Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to integrate $\sqrt{x^2-1}$ How do you integrate the following integral? I have tried u-substitution, but it doesn't work out.
$$\int(x^2-1)^{0.5}dx$$
| $I=\int 1\cdot\sqrt{x^2-1}dx$
Put $x = \sec y$
then $dx = \sec y.\tan y dy$
$= \int \sqrt{\sec^2 y-1}. \sec y.\tan y dy$
$= \int \sqrt{\tan^2 y}. \sec y.\tan y dy$
$= \int \tan y. \sec y. \tan y dy$
$= \int \tan^2 y. \sec y dy$
$= \int (\sec^2 y - 1). \sec y dy$
$= \int \sec^3 y dy - \int \sec y dy$
$= \int \sec^3 y dy + log|\sec y + \tan y|$ .......(1)
Now,
$\int \sec^3 y dy=\int \sec y (\sec^2 y dy)$
Put $\tan y = t$
$\sec^2 y dy = dt$
and $\sec y = \sqrt{1+t^2}$
$\int \sqrt{1 + t^2}dt$
$\frac{t\sqrt{1+t^2}}{2} + \frac{\log(t+\sqrt{1+t^2})}{2}$
$t = \tan y$
$\frac{\tan y\sqrt{1 + \tan^2 y}}{2} + \frac{\log(\tan y + \sqrt{1 + \tan^2 y})}{2}$
Now from equation (1)
$= \frac{\tan y \sqrt{1 + \tan^2 y}}{2} + \frac{\log(\tan y + \sqrt{1 + \tan^2 y})}{2} + log|\sec y + \tan y|$
And replace y = $\sec^{-1}x$ and you got answer.
| {
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"url": "https://math.stackexchange.com/questions/2068646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is every parallelogram a rectangle ??
Let's say we have a parallelogram $\text{ABCD}$.
$\triangle \text{ADC}$ and $\triangle \text{BCD}$ are on the same base and between two parallel lines $\text{AB}$ and $\text{CD}$, So, $$ar\triangle \text{ADC}=ar\triangle \text{BCD}$$
Now the things those should be noticed are that:
In $\triangle \text{ADC}$ and $\triangle \text{BCD}$:
$$\text{AD}=\text{BC}$$ $$\text{DC}=\text{DC}$$ $$ar\triangle \text{ADC}=ar\triangle \text{BCD}.$$
Now in two different triangles, two sides are equal and their areas are also equal, so the third side is also equal or $\text{AC}=\text{BD}$. Which make this parallelogram a rectangle.
Isn't it a claim that every parallelogram is a rectangle or a parallelogram does not exist?
| As already hinted in comments and answers, you appear to have
applied Heron's formula
to the areas of the triangles $\triangle ADC$ and $\triangle BCD$.
Then observing that the only variables in Heron's formula are
the three sides of the triangle and its area,
and that three of these quantities in the formula for $\triangle ADC$
are equal to the corresponding quantities in the formula
for $\triangle BCD$, you concluded that the fourth quantity
must also be the same.
I will try to show that Heron's formula not only allows for multiple
solutions but (in the case of a non-rectangular parallelogram)
gives exactly the two distinct lengths of the diagonals.
One usually sees Heron's formula in a symmetric form such as this:
\begin{align}
s &= \tfrac12(a+b+c),\\
\triangle &= \sqrt{s(s-a)(s-b)(s-c)}.
\end{align}
This format is slightly deceptive since the second line shows
only one explicit occurrence of the variable $c,$ but there are
four implicit occurrences of $c$, one in each place where $s$ occurs.
A more explicit formula is
$$
\triangle = \tfrac14 \sqrt{(a+b+c)(b+c-a)(c+a-b)(a+b-c)}.
$$
Algebra shows that the quantity inside the radical can be regrouped
as follows:
$$
\triangle = \tfrac14 \sqrt{(2ab)^2 - (a^2+b^2-c^2)^2}. \tag1
$$
Now, given
$\triangle = \text{Area}(\triangle ADC) = \text{Area}(\triangle BCD),$
$a = AD = BC,$ and $b = CD,$
suppose that $c$ is a solution to Equation $1.$
In order for the quantity under the radical in Equation $1$
to be positive, it must be true that $c^2\leq 2(a^2+b^2).$
Let
$$
c' = \sqrt{2(a^2+b^2) - c^2}. \tag2
$$
Squaring both sides of Equation $2$ and rearranging terms,
$$
a^2 + b^2 - c'^2 = -(a^2+b^2-c^2) \tag3
$$
and the squares of both sides of Equation $3$ are equal.
Hence we can substitute $(a^2 + b^2 - c'^2)^2$ for
$(a^2 + b^2 - c^2)^2$ in Equation $1$,
that is, $c'$ is also a solution of that equation.
It is still possible that $c'$ is the same as $c.$
This happens precisely when $c^2 = a^2 + b^2,$
that is, when $\triangle ADC$ and $\triangle BCD$ are right triangles.
In that case the parallelogram is a rectangle, its diagonals are equal,
and the two triangles are congruent.
But any value of $c$ such that $\lvert a - b \rvert < c < a+b$
may be a diagonal of a parallelogram with sides $a$ and $b,$
and for every such value such that $c \neq \sqrt{a^2 + b^2}$
there is a distinct value $c'$ that is the length of the other diagonal.
| {
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"timestamp": "2023-03-29T00:00:00",
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The function $f (n) = (1 + 1 / n) ^ {n+1}$ is decreasing I cannot prove that the function $$f (n) = \left(1 + \frac1n\right) ^ {n + 1},$$ defined for every positive integer $n$, is strictly decreasing in $n$. I already tried to prove by induction and also tried to prove by calculating the difference between $f (n + 1)$ and $f (n)$.
I need help.
| Take the fraction
$$
\frac{f(n)}{f(n-1)}=\left(\frac{n+1}{n}\right)^{n+1}\bigg/\left(\frac{n}{n-1}\right)^n=\left(\frac{(n-1)(n+1)}{n^2}\right)^n\frac{n+1}{n}=\frac1{\left(1+\frac{1}{n^2-1}\right)^n}\frac{n+1}{n}.
$$
Using the induction, show that for $n\ge2$ and $x>-1$, $x\ne0$ (if $x=0$ then there would be a trivial equality) $(1+x)^n>1+nx$. For $n=2$ we have $(1+x)^2=1+2x+x^2>1+2x$. For $n+1$: $(1+x)^{n+1}=(1+x)^n(1+x)>(1+nx)(1+x)=1+(n+1)x+nx^2>1+(n+1)x$.
Using this inequality one can conclude that
$$
\left(1+\frac{1}{n^2-1}\right)^n>1+n\frac1{n^2-1}=\frac{n^2+n-1}{n^2-1},
$$
and
$$
\frac{f(n)}{f(n-1)}=\frac1{\left(1+\frac{1}{n^2-1}\right)^n}\frac{n+1}{n}<\frac{n^2-1}{n^2+n-1}\frac{n+1}{n}=\frac{n^3+n^2-n-1}{n^3+n^2-n}<1.
$$
Therefore, $f(n)$ is decreasing in $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$t^3$ is never equal to
Let $t$ be a real number such that $t^2=at+b$ for some positive integers $a$ and $b$. Then for any choice of positive integers $a$ and $b$, $t^3$ is never equal to:
(A). $4t+3$
(B). $8t+5$
(C). $10t+3$
(D). $6t+5$
We have $t^3 = at^2+bt$ and so for (A) we have $t^3 = 4t+3 = 4(at^2+bt)+3$. I don't see how to continue since this is a cubic equation. Is there an easier way of finding a contradiction?
| $t^2 = ax + b$
$t^3 = cx +d$
where $a$, $b$, $c$, $d$ are integers.
See 2 cases:
*
*$t^2=at+b$ has an integer root $\implies$ t is integer
*$t^2=at+b$ has no integer root $\implies$ roots are $p+\sqrt{q}$ and
$p−\sqrt{q}$, where $p$, $q$ are rational. You can see that if one
of these roots is a root of $x^3=cx+d$, the second is the root of
$x^3=cx+d$ too, and the third root must be an integer
$\implies$ $t^3-cx-d = 0$ has an integer root
A) $t = -1$
B) No integer root
C) $t= -3$
D) $t= -1$
So, answer is (B)
| {
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AM-GM inequality: $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d$
Let $a, b, c, d > 0$. Prove that $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{d} + \frac{d^2}{a} \geq a + b + c + d$.
I'm supposed to prove this by AM-GM, but I can't see how. Any help would be appreciated.
| Using this inequality is probably the quickest way
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{d}+\frac{d^2}{a}\geq \frac{(a+b+c+d)^2}{b+c+d+a}=a+b+c+d$$
Otherwise, using AM-GM we have:
$$\frac{a^2}{b}+b+\frac{b^2}{c}+c+\frac{c^2}{d}+d+\frac{d^2}{a}+a \geq 2\sqrt{\frac{a^2}{b}b}+2\sqrt{\frac{b^2}{c}c}+2\sqrt{\frac{c^2}{d}d}+2\sqrt{\frac{d^2}{a}a}=2(a+b+c+d)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Curious relation between $e$ and $\pi$ that produces almost integers I have seen in an article, without proof, that the following expression involving $e$ and $\pi$ is an almost integer very close to 1:
$ e^{-\frac{\pi}{9}} + e^{-4\frac{\pi}{9}} + e^{-9\frac{\pi}{9}} + e^{-16\frac{\pi}{9}} + e^{-25\frac{\pi}{9}} + e^{-36\frac{\pi}{9}} + e^{-49\frac{\pi}{9}} + e^{-64\frac{\pi}{9}} = 1.0000000000010504... $
Furthermore, I have read that the following "approximate theorem" holds
If $n$ is an odd square, then
$\sum\limits_{k = 1}^{n-1} e^{-k^2\frac{\pi}{n}}$
is an almost integer. By some experiments I see easily that the integer being approximated is
($\sqrt{n} - 1)/2$
How to prove that this curious relation holds?
| Using the identity provided by Robert Israel (which can be derived using the Poisson summation formula), we have $$\sum_{k = -\infty}^{\infty}e^{-k^2\tfrac{\pi}{n}} = \sum_{k = -\infty}^{\infty}\sqrt{n}e^{-n\pi k^2}.$$
We can bound the second summation by: $$\sqrt{n} \le \displaystyle\sum_{k = -\infty}^{\infty}\sqrt{n}e^{-n\pi k^2} = \sqrt{n}+2\sqrt{n}\displaystyle\sum_{k = 1}^{\infty}e^{-n\pi k^2} \le \sqrt{n}+2\sqrt{n}\displaystyle\sum_{k = 1}^{\infty}e^{-n\pi k} = \sqrt{n} + \dfrac{2\sqrt{n}}{e^{n\pi}-1}.$$
Also, we can bound the tails of the first summation by:
$$2e^{-n\pi} \le \displaystyle\sum_{|k| \ge n}e^{-k^2\tfrac{\pi}{n}} = 2\displaystyle\sum_{k = n}^{\infty}e^{-k^2 \tfrac{\pi}{n}} \le 2\displaystyle\sum_{k = n}^{\infty}e^{-nk \tfrac{\pi}{n}} = \dfrac{2e^{-n\pi}}{1-e^{-\pi}}.$$
Hence, $$\displaystyle\sum_{k = -(n-1)}^{n-1}e^{-k^2\tfrac{\pi}{n}} = \sum_{k = -\infty}^{\infty}\sqrt{n}e^{-n\pi k^2} - \displaystyle\sum_{|k| \ge n}e^{-k^2\tfrac{\pi}{n}}$$ can be bounded by $$\sqrt{n}-\dfrac{2e^{-n\pi}}{1-e^{-\pi}} \le \displaystyle\sum_{k = -(n-1)}^{n-1}e^{-k^2\tfrac{\pi}{n}} \le \sqrt{n} + \dfrac{2\sqrt{n}}{e^{n\pi}-1}-2e^{-n\pi}.$$
To bound $\displaystyle\sum_{k = 1}^{n-1}e^{-k^2\tfrac{\pi}{n}}$, simply subtract $1$ (for the $k = 0$ term) and divide by $2$ (since the $-k$-th term and the $k$-th term are equal) to get
$$\dfrac{\sqrt{n}-1}{2}-\dfrac{e^{-n\pi}}{1-e^{-\pi}} \le \displaystyle\sum_{k = 1}^{n-1}e^{-k^2\tfrac{\pi}{n}} \le \dfrac{\sqrt{n}-1}{2} + \dfrac{\sqrt{n}}{e^{n\pi}-1}-e^{-n\pi}.$$
Thus, the error in approximating $\dfrac{\sqrt{n}-1}{2}$ is at most $\max\left\{\dfrac{e^{-n\pi}}{1-e^{-\pi}},\dfrac{\sqrt{n}}{e^{n\pi}-1}-e^{-n\pi}\right\}$. For $n = 9$, this is roughly $1.051\times 10^{-12}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that if a prime $p$ divides $5^n-2$ and $2^n-5$, then $p = 3$ The result below has been disproven.
Let $n \in \mathbb{N}$. Prove that if a prime $p$ divides $5^n-2$ and $2^n-5$, then $p = 3$.
We know that $p \neq 2,5$. We need to have \begin{align*}5^n &\equiv 2 \pmod{p}\\2^n &\equiv 5 \pmod{p}.\end{align*} This gives us $10^{n-1} \equiv 1 \pmod{p}$. Thus $\text{ord}_{p}(10) \mid (n-1)$. How do we continue?
| $\gcd(5^{65} - 2, 2^{65} - 5) = 12681 = 3^2 \cdot 1409$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve equation in determinant Let $ a,b,c,m,n,p\in \mathbb{R}^{*} $, $ a+m+n=p+b+c $. Solve the equation:
$$
\begin{vmatrix}
x & a & b &c \\
a & x & b &c \\
m &n & x &p \\
m& n& p& x
\end{vmatrix}
=0
$$
I had used the Schur complement ($\det(M)=\det(A)\cdot (D-C\cdot A^{-1}\cdot B)$, for $ M= \begin{bmatrix}
A &B \\
C & D
\end{bmatrix}) $ but it didn't help me.
| The equation
$$\begin{vmatrix}
x & a & b &c \\
a & x & b &c \\
m &n & x &p \\
m& n& p& x
\end{vmatrix}
=0$$
is equivalent to $p_A(-x)=0$, where $p_A$ is the characteristic polynomial of
$$A=\begin{pmatrix}
0 & a & b & c \\
a & 0 & b & c \\
m &n & 0 &p \\
m& n& p& 0
\end{pmatrix};$$hence the roots of your equation must be the opposite of the eigenvalues of $A$. The condition that $a+m+n=b+c+p$ is equivalent to $a-b-c=-m-n+p$, which tells you that $(1,1,-1,-1)$ is an eigenvector, with associated eigenvalue $a-b-c$. Moreover $-a$ and $-p$ are obviously eigenvalues, and the trace of the matrix is $0$; hence the sum of the eigenvalues is zero, which means that the last eigenvalue must be $b+c-a+a+p=b+c+p$.
So the solutions to your equation are $a$, $p$, $-(b+c+p)$ and $b+c-a$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$a \cot A + b \cot B + c \cot C = 2(R+ r)$ With usual notations, prove that in a triangle $ABC$:
$$a \cot A + b \cot B + c \cot C = 2(R+ r)$$
I know $R = (abc)/(4\Delta)$ and $r = \Delta/s$.
I could not understand how to start it.
| \begin{align}
a \cot A + b \cot B + c \cot C &= 2R\Bigg[\sin A \frac{\cos A}{\sin A} + \sin B \frac{\cos B}{\sin B} + \sin C \frac{\cos C}{\sin C}\Bigg]\\
&= 2R\big[\cos A + \cos B +\cos C\big]\\
&= 2R\Bigg[2\cos \frac{A+B}{2}\cos \frac{A-B}{2} + 1-2\sin^2 \frac{C}{2}\Bigg]\\
&=2R\Bigg[1+ 2\sin \frac{A}{2}\sin \frac{B}{2} \sin \frac{C}{2}\Bigg]\\ &=2R\Big[1+\frac{r}{R}\Big] =2(R+r).
\end{align}
Hope it helps.
| {
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Show $\forall a,b\in\mathbb{R}^*\quad 3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2 (\frac{a}{b}+\frac{b}{a})$ Let $a,b \in \mathbb{R}^*$ Show that :
$$3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2\left(\frac{a}{b}+\frac{b}{a}\right)$$
Let $a,b\in\mathbb{R}^*$
let $t=\dfrac{a}{b}+\dfrac{b}{a}$ , we've :
$$
\begin{aligned}
3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2\left(\frac{a}{b}+\frac{b}{a}\right) &\iff
3+\frac{a^2}{b^2}+\frac{b^2}{a^2}-2\left(\frac{a}{b}+\frac{b}{a}\right)\\
&\iff t^{2}-2t+1\geq 0\\
&\iff \left(t-1\right)^{2}\geq 0
\end{aligned}$$
since $\forall t\in \mathbb{R}\quad (t-1)^{2}\geq 0$ holds then also $$\forall a,b\in \mathbb{R}^*\quad 3+\frac{a^2}{b^2}+\frac{b^2}{a^2}\ge 2\left(\frac{a}{b}+\frac{b}{a}\right) $$ holds
Am i right ? is there any other ways
| We have: $1+\dfrac{a^2}{b^2} \ge \dfrac{2a}{b}, 1+\dfrac{b^2}{a^2} \ge \dfrac{2b}{a}$. Adding these we have the desire inequality. Thus the inequality is strict.
| {
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finding value of indefinite integration of $\frac{y^7-y^5+y^3-y}{y^{10}+1}dy$ finding value of indefinite integration $\displaystyle \int\frac{y^7-y^5+y^3-y}{y^{10}+1}dy$
$\displaystyle \int\frac{y^5(y^2-1)+y(y^2-1)}{y^{10}+1}dy = \int\frac{(y^5+y)(y^2-1)}{y^{10}+1}dy = \int\frac{(y^5+y)(y^2+1-2)}{y^{10}+1}dy$
$\displaystyle =\int\frac{y^5+y}{y^8-y^6+y^4-y^2+1}-2\int\frac{y^5+y}{y^{10}+1}dy$
i wan,t be able to proceed after that, could some help me with this
| HINT: $$y^{10}+1= \left( {y}^{2}+1 \right) \left( {y}^{8}-{y}^{6}+{y}^{4}-{y}^{2}+1
\right)
$$
further we get
$$\frac{y^7-y^5+y^3-y}{y^{10}+1}=1/5\,{\frac {y \left( 4\,{y}^{6}-3\,{y}^{4}+2\,{y}^{2}-1 \right) }{{y}
^{8}-{y}^{6}+{y}^{4}-{y}^{2}+1}}-4/5\,{\frac {y}{{y}^{2}+1}}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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"answer_id": 2
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If $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle $ABC$ is equilateral In a triangle $ABC,$ if $2(bc^2+ca^2+ab^2) = b^2c+c^2a+a^2b+3abc, $ then prove that triangle
$ABC$ is equilateral
$\displaystyle \frac{2(bc^2+ca^2+ab^2)}{abc} = \frac{b^2c+c^2a+a^2b+3abc}{abc}$
$\displaystyle 2\left(\frac{c}{a}+\frac{a}{b}+\frac{b}{c}\right) = \left(\frac{b}{a}+\frac{c}{b}+\frac{a}{c}\right)+3$
let $\displaystyle \frac{a}{b} = x,\frac{b}{c}=y,\frac{c}{a}=z$
so $\displaystyle 2(x+y+z) = \frac{1}{x}+\frac{1}{y}+\frac{1}{z}+3$
wan,t be able to proceed after that, could some help me
| Since $a,b,c$ are sides of a triangle we may write $(a,b,c) = (y+z,z+x,x+y)$
with $x,y,z>0$. The difference between the two sides is then
$$
x^3 + y^3 + z^3 \; + \; x^2 y + y^2 z + z^2 x
- 2 (xy^2 + yz^2 + zx^2).
$$
By cyclic symmetry we may assume $x \geq z$ and $y \geq z$, and write
$x = (1 + \alpha) z$ and $y = (1 + \beta) z$ with $\alpha,\beta \geq 0$
to find $z^3$ times
$$
2(\alpha^2 - \alpha\beta + \beta^2) + \beta(\alpha-\beta)^2 + \alpha^3,
$$
in which each term is nonnegative, and all are zero iff
$\alpha = \beta = 0$ $-$ which in turn is equivalent to $x=y=z$,
and thus $a=b=c$, so triangle $ABC$ is equilateral as claimed.
$\Box$
[Added later: The solution above applies a general technique for such
problems; but I see that for the present question
Hari Shankar posted (in a comment) a link to an
AoPS item
that gives an even simpler conclusion that retains the cyclic symmetry:
$$
x^3 + y^3 + z^3 \; + \; x^2 y + y^2 z + z^2 x
- 2 (xy^2 + yz^2 + zx^2) = x(x-z)^2 + y(y-x)^2 + z (z-y)^2,
$$
again with all terms nonnegative; so equality implies that
they all vanish and the rest follows as before. That link also
gives the example $(a:b:c) = (10:3:24)$ of a rational point on the cubic
$2(bc^2 + ca^2 + ab^2) = b^2c + c^2a + a^2b + 3abc$ with positive variables
that do not satisfy the triangle inequality.]
| {
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Prove that $\ln(x+\sqrt{x^2 + 4}) - \ln2$ is odd I know that a function is odd when
$$f(-x) = -f(x)$$
Therefore I can say that if for a function $$-f(x) + f(x) = f(-x) + f(x) = 0$$
Then the function is odd!
I tried to use this trick to prove that $f(x) = \ln\left(x+\sqrt{x^2 + 4}\right) - \ln2$ is odd.
However, I would want to prove directly that $$f(-x) = -f(x)$$
In other words, I want to solve $$\ln\left(-x+\sqrt{(-x)^2 + 4}\right) - \ln2$$
and to come at the end to this:
$$-\ln\left(x+\sqrt{x^2 + 4}\right) + \ln2$$
This was my approach:
$$\ln\left(-x+\sqrt{(-x)^2 + 4}\right) - \ln2$$
$$\ln\left(-x+\sqrt{x^2 + 4}\right) - \ln2$$
$$\ln\left(\frac{-x+\sqrt{x^2 + 4}}{2}\right)$$
$$\ln\left(\left(\frac{2}{-x+\sqrt{x^2 + 4}}\right)^{-1}\right)$$
$$-\ln\left(\frac{2}{-x+\sqrt{x^2 + 4}}\right)$$
Here I got stuck. I want to get to $-\ln\left(x+\sqrt{x^2 + 4}\right) + \ln2$ but if I use $\ln\left(\frac ab\right) = \ln a - \ln b$ then I will get back to $f(-x)$ and not to $-f(x)$.
Any help?
| Hint:
$$\frac{2}{\sqrt{x^2+4}-x}=\frac{2(\sqrt{x^2+4}+x)}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2080644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx={1\over 2}$ using an alternative method Prove that
$$\int_0^1 {3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \,dx={1\over 2}\tag1$$
My try:
$u=1+\sqrt{5}x$ then $du=\sqrt{5} \, dx$
$${1\over \sqrt 5}\int_1^{1+\sqrt{5}}(4u^{-3}-u^{-2}) \, du$$
$$\left. {1\over \sqrt{5}}(-2u^{-2}+u^{-1}) \right|_1^{1+\sqrt{5}}={1\over 2}$$
Prove $(1)$ using an alternative method other than substitution method.
| Set $x=\frac{\sinh^2u}{\sqrt{5}}$ and $dx=\frac{2\sinh u\cosh u}{\sqrt{5}}$
So $$\int{3-\sqrt{5}x\over (1+\sqrt{5}x)^3} \, dx=\int \frac{6\sinh u\cosh u}{\sqrt{5}\cosh^6 u}du -\int\frac{2\sinh^3u \cosh u}{\sqrt{5}\cosh^6 u} \, du = \int \frac{6\sinh u}{\sqrt{5}\cosh^5 u} \, du -\int \frac{2(1+\cosh^2 u ) \sinh u}{\sqrt{5}\cosh^5 u} \, du$$
And you can calculate these integrals using the substitution $t=\cosh u$ and $dt=\sinh u \, du$
Another way using partial fractions:
$$\int{3-\sqrt{5}x\over (1+\sqrt{5}x)^3} dx= \int \frac{20\sqrt{5}}{(5x+\sqrt{5})^3}- \frac{5}{(5x+\sqrt{5})^2}dx $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there any double series that cannot exchange the sum? I want to find a sequence $(u_{n,p})_{(n,p)\in\mathbb{N}^2}$ that satisfied:
$$
\sum_{n=0}^{\infty}\sum_{p=0}^{\infty}u_{n,p} ~\text{is convergent}
$$
$$
\sum_{p=0}^{\infty}\sum_{n=0}^{\infty}u_{n,p} ~\text{is convergent too}
$$
but
$$
\sum_{p=0}^{\infty}\sum_{n=0}^{\infty}u_{n,p} \neq \sum_{n=0}^{\infty}\sum_{p=0}^{\infty}u_{n,p}
$$
| Let $u_{mn} = 1/(m^2 - n^2)$ if $m \neq n$ and $u_{mn} =0$ if $m = n.$
Note that
$$\frac{\pi^2}{12} = \sum_{n=1}^\infty \sum_{m=1, m \neq n}^\infty \frac{1}{m^2 - n^2} \neq \sum_{m=1}^\infty \sum_{n=1, n \neq m}^\infty \frac{1}{m^2 - n^2} = - \frac{\pi^2}{12}.$$
By anti-symmetry, the double sum changes sign with an interchange of indices.
To find the sum, use
$$\begin{align}\sum_{m=1, m \neq n}^\infty \frac{1}{m^2 - n^2} &= \lim_{M \to \infty} \frac{1}{2n} \sum_{m=1, m \neq n}^M \left(\frac{1}{m-n} - \frac{1}{m+n} \right) \\ &= \lim_{M \to \infty} \frac{1}{2n} \left(\sum_{k=1}^{M-n} \frac{1}{k} - \sum_{k=1}^{n-1} \frac{1}{k} - \sum_{k=n+1}^{M+n} \frac{1}{k}\right)\\ &= \lim_{M \to \infty} \frac{1}{2n} \left(\frac{1}{n}- \frac{1}{M-n+1} - \ldots - \frac{1}{M+n} \right) \\ &= \frac{1}{2n^2} \end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If, in a triangle, $\cos(A) + \cos(B) + 2\cos(C) = 2$ prove that the sides of the triangle are in AP By using the formula :
$$
\cos(A)+\cos(B)+\cos(C) = 1 + 4 \sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right) \sin\left(\frac{C}{2}\right)
$$
I've managed to simplify it to :
$$
2\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)=\sin\left(\frac{C}{2}\right)$$
But I have no idea how to proceed.
| using $$\cos(\alpha)=\frac{b^2+c^2-a^2}{2bc}$$ and so on and plugging these equations in your equation and factorizing we get
$$-1/2\,{\frac { \left( c+a-b \right) \left( -c+a-b \right) \left( -2
\,c+a+b \right) }{bca}}
=0$$
can you proceed?
| {
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Integral of a rational function I need a closed form for the following integral
$$\int^\infty_0\frac{t+b}{(t^2+a)(k^2+(t+b)^2)}dt$$
where $a,b \in \mathbb{C}$ and $k \in \mathbb{N}$. That $t+b$ makes the integral complicated for me.
Any simple approaches for that ?
I tried wolfram |Alpha but it fails. A final solution using CAS is acceptable for me.
I guess a closed form can be obtained using Incomplete Beta function, But I need a simpler one ?
| Use partial fractions to do the calculation. Let
$$ \frac{t+b}{(t^2+a)(k^2+(t+b)^2)}=\frac{At+B}{t^2+a}+\frac{Ct+D}{(t+b)^2+k^2}$$
and then one has
$$ (At+B)((t+b)^2+k^2)+(Ct+D)(t^2+a)=t+b.$$
It is not hard to obtain
\begin{eqnarray}
A&=&\frac{k^2-a-b^2}{a^2+2 a b^2-2 ak^2+b^4+2 b^2 k^2+k^4},\\
B&=&\frac{a b+b^3+b k^2}{a^2+2 a b^2-2 a
k^2+b^4+2 b^2 k^2+k^4},\\
C&=&-A,\\
D&=&\frac{a b+b^3-3 b k^2}{a^2+2 a b^2-2 a k^2+b^4+2 b^2
k^2+k^4}.
\end{eqnarray}
(Suppose $a,b,k>0$.) It is not hard to get
\begin{eqnarray}
&&\int^\infty_0\frac{t+b}{(t^2+a)(k^2+(t+b)^2)}dt\\
&=&\int^\infty_0\left(\frac{At+B}{t^2+a}+\frac{-At+D}{(t+b)^2+k^2}\right)\\
&=&\frac A2\ln\frac{t^2+a}{k^2+(t+b)^2}+\frac{B}{\sqrt a}\arctan\frac{t}{\sqrt a}+\frac{-bC+D}{k}\arctan\frac{b+t}{k}\bigg|_0^\infty\\
&=&(\frac{B}{\sqrt a}+\frac{-bC+D}{k})\frac{\pi}{2}-\frac{A}{2}\ln\frac{a}{k^2+b^2}-\frac{-bC+D}{k}\arctan\frac{b}{k}.
\end{eqnarray}
| {
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Help in understanding a simple proof. Let $Ax + By + C = 0$ be a general equation of a line and $x\cos \alpha + y\sin \alpha - p = 0$ be the normal form of the equation.
Then,
$${-p\over C } = { \cos \alpha\over A} = { \sin\alpha\over B}\tag{1}$$
$${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} \tag{2}$$
$${p\over C } = { \cos \alpha\over -A} = { \sin\alpha\over -B} = {\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}} = {1\over \sqrt{A^2 + B^2}} \tag{3}$$
$$\therefore \bbox[ #FFFDD0, 10px, Border:2px solid #DC143C]{p = {C\over \sqrt{A^2 + B^2}}, \cos \alpha = {-A\over \sqrt{A^2 + B^2}},\sin\alpha = {-B\over \sqrt{A^2 + B^2}}} $$
I did not get the $(3)$ part. Where does $\displaystyle{\sqrt{\sin^2 \alpha + \cos^2 \alpha}\over \sqrt{A^2 + B^2}}$ come from ?
| I'd do it differently: since $A$ and $B$ are not both zero, we can divide by $-\sqrt{A^2+B^2}$, getting
$$
\frac{-A}{\sqrt{A^2+B^2}}x+\frac{-B}{\sqrt{A^2+B^2}}-\frac{C}{\sqrt{A^2+B^2}}=0
$$
and we can choose $\alpha$ so that
$$
\begin{cases}
\cos\alpha=-\dfrac{A}{\sqrt{A^2+B^2}}\\[6px]
\sin\alpha=-\dfrac{B}{\sqrt{A^2+B^2}}
\end{cases}
$$
and set
$$
p=\frac{C}{\sqrt{A^2+B^2}}
$$
For the mysterious formula, observe that, if
$$
k=\frac{\cos\alpha}{-A}=\frac{\sin\alpha}{-B}
$$
then
$$
A^2k^2=\cos^2\alpha,\quad B^2k^2=\sin^2\alpha
$$
so also
$$
(A^2+B^2)k^2=1
$$
and
$$
|k|=\frac{1}{\sqrt{A^2+B^2}}
$$
Of course one needs to be very precise about the choice of $p$, using this method. It is implied that $p<0$ if $C<0$ and $p>0$ if $C>0$; a special case should be made when $C=0$.
The method above is independent of these considerations.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $a^3-b^3=2 \ \text{ and }\ a^5-b^5 \ge 4$, then $a^2+b^2 \ge 2$ Prove that if
$$a^3-b^3=2 \ \text{ and }\ a^5-b^5 \ge 4$$
then $$a^2+b^2 \ge 2.$$
| From the conditions we have $\frac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^2+ab+b^2}\geq2$.
Thus, it remains to prove that $a^2+b^2\geq\frac{a^4+a^3b+a^2b^2+ab^3+b^4}{a^2+ab+b^2}$, which is $a^2b^2\geq0$.
Done!
| {
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Computing $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$
What is $\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k}$ ?
Here are a few remarks:
*
*Since $x\mapsto \frac{2^x}{x}$ is increasing when $x\geq 2$, one might be tempted to use the integral test. This fails: when doing so, one gets $a_n\leq \sum_{k=1}^n \frac{2^k}{k}\leq b_n$ where $a_n\sim \frac{2^n}{\ln (2)n}$ and $b_n\sim \frac{2^{n+1}}{\ln (2)n}$.
Unfortunately $b_n$ is too big and this estimate doesn't yield the limit.
*Here's my solution: since it's easy to sum $2^k$ and the difference $\frac{1}{k}-\frac{1}{k+1}$ is small, it's natural to try summation by parts: $$\begin{align} \sum_{k=1}^n \frac{2^k}{k}
&=\frac{S_n}{n+1}-1+\sum_{k=1}^n S_k \left(\frac{1}{k}-\frac{1}{k+1} \right)\quad \text{where} \; S_n=\sum_{k=0}^n 2^k\\
&= \frac{2^{n+1}}{n+1} + \sum_{k=1}^n \frac{2^{k+1}}{k(k+1)} - \underbrace{1 - \sum_{k=1}^n\left(\frac{1}{k(k+1)}\right) - \frac{1}{n+1}}_{\text{bounded}}\\
\end{align}$$
Intuition suggests $\displaystyle \sum_{k=1}^n \frac{2^{k+1}}{k(k+1)}=o\left(\frac{2^n}n \right)$ but it's not immediate to prove. I had to resort to another summation by parts! Indeed
$$\begin{align}\small\sum_{k=1}^n \frac{2^{k+1}}{k(k+1)}&= \small 2\left[ \frac{2^{n+1}}{n(n+1)} + 2\sum_{k=1}^n \left(\frac{2^{k+1}}{k(k+1)(k+2)}\right)-\frac 12 -2\sum_{k=1}^n \left(\frac{1}{k(k+1)(k+2)}\right) - \frac{1}{n(n+1)}\right]\\
&\small\leq \frac{2^{n+2}}{n(n+1)}+\frac{2^{n+2}}{n(n+1)(n+2)}\cdot n \\
&\small= o\left(\frac{2^n}n \right)
\end{align}$$
Hence $$\sum_{k=1}^n \frac{2^k}{k} = \frac{2^{n+1}}{n+1} + o\left(\frac{2^n}n \right)$$ and $$\lim_n \frac{n}{2^n}\sum_{k=1}^n \frac{2^k}{k} = 2$$
This solution is quite tedious and computational... That's why I'm looking for a shorter or smarter solution that avoids summation by parts (integration by parts is easy to perform on functions, it just gets quite heavy with series).
| Let $f(x)=\sum_{k=1}^n\frac{x^k}{k}$. Then, we can write
$$f(x)=\sum_{k=1}^n \int_0^x t^{k-1}\,dk=\int_0^x \frac{1-t^n}{1-t}\,dt$$
so that $f(2)=\sum_{k=1}^n\frac{2^k}{k}=\int_0^2 \frac{1-t^n}{1-t}\,dt$.
Next, we choose $0<\delta<1$ and write
$$\begin{align}
\frac{n}{2^n}\int_0^2 \frac{1-t^n}{1-t}\,dt&=\frac{n}{2^n}\int_0^{2-\delta}\frac{1-t^n}{1-t}\,dt+\frac{n}{2^n}\int_{2-\delta}^2\frac{t^n-1}{t-1}\,dt \tag 1
\end{align}$$
We can apply the mean-value theorem for the second term on the right-hand side of $(1)$ to reveal that for some $\xi_n\in [2-\delta,2]$
$$\begin{align}
\frac{n}{2^n}\int_{2-\delta}^2 \frac{t^n-1}{t-1}\,dt&=\frac{n}{2^n(\xi_n-1)}\int_{2-\delta}^2(t^n-1)\,dt\\\\
&=\frac{n}{2^n(\xi-1)}\left(\frac{2^{n+1}-(2-\delta)^{n+1}}{n+1}+\delta\right) \tag 2
\end{align}$$
Letting $n\to \infty$ in $(2)$ yields
$$\lim_{n\to \infty}\frac{n}{2^n}\int_{2-\delta}^2 \frac{t^n-1}{t-1}\,dt=\frac{2}{\xi-1}$$
For any given $\epsilon>0$, we can choose $0<\delta<1$ so small that $\left|\frac{2}{\xi_n-1}-2\right|<\epsilon$. Now, we proceed with that $0<\delta<1$ fixed.
Noting that $\frac{1-t^n}{1-t}$ is positive and monotonically increasing on $[0,2]$, the first term on the right-hand side of $(1)$ is bounded above by $\frac{n}{2^n}\frac{(2-\delta)^n-1}{1-\delta}\to 0$ as $n\to \infty$.
Putting it all together, we obtain the coveted limit
$$\lim_{n\to \infty}\frac{n}{2^n}\sum_{k=1}^n\frac{2^k}{k}=2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove that $(x+ {\sqrt{1+x^2}} ) ( y+ {\sqrt {1+y^2}}) = 1$ if $(x+ {\sqrt{1+y^2}} ) ( y+ {\sqrt {1+x^2}}) = 1$ Let $x,y$ be real numbers such that :
$(x+ {\sqrt{1+y^2}} ) ( y+ {\sqrt {1+x^2}}) = 1$.
Prove that :
$(x+ {\sqrt{1+x^2}} ) ( y+ {\sqrt {1+y^2}}) = 1$.
I tried taking $x=y$. It simplifies everything a lot. But I'm not able to progress when both $x$ and $y$ are in the same equation.
| Let $x =\sinh(a), y=\sinh(b)$.
Then
$$(\sinh(a)+\cosh(b)) (\cosh(a) +\sinh(b))=1 \\
\sinh(a)\cosh(a)+\sinh(b)\cosh(b)+\cosh(a)\cosh(b)+\sinh(a)\sinh(b)=1 \\
\sinh(a+b)+\cosh(a+b)=1
$$
Now,
$$1=\cosh^2(a+b)-\sinh^2(a+b)=\left( \sinh(a+b)+\cosh(a+b)\right)\left( \cosh(a+b)-\sinh(a+b)\right) \\=\left( \cosh(a+b)-\sinh(a+b)\right)$$
Therefore
$$\sinh(a+b)+\cosh(a+b)=1\\
\cosh(a+b)-\sinh(a+b)=1$$
and hence
$$\cosh(a+b)=1 \\
\sinh(a+b)=0$$
which proves that $a+b=0$.
Therefore $y=-x$.
From here it is trivial.
| {
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If $ \sin \alpha = \frac 45 $ and $ \cos \beta = \frac{5}{13} $, prove that $ \cos \frac{\alpha-\beta}{2} = \frac{8}{\sqrt {65}} $ I can solve it easily if I assume that $ 0 < \alpha, \beta < \frac{\pi}{2}$
But there is no mention of the quadrants in which $ \alpha $ and $ \beta $ lie in.
Is the question wrong ?
| HINT:
\begin{align}
\cos(\alpha/2-\beta/2)&=\sqrt{\frac{1+\cos(\alpha-\beta)}{2}}=\sqrt{(1+\cos\alpha\cos\beta+\sin\alpha\sin\beta)/2}\\
\end{align}
$$ \cos\alpha =\frac35, \cos\beta = \frac{5}{13},\, \sin\alpha=\frac45,\, \sin\beta =\frac{12}{13}, $$
plug in to get $ 8/\sqrt{65}$
For other quadrants calculate other three values also which are possible with inverse trig functions.
$$\sqrt{(1\pm\cos\alpha\cos\beta\pm\sin\alpha\sin\beta)/2}$$
| {
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Prove that $37$ is a factor of $(10^{2014} + 10^{2015} + 2)^{2016} − 1$ How can I prove that $37$ is a factor of $(10^{2014} + 10^{2015} + 2)^{2016} − 1$, probably using basic number theory?
| Below is a somewhat not so elementary approach:
Notice $2016=36\times56$, and by Fermat's little theorem, for any $37\not\mid a$, we have $a^{36}\equiv1\pmod{37}$.
So $(10^{2014} + 10^{2015} + 2)^{2016} − 1\equiv-1\text{ or } 0\pmod{37}$.
But we know $10^{2016}\equiv1\pmod{37}$ so $10^{2015}\equiv10^{-1}\equiv2^{-1}\times5^{-1}\equiv19\times15\equiv26\pmod{37}$ and $10^{2015}+2\equiv28\pmod{37}$.
Moreover we see that if $10^{2014} + 10^{2015} + 2\equiv10^{2014}+28\equiv0\pmod{37}$, then $10^{2014}\equiv9\pmod{37}$.
However this is impossible: $10^1\equiv10\pmod{37}, 10^{2}\equiv26\pmod{37}, 10^{3}\equiv1\pmod{37}$.
Hence $10^{2014} + 10^{2015} + 2\not\equiv0\pmod{37}$ and $(10^{2014} + 10^{2015} + 2)^{2016}\equiv1\pmod{37}$.
Hope this helps.
| {
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The quadratic equation $x^2+x=3kx-k^2$ has two different real roots. Find the range of $k$
The quadratic equation $x^2+x=3kx-k^2$ has two different real roots. Find the range of $k$.
My answer is $k<1$ or $k<\frac{1}{5}$, but the answer sheet says $k<\frac{1}{5}$ or $k>1$.
What have I done wrong? Please help.
What I've done
$(1-3k)^2-4(k^2)>0$
$1-6k+9k^2-4k^2>0$
$\frac{6 \pm 4}{10}>k$
$k<1$ or $k<\frac{1}{5}$
| We have that the discriminant $$\Delta _{f (x)} =(1-3k)^2-4k^2 =(5k-1)(k-1) $$ We can analyse this into three cases:
Case $(1) $: If $k <\frac {1}{5} $, then surely $(5k-1) $ and $(k-1) $ are negative. So, $\Delta $ is positive. (Acceptable)
Case $(2) $: If $k\in (\frac {1}{5},1) $, then $(5k-1) $ is positive but $(k-1) $ is negative. (Not acceptable)
Case $(3) $: If $k>1$, then both $(5k-1) $ and $(k-1) $ are positive. (Acceptable)
Using these cases, the result follows. Hope it helps.
| {
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Maximum of $\frac{1+3a^2}{(a^2+1)^2}$ What is the maximum value of
$\displaystyle{{1 + 3a^{2} \over \left(a^{2} + 1\right)^{2}}}$, given that $a$ is a real number, and for what values of $a$ does it occur ?.
| Let $a^2=x$.
Hence, $\frac{1+3x}{(1+x)^2}\leq\frac{9}{8}$ it's $(3x-1)^2\geq0$.
The equality occurs for $x=\frac{1}{3}$, which says that the answer is $\frac{9}{8}$.
| {
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Determine if function is convex I want to determine if the function
$$f(x,y)=e^x+e^y+x^2-2xy+4y^2+2x+3y-8$$
is convex on $\mathbb{R}^2$. Any ideas on how to prove convexity for two variable functions?
| $$f(x,y)=e^x+e^y+x^2-2xy+4y^2+2x+3y-8$$
$$\nabla f=\begin{bmatrix} e^x+2x-2y+2 \\ e^y-2x+8y+3\end{bmatrix} $$
$$\nabla^2 f=\begin{bmatrix} e^x+2 & -2 \\ -2 & e^y+8\end{bmatrix} $$
Since
$$e^x+2>0$$
and $$(e^x+2)(e^y+8)-2^2=8e^x+2e^y+16-4=8e^x+2e^y+12>0$$
The function is convex.
Another way to see this is to observe that since
$$\nabla^2 f=\begin{bmatrix} e^x+2 & -2 \\ -2 & e^y+8\end{bmatrix} =\begin{bmatrix} e^x & 0 \\ 0 & e^y\end{bmatrix}+ \begin{bmatrix} 2 & -2 \\ -2 & 8\end{bmatrix}.$$
Observe that
$\begin{bmatrix} 2 & -2 \\ -2 & 8\end{bmatrix}$ is positive semidefinite from Gershgorin circle theorem, the eigenvalues lie between $0$ and $10$.
Also, $\begin{bmatrix} e^x & 0 \\ 0 & e^y\end{bmatrix}$ is positive definite.
Hence $\nabla^2f$ is positive definite.
| {
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If $-1 < a < 1$ then $\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a} < 3$?
Prove that if $-1<a<1$, then :
$\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a} < 3 $
I do not know how to even approach this problem. Those fourth roots confuse me a lot.
Any help would be appreciated.
Also I would like to know how the left hand side of the inequality behaves if there are no given bounds for the value of $a$.
Thanks in advance :) .
| It's wrong of course. Try $a=0$.
$\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a}\leq3$ is true.
Indeed, by P-M $(x+y+z)^4\leq27(x^4+y^4+z^4)$.
This inequality follows also from Holder and more:
$$(1+1+1)^3(x^4+y^4+z^4)\geq(x+y+z)^4$$
Hence, $$\left(\sqrt[4]{1-a^2} + \sqrt[4]{1-a} + \sqrt[4]{1+a}\right)^4\leq27(1-a^2+1-a+1+a)=27(3-a^2)\leq81$$
and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Factorizing $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ The intermediate factorization calculation $z^2-z\frac{a^2 +b^2}{ab}+1 = (z-a/b)(z-b/a)$ was coined straightforward by Riley's Mathematical Methods for Physics and Engeneering (eq.24.66).
I've tried completing the square,
$$
z^2-z\frac{a^2 +b^2}{ab}+1 = 0 \\
\bigg( z-\frac{a^2 +b^2}{2ab} \bigg)^2 +1 -\frac{(a^2+b^2)^2}{(2ab)^2} = 0 \\
\bigg( z-\frac{a^2 +b^2}{2ab} \bigg)^2 = \frac{(a^2+b^2)^2}{(2ab)^2}-1 \\
z= \frac{a^2 +b^2}{2ab} \pm \sqrt{\frac{(a^2+b^2)^2}{(2ab)^2}-1} \ , \\
$$
which doesn't nearly suggest the simple end result.
I'm sure it will roll out after reorganizing terms, but I'm interested in an alternative approach which more naturally suggests the end result of $(z-a/b)(z-b/a)$, reflecting the word-use of straightforward in the text.
Context.
Using contour integration to evaluate $I = \int_{0}^{2\pi} \frac{\cos(2\theta)}{a^2+b^2-2ab\cos(\theta)}d\theta$
| You have one mistake.
$\frac{(a^2+b^2)^2}{(ab)^2}$ this term should be
$\frac{(a^2+b^2)^2}{(2ab)^2}$ then its easily solve.
| {
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Find all prime numbers $y$ and $z$ if $x$ is an integer and $x^2-y^2-z^2=2017.$ Find prime numbers $y$ and $z$ if $x$ is an integer and $x^2-y^2-z^2=2017.$
$x^2-y^2-z^2$ is not factorable, so what should I do? My mind is blank here. Any help is appreciated.
| We have that $$a^2\equiv 1\lor 0\pmod{4}$$
We also have that $$2017\equiv 1\pmod{4}$$.Now try to choose $x^2,y^2,z^2$ so that you get $1\pmod{4}$.You can see that the only solution is if $y^2\equiv 0\pmod{4}$ and $z^2\equiv 0\pmod{4}$ but since $z,y$ are prime we have that $z=y=2$ so the only solution is
$x^2-8=2017,x^2=2025,x=45$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solutions to $|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|$ Find all the complex numbers $z$ satisfying
\begin{equation}
|z+1-i\sqrt{3}|=|z-1+i\sqrt{3}|
\end{equation}
I tried using $z=a+bi$, then using the formula for absolute value:
\begin{equation}
|(a+1)+i(b-\sqrt{3})|=|(a-1)+i(b+\sqrt{3})|\\
\sqrt{(a+1)^2+(b-\sqrt{3})^2}=\sqrt{(a-1)^2+(b+\sqrt{3})^2}
\end{equation}
Is this the right way to go? Should I solve for b or a, and in that case what does that tell me about possible solutions? Any help would be appreciated.
Update:
Solved the equation for $y$, got
\begin{equation}
y=\frac{x}{\sqrt{3}}
\end{equation}
Does this mean that the equation is satisfied for all complex numbers on the line $y=\frac{x}{\sqrt{3}}$?
| Write
$|z-(-1+i\sqrt{3})|=|z-(1-i\sqrt{3})|$
This is the locus of all $z$ which thier diatance from two points $A=-1+i\sqrt{3}$ and $B=1-i\sqrt{3}$ is equal. It halfs the fragment $AB$ which should be $y=\sqrt{3}x$.
| {
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Integral of a two variable function $e^{-(5x^2-6xy+5y^2)}$ How to compute $$\int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(5x^2-6xy+5y^2)} \, dx \, dy \;?$$
I have seen the integral first time. I can't find any hint how to even proceed? Kindly HELP. Thank You!
| Try to use the fact that the integral of the density of the normal distribution is 1:
$$\frac{1}{\sigma\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{(x-\mu)^2}{2\sigma^2}} \, dx=1.$$
Here $-(5x^2-6xy+5y^2)=-5(x-\frac{3}{5}y)^2-\frac{16}{5}y^2$, so our integral is
$$\int_{-\infty}^\infty e^{-\frac{16}{5}y^2}\int_{-\infty}^\infty e^{-5(x-\frac{3}{5}y)^2} \, dx\,dy=\int_{-\infty}^\infty e^{-\frac{16}{5}y^2}\frac{\sqrt{\pi}}{\sqrt{5}}\frac{1}{\frac{1}{\sqrt{10}}\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{(x-\frac{3}{5}y)^2}{2\cdot(\frac{1}{\sqrt{10}})^2}}\,dx\,dy.$$
Here $$\frac{1}{\frac{1}{\sqrt{10}}\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{(x-\frac{3}{5}y)^2}{2\cdot(\frac{1}{\sqrt{10}})^2}}dx=1,$$ so our integral is
$$\frac{\sqrt{\pi}}{\sqrt{5}}\int_{-\infty}^\infty e^{-\frac{16}{5}y^2}dy=\frac{\sqrt{\pi}}{\sqrt{5}}\frac{\sqrt{2\pi}}{\frac{\sqrt{32}}{\sqrt{5}}}\frac{1}{\frac{\sqrt{5}}{\sqrt{32}}\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{y^2}{2\cdot(\frac{\sqrt{5}}{\sqrt{32}})^2}}dy.$$
Here, again, $$\frac{1}{\frac{\sqrt{5}}{\sqrt{32}}\sqrt{2\pi}}\int_{-\infty}^\infty e^{-\frac{y^2}{2\cdot(\frac{\sqrt{5}}{\sqrt{32}})^2}}dy=1,$$ so our integral is
$$\frac{\sqrt{\pi}}{\sqrt{5}}\frac{\sqrt{2\pi}}{\frac{\sqrt{32}}{\sqrt{5}}}=\frac{\pi}{4}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim _{x\to \infty }\left(\cos\sqrt{x}-\cos\sqrt{x-1}\right)$ How should I determine the following limit?
$\lim _{x\to \infty }\left(\cos\sqrt{x}-\cos\sqrt{x-1}\right)$
| Direct approach:
Let $f(x) := \cos\sqrt{x} - \cos\sqrt{x - 1}$.
By the mean value theorem, for two values $a, b \in \mathbb R$, there exists $t\in\mathbb [a, b]$ such that $\cos(b) - \cos(a) = \cos'(t)(b - a)$. Thus,$$
|f(x)|=|\cos\sqrt{x} - \cos\sqrt{x - 1}| \\
=|\cos'(t)(\sqrt{x} - \sqrt{x - 1})| \\
=|-\sin(t)|\cdot|\sqrt{x} - \sqrt{x - 1}|\\
\le|\sqrt{x} - \sqrt{x - 1}| = \sqrt{x} - \sqrt{x - 1} =: g(x)$$
for all $x>1$.
Assertion: $\lim_{x\rightarrow\infty} g(x) = 0$.
Proof: Let $\epsilon > 0$. Then
$$g(x) \le \epsilon\\
\Leftrightarrow\sqrt{x} \leq \epsilon + \sqrt{x - 1}\\
\Leftrightarrow x \leq \epsilon^2 + 2\sqrt{x - 1} + x -1\\
\Leftrightarrow x \geq 1+(1-\epsilon^2)^2/4 =: x_0$$
Thus, for all $x> x_0$, we have $|g(x)| = g(x) \le \epsilon$.
Together with $|f(x)|\le g(x)$, this yields $\lim_{x\rightarrow\infty} f(x) = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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series: $\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots$
We have the infinite series:$$\frac{1}{2\cdot 3\cdot 4}+\frac{1}{4\cdot 5\cdot 6}+\frac{1}{6\cdot 7\cdot 8}+\cdots$$
This is not my series: $\frac{1}{1\cdot 2\cdot 3}+\frac{1}{2\cdot 3\cdot 4}+\frac{1}{3\cdot 4\cdot 5}+\frac{1}{4\cdot 5\cdot 6}+\cdots$ so I cannot use $\sum_{k=1}^n \frac{1}{k(k+1)(k+2)}$
My attempt:
I know that this type of series solved by making it telescoping series but here, I am unable find general term of the series. Thank you.
| $$\dfrac2{2n(2n+1)(2n+2)}=\dfrac{2n+2-2n}{2n(2n+1)(2n+2)}=\dfrac1{2n(2n+1)}-\dfrac1{(2n+1)(2n+2)}$$
$$=\dfrac1{2n}-\dfrac1{2n+1}-\left(\dfrac1{2n+1}-\dfrac1{2n+2}\right)$$
$$\sum_{n=2}^\infty\left(\dfrac1{2n}-\dfrac1{2n+1}-\left(\dfrac1{2n+1}-\dfrac1{2n+2}\right)\right)$$
$$=\left(\dfrac12-\dfrac13+\dfrac14-\dfrac15+\cdots\right)-\left(\dfrac13-\dfrac14+\dfrac15+\cdots\right)$$
Now $\ln2=1-\dfrac12+\dfrac13-\dfrac14+\cdots$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a Reduction Formula I'm trying to find a reduction formula for:
$${I}_n=\int \frac{x^{n}}{\sqrt{ax+b}}dx$$
This is where I have gotten to so far:
$${I}_n= \frac{2x^{n}\sqrt{ax+b}}{a}-\frac{2n}{a} \int \frac{x^{n-1}(ax+b)}{\sqrt{ax+b}}dx$$
I'd appreciate a little nudge in the right direction :)
| You're almost there. Simply note that for $I_n=\int \frac{x^n}{\sqrt{ax+b}}\,dx$, we have
$$\begin{align}
I_n&=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2n}{a}\int x^{n-1}\sqrt{ax+b}\,dx\\\\
&=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2n}{a}\int x^{n-1}\frac{ax+b}{\sqrt{ax+b}}\,dx\\\\
&=\frac{2x^n\sqrt{ax+b}}{a}-\frac{2n}{a}\left(a\int \frac{x^n}{\sqrt{ax+b}}\,dx+b\int \frac{x^{n-1}}{\sqrt{ax+b}}\right)\\\\
&=\frac{2x^n\sqrt{ax+b}}{a}-2nI_n-\frac{2nb}{a}I_{n-1}\\\\
I_n&=\frac{2x^n\sqrt{ax+b}}{a(1+2n)}-\frac{2nb}{a(2n+1)}I_{n-1}
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $z_1^2+z_2^2$ is real, $z_1(z_1^2-3z_2^2)=2$, and $z_2(3z_1^2-z_2^2)=11$, then find $(z_1^2+z_2^2)^2$
If $z_1$ and $z_2$ are complex numbers such that $z_1^2+z_2^2 \in\mathbb R$ and $$z_1(z_1^2-3z_2^2)=2,\qquad z_2(3z_1^2-z_2^2)=11,$$ then find the value of $(z_1^2+z_2^2)^2$. Given answer is $25$.
I have tried many things but I am not getting the answer.
I subtracted two equations to observe that $11/z_2 - 2/z_1$ must be real. Also if $z_1=x_1+iy_1$ and $z_2=x_2+i y_2$, then using the fact that $z_1^2+z_2^2 \in\mathbb R$, we get $x_1 y_1+x_2y_2=0$ but I am not able to compile these results to get the desired value.
| $$z_{1}(z^2_{1}-3z^2_{2}) = 2\Rightarrow z^3_{1}-3z_{1}z^2_{2} = 2\cdots \cdots (1)$$
Similarly $$z_{2}(3z^2_{1}-z_{2}^2) = 11\Rightarrow 3z^2_{1}z_{2}-z^3_{2} = 11\cdots (2)\times i$$
Now Add and Subtract these two equation, we get
$$(z_{1}+iz_{2})^3 = 2+11i$$
$$(z_{1}-iz_{2})^3 = 2-11i$$
So $$(z^2_{1}+z^2_{2})^3 = (2+11i)(2-11i) = 125\Rightarrow z^2_{1}+z^2_{2}=5\Rightarrow (z^2_{1}+z^2_{2}) = 25$$
| {
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Using mathematical induction prove that $ 11 \cdot 3^n + 3 \cdot 7^n - 6$ is divisible by 8 Prove that $ \phi(n) =11 \cdot 3^n + 3 \cdot 7^n - 6 $ is divisible by 8 for all $n \in N$.
Base: $ n = 0 $
$ 8 | 11 + 3 - 6 $ is obvious.
Now let $\phi(n)$ be true we now prove that is also true for $ \phi(n+1)$.
So we get $ 11 \cdot 3^{n+1} + 3 \cdot 7^{n+1} - 6$ and I am stuck here, just can't find the way to rewrite this expression so that I can use inductive hypothesis or to get that one part of this sum is divisible by 8 and just prove by one more induction that the other part is divisible by 8.
For instance, in the last problem I had to prove that some expression a + b + c is divisible by 9. In inductive step b was divisible by 9 only thing I had to do is show that a + c is divisible by 9 and I did that with another induction, and I don't see if I can do the same thin here.
| Setup the same as your current work:
$\dots$
$\dots = 11\cdot 3^{n+1}+3\cdot 7^{n+1}-6 = 11\cdot 3\cdot 3^{n}+ 3\cdot 7\cdot 7^n - 6$
$=33\cdot 3^n + 21\cdot 7^n - 6 = (11+22)\cdot 3^n + (3 + 18)\cdot 7^n - 6$
$=\underbrace{11\cdot 3^n + 3\cdot 7^n - 6}_{\text{should be familiar}} + \underbrace{22\cdot 3^n + 18\cdot 7^n}_{\text{unknown}}$
Now, what can we say about $22\cdot 3^n+18\cdot 7^n$? Anything? You say in a previous example, you had to run a second induction proof to finish, might that be useful here?
| {
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If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, the If the polynomial $x^4-6x^3+16x^2-25x+10$ is divided by another polynomial $x^2-2x+k$, theremainder is $x+a$, find $k$ and $a$.
My Attempt,
$f(x)=x^4-6x^3+16x^2-25x+10$
$g(x)=x^2-2x+k$
$R=x+a$
Here, the divisor is in the quadratic form. so how do I use the synthetic division
| To find the remainder of the long division by $x^2-2x+k$ you can keep replacing $x^2$ with $2x-k$ repeatedly, until getting the remainder of degree $1\,$:
$$
\begin{align}
x^4-6x^3+16x^2-25x+10 & = (2x-k)^2 - 6x(2x-k)+16(2x-k)-25x+ 10 \\
& = -8x^2 + (-4k +6k +32-25)x+k^2-16k+10 \\
& = -8(2x-k) + (2k+7)x+k^2-16k+10 \\
& = (2k-9)x + k^2-8k+10
\end{align}
$$
Identifying coefficients between the calculated remainder and $x+a$ it follows that $2k-9=1$ so $k=5\,$, and $a=5^2-8 \cdot 5+10=-5\,$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the equation $ (100 a+10b+c)^2 =(a+b+c)^5.$ Find a three-digits number $\overline{abc}$ such that $\overline{abc}^2=(a+b+c)^5.$
It is easy to see that
$$
(a+b+c)^5 \leq 999^2 \implies a+b+c< \sqrt[5]{999^2}\leq 15
$$
and
$$
(100 a+10b+c)^2<15^5 \implies 100 a+10b+c>\sqrt{15^5} \leq 871.
$$
Also
$$
(100 a+10b+c)^2 =(a+b+c) \mod 2
$$
implies $a+b=0 \mod 2$.
Similarly
$$
(100 a+10b+c)^2 = c^2 = (a+b+c) \mod 5.
$$
But it not enough to find the solution $243$.
No more ideas.
| $(a+b+c)$ must be a square, so it can only be $1,4,9,16$ or $25$.
By your inequalities we have $a+b+c=1,4$ or $9$.
notice that $(100a+10b+c)^2\geq 100^2$.On the other hand $1^5$ and $4^5$ are too small.
We conclude that $a+b+c=9$.
So now we must have $(100a+10b+c)=\sqrt{9^5}=3^5=243$
| {
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"url": "https://math.stackexchange.com/questions/2121275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why does the discriminant tell us how many zeroes a quadratic equation has? The quadratic formula states that:
$$x = \frac {-b \pm \sqrt{b^2 - 4ac}}{2a}$$
The part we're interested in is $b^2 - 4ac$ this is called the discriminant.
I know from school that we can use the discriminant to figure out how many zeroes a quadratic equation has (or rather, if it has complex, real, or repeating zeroes).
If $b^2-4ac > 0$ then the equation has 2 real zeroes.
If $b^2-4ac < 0$ then the equation has 2 complex zeroes.
If $b^2-4ac = 0$ then the equation has repeating zeroes.
But I don't uderstand why this works.
| Well if $a \ne 0$
$ax^2 + bx + c = 0 \iff$
$x^2 + \frac bax = -\frac ca \iff$
$x^2 + \frac bax + \frac {b^2}{4a^2} = \frac {b^2}{4a^2}- \frac ca \iff $
$x^2 + 2*\frac {b}{2a}x +(\frac {b}{2a})^2 = \frac {b^2 - 4ac}{4a^2} \iff $
$(x + \frac b{2a})^2 = \frac {b^2 - 4ac}{4a^2} \iff $
$x + \frac b{2a} = \pm \sqrt{\frac {b^2 - 4ac}{4a^2}} \iff $
$x + \frac b{2a} = \pm \frac {\sqrt{b^2 - 4ac}}{\pm 2a} \iff $
$x + \frac b{2a} = \pm \frac {\sqrt{b^2 - 4ac}}{2a}\iff $
$x = -\frac b{2a} \pm \frac {\sqrt{b^2 - 4ac}}{2a}\iff $
$x = \frac { -b \pm \sqrt{b^2 - 4ac}}{2a}$
Now three things can happen:
1) $b^2 - 4ac < 0$
If so then $\sqrt{b^2 - 4ac}$ is not a real number. So
$x = \frac { -b \pm \sqrt{b^2 - 4ac}}{2a}$ is never a real number.
So $ax^2 +bx +c = 0$ is impossible for any real x.
There are no real solutions.
2) $b^2 - 4ac = 0$
Then $\sqrt{b^2 - 4ac} = 0$
Then $x = \frac { -b \pm \sqrt{b^2 - 4ac}}{2a}$ would mean $x = \frac {-b}{2a}$.
So $ax^2 + bx + c = 0$ is only possible if (and will be possible if) $ x = \frac {-b}{2a}$. And so $ax^2 + bx + c = 0$ has exactly one solution when $x = \frac {-b}{2a}$.
or 3)
$b^2 - 4ac > 0$
Then $\sqrt{b^2 - 4ac} = k > 0$
then $\frac { -b + \sqrt{b^2 - 4ac}}{2a}$ and $\frac { -b - \sqrt{b^2 - 4ac}}{2a}$ are two different numbers.
And $ax^2 + bx + c = 0 \iff x = \frac { -b + \sqrt{b^2 - 4ac}}{2a}$ or $x = \frac { -b - \sqrt{b^2 - 4ac}}{2a}$.
So $ax^2 +bx + c = 0$ has exactly two solutions. They are $x = \frac { -b + \sqrt{b^2 - 4ac}}{2a}$ or $x = \frac { -b - \sqrt{b^2 - 4ac}}{2a}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Algebraic manipulation: $\sum_{r=1}^n \frac{n-r-1}{n-2} \frac{n-r}{n-1} \frac{1}{n} = \frac{1}{3}$ If you're up for a tedious algebraic manipulation, I'm stuck.
The following is true: $\sum_{r=1}^n \frac{n-r-1}{n-2} \frac{n-r}{n-1} \frac{1}{n} = \frac{1}{3}$.
However, I can't for the life of me figure out the how to show it.
| No tedious algebraic manipulation required!
Consider the following experiment: say that we choose a sequence $x_1,x_2,x_3$ from $\{1,\ldots,n\}$, without replacement. What is the probability that $x_1$ is the smallest of the three numbers?
On the one hand, this probability is $\frac{1}{3}$ by symmetry. Why? Given any three distinct numbers $a,b,c\in\{1,\ldots,n\}$, the sequences $abc$, $acb$, $bac$, $bca$, $cab$, and $cba$ are all equally likely to occur.
On the other hand, we can split into cases based on what that smallest number is. We first choose which number $r$ we assign to $x_1$. This assignment occurs with probability $\frac{1}{n}$. Then, from the remaining $n-1$ elements, we choose any of the $n-r$ numbers bigger than $r$ for $x_2$ (probability $\frac{n-r}{n-1}$); finally, from the remaining $n-2$ elements, we choose any of the $n-r-1$ remaining elements that exceed $r$ (probability $\frac{n-r-1}{n-2}$).
So, all told, we find that the probability should be
$$
\sum_{r=1}^{n-2}\frac{1}{n}\cdot\frac{n-r}{n-1}\cdot\frac{n-r-1}{n-2},
$$
which differs from your sum only in limits of summation; and, you'll notice that the terms $r=n-1$ and $r=n$ that are missing here are both $0$ in your sum.
So, because these two different expressions (the sum and $\frac{1}{3}$) are both the probability of the same event, they must be equal.
Now, if you PREFER tedious algebraic manipulation, it actually isn't too bad here. Note that changing your index of summation from $r$ to $k:=n-r$ yields
$$
\sum_{r=1}^{n}\frac{n-r-1}{n-2}\cdot\frac{n-r}{n-1}\cdot\frac{1}{n}=\frac{1}{n(n-1)(n-2)}\sum_{k=1}^{n-1}k(k-1).
$$
(Again, I've dropped indices of summation that contribute zero.) But, we can write
$$
\sum_{k=1}^{n-1}k(k-1)=\sum_{k=1}^{n-1}k^2-\sum_{k=1}^{n-1}k.
$$
Recall these identities:
$$
\sum_{i=1}^{m}i=\frac{m(m+1)}{2}\qquad\sum_{i=1}^{m}i^2=\frac{m(m+1)(2m+1)}{6}
$$
With these,
$$
\sum_{k=1}^{n-1}k^2-\sum_{k=1}^{n-1}k=\frac{(n-1)n(2n-1)}{6}-\frac{n(n-1)}{2},
$$
so that the original sum is
$$
\begin{align*}
\frac{(n-1)n(2n-1)}{6n(n-1)(n-2)}-\frac{n(n-1)}{2n(n-1)(n-2)}&=\frac{2n-1}{6(n-2)}-\frac{1}{2(n-2)}\\
&=\frac{2n-4}{6(n-2)}\\
&=\frac{1}{3}.
\end{align*}
$$
| {
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"url": "https://math.stackexchange.com/questions/2122635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$ I'm trying to find the derivative of $f=\arcsin\left(\frac{2x}{1+x^2}\right)$. I think I'm mistaken and perhaps using the chain rule incorrectly.
Let $g(x) = \frac{2x}{1+x^2}$ and let $h(x) = \arcsin x$
According to the chain rule -
$$f'(x) = \frac{1}{\sqrt{1-\frac{2x}{1+x^2}}}⋅((2⋅(1+x^2 )-2x⋅2x)/(1+x^2 )^2 ) = \cdots \frac{-2(x^4-1)}{x-1}$$
Is this a correct usage of the chain rule?
| If $\arctan x=y,-\dfrac\pi2\le y\le\dfrac\pi2, x=\tan y$
$\implies\dfrac{2x}{1+x^2}=\sin2y$
$\implies\arcsin\dfrac{2x}{1+x^2}=\begin{cases} 2\arctan x &\mbox{if }-\dfrac\pi2\le2y\le\dfrac\pi2\iff-1\le x\le1 \\\pi-2\arctan x & \mbox{if } 2y>\dfrac\pi2\iff x>1 \\-\pi-2\arctan x & \mbox{if } 2y<-\dfrac\pi2\iff x<-1\end{cases}$
Now $\dfrac{d(\arctan x)}{dx}=?$
Actually, $\sqrt{(1+x^2)^2-4x^2}=\sqrt{(1-x^2)^2}=|1-x^2|= \begin{cases} 1-x^2 &\mbox{if }1-x^2\ge0\iff-1\le x\le1\\
x^2-1& \mbox{if }x^2>1\iff x>1\text{ or }x<-1\end{cases}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Explain how these answers were computed from this graph and equation? this question was given on a previous exam:
Below is a graph of a function y = f(x) with a few key points.
Draw a graph of y = 3f(2x) + 4, with the corresponding points shown in the graph above.
And the answer key is given as the points:
(-1/2,10), (1/2, 10), (1/4), (3/2,10)
for the new points on the new graph; but, I am confused as to how these answers were computed. It would be great if someone can explain to me how these answers were computed.
Thank you!
| Suppose you have a function defined by some equation $y=f(x)$ as in this problem and suppose each $y$ in the equation is replaced by a linear expression $ay+b$ where $a$ and $b$ are constants, $a\ne0$. Suppose also that each $x$ in the equation is replaced by a linear expression $cx+d$ where $c\ne0$ and $d$ are constants.
Then one will have a new functional equation which maps each straight line connecting any two points of the original graph to a straight line connecting two points of the new, transformed graph. Such a transformation is called a linear transformation.
In order to use this information to graph $y=3f(2x)+4$ we must first solve for $f(2x)$ which we find is
$$\frac{1}{3}y-\frac{4}{3}=f(2x)$$
which we compare to
$$y=f(x)$$
to see that the linear transformation which is being applied to the given graph is
$$x\,\to\,2x,\qquad y\,\to\,\frac{1}{3}y-\frac{4}{3}$$
Since the graph of $y=f(x)$ consists of straight line segments and since this is a linear transformation, we just have to find out where the four endpoints of $y=f(x)$ are mapped and connect them with straight line segments.
We can do this with each of the four points as follows:
*
*$2x=-1$ so $x=-\frac{1}{2}.\qquad$ $\frac{1}{3}y-\frac{4}{3}=2$ so $y=10$. Therefore $(-1,2)\,\to\,\left(-\frac{1}{2},10\right)$
*$2x=1$ so $x=\frac{1}{2}.\qquad$ $\frac{1}{3}y-\frac{4}{3}=2$ so $y=10$. Therefore $(1,2)\,\to\,\left(\frac{1}{2},10\right)$
*$2x=2$ so $x=1.\qquad$ $\frac{1}{3}y-\frac{4}{3}=0$ so $y=4$. Therefore $(2,0)\,\to\,(1,4)$
*$2x=3$ so $x=\frac{3}{2}.\qquad$ $\frac{1}{3}y-\frac{4}{3}=2$ so $y=10$. Therefore $(3,2)\,\to\,\left(\frac{3}{2},10\right)$
The original graph is mapped to the new graph shown in red.
Below the graph is a second way to work the problem which requires finding the piece-wise equations for the given graph.
There is a second way to work this kind of problem if you are given the equation. This particular function consists of three line segments "glued" together.
$y=\begin{cases}
2 & \text{for }-1\le x<1\\
-2x+4 & \text{for }\phantom{-}1\le x<2\\
2x-4 & \text{for }\phantom{-}2\le x\le3\\
\end{cases}$
Now apply the transformation $x\,\to\,2x,\quad y\,\to\,\frac{1}{3}y-\frac{4}{3}$
$\frac{1}{3}y-\frac{4}{3}=\begin{cases}
2 & \text{for }-1\le 2x<1\\
-4x+4 & \text{for }\phantom{-}1\le 2x<2\\
4x-4 & \text{for }\phantom{-}2\le 2x\le3\\
\end{cases}$
Remove fractions by multiplying all three equations by $3$ to obtain
$y-4=\begin{cases}
6 & \text{for }-1\le 2x<1\\
-12x+12 & \text{for }\phantom{-}1\le 2x<2\\
12x-12 & \text{for }\phantom{-}2\le 2x\le3\\
\end{cases}$
Then add four to each side of the three equations.
$y=\begin{cases}
10 & \text{for }-1\le 2x<1\\
-12x+16 & \text{for }\phantom{-}1\le 2x<2\\
12x-12 & \text{for }\phantom{-}2\le 2x\le3\\
\end{cases}$
Finally, solve the three inequalities for $x$ to express the domain correctly.
$y=\begin{cases}
10 & \text{for }-\frac{1}{2}\le x<\frac{1}{2}\\
-12x+16 & \text{for }\phantom{-}\frac{1}{2}\le x<1\\
12x-12 & \text{for }\phantom{-}1\le x\le\frac{3}{2}\\
\end{cases}$
This gives the piece-wise equations for the transformed graph.
Note that the actual reason for learning this is so that you can do the process in reverse. That is when the translating, reflecting and scale changes comes into play.
For example, an equation such as
\begin{equation}
y=3+\sqrt{x+1}
\end{equation}
can be re-written as
\begin{equation}
y-3=\sqrt{x+1}
\end{equation}
so that we recognize it as the graph of $y=\sqrt{x}$ transformed by $x\,\to\,x+1,\quad y\,\to\,y-3$ which translates the graph graph of $y=\sqrt{x}$ by $-1$ units horizontally and $+3$ units vertically.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
maximum value of expression $6bc+6abc+2ab+3ac$ If $a,b,c>0$ and $a+2b+3c=15,$ then finding maximum value of $6bc+6abc+2ab+3ac$ is
with the help of AM - GM inequality
$4ab\leq (a+b)^2$ and $4bc \leq (b+c)^2$ and $\displaystyle 4ca \leq (c+a)^2$
and $27(abc)\leq (a+b+c)^3$
could some help me, thanks
| You can proceed as follows:
$$a+2b+3c=15\tag1$$
From $(1)$ , we have by AM-GM inequality,
$$\frac{a+2b+3c}{3} \ge \sqrt[3]{6abc}$$
$$\implies \frac{15}{3} \ge \sqrt[3]{6abc}$$
$$\implies 5^3 \ge 6abc$$
Again from $(1)$ , we have by AM-GM inequality,
$$\frac{a+2b+3c}{2} \ge \frac{a+2b}{2} \ge \sqrt{2ab}$$
$$\frac{15}{2} \ge \sqrt{2ab}$$
$$\frac{225}{4} \ge 2ab$$
Again from $(1)$ , we have by AM-GM inequality,
$$\frac{a+2b+3c}{2} \ge \frac{2b+3c}{2} \ge \sqrt{6bc}$$
$$\frac{15}{2} \ge \sqrt{6bc}$$
And so on .
Hope this helps you and you can complete the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2125800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
If $xy+xz+yz=1+2xyz$ then $\sqrt{x}+\sqrt{y}+\sqrt{z}\geq2$. Let $x$, $y$ and $z$ be non-negative numbers such that $xy+xz+yz=1+2xyz$. Prove that:
$$\sqrt{x}+\sqrt{y}+\sqrt{z}\geq2$$
The equality occurs for $x=y=1$ and $z=0$.
I tried Lagrange Multipliers and more, but I don't see a proof.
| Short proof.
Clearly $xy+yz+zx \ge 1$
We have by AM-GM
$$(\sqrt{x}+\sqrt{y}+\sqrt{z})^2=x+y+z+2(\sqrt{xy}+\sqrt{yz}+\sqrt{zx})$$
$$\ge x+y+z+ \frac{4xy}{x+y}+\frac{4yz}{y+z}+\frac{4zx}{z+x} $$
$$ \ge x+y+z + \frac{4(xy+yz+zx)}{x+y+z} \ge x+y+z+\frac{4}{x+y+z} \ge 4$$
The proof is complete.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2127470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
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} |
value of $2\lambda^2-1 $ in trigonometric expression if $\displaystyle \frac{\sin^3 \beta}{\sin(\beta - 3 \alpha)} = \frac{\cos^3 \beta}{\cos(\beta-3 \alpha)} = \lambda\;,$ then $2\lambda^2-1 = $
options
$(a)\; \lambda \sin \beta$
$(b)\; \lambda \cos \beta$
$(c)\; 2\lambda \cos \beta$
$(d)\; \lambda \tan \beta$
Attempt with ratio and proportion
$\displaystyle \frac{\sin^4 \beta}{\sin \beta \sin(\beta - 3 \alpha)} = \frac{\cos^4 \beta}{\cos \beta \cos(\beta-3 \alpha)} = \frac{\cos^4 \beta - \sin^4 \beta}{\cos \beta \cos(\beta-3 \alpha)-\sin \beta \sin(\beta - 3 \alpha)}$
so $\displaystyle \frac{\sin^4 \beta}{\sin \beta \sin(\beta - 3 \alpha)} = \frac{\cos^4 \beta}{\cos \beta \cos(\beta-3 \alpha)} = \frac{\cos 2 \beta}{\cos (2\beta-3 \alpha)}$
wan,t be able to go further ,could some help me
| HINT:
Clearly we need to eliminate $\alpha$
Expand $\sin(\beta-3\alpha),\cos(\beta-3\alpha)$ to form two simultaneous equations in $\sin(3\alpha),\cos(3\alpha)$
Solve for $\sin(3\alpha),\cos(3\alpha)$
Use $\sin^2(3\alpha)+\cos^2(3\alpha)=1$ to eliminate $\alpha$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2128686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Finding the area between a line and a curve
The two equations are $x+1$ and $4x-x^2-1$.
The answer is $\frac{1}{6}$, but I've done it 4 different times and gotten -$\frac{15}{2}$ each time.
My working:
*
*$x+1$ = $4x-x^2-1$
*$x^2-3x+2 = 0$
*$(x-1)(x-2)$ means $x=1$ or $x=2$
*$\int_1^2$ $3x-x^2$
*$[\frac{3x^2}{2}-\frac{x^3}{3}]_1^2$
*$\frac{3(1)^2}{2}-\frac{(1)^3}{3}$ = $\frac{3}{2}-\frac{1}{3}$
*$\frac{3(2)^2}{2}-\frac{(2)^3}{3}$ = $\frac{12}{2}-\frac{8}{3}$
*($\frac{3}{2}-\frac{1}{3}$)-($\frac{12}{2}-\frac{8}{3}$) = $-\frac{9}{2}-\frac{9}{3}$
*-$\frac{15}{2}$
| You got the two $x$ points right, but you make a mistake in the next step. The correct version should be:
$$\int_1^2[\text{top function}-\text{bottom function}]\ dx$$
Here, the top function is $4x-x^2-1$ and the bottom function is $x+1$, hence
$$\begin{align}\text{Area}&=\int_1^2[(4x-x^2-1)-(x+1)]\ dx\\&=\int_1^2[4x-x^2-1-x-1]\ dx\\&=\int_1^2[3x-x^2-2]\ dx\end{align}$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2131864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
} |
Polynomial with no three distinct rational roots Prove that $x^3-2x^2-2x+a$, where $ a\in \mathbb{R} $, has no three distinct rational roots.
| Suppose
$$f(x) = x^3 - 2x^2 - 2x + a = (x - P)(x - Q)(x - R)$$
where $P,Q,R$ are rational numbers.
Our goal is to derive a contradiction.
If any of $P,Q,R$ is zero, then $a = 0$, hence $f(x) = x(x^2 - 2x - 2)$, contradiction, since $x^2 - 2x - 2$ has no rational roots.
Thus, assume $P,Q,R$ are nonzero.
By Vieta's formulas,
\begin{align*}
P + Q + R &= 2\\[6pt]
PQ + QR + RP &= -2
\end{align*}
Let $d$ be the least common denominator of $P,Q,R$. Then $d$ is a positive integer and
\begin{align*}
P &= \frac{p}{d}\\[6pt]
Q &= \frac{q}{d}\\[6pt]
R & = \frac{r}{d}
\end{align*}
where $p,q,r$ are nonzero integers such that $\text{gcd}(p,q,r) = 1$.
Then Vieta's formulas yield
\begin{align*}
p + q + r &= 2d\\[6pt]
pq + qr + rp &= -2d^2\\[6pt]
\end{align*}
Then
\begin{align*}
p^2 + q^2 + r^2 &= (p + q + r)^2 - 2(pq + qr + rp)\\[6pt]
&=(2d)^2 - 2(-2d^2)\\[6pt]
&=8d^2
\end{align*}
Since $p + q + r$ is even, and $\text{gcd}(p,q,r) = 1$, it follows that exactly one of $p,q,r$ is even.
Without loss of generality, assume $p$ is even, and $q,r$ are odd.
Recall that odd squares are congruent to $1$ mod $8$, and even squares are congruent to
$0$ or $4$ mod $8$. Then
\begin{align*}
&p^2 + q^2 + r^2 = 8d^2\\[6pt]
\implies\; &p^2 + q^2 + r^2 \equiv 0 \pmod 8\\[6pt]
\implies\; &p^2 + 1 + 1 \equiv 0 \pmod 8\\[6pt]
\implies\; &p^2 \equiv 6 \pmod 8
\end{align*}
contradiction.
Therefore there do not exist rational numbers $P,Q,R$ such that
$$f(x) = x^3 - 2x^2 - 2x + a = (x - P)(x - Q)(x - R)$$
| {
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"url": "https://math.stackexchange.com/questions/2133172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $x^2 \equiv -1 \pmod{365}$ solvable? I know that a similar question already exists, but I have a different question to ask.
We want to examine if $x^2 \equiv -1 \pmod{365}$ has a solution.
My thought is: $365=5\cdot 73$. So,The congruence $x^2 \equiv -1 \pmod{365}$ has solution, if and only if, the congrueces $x^2 \equiv -1 \pmod 5$ and $x^2 \equiv -1 \pmod{73}$ has solutions. So, if we use Legendre's Symbol we have
*
*$x^2 \equiv -1 \pmod 5$ has solution $\iff (-1/5)=1 $ (and with simple calculations, indeed)
*$x^2 \equiv -1 \pmod{71}$ has solution $\iff (-1/73)=1 $
Now, can we conclude that the congruence $x^2 \equiv -1 \pmod{365}$ has solution?
And more general: If we have the congruence $x^2 \equiv a \pmod n$ with $n=p_1^{n_1}\cdots p_k^{n_k},\ \gcd(a,n)=1$, which is equivalent with the system $x^2 \equiv a {\pmod p_1^{n_1}},\ldots,x^2 \equiv a \pmod{p_k^{n_k}}$, can we conclude that the first has solution if and only if each one of $x^2\equiv a\pmod{p_i^{n_i}},\ \forall i=1,\ldots,k$ has solution?
Thank you.
| We have $365 = 5 \times 73$. The congruence becomes $x^2 = -1 \mod 5$ and $x^2 = -1 \mod 73$.
We have if $p = 1 \mod 4 \implies x^2 = -1 \mod p$ has exactly $2$ solutions.
Thus $x^2 = -1 \mod 5$ has solutions $x_0,x_1$ and $x^2 = -1 \mod 73$ has solutions $y_0,y_1$.
The original solutions satisfies either:
$x = x_0 \mod 5, x = y_0 \mod 73$; $x = x_0 \mod 5, x = y_1 \mod 73$; $x = x_1 \mod 5, x = y_0 \mod 73$ ; $x = x_1 \mod 5, x = y_1 \mod 73$.
For each pair of congruence , $x$ is uniquely determined $\mod 365$ by the Chinese remainder theorem. Hence the original congruence has $4$ solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2133512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Limit of function which is product of $x^x $ primitives I'm trying to analytically derive the following limit: $$ \lim_{x\to \infty}\left(\frac{1}{8}\right)\left(e^{1-\frac{\sqrt{x}}{2}}\right)\left(\frac{x}{x-\frac{\sqrt{x}}{2}+1}\right)^{x-\frac{\sqrt{x}}{2}+1} $$
I've found it extremely difficult because, as you apply L'Hospital's rule, you need to differentiate $x$ times, thus getting an infinite chain of differentiations (each one invoking the product rule).
WorlframAlpha says the limit is equal to $\frac{1}{8e^{1/8}}$ which seems to match perfectly the numerical estimates I've made with my computer up to $x=2\times10^4$, but it gives no explanation for how it derives this result. How can I analytically evaluate this limit? On it's face, it seems intractable. Yet there appears to be an analytical solution.
The purpose of evaluating this limit is to calculate the following limit of a combinatorial probability:
$$\lim_{N\to\infty}\left(\frac{M^2(N!)}{2N^M(N-M+1)!}\right)$$
Where $M=\frac{N^{1/2}}{2}$.
The original limit above can be found by applying Sterling's Approximation on the second limit. If anyone is aware of any alternative techniques for analytically evaluating this second limit, that would be just as helpful.
| $$\lim _{ x\to \infty } \left( \frac { 1 }{ 8 } \right) \left( e^{ 1-\frac { \sqrt { x } }{ 2 } } \right) \left( \frac { x }{ x-\frac { \sqrt { x } }{ 2 } +1 } \right) ^{ x-\frac { \sqrt { x } }{ 2 } +1 }=\\ =\lim _{ x\to \infty } \left( \frac { 1 }{ 8 } \right) \left( e^{ 1-\frac { \sqrt { x } }{ 2 } } \right) \left( \frac { 1 }{ 1+\frac { 1 }{ x } -\frac { 1 }{ 2\sqrt { x } } } \right) ^{ x-\frac { \sqrt { x } }{ 2 } +1 }=\\ =\lim _{ x\to \infty } \left( \frac { 1 }{ 8 } \right) \left( e^{ 1-\frac { \sqrt { x } }{ 2 } } \right) \left( 1+\frac { 1 }{ x } -\frac { 1 }{ 2\sqrt { x } } \right) ^{ \frac { \sqrt { x } }{ 2 } -x-1 }=\\ =\lim _{ x\to \infty } \left( \frac { 1 }{ 8 } \right) \left( e^{ 1-\frac { \sqrt { x } }{ 2 } } \right) \left( 1+\frac { 2-\sqrt { x } }{ 2x } \right) ^{ \frac { \sqrt { x } }{ 2 } -x-1 }=\\ =\lim _{ x\to \infty } \left( \frac { 1 }{ 8 } \right) \left( e^{ 1-\frac { \sqrt { x } }{ 2 } } \right) { \left[ \left( 1+\frac { 2-\sqrt { x } }{ 2x } \right) ^{ \frac { 1 }{ \frac { 2-\sqrt { x } }{ 2x } } } \right] }^{ \left( \frac { 2-\sqrt { x } }{ 2x } \right) \cdot \left( \frac { \sqrt { x } }{ 2 } -x-1 \right) }=\\ =\lim _{ x\to \infty } \left( \frac { 1 }{ 8 } \right) \cdot e^{ 1-\frac { \sqrt { x } }{ 2 } }\cdot { { e }^{ \left( \frac { 2-\sqrt { x } }{ 2x } \right) \cdot \left( \frac { \sqrt { x } }{ 2 } -x-1 \right) } }=\frac { 1 }{ 8 } { e }^{ \lim _{ x\to \infty } \left( 1-\frac { \sqrt { x } }{ 2 } +\frac { 1 }{ 2\sqrt { x } } -1-\frac { 1 }{ x } -\frac { 1 }{ 4 } +\frac { \sqrt { x } }{ 2 } +\frac { 1 }{ 2\sqrt { x } } \right) }=\\ \\ =\frac { 1 }{ 8{ e }^{ 1/4 } } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find limit of $\lim_{n\to\infty} (\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}$)? I am stuck on this limit.
$$\lim_{n\to\infty} \sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}$$
I couldn't find the limit using the basic properties of limits, since that just yields: $$\infty-\infty$$ which is undefined. Could I get any hints for finding this limit?
| One developes an intuition that as $n^2$ becomes large the $+5$ and $-3$ become negligible and the whole thing becomes but how to prove it...
Well we can is eliminate the difference of square roots by multiplying by the compliment so we can probably do the same for cube roots.
I.e. $(\sqrt[3]a - \sqrt[3]b)(\sqrt[3]a^2 + \sqrt[3]a\sqrt[3]b+\sqrt[3]b^2) = a - b$.
So $\lim (\sqrt[3]{n^2 + 5} - \sqrt[3]{n^2 + 3})=$
$\lim (\sqrt[3]{n^2 + 5} - \sqrt[3]{n^2 + 3})\frac {\sqrt[3]{n^2 + 5}^2 + \sqrt[3]{n^2 + 5}\sqrt[3]{n^2 + 3}+\sqrt[3]{n^2 + 3}^2}{\sqrt[3]{n^2 + 5}^2 + \sqrt[3]{n^2 + 5}\sqrt[3]{n^2 + 3}+\sqrt[3]{n^2 + 3}^2}=$
$\lim \frac{(n^2 + 5)-(n^2 +3)}{\sqrt[3]{n^2 + 5}^2 + \sqrt[3]{n^2 + 5}\sqrt[3]{n^2 + 3}+\sqrt[3]{n^2 + 3}^2}=$
$\lim \frac{2}{\sqrt[3]{n^2 + 5}^2 + \sqrt[3]{n^2 + 5}\sqrt[3]{n^2 + 3}+\sqrt[3]{n^2 + 3}^2}=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2137187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}}dx$ $$I = \int \frac{\sin^{3}x+\cos^{3}x}{\sin{x}\cos{x}}$$
Thought for a while but cannot seem to find how do i procced with this integral.
Another variant of this problem is $$I = \int \frac{\sin^{3}x+\cos^{3}x}{\sin^{2}{x}\cos^{2}{x}}$$ where you just seperate the terms in numerator and it simplifies to $$I=\int \sec{x}\tan{x}dx+\int \csc{x}\cot{x}dx$$ which gives $$\sec{x}-\csc{x}+C$$ on integrating.
But in this case it doesn't simplify into something straightforward.
| Notice that
\begin{eqnarray}
\dfrac{\sin^3x+\cos^3x}{\sin x\cos x}&=&
\dfrac{\sin^3x}{\sin x\cos x}+
\dfrac{\cos^3x}{\sin x\cos x}\\
&=&\dfrac{\sin^2x}{\cos x}+\dfrac{\cos^2x}{\sin x}\\
&=&\dfrac{1-\cos^2x}{\cos x}+\dfrac{1-\sin^2x}{\sin x}\\
&=&\dfrac{1}{\cos x}-\dfrac{\cos^2x}{\cos x}+\dfrac{1}{\sin x}-\dfrac{\sin^2x}{\sin x}\\
&=&\dfrac{1}{\cos x}+\dfrac{1}{\sin x}-\cos x-\sin x
\end{eqnarray}
To compute the two integrals
$$
\int\dfrac{1}{\cos x}\,dx \text{ and } \int\dfrac{1}{\sin x}\,dx
$$
we use the substitution
$$
t=\tan\dfrac{x}{2} \text{ or equivalently } x=2\tan^{-1}t.
$$
Since
\begin{eqnarray}
\cos x&=&\dfrac{\cos^2\dfrac{x}{2}-\sin^2\dfrac{x}{2}}{\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}}=\dfrac{1-t^2}{1+t^2}\\
\sin x&=&\dfrac{2\sin\dfrac{x}{2}\cos\dfrac{x}{2}}{\cos^2\dfrac{x}{2}+\sin^2\dfrac{x}{2}}=\dfrac{2t}{1+t^2}\\
dx&=&\dfrac{2}{1+t^2}\,dt
\end{eqnarray}
we have
\begin{eqnarray}
\int\dfrac{1}{\cos x}\,dx&=&\int\dfrac{1+t^2}{1-t^2}\cdot\dfrac{2}{1+t^2}\,dt=\int\dfrac{2}{1-t^2}\,dt=\int\left(\dfrac{1}{1-t}+\dfrac{1}{1+t}\right)\,dt=\ln\left|\dfrac{1+t}{1-t}\right|+c_1\\
\int\dfrac{1}{\sin x}\,dx&=&\int\dfrac{1+t^2}{2t}\cdot\dfrac{2}{1+t^2}\,dt=\int\dfrac{1}{t}\,dt=\ln|t|+c_2
\end{eqnarray}
Hence
\begin{eqnarray}
\int\dfrac{\sin^3x+\cos^3x}{\sin x\cos x}\,dx&=&\int\left(\dfrac{1}{\cos x}+\dfrac{1}{\sin x}-\cos x-\sin x\right)\,dx\\
&=&\ln\left|\dfrac{1+\tan\dfrac{x}{2}}{1-\tan\dfrac{x}{2}}\right|+\ln\left|\tan\dfrac{x}{2}\right|-\sin x+\cos x+c
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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To find value of $\tan \frac{\theta}{2} (1+\sec \theta)(1+\sec 2\theta)(1+\sec 4\theta)\dotsm(1+\sec 2^n\theta)$ I have to find value of $f_{n} \theta = \tan \frac{\theta}{2} (1+\sec \theta)(1+\sec 2\theta)(1+\sec 4\theta)\dotsm(1+\sec 2^n\theta)$.
I tried to use $1 + \cos\theta = 2 \cos^2 \frac{\theta}{2}$, there were some cancellations but in end I got geometric progression of angles whose cosine were in product form.
How to deal with this?
Thanks
| We have $$f_0 \theta = \tan \frac {\theta}{2}(1+ \sec \theta) = \frac {\sin \theta/2}{\cos \theta/2} \frac {2\cos^2 \theta/2}{\cos \theta} = \frac {2\sin \theta/2 \cos \theta/2}{\cos \theta} = \tan \theta $$ Similarly, $$f_1 \theta = \tan \theta \frac {2\cos^2 \theta}{\cos 2\theta} = \tan 2\theta $$ $$f_2 \theta = \tan 2\theta \frac {2\cos^2 2\theta}{\cos 4\theta} = \tan 4\theta $$ $$\vdots $$
We can thus easily observe that $$f_n \theta = \tan 2^n \theta $$
Hope it helps.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Group of Rotations.
Attempt:
$$\frac{(2^6 + 6 \cdot 2^3 + 3 \cdot 2^4 + 8 \cdot 2^2 + 6 \cdot 2^3)}{24} = 10$$
| $10$ appears correct for distinct cubes allowing rotation mapping. There are:
- $2$ solid-colour cubes $(B^6,R^6)$,
- $2$ cubes with one face different $(B^5R, BR^5)$,
- $2$ (opposite-face and adjacent-face) options for each of the $4{+}2$ colourings $B^4R^2, B^2R^4$
- and $2$ similar options for the evenly-split colouring $B^3R^3$,
which overall gives $1+1+2+2+2+1+1 = 10$
The restriction to rotations (no reflections) makes no difference in this case.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $(xy+2)^2+(x-1)^2+(y-1)^2 > 1$ for real $x,y$ How do I prove that $$(xy+2)^2+(x-1)^2+(y-1)^2 > 1$$ for all real $x,y$? I tried setting the partial derivative to $0$ but I couldn't get it to work.
| Using @quasi 's original comment, consider the function
$$
f(x,y)=(xy+2)^2+(x-1)^2+(y-1)^2.
$$
If we take partial derivatives of $f$, we get
\begin{align*}
\frac{\partial f}{\partial x}&=2(xy+2)y+2(x-1)\\
\frac{\partial f}{\partial y}&=2(xy+2)x+2(y-1).
\end{align*}
At minima, these are equal to zero, so we consider
\begin{align*}
2(xy+2)y+2(x-1)&=0\\
2(xy+2)x+2(y-1)&=0.
\end{align*}
Next, we can subtract the second equation from the first equation to get
$$
2(xy+2)(y-x)+2(x-y)=0.
$$
Factoring this, we have
$$
2(y-x)(xy+2-1)=0
$$
or that
$$
2(y-x)(xy+1)=0.
$$
Therefore, minima occur when, either
1. $x=y$ or
2. $xy+1=0$.
In the first case, $(xy+2)^2$ simplifies to $(x^2+2)^2$, since squares are always positive, the inside of the square is at least $2$, so the LHS is at least $4$.
In the second case, if $xy+1=0$, then $xy=-1$, and, plugging this into the formula for $f$, we see that the LHS is at least $(xy+2)^2=(-1+2)^2=1$. Now, we can't have both $x=1$ and $y=1$ and have their product $-1$, so at least one of $(x-1)^2$ or $(y-1)^2$ is positive, and the LHS is greater than $1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Help with the limit of a function in the $\frac{0}{0}$ case The limit is this one:
$$\lim_{x \rightarrow 0}\frac{(1+2x)^\frac{1}{x}-(1+x)^\frac{2}{x}}{x}$$
I have found that both $(1+2x)^\frac{1}{x}$ and $(1+x)^\frac{2}{x}$ tend to $e^2$, so the numerator tends to 0. I think that the book said that the result of this limit is $-e^2$ if I recall correctly.
| Note that we use here two well known limits $$\lim _{ \quad x\rightarrow 0 }{ { \left( 1+x \right) }^{ \frac { 1 }{ x } } } =e\\ \lim _{ x\rightarrow 0 }{ \frac { { e }^{ x }-1 }{ x } } =1$$
$$\lim _{ x\rightarrow 0 } \frac { (1+2x)^{ \frac { 1 }{ x } }-(1+x)^{ \frac { 2 }{ x } } }{ x } =\lim _{ x\rightarrow 0 } \frac { { e }^{ \frac { 1 }{ x } \ln { \left( 1+2x \right) } }-{ e }^{ \frac { 2 }{ x } \ln { \left( 1+x \right) } } }{ x } =\\ =\lim _{ x\rightarrow 0 } \frac { { e }^{ \frac { 2 }{ x } \ln { \left( 1+x \right) } }\left[ { e }^{ \frac { 1 }{ x } \ln { \left( 1+2x \right) } -\frac { 2 }{ x } \ln { \left( 1+x \right) } }-1 \right] }{ x } =\\\lim _{ x\rightarrow 0 } \frac { \left[ { e }^{ \frac { 1 }{ x } \ln { \left( 1+2x \right) } -\frac { 2 }{ x } \ln { \left( 1+x \right) } }-1 \right] }{ \frac { 1 }{ x } \ln { \left( 1+2x \right) } -\frac { 2 }{ x } \ln { \left( 1+x \right) } } \cdot \frac { { e }^{ \frac { 2 }{ x } \ln { \left( 1+x \right) } } }{ x } \cdot \left[ \frac { 1 }{ x } \ln { \left( 1+2x \right) } -\frac { 2 }{ x } \ln { \left( 1+x \right) } \right] =\\\ =\lim _{ x\rightarrow 0 } \frac { { e }^{ \frac { 2 }{ x } \ln { \left( 1+x \right) } } }{ x } \cdot \left[ \frac { 1 }{ x } \ln { \left( 1+2x \right) } -\frac { 2 }{ x } \ln { \left( 1+x \right) } \right] =\lim _{ x\rightarrow 0 } \frac { (1+x)^{ \frac { 2 }{ x } } }{ { x }^{ 2 } } \cdot \ln { \left( \frac { 1+2x }{ { \left( 1+x \right) }^{ 2 } } \right) } =\\=\lim _{ x\rightarrow 0 } (1+x)^{ \frac { 2 }{ x } }\cdot \ln { { \left( \frac { 1+2x }{ 1+2x+{ x }^{ 2 } } \right) }^{ \frac { 1 }{ { x }^{ 2 } } } } =\\ =\lim _{ x\rightarrow 0 } (1+x)^{ \frac { 2 }{ x } }\cdot \\\ln { { \left( 1+\frac { -{ x }^{ 2 } }{ 1+2x+{ x }^{ 2 } } \right) }^{ \frac { 1 }{ { x }^{ 2 } } } } =\lim _{ x\rightarrow 0 } (1+x)^{ \frac { 2 }{ x } }\cdot \ln { { \left[ { \left( 1+\frac { -{ x }^{ 2 } }{ 1+2x+{ x }^{ 2 } } \right) }^{ \frac { 1+2x+{ x }^{ 2 } }{ -{ x }^{ 2 } } } \right] }^{ \frac { 1 }{ { x }^{ 2 } } \cdot \frac { -{ x }^{ 2 } }{ 1+2x+{ x }^{ 2 } } } } =\lim _{ x\rightarrow 0 } (1+x)^{ \frac { 2 }{ x } }\cdot \frac { 1 }{ { x }^{ 2 } } \cdot \frac { -{ x }^{ 2 } }{ 1+2x+{ x }^{ 2 } } =\color{red}{-{ e }^{ 2 }}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2144537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Maximize $P=a^2+b^2+c^2+ab+ac+bc$
For real numbers $a, b, c$ that satisfy $a + b + c = 6$ and $0\leq a,b,c \leq 4$,
maximize $P=a^2+b^2+c^2+ab+ac+bc$.
My try:
$$\begin{align} \begin{cases} a+b+c=6(1) \\ 0\leq a,b,c\leq4(2) \end{cases} \end{align}$$ $$(1)\Rightarrow \begin{align} \begin{cases} b+c=(6-a) \\b^2+c^2+bc=(6-a)^2-bc \end{cases} \end{align}$$ $$P=a^2+(b^2+c^2+bc)+a(b+c)=a^2+[(6-a)^2-bc]+a(6-a)$$ $$P=(a^2-12a+36)-bc=(a-6)^2-bc (2)\Rightarrow bc\leq 0 \Rightarrow P\geq (a-6)^2$$ When $bc=0 \Rightarrow [{\begin{matrix}b=0\\c=0\end{matrix}}(3)$. From $(1)$ and $(3)$, $\Rightarrow 2\leq a\leq 4(4)$ $P_{max} \implies |a-6|$ max satisfy $(4)$ $\implies a=2$ from $(1)$ and $(3)$ $\implies b=c=4$ $\implies P_{max}(a,b,c)=P(4;2;0)=28$
| Let $f(x)=x^2$ and $a\geq b\geq c$.
Hence, $f$ is a convex function and $(4,2,0)\succ(a,b,c)$.
Thus, by Karamata
$$\sum_{cyc}(a^2+ab)=a^2+b^2+c^2+\frac{36-a^2-b^2-c^2}{2}=\frac{1}{2}(a^2+b^2+c^2)+18\leq$$
$$\leq\frac{1}{2}(4^2+2^2+0^2)+18=28.$$
Done!
| {
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"url": "https://math.stackexchange.com/questions/2145597",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Compute $5!25! \mod 31$ For an exercise, I was asked to compute $5!25! \mod 31$. I noticed that $5! = 120 \equiv -4 \equiv 27 \mod 31$. Therefore we have that
$$5!25! \equiv 27 \cdot 25! \mod 31.$$
Because of the congruence of Wilson, I also know that $30! \equiv -1 \mod 31$.
We have that $30! \equiv 30 \cdot 29 \cdot 28 \cdot 26 \cdot 27 \cdot 25! \equiv -1 \mod 31$, so I computed
$$(30 \cdot 29 \cdot 28 \cdot 26) \equiv (-1 \cdot (-2) \cdot (-3) \cdot (-5)) \equiv 30 \equiv -1 \mod 31.$$ Hence we find that $-1 \cdot (27 \cdot 25!) \equiv -1 \mod 31$ and therefore $27 \cdot 25! \equiv 1 \mod 31$. This proves that $5!25! \equiv 1 \mod 31$.
Is this correct?
| I think your answer is ok, but I would rewrite it as follows
$$
5! = 5 \cdot 4 \cdot 3 \cdot 2 \cdot 1
\equiv
(-26) \cdot (-27) \cdot (-28) \cdot (-29) \cdot (-30)
\pmod{31}.
$$
Thus
$$
5! \cdot 25! \equiv (-1)^{5} \, 30! \equiv (-1) \cdot (-1) = 1 \pmod{31},
$$
using as you did Wilson's Theorem.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Generating function of algebraic equations modulo prime powers Given an integer polynomial $Q(x_1,x_2,...,x_m)$, let $N(p,n)$ denote the number of solutions to the equation $$Q(x_1,...,x_m) \equiv 0$$ modulo $p^n.$
We can consider the ordinary generating functions $$F(p,x) = \sum_{n=0}^{\infty} N(p,n) x^n.$$ I find experimentally that these $F(p,x)$ tend to be rational functions.
Here are a few examples:
(1) If $Q(x) = x^2$, then $$F(2,x) = 1 + x + 2x^2 + 2x^3 + 4x^4 + 4x^5 + ... = \frac{1+x}{1 - 2x^2}.$$ More generally, $F(p,x)$ is $\frac{1+x}{1-px^2}$ for all primes $p$.
(2) If $Q(x) = x^3$, then for any prime $p$, $$F(p,x) = 1 + x + px^2 + p^2 x^3 + p^2 x^4 + p^3 x^5 + ... = \frac{1+x+px^2}{1 - p^2x^3}.$$
(3) If $Q(x,y) = x^2 + xy + 2y^2$, then $$F(2,x) = 1 + 3x + 8x^2 + 20x^3 + 48x^4 + 112x^5 + ... = \frac{1-x}{(1-2x)^2},$$ while $$F(5,x) = 1 + x + 25x^2 + 25x^3 + 625 x^4 + 625x^5 + ... = \frac{1+x}{1-25x^2}$$ and $$F(7,x) = 1 + 7x + 49x^2 + 343x^3 + ... = \frac{1}{1-7x}.$$
(4) If $Q(x,y,z) = -x^2 + 2y^2 + z^2 + xy + xz + yz$ then $$F(2,x) = 1 + 4x + 20x^2 + 80x^3 + 352x^4 + 1408x^5 + ... = \frac{1-4x^2}{(1-4x)(1-8x^2)}$$ and $$F(5,x) = 1 + 25x + 725x^2 + 18125x^3 + ... = \frac{1}{(1 - 25x)(1 - 100x^2)}.$$
Question: Is there a general result that says that this is always a rational function? If so, can we find bounds on the degree of the numerator and denominator? I am particularly interested in the case that $Q$ is a quadratic form.
| Yes, it is always rational. It is related to the Igusa zeta function by the formula $$F(p,x) = \frac{1 - x p^d \zeta_{Igusa}(d-s)}{1 - x p^d},$$ and $\zeta_{Igusa}(s)$ is known to be a rational function in $p^{-s}.$ This is shown in the article here.
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this seqence $a_{n}$ closed form if such Strange condition
Let $\{a_{n}\}$ such
$$a_{0}=a_{1}=a_{2}=1,a_{3}=2,a_{4}=3,a_{5}=4,a_{6}=10,a_{7}=18$$
and such
$$\begin{vmatrix}
a_{n+8}&a_{n+7}&a_{n+6}\\
a_{n+5}&a_{n+4}&a_{n+3}\\
a_{n+2}&a_{n+1}&a_{n}
\end{vmatrix}=2\cdot (n+1)!$$
Find the $a_{n}$ colsed form
or
$$a_{n+8}a_{n+4}a_{n}+a_{n+5}a_{n+1}a_{n+6}+a_{n+2}a_{n+3}a_{n+7}
-a_{n+6}a_{n+4}a_{n+2}-a_{n+7}a_{n+5}a_{n}-a_{n+8}a_{n+1}a_{n+3}=2\cdot(n+1)!$$
if we find $a_{n}$ colsed form,I think easy to mathematical induction solve it?
| The first calculated terms are : $1,1,1,2,3,4,10,18,28,80,162,280,880$
So the closed formula appear to be $\bbox[5px,border:2px solid]{\begin{cases}a_0=1\\a_1=1\\a_2=1\\a_n=(n-1)a_{n−3}\quad n\ge3\end{cases}}$
Let see what the determinant look like under the recurrence hypothesis until $a_{n+7}$.
$D=\begin{array}{l}
+a_{n+8}a_{n+4}a_{n}\\
+a_{n+5}a_{n+1}a_{n+6}\\
+a_{n+2}a_{n+3}a_{n+7}\\
-a_{n+6}a_{n+4}a_{n+2}\\
-a_{n+7}a_{n+5}a_{n}\\
-a_{n+8}a_{n+1}a_{n+3}\end{array} =
\begin{array}{l}
+(n+3)a_{n+8}a_{n+1}a_{n} \\
+(n+4)(n+5)(n+2)a_{n+2}a_{n+1}a_{n} \\
+(n+2)(n+6)(n+3)a_{n+2}a_{n}a_{n+1} \\
-(n+5)(n+2)(n+3)a_{n}a_{n+1}a_{n+2} \\
-(n+6)(n+3)(n+4)a_{n+1}a_{n+2}a_{n} \\
-(n+7)(n+4)(n+2)a_{n+2}a_{n+1}a_{n}
\end{array}$
$=(n+3)a_{n+8}a_{n+1}a_{n}+[-n^3-14n^2-61n-82]\,a_{n+2}a_{n+1}a_{n}$
$=(n+3)a_{n+8}a_{n+1}a_{n}-[(n+3)(n+7)(n+4)-2]\,a_{n+2}a_{n+1}a_{n}$
$=(n+3)a_{n+1}a_{n}\times[a_{n+8}-(n+7)(n+4)a_{n+2}]+2a_{n+2}a_{n+1}a_{n}$
$=(n+3)a_{n+1}a_{n}\times[a_{n+8}-(n+7)a_{n+5}]+2a_{n+2}a_{n+1}a_{n}$
Let's have examine for $n=3k,\ 3k+1,\ 3k+2$ the product $a_{n+2}a_{n+1}a_{n}$
$a_{n+2}a_{n+1}a_{n}=(n+1)(n)(n-1)a_{n-1}a_{n-2}a_{n-3}=\ldots=(n+1)(n+0)(n-1)\ldots(4)(3)(2)a_2a_1a_0=(n+1)!\times 1\times 1\times 1=(n+1)!$
$a_{n+2}a_{n+1}a_{n}=(n+1)(n+0)(n-1)\ldots(5)(4)(3)a_3a_2a_1=\frac{(n+1)!}{2}\times 2\times 1\times 1=(n+1)!$
$a_{n+2}a_{n+1}a_{n}=(n+1)(n+0)(n-1)\ldots(6)(5)(4)a_4a_3a_2=\frac{(n+1)!}{2\times 3}\times 3\times 2\times 1=(n+1)!$
So $D=(n+3)a_{n+1}a_{n}\times[a_{n+8}-(n+7)a_{n+5}]+2\,(n+1)!$
This allows to conclude that $a_{n+8}-(n+7)a_{n+5}=0$ which terminates the recurrence.
| {
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Partial Simplified Proof for the prime version of the Catalan Conjecture I have found an elementary approach which seems to work in proving that Catalan's conjecture is true for specific primes. I am attempting to generalize this approach to see how far it will go.
I am presenting it here with the hopes of finding a mistake or confirming that this approach is worth exploring further.
From Wikipedia, the Catalan Conjecture can be defined as:
*
*Let $a>1,b>1,x>0,y>0$
*Then if:
$x^a - y^b = 1$
*
*The only solution is $x=3, a=2, y=2, b=3$
I will attempt now to prove the case where:
$x=2, y$ is an odd prime
(1) Making this assumption, we have:
$$2^a = 2\left[y(y^{b-2} + y^{b-3} + \dots + 1)(\frac{y-1}{2}) + \frac{y+1}{2}\right]$$
(2) $\frac{y+1}{2}$ is even
If $\frac{y+1}{2}$ is odd, then $\frac{y-1}{2}$ is even and $\left[y(y^{b-2}+\dots+1)(\frac{y-1}{2}) + \frac{y+1}{2}\right]$ is odd.
(3) $b$ is odd
If $\frac{y+1}{2}$ is even, then $\frac{y-1}{2}$ is odd and it follows that $(y^{b-2} + \dots + 1)$ must be even. Therefore, $b-2$ must be odd.
(4) Let $2^u$ be the highest power of $2$ that divides $\frac{y+1}{2}$
(5) $y \equiv -1 \pmod {2^{u+1}}$ since:
There exists $m$ such that $\frac{y+1}{2^{u+1}}=2m+1$ which means $y = 2^{u+1}(2m + 1) - 1$
(6) $2^{u+1} | (y^{b-2} + \dots 1)$ since:
$b-2$ is odd and $(y^{b-2} + y^{b-3} + \dots + y + 1) \equiv (-1 + 1) +\dots + (-1 + 1) \equiv 0 \pmod {2^{u+1}}$
(7) But then we have a contradiction since:
$y(\frac{y^{b-2}+\dots+1}{2^u})(\frac{y-1}{2})$ is even but $\frac{y+1}{2^{u+1}}$ is odd so that:
$$2^a \ne 2^{u+1}\left[y(\frac{^{b-2}+\dots+1}{2^u})(\frac{y-1}{2}) + \frac{y+1}{2^{u+1}}\right]$$
Edit: Attempting to greatly simplify the argument based on feedback received.
| I couldn't find any errors so your approach seems to be correct.
Note, though, that your solution is more general in that it also works for any $x$ being a positive power of $2$, plus for all odd $y \gt 1$, not just odd primes. For the first case, this is because your proof just requires the LHS to be a power of $2$. For the second case, this is due to you only using that $y$ is odd, and in particular not using any particular property of primes, in your solution.
As for "exploring further" with this approach, I'm not quite sure what you mean. However, note you can use a similar approach to show that the only values which work for $y$ being a positive power of $2$ is the only known solution. To show this, you can follow a similar approach to what you did. First, let $y = 2^c$ for an integer $c \ge 1$. Next, the equation can now be rewritten as
$$2^{bc} = x^a - 1 \tag{1}\label{eq1}$$
This shows that $x$ must be odd. First, consider that $a$ is even, i.e., $a = 2d$. Then $x^a - 1 = \left(x^d + 1\right)\left(x^d - 1\right)$, so both $x^d + 1$ and $x^d - 1$ must be powers of $2$. First, $x^d - 1 = 1 \; \Rightarrow \; x^d = 2$, so it's not possible. Next, $x^d - 1 = 2 \; \Rightarrow \; x^d = 3$, so $x = 3$ and $d = 1$, giving the $1$ known solution. No larger values of $x^d - 1$ as a power of $2$ have that $x^d + 1$ is also a power of $2$.
Next, consider that $a$ is odd. Factoring the LHS gives $x^a - 1 = \left(x - 1\right)\left(x^{a-1} + x^{a-2} + \cdots + x + 1\right)$. Since $x$ is odd and there are $a$ terms in the second factor, this factor must be odd and $\gt 1$. However, the LHS of \eqref{eq1} is a power of $2$, so it has no odd factors. Thus, $a$ can't be an odd integer.
Putting this solution together with what you've shown proves that, apart from the one known solution, there are no other ones where $x$ or $y$ is a power of $2$.
You could try to do something similar with odd prime numbers, but I don't believe that just these sorts of basic remainder and number of factors type arguments are sufficient to show there is no solution if either $x$ or $y$ is of a given simple form (e.g., just a prime or a power of a prime). If you haven't already done so, you may wish to read & analyze Preda Mihăilescu's proof, such as discussed & outlined in the AMS article at Catalan's Conjecture: Another Old Diophantine Problem Solved.
| {
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} |
Derive the Laurent series of $\frac{1}{\sqrt{s^2-a^2}}$ So, I was trying to derive the Laplace Tranform of the Modified Bessel Function Io(at) using its summation formula. I arrived at the answer $$\frac{1}{s}+\frac{a^2}{2s^3}+\frac{3a^4}{8s^5}+\frac{5a^6}{16s^2}+\frac{35a^8}{128s^9}+\frac{63a^{10}}{256s^{11}}+\cdots$$
It wasn't the answer yet so I decided to enter the Laplace Transform of the Modified Bessel function which was $\frac{1}{\sqrt{s^2-a^2}}$ to Wolfram Alpha. According to Wolfram Alpha, its Laurent series $\frac{1}{\sqrt{s^2-a^2}}$ is equal to the answer I had.
$$\frac{1}{s}+\frac{a^2}{2s^3}+\frac{3a^4}{8s^5}+\frac{5a^6}{16s^2}+\frac{35a^8}{128s^9}+\frac{63a^{10}}{256s^{11}}+\cdots$$
I don't know anything about Laurent series. Help
https://www.wolframalpha.com/input/?i=1%2Fsqrt(s%5E2-a%5E2)
| Laurent series are quite "similar" to Taylor series.
Considering that you want the Laurent series of $\frac{1}{\sqrt{s^2-a^2}}$, consider that $$\frac{1}{\sqrt{s^2-a^2}}=\frac 1 s \frac{s}{\sqrt{s^2-a^2}}=\frac 1 s\frac 1{\sqrt{1-\frac{a^2}{s^2}}}$$ Now, by Taylor series or generalized binomial theorem $$\frac 1{\sqrt{1-x^2}}=1+\frac{x^2}{2}+\frac{3 x^4}{8}+\frac{5 x^6}{16}+\frac{35
x^8}{128}+O\left(x^9\right)$$ Replace $x$ by $\frac a s$ to get $$\frac 1 s\frac 1{\sqrt{1-\frac{a^2}{s^2}}}=\frac 1 s\left(1+\frac{a^2}{2 s^2}+\frac{3 a^4}{8 s^4}+\frac{5 a^6}{16 s^6}+\frac{35 a^8}{128
s^8}+O\left(\frac{1}{s^9}\right) \right)$$ $$\frac{1}{\sqrt{s^2-a^2}}=\frac{1}{s}+\frac{a^2}{2 s^3}+\frac{3 a^4}{8 s^5}+\frac{5 a^6}{16 s^7}+\frac{35
a^8}{128 s^9}+O\left(\frac{1}{s^{10}}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$ if :
$$\lim_{ x \to 0 }\left( \frac{\sin 3x}{x^3}+\frac{a}{x^2}+b \right)=0$$
then $a+b=?$
Without the use of the L'Hôspital's Rule
My Try :
$$\lim_{ x \to 0 }\left( \frac{ax+bx^3+\sin 3x}{x^3} \right)=0$$
$$\lim_{ x \to 0 }\left( \frac{x(a+bx^2)+\sin 3x}{x^3} \right)=0$$
$$\lim_{ x \to 0 }x(a+bx^2)+\sin 3x=0 $$
now ?
| The MacLaurin expansion of $\sin3x$ up to order $3$ is
$$\sin3x = 3x - \frac{9x^3}{2} + \mathcal O(x^5)$$
Plugging in the limit we obtain
$$\lim_{x \to 0} \frac{(3 + a)x + (b - \frac92)x^3}{x^3} =0$$
If the numerator is not zero then the limit is infinite, because the denominator goes to $0$ as order $3$. Therefore it has to be
$$\left\{\begin{align*}
a &= -3\\
b &= \frac92
\end{align*}\right.$$
| {
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Prove $B(p, q) = \int \limits_0^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx$ Prove the equality $$B(p, q) = \int \limits_0^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx$$ for $p > 0, q > 0$
I think the first step is
$$B(p, q) = \int \limits_0^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx = \int \limits_0^1 \frac{x^{p-1}}{(1+x)^{p+q}}dx + \int \limits_0^1 \frac{x^{q-1}}{(1+x)^{p+q}}dx$$
| Note that
$$\begin {align}
2\int_0^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx &=\int_0^1 \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx+\int_1^\infty \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx\\&=\int_0^\infty \frac{x^{p-1} + x^{q-1}}{(1+x)^{p+q}}dx\\&=\int_0^\infty \frac{x^{p-1} }{(1+x)^{p+q}}dx+\int_0^\infty \frac{x^{q-1}}{(1+x)^{p+q}}dx \\&=B (p,q)+B(q,p)\\&=2 B(p,q)
\end {align}$$
| {
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For $a,b,c<1$. Minimize $M=\frac{a^2\left(1-2b\right)}{b}+\frac{b^2\left(1-2c\right)}{c}+\frac{c^2\left(1-2a\right)}{a}$ For $0<a,b,c<1$ and $ab+bc+ca=1$, minimize $$M=\frac{a^2\left(1-2b\right)}{b}+\frac{b^2\left(1-2c\right)}{c}+\frac{c^2\left(1-2a\right)}{a}$$
My Try:
From $ab+bc+ca=1\Leftrightarrow \sqrt{3}\leq a+b+c<3$
We have: $M=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-2\left(a^2+b^2+c^2\right)$
$\ge\frac{\left(a+b+c\right)^2}{a+b+c}-2\left(a^2+b^2+c^2\right)-4\left(ab+bc+ca\right)+4\left(ab+bc+ca\right)$
$=a+b+c-2(a+b+c)^2+4 (1)$
Let $a+b+c=x (\sqrt {3}\leq x<3)$
We find Min of function $y=-2x^2+x+4$
And $y_{Min}=-2+\sqrt {3}$ when $x=\sqrt {3}$
Right or Wrong ?
| We will prove that $$M=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-2(a^2+b^2+c^2)\geq -2+\sqrt{3} (*)$$
or equivalently
$$\left ( \frac{a^2}{b}-2a+b \right )+\left ( \frac{b^2}{c}-2b+c \right )+\left ( \frac{c^2}{a}-2c+a \right )+(a+b+c-\sqrt{3})-2(a^2+b^2+c^2-1)\geq 0.$$
We have
$$\sum_{cyc} \left (\frac{a^2}{b}-2a+b\right)=\frac{(a-b)^2}{b}+\frac{(b-c)^2}{c}+\frac{(c-a)^2}{a}$$
$$a+b+c-\sqrt{3}=\frac{(a+b+c)^2-3}{a+b+c+\sqrt{3}}=\frac{(a-b)^2+(b-c)^2+(c-a)^2}{2(a+b+c+\sqrt{3})}$$
$$2(a^2+b^2+c^2-1)=2(a^2+b^2+c^2-ab-bc-ac)=(a-b)^2+(b-c)^2+(c-a)^2.$$
Hence
$$(*) \Leftrightarrow (a-b)^2\left [ \frac{1}{b}+\frac{1}{2(a+b+c+\sqrt{3})}-1 \right ]+(b-c)^2\left [ \frac{1}{c}+\frac{1}{2(a+b+c+\sqrt{3})}-1 \right ]+(c-a)^2\left [ \frac{1}{a}+\frac{1}{2(a+b+c+\sqrt{3})}-1 \right ]\geq 0 $$
This true because $\frac{1}{a};\frac{1}{b};\frac{1}{c}>1$ when $a=b=c=\frac{1}{\sqrt 3}$
| {
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The number of 3 digit numbers of the form xyz such that $x
The number of 3 digit numbers of the form $xyz$ such that $x<y$ and $z\leq y$ is N such that n is a 3 digit number of the form $abc$ then find $a+c-b$
My approach:
Case 1:
$x<y$
$z<y$
$x,y,z\neq 0$
Choosing 3 digits out of 9 numbers: $\binom 93$, greatest of them will always be $y$, hence ways of arranging is (exchanging x and z) $2\cdot\binom 93=54$.
Case 2:
$x,y$ can't be $0$. Taking $z=0$, number of ways of selecting the other 2 digits are $\binom 92=36$. There is only one way to arrange them.
case 3:
$y=z$,
_99, values for $x=8$
_88, values for $x=7$
_77, values for $x=6$
.
.
.
_11, values for $x=0$
Total possibilities=$8+7+6+\cdots+1=36$
Hence total permutations=$54+36+36=126$
This gives $a+c-b=5$ which is not the right answer. Why is this logic wrong?
| Your case 1 needs supplementing with case 1a) $x=z$, which gives $\binom 92$ additional cases, with the larger choice $=y$ and the smaller $=x=z$ -- another $36$ cases.
Then $$\overset{\text{case 1}}{2\cdot \binom 93} + \overset{\text{case 1a}}{\binom 92} + \overset{\text{case 2}}{\binom 92} + \overset{\text{case 3}}{\binom 92}
= 2\cdot 84 +3\cdot 36 = 168+108 = 276$$
giving your final calculation of $2+6-7 = 1$
| {
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Find the determinant and possible sizes of a matrix $A$, which satisfies the equation $A^2 - 2A + 4E = 0$ Let $A \in M_n(\mathbb{R})$ satisfy the equation:
$$A^2 - 2A + 4I = 0$$
(a) What is $\det A$? (3 points)
(b) What are possible sizes of a matrix $A$? (6 points)
My solution
We can derive 2 useful equations from the given one:
1) $A^2 = -2(2I - A)\Rightarrow (\det A)^2 = \det (-2(2I - A)) = (-2)^n \cdot \det (2I - A)$
2) $A(2I - A) = 4E \Rightarrow \det A \cdot \det (2I - A) = 4^n = 2^{2n}$
From the first we see that the size $n$ must be even. This is the answer to the question (b).
Now, let's build the system of equations, where $x = \det A$, $y = \det (2I - A)$. Also, we can replace $(-2)^n$ with $2^n$, because we know, that $n$ is even:
$$\begin{cases} x^2 = 2^n \cdot y, \\ xy = 2^{2n} \end{cases} \Rightarrow \begin{cases} y = x^2\cdot2^{-n}, \\ x\cdot(x^2\cdot2^{-n}) = 2^{2n} \end{cases} \Rightarrow x = (2^{2n}\cdot2^n)^{1/3} = 2^n$$
So, $\det A = 2^n$ and it depends on the size of the matrix. This is the answer to the question (a).
The question
Is this solution correct? I have big doubts about it, because the question (b) is worth twice more points than (a), but happened to be much easier and also order of tasks suggests, that we should solve (a) first.
| Using eigenvalues: Let $x$ be an eigenvector of $A$ associated with $\lambda$. Note that
$$
0 = 0x = (A^2 - 2A + 4I)x = (\lambda^2 - 2\lambda + 4)x
$$
so, each $\lambda$ must be a root of $x^2 - 2x + 4$, which is to say that each eigenvalue is either $1 + \sqrt{3}i$ or $1 - \sqrt{3} i$. Since $A$ is real, its eigenvalues come in conjugate pairs. So, $n$ must be even, and the eigenvalues $1+\sqrt{3}i$ and $1 - \sqrt{3}i$ must have the equal multiplicities $n/2$.
The determinant of $A$ is the product of its eigenvalues, so
$$
\det(A) = [(1 + \sqrt{3}i)(1 - \sqrt{3} i)]^{n/2} = 4^{n/2} = 2^n
$$
| {
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Show that $\frac 1x+\frac 1y =(\frac 27)^a$ does not produce integer solutions for $a > 3$. Show that $\frac 1x+\frac 1y =(\frac 27)^a$ does not produce integer solutions for $a > 3$.
I have shown it is not possible for $a=4$, but not for any a greater than $4$.
| We see that
$$\frac{x+y}{xy}=\frac{2^a}{7^a}$$
and that leads to
$$7^a(x+y)=2^a(xy)$$
now let $\gcd(x,y)=d$, and write $x=dx_0$ and $y=dy_0$.
We get:
$$7^a(dx_0+dy_0)=2^a(dx_0dy_0)$$
or, rewritten:
$$7^a(x_0+y_0)=2^a(dx_0y_0)$$
note that since $\gcd(x_0,y_0)=1$, we know $\gcd(x_0y_0,x_0+y_0)=1$; thus, $x_0$ and $y_0$ can only contain factors $7$ (since $x_0y_0|7^a(x_0+y_0)$ implies $x_0y_0|7^a$). Now write $x=dx_0=d\cdot7^n$ and $y=dy_0=d\cdot7^m$. We substitute this into the equation to get
$$7^a(7^n+7^m)=2^a(d\cdot7^{n+m})$$
meaning, $2^a|7^n+7^m$. Now set w.l.o.g. $n\geq m$, so that we can write $7^n+7^m=7^m(7^{n-m}+1)$. We know there's no factors $2$ in $7^m$, so we can say $2^a|7^{n-m}+1$. We can write $n-m=k$ to simplify this to $2^a|7^k+1$. Proving that $7^k+1$ can never contain four factors $2$ would thus be sufficient.
Let's look at this $\mod 16$. We see that $7^2=49\equiv 1\mod 16$, and as such, we know that $7^k+1\equiv 2\mod 16$ if $k$ is even, and if $k$ is odd, $7^k+1=7\cdot 7^{k-1}+1\equiv 7\cdot 1+1\equiv 8\mod 16$. Therefore, $7^k+1$ is never divisible by $16$, and as a result of that, never contains 4 or more prime factors. This proves that the initial equality has no integer solutions if $a>3$.
| {
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What does this series converge to, if anything? $$\arctan{1} + \arctan{\frac{1}{2}} + \arctan{\frac{1}{3}} + \arctan{\frac{1}{4}} ...= ?$$
The infinite series for arctan is
$$\arctan{x} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} ...$$
So I want to sum up the $\arctan{1\over n}$ where $n$ starts at $1$ and goes to infinity.
I originally thought the resulting series can be written this way:
$$(1 + \frac{1}{2} + \frac{1}{3} ...) - \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^3}{3} + \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^5}{5} - \frac{(1 + \frac{1}{2} + \frac{1}{3} ...)^7}{7} ...$$
But that is wrong. The right way to "insert" the series is:
$$1 + \frac{1}{2} + \frac{1}{3} ... - \frac{1}{3} - \frac{(\frac{1}{2})^3}{3} - \frac{(\frac{1}{3})^3}{3} ... + \frac{1}{5} + \frac{(\frac{1}{2})^5}{5} + \frac{(\frac{1}{3})^5}{5} ... ...$$
So it looks like a bunch of harmonic serieses manipulated. Harmonic series diverges, but I remember that doesn't necessarily mean a similar series diverges. I remember from Calculus 2 that, for example, $\lim_{n\to\infty}$ of $\sin(n)\over n$ converges to zero even though $\sin(n)$ does not converge.
So how do I figure out what this series converges to, if anything?
| We have,
$$\lim_{n \to \infty} \frac{\arctan(\frac{1}{n})}{\frac{1}{n}}=1$$
This follows easily by first the change of variables $\frac{1}{n}=h$ then by Taylor series.
We also have that $a_n=\arctan (\frac{1}{n}) \geq 0$ for all $n \geq 1$. Hence by the limit comparison test your series $\sum_n a_n$ diverges with comparison to the harmonic series.
| {
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If coefficients of the Quadratic Equation are in AP find $\alpha+\beta +\alpha\beta$. $ax^2+bx+c=0$ is a quadratic equation with integer roots $\alpha,\beta$ and the positive coefficients are in AP.Find $\alpha+\beta +\alpha\beta$.
My attempt:
$\alpha+\beta +\alpha\beta=\frac {-b}{a}+\frac {c}{a}$ and we know $\frac{a+c}{2}=b$.
But I cannot eliminate the $a,b,c$ terms to get the answer $7$ given.
How to do it?Any quick,short,simple solutions?
| First of all we need to that if the roots are integer then $S=\alpha+\beta+\alpha\cdot \beta$ is also integer then
$$S=\frac{c-b}{a}=\frac{b-a}{a}=\frac{b}{a}-1\to a|b$$
and once $2b=a+c$ we get that $a|c$ so we can write the terms $(a,b,c)$ like $(a,a(1+r),a(1+2r))$ and then
$$S=r$$
We also have the discriminant:
$$\Delta=b^2-4ac=a^2(1+r)^2-4a\cdot a(1+2r)=a^2[(1+r)^2-4(1+2r)]$$
and once the roots are integer then $\Delta$ should be a square and then
$$(1+r)^2-4(1+2r)=k^2\to r^2-6r-(3+k^2)=0\quad (1)$$
and looking this new quadratic equation (in $r$) we have the discriminant
$$\Delta '=36-4[-(3+k^2)]=4(12+k^2)$$
and $\Delta'$ has also to be an integer, then:
$$12+k^2=p^2\to (p-k)(p+k)=12$$
spliting $12$ as a product of two numbers we get the solutions $(k,p)\in\{(2,4),(-2,-4),(2,-4),(-2,4)\}$. So, $k=\pm 2\to k^2=4$ and then backing to $(1)$ we get $r=7$ or $r=-1$.
For $r=-1$ we get the sequence $(a,0,a)$ and the equation $ax^2+a=a(x^2+1)$ what not give us real solution.
For $r=7$ we get the equation $ax^2+8ax+15a=0$ what give us the roots $-3$ and $-5$, then
$$S=r=7$$
| {
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If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $x^3+y^3 \le 2$. As in the title. If $x,y>0$ and $x^2+y^3\ge x^3+y^4$, prove that $$x^3+y^3 \le 2.$$
This seems to be a very tricky one. I tried applying various inequalities like AM-GM, unfortunately, none of techniques I'm familiar with seem to work here.
I'd greatly appreciate any hints.
| By AM-GM $2x^3+1\geq3x^2$ and $3y^4+1\geq4y^3$.
Thus, $$2x^3+3y^4+2\geq3x^2+4y^3$$ or
$$x^3+y^3\leq2+3(x^3+y^4-x^2-y^3)\leq2$$
| {
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If $|a-b| \leq C$ and $|x-y|\leq D$, is there a way to bound $\left|\frac{a}{x} - \frac{b}{y}\right|$? If $|a-b| \leq C$ and $|x-y|\leq D$, is there a nice way to bound $\left|\frac{a}{x} - \frac{b}{y}\right|$?
I made an attempt using Triangle inequality but to no avail.
\begin{align*}\left|\frac{a}{x} - \frac{b}{y}\right| &= \left|\frac{a}{x} - \frac{b}{x} + \frac{b}{x} - \frac{b}{y}\right| \\
&\leq \frac{|a-b|}{|x|} + |b| \left|\frac{y - x}{xy}\right| \\
&\leq \frac{C}{|x|} + \frac{|b|D}{|xy|}
\end{align*}
Is there a nicer one?
| Think about this intuitively. Consider $|\frac{a}{x} - \frac{b}{y}|$. Let's just keep $a$ and $b$ fixed for now. Consider $|x - y| \leq D$, which basically means $-D \leq x-y \leq D$. If we choose $x$ and $y$ from $(-\frac{D}{2}, \frac{D}{2})$, then we will always have $x-y$ in $(-D,D)$ (why?).
Okay, but let's fix $y$ for a moment. We can choose $x$ as close to $0$ as we want. That will make $|\frac{a}{x} - \frac{b}{y}|$ grow unboundedly toward $\infty$ as $x$ gets closer and closer to $0$. So, we've seen that under all of your assumptions, we can find an example where $|\frac{a}{x} - \frac{b}{y}|$ is actually an unbounded expression.
| {
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trigonometry equation - $\sin^3(x)+\sin^3(2x)+\sin^3(3x) = (\sin(x)+\sin(2x)+\sin(3x))^3$ how can i find all possible solutions to this equation?
$\sin^3(x)+\sin^3(2x)+\sin^3(3x) = (\sin(x)+\sin(2x)+\sin(3x))^3$
I've tried writing it all as a sum in the form $\sin^m(x)\cos^n(x) + ... sin(x)$, with just $\sin(x)$ and $\cos(x)$ occuring. For my luck, you can write it in the form $\sin(x)* (...) = 0$ and therefore you can find a few solutions. But inside the brackets there is an expression starting with somthing like $\sin^7(x)$ which i cant handle and dont find any other solutions.
| Let $\sin x=a$, $\sin2x=b$,
and $\sin3x=c$
Expand out
$(a+b+c)^3$
Then subtract $a^3, b^3$ and $c^3$, which yields
$3 a^2 b + 3 a^2 c + 3 a b^2 + 6 a b c + 3 a c^2 + 3 b^2 c + 3 b c^2=0$.
The long expression can be factored nicely into :
$3(a+b)(b+c)(a+c)$, which equal to $0$.
The factor $3$ makes no difference, so solve for:
$\sin(x)+\sin(2x)=0$,
$\sin(2x)+\sin(3x)=0$, and
$\sin(x)+\sin(3x)=0$.
| {
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Find all pairs of prime numbers $p$ and $q$ such that $\,p^2-p-1=q^3.$ I'm preparing a mathematical olympiad and our group is stuck in this problem. Here is all we've done:
First, let's rewrite the equation
$$p^2-p-1=q^3\Rightarrow p(p-1)=(q+1)(q^2-q+1)$$
It's obvious that $q$ must be less than $p$. Then, $p|(q^2-q+1)$ and then equation is of the form
$$p-1=k(q+1)$$
For some integer $k$. Putting that onto the first equation
$$\left(\frac{p-1}{k}-1\right)^3=q^3=p^2-p-1$$
Which is the same as
$$p^2-(2+3k+k^3)p+(3k^2+3k+1)=0$$
Since $p$ is prime $p|3k^2+3k+1$, a possible solution is $p=3k^2+3k+1$. Substituting
\begin{align*}
(3k^2+3k+1)^2-(2+3k+k^3)(3k^2+3k+1)+3k^2+3k+1=-k^2(k-3)(3k^2+3k+1)=0
\end{align*}
$k$ cannot be 0, so it's unique possible integer value is $3$ and $p=37$. Solving, $q=11$ which is a valid solution but we don't know if there are more or how to prove there aren't more, please help us continue.
| You already have
$$q^2-q+1=pk\quad\text{and}\quad p-1=k(q+1)$$
where $k$ is a positive integer.
Eliminating $p$ gives
$$q^2-q+1=(kq+k+1)k,$$
i.e.
$$q^2+(-1-k^2)q-k^2-k+1=0$$
to have
$$q=\frac{k^2+1+\sqrt{k^4+6k^2+4k-3}}{2}$$
Note here that we get, for $k\gt 3$,
$$k^2+3\lt \sqrt{k^4+6k^2+4k-3}\lt k^2+4$$
from which we have to have $k=1,2,3$.
*
*For $k=1$, $\sqrt{k^4+6k^2+4k-3}=\sqrt 8\not\in\mathbb Z$
*For $k=2$, $\sqrt{k^4+6k^2+4k-3}=\sqrt{45}\not\in\mathbb Z$
*For $k=3$, $q=11,p=37$.
Hence, $\color{red}{(p,q)=(37,11)}$ is the only solution.
| {
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"url": "https://math.stackexchange.com/questions/2180688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}$ We know that $$\int_0^1\frac{\ln (1+x)}{1+x^2}dx=\frac{\pi}8\ln 2,$$ but how about $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx?$$Prove that $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}.$$
| Let $$I=\int_0^1\frac{x\ln(1+x)}{1+x^2}dx$$
and $$I(a)=\int_0^1\frac{x\ln(1+ax)}{1+x^2}dx.$$
Note that $I(1)=I$ and $I(0)=0.$
$$I'(a)=\int_0^1\frac{x^2}{(1+x^2)(1+ax)}dx$$
$$=-\frac{\pi}{4}\frac{1}{1+a^2}+\frac12\ln(2)\frac{a}{1+a^2}+\frac{\ln(1+a)}{a}-\frac{a\ln(1+a)}{1+a^2}.$$
$$\Longrightarrow I=-\frac{\pi}{4}\int_0^1\frac{da}{1+a^2}+\frac12\ln(2)\int_0^1\frac{a}{1+a^2}da+\int_0^1\frac{\ln(1+a)}{a}da-\int_0^1\frac{a\ln(1+a)}{1+a^2}da$$
$$=-\frac{\pi}{4}\left(\frac{\pi}{4}\right)+\frac12\ln(2)\left(\frac12\ln(2)\right)+\frac{\pi^2}{12}-I$$
$$\Longrightarrow I=\frac{\pi^2}{96}+\frac18\ln^2(2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Inclusion/Exclusion - Assigned Seating Arrangements Question. Four people sit down for dinner at a round table. They have assigned seating. What is the probability that 1, and only 1, person sits in their assigned seat.
My try:
We fix one of the members to a seat that is not theirs. Suppose the events $A_i, i=1,2,3$ are denoted as someone sitting in their correct seat. Then
$$
\begin{align*}
P(A_1 \cup A_2 \cup A_3) &= P(A_1) + P(A_2) + P(A_3) - P(A_1 \cap A_2) - \dots + P(A_1 \cap A_2 \cap A_3)\\
&= \frac{1}{3} + \frac{1}{3} + \frac{1}{3} - \frac{1}{3} \cdot \frac{1}{2} - \dots + \frac{1}{3} \cdot \frac{1}{2} \\
&= 1 - \frac{1}{2} + \frac{1}{6} \\
&= \frac{2}{3}.
\end{align*}
$$
However, the answer is $\frac{1}{3}$. Meaning we are taking the complement. So, am I wrong in assuming that the fixed person is not in their assigned chair? If we fix him to be in his assigned chair then the complement makes sense (i.e. we need none of the other 3 in the correct spot). What is the basis for fixing the assigned person to their correct spot though?
| Assume that person 1 is in his seat. Then 2,3,4 should not be in their assigned seats and this can be done in 2 ways (derangement on three symbols). Thus the number of ways in which only one is in the assigned seat is $4 \times 2 = 8$. Thus the required probability is $\frac{8}{24} = \frac{1}{3}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Multinomial Verification Here's my question, which is similar to the one I asked a couple weeks ago:
Find the coefficient of $x^4$ in the expansion of $(1+x+x^2+x^3)^7$.
I was suggested to use the Binomial Theorem, but it would become very tedious and just decided to use the Multinomial Theorem instead. So I need to find the terms that have $x^4$ in them. 3 such instances do occur and they are:
$$1^5 \cdot (x)^1 \cdot (x^2)^0 \cdot (x^3)^1$$
$$1^5 \cdot (x)^0 \cdot (x^2)^2 \cdot (x^3)^0$$
$$1^3 \cdot (x)^4 \cdot (x^2)^0 \cdot (x^3)^0$$
Notice that all of the exponents add up to $7$. I'm supposed to get $203$ as my result (verified using Wolfram), but I got $98$ as my answer. I believe I'm missing some other factors, but cannot find any more other than those $3$.
| $$\begin{eqnarray*}[x^4](1+x+x^2+x^3)^7 &=& [x^4](1-x^4)^7\cdot\frac{1}{(1-x)^7}\\&=&[x^4]\left(\sum_{k=0}^{7}(-1)^k \binom{7}{k}x^{4k}\right)\cdot\left(1 + 7 x + 28 x^2 + 84 x^3 + 210 x^4+\ldots\right)\end{eqnarray*}$$
hence the answer is given by $\binom{7}{0}\cdot 210-\binom{7}{1}\cdot 1 = \color{red}{203}$ as predicted by WA.
| {
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"url": "https://math.stackexchange.com/questions/2185257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How Do You Find The Point Of Intersection Of $2$ Vectors? Two lines $\textbf{v} = \begin{pmatrix} 7 \\ -3 \\ 1 \end{pmatrix} + \begin{pmatrix} -2 \\ 5 \\ 1 \end{pmatrix} t$ and $\textbf{w} = \begin{pmatrix} 8 \\ -1 \\ -1 \end{pmatrix} + \begin{pmatrix} 1 \\ -4 \\ 0 \end{pmatrix} u$ intersect. What point do they intersect at?
| We require the $x,y,z$ coordinates to be equal at the point of intersection, so we solve the following set of equations:
$(1)$ $$7-2t = 8+u$$
$(2)$ $$-3+5t = -1-4u$$
$(3)$ $$1+ t = -1$$
Thus from $(3)$ we get $t=-2$
Then from $(1)$ we get $7+4 = 8+u \Rightarrow u = 3$
Then we must check that this satisfies equation $(2)$
$-3+5t = -3 + 5\cdot -2 = -13 $
$-1-4u = -13$ hence there is a point of intsersection
Just plugin $u = 3$ into the equation for $w$ or $t=-2$ into your equation for $v$ to find the point of intersection.
Note the point of intersection you should get is
$\begin{pmatrix}
11\\ -13
\\ -1
\end{pmatrix}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the roots of quadratic polynomial and factor into linear factors I'm looking at Calculus by Adams, 7th Ed. page 43 Example 3b. It shows how to find the roots of $9x^2-6x+1$ by using the quadratic formula to arrive at the double root $\frac{1}{3}$. That I understand.
However, I don't understand how the factoring (?) is carried out in the next line:
$9x^2-6x+1 = 9\left(x - \frac{1}{3}\right)^2 = (3x - 1)^2$
I thought the factor theorem says something along the lines of that if we have roots such as $\frac{1}{3}$, we can write the factors $(x - \frac{1}{3})(x - \frac{1}{3}) = \left(x - \frac{1}{3}\right)^2$. I don't get where the $9$ comes from after the first equal sign. Are they using a different method than completing the square?
| The factor theorem says only that $x-c$ is a factor of a polynomial $p(x)$ if and only if $p(c)=0$. What this exactly means is that there is a polynomial $q(x)$ such that $p(x) = (x-c)q(x)$. The theorem doesn't say, but it's required that $q(x)$ has degree one less than $p(x)$.
Applied to a quadratic polynomial such as your polynomial. We set $p(x) = 9x^2-6x+1$ has the only roots at $x=1/3$. This means that $x-1/3$ is a factor of it which means that is there's a polynomial $q(x)$ such that $p(x) = (x-1/3)q(x)$. Now $q(x)$ must be of degree $1$ that is on the form $q(x) = ax+b = a(x+b/a)$, we see from this that $x+b/a$ must be a factor of $p(x)$. So we know that by the factor theorem that $p(-b/a) = 0$ which we now requires $b/a$ to be $1/3$. We now know that:
$$9x^2-6x+1 = a(x+b/a)(x-1/3) = a(x-1/3)(x-1/3) = a(x-1/3)^2 = a\left(x^2-{2\over3}x + {1\over 9}\right)$$
With $a$ needing to be $9$ in order to the LHS and RHS to be equal.
Another approach would be to complete the square (this is how the quadratic equation is solved without the formula), we know that $9x^2-6x = (3x-1)^2 -1$ (since $(3x-1)^2 = 9x^2-6x+1$, the $3x-1$ is chosen so that the $x^2$ and $x$ terms match). This means that
$$9x^2-6x+1 = (3x-1)^2-1 + 1 = (3x-1)^2$$
If you want to bring that into the other form you just rewrite the factors using $(3x-1) = 3(x-1/3)$
Note that one often shortcuts this reasoning. However you come to the conclusion you must of course end up with a factorization that is correct. You can just rely on that you by expanding $9(x-1/3)^2$ via the square rule come to the correct polynomial.
| {
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Calculating Euler Number limit Please, so far I did
$$\lim_{x\to +\infty}\left(\frac{x^2-x+1}{x+2}\right)^{\frac{1}{x-1}},$$ but I can write
$$\frac{x^2-x+1}{x+2}=1+\frac{x^2-2x-1}{x+2}=1+\frac{1}{\frac{x+2}{x^2-2x-1}}.$$
But
$$\lim_{x\to +\infty}\frac{x+2}{x^2-2x-1}=0,$$ so I can not use
$$e =\lim_{N\to \infty}(1+\frac{1}{N})^N$$
| Notice that for sufficiently large $x$,
\begin{align}
1 < \left(\frac{x^2-x+1}{x+2}\right)^\frac{1}{x-1}
& = \left(x-3+\frac{7}{x+2}\right)^\frac{1}{x-1} \\
& < (x-1)^\frac{1}{x-1}
\end{align}
which has a well-known limit as $x \to +\infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Teacher's proof of sequence convergence that I don't understand. Prove $a_n = \frac{n^2-2}{n^2 +2n + 2}$ converges to 1.
Solution:
WTS $\exists L \in R, \forall \epsilon > 0, \exists N > 0$, such that for all $n \in N$, if $n > N$, then $|a_n - 1| < \epsilon$.
Let $\epsilon > 0$ be arbitrary
Choose N = $\max(\frac{6}{\epsilon}, 1) > 0$
Suppose n > N, then
$$\begin{align}
\left|\frac{n^2-2}{n^2+2n+2} - 1\right|&= \left|\frac{-2n-4}{n^2+2n+2}\right|
\qquad\text{(By algebra)}\\
&=\frac{2|n+2|}{|n^2+2n+2|} \qquad \text{ (abs value properties)}\\
&\leq \frac{2|n+2|}{n^2} \qquad\qquad \text{ (By min denominator)}\\
&= 2\frac{1}{n^2} n \left|1+\frac{2}{n}\right| \qquad \text{ (factoring)}\\
&= 2\frac{1}{n} \left|1+\frac{2}{n}\right| \qquad \text{ (factoring/potential problem)}
\end{align}$$
Helper Assumption:
If $n > 1$ ((ie) $ N \geq 1$)
$$\text{Note 1: } \max\left(\frac{6}{\epsilon}, 1\right) = \frac{1}{6} \implies \frac{1}{n} < 1$$
$$\begin{align}
\text{Note 2: } \max\left(\frac{6}{\epsilon}, 1\right)=1 &\implies\frac{2}{n}<2\\
&\implies 1 + \frac{2}{n} < 3
\end{align}$$
Thus:
$$|a_n-1|\leq \frac{6}{n} < \frac{6}{N} = \epsilon$$
This is the entire proof, I get everything until we get to the potential problem / helper assumption. Could someone explain, or give a easier way because I thought this question could have been much easier.
| Here's a much easier solution:
We have
$$1 > \frac{n^2 - 2}{n^2 + 2n + 2} > \frac{n^2 - 4}{n^2 + 4n + 4} = \frac{(n - 2)(n + 2)}{(n + 2)^2} = \frac{n - 2}{n + 2},$$
thus
$$\left\vert \frac{n^2 - 2}{n^2 + 2n + 2} - 1 \right\vert < \left\vert \frac{n - 2}{n + 2} - 1\right\vert,$$
so it suffices to show that
$$\lim_{n\to\infty}\frac{n-2}{n+2} = 1.$$
Pick $\epsilon > 0$ and choose $N > \tfrac{4}{\epsilon}$. Then if $n > N$, we have
$$\left\vert \frac{n - 2}{n + 2} - 1\right\vert = \left\vert \frac{-4}{n + 2} \right\vert < \frac{4}{n} < \epsilon.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximize $P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}$ For $x,y,z$ are positive real numbers that satisfy $xy+yz+xz=1$. Maximize $$P=\frac{1}{\sqrt{1+x^{2}}}+\frac{1}{\sqrt{1+y^{2}}}+\frac{1}{\sqrt{1+z^{2}}}.$$
I think if we let $x=\tan A;y=\tan B;z=\tan C$, then
$$P\Leftrightarrow \displaystyle \text {sin}A+\text {sin}B+ \text{sin}C\leq \frac{3}{2}. \hspace{2cm}(1)$$
But I can't prove $(1)$.
| For $x=y=z=\frac{1}{\sqrt3}$ we get a value $\frac{3\sqrt3}{2}$.
We'll prove that it's a maximal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{1}{\sqrt{x^2+xy+xz+yz}}\leq\frac{3\sqrt3}{2\sqrt{xy+xz+yz}}$$ or
$$\sum_{cyc}\sqrt{x+y}\leq\frac{3\sqrt{3(x+y)(x+z)(y+z)}}{2\sqrt{xy+xz+yz}}.$$
But by C-S $$\left(\sum_{cyc}\sqrt{x+y}\right)^2\leq(1+1+1)\sum_{cyc}(x+y)=6(x+y+z).$$
Thus, it remains to prove that
$$\sqrt{6(x+y+z)}\leq\frac{3\sqrt{3(x+y)(x+z)(y+z)}}{2\sqrt{xy+xz+yz}}$$ or
$$9(x+y)(x+z)(y+z)\geq8(x+y+z)(xy+xz+yz)$$ or
$$\sum_{cyc}z(x-y)^2\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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integrate $\int\frac{x\cdot dx}{(x^3+1)^2}$ What methods are there to integrate: $$\int\frac{x\cdot dx}{(x^3+1)^2}$$
I know about partial fractions:
$$\int\frac{x\cdot dx}{(x^3+1)^2} $$
$$= \int\frac{x\cdot dx}{((x+1)(x^2-x+1))^2} $$
$$= \int \left(\frac{A}{x+1}+\frac{Bx+C}{(x+1)^2} + \frac{Dx+E}{x^2-x+1} + \frac{Fx^3+Gx^2+H+I}{(x^2-x+1)^2}\right)dx$$ and after this solving is easy, i was trying to do the same many times, but i can't find coefficients because mistakes or something other.
I want to know about another methods to solve it.
| HINT:
First integrate by parts,
$$3I=\int\dfrac1x\cdot\dfrac{3x^2}{(1+x^3)^2}dx=-\dfrac1{x(1+x^3)}+\int\dfrac{dx}{x^2(1+x^3)}$$
Now use Partial fraction $$\dfrac1{x^2(1+x^3)}=\dfrac Ax+\dfrac B{x^2}+\dfrac C{x+1}+\dfrac{Dx+E}{x^2-x+1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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The remainder when $33333\ldots$ ($33$ times) is divided by $19$ $A= 33333\ldots$ ($33$ times). What is the remainder when $A$ is divided by $19$?
I don't know the divisibility rule of $19.$
What I did was
$32\times(33333\times100000)/19$ and my remainder is not zero and this is completely divisible by $19.$
This is a gmat exam question.
| Since $19$ is prime, the first $18$ powers of $10$ (starting with $10^0$) will produce all of the remainders $1, 2, 3, \ldots, 18$ (although not in that order) when each power is divided by $19.$ Hence the sum of these $18$ powers of $10,$ namely $111111111111111111,$ satisfies
\begin{align}
111111111111111111 &\equiv 1 + 2 + 3 + \cdots + 18 \\
&\equiv (1 + 18) + (2 + 17) + \cdots + (9 + 10) \\
&\equiv 0 + 0 + \cdots + 0 \\
&\equiv 0 \pmod{19}.
\end{align}
So $111111111111111111$ is divisible by $19,$ and also any multiple of
$111111111111111111$
(such as $333333333333333333 \times 10^n$ for any non-negative integer $n$)
is also divisible by $19.$
It follows that if a number's decimal representation consists of $165$ digits each of which is $3,$ the first $9 \times 18 = 162$ digits are a number divisible by $19.$ So we just have to deal with the last three digits,
that is, what is the remainder when $333$ is divided by $19$?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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exponent problem
Can anyone answer this question? $$3^{n+2} + (3^{n+3} - 3^{n+1})= ?$$
I really want to know how to solve this one, our solution can't seem to agree with the problem's answer thank you!
| If the answer is $3/8$ probably the question is:
$$\frac{3^{n+2}}{3^{n+3}-3^{n+1}}=\frac{3^2\cdot3^n}{3^3\cdot3^{n}-3\cdot3^{n}}=\frac{3^2\cdot3^n}{3^n(3^3-3)}=\frac{3^2}{27-3}=\frac{3}{8}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the expression of a power series. So I have the infinite sum:
$$\sum_{n=1}^\infty (-1)^{n-1} nx^{2n}$$
I have found in a previous part that the convergence interval is $(-1,1)$.
I know that the sum should equal
$$\frac {x^2}{{(x^2 +1)}^2}$$
when $|x|\lt1$
But I'm not sure how I can show this from the sum.
| \begin{align}
\sum_{n=1}^\infty (-1)^{n-1} nx^{2n} = x^2 - {}2x^4 + 3x^6 - 4x^8 + 5x^{10} - \cdots
\end{align}
$$
\begin{array}{cccccccccc}
n=1 & & n=2 & & n=3 & & n=4 & & n=5 & & \cdots \\[10pt] \hline
x^2 & + & (-x^4) & + & x^6 & + & (-x^8) & + & x^{10} & + & \cdots \\[10pt]
& & (-x^4) & + & x^6 & + & (-x^8) & + & x^{10} & + & \cdots \\[10pt]
& & & & x^6 & + & (-x^8) & + & x^{10} & + & \cdots \\[10pt]
& & & & & & (-x^8) & + & x^{10} & + & \cdots \\[10pt]
& & & & & & & & x^{10} & + & \cdots \\[10pt]
& & & & & & & & & & \cdots
\end{array}
$$
First find the sum of each row separately. That's easy because each one is a geometric series.
Then find the sum of all those resulting sums. That also easy because that is also a geometric series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2199006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove this Generalizing AM-GM inequality
Let $n\ge 2$ and $a_{i} \ge 0,i=1,2,\cdots,n$, show that
$$(n-1)^{n-1}(a^n_{1}+a^n_{2}+\cdots+a^n_{n})+n^na_{1}a_{2}\cdots a_{n}\ge (a_{1}+a_{2}+\cdots+a_{n})^n$$
When $n=2$,
$$a^2_{2}+a^2_{2}+4a_{1}a_{2}=(a_{1}+a_{2})^2+2a_{1}a_{2}\ge (a_{1}+a_{2})^2$$
When $n=3$, it is
$$4(a^3_{1}+a^3_{2}+a^3_{3})+27a_{1}a_{2}a_{3}\ge (a_{1}+a_{2}+a_{3})^3$$
By
$$(a_{1}+a_{2}+a_{3})^3=a^3_{1}+a^3_{2}+a^3_{3}+3a_{1}a_{2}(a_{1}+a_{2})+3a_{1}a_{3}(a_{1}+a_{3})+3a_{2}a_{3}(a_{2}+a_{3})+6a_{1}a_{2}a_{3}$$
so it's enough to prove
$$a^3_{1}+a^3_{2}+a^3_{3}+7a_{1}a_{2}a_{3}\ge a_{1}a_{2}(a_{1}+a_{2})+a_{1}a_{3}(a_{1}+a_{3})+a_{2}a_{3}(a_{2}+a_{3})$$
which is clear by using Schur inequality:
$$a^3+b^3+c^3+3abc\ge ab(a+b)+bc(b+c)+ac(a+c)$$
| Maybe this can help you, it seems very similar.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Asymptotics of $\Sigma x^n/n^p$ I am looking for asymptotics of
$$\sum_{n=1}^N \dfrac{x^n}{n^p}$$
in terms of $N$, where $x>1$ and $p$ is a positive number.
So I am looking for a $O(f(N))$ type result. For example what is the approximate value of
$\sum_{n=1}^N \dfrac{5^n}{n^2}$ for large $N$. One possibility is probably something like $\frac{5^N}{N^2} (a+\frac{b}{N}+\frac{c}{N^2}+\cdots)$.
| We find
\begin{align*}
\sum_{n=1}^N \frac{x^n}{n^p}
&= \sum_{n=1}^\infty \frac{x^n}{n^p}
- \sum_{n=N+1}^\infty \frac{x^n}{n^p} \\
&= \sum_{n=1}^\infty \frac{x^n}{n^p} - x^{N+1}\sum_{k=0}^\infty \frac{x^k}{(k+N+1)^p}
\qquad (\textrm{let } n=N+1+k) \\
&= \mathrm{Li}_p(x) - x^{N+1}\Phi(x,p,N+1)
\qquad \textrm{(polylogarithm and Lerch transcendent)} \\
&\sim \frac{x^{N+1}}{x-1} \frac{1}{(N+1)^p}.
\end{align*}
In the last step we use the known asymptotics for the Lerch transcendent. Higher order corrections can be found.
See, for example,
Chelo Ferreira, José L. López, Asymptotic expansions of the Hurwitz–Lerch zeta function, Journal of Mathematical Analysis and Applications, Volume 298, Issue 1, 2004, Pages 210-224, ISSN 0022-247X, http://dx.doi.org/10.1016/j.jmaa.2004.05.040.
Addendum
As @martycohen notes, the expansion in the paper is for $x\in\mathbb{C}-[1,\infty)$.
It should be possible to make a complex analytic argument that the asymptotic expansion above is valid for $x>1$.
Rather than do that, instead I will present a more elementary approach.
Let $1<k<N$. Then
$$\sum_{n=1}^N\frac{x^n}{n^p}
= \sum_{n=N-k}^N\frac{x^n}{n^p}
+ \sum_{n=1}^{N-k-1}\frac{x^n}{n^p}.$$
But
$$\sum_{n=1}^{N-k-1}\frac{x^n}{n^p}
< \sum_{n=1}^{N-k-1}x^n
= \frac{x^{N-k}-x}{x-1}$$
and
$$\sum_{n=1}^{N-k-1}\frac{x^n}{n^p}
> \frac{1}{N^p}\sum_{n=1}^{N-k-1}x^n
= \frac{1}{N^p}\frac{x^{N-k}-x}{x-1}$$
and so
$$\sum_{n=1}^{N-k-1}\frac{x^n}{n^p} = O(x^{N-k}).$$
Therefore,
$$\sum_{n=1}^N\frac{x^n}{n^p}
= \sum_{n=N-k+1}^N\frac{x^n}{n^p} + O(x^{N-k}).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2201643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
If $999\times \mathrm{abc}=\mathrm{def123}$ in decimal system, then find $\mathrm{a,b,c,d,e\ and\ f}.$ Consider the following multiplication in decimal system:
$$999\times \mathrm{abc}=\mathrm{def123}$$
then find the value of digits $\mathrm{a,b,c,d,e\ and\ f}.$
Here $\mathrm{abc}$ means not $(a\times b\times c$), $\mathrm{abc}$ is a number of 3 digits decimal system (e.g. if $\mathrm{abc}=123$, then $
\mathrm{a=1,\ b=2,\ c=3}$).
| Using the expansion of a natural number into powers of $10$ your equation becomes
\begin{equation*}
999\times 10^2a+999\times 10b+999c=10^5d+10^4e+10^3f+123.\tag{1}
\end{equation*}
This implies that $c=7$ because the final digit of the right-hand side of $(1)$ is $3 $, so should be the left-hand side's; and since we have $9(7)=63$, the unit's digit of $999c $ shall equal $3 $. Plugging this value into $(1)$, simplifying and dividing both sides by $10$, we obtain
\begin{equation*}
999\times 10a+999b+687=10^4d+10^3e+10^2f,\tag{2}
\end{equation*}
which yields $b=7$ because now the final digit of both sides of $(2)$ must equal $0$, and since $9(7)+7=70$, the final digit of $999b+687 $ shall equal $0 $. Proceeding in a similar way to the above, we find that the single solution of $(1)$ is
$a=8$, $b=7$, $c=7$, $d=8$, $e=7$, $f=6$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Prove that $\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$ For $a,b,c$ are positive real number. Prove that
$$\sqrt{\frac{2}{a}}+\sqrt{\frac{2}{b}}+\sqrt{\frac{2}{c}}\le\sqrt{\frac{a+b}{ab}}+\sqrt{\frac{b+c}{bc}}+\sqrt{\frac{c+a}{ca}}$$
Let $\left(\dfrac{1}{a};\dfrac{1}{b};\dfrac{1}{c}\right)\rightarrow\left(x;y;z\right)$
We need prove $\sqrt{2x}+\sqrt{2y}+\sqrt{2z}\le\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}$
We have: $\left(\sqrt{x+y}+\sqrt{y+z}+\sqrt{z+x}\right)^2=2\left(x+y+z\right)+2\left[\sqrt{\left(x+y\right)\left(y+z\right)}+\sqrt{\left(y+z\right)\left(z+x\right)}+\sqrt{\left(z+x\right)\left(x+y\right)}\right]$
By C-S we have: $\sqrt{\left(x+y\right)\left(y+z\right)}\ge\sqrt{xy}+\sqrt{yz}$
$\Rightarrow 2\sum\sqrt{\left(x+y\right)\left(y+z\right)}\ge4\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)$
$\Rightarrow LHS^2\ge 2(x+y+z+2\sqrt {xy}+2\sqrt {yz}+2\sqrt {xz})=2(\sqrt{x}+\sqrt{y}+\sqrt{z})=RHS^2$
Can do other way ?
| Hint: prove the following first (e.g. by squaring), then add up all the combinations thereof:
$$
\sqrt{\dfrac{1}{a}}+\sqrt{\dfrac{1}{b}} \le \sqrt{2\left(\dfrac{1}{a}+\dfrac{1}{b}\right)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2203382",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Given is a linear mapping and a basis. Determine the transformation matrix
Given is $f: \mathbb{R}^{3} \rightarrow \mathbb{R}^{3}$ with
$f:(x,y,z) \rightarrow (x+2y+z, y+z, -x+3y+4z)$. Determine the
transformation matrix in terms of the basis $B= \left\{\begin{pmatrix}
1\\ 0\\ 0 \end{pmatrix}, \begin{pmatrix} 0\\ 1\\ 1
\end{pmatrix},\begin{pmatrix}
-1\\ 0\\ 1 \end{pmatrix}\right\}$
$f(\begin{pmatrix}
1\\
0\\
0
\end{pmatrix})= \begin{pmatrix}
1\\
0\\
-1
\end{pmatrix}= 0 \cdot \begin{pmatrix}
1\\
0\\
0
\end{pmatrix} + 0 \cdot \begin{pmatrix}
0\\
1\\
1
\end{pmatrix}- 1 \cdot \begin{pmatrix}
-1\\
0\\
1
\end{pmatrix}$
$f(\begin{pmatrix}
0\\
1\\
1
\end{pmatrix})= \begin{pmatrix}
3\\
2\\
7
\end{pmatrix}= 8 \cdot \begin{pmatrix}
1\\
0\\
0
\end{pmatrix} + 2 \cdot \begin{pmatrix}
0\\
1\\
1
\end{pmatrix}+5 \cdot \begin{pmatrix}
-1\\
0\\
1
\end{pmatrix}$
$f(\begin{pmatrix}
-1\\
0\\
1
\end{pmatrix})= \begin{pmatrix}
0\\
1\\
5
\end{pmatrix}= 4 \cdot \begin{pmatrix}
1\\
0\\
0
\end{pmatrix} + 1 \cdot \begin{pmatrix}
0\\
1\\
1
\end{pmatrix}+ 4 \cdot \begin{pmatrix}
-1\\
0\\
1
\end{pmatrix}$
I'm not sure how to read the transformation matrix now. Either I will read it correctly or I will accidentally read its transposition :s
$T=\begin{pmatrix}
0 & 8 & 4\\
0 & 2 & 1\\
-1 & 5 & 4
\end{pmatrix}$
Did I do it all correctly? If it's alright, there are maybe faster ways doing this?
| Perhaps a more straightforward way of computing the matrix is by using the relation
\begin{equation} [T]_\mathcal{B} = P^{-1} T P, \end{equation}
where $[T]_\mathcal{B}$ is the matrix that you computed,
$$P = (b_1\quad b_2\quad b_3) = \begin{pmatrix} 1 & 0 & -1 \\
0 & 1 & 0 \\ 0 & 1 & 1\end{pmatrix}, $$
and $$T = \begin{pmatrix} f(e_1) & f(e_2) & f(e_3) \end{pmatrix}
= \begin{pmatrix}1 & 2 & 1\\ 0 & 1 & 1 \\ -1 & 3 & 4 \end{pmatrix},$$
where $e_1, e_2, e_3$ are the standard basis vectors. With this approach, all you have to do is invert $P$, which is easy, and then multiply the matrices. (check to see that it yields the same result!)
Heres a nice diagram that illustrates why the relation is true, from
Linear Algebra Basis Trick.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
To find factor of a polynomial equation
One of the factors of $4x^2+y^2+14x-7y-4xy+12$ is equal to
*
*$2x-y+4$
*$2x-y-3$
*$2x+y-4$
*$2x-y+3$
Step $1$:
$4x^2+y^2-4xy$ can be simplified as $(2x-y)^2$
Step $2$:
$14x-7y$ can be simplified as $7(2x-y)$
and finally
$(2x-y) (2x-y+7) + 12$
I can able to factor to this extent only. however can't able to arrive at the answer.
The answer is given in the book. it states that $4x^2+y^2+14x-7y-4xy+12$ is product of $(2x-y+3)$ and $(2x-y+4)$ I am in need of steps
| We can write the given equation as the product of two smaller polynomials:
\begin{align}
\rlap{4x^2+y^2+14x−7y−4xy+12} \\
&= (ax+by+c)(dx+ey+f) \\
&= adx^2+aexy+afx+bdxy+bey^2+bfy+cdx+cey+cf \\
&= \rlap{adx^2+bey^2+(af+cd)x+(bf+ce)y+(ae+bd)xy+cf}
\end{align}
We can equate coefficients to get the following:
\begin{align}4 &= ad \tag{$x^2$ term} \\
1 &= be \tag{$y^2$ term} \\
14 &= af+cd \tag{$x$ term} \\
-7 &= bf+ce \tag{$y$ term} \\
-4 &= ae+bd \tag{$xy$ term} \\
12 &= cf \tag{constant term}\end{align}
We can solve these simultaneous equations to get
\begin{align}a &= 2 \\
b &= -1 \\
c &= 4 \\
d &= 2 \\
e &= -1 \\
f &= 3\end{align}
Therefore, we can say that $$4 x^2 + y^2 + 14 x − 7 y − 4 x y + 12 = (2 x - y + 4) (2 x - y + 3)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2205323",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
limit of an indetermined sequence I'm trying to find out the limits of a sequence of the type $\frac{\infty}{\infty}$, but I got stuck and am starting to get frustrated. The sequence in question is: $$\lim_{n \to \infty}\frac{\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1}}{\ln(1+n) - \ln(2+n)}$$ I tried looking at its parts and figuring out their own limits to try and help me, but I didn't get anywhere useful. I tried rationalizing the square roots and this is what I have at the moment: $$\lim_{n \to \infty}\frac{\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1}}{\ln(1+n) - \ln(2+n)}\times \frac{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}}{\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1}} = \lim_{n \to \infty}\frac{(n^2+3n+1)-(n^2+3n-1)}{(\ln(1+n) - \ln(2+n))(\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1})}=\lim_{n \to \infty}\frac{2}{{(\ln(1+n) - \ln(2+n))(\sqrt{n^2+3n+1} + \sqrt{n^2+3n-1})}}$$
| I get the limit to be $-1$.
Note that $\frac{\sqrt{n^2+3n+1} - \sqrt{n^2+3n-1}}{\ln(1+n) - \ln(2+n)} =
\frac{ n^2( \sqrt{1+{3 \over n} + {1 \over n^2}} - \sqrt{1+{3 \over n} - {1 \over n^2}})}{n\ln(\frac{1+{1 \over n}}{1 + {2 \over n}})}$.
We have $\lim_{x \to 0} {1 \over x} \ln(\frac{1+x}{1 + 2x}) = -1 $ and
$\lim_{x \to 0} {{ \sqrt{1+{3 x} + x^2} - \sqrt{1+{3 x} - x^2}} \over x^2} = 1$.
To compute the first limit, let $f_0(x) = \ln(\frac{1+x}{1 + 2x})$
and note that $\lim_{x \to 0} {1 \over x} \ln(\frac{1+x}{1 + 2x}) = f_0'(0) = -1$.
To compute the latter limit, let $f_1(x) = \sqrt{1+{3 x} + x^2}-\sqrt{1+{3 x} - x^2}$, we have $f_1'(0) = 0, f_1''(0) = 2$, hence
$\lim_{x \to 0} {{ \sqrt{1+{3 x} + x^2} - \sqrt{1+{3 x} - x^2}} \over x^2} = {f_1''(0) \over 2} = 1$.
Hence the limit is $-1$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Want less brutish proof: if $a+b+c=3abc$ then $\frac1a+\frac1b+\frac1c\geq 3$ In the course of showing for positive $a,b,c\,$ that $a+b+c=3abc$ implies $$\frac1{2a+b} + \frac1{2b+c} + \frac1{2c+a} \geq 1$$ I needed to show that (given that same constraint)
$$
\frac1a+\frac1b+\frac1c\geq 3$$
I have a proof which is human-readable but less elegant than I would like:
$$
3\left( ab-\frac12(a+b)\right)^2 +\frac14(a-b)^2 \geq 0
$$
rearranges to
$$
-3ab(a+b) + (a+b)^2 + \left[ 3a^2b^2-ab\right] \geq 0
$$
and dividing by the positie quantity $ab(a+b)$ this becomes
$$
\frac1a + \frac1b + \frac{3ab-1}{a+b} \geq 3
$$
and the third term on the right is, by the constraint, equal to $\frac1c$.
What I dislike is that to find that first expression, I used a Buffalo-like technique: I solved the constraint for $c$ in terms of $a$ and $b$, substituted in the inequality, multiplied through by the product of denominators $ab(a+b)$, found the first term in the sum of squares by demanding that the $a^2b^2$ and $(a^2b+b^2a)$ terms be covered.
By some lucky chance the remaining terms were a recognizable square (although the Buffalo Way could also handle remaining terms with all positive coefficients).
It seems to me that there should be a better way of showing this, perhaps using Jensen's inequality, valid for convex upward functions $f(t)$:
$$
f\left(\frac{\sum w_ix_i}{\sum w_i} \right)\leq \frac{\sum w_if(x_i)}{\sum w_i}
$$
The usual trick is to use some simple $f(t)$ like $t^{-2}$ and choose the weights and the $x_I$ to make the numerator on the right hand side come out to the business end of the inequality (and the other sums expressions that can be handled using the given constraint). But I can't find the way to do that here.
Can anybody find a proof that is a bit less ad hoc in its starting point?
ADDED LATER
After seeing the excellent proofs offered by Arnaldo and Michael, I realized that although this was nice to prove, it in fact does not help me prove that
$$\frac1{2a+b} + \frac1{2b+c} + \frac1{2c+a} \geq 1$$
Oh well...
| I will use that
$$a+b+c=3abc \Leftrightarrow \frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}=3$$
By AM-GM
$$\left(\frac{1}{a}+\frac{1}{b}\right)^2\ge\frac{4}{ab}\\
\left(\frac{1}{a}+\frac{1}{c}\right)^2\ge\frac{4}{ac}\\
\left(\frac{1}{b}+\frac{1}{c}\right)^2\ge\frac{4}{bc}$$
so
$$\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2\ge 4\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)=12\quad (1)$$
but $x^2+y^2+z^2=(x+y+z)^2-2(xy+xz+yz)$, so
$$\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=4\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+9\right)\quad (2)$$
and also,
$$\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-2\left(\frac{1}{ab}+\frac{1}{ab}+\frac{1}{bc}\right)=\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-6\quad (3)$$
Now use $(3)$ in $(2)$ and get:
$$\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{a}+\frac{1}{c}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-6 \quad (4)$$
Now put $(4)$ in $(1)$:
$$2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2-6\ge 12\to \frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge 3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2207571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
proving $\cos (A+B)>0$, if given angles $A$ and $B$ If $\displaystyle A=3\sin^{-1}\left(\frac{6}{11}\right)$ and $\displaystyle B = 3\cos^{-1}\left(\frac{4}{9}\right),$ then proving $\cos (A+B)>0$
Attempt: $$ 3\sin^{-1}\left(\frac{1}{2}\right)<3\sin^{-1}\left(\frac{6}{11}\right)<3\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)\Rightarrow \frac{\pi}{2}<A<\pi$$
same way $$\displaystyle 3\cos^{-1}\left(0\right)<3\cos^{-1}\left(\frac{4}{9}\right)<3\cos^{-1}\left(\frac{1}{2}\right)\Rightarrow \frac{\pi}{2}<B<\frac{3\pi}{2}$$
so $$\pi<A+B<\frac{5\pi}{2}$$
could some help me how to prove $\cos (A+B)>0,$ thanks
| Let $\displaystyle \alpha=\sin^{-1}\left(\frac{6}{11}\right)=\frac{A}{3}$ and let $\displaystyle\beta=\cos^{-1}\left(\frac{4}{9}\right)=\frac{B}{3}$
Then since $\frac{1}{2}<\frac{6}{11}<\frac{\sqrt{2}}{2}$ we have $\frac{\pi}{6}<\alpha<\frac{\pi}{4}$.
And since $0<\frac{4}{9}<\frac{1}{2}$ we have $\frac{\pi}{3}<\beta<\frac{\pi}{2}$.
Addition of the two inequalities gives $\frac{\pi}{2}<\alpha+\beta<\frac{3\pi}{4}$.
Multiplication by $3$ gives $\frac{3\pi}{2}<3\alpha+3\beta<\frac{9\pi}{4}$.
Thus $\frac{3\pi}{2}<A+B<\frac{9\pi}{4}$ meaning that $A+B$ lies in either quadrant IV or quadrant I.
Thus $\cos(A+B)>0$.
What has been shown here is that, referring to the following diagram, for any standard angle $A$ such that the terminal side of $\dfrac{A}{3}$ lies in the blue region and for any standard angle $B$ such that the terminal side of $\dfrac{B}{3}$ lies in the green region it will be the case that $\cos(A+B)>0$. The "red" lines are the terminal sides of the particular $\dfrac{A}{3}$ and $\dfrac{B}{3}$ given in the question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2208004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Exponential map of Beltrami-Klein model of hyperbolic geometry In the Betrami-Klein model of hyperbolic geometry, geodesics are represented as straight lines. Hence the exponential map of a tangent vector $\mathbf{v}$ at a point $\mathbf{p}$ is $\mathbf{p} + \lambda \mathbf{v}$, where $\lambda$ is a scalar that depends on $\mathbf{p}$ and $\mathbf{v}$. For example, suppose $\mathbf{p} = 0$. Then the exponential map is
$$ \exp_\mathbf{p}(\mathbf{v}) = \hat{\mathbf{v}} \tanh \lvert \mathbf{v} \rvert $$
Notice that $\tanh \lvert \mathbf{v} \rvert < 1$ always, since we cannot reach the circle at infinity with any finite $\mathbf{v}$.
Now consider the diagram below:
where $\mathbf{q} = \mathbf{p} + \lambda \mathbf{v}$. The hyperbolic distance between $\mathbf{p}$ and $\mathbf{q}$ must be equal to the length of the tangent vector, that is, $d(\mathbf{p}, \mathbf{q}) = \lvert \mathbf{v} \rvert$. This hyperbolic distance is
$$ d(\mathbf{p}, \mathbf{q}) = \frac{1}{2} \log \frac{\lvert\mathbf{p} - \mathbf{r_1}\rvert \lvert\mathbf{q} - \mathbf{r_2}\rvert}{\lvert\mathbf{p} - \mathbf{r_2}\rvert \lvert\mathbf{q} - \mathbf{r_1}\rvert}$$
where $\mathbf{r_1}$ and $\mathbf{r_2}$ are the two ideal points intersected by the straight line connecting $\mathbf{p}$ and $\mathbf{q}$. We can find these ideal points as follows:
\begin{align}
1 &= \mathbf{r} \cdot \mathbf{r} \\
&= (\mathbf{p} + \mu \mathbf{v}) \cdot (\mathbf{p} + \mu \mathbf{v}) \\
&= \mathbf{p} \cdot \mathbf{p} + 2 \mu \mathbf{p} \cdot \mathbf{v} + \mu^2 \mathbf{v} \cdot \mathbf{v}
\end{align}
Hence
\begin{align}
\mu &= \frac{-2 \mathbf{p} \cdot \mathbf{v} \pm \sqrt{(2 \mathbf{p} \cdot \mathbf{v})^2 - 4 (\mathbf{v} \cdot \mathbf{v})(\mathbf{p} \cdot \mathbf{p} - 1)}}{2 \mathbf{v} \cdot \mathbf{v}} \\
&= \frac{-\mathbf{p} \cdot \mathbf{v} \pm \sqrt{(\mathbf{p} \cdot \mathbf{v})^2 - (\mathbf{v} \cdot \mathbf{v}) (\mathbf{p} \cdot \mathbf{p} - 1)}}{\mathbf{v} \cdot \mathbf{v}}
\end{align}
so that $\mathbf{r} = \mathbf{p} + \mu \mathbf{v}$ for each root. Hence we have
\begin{align}
\lvert \mathbf{v} \rvert &= \frac{1}{2} \log \frac{\lvert\mathbf{p} - (\mathbf{p} + \mu_1 \mathbf{v})\rvert \lvert(\mathbf{p} + \lambda \mathbf{v}) - (\mathbf{p} + \mu_2 \mathbf{v})\rvert}{\lvert\mathbf{p} - (\mathbf{p} + \mu_2 \mathbf{v})\rvert \lvert(\mathbf{p} + \lambda \mathbf{v}) - (\mathbf{p} + \mu_1 \mathbf{v})\rvert}
\\
&= \frac{1}{2} \log \frac{\lvert\mu_1 \mathbf{v}\rvert \lvert(\lambda - \mu_2) \mathbf{v}\rvert}{\lvert\mu_2 \mathbf{v}\rvert \lvert(\lambda - \mu_1) \mathbf{v}\rvert}
\\
&= \frac{1}{2} \log \frac{\lvert\mu_1\rvert \lvert\lambda - \mu_2\rvert}{\lvert\mu_2\rvert \lvert\lambda - \mu_1\rvert}
\end{align}
or equivalently
\begin{align}
\left\lvert \frac{\lambda - \mu_2}{\lambda - \mu_1} \right\rvert = \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert}
\\
\frac{\lambda - \mu_2}{\lambda - \mu_1} = \pm \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert}
\\
\lambda - \mu_2 = \pm \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert} (\lambda - \mu_1)
\\
\lambda \left( 1 \mp \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert} \right) = \mu_2 \mp \mu_1 \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert}
\\
\lambda = \frac{\mu_2 \mp \mu_1 \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert}}{1 \mp \left\lvert \frac{\mu_2}{\mu_1} \right\rvert \mathrm{e}^{2 \lvert \mathbf{v} \rvert}}
\end{align}
We pick the root $\lambda > 0$. Is this result correct? Can it be simplified? Can anyone find a source for it?
| The hyperbolic plane doesn't have a center. However, the Klein model does. Because of the isotropy of the hyperbolic plane we can do our calculations from the center of the model. All the results will be the same. Let $p=(0,0)$ and let $\|v\|$ be arbitrary, and let $v$ be originated at $p$ (see the blue vector). $r=p+\lambda v$ is the red vector. We can replace the vectors by their lengths in this setup.
I would be surprised if the answer wouldn't be $\lambda=\pm1$.
EDITED
I guess, I understand the question now. Let $v$ be the Euclidean distance of a point from the origin in the Klein model. Also, let $\lambda$ be a real variable. We are looking for a $\lambda$ for which the hyperbolic distance of the point at $\lambda v$ equals $v$.
The hyperbolic distance (from the origin) of the point at $\lambda v$ is $\operatorname {atanh} (\lambda v)$. And we are looking for a $\lambda$ for which
$$\operatorname {atanh} (\lambda v)=v.$$
Take the $\operatorname {tanh}$ of both sides. We have now
$$\lambda =\pm\frac{\operatorname {tanh}(v)}v.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Use substitution of $x =\tan\theta$ to show that Use the substitution $x = \tan\theta$ to show that $$\int\frac {1 - x^2} {(1+x^2)^2} dx = \int \cos2\theta\ d\theta $$
I'm a bit lost on how to handle this question, I have tried subtituting $d\theta/dx = 1 / \sec^2\theta$ but I still don't reach the answer.
| I have done with your substitution yet it is so boring
$$\int \frac { 1-x^{ 2 } }{ (1+x^{ 2 }) } dx=\int { \frac { 1-\tan ^{ 2 }{ \theta } }{ 1+\tan ^{ 2 }{ \theta } } \frac { 1 }{ \cos ^{ 2 }{ \theta } } d\theta =\int { \left( 1-\tan ^{ 2 }{ \theta } \right) d\theta =\int { \frac { \cos ^{ 2 }{ \theta -\sin ^{ 2 }{ \theta } } }{ \cos ^{ 2 }{ \theta } } } d\theta =\int { \frac { \cos { 2\theta } }{ \cos ^{ 2 }{ \theta } } d\theta } = } } \\ =\int { \cos { 2\theta \quad d\left( \tan { \theta } \right) } } $$
the last part we can evaluate by parts
$$\\ \\ \int { \cos { 2\theta \quad d\left( \tan { \theta } \right) } } =\cos { 2\theta \tan { \theta +2\int { \sin { 2\theta } \cdot \tan { \theta = } } } } \cos { 2\theta \tan { \theta +4\int { \sin ^{ 2 }{ \theta d\theta } } } } =\cos { 2\theta \tan { \theta +2\int { \left( 1-\cos { 2\theta } \right) d\theta } = } } \\ =\cos { 2\theta \tan { \theta +2\left( \theta -\frac { \sin { 2\theta } }{ 2 } \right) } } =\cos { 2\theta \tan { \theta +2\theta -\sin { 2\theta +C } } } $$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the Rate of change of f (x^2) given rate of change of f (x) Find the average rate of change of $f(x^2)$ on the interval [1,4] given that the average rate of change of $f (x)$ equals 9
on interval [1,16]?
This question has two different "answers" according to two different teachers
The first give answer of 9 by assuming $y=x^2$ and applying the formula of rate of change as following
Let$ y=x^2 $ with $I$= [1,4] then$ f (1)$=1,$f (4)$=16
$\frac{f(4^2) - f(1^2)}{(4^2 -1^2)}$ =
$\frac{f (y_{2})-f (y_{1})}{y_{2}-y_{1}}$
=
$\frac{f(16) - f(1)}{(16-1)}$ =9
The second give the answer of 45
as following
$\frac{f(b) - f(a)}{(b - a)}$ to $\frac{f(16) - f(1)}{16 - 1}=9$ to
$\frac{f(16) - f(1)}{15}=9$
$f(16) - f(1)=9*15$
Now we find rate of change of $f (x^2)$ following
$\frac{f(16) - f(1)}{4 - 1}=\frac{9*15}{3}=45$
what is
the right answer?
Can we ensure the either answers geometrically?
Thank you for helping
| Let $g(x) = f(x^2)$.
The average rate of change of a function $f$ on an interval $[a,b]$ is $\dfrac{f(b) - f(a)}{b-a}$.
The average rate of change of $f$ on the interval $[1,16]$ is $$\frac{f(16) - f(1)}{16 - 1} = \frac{f(16) - f(1)}{15}.$$
You are told this rate of change equals $9$.
The average rate of change of $g$ on the interval $[1,4]$ is thus equal to $$ \frac{g(4) - g(1)}{4-1} = \frac{f(16) - f(1)}{3} = 5 \cdot \frac{f(16) - f(1)}{15} = 45.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is every integer odd $p$ not divisible by $3$ a sum of difference of powers of $2$ and $3$ or vice versa? Does every odd integer $p$ not divisible by $3$ have the form
$2^a+3^b$, $2^a-3^b$, and or $3^b-2^a$ for integers $a$ and $b$?
Please show proofs if possible. Thanks for help.
| It can be verified by a finite if tedious computation that none of the equations
\begin{align}
2^a + 3^b \equiv 133 \pmod {683} \\
2^a - 3^b \equiv 133 \pmod {683} \\
3^b - 2^a \equiv 133 \pmod {683}
\end{align}
have a solution $(a,b)$, and therefore $133$ cannot be written as sum or difference of a power of $2$ and a power of $3$.
(I chose $683$ because both $2$ and $3$ have relatively small multiplicative orders modulo $683$: $2^{22} \equiv 3^{31} \equiv 1 \pmod{683}$. Then I tested the above congruences by brute force for $0 \le a < 22$ and $0 \le b < 31$. I tried this for several primes, but $683$ was the first that worked.)
| {
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If $a^3+a^2+a=9b^3+b^2+b$ and $a,b$ are integers then show $a-b$ is a perfect cube. If $a^3+a^2+a=9b^3+b^2+b$ and $a,b$ are integers then show $a-b$ is a perfect cube.
My attempt:I factorized it like below:
$(a-b)(a^2+b^2+ab+a+b+1)=8b^3=(2b)^3$
I take $gcd(a-b,a^2+b^2+ab+a+b+1)=d$
If $d=1$ then it is clear that $a-b$ is a perfect cube then consider $d>1$ then there is a $p$ that is prime and $p \mid d$.We have :
$p\mid a-b \Rightarrow p \mid (2b)^3 \Rightarrow p \mid 2b \Rightarrow p\mid 2$ or $p\mid b$
If $p\mid b$ then also $p\mid a$ (as $p\mid a-b$ holds). Then we will get to $p \mid 1$ because:
$ p \mid a^2+b^2+ab+a+b,p \mid a^2+b^2+ab+a+b+1 \Rightarrow p \mid 1$
which is clearly wrong then we have $p\mid 2$ so $p=2$ means $d=2^k$ where $k$ is a natural number including $0$.In the case $d=1$ we have the right result.So assemble $k \ge 1$.Because $2 \mid a-b$ we can conclude that $a,b$ have the same parity @Ghartal showed in his answer that if $a,b$ are both even we have a right result but if $a,b$ are both odd we don't.So maybe we have to prove $a,b$ can,t be both odd.
| Revised proof:
We have $(a-b)(a^2 + b^2 + ab + a + b + 1) = (2b)^3$
From OP, we see that if $p$ divides $d$ then $p$ divides $2b \implies p \mid 2$ or $p \mid b$.
Assume $p$ doesn't divide $2$.
$p \mid b \implies p \mid a$ but then $p$ does not divide $a^2+b^2+ab +a+b+1$.
Therefore $p=1$ or $p=2$. If $p=1$ then $a-b$ is a cube.
If $p=2$ then since $p$ does not divide $b$ both $a$ and $b$ are odd.
Now $a^2 + b^2 + ab + a + b + 1 \ne 0 \mod 4$. Therefore largest power
of $2$ in $a^2 + b^2 + ab + a + b + 1$ is $2$.
Largest power of $2$ in $a-b$ should be $4$ (as $b$ is odd)
So one possibility is $a-b = 4m^3$ and $a^2 + b^2 + ab + a + b + 1 = 2n^3$ where $b = mn$
I guess I am stuck at this point.
| {
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Trig identities - stuck solving $\tan^2\theta = -\frac 32 \sec\theta$ Solve the equation on the interval $0\leq \theta < 2\pi$
$$\tan^2 \theta = -\frac{3}{2}\sec \theta $$
Here are the steps I have so far:
Identity: $\tan^2 \theta = \sec^2 \theta -1 $
Substitute:
$$\sec^2 \theta -1 = -\frac{3}{2}\sec \theta $$
$$2\sec^2 \theta -2 = {-3}\sec \theta $$
$$2\sec^2 \theta +3\sec \theta - 2 = 0 $$
Is this factoring correct?:
$$(2\sec\theta+4)(\sec\theta-1) = 0 $$
$$2\sec\theta+4 =0$$
$$2\sec\theta = -4 $$
$$\sec \theta = -2$$
$$(2\pi/3), (4\pi/3) $$
$$\sec\theta - 1 = 0$$
$$\sec\theta = 1 $$
$\sec\theta=1$ would evaluate to $0$, for some reason that is an invalid answer? (according to my assignment)
So is $\ (2\pi/3), (4\pi/3) $ the full answer?
| No, the factoring is not correct - it should be $\left(2\sec\theta +4\right)\left(\sec\theta- \frac12\right)$
| {
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