1)$ using properties of hypergeometric functions.
Using the duplication formula for the gamma function, along with the fact that $$\frac{1}{(m-n)!} = \frac{(-1)^{n}(-m)_{n}}{m!}, $$ we can express the left side of $(1)$ as the hypergeometric series $$_{3}F_{2} \left(1,-m,m+1;\frac{3}{2}, \frac{3}{2}; 1 \right). $$
Then using Euler's integral representation of the generalized hypergeometric function, we have $$_{3}F_{2} \left(1,-m,m+1;\frac{3}{2}, \frac{3}{2}; 1 \right) = \frac{1}{2} \int_{0}^{1} (1-t)^{-1/2} \, _2F_{1} \left(-m, m+1; \frac{3}{2} ; t \right) \mathrm dt, $$
where $$_2F_{1} \left(-m, m+1; \frac{3}{2} ; t \right) = \frac{\sin \left[(2m+1)\arcsin (\sqrt{t}) \right]}{(2m+1)\sqrt{t}}. $$
(See my answer here to a previous question.)
Therefore, $$ \begin{align} \sum_{n=0}^{m} (-1)^{n} \, \frac{(m+n)!}{(m-n)!} \left( \frac{n!2^{2n}}{(2n+1)!} \right)^{2} &= \frac{1}{2(2m+1)} \int_{0}^{1} (1-t)^{-1/2} \, \frac{\sin \left[(2m+1)\arcsin (\sqrt{t}) \right]}{\sqrt{t}} \, \mathrm dt \\ &= \frac{1}{(2m+1)^{2}} \int_{0}^{(2m+1)\pi /2} \sin (u) \, \mathrm du \\ &= \frac{1}{(2m+1)^{2}}. \end{align}$$
| {
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Sums of squares
Let $R(n)$ be the number of ordered pairs of integers $(x,y)$ with $x^2+y^2 = n$. Using the identity $$(a+b)^2+(a-b)^2 =2(a^2 +b^2),$$
prove that for all integers $n$, $R(2n) = R(n)$.
I'm not really sure how to start this question. Any help would be appreciated.
| If $n$ is even and $$n = x^2 + y^2$$ for integers $x$ and $y$, then $$\frac{n}{2} = (\frac{x+y}{2})^2 + (\frac{x-y}{2})^2,$$ (using the mentioned identity) and both these terms are integers, as $x$ and $y$ must have the same parity.
On the other hand, if $n$ is an integer with $$n = a^2 +b^2$$ for integers $a$ and $b,$ then $$2n = (a+b)^2 + (a-b)^2.$$ Hence if $n$ is even, we can replace $n$ by $\frac{n}{2},$ and $R(n)$ doesn't change.
| {
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Find $(a,b)$ such that in $x^2+ax+b$, whenever $v$ is a root, then $v^2 - 2$ is also a root Find $(a,b)$ such that in $x^2+ax+b$, whenever $v$ is a root, then $v^2 - 2$ is also a root
$a,b$ are real numbers. Roots may or may not be real.
In this question, the aim is to find values of and b ,eg. 2,4.
| Continuing toufik's solution, one can state the following :
Theorem Let $P=X^2+aX+b$ where $a$ and $b$ are real. Then $P$
factorizes as $(X-v)(X-(v^2-2))$ for some $v\in{\mathbb C}$ if and only if
$9-4a \geq 0$, and $b=\frac{3a-5\pm(a-1)\sqrt{9-4a}}{2}$.
Proof of theorem. In one direction, suppose that $P$
factorizes as $(X-v)(X-(v^2-2))$ for some $v\in{\mathbb C}$. Then
$X^2+aX+b=(X-v)(X-(v^2-2))$, so
$$
a=-v^2-v+2, \ b=v^3-2v \tag{1}
$$
We deduce $(v-1)a=-v^3+3v-2$ and hence $b+(v-1)a=v-2$.
So if $a=1$, we must have $b=(-1)$. Otherwise we can write
$$
v=\frac{a-(b+2)}{a-1} \tag{2}
$$
Reinjecting this into (1) we see that
$$
a=-\big(\frac{a-(b+2)}{a-1}\big)^2+\big(\frac{a-(b+2)}{a-1}\big)+2 \tag{3}
$$
Expanding and clearing out denominators in (3), we obtain
$$
b^2 + (-3a + 5)b + (a^3 - 2a^2 - 2a + 4)=0 \tag{4}
$$
The discriminant is
$$(-3a+5)^2-4(a^3 - 2a^2 - 2a + 4)=-4a^3 + 17a^2 - 22a + 9
=(9-4a)((a-1)^2)
$$
So that
$$
b=\frac{3a-5\pm(a-1)\sqrt{9-4a}}{2} \tag{5}
$$
Note that (5) still holds when $(a,b)=(1,-1)$.
Conversely, suppose that (5) holds, so that
$b=\frac{3a-5+\varepsilon(a-1)\sqrt{9-4a}}{2}$ for some
$\varepsilon=\pm 1$. Inspired by (2), let us put
$$
v=\frac{-1-\varepsilon\sqrt{9-4a}}{2} \tag{6}
$$
Then $v^2=\frac{5-2a-\varepsilon\sqrt{9-4a}}{2}$, so
$$
v^2-2=\frac{1-2a+\varepsilon\sqrt{9-4a}}{2} \tag{7}
$$
Multiplying (6) and (7), we see that
$$
v(v^2-2)=\frac{2a-1-(9-4a)+\varepsilon\sqrt{9-4a}(2a-1-1)}{4}=
\frac{6a-10+\varepsilon\sqrt{9-4a}(2a-2)}{4}=
b \tag{8}
$$
and adding (6) and (7), we see that
$$
v+(v^2-2)=-a \tag{9}
$$
Then (7) and (8) together imply that $X^2+aX+b=(X-v)(X-(v^2-2))$.
This concludes the proof.
| {
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How do we solve this system of equations? $a,b \in \Bbb R$ and $$\frac{a^5b-b^5a}{a-b}=30$$ and $$a^5+b^5 = 33$$
I get that $a^6-b^6=(a-b)63$ But I have no idea how to solve after that. Someone could help me?
| $$ a=2, b=1 $$
works. So does $$ a=1, b=2. $$
If you draw a graph of $x^5 + y^5 = C > 0$ you find that $x+y > 0.$ Thus, with $$ ab(a+b)(a^2 + b^2) = 30 $$ we find $ab > 0.$ Then, with $a^5 + b^5 = 33,$ we have both positive.
Writing $$ a = r \cos \theta, \; b = r \sin \theta $$ and pulling out $r^5,$
we get
$$ 33 \cos \theta \sin \theta (\cos \theta + \sin \theta) = 30 (\cos^5 \theta + \sin^5 \theta). $$
Taking second derivatives, the left hand side has negative second derivative on $0 < \theta < \pi / 2.$ On the same range, the right hand side has negative first derivative until $\theta = \pi / 4,$ after which it has positive first derivative. As a result, there are at most two points of equality, and we have already found those.
| {
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Solve: $2\log_{3}(x)-\log_{3}(x+6)=1$ just getting going with logarithms, having trouble with this question.
$$2\log_{3}(x)-\log_{3}(x+6)=1$$
$$\log_{3}x^2-\log_{3}(x+6)=1$$
Stuck at this point: What do I do next?
$$\log_{3}\bigg({\dfrac{x^2}{x+6}}\bigg)=1$$
Edit: Got it!
$$\dfrac{x^2}{x+6}=3^1$$
$$\dfrac{x^2}{x+6}=3$$
$${x^2}=3(x+6)$$
$$x^2=3x+18$$
$$x^2-3x-18=0$$
$$(x-6)(x+3)=0$$
$x=6 $ and cannot equal $\ne3$ as $2\log_{3}(-3) < 0$.
| Good so far. Now use the property that
$$\log_3 a = b \iff a = 3^b$$
This leads to
$$\frac{x^2}{x + 6} = 3$$
Rearranging, this is a quadratic equation.
| {
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Calculate $\lim_{n\to\infty}\sqrt{n^2+n} - n$. Calculate $\lim_{n\to\infty}\hspace{2 pt}a_n$, where $a_n = \sqrt{n^2+n} - n$ is a sequence of complex numbers.
I performed the ratio test:
$$\lim_{n\to\infty}\frac{a_{n+1}}{a_n} = \lim_{n\to\infty}\frac{\sqrt{(n+1)^2 + (n+1)} - (n+1)}{\sqrt{n^2+n} - n} = \lim_{n\to\infty}\frac{\sqrt{n^2+3n+2} - (n+1)}{\sqrt{n^2+n} - n}$$
which ultimately leads to, when multiplied by $\frac{\frac{1}{n}}{\frac{1}{n}}$,
$$\lim_{n\to\infty}\frac{\sqrt{1+\frac{3}{n}+\frac{2}{n^2}} - (1+\frac{1}{n})}{\sqrt{1+\frac{1}{n}} - 1}$$
Should I multiply by the complex conjugate? It leads to a rather messy expression. I am stuck.
How can one perform the root test here? I attempted: $$\lim_{n\to\infty}\sqrt[n]{\sqrt{n^2+n} - n} \Longleftrightarrow \lim_{n\to\infty}((n^2 + n)^{\frac{1}{2}} - n)^{\frac{1}{n}}$$
I have also no idea how to proceed.
| $$a_n = \sqrt{n^2+n} - n = \dfrac{\sqrt{n^2+n}+n}{\sqrt{n^2+n}+n} \cdot (\sqrt{n^2+n}-n) = \dfrac{n^2+n - n^2}{\sqrt{n^2+n}+n} = \dfrac{n}{\sqrt{n^2+n} + n} = \dfrac1{\sqrt{1+1/n}+1}$$
Now let $n \to \infty$ to conclude the limit.
| {
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Evaluating Definite Integral $\int_1^2\arcsin\left(\frac{4-3\sqrt{x^2-1}}{5x}\right)dx$ How can I prove that
$$\int_1^2\arcsin\left(\frac{4-3\sqrt{x^2-1}}{5x}\right)dx=\frac{\pi}{3}-2\text{arcsec}(\sqrt{5})+\log(2+\sqrt{3})$$
Can someone help me, please?
| Using Trigonometric substitution
let $x=\sec\theta$
$\displaystyle\implies (i) 0\le \theta\le\pi$ using the definition of principal value
and $\displaystyle(ii)\frac{4-3\sqrt{x^2-1}}{5x}=\frac{4-3|\tan\theta|}{5\sec\theta}$
Method $1:$ If $\displaystyle0\le \theta\le\frac\pi2,$
when $x=2, \theta=\text{arcsec}2=\frac\pi3$
when $x=1, \theta=\text{arcsec}1=0$
$\displaystyle\frac{4-3\sqrt{x^2-1}}{5x}=\frac45\cos\theta-\frac35\sin\theta=\sin\left(\arcsin\frac45-\theta\right)$ (using this)
$$\implies\int_1^2\arcsin\left(\frac{4-3\sqrt{x^2-1}}{5x}\right)dx=\int_0^{\frac\pi3}\left(\arcsin\frac45-\theta\right)\sec\theta\cdot\tan\theta d\theta$$
Now integrating by parts,
$$\int \left(\arcsin\frac45-\theta\right)\sec\theta\cdot\tan\theta d\theta$$
$$=\left(\arcsin\frac45-\theta\right)\int\sec\theta\cdot\tan\theta d\theta-\int\left(\frac{d(\arcsin\frac45-\theta)}{d\theta}\cdot\int\sec\theta\cdot\tan\theta d\theta\right)d\theta$$
$$=\left(\arcsin\frac45-\theta\right)\sec\theta-\int\left(-1\cdot\sec\theta\right)d\theta$$
$$=\left(\arcsin\frac45-\theta\right)\sec\theta+\ln|\sec\theta+\tan\theta|+C$$
Observe that $\displaystyle 2\text{arcsec}(\sqrt5) =2\arccos\frac1{\sqrt5}=\cos\left(\frac25-1\right)$
$\displaystyle=\cos\left(-\frac35\right)=\pi-\arccos\frac35=\pi-\arcsin\frac45$ as $\arccos x=+\arcsin\sqrt{1-x^2}\forall $ real $x$ based on the Principal value of inverse cosine ratio
Method $2:$
If $\displaystyle \frac\pi2\le \theta\le\pi,\text{arcsec}(2)=?$
$$\frac{4-3\sqrt{x^2-1}}{5x}=\frac45\cos\theta+\frac35\sin\theta=\sin\left(\arcsin\frac45+\theta\right)$$
| {
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Proving by induction inequalities that lack the variable on the right side: $\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$ Doing proof by induction exercises with inequalities, I got stuck on one that is a bit different from the others. There is no $n$ term on the rightmost part of the inequality:
Prove that the following holds for all $n \ge 1$:
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$$
All the other proofs I did before had some $n$ involved there, but now there is none. I wonder how will this difference affect my standard attempt:
Test for $n = 1$:
$$\frac{1}{3} \le \frac{5}{6} \implies 6 \le 15$$
We have to prove that it holds for $n + 1$, that is:
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
Asssume
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}\le \frac{5}{6}$$
We start off with
$$\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
And now I see the problem. Normally, I apply the hypothesis first on this part:
$$\color{blue}{\frac{1}{n + 1}+\frac{1}{n + 2}+...+\frac{1}{n+(n+1)}}+\frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
But all it yields is this:
$$\color{blue}{\left ( \textrm{something smaller or equal than } \frac{5}{6} \right )} + \frac{1}{(n + 1)+(n+2)}\le \frac{5}{6}$$
If I move $\frac{1}{(n + 1)+(n+2)}$ to the other side, all I get is
$$\color{blue}{\left ( \textrm{something smaller or equal than } \frac{5}{6} \right )} \le \frac{5}{6} - \frac{1}{(n + 1)+(n+2)}$$
Which cannot be guaranteed.
How can I prove this, then?
| Let $$a_n=\sum_{k=n+1}^{2n+1}\frac1k\;,$$ and show that the sequence $\langle a_n:n\ge 1\rangle$ is decreasing. There’s more in the spoiler-protected block below.
$\displaystyle\sum_{k=n+1}^{2n+1}\frac1k-\sum_{k=n+2}^{2n+3}\frac1k=\frac1{n+1}-\frac1{2n+2}-\frac1{2n+3}=\frac1{2n+2}-\frac1{2n+3}=\ldots$
| {
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Value of $\frac{1}{\sqrt{3}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{11}}+\frac{1}{\sqrt{11}+\sqrt{15}}+\cdots$ ($n$ terms) Sum $$\frac{1}{\sqrt{3}+\sqrt{7}}+\frac{1}{\sqrt{7}+\sqrt{11}}+\frac{1}{\sqrt{11}+\sqrt{15}}+\cdots \text{ ($n$ terms)}$$
I know how to use the telescoping series method when the terms are in product. Here i tried to make $f(n+1)-f(n)$ form but failed to do so. How do we go about with this one?
Any help is welcome,
Thanks.
| Multiplying top and bottom by the denominator with a $-$ sign, we can rewrite this as
$$\frac{\sqrt 7 - \sqrt 3}{7 - 3} + \frac{\sqrt{11} - \sqrt 7}{11 - 7} + \frac{\sqrt{15} - \sqrt{11}}{15 - 11} + \dots + \text{ final term}$$
This does telescope.
| {
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How to integrate $\frac{4x}{x^3+x^2+x+1}$ I'd like to know how to integrate $$\frac{4x}{x^3+x^2+x+1}$$
Please could anyone help me
Thanks all
| Note that $x^3+x^2+x+1=(x+1)(x^2+1)$. We use partial fracions. So we try to find constants $A,B,C$ such that
$$\frac{4x}{(x+1)(x^2+1)}=\frac{A}{x+1}+\frac{Bx+C}{x^2+1}.$$
Bring the right-hand side to the common denominator $(x+1)(x^2+1)$. The numerators must be identically equal. It follows that
$$4x=A(x^2+1)+(Bx+C)(x+1).$$
Set $x=-1$. We get $-4=2A$, and therefore $A=-2$.
On the right, the coefficient of $x^2$ is $-2+B$. On the left it is $0$. It follows that $B=2$.
On the right, the coefficient of $x$ is therefore $2+C$. Thus $C=2$. We conclude that
$$\frac{4x}{(x+1)(x^2+1)}=-\frac{2}{x+1}+\frac{2x+2}{x^2+1}.$$
So we want to integrate $-\frac{2}{x+1}+\frac{2x}{x^2+1}+\frac{2}{x^2+1}$.
The rest is straightforward. To integrate $\frac{2x}{x^2+1}$ let $u=x^2+1$.
| {
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Find $x$ such that $\sqrt{x+\sqrt{x+7}}\in \mathbb{N}$ Find $x$ such that $$\sqrt{x+\sqrt{x+7}}\in \mathbb{N}$$
I tried many ways: $$\sqrt{x+\sqrt{x+7}}=n$$
$$\sqrt{x+\sqrt{x+7}}^2=n^2$$
$$x+\sqrt{x+7}=n^2$$
then solve for $x$ but didn't do with success.
I think this is the most difficult problem in my lifetime
Also $x$ must be made
of $2$ digit.
Thanks to everybody for helping me understand this problem and its solution!
| $$x + 7 = ((n^2)-x)^2 = n^4 - 2n^2x + x^2$$
$$x^2-(2n^2+1)x+n^4-7=0$$
$$x_{1,2}=\frac{2n^2+1 \pm \sqrt{(2n^2+1)^2-4(n^4+7)}}{2}\\
=\frac{2n^2+1 \pm \sqrt{4n^4+4n^2+1-4n^4+28}}{2}\\
=\frac{2n^2+1 \pm \sqrt{4n^2+29}}{2}\\$$
This is an integer if and only if $ \sqrt{4n^2+29}$ is an integer (the converse follows immediately from the observation that if $\sqrt{4n^2+29}$ is an integer, it must be odd.
Claim 1: If $\sqrt{4n^2+29}$ is an integer, then $n \leq 7$.
Proof:
$$\sqrt{4n^2+29} > \sqrt{4n^2}=2n \,.$$
Thus
$$\sqrt{4n^2+29} \geq 2n+1 \,.$$
Hence
$$4n^2+29 \geq 4n^2+4n+1 \,.$$
This implies that $n \leq 7$, with equality if and only if $n=7$.
Claim 2: If $\sqrt{4n^2+29}$ is an integer, and $n \neq 7$ then $n \leq 1$.
Proof: Exactly like in Claim $1$
$$\sqrt{4n^2+29} \geq 2n+1 \,.$$
But we proved that we only have equality for $n=7$. Thus
$$\sqrt{4n^2+29} > 2n+1 \,.$$
As $\sqrt{4n^2+29}$ is odd, we get
$$\sqrt{4n^2+29} \geq 2n+3 \,.$$
$$4n^2+29 \geq 4n^2+12n+9 \,.$$
$$29 \geq 12n+9 \,.$$
Thus $n \leq 1$.
Thus we showed that the only $n$ which can work are $n = 0, n = 1$ and $n = 7$.
If
$$n=0 \Rightarrow x_{1,2}
=\frac{1 \pm \sqrt{29}}{2} \notin \mathbb Z$$
$$n=1 \Rightarrow x_{1,2}
=\frac{2+1 \pm \sqrt{4+29}}{2}\notin \mathbb Z$$
$$n=7 \Rightarrow x_{1,2}
=\frac{99 \pm 15}{2}$$
P.S. Don't forget that we squared couple times, which means that the solutions we got are solutions to $$x + 7 = ((n^2)-x)^2 = n^4 - 2n^2x + x^2$$
but not necessarily to the original question.
One needs to check them in the original equation, and only one works.
The reason why the other solution doesn't work is because it appears as an extra solution when we square:
$$\sqrt{x + 7 }= (n^2)-x$$
If we observe in the original equation that $n^2 \geq x$, this eliminates the wrong extra solution .
| {
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Series $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ I want to check, whether $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ converges or diverges.
I tried to use the Ratio test:
$|\frac{a_{n+1}}{a_n}|= \frac{n+2}{2^{n+1}} \frac{2^n}{n+1} = \frac{1}{2} \frac{n+1+1}{n+1} = \frac{1}{2} (1+ \frac{1}{n+1})$
$\lim\limits_{n \rightarrow \infty}{{(\frac{1}{2}(1+ \frac{1}{n+1})) = \frac{1}{2}}} \leq 1$
So $\sum\limits_{n=0}^{\infty}\frac{n+1}{2^n}$ converges.
Could somebody please check my solution?
| $\sum_{n=0}^\infty (n+1) 2^{-n}$
$ = \sum_{n=0}^\infty \left( n 2^{-(n-1)} - (n+1) 2^{-n} + 2 \cdot 2^{-n} \right)$
$ = \left( 0 \cdot 2^{-(0-1)} - \lim_{n \to \infty} (n+1)2^{-n} \right) + 2 \sum_{n=0}^\infty { 2^{-n} }$
$ = 4$
The first step can be found be guesswork in simple cases, or be systematically doing "anti-difference by parts" in almost the same manner as "integration by parts", where the "by-parts" formula can be derived in exactly the same way from the "product rule" for the difference operator. Either of the forward or backwards difference operators work. This can easily handle things like $\sum_{n=0}^\infty n^3 3^{-n}$.
| {
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Equation with seven unknowns Let $a,b,c,d,e,f,g$ be positive integers greater than or equal to $2$. What values of these numbers satisfy the equations $$a+b+c+d+e+f+g =18 \tag 1$$
$$a(b+c+d+e+f+g+3) + b(c+d+e+f+g+3) + c(d+e+f+g+3) + d(e+f+g+3) + e(f+g+3) + f(g+3) + 3g = 188 \tag 2$$
| Equation (2) is just $\displaystyle \sum_{\text{sym}} ab + 3\sum_{sym} a = 188$. Using (1) we then get $\displaystyle \sum_{\text{sym}} ab = 134$.
Now squaring (1) and subtracting twice the above we get $\displaystyle \sum_{\text{sym}} a^2 = 56$.
Now, at least $3$ of the variables are $2$, because otherwise their sum is greater than $18$. Suppose $e,f,g$ are $2$, then we get $a+b+c+d = 12$ and $a^2+b^2+c^2+d^2 = 44$.
Now $4\cdot 3^2 < 44$, so we must have another variable equal to $2$ (say $d$). Hence we get $a+b+c = 10$ and $a^2+b^2+c^2 = 40$. $(a,b,c) = (3,3,4)$ still has $a^2+b^2+c^2$ too small, so another variable must be $2$, say $c$. So we get $a+b=8$, $a^2+b^2 = 36$. $4^2+4^2$ is too small, as is $3^2 + 5^2$, while $2^2 + 6^2$ is too big.
Thus there are no solutions.
| {
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Solve$(log_{2}(x+1))^2=4$ $$(log_{2}(x+1))^2=4$$
$$log_{2}(x+1)*log_{2}(x+1)=log_{2}16$$
$$x^{2}+2x-15=0$$
$$(x+1)*(x+1)=16$$
$$x^{2}+2x+1=16$$
$$x^{2}+2x-15=0$$
$$(x+5)(x-3)=0$$
$$x_1=-5; x_2=3$$
The solution is only $x_1=3$. Is this correct?
| $$(\log_{2}(x+1))^2=4\iff\log_{2}(x+1)=\pm2$$
[1]$$\log_{2}(x+1)=2,x+1=2^2,x_1=3$$
[2]$$\log_{2}(x+1)=-2,x+1=2^{-2},x_2=-3/4$$
| {
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Trigonometric equation: $2(\sin^6 x+\cos^6 x)-3(\sin^4 x+\cos^4 x)+1=0$ I'm new here, but I need your help so much to solve an equation:
$$2(\sin^6 x+\cos^6 x)-3(\sin^4 x+\cos^4 x)+1=0$$
I tried a lot like making $2[(\sin^2 x)^3 + (\cos^2 x)^3$
| $$2(\sin^6 x+\cos^6 x)-3(\sin^4 x+\cos^4 x)=0$$
$\sin x=t,\cos^2x=1-t^2$
$$2(t^6+(1-t^2)^3)-3(t^4+(1-t^2)^2)=0$$
$$2(t^6+1-3t^2+3t^4-t^6)-3(t^4+1-2t^2+t^4)=0$$
$$2(1-3t^2+3t^4)-3(2t^4+1-2t^2)=0$$
$$(2-6t^2+6t^4)-(6t^4+3-6t^2)=0$$
$$-1=0$$ CONTRADICTION
| {
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Limit of $\lim_{x \rightarrow 0}\frac{9}{x}\bigl(\frac{3}{(x+3)^3}-\frac{1}{9}\bigr)$ I have to determine the following:
$$\lim_{x \rightarrow 0}\frac{9}{x}\left(\frac{3}{(x+3)^3}-\frac{1}{9}\right)$$
I've got so far:
$$\lim_{x \rightarrow 0}\frac{9}{x}\left(\frac{3}{(x+3)^3}-\frac{1}{9}\right)= \lim_{x \rightarrow 0}\left(\frac{27}{x(x+3)^3}-\frac{1}{x}\right)=\lim_{x \rightarrow 0} \left(\frac{27-(x+3)^3}{x(x+3)^3}\right)=\cdots$$
How to go on? I've got $\frac{\infty}{0}...$
| When you plug in $0$ to $x$, you see that the answer is $0/0$. You have to use L'Hospital's Rule, which says
$$\lim_{x\to a} \frac{f(x)}{g(x)} = \lim_{x\to a} \frac{f'(x)}{g'(x)}.$$
This applies only to $0/0$ or $\infty/\infty$.
Hence, you just need to take the derivative of the top and bottom until you get an answer that is either the answer or you can't use L'Hospital's anymore.
So let's take the derivatives of the numerator and denominator:
$$\frac{d}{dx}\left(27-(x+3)^3\right)= -3(x+3)^2$$
$$\frac{d}{dx}\left(x(x+3)^3\right) = x(3(x+3)^2) + (x+3)^3$$
So now we can just plug zero in!
$$\lim_{x\to 0} \frac{-3(x+3)^2}{3x(x+3)^2 + (x+3)^3} = -27/27 = -1$$
| {
"language": "en",
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Solve inverse tangents How do I solve the following equation:
$$ \tan^{-1}\frac{x}{10^6}+\tan^{-1}\frac{x}{10^7}=90^{\circ}$$
WA
Step by step solution from wolframalpha is unavailable.
| To solve this equation, what you'd first think to do is apply $\tan$ to both sides, to get
$$\tan \left(\tan^{-1} \dfrac{x}{10^6} + \tan^{-1} \dfrac{x}{10^7}\right) = \tan(90^{\circ})$$
but this doesn't work since $\tan(90^{\circ})$ is undefined.
But look at the left-hand side and compare it to the identity
$$\tan(A+B) = \dfrac{\tan A + \tan B}{1-\tan A\tan B}$$
Here we'd set $A = \tan^{-1}\dfrac{x}{10^6}$ and $B = \tan^{-1}\dfrac{x}{10^7}$.
For this to be undefined in this case, the denominator must be equal to zero.
So
$$\tan^{-1} \dfrac{x}{10^6} + \tan^{-1} \dfrac{x}{10^7} = 90^{\circ} \quad \Leftrightarrow \quad 1- \left(\dfrac{x}{10^6}\right) \left( \dfrac{x}{10^7} \right) = 0$$
I leave the rest to you.
| {
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How to integrate $\int{\frac{6x}{x^3+8}dx}$ I'm having some trouble solving this integral using partial fraction method: $$\int{\frac{6x}{x^3+8}dx}.$$
After expanding $x^3+8$ into $(x-2)(x^2+2x+4)$ and expanding the original integral into $$\int{\frac{A}{x-2}+\frac{Bx+C}{x^2+2x+4}}dx,$$ I got $1$,$-1$, and $2$ for $A$, $B$, and $C$, respectively, yielding $$\int{\frac{1}{x-2}+\frac{-x+2}{x^2+2x+4}}dx.$$
This simplifies to $$\ln{|x-2|}-\int{\frac{x-2}{(x+1)^2+3}dx}.$$
How do I solve this resultant integral? That's where I'm stuck.
| Write the numerator of the last integral as $(x+1)-3,$ and break the integral up into the two parts. The first integral will be a logarithm the second an arctan but do the calculations yourself.
| {
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negative exponent problem $$\sqrt{\frac{1}{3^0 + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}}}$$
Does this equal =
$$
\begin{align*}
& \sqrt{3^0 + 3^1 + 3^2 + 3^3 + 3^4} \\
=&\sqrt{1 + 3 + 9 + 27 + 81} \\
=&\sqrt{121} \\
=&11.
\end{align*}
$$
The answer is apparently $\frac{9}{11}$ and I'm not sure what rule of negative exponents I got wrong.
The rule I'm using, incorrectly, is this:
$$\frac{1}{3^{-2}} = 3^2 = 9.$$
| Note that $3^{-n} = 1/3^{n}$ for all $n \geq 0$, where $3^{0} = 1$, by convention. First, simplify the denominator utilizing the last fact:
$$3^{0} + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4} = 1 + 1/3 + 1/9 + 1/27 + 1/81 \\ = 81/81 + 27/81 + 9/81 + 3/81 + 1/81 \\ = 121/81$$
Thus, we have simplified the original expression to:
$$\sqrt{1/(121/81)} = \sqrt{81/121} = 9/11$$
where the last equality comes from knowing perfect squares for natural numbers less than $20$ (recommended for any student of mathematics). Observe from the above that you had the correct "fact" all along, you simply need to recognize that each term of the denominator cannot be inverted separately. Explicitly,
$$1/(3^{0} + 3^{-1} + 3^{-2} + 3^{-3} + 3^{-4}) \neq 1/3^{0} + 1/3^{-1} + 1/3^{-2} + 1/3^{-3} + 1/3^{-4}$$
It might also be useful for you multiply the numerator and denominator by $\sqrt{3^{4}}$ as was mentioned in some of the other answers, but it is not too difficult to think about it without doing this in my opinion. Let me know if you have any further questions.
| {
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Algebra Inequality Proof Let $a, b, c, d$ be positive numbers. Prove that $\frac{a+b+c+d}{4} \ge \sqrt{\frac{ab+ac+ad+bc+bd+cd}{6}}$.
I was told to rewrite the sum on the right side in terms of $a^2 + b^2 + c^2 + d^2$ and $a + b + c + d$ but I am unsure how to combine the terms.
| Hint:
$$(a+b+c+d)^2=(a^2+b^2+c^2+d^2)+2(ab+ac+ad+bc+bd+cd)$$
Let
$$A=a+b+c+d$$
$$B=a^2+b^2+c^2+d^2$$
Now, this amounts to prove,
$$\frac{A^2}{16} \geq \frac{A^2-B}{12}$$
$$4B \geq A^2$$
You can also write it
$$ \sqrt{a^2+b^2+c^2+d^2} \geq \frac{a+b+c+d}{2}$$
Or
$$ \sqrt{\frac{a^2+b^2+c^2+d^2}{4}} \geq \frac{a+b+c+d}{4}$$
And you should recognize this as a standard inequality between quadratic and arithmetic mean. You can prove it using Cauchy-Schwarz inequality, which states that
$$\left| \sum_{i=1}^n a_ib_i \right| \leq \sqrt{\sum_{i=1}^n a_i^2} \sqrt{\sum_{i=1}^n b_i^2}$$
With $b_i=1$ and $a_i \geq 0$, you get
$$ \sum_{i=1}^n a_i \leq \sqrt{\sum_{i=1}^n a_i^2} \sqrt{n}$$
Or
$$ \frac{1}{n}\sum_{i=1}^n a_i \leq \sqrt{\frac{1}{n}\sum_{i=1}^n a_i^2}$$
In case you don't know Cauchy-Schwarz inequality, here is a standard proof.
The following inequality is obviously true because it's a sum of squares:
$$\sum_{i=1}^{n} (a_i - \lambda b_i)^2 \geq 0$$
Now, develop w.r.t. $\lambda$, to get
$$ \left(\sum_{i=1}^n b_i^2\right) \lambda^2 - 2 \left(\sum_{i=1}^n a_ib_i\right) \lambda + \left(\sum_{i=1}^n a_i^2\right) \geq 0$$
It's a positive trinomial (in $\lambda$), thus its discriminant must be $\leq 0$ (otherwise there would be two roots, and it would becom negative between them, which is not possible).
Write the inequality on the discriminant:
$$4 \left(\sum_{i=1}^n a_ib_i\right)^2 - 4\left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right) \leq 0$$
And finally,
$$ \left(\sum_{i=1}^n a_ib_i\right)^2 \leq \left(\sum_{i=1}^n a_i^2\right)\left(\sum_{i=1}^n b_i^2\right)$$
Or
$$\left| \sum_{i=1}^n a_ib_i \right| \leq \sqrt{\sum_{i=1}^n a_i^2} \sqrt{\sum_{i=1}^n b_i^2}$$
| {
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Convergence of $\int_1^\infty\frac{x^2+kx}{x^4+k^px^2+k^2}dm(x)$; $k\in\mathbb{N},1\le p<\infty$ Consider the integrals
$$I(k,p)=\int_1^\infty\frac{x^2+kx}{x^4+k^px^2+k^2}dm(x),$$ with $k\in\mathbb{N}$ and $1\le p<\infty$.
*
*For which $p$ does the integrand have an integrable majorant?
*For which $p$ do the integrals tend to $0$?
I'm thinking that the integrals have an integrable majorant when $p\ge 1$, but I'm not sure if it's true or how to show it.
| Here we obtain an integral majorant ( independent from $k$) and convergence for $p>1$. For $p=1$ we obtain convergence but no integrable majorant ( independent from $k$)
Case $p>1$
$$ \frac{kx}{x^4+k^px^2+k^2}\leq \frac{kx}{x^4+k^px^2}=\frac{k}{x(x^2+ k^p)}=:\phi(x,k)
$$
$\partial_k\phi = \frac{(x^2+k^p)-pk^p}{x(x^2+k^p)^2}=\frac{x^2-(p-1)k^p}{x(x^2+k^p)^2}=0$ has solution $k_*=\Big(\frac{x^2}{p-1}\Big)^{1/p}$, and $\partial^2_{kk}\phi(k_*)<0$. Since $\phi(x,0)=0=\lim_{k\rightarrow\infty}\phi(x,k)$, we have that $\phi(x,k)\leq \phi(x,k^*)$ for all $k$ and so
$$
\phi(x,k)=\frac{k}{x(x^2+ k^p)}\leq C_p\frac{x^{2/p}}{x^3}=C_px^{\tfrac2p -3}
$$
for all $x\geq1$ and some constant $C_p$ depending only on $p$. Then, for $p>1$
$$
f_{p,k}(x):=\frac{x^2+kx}{x^4+k^px^2+k^2}\leq \frac{1}{x^2}+C_px^{\tfrac2p -3}\in L_1([1,\infty),m)
$$
For each $x\geq1$
$$
\lim_{k\rightarrow\infty}\frac{x^2+kx}{x^4+k^px^2+k^2}=0
$$
Hence, if $p>1$, $\lim_{k\rightarrow\infty}\int^\infty_1 f_{p,k}\,dm =0$ by dominated convergence.
Case $p=1$:
Convergence can be addressed by direct computation:
$$\int^\infty_1 f_{1,k}=\int^\infty_1\frac{x^2\,dx}{x^4+kx^2+1} +k\int^\infty_1\frac{x\, dx}{x^4+kx^2+k^2}\,dx$$
Since $g_k(x):=\frac{x^2}{x^4+kx^2+k^2}\leq \frac{1}{x^2}$ and $0<g_k(x)\xrightarrow{k\rightarrow\infty}0$, $$I_k:=\int^\infty_1\frac{x^2\,dx}{x^4+kx^2+1} \xrightarrow{k\rightarrow\infty}0$$
The second intergral can be estimated directly:
\begin{aligned}
J_k=k\int^\infty_1\frac{x}{x^4+kx^2+k^2}\,dx&=k\int^\infty_1\frac{x\,dx}{(x^2+\tfrac{k}{2})^2+\tfrac34k^2}\\
&=\frac{k}{2}\int^\infty_{1+\tfrac{k}{2}}\frac{du}{u^2+\tfrac{3}{4}k^2}\\
&=\frac{1}{\sqrt{3}}\Big(\frac{\pi}{2}-\arctan\big(\tfrac{2(k+\tfrac12)}{\sqrt{3}k}\big)\Big)\xrightarrow{k\rightarrow\infty}\frac{1}{\sqrt{3}}\Big(\frac{\pi}{2}-\arctan\big(\tfrac{2}{\sqrt{3}}\big)\Big)
\end{aligned}
It follow from this that there is not integrable majorant for $f_{1,k}$ (that is independent of $k$)
Alternative method: only $p\geq2$
From
$$\sqrt{x^4k^px^2}=x^3 k^{p/2}\leq\frac{x^4 + k^px^2}{2}$$
we obtain
\begin{aligned}
f_{p,k}(x)&=\frac{x^2+kx}{x^4+k^px^2+k^2}\leq \frac{1}{x^2}+\frac{kx}{x^4+k^px^2}\leq \frac{1}{x^2}+ \frac{1}{2x^2}k^{1-p/2}
\end{aligned}
For $p\geq2$
$$0<f_{p,k}(x)\leq \frac{3}{2x^2}$$
since $x\geq1$ and $k\geq1$.
| {
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Prove an trigonometric identity. Can someone help me by solving it? $$1-\frac{\sin^2 x}{1+\cot x}-\frac{\cos^2 x}{1+\tan x}=\sin x \cos x.$$
I cannot come to the result. I make that like $1-\dfrac{\sin^2 x}{1+\frac{\cos x}{\sin x}}$
| We can transform the left side to the right.
$$\underbrace{\color{white}{\underline{\color{black}{1-\dfrac{\sin^2x}{1+\cot x}-\dfrac{\cos^2 x}{1+\tan x}}}}}_{\displaystyle\cal I}=\sin x\cos x. \\ \eqalign{
{\cal I} \ : & \; \ 1-\dfrac{\sin^2x}{\dfrac{\sin x+\cos x}{\sin x}}-\dfrac{\cos^2x}{\dfrac{\cos x+\sin x}{\cos x}} \\
&= \dfrac{\sin x+\cos x-\sin^3 x-\cos^3x}{\cos x+\sin x} \\
&= \dfrac{\sin x+\cos x-(\sin x+\cos x)\left(\sin^2x-\sin x\cos x+\cos^2x \right)}{\cos x+\sin x} \\
&=\require{cancel}\dfrac{(\cancel{\cos x+\sin x })\left(1-\sin^2x-\cos^2x+\sin x\cos x\right)}{\cancel{\cos x+\sin x }}\\
&=1-1+\sin x\cos x=\color{lightblue}{\underline{\color{black}{\sin x\cos x}}}
\;\checkmark}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Roots of $f(x)=\sin(x)-ax$ How many roots are there of the function $f(x)=\sin(x)-ax$, where $a$ is a positive number? Clearly for all $a$, $x=0$ is a root; if $a>1$ that is the only root. The roots will also be symmetric about the origin. I conjecture that there are $2\lfloor 1/(a\pi)\rfloor+1$, but I'm not sure how to prove this.
| Let $\#(a)$ be the number of real solutions to the equation $\sin x = ax$.
For $a \lesssim 1$ we have $\#(a) = 3$, and this remains constant as $a$ decreases until $a = a_0$, when the line $y = ax$ is tangent to the curve $y = \sin x$. This point of tangency occurs just to the left of $x = 2\pi + \tfrac{\pi}{2}$. Here $\#(a_0) = 5$, and then for $a \lesssim a_0$ we have $\#(a) = 7$.
The function $\#(a)$ continues in this manner, increasing by $2$ when $a$ decreases to a value for which the graphs are tangent, then increases by $2$ again when $a$ becomes smaller than that.
The value of $a$ for which $\#(a) = 4n+1$, which is when the line $y = ax$ is tangent to the graph $\sin x$, is the unique real solution to the equation
$$
\sqrt{1-a^2} = a (2\pi n + \arccos a). \tag{$*$}
$$
We can express this solution as
$$
a = \frac{1}{2\pi n + \frac{\pi}{2} - \epsilon_n},
$$
where $\epsilon_n > 0$ and $\epsilon_n \to 0$ as $n \to \infty$. Further,
$$
\#\left(\frac{1}{2\pi n + \frac{\pi}{2}}\right) = 4n+3.
$$
In this sense,
$$
\begin{align}
\#(a) &= 3 + 4\sum_{n=1}^{\infty} \left[1 - H\left(a - \frac{1}{2\pi n + \frac{\pi}{2} - \epsilon_n}\right)\right] \\
&\approx 3 + 4\sum_{n=1}^{\infty} \left[1 - H\left(a - \frac{1}{2\pi n + \frac{\pi}{2}}\right)\right] \tag{$**$}
\end{align}
$$
for small $a$, where $H$ is the Heaviside step function. Here's a plot of this approximation:
Of course we don't need every term of the infinite sum to calculate $\#(a)$, we only need to sum to the largest $n$ such that
$$
a \leq \frac{1}{2\pi n+ \frac{\pi}{2} - \epsilon_n} \approx \frac{1}{2\pi n+ \frac{\pi}{2}},
$$
which is
$$
n \approx \left\lfloor \frac{1}{2\pi a} - \frac{1}{4}\right\rfloor.
$$
Thus we get the approximation
$$
\#(a) \approx 3 + 4 \left\lfloor \frac{1}{2\pi a} - \frac{1}{4}\right\rfloor, \tag{$***$}
$$
which, unfortunately, skips the values of $\#(a)$ where $ax$ is tangent to $\sin x$.
I would expect that finding a closed form for the value of $a$ which solves $(*)$ would be difficult, if not impossible.
| {
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.