Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
What's the limit of the string?
Find $$\lim_{n\to\infty} (x_n\sqrt{n})^{\sqrt{n^2-1}},$$ where $ x_{n+1} = \frac{x_n}{\sqrt{1+x_n^2}}$ and $x_1 = 2$.
I showed $x_n \to 0$, $x_n\sqrt{n} \to 1$, but i don't know how to solve limit properly.
| Later EDIT: In the initial post there was an error, the old computation was using $y_0=2$, instead of $y_{\color{red}1}={\color{red}4}$.
Let the sequence $(y_n)$ be given by $y_n=x_n^2$. Then we have the Möbius transformation action giving the recursion:
$$
y_n =\begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}\cdot y_{n-1}\ .
$$
Here, the action of the $2\times 2$-matrix with entries $a,b,c,d$ on an element in $z\in \Bbb R$ (or $\Bbb C$) is given by
$$
\begin{bmatrix} a & b\\ c&d \end{bmatrix}\cdot z
=\frac {az+b}{cz+d}\ .
$$
This gives
$$
\begin{aligned}
y_n &=
\begin{bmatrix} 1 & 0\\ 1 & 1 \end{bmatrix}^{n-1}\cdot y_1
=
\begin{bmatrix} 1 & 0\\ n-1 & 1 \end{bmatrix}\cdot 4
\\
&=
\frac {4}{4n-3}\text{ . So we have:}
\\
x_n &=
\sqrt{\frac {4}{4n-3}}\ .
\end{aligned}
$$
(In case the argument using Möbius transformations is to heavy, just use induction to show the above formula.)
So we have to compute the limit of
$$
(x_n\sqrt{n})^{\sqrt{n^2-1}}
=
\left(\sqrt{\frac {4n}{4n-3}}\right)^{\sqrt{n^2-1}}
=
\left(1+\frac 3{4n-3}\right)^{\frac 12\sqrt{n^2-1}}
=
\left(1+\frac 3{4n-3}\right)^{(4n-3)\frac 1{2(4n-3)}\sqrt{n^2-1}}
=
\left[\ \left(1+\frac 3{4n-3}\right)^{(4n-3)}\ \right]^{\frac 1{2(4n-3)}\sqrt{n^2-1}}
\ .
$$
The limit is now easy to compute,
the $[\dots]$ expression goes to $e^3$, and its exponent to $\frac 18$, so
the limit is $\exp\frac 38$.
| {
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"timestamp": "2023-03-29T00:00:00",
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solve for real x and y: $x(x^2-3y^2)=2, y(3x^2-y^2)=11$ Solve for real x and y:
$x(x^2-3y^2)=2$
$y(3x^2-y^2)=11$
My attempt: I got $(x-y)^3=13$ but this doesn't always hold, I got a solution $(2,1)$.
How to proceed?
| NOTE: You made a mistake in getting $(x-y)^3=13$. Once you add the two equations, the left side is not $(x-y)^3$.
Idea:
\begin{align*}
\frac{x(x^2-3y^2)}{y(3x^2-y^2)}&=\frac{2}{11}.
\end{align*}
Let $y=mx$, then we have
\begin{align*}
\frac{(1-3m^2)}{m(3-m^2)}&=\frac{2}{11}.
\end{align*}
This gives
$$2m^3-33m^2-6m+11=0 \implies (2m-1)(m^2-16m-11)=0.$$
One of the solutions for this is $m=1/2$, now others can be found as well and they are $8 \pm 5\sqrt{3}$.
| {
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How to prove $x^2 + y^2 + z^2\geq xy + xz + yz$ Result: Let $, , ∈ ℝ$. Then we have $^2 + ^2 + ^2 ≥ + + $
Need some help proving this, just a few steps with work.
Was thinking you start with $x^2+y^2+z^2−xy−xz−yz$
then factor?
can anyone show me how to solve this?
$x^2+y^2+z^2≥xy+yz+zx⇔ $
$⇔2(x^2+y^2+z^2)≥2(xy+yz+zx)⇔$
$⇔x^2−2xy+y^2+y^2−2yz+z^2+z^2−2xz+x^2≥0⇔$
$⇔(x−y)^2+(y−z)^2+(z−x)^2≥0$
Should I do anything else? Or is this right?
| Your proof is right, but maybe it looks a bit of better by using a cyclic sum:
$$\sum_{cyc}(x^2-xy)=\frac{1}{2}\sum_{cyc}(2x^2-2xy)=\frac{1}{2}\sum_{cyc}(x^2-2xy+y^2)=\frac{1}{2}\sum_{cyc}(x-y)^2\geq0.$$
We can use also the following.
$$x^2+y^2+z^2-xy-xz-yz=x^2-(y+z)x+y^2-yz+z^2=$$
$$=\left(x-\frac{y+z}{2}\right)^2+y^2-yz+z^2-\frac{(y+z)^2}{4}=\left(x-\frac{y+z}{2}\right)^2+\frac{3(y-z)^2}{4}\geq0.$$
| {
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Find the value of $a_0a_1a_2\cdots a_n\left(\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)$
Given that the sequence $\left\{a_n\right\}$ satisfies $a_0 \ne 0,1$ and $$a_{n+1}=1-a_n(1-a_n)$$ $$a_1=1-a_0$$
Find the value of $$a_0a_1a_2\cdots a_n\left(\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}\right)$$
We actually get the terms of the sequence as:
$$a_1=1-a_0$$
$$a_2=1-a_1a_0$$
$$a_3=1-a_2a_1a_0$$
$$\vdots$$
$$a_{n+1}=1-a_na_{n-1}a_{n-2}\cdots a_0$$
Any way from here?
| For $n\ge 1$, we have $a_{n+1}-1=a_n(a_n-1)$. It follows that if $a_n\notin\{0,1\}$, then $a_{n+1}\notin\{0,1\}$, but since $a_0\notin\{0,1\}$, we see that $a_n\notin\{0,1\}$ for all $n$. Now $$\frac1{a_{n+1}-1}=\frac{1}{a_n(a_n-1)}=\frac{1}{a_n-1}-\frac{1}{a_n}$$
for all $n\ge 1$. That is
$$\frac1{a_n}=\frac{1}{a_n-1}-\frac{1}{a_{n+1}-1}$$
if $n\ge 1$, so
$$\frac1{a_1}+\frac{1}{a_2}+\ldots+\frac1{a_{n}}=\frac1{a_1-1}-\frac{1}{a_{n+1}-1}$$
Since $a_1=1-a_0$, we get $a_1-1=-a_0$. Thus $\frac{1}{a_1-1}=-\frac1{a_0}$ and
$$\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac1{a_{n}}=-\frac{1}{a_{n+1}-1}.$$
From your calculation, $a_{n+1}-1=-a_0a_1a_2\cdots a_n$, so
$$a_0a_1a_2\cdots a_n\left(\frac{1}{a_0}+\frac{1}{a_1}+\frac{1}{a_2}+\ldots+\frac1{a_{n}}\right)=\frac{a_{n+1}-1}{a_{n+1}-1}=1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Solve $\sin 84^\circ \sin(54^\circ-x)=\sin 126^\circ \sin x$.
Find $x$ in degrees, where $$\sin 84^\circ\cdot \sin(54^\circ-x)=\sin126^\circ\cdot \sin x\,.$$
I tried to use trigonometry identities to transform the product in sums, but I can't simplify moreover. I know the result is 30° since I solved it in a calculator, but there must be an algebraic way.
For context, this equation comes from solving this problem:
How can I solve this geometry problem without trigonometry?
| Use
$$\cos36 = \frac {\sin108}{2\sin36}
= \frac {\sin36+2\sin36\cos72}{2\sin36}=\frac12+\cos72$$
to factorize the equation as follows
$$\begin{align}
& \sin 84\sin(54-x)-\sin 54 \sin x \\
& =\cos 6 \cos 36 \cos x - ( \cos 36+\cos 6 \sin 36 )\sin x \\
& =\frac12(\cos30+\cos 42 ) \cos x
- \left(\frac12 + \cos 72 + \frac12(\sin 42+ \sin30 )\right)\sin x \\
& =\frac12\left(\frac{\sqrt3}2+\cos 42 \right) \cos x
- \left(\frac34 +\cos (42+30) + \sin30\sin 42\right)\sin x \\
& =\frac12\left(\frac{\sqrt3}2+\cos 42 \right) \cos x
- \left(\frac34 + \frac{\sqrt3}2\cos42 \right)\sin x \\
& =\frac12\left(\frac{\sqrt3}2+\cos 42 \right) (\cos x -\sqrt3 \sin x) = 0
\end{align}$$
Thus,
$$\tan x = \frac1{\sqrt3}$$
and the angle in the source problem is $30^\circ$.
| {
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"url": "https://math.stackexchange.com/questions/3612928",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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An inequality on sides of triangle
Let $a,b,c$ denote the sides of a triangle. Then prove that the quantity $a\over b+c$$+$$b\over c+a$$+$$c\over a+b$ lies between $3\over 2$ and $2$.
My attempt :
Let $P=a+b+c$
With a little manipulation it becomes
$-3 + {P\over P-a} + {P\over P-b} + {P\over P-c} $
Applying $AM>HM$ for the three latter terms , I got the lower limit , but I am not able to find the upper limit.
Could someone please help with the upper limit ?
Thanks.
| $$2>F=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} \ge \frac{3}{2}~~~(1)$$
$$F+3=(a+b+c)(\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{a+b})~~~~(2)$$
Using AM_HM:
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge \frac{9}{x+y+z}~~~~(3)$$
Using (3) in (2), we prove that
$$F+3 \ge \frac{9(a+b+c)}{2(a+b+c)} \implies F \ge 9/2-3= \frac{3}{2}.$$
The equality holds when the triangle is equilateral.
For LHS, in a triangle $$\frac{a}{b+c}<1 \implies \frac{a}{b+c} <\frac{2a}{a+b+c}$$
When we add the same positive number up and down in a proper fraction we get a bigger
fraction:
If $$\frac{P}{Q}<1 \implies \frac{P}{Q}< \frac{P+R}{Q+R},~~ P,Q,R>0$$
Similarly we get two more results:
$$\frac{b}{c+a} <\frac{2b}{a+b+c},~~~\frac{c}{a+b} < \frac{2c}{a+b+c}$$
Adding last three results we get
$$F=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} <2$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Recursive formula, eigenvalue problem A sequence $ \{X_i \}_{i \geq 0} $ is defined recursive by $ X_{n+1} = 3 X_{n} - 2X_{n-1}, \qquad n \geq1. $
and $ X_0 = \begin{pmatrix}
1 & 0 \newline
1 & 1
\end{pmatrix},X_1 = \begin{pmatrix}
1 & 1 \newline
1 & 1
\end{pmatrix}. $ Find an explicit formula for $ X_n $.
My attempt;
We can write $ = \begin{pmatrix}
X_{n+1} \newline X_{n} \end{pmatrix} = \begin{pmatrix} 3 & -2 \newline 1 & 0 \end{pmatrix} \begin{pmatrix} X_{n} \newline X_{n-1} \end{pmatrix} $ for $ n \geq 1. $ Diagonalizing the matrix yields
$\begin{pmatrix} X_{n+1} \newline X_{n} \end{pmatrix} = \begin{pmatrix} 2 & 1 \newline 1 & 1 \end{pmatrix} \begin{pmatrix} 2^n & 0 \newline 0 & 1 \end{pmatrix} \begin{pmatrix} 1 & -1 \newline -1 & 2 \end{pmatrix} \begin{pmatrix} X_{n} \newline X_{n-1} \end{pmatrix}$
This is and old exam; now the answer concludes that $X_n $ can be written as $ X_n = A + 2^n B $, for some matrices A and B, which can be found by plugging in $ X_1 $ and $ X_0.$ My question is how did he draw this conclusion, is it only from the eigenvalues? How would one approach questions on this form if the matrix couldnt be diagonalized?
| Note that $X_{n+1}-2 X_n = X_n - 2 X_{n-1} = \Delta = X_1 - 2 X_0$.
Hence $X_{n+1} = 2 X_n + \Delta$.
The general solution is
$X_n = 2^n X_0 + \sum_{k=0}^{n-1} 2^{n-k-1} \Delta = 2^n X_0+(2^{n}-1)\Delta = 2^n(X_1-X_0)-X_1+2X_0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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${\displaystyle \int_0^1 \dfrac{1}{x} \cdot \dfrac{1}{1-x} \cdot \left(1-\dfrac{1}{x} \right) \mathrm dx}$
Evaluate
$$\displaystyle \int_0^1 \left\{\dfrac{1}{x}\right\} \cdot \left\{\dfrac{1}{1-x}\right\} \cdot \left\{1-\dfrac{1}{x} \right\} \mathrm dx$$
where $\{x\}$ denotes the fractional part of $x$.
| Since $\{1 - \frac{1}{x}\} = \{-\frac{1}{x}\} = 1 - \{\frac{1}{x}\}$, we have
\begin{align}
I &= \int_0^1 \{\tfrac{1}{x}\}\{\tfrac{1}{1-x}\} (1 - \{\tfrac{1}{x}\})\mathrm{d}x \\
&= \int_0^1 \{\tfrac{1}{x}\}\{\tfrac{1}{1-x}\} \mathrm{d}x
- \int_0^1 \{\tfrac{1}{x}\}^2\{\tfrac{1}{1-x}\} \mathrm{d}x.
\end{align}
From 2.10 and 2.12 (page 101) in [1], we have
$$\int_0^1 \{\tfrac{1}{x}\}\{\tfrac{1}{1-x}\} \mathrm{d}x = 2\gamma - 1$$
and
$$\int_0^1 \{\tfrac{1}{x}\}^2\{\tfrac{1}{1-x}\} \mathrm{d}x = \frac{5}{2} - \gamma - \ln (2\pi)$$
where $\gamma $ is Euler-Mascheroni constant.
Thus, we have $I = 3\gamma - \frac{7}{2} + \ln (2\pi) \approx 0.069524062$.
Also, I approximated the integral numerically in Maple which resembles the analytic result above.
Reference:
[1] Ovidiu Furdui, "Limits, Series, and Fractional Part Integrals: Problems in Mathematical Analysis".
| {
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Solution of a repeating polynomial I have a polynomial series
$$ P_n(x) = x(1-P_{n-1}(x)) $$
With initial value $ P_1 = x $, how can I solve this as a sum?
$$ \Sigma_{n=1}^\infty nP_n(x)$$
| Let's put first
$$
\eqalign{
& P_{\,n} (x) = x\left( {1 - P_{\,n - 1} (x)} \right)\quad \left| {\;P_{\,1} (x) = x} \right.\quad \Rightarrow \cr
& \Rightarrow \quad Q_{\,n} (x) = x\left( {1 - Q_{\,n - 1} (x)} \right)\quad \left| \matrix{
\;P_{\,n + 1} (x) = Q_{\,n} (x) \hfill \cr
\;Q_{\,0} (x) = x \hfill \cr} \right. \cr}
$$
Then let's take the z-Transform of both sides
$$
\eqalign{
& F(x,z) = \sum\limits_{0\, \le \,n} {Q_{\,n} (x)z^{\,n} } = \sum\limits_{0\, \le \,n} {x\left( {1 - Q_{\,n - 1} (x)} \right)z^{\,n} } = \cr
& = x\sum\limits_{0\, \le \,n} {z^{\,n} } - xz\sum\limits_{0\, \le \,n} {Q_{\,n - 1} (x)z^{\,n - 1} } = \cr
& = {x \over {1 - z}} - xzF(x,z)\quad \Rightarrow \cr
& \Rightarrow \quad F(x,z) = {x \over {\left( {1 - z} \right)\left( {1 + xz} \right)}}
= {{x^{\,2} } \over {\left( {1 + x} \right)\left( {1 + xz} \right)}} + {x \over {\left( {1 + x} \right)\left( {1 - z} \right)}} \cr}
$$
That means that
$$
Q_{\,n} (x) = {x \over {1 + x}}\left( {x\left( { - 1} \right)^{\,n} x^{\,n} + 1} \right)
= \left( { - 1} \right)^{\,n} {x \over {1 + x}}\left( {x^{\,n + 1} - \left( { - 1} \right)^{\,n + 1} } \right)
$$
And finally
$$
\eqalign{
& Q_{\,2n} (x) = x{{x^{\,2n + 1} + 1} \over {1 + x}} = x{{1 - \left( { - x} \right)^{\,2n + 1} } \over {1 - \left( { - x} \right)}} = \cr
& = x\sum\limits_{k = 0}^{2n} {\left( { - x} \right)^k } = \sum\limits_{k = 0}^{2n} {\left( { - 1} \right)^k x^{k + 1} } \cr
& Q_{\,2n + 1} (x) = - x{{x^{\,2n + 2} - 1} \over {1 + x}} = - x{{\left( {x^{\,n + 1} + 1} \right)\left( {x^{\,n + 1} - 1} \right)} \over {1 + x}} = \cr
& = - x\left( {x^{\,n + 1} - 1} \right){{\left( {1 - \left( { - x} \right)^{\,n + 1} } \right)} \over {1 - \left( { - x} \right)}}
= \left( {1 - x^{\,n + 1} } \right)\sum\limits_{k = 0}^n {\left( { - 1} \right)^k x^{k + 1} } \cr}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to bring elliptical equation $2x^2+2y^2+3xy-x-y=0$ into canonical form I have this ellipse: $$2x^2+2y^2+3xy-x-y=0$$
Canonical form is $$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$$
How can I bring my ellipse to that canonical form? It seems like I need some substitution.
| Rewrite the equation,
$$\begin{align}
& 2x^2+2y^2+3xy-x-y\\
& = 2(x+y)^2 -xy -(x+y)\\
& = 2(x+y)^2 -\frac14[(x+y)^2-(x-y)^2] -(x+y)\\
& =\frac74 \left(x+y-\frac27\right)^2+\frac14 (x-y)^2 - \frac17=0 \\
\end{align}$$
Then, let $u=\frac 1{\sqrt2}(x+y-\frac27)$ and $u=\frac 1{\sqrt2}(x-y)$ to get the canonical form
$$\frac{u^2}{\left(\frac{\sqrt2}7\right)^2} + \frac{v^2}{\left(\sqrt{\frac{2}7}\right)^2} =1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How prove this inequality $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}<6$ let $a,b,c\ge\dfrac{1}{3}$,and such $$a^2+b^2+c^2=a+b+c$$
show that
$$\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}<6$$
I try:$$a-a^2=(b^2-b)+(c^2-c)\ge-\dfrac{1}{4}-\dfrac{1}{4}=-\dfrac{1}{2}$$
so we have
$$a^2-a-\dfrac{1}{2}\le 0\Longrightarrow a\le\dfrac{1+\sqrt{3}}{2}$$
so we have
$$a,b,c\in [\dfrac{1}{3},\dfrac{1+\sqrt{3}}{2}]$$
let $f(a)=\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}$,and $f''(a)>0$. so $LHS$ is maximum when $a,b,c=\{\dfrac{1}{3},\dfrac{1+\sqrt{3}}{2}\}$.but I found all case this value is big $6$
| Actually, this problem could be solved analytically or using Lagrange multipliers, but here the (Analytical solution).
show :
$\frac{a²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${< 6}$
s.t:
${a, b, c ≥ ⅓}$
${a² + b² + c² = a + b + c}$
Changing the problem a little
${a, b, c ≥ 1}$
$\frac{a²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${< 2}$
rearranging the function
${a² = a + b + c - b² - c²}$
$\frac{a + b + c - b² - c²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${< 2}$
then
$\frac{a}{b}$ + ${1}$ + $\frac{c}{b}$ - ${b}$ - $\frac{c²}{b}$ + $\frac{b²}{c}$ + $\frac{c²}{a}$ ${< 2}$
*
*${b ≥ 1}$ ⇒ ${1-b}$ ≤ 0
*$\frac{c}{b}$ - $\frac{c²}{b}$ = $\frac{c-c²}{b}$ ⇒ $\frac{c-c²}{b}$ ${≤ 0}$
*$\frac{a}{b}$ + $\frac{c²}{a}$ = $\frac{a² + bc²}{bc}$ = $\frac{(a + b + c - b² - c² ) + bc² }{ba}$
*$\frac{1}{b}$ + $\frac{b-b²}{ba}$ + $\frac{c - c²}{ba}$ + $\frac{c²}{a}$
*$\frac{1}{b}$ ${≤1}$ and $\frac{b-b²}{ba}$ ${≤ 0}$ and $\frac{c - c²}{ ba}$ ${≤ 0}$
then we need to proof that $\frac{c²}{a}$ ${≤ 1}$
*$\frac{c² + a² - a² }{a}$
= $\frac{c² + a²}{a}$ - ${a}$
= $\frac{c² + (a + b + c - b² - c²)}{a}$ - ${a}$
= ${1 - a}$ + $\frac{2c - c²}{a}$ + $\frac{b-b²}{a}$
${1-a}$ ${≤ 0}$ and
$\frac{b-b²}{a}$ ${≤ 0}$
we need to proof that $\frac{2c - c²}{a}$ ${< 1}$ or ${2c ≤ c²}$ but ${c≥1}$
then ${2c ≤ c²}$ for all ${c≥1}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Framing problem : Show that $\frac{x^2}{yz+1}+\frac{y^2}{zx+1}+\frac{z^2}{xy+1}\leq\frac{x^2+y^2+z^2}{4} +1$ Let $x,y,z\gt 0$ reals and let $x^3+y^3+z^3=4xyz$
Show that :
$$\frac{x^2}{yz+1}+\frac{y^2}{zx+1}+\frac{z^2}{xy+1}\leq\frac{x^2+y^2+z^2}{4} +1$$
Can I use Cauchy inequality to solve it ?
| Using Cauchy-Inequality:
$$\left(\frac{x^2}{yz}+\frac{x^2}{1}\right)\left(yz+1\right)\geq (x+x)^2\implies \frac{x^2}{yz+1}\leq \frac{x^2}{4yz}+\frac{x^2}{4}$$
Therefore
$$\frac{x^2}{yz+1}+\frac{y^2}{zx+1}+\frac{z^2}{xy+1}\leq\frac{x^2}{4yz}+\frac{y^2}{4zx}+\frac{z^2}{4xy}+ \frac{x^2+y^2+z^2}{4}=1+ \frac{x^2+y^2+z^2}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve the following expression I have the following expression: $2^{n-3}+\sum_{k=1}^{n-3} 2^{k-1}(n-k+1)^2$ and I have no idea how to solve it. I tried plugging it in WolframAlpha to get an idea but it gives me $-n^2 - 4 n + 7 \cdot 2^{n - 1} - 6$ and I have even less of an idea how I would get there.
I don't need the complete solution done. I just would like pointers on how to solve it.
| You'll just to simplify the expression
$f(n) = 2^{n-3}+\sum_{k=1}^{n-3} 2^{k-1}*(n-k+1)^2$
Now let's expand
$f(n) = 2^{n-3}+\sum_{k=1}^{n-3} 2^{k-1}*(n^2-2*k*n+2*n+k^2-2*k+1)$
Let's reduce it
$f(n) = 2^{n-3}+\sum_{k=1}^{n-3} 2^{k-1}*(n^2+2*n+1) +\sum_{k=1}^{n-3} 2^{k-1}*(-2*k*n-2*k) + \sum_{k=1}^{n-3} 2^{k-1}*(k^2)$
Simplify more
$f(n) = 2^{n-3}+ (n^2+2*n+1)*\sum_{k=1}^{n-3} 2^{k-1} -2*(n+1)*\sum_{k=1}^{n-3} k*2^{k-1}+\sum_{k=1}^{n-3} (k^2)*2^{k-1}$
Remember that
$\sum_{k=1}^{n-3} 2^{k-1} = 1+2+2^2+2^3+...........+2^{n-3-1} = 2^{n-3}-1$
It's becomes difficult here
$\sum_{k=1}^{n-3} k*2^{k-1} = h(x) = k*2^{k-1}+(k-1)*2^{k-2}+(k-2)*2^{k-3}+........+3*2^2+2*2^1+1*2^0$
But It's not so hard to see that
$h(x)*2-k*2^k = 2*h(x-1)$
You'll need to solve some recurring equation now
$h(x) = 2^x*(x-1)+1$
$\sum_{k=1}^{n-3} (k^2)*2^{k-1} = g(x) = (k^2)*2^{k-1}+(k-1)^2*2^{k-2}+(k-2)^2*2^{k-3}+........+ 3^2*2^2+2^2*2+1$
Same process involved we can also see that
$g(x)*2 -x^2*2^x = 2*g(x-1)$
So that $g(x) = 2^x*(x^2-2*x+3)-3$
Don't forget that the plugin value is $h(n-3)$ and $g(n-3)$
Therefore
$f(n) = 2^{n-3}+(n^2+2*n+1)*(2^{n-3}-1)-2*(n+1)*(2^{n-3}*(n-3-1)+1)+2^{n-3}*((n-3)^2-2*(n-3)+3)-3$
Expand and simplify this to get our solution
$f(n) = 7*2^{n-1}-n^2-4*n-6$
So you see that the trick is recurring equation, which Wolfram automatically solves
| {
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How to evaluate $ \int_{0}^{\infty} \frac{dx}{1+x+x^2+x^3......+x^n} ; n> 1$ I have found like ;
1.$ \int_{0}^{\infty} \frac{dx}{1+x+x^2} = \frac{2\pi}{3\sqrt3}$
2.$ \int_{0}^{\infty} \frac{dx}{1+x+x^2+x^3} = \frac{\pi}{4}$
3.$ \int_{0}^{\infty} \frac{dx}{1+x+x^2+x^3+x^4} = \frac{\pi}{5} \cdot \sqrt{2-\frac{2}{\sqrt5}}$
4.$ \int_{0}^{\infty} \frac{dx}{1+x+x^2+x^3+x^4+x^5} = \frac{2\pi}{6\sqrt3}$
5.$ \int_{0}^{\infty} \frac{dx}{1+x+x^2+x^3+x^4+x^5+x^6+x^7} = \frac{2\pi}{8\sqrt3}$
In that way i'm trying to find $ \int_{0}^{\infty} \frac{dx}{1+x+x^2+x^3......+x^n} ; n> 1;$ But get stuck , can anyone help me by showing the process how to evaluate it .
| Note that $\sum_{0 \le k \le n} x^k = \frac{1 - x^{n +1}}{1 - x}$. The numerator of this can be factored into linear (complex) factors, partial fractions should take care of your general integral.
| {
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How many cycles in this adjacency matrix? Consider the $N\times N$ adjacency matrix $A$ such that
$$A_{ij}=\begin{cases} 1, &\; \; (i\leq n \;\text{ or }\; j \leq n), \;\; i\neq j \\ 0, & \; \; \text{otherwise}\end{cases} $$
for some integer $0\leq n \leq N$
Example matrix, $N=5, n=3$
$$\begin{pmatrix}0&1&1&1&1 \\ 1&0&1&1&1 \\ 1&1&0&1&1 \\ 1&1&1&0&0 \\ 1&1&1&0&0 \end{pmatrix}$$
Example matrix, $N=7, n=4$
$$\begin{pmatrix}0&1&1&1&1&1&1 \\ 1&0&1&1&1&1&1 \\ 1&1&0&1&1&1&1 \\ 1&1&1&0&1&1&1 \\ 1&1&1&1&0&0&0 \\ 1&1&1&1&0&0&0 \\ 1&1&1&1&0&0&0\end{pmatrix}$$
Question: How many cycles are in $A(N,n)$?
I only need the answers for $A(30,5)$ and $A(30,7)$ but I am curious about the general case.
| Hint: Interpret the graph as 1) Complete graph on $n$ elements, 2) $N-n$ elements that are connected only to the vertices of the complete graph.
Hint: View a $k-$cycle as a $k-$permutation of $N$. What are the restrictions?
For a $k-$ cycle to be valid, the only restriction is that elements that are $ > n$ cannot be located next to each other. Hence, we can
1. Pick $\frac{k}{2} \leq l \leq k$ elements from $1$ to $n$,
2. Pick $k - l $ elements from $n+1$ to $N$,
3. Pick $k-l$ spots from $l$ spots to insert the large elements
4. Divide out by $k$ if desired to count unique cycles.
5. Sum over all $l$ possible values.
Sum up over all $k$. Note that $k \leq 2n$.
| {
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Find locus of $S$ denoting set of complex numbers $\frac{z+1}{z-3}$, where $z$ varies over set of $|z|=1$.
Question: Let $S$ denote the set of all complex numbers of the form $\frac{z+1}{z-3}$, where $z$ varies over the set of all complex numbers with $|z|=1.$ Find the locus of the points in set $S$.
My approach:
Let $z=x+iy$, with $x,y\in\mathbb{R}$ such that $|z|=1\implies |z|^2=1$. This implies that we must have $x^2+y^2=1$. Now $z+1=(x+1)+iy$ and $z-3=(x-3)+iy$.
Thus
\begin{align*}
\frac{z+1}{z-3}&=\frac{(z+1)(\overline{z-3})}{|z-3|^2}\\
&=\frac{(z+1)(\overline{z}-3)}{|z-3|^2}\\
&=\frac{x^2-2x-3+y^2-4iy}{(x-3)^2+y^2}\\
&=\frac{-2x-2-4iy}{10-6x}\\
&=\frac{-x-1-2iy}{5-3x}.
\end{align*}
Thus we have
$$
\Re\left(\frac{z+1}{z-3}\right)=\frac{x+1}{3x-5}
$$
and
$$
\Im\left(\frac{z+1}{z-3}\right)=\frac{2y}{3x-5},$$ and our task is to find a relationship between these two given that $|z|=1$.
For our ease let us have $\alpha=\frac{z+1}{z-3}\,\, \forall z$ satisfying $|z|=1$.
Now we obtain two useful information using the triangle inequality. We have
$$
|z+1|\le |z|+1=2
$$
and
$$
|z-3|\le |z|+3=4.
$$
From here we can conclude that $$|z+1|^2=(x+1)^2+y^2=2+2x\le 4 \iff 1+x\le 2\text{ and } |z-3|^2=5-3x\le 16.$$
Thus we have $$0\le 1+x\le 2 \text{ and } 0\le 5-3x\le 16\implies -1\le x\le 1.$$
Observe that even from $x^2+y^2=1$, we can directly conclude that $-1\le x,y\le 1$.
But this doesn't help much to find a relationship between $\Re(\alpha)$ and $\Im(\alpha)$.
How to proceed?
| As @Richard said, write $w=1+ \frac{4}{z-3}$
$\Rightarrow z-3=\frac {4}{w-1}$
$\Rightarrow z=\frac {3w+1}{w-1}$
$|z|=|\frac {3w+1}{w-1}|=1$
$|\frac {w+\frac{1}{3}}{w-1}|=\frac{1}{3} \ne1$
Thus, w lies on a circle
| {
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Get the coefficient of $x^n$ in the binomial expansion I have this statement:
Get the coefficient of $x^n$ in $(1-x+x^2)(1+x)^{2n-1}$
My development was:
Note that all exponents of $(1+x)^{2n-1}$ will be modified by $+2$ by $x^2$ and $+1$ by $x$
Then the to get $x^n$ i need the $x^{n-2}$ term in the binomial expansion of $(1+x)^{2n-1}$.
The $x^{n-2}$ term is given by $1^{2n-1-c} \cdot x^{c} \cdot \binom{2n-1}{c}$ And $c$ need to take the value of $x-2$, then i will get: $\binom{2n-1}{x-2}x^{n-2}\cdot(x^2-x+1) = \binom{2n-1}{x-2}( x^n - x^{n-1} +1)$, Therefore the sign of $x^n$ is $\binom{2n-1}{x-2}$.
But according to the guide, my answer is incorrect and i don't know what is wrong with my development. Thanks in advance.
| You have the right basic idea. However, in your case, $c$ would take the value of $n - 2$, not $x - 2$. Also, note with the $1$ term in the left factor, you will use the coefficient of $x^n$ term in $(1 + x)^{2n-1}$. Similarly, with the $-x$ term, you will subtract the coefficient of $x^{n-1}$ term in $(1 + x)^{2n-1}$, and finally with the $x^2$ term, you will add the coefficient of the $x^{n-2}$ term of $(1 + x)^{2n-1}$. Putting this together gives that the coefficient of $x^n$ is
$$\binom{2n-1}{n} - \binom{2n-1}{n - 1} + \binom{2n-1}{n - 2} \tag{1}\label{eq1A}$$
| {
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Doubt regarding the quadrant of angles This question is given in a text book.
and the answer given is
My doubt is that when $A > 0$ and $sin$ $A$ is positive, the angle $A$ may be in the first or second quadrant. It results in two values of $A$.
Then, how can we determine the sign of $cos$ $A$?
Similarly, angle $B$ is less than $\pi/2$ and $cos A$ is positive, it may be in first or 4th quadrant.
How can we determine the sign of $sin$ $B$?
We have to use the formula of $sin$$(A+B)$.
| Since it's given that $0 \lt A, B \lt \frac{\pi}{2}$, you have as Andrew Chin's question comment states, i.e., $0 \lt A \lt \frac{\pi}{2}$ and $0 \lt B \lt \frac{\pi}{2}$. This means $0 \lt \sin(A), \cos(A), \sin(B), \cos(B) \lt 1$, i.e., all of these values are always positive. Since the $4$ formulas used in the question can be expressed using just those $4$ trig. values (i.e., $\sin(A), \cos(A), \sin(B), \cos(B)$), you don't have to worry about the signs of each component as it's always positive. Thus, from $\sin^2(x) + \cos^2(x) = 1$ for all $x$, you have $\sin(A) = \frac{4}{5} \implies \cos(A) = \frac{3}{5}$ and $\cos(B) = \frac{5}{13} \implies \sin(B) = \frac{12}{13}$, so for example with part (i) of your question you get
$$\begin{equation}\begin{aligned}
\sin(A + B) & = \sin(A)\cos(B) + \cos(A)\sin(B) \\
& = \left(\frac{4}{5}\right)\left(\frac{5}{13}\right) + \left(\frac{3}{5}\right)\left(\frac{12}{13}\right) \\
& = \frac{20}{65} + \frac{36}{65} \\
& = \frac{56}{65}
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
| {
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Solve $f(x) = |x+1|−|x|+3|x−1|−2|x−2|−(x+2)=0$ $f(x) = |x+1|−|x|+3|x−1|−2|x−2|−(x+2)=0$
I saw this problem, and I couldn’t think of a good way to do this (initially thought about isolating a term with the modulus sign and squaring, but that seemed to take too long). The solution said something along the lines of:
‘ We have to solve f (x) = 0 in the five regions of the x-axis determined by the modulus functions’. Could someone please elaborate on what this means. Many thanks.
| We have $|x-a| = \begin{cases}x-a,& \text{if } x\ge a \\ -(x-a), & \text{if } x< a\end{cases}$
So, $f(x) = \begin{cases}-(x+1) +x-3(x-1)+2(x-2)-(x+2) & \text{for } x <-1 \\
(x+1) +x-3(x-1)+2(x-2)-(x+2) &\text{for }-1\le x <0 \\ (x+1) -x -3(x-1)+2(x-2)-(x+2) & \text{for } 0\le x<1 \\
(x+1)-x+3(x-1)+2(x-2)-(x+2) &\text{for }1\le x<2 \\
(x+1)-x+3(x-1)-2(x-2)-(x+2) & \text{for } x\ge2\end{cases}, $
Simplify these and equate to $0$ to get the values of $x$
| {
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Calculus A Level Line Tangent to Circle How can you find values of $k$ such that $y = kx + 1$ is tangent to the circle $(y-1)^2 + (x-5)^2 = 9 $?
I first rewrote the circle equation in terms of y:
$$ (y-1)^2 = -(x-5)^2 + 9 \\y-1 = \pm\sqrt{-x^2+10x -16} \\ y = \pm\sqrt{-x^2+10x -16} + 1 $$
Then I took two derivatives, one for the positive equation and one for the negative equation:
$$y\prime_{_{+}} = \frac{1}{2}(-x^2+10x-16)^\frac{-1}{2}(-2x+10) \\=-(x-5) (-x^2+10x-16)^\frac{-1}{2}\\y\prime_{_{-}} = -\frac{1}{2}(-x^2+10x-16)^\frac{-1}{2}(-2x+10) \\=(x-5) (-x^2+10x-16)^\frac{-1}{2}$$
So if I look at just the positive semicircle, I have a line $y = kx + 1$ which intersects $y = \sqrt{-x^2+10x-16} + 1 $ at one spot and tangent at that spot $ k = -(x-5)(-x^2+10x-16)^\frac{-1}{2} $.
How do I actually find the value of $k$?
| You can use implicit differentiation to find the slope $y'$ at any $(x,y) $ in the circle with $y \neq 1$:
$$ 2(y-1)y' + 2(x-5) = 0 \implies y' = \dfrac{5-x}{y-1} $$
if $y' = k$, then we have $5-x = k(y-1)$ and together with the equation of the circle and the equation of the line, we obtain the following system of three equations and three unknowns:
$$ \begin{cases} 5-x = k(y-1) \\ y = kx + 1 \\ (y-1)^2 + (x-5)^2 = 9 \end{cases} $$
Solving this will find your k
| {
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Prove that sum of matrices equals zero
The matrix $A$ has size $3 \times 3$ and we know that for any column
vector $v\in \mathbb{R}^{3}$ the vectors $Av$ and $v$ are orthogonal. Prove that
$A^{T} + A = 0$, where $A^{T}$ is the transposed matrix $A$.
So if
$$A = \begin{pmatrix}
a_{11} & a_{12} & a_{13}\\
a_{21} & a_{22} & a_{23}\\
a_{31} & a_{32} & a_{33}
\end{pmatrix}$$
and
$$v = \begin{pmatrix}
x\\
y\\
z
\end{pmatrix}$$
orthogonality of $Av$ and $v$ brought me to the equation
$x \cdot y \cdot (a_{12}+a_{21}) + x \cdot z \cdot (a_{13}+a_{31}) + y \cdot z \cdot (a_{23}+a_{32}) + a_{11} \cdot x^{2} + a_{22} \cdot y^{22} + a_{33} \cdot z^{2} = 0$
and the matrix $A^{T} + A$ equals
$$A^{T} + A = \begin{pmatrix}
2a_{11} & a_{12} + a_{21} & a_{13} + a_{31}\\
a_{12} + a_{21} & 2a_{22} & a_{23} + a_{32}\\
a_{13} + a_{31} & a_{23} + a_{32} & 2a_{33}
\end{pmatrix}$$
However I don't see how then prove that $A^{T} + A = 0$.
| Following your working,
$x \cdot y \cdot (a_{12}+a_{21}) + x \cdot z \cdot (a_{13}+a_{31}) + y \cdot z \cdot (a_{23}+a_{32}) + a_{11} \cdot x^{2} + a_{22} \cdot y^{22} + a_{33} \cdot z^{2} = 0 \tag{1}$
Note that equation $(1)$ holds for any choice of $v=\begin{pmatrix}x \\ y \\ z \end{pmatrix}$. That is we can choose $x, y, z$.
If you substitute $x=1, y=0, z=0$, we conclude that $a_{11}=0$.
If you substitute $x=0, y=1, z=0$, we conclude that $a_{22}=0$.
If you substitute $x=0, y = 0, z=1$, we conclude that $a_{33}=0$.
By now, we can conclude that the diagonal entries of $A+A^T$ are all zeros.
Hence equation $(1)$ now reduces to
$x \cdot y \cdot (a_{12}+a_{21}) + x \cdot z \cdot (a_{13}+a_{31}) + y \cdot z \cdot (a_{23}+a_{32}) = 0 \tag{2}$
If we substitute $x=1,y=1, z=0$, we conclude that $a_{12}+a_{21}=0$.
If we substitute $x=1, y=0, z=1$, we conclude that $a_{12}+a_{31}=0$.
If we substitue $x=0, y=1, z=1$, we conclude that $a_{23}+a_{32}=0$
Hence we can conclude that the off diagonal of $A+A^T$ are zeros as well.
Hence we have $A+A^T=0$.
The result actually hold for square matrices $A$ of general size $n \times n$. That is if $A \in \mathbb{R}^{n \times n}$ and for any $v \in \mathbb{R}^n$, if $v$ and $Av$ are orthogonal, then $A+A^T=0$.
From the condition that $v$ and $Av$ are orthogonal, we have $v^TAv=0$.
We let $e_i$ be the standard unit basis with the $i$-th position being $1$ and the other positions being $0$. For example in the case of $n=3$, we have $e_2 = \begin{pmatrix} 0 \\ 1\\ 0\end{pmatrix}$. These matrices are very useful, we have $$e_i^TAe_j = A_{ij}$$
because you can check that $Ae_j$ would be just the $j$-th column of $A$ and then $e_i$ would select the $i$-th row.
We first handle the diagonal entries. We know that $e_iAe_i=0$, this is just saying that $A_{ii}=0$, hence the diagonal part s of $A+A^T=0$.
Also, we know that \begin{align}(e_i+e_j)^TA(e_i+e_j)&=e_i^TAe_i+e_i^TAe_j+e_j^T Ae_i + e_j^TAe_j \\
&= A_{ii} + A_{ij}+A_{ji}+A_{jj}\\
&= A_{ij}+A_{ji}\end{align}
and we know that it is equal to $0$.
Hence $A+A^T=0$.
| {
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If $x=y^2+z^2$, $y=z^2+x^2$, and $z=x^2+y^2$, prove $\frac{x}{x+1} + \frac{y}{y+1} + \frac{z}{z+1} = 1$ If $x=y^2+z^2$, $y=z^2+x^2$, and $z=x^2+y^2$,
prove
$$\frac{x}{x+1} + \frac{y}{y+1} + \frac{z}{z+1} = 1$$
| Hint:
From the first two,
$$x-y=y^2-x^2$$ factors as $$(x-y)(x+y+1)=0$$
which defines a pair of planes, and similarly
$$(z-x)(z+x+1)=0.$$
Eliminate $y,z$ and plug in the last identity.
Note that
$$\frac{-(x+1)}{-(x+1)+1}=\frac{x+1}{x}$$
and the last identity has one of the forms
$$3t=1,\\2t+t^{-1}=1,\\t+2t^{-1}=1.$$
| {
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Prove that $3x^3-41x+48\leq 0$ for $x \in [\sqrt 3, \sqrt 6]$ Prove that $3x^3-41x+48\leq 0$ for $x \in [\sqrt 3, \sqrt 6]$.
This is from an inequality in one of Titu Andreescu’s inequality books. More exactly, $2(a+b+c)\geq 3+\frac38(a+b)(b+c)(c+a)$ for positive numbers with $a^2+b^2+c^2=3$. You get the inequality in the op by using the pqr method.
Please don’t use polynomial roots or continuity (I know that you can consider it a polynomial $f$ and then simply checking some values for which $f(x_0)$ is positive and negative will prove the inequality). I’m struggling to find an algebraic proof. Please help. Thank you!
Polynomial proof:
Let $f(x)=3x^3-41x+48$. Then $f(-1)>0$, $f(\sqrt 3)<0$, $f(\sqrt 6)<0$ and $f(3)>0$. Thus, $f$ can’t have any roots in $[\sqrt 3, \sqrt 6] $ and we’re done.
| We can prove this inequality by the following way.
Let $a+b+c=3u$, $ab+ac+bc=3v^2$, $abc=w^3$ and $u^2=tv^2$.
Thus, $t\geq1$, the condition gives $3u^2-2v^2=1$ and we need to prove
$$6u(3u^2-2v^2)\geq3\sqrt{(3u^2-v^2)^3}+\frac{3}{8}(9uv^2-w^3)$$ or
$$48u^3-41uv^2+w^3\geq8\sqrt{(3u^2-2v^2)^3}.$$
Now, by Schur $$w^3\geq4uv^2-3u^3,$$ which says that it's enough to prove
$$45u^3-37uv^2\geq8\sqrt{(3u^2-2v^2)^3}$$ or
$$t(45t-37)^2\geq64(3t-2)^3$$ or
$$(t-1)(297t^2+423t-512)\geq0,$$ which is obvious.
Now, you know that your polynomial inequality is true
because we solved the problem by another way (just by using Schur only).
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what is the value of $\binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\binom{n}{10}+\binom{n}{13}+\dots$ what is the value of $$\binom{n}{1}+\binom{n}{4}+\binom{n}{7}+\binom{n}{10}+\binom{n}{13}+\dots$$ in the form of number,
cos, sin
attempts : I can calculate the value of $$\binom{n}{0}+\binom{n}{3}+\binom{n}{6}+\binom{n}{9}+\binom{n}{12}+\dots=\frac{1}{3}\left(2^n+2\cos \frac{n\pi}{3}\right)$$ by use primitive $3^\text{rd}$ root of the unity
but this problem i cant solve it.
| As we are interested in every third term
Let $$\sum_{r=0}^{3r+1\le n}\binom n{3r+1}=\sum_{k=0}^2a_k(1+w_k)^n$$
where $w_k=w^k;k=0,1,2$ and $w$ is a complex cube root of unity so that $$1+w+w^2=0$$
Set $n=0,1,2$ to find $a_k;k=0,1,2$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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If $x+y+2xy=83$, find the value of $x+y$.
Let $x$ and $y$ be integers. If $x+y+2xy=83$, find the value of $x+y$.
I tried to multiply both sides by $x+y-2xy$ but I could never manage to simplify it. Is there a better way to solve this question?
| Let's assume we have a solution $(x,y)$. First we make a substitution $y = x + a$ for some integer $a$. Substituting gives $2x^2 +2x(a+1) + (a - 83) = 0$ which we can think of as a quadratic in $x$.
If a general quadratic $ax^2 + bx + c = 0$ has integer roots then we have that it's discriminant $b^2 - 4ac$ is a square because it appears under the squareroot in the quadratic formula.
So in our case, we have that $4(a+1)^2 - 4 \cdot 2 \cdot (a - 83)$ is a square. Simplifying gives us that $4(a^2 + 167)$ is a square. Suppose $a^2 + 167 = k^2$ for some integer $k$, then $(k-a)(k+a) = 167$. Since $167$ is prime we have $k = \pm 84 $ and $a = \pm 83$. Note that $2k$ is the discriminant of our quadratic.
So to find $x$ we plug $a$ into the quadratic formula.
$$
x = \frac{-2(a+1) \pm 2k}{2 \cdot 2} = \frac{-2(\pm 83 + 1) \pm 2 \cdot 84}{4} = 0, 83, -84 \textrm{ or } -1.
$$
So now we check the possible solutions $(x,y)$. A trick we can use here it to note that the equation $x+y+2xy = 83$ is symmetric in $x$ and $y$ so the possible values for $y$ are also $0, 83, -84$ and $-1$. Going through the options gives the solutions $(0,83), (83, 0), (-84,-1)$ and $(-1,84)$.
| {
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Prove $\sum_{n\geq1}\frac{2^n (1-\cos\frac{x}{2^n})^2}{\sin\frac{x}{2^{n-1}}}=\tan\frac{x}{2}-\frac{x}{2}$ How to prove for $|x|<\pi$:
*
*$\sum_{n\geq1}\frac{2(1-\cos(\frac{x}{2^n}))}{\sin(\frac{x}{2^{n-1}})}=\tan(\frac{x}{2})$
*$\sum_{n\geq1}\frac{2^n (1-\cos(\frac{x}{2^n}))^2}{\sin(\frac{x}{2^{n-1}})}=\tan(\frac{x}{2})-\frac{x}{2}$
Any help will be appreciated.
| Hint:
If $\sin4y\ne0,$
$$\dfrac{(1-\cos2y)^2}{\sin4y}=\dfrac{\sin^3y}{\cos y\cos2y}$$
$$=\dfrac{\sin y(1-\cos^2y)}{\cos y\cos2y}$$
$$=\dfrac{\sin(2y-y)}{\cos y\cos2y}-\dfrac{\tan2y}2$$
$$=\dfrac{\tan2y}2-\tan y$$
Set $2y=\dfrac x{2^n}$
$$\sum_{n=1}^\infty\dfrac{2^n\left(1-\cos\dfrac x{2^n}\right)^2}{\sin\dfrac x{2^{n-1}}}$$ $$=\sum_{n=1}^\infty2^n\left(\dfrac{\tan\dfrac x{2^n}}2-\tan\dfrac x{2^{n+1}}\right)$$
$$=\sum_{n=1}^\infty\left(f(n)-f(n+1)\right)\text{ (Telescoping series)}$$
$$=f(1)-\lim_{n\to\infty}f(n+1)$$
where $f(m)=2^{m-1}\tan\dfrac x{2^m}$
Now $$\lim_{n\to\infty}f(n+1)=\dfrac x2\cdot\lim_{n\to\infty}\dfrac{\tan\dfrac x{2^{n+1}}}{\dfrac x{2^{n+1}}}=?$$
Second part is much simpler
Set $2y=\dfrac x{2^n}$
$$\dfrac{1-\cos2y}{\sin4y}=\dfrac{2\sin^2y}{4\sin y\cos y\cos2y}=\dfrac{\sin(2y-y)}{2\cos y\cos2y}=?$$
| {
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Show a parallelogram with angle $60^\circ$ is a rhombus
If $ABCD$ is a parallelogram with $\angle BAD=60 ^\circ$ and $\dfrac{AC^2}{BD^2}=\dfrac31$, show $ABCD$ is a rhombus.
We have the squares of $AC$ and $BD$ so MAYBE it is a good idea to construct right triangles. Let $DD_1\perp AB$ and $CC_1 \perp AB$. Now we have the right triangles $BD_1D$ and $AC_1C$ with hypotenuses $BD$ and $AC$, respectively. By the Pythagorean theorem we can get $AC^2=AC_1^2+CC_1^2$ and $BD^2=BD_1^2+DD_1^2$. This does not seem to help. Can you give me some hints? Thank you in advane! :)
I am trying to solve it by the Pythagorean theorem.
| You should be right on target.
Let $AD=1$ be our unit length.
We want to prove $AB = 1$.
But that would make $\triangle ADD_1$ an isosceles triangle with a $60$ degree vertex, or in other words and equilateral triangle, so $AB=BD=1$.
Let $BD = x$.
We have:
$AD_1 = \frac 12$. $DD_1= \frac{\sqrt 3}2$. (Because $\triangle ADD_1$ is a $30-60-90$ triangle.)
$D_1B = \sqrt{x^2-\frac 34}$. (By Pythagorean theorem)
$BC_1 = AD_1 =\frac 12$. (Because $ABCD$ is a parallelogram.)
And $AC_1 =AD_1 + D_1B+BC_1 = \sqrt{3x^2 -\frac 34}$.(By Pythagorean theorem)
So we have $\frac 12 + \sqrt{x^2-\frac 34} + \frac 12=\sqrt{x^2-\frac34}+1 = \sqrt{3x^2 -\frac 34}$
Which has solution $x = 1$
And so $D_1B= \sqrt{x^2-\frac 34}=\sqrt{1^2 -\frac 34}=\frac 12$ and $AB = AD_1 + D_1B =1$.
| {
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find the values of the real numbers A and B such that...
i expanded and simplified the RHS to $x^4+Bx^3+4x^2+Ax^3+ABx^2+2Ax+2Bx+4$
i don’t know where to go from here.
| After your expansion
$$x^4+Bx^3+2x^2+Ax^3+ABx^2+2Ax+2x^2+2Bx+4$$
Just like how you grouped the $2x^2+2x^2$, we can group like terms (same $x^i$)
$$x^4+(A+B)x^3+(AB+4)x^2+(2A+2B)x+4\equiv x^4+4$$
Now since this is true for all $x$, all the matching coefficients must be equal! (Might not be obvious)
$$x^4+(A+B)x^3+(AB+4)x^2+(2A+2B)x+4\equiv x^4+0x^3+0x^2+0x+4$$
$$A+B=0,AB+4=0,2A+2B=0$$
$$\implies A=-B,AB+4=0$$
$$\implies -B^2+4=0$$
$$A,B=2,-2$$
$$\therefore x^4+4=(x^2+2x+2)(x^2-2x+2)$$
| {
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Using different pdfs to calculate the same thing I have a question that is asking me to evaluate the following: $P(|X-\frac{1}{2}|<\frac{1}{8})$. Before I get into any other details in this particular question $Y = X(1-X)=-(X-\frac{1}{2})^2+\frac{1}{4}$, this will come in handy when computing using the pdf of $Y$.
I have two pdfs, one of $X$ which is $1$ and is from $0<x<1$ and $0$ otherwise.
The other one is of $Y$, $\frac{1}{\sqrt{\frac{1}{4}-y}}$ from $0<y<\frac{1}{4}$ and $0$ otherwise.
The question is asking me to evaluate the expression above, using both the pdfs of $X$ and $Y$. I first did the pdf of $X$ and this is what I did:
$P(|X-\frac{1}{2}|<\frac{1}{8})=P(\frac{3}{8}<x<\frac{5}{8})=\int_\frac{3}{8}^\frac{5}{8}1\space dx = \frac{1}{4}$.
Meanwhile using the pdf of $Y$ I did:
$P(|X-\frac{1}{2}|<\frac{1}{8})=P(|X-\frac{1}{2}|^2<(\frac{1}{8})^2)=P((X-\frac{1}{2})^2+\frac{1}{4}<\frac{1}{64}+\frac{1}{4})=P(Y<\frac{17}{64})=\int_0^\frac{1}{4}1\space dy + \int_\frac{1}{4}^\frac{17}{64}0\space dy = 1$.
My answers are different so I was wondering can see if my logic is correct and where I went wrong...
| Why did you use $1$ in the integral for $P(Y< \frac{17}{64})$? You should be using the pdf for $Y$. There is another small error: you say that $P((X- \frac12)^2 + \frac14 < \frac 1{64} + \frac14) = P(Y< \frac {17}{64})$ but this is not true. $Y= -(X-\frac12)^2 + \frac14$. Notice the extra minus sign out front.
You were definitely extremely close with your attempt. Start over from $P(|X-\frac12|^2 < (\frac18)^2)$, as that was your last correct step, and then multiply both sides of the inequality by $-1$ and see where that takes you. And remember to use the right pdf for $Y$ !
| {
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Unconstrained optimization - parameters' values for global maximum Consider the following function
$f(x,y)=p \cdot x^a \cdot y^b - w_xx-w_yy$
with $(p,x,y,w_x,w_y) \in \mathbb{R^5_+} $ (this is because $f(x,y)$ is a profit function)
(a) Solve the first-order conditions in order to find $(x,y)$ that maximizes $f(x,y)$.
Here's my answer (it's a bit long, so I'll skip some steps).
$$\begin{cases} f_x = 0 \therefore p \cdot a \cdot x^{a-1} \cdot y^b = w_x \,\, \dots I \\ f_y = 0 \therefore p \cdot b \cdot x^a \cdot y^{b-1} = w_y \,\, \dots II\end{cases}$$
Since $x \neq 0, y \neq 0$, we can divide the equations and get:
$$ \frac{a}{b} \cdot x^{-1} \cdot y = \frac{w_x}{w_y} \therefore y = \frac{b}{a} \cdot \frac{w_x}{w_y} \cdot x \,\, \dots III$$
Substituting back in equation (I), we find:
$$ \large{x^* = \left( \frac{1}{p} \cdot a^{b-1} \cdot b^{-b} \cdot w_2^b \cdot w_1^{1-b} \right)^\frac{1}{a+b-1}}$$
and
$$ \large{y^* = \frac{b}{a} \cdot \frac{w_1}{w_2} \cdot \left( \frac{1}{p} \cdot a^{b-1} \cdot b^{-b} \cdot w_2^b \cdot w_1^{1-b} \right)^\frac{1}{a+b-1} }$$
(b) Use the second-order conditions to determine the parameters' $a,b,p,w_x,w_y$ values such that the solution for the optimization problem is a global maximum.
The Hessian will be:
$$ H_{f(x,y)}= \begin{bmatrix} p \cdot a \cdot (a-1) \cdot x^{a-2} \cdot y^b & p \cdot a \cdot b \cdot x^{a-1} \cdot y^{b-1} \\ p \cdot a \cdot b \cdot x^{a-1} \cdot y^{b-1} & p \cdot b \cdot (b-1) \cdot x^a \cdot y^{b-2}\end{bmatrix} $$
In order for $f(x,y)$ to be strictly concave, $|H_2| < 0, |H| > 0$ where $H_2$ is the sub-matrix of $H$ that is found by removing the second column and second row.
For $|H|$, we have:
$$ |H| = p^2 \cdot a \cdot (a-1) \cdot b \cdot (b-1) \cdot (x^{a-1})^2 \cdot (y^{b-1})^2 - p^2\cdot a^2 \cdot b^2 \cdot (x^{a-1})^2 \cdot (y^{b-1})^2 \therefore \\\\ |H| = p^2 \cdot a \cdot b \cdot (x^{a-1})^2 \cdot (y^{b-1})^2 \cdot [ (a-1) \cdot (b-1)-ab ]$$
Now if we make $p^2 \cdot (x^{a-1})^2 \cdot (y^{b-1})^2 = k > 0$, all we have to worry about is:
$a \cdot b \cdot [(a-1)(b-1) -ab] = ab \cdot [ab - a -b + 1 - ab] = ab \cdot (1-a-b)$
So our first inequality is $ab \cdot (1-a-b) > 0$.
Now, substituting our optimal values into $|H_2|$ we have:
$|H_2(x^*,y^*)| = p \cdot a \cdot (a-1) \cdot \left( \left( \frac{1}{p} \cdot a^{b-1} \cdot b^{-b} \cdot w_2^b \cdot w_1^{1-b} \right)^\frac{1}{a+b-1} \right)^{a-2} \cdot \left( \frac{b}{a} \cdot \frac{w_1}{w_2} \cdot \left( \frac{1}{p} \cdot a^{b-1} \cdot b^{-b} \cdot w_2^b \cdot w_1^{1-b} \right)^\frac{1}{a+b-1} \right)^b \\\\ \therefore |H_2(x^*,y^*)| = p \cdot a \cdot (a-1) \cdot \left( \frac{b}{a} \cdot \frac{w_x}{w_y} \right)^b \cdot \left( \frac{1}{p} \cdot a^{b-1} \cdot b^{-b} \cdot w_y^b \cdot w_x^{1-b}\right)^{\frac{a+b-2}{a+b-1}} $
Now since $p > 0, w_x > 0, w_y > 0$, I'll remove them from the equation to declutter it a bit and I'll also drop the $(x^*,y^*)$.
$$ \large{|H_2| = a \cdot (a-1) \cdot b^{-b} \cdot a^{-b} \cdot a^{\frac{(b-1)(a+b-2)}{a+b-1}} \cdot b^{\frac{b \cdot (a+b-2)}{a+b-1}} \therefore \\ |H_2| = (a-1) \cdot a^{\frac{(1-b)(a+b-1) + (b-1)(a+b-2)}{a+b-1}} \cdot b^{\frac{-b(a+b-1) + b(a+b+2)}{a+b-1}} \therefore \\ |H_2| = (a-1) \cdot a^{\frac{1-b}{a+b-1}} \cdot b^{\frac{3b}{a+b-1}} \therefore \\\\ |H_2| = (a-1) \cdot a^{\frac{1-b}{a+b-1}} \cdot b^{\frac{b}{a+b-1}} \cdot \left( b^{\frac{b}{a+b-1}} \right)^2}$$
Now, because the last term is always positive, all we need is:
$$ \large{(a-1) \cdot a^{\frac{1-b}{a+b-1}} \cdot b^{\frac{b}{a+b-1}} < 0 } $$
Without imposing further restrictions I have no idea how to continue. If we impose $a>0,b>0$, than our problem can be reduced to
$$ \begin{cases} -a-b+1 >0 \\ (a-1) \cdot a < 0 \end{cases} $$
and our final solution is $$ \boxed{\boxed{0 < a < 1, \\ 0 < b < 1-a, \\(p,w_x,w_y) \in \mathbb{R_+^3} }} $$
Is my solution correct? Did I miss something that would make the calculations simpler or at least feasible?
Thanks in advance!
| For the second part, you only need the hessian to be negative semidefinite everywhere and strictly negative definite at the candidate point found from the first part. This means that you have a concave function with only one local maximum, that is, this local minimum is unique in a small neighborhood. Thus, it is the unique global maximum of the function due to its concavity.
Also, the determinant of the symmetric matrix $H$ must be nonpositive as a necessary condition for negative semidefiniteness.
| {
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Proving $\left|\begin{smallmatrix} b^2+c^2 & ab & ac \\ ba & c^2 +a^2 & bc\\ ca & cb & a^2+b^2\end{smallmatrix}\right|=4a^2b^2c^2$ I am attempting to prove the following expression, using elementary row and column operations. I have included the attempted solution.
$$\det \begin{bmatrix} b^2+c^2 & ab & ac \\ ba & c^2 +a^2 & bc\\ ca & cb & a^2+b^2\end{bmatrix}=4a^2b^2c^2$$
Attempted Solution:
$$\begin{matrix} R_{1}\to aR_{1}\\ R_{2}\to bR_{2}\\ R_{3}\to cR_{3}\end{matrix}\mapsto \det\begin{bmatrix}b^2+c^2 & a^2 & a^2\\ b^2 &c^2+a^2 & b^2 \\ c^2 &c^2 &a^2+b^2\end{bmatrix} \tag1$$ $$\begin{matrix}R_{1}\to R_{1}-(R_{2}+R_{3})\mapsto \det\begin{bmatrix}0 &-2c^2 &-2b^2 \\ b^2 &c^2+a^2 &b^2 \\ c^2 & c^2 &a^2+b^2\end{bmatrix}\end{matrix}\tag2$$ $$\begin{matrix}R_{2}\to c^2R_{2}-b^2R_{3}\end{matrix}\mapsto \det\begin{bmatrix}0 &-2c^2&-2b^2\\ 0 & c^4+a^2c^2-b^2c^2 & c^2b^2-a^2b^2-b^4\\ c^2 &c^2 &a^2+b^2\end{bmatrix}\tag3$$
I seem to have made some mistake in the transformation $(3)$. Any pointers as to why this transformation is invalid are appreciated. Thanks.
| The transformation you did in step $(3)$ actually involves two steps: $$R_2 \mapsto c^2 R_2 \\ R_2 \mapsto R_2-b^2 R_3$$
So, you have to divide by $c^2$ to preserve the identity of the determinant.
| {
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How to prove that $\ln (N+1) - \ln (N) \geq \frac{1}{N+1}$ How to prove that for each natural number ($N$) the following is true:
$$\ln(N+1)-\ln(N) \geq \frac{1}{N+1}$$
I tried using induction but it seems like the end of a road.
The first case on ($N=0$) was easy but the rest not.
Note: I'm looking for solution using induction
| This way is a bit long-winded but it does work. Note: I started writing this before the request for a proof by induction was made.
Consider $$\ln(N+1)-\ln(N)$$
$$=\ln\bigg(\frac{N+1}{N}\bigg)$$
$$=\ln\bigg(1 +\frac{1}{N}\bigg)$$
$$=\sum_{j=1}^{\infty} \frac{(-1)^{j+1}\Big(\frac{1}{N}\Big)^j}{j}$$
The Taylor series converges if $|\frac{1}{N}|< 1$, so this works for $N\ge 2$. However, the cases for $N = 0,1$ are easy to verify by calculation.
For $N \ge 2$,
$$\ln\bigg(1 +\frac{1}{N}\bigg) = \frac{1}{(1)N} - \frac{1}{(2)N^2} + \frac{1}{(3)N^3} -\frac{1}{(4)N^4} + \cdots$$
$$= \Bigg(\frac{1}{(1)N} - \frac{1}{(2)N^2}\Bigg) + \Bigg(\frac{1}{(3)N^3} -\frac{1}{(4)N^4}\Bigg) + \cdots$$
It should be clear that each bracketed term is positive; thus,
$$\ln\bigg(1 +\frac{1}{N}\bigg) > \frac{1}{N} - \frac{1}{2N^2}$$
Now consider
$$\Bigg(\frac{1}{N} - \frac{1}{2N^2}\Bigg) - \Bigg(\frac{1}{N+1}\Bigg)$$
$$ = \Bigg(\frac{2N-1}{2N^2}\Bigg) - \Bigg(\frac{1}{N+1}\Bigg)$$
$$ = \frac{N-1}{(2N^2)(N+1)} $$
$$ > 0 $$ since $N \ge 2$. Therefore,
$$\frac{1}{N} - \frac{1}{2N^2} > \frac{1}{N+1} $$
$$\implies \ln\bigg(1 +\frac{1}{N}\bigg) \ge \frac{1}{N+1}$$
| {
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On integer solutions to $x\sqrt{y}+y\sqrt{x} = a, x+y = b $ A question in quora
asked to
find real solution(s) to
$x\sqrt{y}+y\sqrt{x} = 6,
x+y = 5.
$
I showed that the solution
with $x \le y$ is
$x = 1, y = 4
$.
This naturally
brings up the question
for
which positive integers $a$ and $b$
do the equations
$x\sqrt{y}+y\sqrt{x} = a,
x+y = b
$
have integer $x$ and $y$
as solutions.
It is clear that
for any integers
$1 \le p \le q$,
$a = pq(p+q), b = p^2+q^2$
has the solution
(with $x \le y$) of
$x = p^2, y = q^2$.
The original question has
$p=1, q=2$.
My question is:
are there any other
integral $a$ and $b$
for which the
equations have
integral solutions?
Note:
If we just
try to solve for $x$,
this happens
(with the help of Wolfy):
$\begin{array}\\
y
&= b-x\\
a
&=x\sqrt{y}+y\sqrt{x}\\
&=x\sqrt{b-x}+(b-x)\sqrt{x}\\
a-x\sqrt{b-x}
&=(b-x)\sqrt{x}\\
a^2-2ax\sqrt{b-x}+x^2(b-x)
&=x(b-x)^2\\
&=x(b^2-2bx+x^2)\\
2ax\sqrt{b-x}
&=x^3-2bx^2+b^2x
-a^2-(bx^2-x^3)\\
&=2x^3-3bx^2+b^2x-a^2\\
4a^2x^2(b-x)
&=(2x^3-3bx^2+b^2x-a^2)^2\\
0
&=a^4 - 2 a^2 b^2 x + 2 a^2 b x^2 + b^4 x^2 - 6 b^3 x^3 + 13 b^2 x^4 - 12 b x^5 + 4 x^6\\
\text{with real roots}\\
x
&= \dfrac12 \left(b \pm \sqrt{b^2 - 4 \left(-\dfrac{r}{12} + \dfrac{(24 a^2 b - b^4)}{12 r} + \dfrac{b^2}{12}\right)}\right)\\
\text{where}\\
r
&=\left(-216 a^4 + 36 a^2 b^3 + 24 \sqrt{3} \sqrt{27 a^8 - a^6 b^3} - b^6\right)^{1/3}\\
\end{array}
$
I don't know
how much help
this is.
| If $$x\sqrt{y}+y\sqrt{x} = a,\tag{1}$$ with $x,y,a\in \mathbb{N},$ then $$x^2y+2xy\sqrt{xy}+xy^2=a^2,$$ so that $\sqrt{xy}$ is rational, and $xy$ is a perfect square. Let $g=\gcd(x,y),$ so that $x=gm, y=gn$ where $m$ and $n$ are co-prime integers. We have that $\sqrt{nm}$ is rational, so that $nm$ is a perfect square, and since $\gcd(n,m)=1$ $m$ and $n$ are each perfect squares, say $x=gr^2,\ y=gs^2,\ r,s\in\mathbb{N}$.
Substituting these values into $(1)$ we get that $\sqrt{g}$ is rational, so that $g$ is a perfect square, and finally, that $x$ and $y$ are perfect squares.
In short, you have found all the possibilities.
| {
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} |
Prove that $\frac{1}{1 - \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$ Given that $0 < a , b , c < 1$.
Prove that $\frac{1}{1 - \sqrt{ab}} + \frac{1}{1 - \sqrt{bc}} + \frac{1}{1 - \sqrt{ca}} \leq \frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}$.
I tried using modified C.S. and brute-force. But , it demands a lot of calculation. So , I want some better solution than this. Thank you.
| This is very similar to N.Quy's approach, but bringing out the monotonicity and convexity of $\frac1{1-x}$ made this inequality clearer for me.
Since $\frac1{1-x}$ is monotonically increasing on $[0,1)$, the AM-GM says that
$$
\frac1{1-\sqrt{xy}}\le\frac1{1-\frac{x+y}2}\tag1
$$
Since $\frac1{1-x}$ is convex on $[0,1)$, we have
$$
\frac1{1-\frac{x+y}2}\le\frac12\left(\frac1{1-x}+\frac1{1-y}\right)\tag2
$$
Therefore,
$$
\frac1{1-\sqrt{xy}}\le\frac12\left(\frac1{1-x}+\frac1{1-y}\right)\tag3
$$
Adding $(3)$ for all $3$ pairs gives
$$
\frac1{1-\sqrt{xy}}+\frac1{1-\sqrt{yz}}+\frac1{1-\sqrt{zx}}\le\frac1{1-x}+\frac1{1-y}+\frac1{1-z}\tag4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3664739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Diophantine equation $1+5^k=2^y+2^z\cdot 5^t$ Find all positive integers solutions of the equation $1+5^k=2^y+2^z\cdot 5^t$.
My idea is, let $y,z\geq 2$. Then $4\mid RHS$. But $LHS\equiv 2(mod4)$ which is contradiction.
Let $y=1$. Then the equation becomes $1+5^k=2+2^z\cdot 5^t\implies 5^k-2^z\cdot5^t=1$ which is false by $(mod5)$.
And I don't know how to continue this solution if $z=1$. I know that $(k,y,z,t)=(2,4,1,1)$ is a solution and I'm trying to prove $k\leq2$. Any hint on this problem?
| COMMENT.-$1+5^k=2^y+2^z\cdot 5^t\Rightarrow6\equiv2^y\pmod{10}$ then $y=4n$ with $n\ge1$. It follows
$$1+5^k=16^n+2^z\cdot5^t$$
►$n=1\iff y=4\Rightarrow 5^{k-1}=3+2^z\cdot5^{t-1}\Rightarrow k=2,z=1,t=1$ so
$(k,y,z,t)=(2,4,1,1)$ is a solution.
►$n=2\iff y=8\Rightarrow5^{k-1}=51+2^z\cdot5^{t-1}$.Therefore if $t\ge2$ we have
$5\equiv1\pmod{10}$ so $t=1$ but in this case we have $1\equiv3\pmod4$, absurde.
►$n=3\iff y= 12\Rightarrow5^{k-1}=819+2^z\cdot5^{t-1}=3^2\cdot7\cdot13+2^z\cdot5^{t-1}$. Like above $t=1$ but in this case we would have $1\equiv3\pmod4$, absurde.
I stop here. This is not an answer, it is a comment.
| {
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Counter-intuitive inequality : $a^{\frac{a^2}{a^2+b^2}}b^{\frac{b^2}{a^2+b^2}}+a^{\frac{a}{a+b}}b^{\frac{b}{a+b}}\leq \frac{2(a^2+b^2)}{a+b}$ I'm studing the following inequality :
Let $a,b>0$ then we have :$$a^{\frac{a^2}{a^2+b^2}}b^{\frac{b^2}{a^2+b^2}}+a^{\frac{a}{a+b}}b^{\frac{b}{a+b}}\leq \frac{2(a^2+b^2)}{a+b}$$
It's related to the function :
Let $x>0$ then we have :$$x^{\frac{x^2}{x^2+1^2}}+x^{\frac{x}{x+1}}\leq \frac{2(x^2+1)}{x+1}$$
Where $x=\frac{a}{b}$
It (RHS-LHS) increases slowly ,equivalently to the logarithmic function so I try to find a refinement in this sense :
$\exists\,x>0$ such that :
$$x^{\frac{x^2}{x^2+1^2}}+x^{\frac{x}{x+1}}+\ln(x)\leq \frac{2(x^2+1)}{x+1}+C$$ Where $C$ is a constant .
To prove it I have tried to use power series without success.
I was thinking of asymptotic evaluation.
Finally we have :
$$\lim_{x \to +\infty} x^{\frac{x^2}{x^2+1^2}}+x^{\frac{x}{x+1}}+\ln(x)-\frac{2(x^2+1)}{x+1}=2$$
My question :
How to prove the initial inequality ?
So if you have an idea to prove it or hints(I prefer it).
Thanks a lot for all your contributions .
P.S. We don't have :$$\lim_{x \to +\infty} x^{\frac{x^2}{x^2+1^2}}+x^{\frac{x}{x+1}}-\frac{2(x^2+1)}{x+1}=0$$
| Suppose that you compose Taylor series for large values of $x$. Then
$$x^{\frac{x^2}{x^2+1}}=x-\frac{\log (x)}{x}+O\left(\frac{1}{x^3}\right)$$
$$x^{\frac{x}{x+1}}=x-\log (x)+\frac{\log (x) (\log (x)+2)}{2 x}+O\left(\frac{1}{x^2}\right)$$
$$\frac{2(x^2+1)}{x+1}=2 x-2+\frac{4}{x}+O\left(\frac{1}{x^2}\right)$$
So
$$RHS-LHS=\log (x)-2+\frac{8-{\log ^2(x)}}{2x}+O\left(\frac{1}{x^2}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solution verification: Given that $x=\cos \theta$ I want to find $\frac{d^2y}{dx^2}$ in terms of $\frac{dy}{d\theta}$ Given that $x=\cos \theta$ I want to find $\frac{d^2y}{dx^2}$ in terms of $\frac{dy}{d\theta}$.
My work $\frac{dx}{dy}=-\sin {\theta} \frac{d\theta}{dy}$, rearanging gives $\frac{dy}{dx}=\frac{dy}{d\theta}\frac{1}{-\sin{\theta}}$. Differentiating again with respect to x gives
$\frac{d^2y}{dx^2}=\frac{d^2y}{d\theta^2}\frac{dy}{dx}\frac{1}{-\sin\theta}+\frac{dy}{d\theta}\frac{\cos\theta}{\sin^2\theta}\frac{d\theta}{dx}=\frac{d^2y}{d\theta^2}(\frac{dy}{d\theta}\frac{1}{-\sin\theta})\frac{1}{-\sin\theta}+-\frac{dy}{d\theta}\frac{\cos\theta}{\sin^3\theta}=\frac{d^2y}{d\theta^2}\frac{dy}{d\theta}\frac{1}{\sin^2\theta}-\frac{dy}{d\theta}\frac{\cos\theta}{\sin^3\theta}$. However, when I checked the answer on the student room it did not contain $\frac{dy}{d\theta}$ in the $\frac{d^2y}{d\theta^2}\frac{dy}{d\theta}\frac{1}{\sin^2\theta}$ bit. The way I get it is just by directly substitution what I have for $\frac{dy}{dx}$ in the previous part. Could someone explain to me why this is wrong?
| $\frac {dy}{dx} = \frac {dy}{d\theta}\frac {d\theta}{dx}\\
\frac {d}{dx} \frac {dy}{dx} = \frac {d}{dx}(\frac {dy}{d\theta}) \frac {d\theta}{dx} + \frac {dy}{d\theta}\frac {d}{dx}\frac {d\theta}{dx}\\
\frac {d^2y}{dx^2} = (\frac {d^2y}{d\theta^2}\frac {d\theta}{dx}) \frac {d\theta}{dx} + \frac {dy}{d\theta}(\frac {d^2\theta}{dx^2})\\
\frac {d^2y}{dx^2} = \frac {d^2y}{d\theta^2}(\frac {d\theta}{dx})^2 + \frac {dy}{d\theta}\frac {d^2\theta}{dx^2}\\
$
$\frac {d\theta}{dx} = -\frac {1}{\sqrt{1-x^2}}\\
\frac {d^2\theta}{dx^2} = -\frac {x}{(1-x^2)^\frac 32}$
$\frac {d^2y}{dx^2} = \frac {d^2y}{d\theta^2}\left(\frac {1}{1-x^2}\right) - \frac {dy}{d\theta}\left(\frac {x}{(1-x^2)^{\frac 32}}\right)$
Or
$\frac {dx}{d\theta} = -\sin\theta\\
\frac {d\theta}{dx} = -\csc\theta\\
\frac {d}{dx}\frac {d\theta}{dx} = \frac {d}{d\theta}(-\csc\theta\frac) {d\theta}{dx}\\
\frac {d}{dx}\frac {d\theta}{dx} = -\csc^2\theta\cot\theta\\
\frac {d^2y}{dx^2} = \frac {d^2y}{d\theta^2}\csc^2\theta - \frac {dy}{d\theta}\csc^2\theta\cot\theta$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3670109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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When $ a^5 < 5 $ show that there exists b such that $ aHere's my approach.
Since $ a^5 < 5, a<\sqrt[5]{5} $
By density of rational number, there must be integer $m$ and natural number $n$ such that
$ a< \frac{m}{n} < \sqrt[5]{5}$
If I let $b= \frac{m}{n} $, then $ a<b$ and $ b<\sqrt[5]{5} $
Since, $ b< \sqrt[5]{5}$, $ b^5 < 5 $,
therefore there exists $b$ such that $a<b$ and $b^5<5$
Is there anything that I should fix or add?
I wonder if I have to show that there exists real number $a$ such that $a^5<5$.
| Let $b=\frac{a+\sqrt[5]5}{2}.$
Thus, since $a<\sqrt[5]5,$ we see that $$b<\frac{\sqrt[5]5+\sqrt[5]5}{2}=\sqrt[5]5,$$ which gives $$b^5<5.$$
Also, $$b-a=\frac{a+\sqrt[5]5}{2}-a=\frac{\sqrt[5]5-a}{2}>0,$$ which says $a<b$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding expected value of X By rolling 6-sided fair dice, and if the outcome equals to 1 or 2, then X = 3; if it is 3 or 4, then X = 4, and if the outcome is 5 or 6, then X = 10. Find E[X].
I solved it by constructing a table of a die probability and got this:
P(X=3) = 1/36 + 2/36 = 3/36;
P(X=4) = 3/36 + 4/36 = 7/36;
P(X=5) = 5/36 + 6/36 = 11/36;
E[X] = 3*3/36 + 4*7/36 + 5*11/36 = 4.08
Is this solution correct or has any mistakes?
| Denote $D$ to mean the value of the die roll. We observe that $P(X = 3) = P(D = 1) + P(D = 2) = \frac{1}{3}$. Similarly, $P(X=4)=P(X=10)=\frac{1}{3}$.
Thus
$$
E[X] = \frac{1}{3}(3+4+10) = \frac{17}{3}.
$$
You could also just observe that the process is really rolling a fair "three sided" die to get the probabilities.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{a(b+c)}{b^2+bc+c^2} \geqslant \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)}.$ Let $a,\,b,\,c$ are non-negative such that $ab+bc+ca>0.$ Prove that
$$\frac{a(b+c)}{b^2+bc+c^2} \geqslant \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)}. \quad (1)$$
Note. Because
$$\sum \frac{8a^2+2a(b+c)-(b-c)^2}{3(a^2+b^2+c^2+ab+bc+ca)} = 2.$$
So, from $(1)$ we get know inequality of Darij Grinberg
$$\frac{a(b+c)}{b^2+bc+c^2}+\frac{b(c+a)}{c^2+ca+a^2}+\frac{c(a+b)}{a^2+ab+b^2} \geqslant 2.$$
My proof use sum of squares method.
| We have$:$
$$48\, \left( b+c \right) ^{2} \left( {b}^{2}+bc+{c}^{2} \right)
\left( {a}^{2}+ab+ac+{b}^{2}+bc+{c}^{2} \right)
\cdot (\text{LHS}-\text{RHS})$$
$$=48\,a ( b+c ) \Big[b^2+c^2-a(b+c)\Big] ^{2}+4\,(b-c)^2 \Big[ 2\, a(b+c)-(2b^2+bc+2c^2) \Big] ^{2}$$
$$+4\,bc \left( 8\,{b}^{2}+7\,bc+8\,{c}^{2} \right) \left( b-c \right) ^
{2} \geqq 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$ For $a,b,c>0$. Prove that$:$
$$\frac{ab}{c}+\frac{bc}{a}+\frac{ca}{b} +\frac{81abc}{4(a+b+c)^2} \geqq \frac{7}{4} (a+b+c)$$
My proof:
We have$:$ $$\text{LHS}-\text{RHS} =\frac{g(a,b,c)}{4abc(a+b+c)^2} \geqq 0$$
Where
$g(a,b,c) =\frac{1}{16} \left( a+b \right) ^{2} \left( 2\,a+2\,b-c \right) ^{2} \left(
a+b-2\,c \right) ^{2}$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\frac{1}{64} \left( a-b \right) ^{2} \cdot \Big[ \left( 2\,c-a-b \right) ^{3} \left(
119\,a+119\,b+30\,c \right)$
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+\left( a+b-2\,c \right) ^{2} \left( 343\,{a}^{2}+346\,ab+343\,{b}^{2}
\right) $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+24\, \left( 2\,c-a-b \right) \left( a+b \right) \left( 16\,{a}^{2}+a
b+16\,{b}^{2} \right) $
$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,+36\, \left( 4\,{a}^{2}-5\,ab+4\,{b}^{2} \right) \left( a+b \right) ^{
2} \Big] \geqq 0$
which is clearly true for $c=\max\{a,b,c\}$
I wish to see another proof without $uvw$! Thanks for a real lot!
You can see also here.
|
here you go .
| {
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Does it follow for $x \ge 785$, that Gautschi's Inequality implies that $\frac{\Gamma(2x + 3 - \frac{1.25006}{\ln n})}{\Gamma(2x+1)} > x^2$ Does it follow for $x \ge 785$, that Gautschi's Inequality implies that $\frac{\Gamma(2x + 3 - \frac{1.25006}{\ln n})}{\Gamma(2x+1)} > x^2$
Here's my reasoning. Please let me know if I made any mistakes or made any jumps in my logic.
(1) From Gautschi's Inequality, from any real $z$ and any real $s$ where $0 < s < 1$, it follows that:
$$z^s > \frac{\Gamma(z+s)}{\Gamma(z)} > (z)(z+1)^{s-1}$$
(2) Setting $z = 2x+2$ gives us:
$$(2x+2)^s > \frac{\Gamma(2x+2+s)}{\Gamma(2x+2)} > (2x+2)(2x+3)^{s-1}$$
(3) Multiplying $2x+1$ to both sides:
$$(2x+1)(2x+2)^s > \frac{\Gamma(2x+2+s)}{\Gamma(2x+1)} > (4x^2+6x+2)(2x+3)^{s-1}$$
(4) Since $\dfrac{1.25506}{\ln x} < 1$ for $x \ge 4$, setting $s = 1 - \dfrac{1.25506}{\ln x}$ gives us:
$$\frac{\Gamma(2x+3-\frac{1.25506}{\ln x})}{\Gamma(2x+1)} > (4x^2+6x+2)(2x+3)^{-\frac{1.25506}{\ln x}} = \frac{4x^2+6x+2}{(2x+3)^{\frac{1.25506}{\ln x}}}$$
(5) Since for $x \ge 785$ (see here for details), $(2x+3)^{\frac{1.25506}{\ln x}} < 4$, it follows that for $x \ge 785$:
$$\frac{\Gamma(2x+3-\frac{1.25506}{\ln x})}{\Gamma(2x+1)} > x^2$$
| It seems to me that the inequality holds even for smaller values of $x$.
Considering that we look for the zero of function
$$f(x)=\log \left(\frac{\Gamma \left(2x+3-\frac{a}{\log (x)}\right)}{\Gamma (2
x+1)}\right)-2\log(x)$$ and using Stirling approximation plus Taylor series
$$f(x)=-\left(\frac{a \log (2)}{\log (x)}+a-2\log (2)\right)+\frac{(3 \log (x)-a) (2 \log
(x)-a )}{4 x \log ^2(x)}+O\left(\frac{1}{x^2}\right)$$ an overestimate of the solution is given by
$$2^{-\frac{a}{a-2 \log (2)}}$$ which, for $a=1.25506$ gives $756.66$.
Newton iterates are
$$\left(
\begin{array}{cc}
n & x_n \\
0 & 756.000 \\
1 & 455.485 \\
2 & 512.524 \\
3 & 517.233 \\
4 & 517.260
\end{array}
\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve $\int_0^2 \int_0^\sqrt{4-x^{2}} \int_0^\sqrt{4-x^2 -y^2} z \sqrt{4-x^2 -y^2} \, dz \, dy \, dx$ in spherical coordinate $$\int_0^2 \int_0^\sqrt{4-x^{2}} \int_0^\sqrt{4-x^2 -y^2} z \sqrt{4-x^2 -y^2} \, dz \, dy \, dx$$
The task is to solve this integral using spherical coordinate. After I tried to change the variable, I got
$$ \int _0^{\frac{\pi }{2}}\int _0^{\frac{\pi }{2}}\int _0^2\left(\rho \:\cos\left(\phi \right)\sqrt{4-\rho ^2\left(\sin\left(\phi \right)\right)^2}\right)\:\rho ^2\sin\left(\phi \right)d\rho \:d\theta \:d\phi
$$
Which I think pretty ugly with $\sqrt{4-\rho ^2\left(\sin\left(\phi \right)\right)^2}$ . Is there anything I did wrong on the variable changing process? If it's not, what are the approaches to solve this integral?
| The integral simplifies like so
$$\int_0^{2}\int_0^{\frac{\pi}{2}}\int_0^{\frac{\pi}{2}}\rho^3\sin\phi\cos\phi\sqrt{4-\rho^2\sin^2\phi}\:d\theta\:d\phi\:d\rho = \frac{\pi}{4}\int_0^{2}\int_0^{\frac{\pi}{2}} \rho\sqrt{4-\rho^2\sin^2\phi} \:d(\rho^2\sin^2\phi)\:d\rho$$
$$= \frac{\pi}{6}\int_0^2 -\rho \left[4-\rho^2\sin^2\phi\right]^{\frac{3}{2}}\biggr|_0^{\frac{\pi}{2}}\:d\rho = \frac{\pi}{6}\int_0^2 8\rho-\rho(4-\rho^2)^{\frac{3}{2}}\:d\rho = \frac{8\pi}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3683415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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locus problem in analytical geometry asking about a constant sum of two tangents to two identical circles yielding an ellipse You are given two circles:
Circle G: $(x-3)^2 + y^2 = 9$
Circle H: $(x+3)^2 + y^2 = 9$
Two lines that are tangents to the circles at point $A$ and $B$ respectively intersect at a point $P$ such that $AP + BP = 10$
Find the locus of all points $P$.
This problem is solvable if we set point $P = (x,y)$ and solve the equation $AP + BP = 10$. After substituting $GP^2 = AP^2 + 3^2$ and $HP^2 = BP^2 + 3^2$ and getting the following equation for an ellipse
$16x^2 +25y^2 = 625$
That's a lot of math and algebra to do, so my question is: What is the geometric reasoning behind why is the locus an ellipse (without using analytical geometry) or is there any other elegant proofs that lack heavy calculations?
| Not answering the question but giving further observation,
\begin{align}
\sqrt{(x-r)^2+y^2-r^2} \pm \sqrt{(x+r)^2+y^2-r^2} &= 2s \\
\sqrt{x^2-2rx+y^2} \pm \sqrt{x^2+2rx+y^2} &= 2s \\
2(x^2+y^2) \pm 2\sqrt{(x^2+y^2)^2-4r^2x^2} &= 4s^2 \\
(x^2+y^2)^2-4r^2x^2 &= 4s^4-4s^2(x^2+y^2)+(x^2+y^2)^2 \\
(s^2-r^2)x^2+s^2y^2 &= s^4 \\
\end{align}
*
*Positive sign is taken when $s^2>r^2$ giving an ellipse.
*Negative sign is taken for constant difference instead, the locus can be two horizontal lines $(s^2=r^2)$, a hyperbola $(s^2<r^2)$ or a vertical line $(s=0)$.
*The loci always pass through the point $(0,\pm s)$.
*The loci don't have contact with two circles when $s^2>2r^2$.
*The eccentricity is given by $e=\dfrac{r}{|s|}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculating Inverse Proportion of 5 or more variables I am trying to create a formula to calculate a wine blend based on the relative boldness (on a scale of 0-100) of each of the five blending components. At this stage, I am assuming that each component is equally liked by the taster and in order to give equal weight to the components in the blend, relative boldness is inversely proportional to the quantity of each component added.
If:
$$\ a∝b∝c∝d∝e$$
I first calculated the percentage of each component assuming they were proportional:
$$\frac{100a}{a+b+c+d+e} +\frac{100b}{a+b+c+d+e}+\frac{100c}{a+b+c+d+e}+\frac{100d}{a+b+c+d+e}+\frac{100e}{a+b+c+d+e}$$
I then calculated the inverse of each component:
$$\frac{a+b+c+d+e}{100a}+...+\frac{a+b+c+d+e}{100e}$$
and divided each by the sum of the inverse of the percentage of each component:
$$\frac{a+b+c+d+e}{100a(\frac{a+b+c+d+e}{100a}...\frac{a+b+c+d+e}{100e})}+...+\frac{a+b+c+d+e}{100e(\frac{a+b+c+d+e}{100a}...\frac{a+b+c+d+e}{100e})}$$
I then simplified the formula:
$$\frac{1}{a(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e})}+...+\frac{1}{b(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e})}$$
I tested the above method with three variables against the following formula for inverse proportions for three variables and got the same answer so I am assuming that my method works:
Given:
$$\ a∝b∝c$$
Inverse percentage of a,b,c with this formula is:
$$\frac{100bc}{ab+ac+bc}+\frac{100ac}{ab+ac+bc}+\frac{100ab}{ab+ac+bc}$$
I've been out of college for 30 years and am a bit rusty with some math. I would appreciate anyone validating my method and providing a simpler formula for calculating inverse proportions with 5 or more variables.
Thank you.
| I am not too sure precisely what you are trying to do, but it seems you want to extend $100\dfrac{\frac{1}{a}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=\dfrac{100bc}{ab+ac+bc}$ and $\dfrac{100bc}{ab+ac+bc}+\dfrac{100ac}{ab+ac+bc}+\dfrac{100ab}{ab+ac+bc}=100$
For five terms this would be
$$ 100\dfrac{\frac{1}{a}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{1}{e}}= \dfrac{100bcde}{abcd+abce+abde+acde+bcde}$$
and
$$\dfrac{100bcde}{abcd+abce+abde+acde+bcde} + \dfrac{100acde}{abcd+abce+abde+acde+bcde} \\+ \dfrac{100abde}{abcd+abce+abde+acde+bcde} + \dfrac{100abce}{abcd+abce+abde+acde+bcde} \\+ \dfrac{100abcd}{abcd+abce+abde+acde+bcde}=100$$
It could be wise not to allow scores of $0$
| {
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If $a+b+c=0$ find $\frac{a^2+b^2+c^2}{b^2-ca}$ I tried finding the value of $a^2+b^2+c^2$ and it is $-2(ab+bc+ca)$ but that doesn't have any common factor with $b^2-ca$ so that didn't help.
I squared on both sides of $a+b=-c$ to get $a^2+b^2-c^2+2ab=0$. But I didn't proceed any further.
| $0 = 0^2 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$
$a^2 + b^2 + c^2 = -2(ab + ac + bc)$
$b² - ac = -b(a + c) - ac = -(ab + ac + bc)$ as $b = -(a + c)$
Hence, the answer is 2.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum of two cubes equal to prime square If $a,b\in \mathbb{N}$ find all primes $p$ such that $a^3+b^3=p^2$
My approach-
$a^3+b^3=(a+b)(a^2-ab+b^2)=p^2$ suppose $a+b=x$ and $a^2-ab+b^2=y$ then there are two cases- $(x,y)=(p^2,1),(p,p)$
Now I am struggling for case 02 where $(x,y)=(p,p)$
| In the case $(x,y)=(p,p)$, we have $a+b=(a+b)^2-3ab=p$, hence $p^2\equiv p\pmod 3$.
If $p\neq 3$, then $p\equiv 1\pmod 3$, hence
$$ab=\frac{p-1}3p$$
Then $p|(ab)$ but since $0<a,b<a+b=p$ this is impossible.
| {
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"timestamp": "2023-03-29T00:00:00",
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Given that $n^4-4n^3+14n^2-20n+10$ is a perfect square, find all integers n that satisfy the condition So, I tried solving that by
$$n^4-4n^3+14n^2-20n+10=x^2\\10=x^2-a^2, a^2=n^4-4n^3+14n^2-20n+10\\10=(x+a)(x-a)$$ but I couldn't find any integers when I solved it
| It always helps to form squares from the biggest power and is a good strategy:
$$n^4-4n^3+14n^2-20n+10=n^2(n^2-4n+4)+10n^2-20n+10=\\
n^2(n-2)^2+10(n-1)^2=(n(n-2))^2+10(n-1)^2=((n-1)^2-1)^2+10(n-1)^2=\\
(n-1)^4-2(n-1)^2+1+10(n-1)^2=(n-1)^4+8(n-1)^2+1=\\
((n-1)^2+4)^2-15=x^2$$
I think this quite large hint will make it a bit easier to solve it.
Just as a note: it just happens we can make a nice square, that is not always the case!
| {
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Find all possible positive integers $x$ and $y$ such that the equation: $(x+y)(x-y)=\frac{(y+1)(y-1)}{24}$ is satisfied. My approach so far:
The given equation can be rewritten as: $x^2 -y^2=\frac{y^2 -1}{24}.$ This gives
$24x^2 +1=25y^2=(5y)^2.$ So $(24x^2+1)$ must also be a perfect square. This implies $x=0, 1$ is two such possible values for $x$. As $x>0$, so $x=1$ is a possible solution. Corresponding to this, we get $y=1$. But how to check does there exist any other solutions or not? Please suggest.. Thanks in advance.
| $x^2-Dy^2=1$
Let $(x_0,y_0)$ is the smallest nontrivial positive solution.
Then general solution is given as follows.
$x_n + y_n\sqrt{D} = (x_0 + y_0\sqrt{D})^n$
Let $X=5y$ and $Y=x$ then $24x^2 - 25y^2 =- 1$ is transformed to $X^2-24Y^2=1.$
Smallest nontrivial positive solution is $(X0,Y0)=(5,1).$
Then all solutions are given $X_n + Y_n\sqrt{D} = (5 + 2\sqrt{6})^n.$
We show ten solutions with $n=1\cdots 10.$
$[5+2\sqrt{6}], [49+20\sqrt{6}], [485+198\sqrt{6}], [4801+1960\sqrt{6}], [47525+19402\sqrt{6}], [470449+192060\sqrt{6}], [4656965+1901198\sqrt{6}], [46099201+18819920\sqrt{6}], [456335045+186298002\sqrt{6}], [4517251249+1844160100\sqrt{6}]$
Since $X_n$ must be divisible by $5$, we know $[485+198sqrt{6}]$ is second solution.
Thus $y=\frac{485}{5}=97, x=\frac{198}{2}=99.$
Third solution is $y=\frac{47525}{5}=9505, x=\frac{19402}{2}=9701.$
| {
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Find radius of largest circle within ellipse $\frac{x^2}{9} + \frac{y^2}4 = 1$ with their intersection only at $(3,0)$ An ellipse is defined by the equation $$\frac{x^2}{9} + \frac{y^2}4 = 1$$
Compute the radius of the largest circle that is internally tangent to the ellipse at $(3,0),$ and intersects the ellipse only at $(3,0).$
How can I write an equation for the largest circle within this ellipse if its equation is given? Is there a property or theorem I'm missing? Is there another way to solve?
| Let the equation of the circle $(x-a)^2+y^2 =(3-a)^2$ that passes the point $(3,0)$. Then, substitute $y^2$ into $\frac{x^2}{9} + \frac{y^2}4 = 1$ to get
$$\frac59 x^2 -2ax +6a-5=0$$
Since the two shapes has only one common point $(3,0)$, the discriminate of above quadratic equation is zero, which yields $a= \frac53$ and hence the radius $3-a= \frac43$.
| {
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Find $\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \to \min$ I'm trying to solve
\begin{align*}
&\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \to \min \\
&x + y + z = 1 \\
&x, y, z > 0,
\end{align*}
using only inequalities.
How can i solve it? I used am-gm, tried to "break" terms, but got only after all $(xy + yz + xz) = \frac{1}{2}$
| By Cauchy-Schwarz we get$$\frac{x^2}{y} + \frac{y^2}{z} + \frac{z^2}{x} \geq \frac{(x+y+z)^2}{x+y+z}=1$$
| {
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Show that $f(x) = |1-x^2|^3$ is a differentiable function. $f: \mathbb{R} \rightarrow \mathbb{R}$
$x \rightarrow |1-x^2|^3$
Show that f is a differentiable function and calculate its derivative.
Check whether f if is a continuous function.
$f'(x_0)=lim_{x\rightarrow x_0} \frac{f(x) - f(x_0}{x-x_0} = lim_{x\rightarrow x_0}\frac{|1-x^2|^3-|1-x_0^2|^3}{x-x_0} = lim_{x\rightarrow x_0} \frac{\sqrt{(1-x^2)^2}^3- \sqrt{(1-x_0^2)^2}^3}{x-x_0} = lim_{x\rightarrow x_0} \frac{\sqrt{(1-x^2)^6}- \sqrt{(1-x_0^2)^6}}{x-x_0} = lim_{x\rightarrow x_0} \frac{(1-x^2)^3- (1-x_0^2)^3}{x-x_0} = lim_{x\rightarrow x_0} \frac{(1-x^2)^3- (1-x_0^2)^3}{x-x_0}$
But I didn't get any further. How to get the $x-x_0$ out? I tried multiplying out the numerator and then using polynomial division but I didn't get a solution.
So I tried calculating the derivative directly, but that doesn't show it generally.
$f(x) = |1-x^2|^3 = (\sqrt{(1-x^2)^2})^3$
$f'(x) \\= 3 \cdot (\sqrt{(1-x^2)^2})^2 \cdot \frac{1}{2} \cdot ((1-x^2)^2)^{-\frac{1}{2}} \cdot 4x\cdot (1-x^2) \\ = 3 \cdot |1-x^2|^2 \cdot \frac{1}{|1-x^2|} \cdot 2x \cdot (1-x^2) \\= |1-x^2| \cdot 6x \cdot (1-x^2)$
I looked at the graph. The graph looks like a "W" but with "soft turns". If I can prove that the function is differentiable in every point then it is a continuous function.
So how do I prove it using this definition: $f'(x_0)=lim_{x\rightarrow x_0} \frac{f(x) - f(x_0}{x-x_0}$ ?
| Two cases: $|x_0|=1$ and $|x_0|\ne 1$ - that's where $|1-x^2|$ does not have a derivative.
For the first one (let us take $x_0=1$; the case $x_0=-1$ is done likewise) you get by definition
$$\lim_{x\to 1}\frac{|1-x^2|^3}{x-1} = \lim_{h\to 0}\frac{|1-(1+h)^2|^3}{h} = \lim_{h\to 0}\frac{|h^2+2h|^3}{h} = \lim_{h\to 0}\frac{|h|^3|h+2|^3}{h} = \lim_{h\to 0}\frac{|h|^3|h+2|^3}{h} = \lim_{h\to 0} \left(sgn (h)\cdot h^2|h+2|^3\right) = 0.$$
In the case where $x_0\ne \pm 1$ you can say that for $x$ sufficiently close to $x_0$ you can expand absolute value; for example, for $|x|<1$ you get $|1-x^2| = 1 - x^2$. After this transformation proceed as you would when proving that polynomials are differentiable.
| {
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Closed form of $\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx$
Can a closed form solution for the following integral be found:
$$\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx\,?$$
I have tried all the standard tricks such as integration by parts, various substitutions, and parametric differentiation (Feynman's trick), but all to no avail.
An attempt is letting
$$f(t):=\int_0^\infty\,\arctan^2\left(\frac{2tx}{1+x^2}\right)\,\text{d}x\,.$$
Therefore,
$$f'(t)=\int_0^\infty\,\frac{8x^2(x^2+1)}{\big(x^4+2(2t^2+1)x^2+1\big)^2}\,\left(1+x^2-4tx\arctan\left(\frac{2tx}{1+x^2}\right)^{\vphantom{a^2}}\right)\,\text{d}x\,.$$
This doesn't seem to go anywhere. Help!
| Here is a solution based on Fubini's theorem.
According to an addition formula
\begin{equation*}
\arctan\left(\dfrac{2x}{1+x^2}\right) = \arctan((\sqrt{2}+1)x)-\arctan((\sqrt{2}-1)x) .
\end{equation*}
Furthermore
\begin{equation*}
\arctan x=\mathrm{sign}(x)\dfrac{\pi}{2}-\arctan\dfrac{1}{x} .
\end{equation*}
Consequently
\begin{equation*}
\arctan\left(\dfrac{2x}{1+x^2}\right) = \arctan\dfrac{\sqrt{2}+1}{x}-\arctan\dfrac{\sqrt{2}-1}{x}=\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{x}{x^2+s^2}\, ds .
\end{equation*}
Via Fubini's theorem we get
\begin{gather*}
\int_{0}^{\infty}\arctan^2\left(\dfrac{2x}{1+x^2}\right)\, dx = \int_{0}^{\infty}\left(\arctan\dfrac{\sqrt{2}+1}{x}-\arctan\dfrac{\sqrt{2}-1}{x}\right)^2\, dx=\\[2ex]
\int_{0}^{\infty}\left(\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{x}{x^2+s^2}\, ds\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{x}{x^2+t^2}\, dt\right)\, dx=\\[2ex]
\int_{\sqrt{2}-1}^{\sqrt{2}+1}\left(\int_{\sqrt{2}-1}^{\sqrt{2}+1}\left(\int_{0}^{\infty}\dfrac{x^2}{(x^2+s^2)(x^2+t^2)}\, dx\right)\, ds\right)\, dt=\\[2ex]
\dfrac{\pi}{2}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\left(\int_{\sqrt{2}-1}^{\sqrt{2}+1}\dfrac{1}{s+t}\, ds\right)\, dt=\\[2ex]
\dfrac{\pi}{2}\int_{\sqrt{2}-1}^{\sqrt{2}+1}\left(\ln(t+\sqrt{2}+1)-\ln(t+\sqrt{2}-1)\right)\, dt=\\[2ex]
2\pi\ln(\sqrt{2}+1)-\sqrt{2}\pi\ln 2.
\end{gather*}
Remark.
Since
\begin{equation*}
\arctan\left(\dfrac{2x\sinh\alpha}{1+x^2}\right)=\arctan\left(\dfrac{e^{\alpha}}{x}\right)-\arctan\left(\dfrac{e^{-\alpha}}{x}\right) = \int_{e^{-\alpha}}^{e^{\alpha}}\dfrac{x}{x^2+s^2}\, ds
\end{equation*}
the $@$Sangchul Lee's generalization can be proved in the same way.
| {
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"timestamp": "2023-03-29T00:00:00",
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Algebra/number theory solution check, number of 0's at end of integer As part of a larger problem, I wish to calculate the value of $\frac{1993^2+1993}{2} \pmod {2000}$. The top reduces to $42$. However, $\gcd(2,2000)>1$, so the solution is not $21$, and carrying out the division would require changing the modulus.
Is there a method to divide in this case (or in a general case entirely applicable to this case), such that doing this division will not change the modulus from 2000, i.e. the solution $x$ is $\frac{1993^2+1993}{2} \equiv x \pmod {2000}$.
I understand that I can just do the arithmetic, but I wonder if it is possible to keep the modulus at $2000$ after the division.
| $1993^2 + 1993 \equiv (-7)^2 +(-7) \equiv 42\pmod {2000}$.
So $\frac {1993^2+1993}2 \equiv \frac {42}2 \pmod {\frac {2000}{\gcd(2,2000)}}$
So $x \equiv 21\pmod{1000}$ so $x \equiv 21, 1021 \pmod {2000}$
$2000 = 125*16$ and if we consider the chinese remainder theorem we get that $x \equiv 21 \pmod{125}$
But to solve $x \pmod{16}$ we solve $1993^2 + 1993\pmod{32}\equiv 9^2 +9\equiv 26\pmod {32}$.
So $\frac{1993^2 + 1993}2 \equiv \frac{32k +26}2\equiv 16k + 13 \equiv 13 \equiv -3\pmod {16}$.
And as $21 \not \equiv -3\pmod{16}$ then $x\equiv 21 \pmod {2000}$ can't be the solution.
So it must be $x \equiv 1021 \pmod{2000}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Increasing function $ f : \mathbb R ^ + \to \mathbb R $ with $ x f ( x ) + 2 > 0 $ and $ f ( x ) f \left( f ( x ) + \frac 4 x \right) = 1 $ Let $ f : \mathbb R ^ + \to \mathbb R $ be an increasing function, such that $ x f ( x ) + 2 > 0 $ and $ f ( x ) f \left( \frac { x f ( x ) + 4 } x \right) = 1 $, then find the value of $ \lfloor f ' ( 1 ) \rfloor $ (where $ \lfloor \cdot \rfloor $ represents greatest integer function).
My approach is as follow $ f ( x ) f ( y ) = 1 $ as $ y = \frac { x f ( x ) + 4 } x $, I am trying to use the function given as $ x f ( x ) + 2 > 0 $ in $ y $ but I'm not able to proceed.
| Let $g(x)=\frac{xf(x)+4}{x}=f(x)+\frac{4}{x}$. We have
$$ f(x)f(g(x)) =1$$
Since $xf(x)+2>0$, we also have $g(x)>\frac{2}{x}>0$, which means that we also have
$$ f(g(x))f(g(g(x))) =1$$
which means that
$$ f(x) = f(g(g(x)))$$
since $f$ is an increasing function, it means that
$$ x=g(g(x))=f(g(x))+\frac{4}{g(x)}= \frac{1}{f(x)}+\frac{4}{f(x)+\frac{4}{x}}$$
Solving for $f(x)$, we can find
$$ f(x) = \frac{1\pm\sqrt{17}}{2x} $$
Since we know that $f$ is increasing, we must conclude that the minus sign is the correct one
$$ f(x) = \frac{1-\sqrt{17}}{2x} $$
Form this, we find
$$ f'(1) = \frac{\sqrt{17}-1}{2} $$
$$ \lfloor f'(1)\rfloor = 1$$
| {
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Inapplicability of Gauss' mean value theorem in an integration What is the value of $$\int_0^{2\pi}\ln|i+2e^{ix}|dx$$
At first glance I thought of applying Gauss mean value theorem, but this is not applicable here as $\ln|z|$ is not analytic at $0$, which belongs in the region $|z-i|<2$.
How one can approach this one? Any help is appreciated.
| Too long for a comment, but using $e^{i\theta}=\cos\theta+i\sin\theta$ we have
$$|i+2\cos(x)+2i\sin(x)|=\sqrt{4\cos^2 x+(1+2\sin x)^2}=\sqrt{4\cos^2x+1+4\sin^2x+4\sin x}$$
so that
$$I=\frac{1}{2}\int_0^{2\pi}\ln(5+4\sin x)dx=\frac{1}{2}\int_0^{2\pi}\ln(4)dx+\frac{1}{2}\int_0^{2\pi}\ln\left(\frac{5}{4}+\sin x\right)dx.$$
Hence
$$I=2\pi\log(2)+\frac{1}{2}\int_0^{2\pi}\ln\left(\frac{5}{4}+\sin x\right)dx=2\pi\log 2.$$
The only way I could show the integral to vanish was using the power series for $\log(1+z)$,
$$\int_0^{2\pi}\ln\left(\frac{5}{4}\left(1+\frac{4}{5}\sin x\right)\right)dx=2\pi\ln(5/4)+\int_0^{2\pi}\ln\left(1+\frac{4}{5}\sin x\right)dx=J.$$
Assuming we can switch integral and infinite sum, and since $4/5|\sin x|<1$, then the integral becomes
$$-\sum_{k=1}^\infty\frac{(-1)^k}{k}\frac{4^k}{5^k}\int_0^{2\pi}\sin^k xdx=-\frac{1}{2}\sum_{k=1}^\infty\frac{1}{k}\frac{4^{2k}}{5^{2k}}\frac{\pi \cdot\left(\frac{3}{2}\right)_{k-1}}{\Gamma (k+1)}=-2\pi\log(5/4),$$
hence $J=0$.
| {
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Sequence problem : Find $|a_1|$
Given a sequence such that
$$a_1a_2=1\ ; \ a_2a_3=2 \ ; \ a_3a_4=3 \ ; \ \dots$$ and
$$\lim_{n\to \infty} \frac{a_n}{a_{n+1}}=1$$
find $|a_1|$.
My attempts :
We can deduce from this $a_1a_2=1 \ ; \ a_2a_3=2 \ ; \ a_3a_4=3 \ ; \ \dots$ that :
$$\prod_{n=k}^{n+1} a_n=n$$
Thus :
$$\begin{align}a_n \times a_{n+1}&=n \\ a_{n+1}&=\frac{n}{a_n} \end{align}$$
Therefore :
$$\begin{align} \lim_{n\to \infty} \frac{a_n}{a_{n+1}}&=\lim_{n\to \infty} \frac{a_n}{\frac{n}{a_n}} \\ &=\lim_{n\to \infty} \frac{a_n^2}{n} \\&=1 \end{align} $$
Is this can lead us to say that :
$$\begin{align} |a_n^2|&\sim n \\ |a_n|&\sim \sqrt{n} \end{align} $$
If this is true we can say that :
$$|a_n| =1$$
But I don't know if this is true or not. Any tips or hints ?
Thanks in advance !
| Since $a_{n+1} = n/a_n$ the sequence $a_n$ is fully determined from the starting value $a_1$.
In fact it turns out that
$$a_n = \frac{n-1}{a_{n-1}} =\frac{n-1}{n-2} a_{n-2} = \cdots = \frac{(n-1)!!}{(n-2)!!} a_1^{(-1)^{n+1}}$$
Where $N!!$ denotes the semifactorial. From this you get that
$$\frac{a_{n}}{a_{n+1}} = \frac{((n-1)!!)^2}{(n-2)!! n !!} (a_1)^{2(-1)^{n+1}}$$
In order to analyze asymptotic behaviour you may want distinguish even and odd $n$
$$\frac{a_{2n}}{a_{2n+1}} = \frac{((2n-1)!!)^2}{(2n)!!(2n-2)!!} a_1^{-2} = \left(\frac{(2n-1)!!}{(2n)!!}\right)^2 \frac{2n}{a_1^2} = \left(\frac{(2n)!}{(2n )!!^2}\right)^2\frac{2n}{a_1^2}$$
$$\frac{a_{2n+1}}{a_{2n+2}} = \frac{((2n)!!)^2}{(2n+1)!!(2n-1)!!} a_1^{2} = \left(\frac{(2n)!!}{(2n-1)!!}\right)^2 \frac{a_1^2}{2n+1} = \left(\frac{(2 n)!!^2}{(2n)!}\right)^2\frac{a_1^2}{2n+1}$$
Passing to logarithms you get
$$\log\left(\frac{a_{2n}}{a_{2n+1}}\right) = 2 \big( \log( (2n)! ) - 2\log( (2n)!!) \big) +\log(2n) - 2\log(a_1)$$
$$\log\left(\frac{a_{2n+1}}{a_{2n}}\right) = 2 \big( 2\log( (2n)!!) - \log((2n)!)\big) - \log(2n+1) + 2\log(a_1)$$
Recall the Stirling approximation
$$\log(n!)=n\log(n)-n+\frac{1}{2} \log(n) + \log(\sqrt{2\pi}) +o(1)$$
from this you get approximations
$$\log((2n)!!) = \log(2^n n!) = n\log(n) + (\log(2)-1) n + \frac{1}{2} \log(n) + \log(\sqrt{2\pi}) +o(1) $$
$$\log((2n)!) = 2n \log(n) + (2\log(2)-2) n + \frac{1}{2} \log(n) + \log(2\sqrt{\pi})+o(1)$$
Subtracting you have $2\log((2n)!!) - \log((2n)!) = \frac{1}{2} \log(n) + \log( \sqrt{\pi}) +o(1)$, so
$$\log\left(\frac{a_{2n}}{a_{2n+1}}\right) = \log(2) - \log(\pi) -2 \log(a_1) +o(1)$$
$$\log\left(\frac{a_{2n+1}}{a_{2n}}\right) = \log(\pi) - \log(2) + 2\log(a_1) +o(1)$$
Hence there is a value of $a_1$ for which $a_n/a_{n+1} \to 1$ and it has to satisfy $2\log(a_1)= \log(2/\pi)$, that is $a_1= \sqrt{2/\pi}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\lim_{x\to\infty} \frac{(x^2+1)^{1/2}-(x^2+1)^{1/3}}{(x^4+1)^{1/4}-(x^4+1)^{1/5}}$ $$\lim_{x\to\infty} \frac{(x^2+1)^{1/2}-(x^2+1)^{1/3}}{(x^4+1)^{1/4}-(x^4+1)^{1/5}}$$
What I did to solve this problem is here
I tried to multiply by $1/x$ the numerator and denominator and I got a limit of $1$ at the end, but I am not sure if what I got is right.
| Your answer $1$ is correct.
$$\lim_{x\to\infty} \frac{(x^2+1)^{1/2}-(x^2+1)^{1/3}}{(x^4+1)^{1/4}-(x^4+1)^{1/5}}$$
$$=\lim_{x\to\infty} \frac{x\left(1+\frac{1}{x^2}\right)^{1/2}-x^{2/3}\left(1+\frac{1}{x^2}\right)^{1/3}}{x\left(1+\frac{1}{x^4}\right)^{1/4}-x^{4/5}\left(1+\frac{1}{x^4}\right)^{1/5}}$$
$$=\lim_{x\to\infty} \frac{\left(1+\frac{1}{x^2}\right)^{1/2}-\frac{1}{x^{1/3}}\left(1+\frac{1}{x^2}\right)^{1/3}}{\left(1+\frac{1}{x^4}\right)^{1/4}-\frac{1}{x^{1/5}}\left(1+\frac{1}{x^4}\right)^{1/5}}$$
$$=\frac{1-0}{1-0}=1$$
| {
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Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$ Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$
My try:
we have $$\frac{13}{32}=\frac{2^2+3^2}{2^5}=\frac{1}{8}\left(1+(1.5)^2)\right)$$
Let $x=1.5$
Now consider the function $$f(x)=\frac{1+x^2}{8}-\ln x$$
$$f'(x)=\frac{x}{4}-\frac{1}{x}$$ So $f$ is Decreasing in $(0,2)$
any help here?
| The difference is so small that I see no other way than to do the computation. Note $$e^x = \sum_{k=0}^\infty \frac{x^k}{k!}$$ implies $$e^{13/32} > 1 + \frac{13}{32} + \frac{(13/32)^2}{2!} + \frac{(13/32)^3}{3!} + \frac{(13/32)^4}{4!} = \frac{12591963}{8388608} > \frac{3}{2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Can anyone suggest other than Stolz-Cesaro Theorem to solve this? $$\lim _{n \rightarrow \infty} \frac{1}{n^{2020}} \sum_{k=1}^{n}(k+2019)^{2019}$$
My approach:
using Stolz-Cesaro Theorem
let,\begin{aligned}
&a_{n}=\sum_{k=1}^{n}\left(k+2019\right)^{2019}, b_{n}=n^{2020}\\
&a_{n+1}-a_{n}=(n+2020)^{2019}=n^{2019 \ldots}\\
&\begin{array}{l}
b_{n +1}-b_{n}=(n+1)^{2020}-n^{2020} \\
then,
\end{array}\\
&\lim _{n \rightarrow \infty} \frac{a_{n+1}-a_{n}}{b_{n+1}-b_{n}}=\frac{1+\cdots}{2020+\cdots}=\left(\frac{1}{2020}\right)
\end{aligned}
| Another approach using Riemann sums.
For integer $a\geq 1$, we have
$$
\frac{1}{n^{a+1}}\sum_{k=1}^n \left(k+a\right)^a
= \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}+\frac{a}{n}\right)^a
\geq \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}\right)^a \xrightarrow[n\to\infty]{}\int_0^1 x^a dx = \frac{1}{a+1}
$$
On the other hand, for any fixed $\varepsilon > 0$, we have for $n\geq a/\varepsilon$
$$
\frac{1}{n^{a+1}}\sum_{k=1}^n \left(k+a\right)^a
= \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}+\frac{a}{n}\right)^a
\leq \frac{1}{n}\sum_{k=1}^n \left(\frac{k}{n}+\varepsilon\right)^a \xrightarrow[n\to\infty]{}\int_0^1 (x+\varepsilon)^a dx = \frac{(1+\varepsilon)^{a+1} - \varepsilon^{a+1}}{a+1}
$$
Therefore, for any fixed $\varepsilon > 0$,
$$
\frac{1}{a+1}
\leq \lim\inf_{n\to\infty} \frac{1}{n^{a+1}}\sum_{k=1}^n \left(k+a\right)^a
\leq \lim\sup_{n\to\infty} \frac{1}{n^{a+1}}\sum_{k=1}^n \left(k+a\right)^a
\leq \frac{(1+\varepsilon)^{a+1} - \varepsilon^{a+1}}{a+1}
$$
Now take the limit as $\varepsilon \to 0^+$ to conclude.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $\int \frac{x}{\sin^2x-3}dx$ I started like this
\begin{align*}
\int \frac{x}{\sin^2x-3}dx & =\int \frac{x\sec^2x}{\tan^2x-3\sec^2x}dx\\
& =-\int \frac{x\sec^2x}{2\tan^2x+3}dx\\
& = -\left [ \frac {x\tan^{-1}\left (\frac {\sqrt {2}\tan x}{\sqrt {3}}\right)}{\sqrt {6}}-\int \frac {\tan^{-1}\left (\frac {\sqrt {2}\tan x}{\sqrt {3}}\right)}{\sqrt {6}}dx\right]
\end{align*}
Now i am unable to solve this further.
| As said in comments, after integration by parts, you are left with
$$I=\int \tan ^{-1}\left(\sqrt{\frac{2}{3}} \tan (x)\right)\,dx$$ which is a monster that I should try to avoid.
In order to compute it, I should rather consider series expansions built around $x=k \pi$ and use for
$$J_k=\int_{k\pi}^{(2k+1)\frac \pi 2} \tan ^{-1}\left(\sqrt{\frac{2}{3}} \tan (x)\right)\,dx$$ the series expansion of the integrand. This is
$$\tan ^{-1}\left(\sqrt{\frac{2}{3}} \tan (x)\right)=t+\frac{t^3}{6}-\frac{3 t^7}{280}-\frac{t^9}{504}+O\left(t^{11}\right)$$ where $t=\sqrt{\frac{2}{3}} (x-\pi k)$.
For illustration, this would give
$$J_0=\frac{\pi ^2}{4 \sqrt{6}}\left(1+\frac{\pi ^2}{72}-\frac{\pi ^6}{80640}-\frac{\pi ^8}{3265920} \right)\approx 1.1305$$ while the monster would give as an exact solution
$$\frac{1}{2 \sqrt{6}}\Phi \left(\frac{2}{3},2,\frac{1}{2}\right)+\frac{1}{2} \log
\left(\frac{3}{2}\right) \tanh ^{-1}\left(\sqrt{\frac{2}{3}}\right)\approx 1.1326$$ For sure, we could have better results pushing the expansion to higher orders.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove derivative by induction $f:(0, \infty) \rightarrow \mathbb{R}$
$f(x) = \sqrt{x}$
a) Calculate the first four derivatives
$f'(x) = \frac{1}{2}\cdot \frac{1}{\sqrt{x}}$
$f''(x) = -\frac{1}{4}\cdot \frac{1}{\sqrt{x^3}}$
$f'''(x) = \frac{3}{8}\cdot \frac{1}{\sqrt{x^5}}$
$f''''(x) = -\frac{15}{16}\cdot \frac{1}{\sqrt{x^7}}$
b) Prove by induction that the following formula holds true:
$f^{(k)}(x) = \frac{(-1)^{k+1}}{2^k}\cdot\prod^{k-1}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k-1}}}$
Base Case: $k=1$:
$f'(x) = \frac{(1)}{2}\cdot\prod^{0}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x}} = \frac{1}{2}\cdot\frac{1}{\sqrt{x}}$
Inductive Hypothesis(IH):
Assumption holds true for some k.
Inductive step:
$k \rightarrow k+1$
to show:
$f^{(k+1)}(x) = \frac{(-1)^{k+2}}{2^k+1}\cdot\prod^{k}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k+1}}}$
$\frac{(-1)^{k+2}}{2^k+1}\cdot\prod^{k}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k+1}}} \\
= \frac{(-1)^{k+1}}{2^k}\cdot\frac{-1}{2} \cdot(2k-1)\cdot\prod^{k-1}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k-1}}}\cdot \frac{1}{\sqrt{x^2}} \\
= [\frac{(-1)^{k+1}}{2^k}\cdot\prod^{k-1}_{j=1}(2j-1)\cdot\frac{1}{\sqrt{x^{2k-1}}}] \cdot\frac{-1}{2}\cdot(2k-1)\frac{1}{\sqrt{x^2}} \\
=^{IH} f^{(k)}(x) \cdot \frac{-1}{2}\cdot(2k-1)\frac{1}{\sqrt{x^2}} \\
= f^{(k)}(x) \cdot (-k+\frac{1}{2})\cdot \frac{1}{|x|} \text{since x $\in$ $(0, \infty)$} \\
= f^{(k)}(x) \cdot [(-k+\frac{1}{2})\cdot \frac{1}{x}]$
This means that to get from one derivative to the next, the factor at the end will be multiplied.
So how do I go on from here?
Usually I would start from the left side. But if I would start here from the left side I would have to transform $f^{(k+1)}(x)$. And I would have to calculate the derivative over the product symbol.
$f^{(k+1)}(x) = f^{(k)'}(x)$
| You start with letting $A=\frac{(-1)^{k+1}}{2^k}\cdot\prod^{k-1}_{j=1}(2j-1)$ (a constant), so
\begin{align*}
f^{(k)}(x) & = A\cdot\frac{1}{\sqrt{x^{2k-1}}}\\
f^{(k+1)}(x) & = A\cdot\frac{d}{dx}\frac{1}{\sqrt{x^{2k-1}}}\\
& = A\cdot\frac{d}{dx}x^{\frac{1-2k}{2}}\\
& = A\cdot \left(\frac{1-2k}{2}\right) \cdot x^{\frac{1-2k-2}{2}}\\
& = \color{red}{A}\cdot \left(\frac{1-2k}{2}\right) \cdot \frac{1}{\sqrt{x^{2(k+1)-1}}}\\
&= \color{red}{\frac{(-1)^{k+1}}{2^k}\cdot\prod^{k-1}_{j=1}(2j-1)} \left(\frac{1-2k}{2}\right) \cdot \frac{1}{\sqrt{x^{2(k+1)-1}}}\\
&= \frac{(-1)^{k+1}}{2^k}\cdot\prod^{k-1}_{j=1}(2j-1) \color{red}{(-1)\left(\frac{2k-1}{2}\right)} \cdot \frac{1}{\sqrt{x^{2(k+1)-1}}}\\
&= \color{blue}{\frac{(-1)^{k+2}}{2^{k+1}}\cdot\prod^{k}_{j=1}(2j-1)} \cdot \frac{1}{\sqrt{x^{2(k+1)-1}}}.
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve for $x, y \in \mathbb R$ - $\left\{ \begin{align} \sqrt x + \sqrt {y + 3} = 3\\ \sqrt{5 - x^2} + \sqrt y = 3 \end{align} \right.$.
Solve the following system of equations for $x, y \in \mathbb R$ - $\left\{ \begin{align} \sqrt x + \sqrt {y + 3} = 3\\ \sqrt{5 - x^2} + \sqrt y = 3 \end{align} \right.$.
$(x \in [0, \sqrt 5], y \ge 0)$
Of course, we could go further and have that $$\sqrt x + \sqrt {y + 3} = \sqrt{5 - x^2} + \sqrt y \iff \sqrt{5 - x^2} - \sqrt x = \sqrt{y + 3} - \sqrt y \le \sqrt 3$$
$$\implies x \in \left[-\frac{1}{2} + \sqrt 3 - \frac{1}{2}\sqrt{4\sqrt 3 - 3}, \sqrt 5\right]$$
Notwithstanding, it wouldn't help much.
Furthermore, we have that $$[(\sqrt{y + 3})^2 - (\sqrt y)^2] - (\sqrt{y + 3} + \sqrt y)(\sqrt{y + 3} - \sqrt y) = 0$$
$$\implies (3 - \sqrt x)^2 - (3 - \sqrt{5 - x^2})^2 - (6 - \sqrt x - \sqrt{5 - x^2})(\sqrt{5 - x^2} - x) = 0$$
$$\iff (\sqrt x - 1)\sqrt x(6 - \sqrt x - \sqrt{5 - x^2}) = 0 \iff \left[ \begin{align} \sqrt x - 1 = 0\\ \sqrt x = 0\\ 6 - \sqrt x - \sqrt{5 - x^2} = 0 \end{align} \right.$$
It needs to be proven that $6 - \sqrt x - \sqrt{5 - x^2} > 0$ for $\forall x \in [0, \sqrt 5]$, which I couldn't have in the three hour I was given in the competition.
$\implies \left[ \begin{align} x = 1\\ x = 0\\ \end{align} \right.$.
For $x = 1$, we have $y = 1$, and for $x = 0$, we have $y = 6$ and $y = \dfrac{\sqrt{10} - \sqrt 2}{2}$, which is, of course, illogical.
The only solution to the above system of equations is $x = y = 1$.
Is the above solution correct? How could I prove that $6 - \sqrt x - \sqrt{5 - x^2} > 0$ for $\forall x \in [0, \sqrt 5]$? And should you come up with any better solutions, please write them down below. Thanks for your attention.
|
Is the above solution correct?
No, it isn't.
Note that $\sqrt{y+3}-\sqrt y$ is equal to $\sqrt{5-x^2}-\color{red}{\sqrt x}$, not $\sqrt{5-x^2}-x$.
More importantly,
$$(3 - \sqrt x)^2 - (3 - \sqrt{5 - x^2})^2 - (6 - \sqrt x - \sqrt{5 - x^2})(\sqrt{5 - x^2} - \color{red}{\sqrt x}) = 0$$
holds for every $x$ such that $x\in[0,\sqrt 5]$. (You can see by expanding LHS that LHS is a constant function which is equal to $0$.)
So, your method does not work.
Eliminating $y$, we get
$$(3-\sqrt x)^2-(3-\sqrt{5-x^2})^2=3$$
which is equivalent to
$$ -(5-x^2)+6\sqrt{5-x^2}-8+(x-6\sqrt x+5)=0$$
which can be written as
$$ (-\sqrt{5-x^2}+2)(\sqrt{5-x^2}-4)+(\sqrt x-1)(\sqrt x-5)=0$$
which is equivalent to
$$\frac{(x-1)(x+1)}{2+\sqrt{5-x^2}}(\sqrt{5-x^2}-4)+\frac{x-1}{\sqrt x+1}(\sqrt x-5)=0,$$
i.e.
$$(x-1)\bigg(\underbrace{\frac{x+1}{2+\sqrt{5-x^2}}}_{\text{positive}}(\underbrace{\sqrt{5-x^2}-4}_{\text{negative}})+\underbrace{\frac{1}{\sqrt x+1}}_{\text{positive}}(\underbrace{\sqrt x-5}_{\text{negative}})\bigg)=0$$
from which we get $x=1$ and $y=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find integer solution $|x^2-y|=8|y^2-1|$ Find integer solution $|x^2-y|=8|y^2-1| (1) $
I try to: If $y^2-1=0$ we have $y\in\{1,-1\}$
With $y=1\Longrightarrow x=\pm 1$
$y=-1\Longrightarrow x^2+1=0$ not satisfied.
In other case: $\left| \dfrac{x^2-y}{y^2-1}\right|=8$.
But I can't to solve it.
| As pointed out by several commenters, because of the absolute values, there are only two possible cases.
Case 1: $x^2-y=8(1-y^2)$. Then $8y^2-y=8-x^2$. The left-hand side is greater than or equal to $7$ for all $y ≠ 0$, while the right-hand side is less than or equal to $8$ for all $x$. The only possible equalities are $7$ or $8$, and it is easily seen that $(x,y)=(\pm 1,1)$ is the only solution.
Case 2: $x^2-y=8(y^2-1)$. This has an infinite number of solutions, which can be found using standard quadratic methods.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplification of $\sqrt{2\zeta^2-1+2\zeta\sqrt{\zeta^2-1}}+\sqrt{2\zeta^2-1-2\zeta\sqrt{\zeta^2-1}}$ If I try to evaluate $\sqrt{2\zeta^2-1+2\zeta\sqrt{\zeta^2-1}}+\sqrt{2\zeta^2-1-2\zeta\sqrt{\zeta^2-1}}$ numerically for real $\zeta$, it looks like it is just equal to $2|\zeta|$ for $\zeta \ne 0$ and $2j$ for $\zeta=0$, but I can't figure out how to simplify to get there...
It's of the form $\sqrt{b+c} + \sqrt{b-c}$ with $b=2\zeta^2-1$ and $c=2\zeta\sqrt{\zeta^2-1}$. I can write:
$$\sqrt{b+c} + \sqrt{b-c} = \frac{(b+c) - (b-c)}{\sqrt{b+c} - \sqrt{b-c}}$$
but that doesn't seem to help either....
| Note
\begin{align}
& \sqrt{2\zeta^2-1+2\zeta\sqrt{\zeta^2-1}}+\sqrt{2\zeta^2-1-2\zeta\sqrt{\zeta^2-1}}\\
= & \sqrt{\left(\zeta+\sqrt{\zeta^2-1}\right)^2}
+ \sqrt{\left(\zeta-\sqrt{\zeta^2-1}\right)^2}\\
= & \left|\zeta+\sqrt{\zeta^2-1}\right|
+\left|\zeta-\sqrt{\zeta^2-1}\right|
=2|\zeta|\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all positive integers $x$, that satisfy $29x^{33} \equiv 27\pmod {11} $
Find all positive integers $x$, that satisfy $29x^{33} \equiv 27 \pmod {11}$.
I approached this the following way:
Since from $29x^{33} \equiv 27 \pmod {11}$ we get that $7x^{33} \equiv 5 \pmod {11}$ and since $\gcd(7,5)=1$ we would get that $\phi(11)=10$ which would imply that $7x^{10} \equiv 5 \pmod {11}$.
How should I continue from here, it doesn't seem to be quite clear.
| Another path: Subtract $11$ from $29$ to get
$$18x^3\equiv 27 \pmod{11}.$$
Divide by $9$
$$2x^3 \equiv 3 \equiv 14 \pmod{11}.$$
Divide by $2$
$$x^3 \equiv 7 \pmod{11}.$$
Cube both sides
$$x^9 \equiv x^{-1} \equiv 49\cdot 7 \equiv 2 \pmod{11}.$$
Solve $2x\equiv 1 \pmod{11}$ to get $x=6$.
| {
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"url": "https://math.stackexchange.com/questions/3727589",
"timestamp": "2023-03-29T00:00:00",
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Proving that $0I need to prove that $0<e-S_{n}<\frac{4}{(n+1)!}$ $\quad\forall n\in\mathbb{N}\quad$ for $\quad S_{n}=1+1+\frac{1}{2!}+\ldots+\frac{1}{n!}$
Previously, I proved that $2<e<4$, and that $S_n$ is taylor polynomial of $e$.
I thought on proving it using induction:
Base: $S_1=1<2<e<4$
Step - for $n+1$:
$e-S_{n+1}=e-S_{n}-\frac{1}{(n+1)!}$
What am I missing here? It's supposed to be simple.
Thank you!
| Let's consider inequality
$$x_n = \left( 1+ \frac{1}{n} \right)^n > 1+ \frac{n}{n} + \frac{n(n-1)}{2!} \frac{1}{n^2} + \cdots + \frac{n(n-1) \dots (n-k+1) }{k!} \frac{1}{n^k}$$
If we now consider limit $n \to \infty$, then we will have
$$e \ge 1+1 +\frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{k!} = S_k$$
Also we have
$$x_n =\left( 1+ \frac{1}{n} \right)^n < 1+1 +\frac{1}{2!} + \frac{1}{3!} + \cdots + \frac{1}{n!} = S_n$$
so, there is
$$x_n < S_n < e$$
which, after taking limit, finishes proof.
For estimation approximation let's consider
$$S_{n+k}-S_n= \frac{1}{(n+1)!}+ \cdots + \frac{1}{(n+k)!}< \frac{1}{(n+1)!}\left( \frac{1}{(n+2)} + \frac{1}{(n+2)^2} +\cdots \right) = \\= \frac{1}{(n+1)!} \frac{n+2}{(n+1)}< \frac{1}{n \cdot n!}$$
Fixing $n$ and taking limit $k \to \infty$ gives
$$0<e-S_n<\frac{1}{n \cdot n!}<\frac{4}{(n+1)!}$$
| {
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Is it possible to find the limit of $\frac{\sin^2ax}{x\sqrt{1-\cos ax}}$? The problem is as follows:
A certain tv signal is modeled by the function shown below:
$f(x)=\frac{\sin^2ax}{x\sqrt{1-\cos ax}}$
where $a>0$
Find the $\lim_{x\rightarrow 0^+} f(x)$.
The alternatives given in this problem are as follows:
$\begin{array}{ll}
1.&-a\sqrt{2}\\
2.&\frac{\sqrt{a}}{2}\\
3.&\sqrt{2}\sqrt{a}\\
4.&\sqrt{2}\\
5.&-\sqrt{a}\\
\end{array}$
How exactly should I assess this problem?.
I'm confused about the simbol used in the limit but I think the intended meaning is to find the limit of the function where $x$ approaches to positive?.
Attempting to insert the zero in the function as it is given would yield an infinite value. Thus I thought to reduce the trigonometric expression by doing this:
$f(x)=\frac{\sin^2ax}{x\sqrt{1-\cos ax}}$
$\frac{1-\cos^2 ax}{x\sqrt{1-\cos ax}}\times\frac{\sqrt{1-\cos ax}}{\sqrt{1-\cos ax}}$
$\frac{(1-\cos ax)(1+\cos ax )(\sqrt{1-\cos ax})}{x(1-\cos ax)}$
Simplifying terms in both denominator and numerator it yields
$\frac{(1+\cos ax )(\sqrt{1-\cos ax})}{x}$
By inserting the expression in the numerator inside the square root I'm getting:
$\frac{\sqrt{(1+\cos ax)^2(1-\cos ax})}{x}$
Expanding the whole expression I'm getting:
$\frac{\sqrt{(1^2+2\cos ax+\cos^2ax)(1-\cos ax})}{x}$
$\frac{\sqrt{1+2\cos ax+\cos^2ax-\cos ax-2\cos^2ax-\cos^3 ax}}{x}$
$\frac{\sqrt{1+\cos ax-\cos^2ax-\cos^3 ax}}{x}$
and that's how far I went. What exactly should be done here?. Can someone help me here?.
| Since $$\lim_{x\rightarrow0}\frac{\sin{x}}{x}=1,$$ we obtain: $$\frac{\sin^2ax}{x\sqrt{1-\cos ax}}=\frac{\sin^2ax}{x\sqrt2\sin\frac{ax}{2}}\rightarrow\frac{a^2}{\sqrt2\cdot\frac{a}{2}}=a\sqrt2.$$
| {
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Easier approach to $\int_0^{\infty} \frac{\mathrm{e}^{-x} \cosh(2x/5)}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x$? After playing with this integral for a little bit last night, I eventually resorted to complex analysis to solve it.
Can this be solved without complex analysis? It feels like there ought to be a way. If not, is there an easier way with complex analysis? (I'm still pretty beginner tier at this sort of thing.)
My solution is somewhat involved and it is as follows:
First, get rid of the cosh.
$$
\begin{split}
I &= \int_0^{\infty} \frac{\mathrm{e}^{-x} \cosh(2x/5)}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x \\
&= \frac1{2} \int_0^{\infty} \frac{\mathrm{e}^{-3/5 x} + \mathrm{e}^{-7/5 x}}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x \\
\frac{\mathrm{e}^{-7/5 x}}{1 + \mathrm{e}^{-2x}} &= \frac{\mathrm{e}^{3/5 x}}{1 + \mathrm{e}^{2x}} \\
2I &= \int_0^{\infty} \frac{\mathrm{e}^{-3/5 x}}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x + \int_0^{\infty} \frac{\mathrm{e}^{3/5 x}}{1 + \mathrm{e}^{2x}} \, \mathrm{d}x
\end{split}
$$
Next, do some u-subs to make it nicer.
$$\begin{split}
u = \mathrm{e}^{-x} & \qquad \mathrm{d}u = - \mathrm{e}^{-x} \, \mathrm{d}x \\
\int_0^{\infty} \frac{\mathrm{e}^{-3/5 x}}{1 + \mathrm{e}^{-2x}} \, \mathrm{d}x &= \int_{0}^{1} \frac{u^{-2/5}}{1 + u^2} \, \mathrm{d}u \\
\\
u = \mathrm{e}^{x} & \qquad \mathrm{d}u = \mathrm{e}^{x} \, \mathrm{d}x \\
\int_0^{\infty} \frac{\mathrm{e}^{3/5 x}}{1 + \mathrm{e}^{2x}} \, \mathrm{d}x &= \int_{1}^{\infty} \frac{u^{-2/5}}{1 + u^2} \, \mathrm{d}u \\
\\
2I &= \int_{0}^{\infty} \frac{u^{-2/5}}{1 + u^2} \, \mathrm{d}u \\
\end{split}
$$
Our contour is the counterclockwise semicircular arc of radius $R > 1$ in the upper half of the complex plane.
$$
\begin{split}
\oint_C \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \int_{-R}^0 \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z + \int_0^{R} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z + \int_0^{\pi} \frac{{\left(R \mathrm{e}^{i \phi}\right)}^{-2/5}}{1 + {\left(R \mathrm{e}^{i \phi}\right)}^2} \, iR\mathrm{e}^{i \phi} \, \mathrm{d}\phi \\
\lim_{R \rightarrow \infty} \int_0^{\pi} \frac{{\left(R \mathrm{e}^{i \phi}\right)}^{-2/5}}{1 + {\left(R \mathrm{e}^{i \phi}\right)}^2} \, iR\mathrm{e}^{i \phi} \, \mathrm{d}\phi &= 0 \\
\oint_{C, R \rightarrow \infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \int_{-\infty}^0 \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z + \int_0^{\infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z \\
\int_{-\infty}^0 \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= - \int_0^{\infty} \frac{(-z)^{-2/5}}{1 + (-z)^2} \, \mathrm{d}(-z) \\
\oint_{C, R \rightarrow \infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \left(1 + \mathrm{e}^{-2\pi i/5}\right) \int_0^{\infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z \\
\end{split}
$$
Finally, take the residue and solve for the original integral.
$$
\begin{split}
\oint_{C, R \rightarrow \infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= 2 \pi i \operatorname{Res}_{z = i} \left( \frac{z^{-2/5}}{1+z^2} \right) \\
&= 2 \pi i \left( \frac{i^{-2/5}}{2 i} \right) \\
&= \pi i^{-2/5} \\
\int_0^{\infty} \frac{z^{-2/5}}{1 + z^2} \, \mathrm{d}z &= \pi \left(\frac{i^{-2/5}}{1 + \mathrm{e}^{-2 \pi i / 5}}\right) \\
&= \frac{\pi}{2} \left(\sqrt{5} - 1 \right) \\
2I &= \frac{\pi}{2} \left(\sqrt{5} - 1 \right) \\
I &= \frac{\pi}{4} \left(\sqrt{5} - 1 \right)
\end{split}
$$
| Yet another special-function solution, this time using beta-integrals: $$\mathrm{B}(a,b)=\int_0^1 x^{a-1}(1-x)^{b-1}\,dx=\int_0^\infty\frac{y^{a-1}\,dy}{(1+y)^{a+b}}\color{blue}{=\int_0^1\frac{z^{a-1}+z^{b-1}}{(1+z)^{a+b}}\,dz},$$ and your integral reduces to a particular case after $z=\mathrm{e}^{-2x}$: $$I=\frac14\int_0^1\frac{z^{-3/10}+z^{-7/10}}{1+z}\,dz=\frac14\mathrm{B}\left(\frac{7}{10},\frac{3}{10}\right)=\frac{\pi}{4\sin(3\pi/10)}=\frac{\pi}{4\cos(\pi/5)}=\ldots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3735801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Three boxes with balls Box $U_1$ contains $1$ white ball and $2$ black balls. Box $U_2$ contains $2$ white balls and $2$ black balls. We extract without reinsertion two balls from every boxes. The four balls are put in a third box $U_3$ initially empty. We randomly extract a ball from $U_3$. Find the probability that the ball is white.
Well, I reasoned in this way. The possible combinations that ensure that $U_3$ contains at least one white ball are BNBB, NBBB, NNBB, BNBN, BNNB, BNNN, NBBN, NBNB, NBNN, NNBN, NNNB. Thus:
*
*$\mathbb{P}$($U_3$ contains $3$ white balls)$=\mathbb{P}($(BNBB)$\cap$(NBBB)$)=(\frac{1}{3}\cdot1 \cdot\frac{1}{2}\cdot\frac{1}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2}\cdot \frac{1}{3})=0,11$
*$\mathbb{P}(U_3$ contains $2$ white balls)$=\mathbb{P}($(NNBB)$\cap$(BNBN)$\cap$(BNNB)$\cap$(NBBN)$\cap$(NBNB)$)=(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})+(\frac{1}{3}\cdot 1\cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{1}{3}\cdot 1 \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{3}\cdot \frac{1}{2} \cdot \frac{2}{3})=0,46$
*$\mathbb{P}(U_3$ contains $1$ white ball)$=\mathbb{P}($(BNNN)$\cap$(NBNN)$\cap$(NNBN)$\cap$(NNNB)$)=(\frac{1}{3}\cdot 1 \cdot \frac{1}{2}\cdot \frac{1}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})+(\frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{2}{3})=0,33$
*$\mathbb{P}(U_3$ doesn't contain any white balls)$=2\mathbb{P}($(NNNN)$)=2(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})=0,1$
Thus $\mathbb{P}($one white ball from $U_3)=\frac{3}{4}\cdot 0,11+\frac{2}{4}\cdot 0,46+\frac{1}{4}\cdot 0,33+\frac{0}{4}\cdot 0,11=0,395$
Is it correct? Particularly I'm interested in reasoning. Thanks in advance.
| Let $X_1, X_2$ be the count of white balls drawn from the first two urn, $Y$ be the sum of these, and $E$ the event that a white ball is then drawn from the third urn.
Your approach was correct. $\mathsf P(E)=\sum_{k=0}^4 \mathsf P(E\mid Y=k)\mathsf P(Y=k)$
However, your evaluations were a bit off.
$$\begin{align}\mathsf P(Y=3)&=\mathsf P(X_1=1)\mathsf P(X_2=2)\\&=\dfrac{\binom 11\binom21}{\binom 32}\dfrac{\binom 22\binom 20}{\binom 42}\\&=\dfrac{1}{9}\\[2ex]\mathsf P(Y=2)&=\mathsf P(X_1=1)\mathsf P(X_2=1)+\mathsf P(X_1=0)\mathsf P(X_2=2)\\&=\dfrac{\binom 11\binom21}{\binom 32}\dfrac{\binom 21\binom 21}{\binom 42}+\dfrac{\binom 10\binom22}{\binom 32}\dfrac{\binom 22\binom 20}{\binom 42}\\&=\dfrac 12\\[2ex]\mathsf P(Y=1)&=\mathsf P(X_1=1)\mathsf P(X_2=0)+\mathsf P(X_1=0)\mathsf P(X_2=1)\\&=\dfrac{\binom 11\binom21}{\binom 32}\dfrac{\binom 20\binom 22}{\binom 42}+\dfrac{\binom 10\binom22}{\binom 32}\dfrac{\binom 21\binom 21}{\binom 42}\\&=\dfrac 13\\[2ex]\mathsf P(Y=0)&=\mathsf P(X_1=0)\mathsf P(X_2=0)\\&=\dfrac{\binom 10\binom22}{\binom 32}\dfrac{\binom 20\binom 22}{\binom 42}\\&=\dfrac 1{18}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find all $x > 0$ s/t $\sqrt{x} + 1/\sqrt{x}$ and $x^{1/3} + 1/x^{1/3}$ are integers. Why doesn't this get all solutions? This is a math puzzle I've been working on, and I'm not sure why my approach isn't yielding all the solutions for $x$. As in the title, the question is to find all positive reals $x$ where $\sqrt{x} + 1/\sqrt{x}$ and $x^{1/3} + 1/x^{1/3}$ are both integers. Here's my approach--I'm not sure where the error is. Where does the following approach go wrong?
Suppose that $\sqrt{x} + 1/\sqrt{x} = m$ and $x^{1/3} + 1/x^{1/3} = n$, where $m,n \in \mathbb{Z}$. We let $m, n > 0$ since $\sqrt{x} + 1/\sqrt{x} > 0$ and $x^{1/3} + 1/x^{1/3} > 0$ for $x > 0$.
We equivalently have $x - m\sqrt{x} + 1 = 0$ and $x^{2/3} -nx^{1/3} + 1 = 0$. Let $u = \sqrt{x}$, so that we have $u^2 - mu + 1 = 0$ and $u^{4/3} - nu^{2/3} + 1 = 0$. It suffices to find all $u$ satisfying both of these equations.
Applying the quadratic formula, $u = \frac{m \pm \sqrt{m^2 - 4}}{2}$ and $u^{2/3} = \frac{n \pm \sqrt{n^2-4}}{2}$. Squaring the first equation gets $u^2 = (\frac{m \pm \sqrt{m^2 - 4}}{2})^2$.Cubing the second equation gets $u^2 = (\frac{n \pm \sqrt{n^2-4}}{2})^3$.
So there are 4 sign combinations setting these two expressions for $u^2$ equal. Since $m,n$ are integers, I look for integer solutions only, finding only $m = 2, n = 2$ as the viable solution in all 4 cases. I checked this part with WolframAlpha--the general idea was to isolate radicals to one side. If we have a nonzero sum of surds, it ends up being irrational while the other side is rational. That leaves integer solutions to the radicals $\sqrt{m^2-4},\sqrt{n^2-4}$, the only positive solutions to which are $m = 2$ and $n = 2$.
This yields $u = 1$ as the only viable solution. However, the solution states that there are infinitely many such $x$.
| Since you don't show certain details of your work, it's hard for me to be able to tell what the issue is. However, I believe the main limitation of your approach is that a non-zero sum of surds doesn't necessarily need to have each term being rational for its sum to be rational, such as what Zerox's question comment states about $\sqrt{m^2 - 4}$ being some rational multiplier of $\sqrt{n^2 - 4}$ can cause the radical side to vanish.
Instead, here's how I show there are an infinite # of solutions for $x$. First, as you did, have
$$\sqrt{x} + \frac{1}{\sqrt{x}} = m \tag{1}\label{eq1A}$$
$$\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}} = n \tag{2}\label{eq2A}$$
for some integers $m$ and $n$. Next, square both sides of \eqref{eq1A} to get
$$x + 2 + \frac{1}{x} = m^2 \tag{3}\label{eq3A}$$
and cube both sides of \eqref{eq2A}, plus simplify, to get
$$\begin{equation}\begin{aligned}
x + 3(\sqrt[3]{x})^2\left(\frac{1}{\sqrt[3]{x}}\right) + 3(\sqrt[3]{x})\left(\frac{1}{\sqrt[3]{x}}\right)^2 + \frac{1}{x} & = n^3 \\
x + 3\left(\sqrt[3]{x} + \frac{1}{\sqrt[3]{x}}\right) + \frac{1}{x} & = n^3 \\
x + 3n + \frac{1}{x} & = n^3
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Next, \eqref{eq4A} minus \eqref{eq3A}, plus manipulations including factoring, gives
$$\begin{equation}\begin{aligned}
3n - 2 & = n^3 - m^2 \\
m^2 & = n^3 - 3n + 2 \\
m^2 & = (n - 1)^2(n + 2)
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
This shows any integer $n \ge 2$, where $n + 2$ is a perfect square, will work. Since $x \gt 0$ and all of the quantities are positive, all of the above steps are reversible, meaning there will not be any extraneous solutions. This proves there are an infinite # of $n$ plus $m$, and thus $x$, which work.
To get the values of $x$ (note for any $x$, you get $\frac{1}{x}$ is also a solution, so apart from $x = 1$, there's always $2$ of them), you could use your $u$ or, from \eqref{eq3A}, you also can get
$$\begin{equation}\begin{aligned}
x + \frac{1}{x} & = m^2 - 2 \\
x^2 + 1 & = (m^2 - 2)x \\
x^2 - (m^2 - 2)x + 1 & = 0
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
Thus, the quadratic formula gives
$$\begin{equation}\begin{aligned}
x & = \frac{m^2 - 2 \pm \sqrt{(m^2 - 2)^2 - 4}}{2} \\
& = \frac{m^2 - 2 \pm \sqrt{(m^4 - 4m^2 + 4) - 4}}{2} \\
& = \frac{m^2 - 2 \pm m\sqrt{m^2 - 4}}{2}
\end{aligned}\end{equation}\tag{7}\label{eq7A}$$
Note $n = 2$ gives $m = 2$, which is your one solution. An example of the other solutions is $n = 7$ which gives $m = 18$, and with \eqref{eq7A} showing $x = 161 \pm 72\sqrt{5}$.
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
} |
If $S=1+\frac{1}{2}+\frac{1}{3}-\frac{3}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{3}{8}+...$ what is the closest integer to $e^S$ If $S=1+\frac{1}{2}+\frac{1}{3}-\frac{3}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{3}{8}+...$ what is the closest integer to $e^S$
I thought that this series could be represented as $\ln{2}$ but it is $\ln{4}$ somehow? Any suggestions please send.
| Starting from @vonbrand's answer,
$$\begin{align*}
S_n
&= H_n - H_{\lfloor n / 4 \rfloor} \implies e^{S_n}=\exp\left( H_n - H_{\lfloor n / 4 \rfloor} \right)=\frac {\exp\left( H_n\right)}{\exp\left(H_{\lfloor n / 4 \rfloor}\right) }
\end{align*}$$ Using the asymptotics of harmonic numbers, an approximation is
$$e^{S_n}\sim\frac{n+\frac{1}{2}+\frac{1}{24 n}-\frac{1}{48 n^2} }{m+\frac{1}{2}+\frac{1}{24 m}-\frac{1}{48 m^2}}\qquad \text{where} \qquad m=\lfloor n / 4 \rfloor$$
Disregarding the floors, this would give for large values of $n$
$$e^{S_n}=4-\frac{6}{n}+\frac{19}{2 n^2}-\frac{39}{4
n^3}+O\left(\frac{1}{n^4}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3746679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Percent sign after a fraction? Convert to percent Isn't it wrong to write the following with only the percent sign? Instead of $100 \%$?
The change in height as a percentage is
$$
\frac{a - b}{a} \% \tag 1
$$
where $a$ is the initial height and $b$ is the final height.
Because if $a=10$, $b=5$ we have
$$
\frac{10-5}{10}\%=\frac{1}{2} \% = 0.5 \frac{1}{100} = 0.005
\quad \text{what?!}
\tag 2
$$
If we convert a decimal number to percent we multiply it by $100$ and add the percent sign. We have $1\%=\frac{1}{100}$, so with $100 \%$ we multiply the number by $1$, i.e.
\begin{align}
\frac{10-5}{10} \cdot \color{blue}{1}
&=
\frac{10-5}{10} \cdot \color{blue}{100 \%}
=
\frac{5}{10} \cdot
\color{blue}{100 \frac{1}{100}}
=\frac{1}{2} \cdot
\color{blue}{100 \frac{1}{100}} \tag 3
\\
&=0.5\cdot \color{blue}{100 \frac{1}{100}}
=50 \color{blue}{\frac{1}{100}} = 50 \color{blue}{\%} \tag 4
\end{align}
So, shouldn't we instead write $(1)$ as
$$
\frac{a - b}{a} 100 \% \quad ? \tag 5
$$
| In what concerns exclusively the use of the percentage sign $\%$, it is simply a shorthand notation for $1/100$, so e.g. five percent of a population of $n$ elements is equivalently expressed as the amount $$n \cdot 5\%=n\cdot \frac{5}{100} =\frac{n}{20}= 0.05 n$$
So it is true that
$$\frac{10-5}{10}\%=\frac{1}{2}\%=0.005$$
although it certainly doesn't correspond to the percentage representation of the ratio $\frac{1}{2}$, which would of course be $50\%$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
simple method for expanding binomial with 3 or more terms? I've seen
Binomial Theorem Question (Expansion of Three Terms)
Binomial Theorem with Three Terms
Expanding Equation with Binomial Theorem
but I'm not such a math expert, I need things explained in simple terms.
Basically I've heard that the solve
(x + y)^
it's essentially
(x + y)(x + y)
and then multiply the first terms, (leftmost), then first term of one with last term of the other (outer) then second term of one with first term of other (inner) then the two last terms of each (right most), and add them up, so
x * x + x * y + y * x + y * y =
x^2 + 2xy + y^2
that is pretty much all I know, now if I want to solve a more complicated binomial, with three or more terms, for example
(x + y + z)(x + y + z)
would I use a similar method? Meaning do I start with the left most terms, then x * y, then, what? then do I do x * z and then move on to the next right term, y, and do y * x + y * y + y * z, and then do the same for z, meaning z * x + z * y + z * z ? Am I missing something here, or is that it?
| You do not need any theorems or results other than the distributive property: $a(b+c)=ab+ac$. Applying this once, we have $(x+y)(x+y)=x(x+y)+y(x+y)$. Here, the left $x+y$ is playing the role of $a$, and the right $x$ and $y$ are playing the roles of $b$ and $c$ respectively. Then we can use the same property again to see $x(x+y)=x^2+xy$ and $y(x+y)=yx+y^2$. Since $xy=yx$, we have
\begin{align}
(x+y)(x+y)&=x(x+y)+y(x+y)\\
&=x^2+xy+yx+y^2\\
&=x^2+xy+xy+y^2\\
&=x^2+2xy+y^2.
\end{align}
For three terms we can do the same thing:
\begin{align}
(x+y+z)(x+y+z)&=x(x+y+z)+y(x+y+z)+z(x+y+z)\\
&=x^2+xy+xz+xy+y^2+yz+xz+yz+z^2\\
&=x^2+y^2+z^2+2xy+2xz+2yz.
\end{align}
| {
"language": "en",
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How do I find the integer solutions that satisfy $xyz = 288$ and $xy + xz + yz = 144$?
Find all integers $x$, $y$, and $z$ such that $$xyz = 288$$ and $$xy + xz + yz = 144\,.$$
I did this using brute force, where $$288 = 12 \times 24 = 12 \times 6 \times 4$$ and found that these set of integers satisfy the equation. How do I solve this without using brute force?
| Without loss of generality, suppose that $x\geq y\geq z$. From the given system of Diophantine equations, we obtain an Egyptian fraction problem:
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{yz+zx+xy}{xyz}=\frac{144}{288}=\frac12\,.\tag{*}$$
Since $xyz=288>0$, the number of variables with negative values among $x$, $y$, and $z$ is either $0$ or $2$. We consider two cases.
Case I: $x>0>y\geq z$. Let $u:=-y$ and $v:=-z$. Then,
$$\frac{1}{x}-\frac1{u}-\frac1{v}=\frac{1}{2}\,.$$
Thus, $\dfrac{1}{x}>\dfrac12$, making $x<2$. Therefore, $x=1$. This implies $$yz=xyz=288$$ and $$y+z=x(y+z)=144-yz=144-288=-144\,.$$
Consequently, the polynomial
$$q(t):=t^2+144t+288$$
has two roots $y$ and $z$. It is easily seen that $q(t)$ has no integer roots, so this case is invalid.
Case II: $x\geq y\geq z>0$. Then,
$$\frac{3}{z}\geq \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac12\,.\tag{#}$$
This shows that $z\leq 6$. Furthermore, it is clear that $z>2$. Hence, there are four possible values of $z$, which are $3$, $4$, $5$, and $6$.
*
*If $z=6$, then by (#), we conclude that $x=6$ and $y=6$. However, $xyz\neq 288$. This subcase yields no solutions.
*If $z=5$, then this is impossible, as $xyz=288$ implies that $z$ divides $288$. This subcase is eliminated.
*If $z=4$, then $$xy=\dfrac{288}{z}=\dfrac{288}{4}=72$$ and $$x+y=\dfrac{144-xy}{z}=\dfrac{144-72}{4}=18\,.$$ Thus, $t=x$ and $t=y$ are the roots of the quadratic polynomial $$t^2-18t+72=(t-6)(t-12)\,.$$ This means $x=12$ and $y=6$.
*If $z=3$, then $$xy=\dfrac{288}{z}=\dfrac{288}{3}=96$$ and $$x+y=\dfrac{144-xy}{z}=\frac{144-96}{3}=16\,.$$ Thus, $t=x$ and $t=y$ are the roots of the quadratic polynomial $t^2-16t+96$, but this polynomial has no real roots.
In conclusion, all integer solutions $(x,y,z)$ to the required system of Diophantine equations are permutations of $(4,6,12)$.
Remark. Note that all $(x,y,z)\in\mathbb{Z}^3$ that satisfy (*) are permutations of the triples listed below.
$$(1,-3,-6)\,,\,\,(1,-4,-4)\,,\,\,(k,2,-k)\,,\,\,(4,3,-12)\,,\,\,(5,3,-30)\,,$$
$$(6,6,6)\,,\,\,(10,5,5)\,,\,\,(20,5,4)\,,\,\,(12,6,4)\,,\,\,(8,8,4)\,,$$
$$(42,7,3)\,,\,\,(24,8,3)\,,\,\,(18,9,3)\,,\text{ and }(12,12,3)\,,$$
where $k$ is any positive integer.
| {
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If $ \gcd(a,b) = 1$ prove that $ \gcd(2a+b, a+2b) = 1$ or $3$? I have seen this question, some other related questions and answers for solving this problem. However, I tried to solve it using a different approach.
Let, $ \gcd(2a+b, a+2b) = d$
Assume $2a+b = qd\tag{1}$
and so $b = qd - 2a$
If we replace b with this in $a+2b$, we get $a+2b = 2qd - 3a$
We know, $\gcd(a,b) = 1 $. Let, $\gcd(a,qd)=m$. So $a=mn$ and $qd=lm$. In equation (1), $2a+b=qd$ or $b=qd−2a=lm−2mn=m(l−2n)$. That means $m|b$ and so $\gcd(a,b)=m$ which is not true. So, $ \gcd(a, qd) = 1$.
Thus $\gcd(2a+b,a+2b)$ = $\gcd(qd,2qd-3a)$ = $\gcd(qd,2qd-3)$. Since $\gcd(2,3) = 1, \gcd(qd,3) = 1$ or $3$ will be the answer.
Is this correct?
| Choose integers $r,s$ such that $ar+bs=1$. Then
$$ (2a+b)(2r-s) + (a+2b)(2s-r) = 3(ar+bs) = 3. $$
So if $g=\gcd(2a+b,a+2b)$, then $g \mid 3$. Hence, $g=1$ or $3$.
Moreover, from $3 \mid \big((2a+b)+(a+2b)\big)$, we have $3 \mid (2a+b)$ if and only if $3 \mid (a+2b)$. Further, note that $3 \mid (2a+b)$ if and only if $3 \mid \big(3a-(2a+b)\big)=a-b$. Therefore,
$$ \gcd(2a+b,a+2b) = \begin{cases} 1 & \:\mbox{if}\: 3 \nmid (a-b); \\ 3 & \:\mbox{if}\: 3 \mid (a-b). \end{cases} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Interview Question -- confused about the answer If $a$, $b$ and $c$ are positive integers then how many possible solutions exist to the equation $a! = 5b! + 27c!$
A. No Solution
B. $1$ Solution
C. $2$
D. $3$
E. Infinite Solution
| The only solution is $32!=5.31!+27.31!$.
Otherwise if $b\ne c$, then either
(i) $b<c<a$, so $$a(a-1)\cdots(b+1)=5+27c(c-1)\cdots(b+1)$$ In this case $b+1=5$ since it is a common factor. There can be no other common factors, so $c=b+1=5$. But $a!=5.4!+27.5!$ has no solutions.
(ii) $c<b<a$, so $$a(a-1)\cdots(c+1)=5b\cdots(c+1)+27$$. In this case $c+1$, $c+2$,$\ldots$, must divide $27$. Since consecutive numbers cannot be multiples of $3$, $b=c+1=3,9,27$. But $a!=5.3!+27.2!=14.3!$, $a!=5.9!+27.8!=8.9!$, and $a!=5.27!+27.26!= 6.27!$ have no solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $y=f(x)=\frac{3x-5}{2x-m}$ find $m$ so that $f(y)=x$. Question: If $y=f(x)=\dfrac{3x-5}{2x-m}$ find $m$ so that $f(y)=x$.
We have $y=\dfrac{3\left(\frac{3x-5}{2x-m}\right)-5}{2\left(\frac{3x-5}{2x-m}\right)-m} $
How can I find $m$? It is given than $m=3$.
| 1-method. The hyperbola has assymptotes: $y=\frac32$ and $x=\frac m2$. The inverse function must be symmetric with respect to the line $y=x$. For the hyperbola to be self inverse, the assymptotes must intersect at the line $y=x$, consequently, $m=3$.
2-method. Find the inverse function and equate to the original:
$$y=\frac{3x-5}{2x-m}\Rightarrow x=\frac{mx-5}{2x-3}\Rightarrow \\
y^{-1}=\frac{mx-5}{2x-3}=\frac{3x-5}{2x-m}\Rightarrow m=3.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3754079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int_{1}^{\sqrt{2}} \frac{x^4}{(x^2-1)^2+1}\,dx$ Evaluate the integral:
$$\int_{1}^{\sqrt{2}} \frac{x^4}{(x^2-1)^2+1}\,dx$$
The denominator is irreducible, if I want to factorize and use partial fractions, it has to be in complex numbers and then as an indefinite integral, we get
$$x + \frac{\tan^{-1}\left(\displaystyle\frac{x}{\sqrt{-1 - i}}\right)}{\sqrt{-1 - i}} + \frac{\tan^{-1}\left(\displaystyle\frac{x}{\sqrt{-1 + i}}\right)}{\sqrt{-1 + i}}+C$$
But evaluating this from $1$ to $\sqrt{2}$ is another mess, keeping in mind the principal values. I also tried the substitution $x \mapsto \sqrt{x+1}$, which then becomes
$$\frac{1}{2}\int_{0}^1 \frac{(x+1)^{3/2}}{x^2+1}\,dx$$
I don't see where I can go from here. Another substitution of $x\mapsto \tan x$ also leads me nowhere.
Should I approach the problem in some other way?
| Start by writing $x^4 = (x^2 - 1 + 1)^2$
$\implies x^4 = (x^2-1)^2 + 1 + 2(x^2-1)$
So the our integral becomes:
$$\int_1^{\sqrt2}\frac{(x^2-1)^2 + 1 + 2(x^2-1)}{(x^2-1)^2 + 1}\,dx$$
$$ = \int_1^{\sqrt2}\frac{(x^2-1)^2 + 1}{(x^2-1)^2 + 1}\,dx + 2\int_1^{\sqrt2}\frac{(x^2-1)}{(x^2-1)^2 + 1}\,dx$$
$$ = \sqrt2 - 1 + 2\int_1^{\sqrt2}\frac{(x^2-\sqrt2 + 1-\sqrt2)}{(x^4 - 2x^2 + 2)}\,dx$$
$$ = \sqrt2 - 1 + 2\int_1^{\sqrt2}\frac{(1-\sqrt2/x^2 )}{((x + \sqrt2/x)^2 - 2 - 2\sqrt2)}\,dx + 2(1-\sqrt2)\int_1^{\sqrt2}\frac{1}{(x^2-1)^2 + 1}\,dx$$
Here, I've split the integral so I can use the fact that after dividing the numerator and denominator by $x^2$, I can complete a square (The square of $(x + \sqrt2/x))$ and I'll have its derivative in the numerator for an easy substitution.
Put $(x + \sqrt2/x) \rightarrow t$ in the first integral, you can see that the upper and lower limits become the same $(1+ \sqrt2)$ So the first integral becomes $0$ and you're left with:
$$\sqrt2 - 1 + 2(1-\sqrt2)\int_1^{\sqrt2}\frac{1}{(x^2-1)^2 + 1}\,dx$$
I was trying to avoid using complex numbers, but this integral becomes so much easier if you write: $(x^2-1)^2 + 1 = (x^2 - 1 + i)(x^2 - 1 + i)$ and use partial fractions.
$$ =\sqrt2 - 1 + \frac{1 - \sqrt2}{i}\int_1^{\sqrt2}\left(\frac{1}{x^2-1-i} + \frac{1}{x^2-1+i}\right)\,dx$$
$$=\sqrt2 - 1 + \frac{1 - \sqrt2}{i}\left(\frac{\tan^{-1}\left(\displaystyle\frac{x}{\sqrt{-1 - i}}\right)}{\sqrt{-1 - i}} + \frac{\tan^{-1}\left(\displaystyle\frac{x}{\sqrt{-1 + i}}\right)}{\sqrt{-1 + i}}\right)\Bigg|_{x=1}^{x=\sqrt2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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An stronger inequality than in AoPS. For $x,y,z >0.$ Prove$:$
$$\sum {\frac {y+z}{x}}+{\frac {1728 {x}^{ 3}{y}^{3}{z}^{3}}{ \left( x+y \right) ^{2} \left( y+z \right) ^{2} \left( z+x \right) ^{2} \left( x+y+z \right) ^{3}}} \geqslant 4\sum {\frac {x}{y+z }}+1$$
I check when $xyz=0$ and $x=y$ and see it's true. So I guess it's true.
So I try to get it in $uvw$ form as follow$:$
$$-26244{u}^{7}{v}^{2}{w}^{3}+19683{u}^{6}{v}^{6}+2916{u}^{6}{w}^{6}+4374{u}^{5}{v}^{4}{w}^{3}-2673{u}^{4}{v}^{2}{w}^{6}+216{u}^{3}{w}^{9}+1728{w}^{12} \geqslant 0$$
Then I don't know how to end proof for it. BW does not help here.
I'm not sure about this inequality. I found it when I prove this inequality.
Help me please. Thanks for a real lot!
| $uvw$ helps!
Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v>0$, and $xyz=w^3$.
Thus, we need to prove that:
$$\frac{9uv^2-3w^3}{w^3}+\frac{1728w^9}{(9uv^2-w^3)^2\cdot27u^3}\geq\frac{4(27u^3-27uv^2+3w^3+9uv^2-3w^3+3w^3)}{9uv^2-w^3}+1$$ or
$$\frac{9uv^2}{w^3}+\frac{64w^9}{u^3(9uv^2-w^3)^2}\geq\frac{4(27u^3-9uv^2+2w^3)}{9uv^2-w^3}$$ or $f(w^3)\geq0,$ where
$$f(w^3)=64w^{12}+8u^3w^9+108u^6w^6-99u^2v^4w^6-972u^7v^2w^3+162u^5v^4w^3+729u^6v^6.$$
But since by Maclaurin $$u\geq v\geq w,$$ we obtain: $$f'(w^3)=256w^9+24u^3w^6+216u^6w^3-198u^2v^4w^3-972u^7v^2+162u^5v^4<0,$$
which says that $f$ decreases and it's enough to prove our inequality for a maximal value of $w^3$,
which happens for equality case of two variables.
Since our inequality is homogeneous and symmetric, it's enough to assume $y=z=1,$ which gives:
$$(x-1)^2(x-2)^2(x^4+10x^3+37x^2+20x+4)\geq0$$ and we are done!
Now we see, why BW does not help: the equality occurs also for $(x,y,z)=(2,1,1).$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Let $n$ be an odd positive integer. Show that $24 \vert(n^3-n).$
Let $n$ be an odd positive integer. Show that $24 \vert(n^3-n).$
So since $n^3-n=(n-1)n(n+1)$, we have that $3\vert(n^3-n).$
Also since we have that $n$ is odd we can say that $n=2k+1$, for some $k \in \mathbb{Z^+}.$
This implies that $(n+1) = 2k +2 = 2(k+1)$, hence $2\vert(n+1).$ Similar argument can be made for $(n-1)$.
Now $24 = 2 \cdot 3 \cdot 4$ so essentially I'm only missing the part where I would have to show that $4$ divides some of the terms. How can I find that?
| Only remaining to show $8\mid (n-1)n(n+1)$.
Observe, for an odd $n$, one of $n-1$ and $n+1$ is divisible by $2$ and another one is divisible by $4$.
So $(n-1)(n+1)$ is divisible by $8$, if $n$ is odd.
| {
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"timestamp": "2023-03-29T00:00:00",
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How do I write the Laurent series for $\frac{1}{z^2(z-i)}$ for $1<|z-1|<\sqrt2$? How do I write the Laurent series for $\frac{1}{z^2(z-i)}$ for $1<|z-1|<\sqrt2$?
I know that I need to rewrite it somehow to fit the geometric series form of $\frac{1}{1-r}$, but I'm stuck on getting here. I'm also aware that $z_0=1$, so then my "r" should have $z-1$ in it correct? After that I think I can do the rest from there.
|
I know that I need to rewrite it somehow to fit the geometric series form of $\frac{1}{1-r}$, but I'm stuck on getting here.
If you know already that you should use partial fractions, then you can jump to Equation $(2)$ and from there on. In B. and C. the geometric series you need are indicated and motivated, I hope.
I'm also aware that $z_0=1$, so then my "r" should have $z-1$ in it correct?
Yes.
A. When the given function is of the form $f(z)=\frac{p(z)}{q(z)}$, with $p(z)$ and $q(z)$ being polynomials in $z$, the first step is to expand it into partial fractions. Due to the form of $f(z)$ this means that
\begin{equation}
f(z)\equiv \frac{1}{z^{2}\left( z-i\right) }=\frac{A}{z^{2}}+\frac{B}{z}+\frac{C}{z-i}. \tag{1}
\end{equation}
To find the coefficients we can use the Heaviside cover-up method.
*
*To determine $C$, multiply by $\left( z-i\right) $ and use the root $z=i$ of the denominator $q(z)= z^{2}\left( z-i\right)$ and evaluate the limit
\begin{equation*}
C=\lim_{z\rightarrow i}f(z)\left( z-i\right) =\lim_{z\rightarrow i}\frac{1}{z^{2}}=\frac{1}{i^{2}}=-1.
\end{equation*}
*To find $A$, multiply by $z^{2}$ and use the root $z=0$ of $q(z)$:
\begin{equation*}
A=\lim_{z\rightarrow 0}f(z)z^{2}=\lim_{z\rightarrow 0}\frac{1}{z-i}=i.
\end{equation*}
*To determine $B$, substitute $C$ and $A$ in one of the equations resulting from $(1)$ after being multiplied by $\left( z-i\right) $ or $z^{2}$ , and pick a meaningful $z$, e.g. $z=1$:
\begin{equation}
f(z)\left( z-i\right) =\frac{1}{z^{2}}=\frac{A}{z^{2}}\left( z-i\right) +\frac{B}{z}\left( z-i\right) +C,
\end{equation}
\begin{equation}
z=1\implies 1=i\left( 1-i\right) +B\left( 1-i\right) -1\implies B=1.
\end{equation}
Then
\begin{equation}
f(z)\equiv \frac{1}{z^{2}\left( z-i\right) }=\frac{i}{z^{2}}+\frac{1}{z}-
\frac{1}{z-i}. \tag{2}
\end{equation}
B. To make some algebraic manipulations easier we now use the substitution $w=z-1$. Then the annulus $1<\left\vert z-1\right\vert <\sqrt{2}$ becomes the new annulus $1<\left\vert w\right\vert <\sqrt{2}$, centered at $w=0$, and $\frac{1}{z^{2}\left( z-i\right) }$ becomes
\begin{equation}
\frac{1}{z^{2}\left( z-i\right) }=\frac{1}{\left( w+1\right) ^{2}\left[
w+\left( 1-i\right) \right] }\equiv g(w). \tag{3}
\end{equation}
By $(2)$ the function $g(w)$ can be expanded as
\begin{equation}
g(w)=\frac{1}{w+1}+\frac{i}{\left( w+1\right) ^{2}}-\frac{1}{w+\left(
1-i\right) }. \tag{4}
\end{equation}
C. Each term can be expanded into a particular geometric series as follows:
*
*For $\color{blue}{1<\left\vert w\right\vert }$ and using the sum of the complex geometric series $\displaystyle\sum_{n\geq 0}\dfrac{\left( -1\right) ^{n}}{w^{n}}=\frac{1}{1-\left(-1/w\right) }$, the first term can be written as
\begin{equation*}
g_{1}(w)\equiv \frac{1}{w+1}=\frac{1}{w}\frac{1}{1+1/w}=\frac{1}{w}\frac{1}{
1-\left( -1/w\right) }
\end{equation*}
and expanded into
\begin{equation}
g_{1}(w)=\frac{1}{w}\sum_{n\geq 0}\frac{\left( -1\right) ^{n}}{w^{n}}
=\sum_{n\geq 0}\frac{\left( -1\right) ^{n}}{w^{n+1}}, \text{ for }
\color{blue}{1<\left\vert w\right\vert} . \tag{5}
\end{equation}
*As for the second term, for $\color{blue}{\left\vert w\right\vert >1}$ as well, since
$$\frac{1}{\left( w+1\right) ^{2}}=-\frac{d}{dw}\left( \frac{1}{w+1}\right) =-\frac{d}{dw}g_{1}(w),$$
it can be expanded into
\begin{align}
g_{2}(w) &\equiv \frac{i}{\left( w+1\right) ^{2}}=-i\frac{d}{dw}g_{1}(w)=-i\frac{d}{dw}\sum_{n\geq 0}\frac{\left( -1\right) ^{n}}{w^{n+1}} \\
&=i\sum_{n\geq 0}\left( -1\right) ^{n}\frac{n+1}{w^{n+2}}=-i\sum_{n\geq
1}\left( -1\right) ^{n}\frac{n}{w^{n+1}},\text{
for }\color{blue}{1<\left\vert w\right\vert }. \tag{6}
\end{align}
*As for the third term, if $\left\vert - \dfrac{w}{1-i}\right\vert =\color{green}{\dfrac{\left\vert w\right\vert }{\sqrt{2}}<1}$, we have that
\begin{align}
g_{3}(w) &\equiv \frac{1}{w+\left( 1-i\right) }=\frac{1}{1-i}\frac{1}{1-\left( -\frac{w}{1-i}\right) } \\
&=\frac{1}{1-i}\sum_{n\geq 0}\frac{\left( -1\right) ^{n}w^{n}}{\left(
1-i\right) ^{n}}=\sum_{n\geq 0}\frac{\left( -1\right) ^{n}w^{n}}{\left(1-i\right) ^{n+1}}\text{ for }\color{green}{\left\vert w\right\vert <\sqrt{2}}. \tag{7}
\end{align}
D. From $(5)-(7)$ it follows that for $\color{blue}{1<}\left\vert
w\right\vert \color{green}{<\sqrt{2}}, $
\begin{align}
g(w) &=g_{1}(w)+g_{2}(w)+g_{3}(w) \\
&=\sum_{n\geq 0}\left( -1\right) ^{n}\left[ \frac{1-in}{w^{n+1}}+\frac{w^{n}}{\left( 1-i\right) ^{n+1}}\right] \text{ for }\color{blue}{1<}\left\vert
w\right\vert \color{green}{<\sqrt{2}}. \tag{8}
\end{align}
In terms of the given function $f(z)$, we thus have the following expansion for $\color{blue}{1<}\left\vert z-1\right\vert \color{green}{<\sqrt{2}} $:
\begin{equation}
f(z)=\sum_{n\geq 0}\left( -1\right) ^{n}\left[ \frac{1-in}{\left( z-1\right)
^{n+1}}+\frac{\left( z-1\right) ^{n}}{\left( 1-i\right) ^{n+1}}\right]
\text{ for }\color{blue}{1<}\left\vert z-1\right\vert \color{green}{<\sqrt{2}} . \tag{9}
\end{equation}
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate limit integrate: $\lim_{n \rightarrow +\infty}{\int_{0}^{\frac {\pi}{2}} \sin^{n} x dx}$ I know,
when n is even,
$$\int_{0}^{\frac {\pi}{2}} \sin^{n} x dx = \frac {(2m-1)!!}{(2m)!!}\frac {\pi}{2} $$
and n is odd,
$$\int_{0}^{\frac {\pi}{2}} \sin^{n} x dx = \frac {(2m)!!}{(2m+1)!!} $$
But when let limit n to infinite, how can I get the limit, or if the limit exists.
--
I find a good way to solve the limit:$$\lim_{x \to +\infty} \frac{(2n-1)!!}{(2n)!!}
$$
as we know,
$2 =\frac{1+3}{2} > \sqrt{1 \cdot 3}$ , so $2n = \frac{(2n-1)+(2n+1)}{2} > \sqrt{(2n-1)(2n+1)}$
so,
$$\lim_{x \to +\infty} \frac{(2n-1)!!}{(2n)!!} = \frac{\sqrt{1 \cdot 3}\cdots \sqrt{(2n-3)(2n-1)}\sqrt{2n-1}}{\sqrt{1\cdot 3} \cdots \sqrt{(2n-3)(2n-1)} \sqrt{(2n-1)(2n+1)}} = \frac{1}{\sqrt{2n+1}} \to 0 (n \to \infty)$$
| It is obviously zero. Think of the graph of $\sin^nx$ with $n$ going to infinity. How the area under the $y=\sin^nx $ changes accordingly
| {
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Solve for $x$, $(1+2i)^{(2x+6)}=(-11-2i)^{(x+1)} $ $$(1+2i)^{(2x+6)}=(-11-2i)^{(x+1)} $$
Solve for $x$
My Work:
Take ln at the both sides:
$$(2x+6)\ln(1+2i)=(x+1)\ln(-11-2i)$$
and$$\ln(1+2i)=\ln(\sqrt{5})+i\arctan(2)$$
$$\ln(-11-2i)=\ln(5\sqrt{5})+i\left[\arctan\left(\frac{2}{11}\right)+\pi\right]\\=\ln(5)+\ln(\sqrt{5})+i\left[\arctan\left(\frac{2}{11}\right)+\pi\right]$$
And things get dirty here because of $\arctan$ and $\ln$ ...
But
$$(2x+6)\ln(1+2i)=(x+1)\ln(-11-2i)$$ from here
$$2x+6=(x+1)\dfrac{\ln(-11-2i)}{\ln(1+2i)}$$
then
$$x\left(2-\dfrac{\ln(-11-2i)}{\ln(1+2i)}\right)=-6+\dfrac{\ln(-11-2i)}{\ln(1+2i)}$$
then
$$x=\dfrac{-6+\dfrac{\ln(-11-2i)}{\ln(1+2i)}}{2-\dfrac{\ln(-11-2i)}{\ln(1+2i)}}$$
But specifically I cannot find x here.
Any elegant way to solve it, or not elegant but proper, any hint any help, would be appreciated. Thank you.
| Hint:
Observe that $11^2+2^2=\cdots=5^3, 5=1^2+2^2$
which encouraged me to find $$(1+2i)^3=1-12+(6-8)i$$
$$\implies(-11-2i)^{(x+1)}=((1+2i)^3)^{x+1}=?$$
So we have $$(1+2i)^{3(x+1)-(2x+6)}=1$$
Taking logarithm in both sides, $$(x-3)\ln(1+2i)=2n\pi i$$ where $n$ is any integer
Now $1+2i=\sqrt{1^2+2^2}\cdot e^{i\arctan2}$
$\implies\ln(1+2i)=\dfrac{\ln5}2+i\arctan2+2m\pi i$ where $m$ is any integer
See also: Find all real numbers $x$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac76$
| {
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"url": "https://math.stackexchange.com/questions/3761033",
"timestamp": "2023-03-29T00:00:00",
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Showing $\sum_{\alpha=0}^k \binom{k+15}{\alpha + 10}\binom{k}{\alpha}=\frac{2^{2k+15}(k+6)(k+7)\left(\frac{2k+15}{2}\right)!}{\sqrtπ (k+10)!}$ I want to simplify $\sum_{\alpha=0}^k \binom{k+15}{\alpha + 10}\binom{k}{\alpha}$ but this can't be directly simplified using Vandermonde Chu Identity. Wolfram shows a nice closed form expression but I can't get to it.
My attempt
$\sum_{k=0}^n \binom{k+15}{5+k-\alpha}\binom{k}{\alpha}=\binom{2k+15}{5+k}$ by thinking of this sum as choosing a set of 5+k objects from 2k+15 objects in total but I feel that this isn't correct as some terms are missing
|
Prove that $$\binom{2k+15}{k+5}=\frac{2^{2k+15}(k+6)(k+7)\left(\frac{2k+15}{2}\right)!}{\sqrtπ (k+10)!}$$
Number of $2$'s in $\left(\frac{2k+15}{2}\right)!=\frac{2k+15+1}{2}=k+8$. So, the RHS becomes
\begin{align*}
&\Rightarrow\frac{2^{k+7}(k+6)(k+7)(2k+15)!!\ \Gamma\left(\frac12\right)}{\sqrtπ(k+10)!}\\
&=\frac{2^{k+5}(k+5)!(2k+12)(2k+14)(2k+15)!!}{(k+10)!(k+5)!}&\left(\because \Gamma\left(\frac12\right)=\sqrt{\pi}\right)\\
&=\frac{(2k+10)!!(2k+12)(2k+14)}{(k+10)!(k+5)!}\frac{(2k+15)!}{(2k+14)!!}\\
&=\frac{(2k+15)!}{(k+10)!(k+5)!}\\
\end{align*}
PS: I think that Wolfram Alpha's algorithm doesn't apply the common logic of choosing $k+5$ people from $2k+15$.
| {
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Distance between set and point, confused of partial derivatives. Let $H = \{(x,y,z)\ \in \mathbb{R}^{3}: x^2+y^2 - z^2 + 4 = 0$
Compute the shortest distance between H and point $p=(2,4,0)$.
I am a bit confused because I tried a direct approach.
$$ x^2+y^2 + 4 = z^2$$
Let $D(H,p) = \sqrt{(2-x)^{2}+(4-y)^{2} + x^{2} + y^{2}+4}$
So I tried compute $$\frac{\partial D}{\partial x} = (\sqrt{2} (-1 + x))/\sqrt{12 - 2 x + x^2 - 4 y + y^2}$$
$$\frac{\partial D}{\partial y} = (\sqrt{2} (-2 + y))/\sqrt{12 - 2 x + x^2 - 4 y + y^2}$$
It seems not nice to compare with zero.
Do you have another idea?
|
Let $H = \{(x,y,z)\ \in \mathbb{R}^{3}: x^2+y^2 - z^2 + 4 = 0\}$.
Compute the shortest distance between $H$ and the point $P\equiv(2,4,0)$.
Approach other than Lagrange's multipliers:
Since the graph is symmetric in $\phi$, you can also minimize $(\sqrt{2^2+4^2}-r)^2+(r^2+4)=2r^2-2\sqrt{20}r+24$ for some $r$, where $\sqrt{(\sqrt{2^2+4^2}-r)^2+(r^2+4)}$ is the distance between the point $(\sqrt{2^2+4^2},\phi,0)$ and $(r,\phi,z)$.
| {
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If $\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\le 1$, prove that $(1+a^2)(1+b^2)(1+c^2)\ge 125$.
QUESTION: Let $a,b,c$ be positive real numbers such that
$$\cfrac{1}{1+a}+\cfrac{1}{1+b}+\cfrac{1}{1+c}\le 1$$
Prove that $$(1+a^2)(1+b^2)(1+c^2)\ge 125$$ When does equality hold?
MY APPROACH: Firstly, let's try to squeeze out all the information we can from what is given. $$\frac{(1+b)(1+c)+(1+a)(1+c)+(1+a)(1+b)}{(1+a)(1+b)(1+c)}≤1$$ Multiplying this out, we get $$3+2(a+b+c)+(ab+bc+ca)≤1+(a+b+c)+(ab+bc+ca)+abc$$ $$\implies 2+(a+b+c)≤abc$$ Also, since $$1≥\sum_{cyc}\frac{1}{1+a}$$ Therefore by AM-GM,
$$1≥\sum_{cyc}\frac{3}{\sqrt[3]{(1+a)(1+b)(1+c)}}$$
$$\implies (1+a)(1+b)(1+c)≥27$$
That's all I ended up in.. At first, I thought Hölder's inequality could be employed, but that too requires the sum of the powers to be $=1$.. and that is not going to be useful in $(1+a^2)(1+b^2)(1+c^2)$ , since here the sum of powers add up to $3$..
I don't know what to do next.. Any help will be much appreciated..
| As $$abc\geq a+b+c+2\overset{AM-GM}{\geq} 4\sqrt[4]{2abc}\implies abc\geq 8$$Now,$$\prod_{cyc} (1+a^2)=\prod_{cyc} \left(1+\frac{a^2} 4+\frac{a^2} 4+\frac{a^2} 4+\frac{a^2} 4\right)\overset{AM-GM}{\geq}\prod_{cyc} 5\times \left(\frac{a}{2}\right)^{8/5}=125\times(abc/8)^{8/5}\geq 125$$
| {
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In △ABC, ∠A= 60$^∘ $, BC=12, BD⊥AC, CE⊥AB and∠DBC = 3∠ECB. Find EC in the form $a(\sqrt{b}+\sqrt{c})$ Given that $m \angle A= 60^\circ$, $BC=12$ units, $\overline{BD}\perp\overline{AC}$, $\overline{CE} \perp \overline{AB}$, and $m \angle DBC = 3m \angle ECB$, the length of segment $EC$ can be expressed in the form $a(\sqrt{b}+\sqrt{c})$ units where $b$ and $c$ have no perfect-square factors. What is the value of $a+b+c$?
Let the intersection of $EC$ and $BD$ be $F$. I have figured out that $\triangle EFB$ and $\triangle DFC$ are both $30-60-90$ triangles, and that $\triangle BDC$ is a $45-45-90$ triangle. Along with this, I have figured out all other angles. However, I can't figure out the next step. Can someone help? Thanks!
| Let $\angle ECB = y$, so $\angle DBC=3y$. From $\triangle AEC$ we have $\angle ACE = 180^\circ-60^\circ-90^\circ= 30^\circ$.
Now let $EC$ and $BD$ intersect at $F$. $\angle BFE=\angle DFC$ by vertical angles and $\angle BEF=\angle CDF=90^\circ$, so $\angle FBE=\angle FCD$, which is equal to 30 degrees. Now summing the angles in $\triangle ABC$, we have $60^\circ+30^\circ+3y+y+30^\circ=180$, solving yields $4y=60$ so $y=15$ and we see $\triangle BDC$ is a 45-45-90 triangle. Also, $\triangle ABD$ is a 30-60-90 triangle.
Let $ AD = x$, so $AB = 2x$ and $DB = DC = x\sqrt{3}$. $BC = x\sqrt{3}\sqrt{2} = x\sqrt{6}$. We are given that this equals 12, so we find $x = 12/\sqrt{6} = 2\sqrt{6}$. It follows that the area of $\triangle ABC$ can be found via[(1/2)(AC)(BD)=(1/2)(x+x\sqrt{3})(x\sqrt{3})=12\sqrt{3}+36.]To find $EC$, notice that the area of $\triangle ABC$ can also be written as $(1/2)(AB)(EC)$. Thus,[(1/2)(4\sqrt{6})(EC)=12\sqrt{3}+36 \Rightarrow EC = 3(\sqrt{2}+\sqrt{6}).]Hence $a=3$, $b=2$, and $c=6$, so $a+b+c=\boxed{11}$.
| {
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Finding the volume when a parabola is rotated about the line $y = 4$. Problem:
Find the volume generated by revolving the region bounded below by the
parabola $y = 3x^2 + 1$ and above by the line $y = 4$ about the line $y = 4$.
Answer:
Let $V$ be the volume we are trying to find. We want to find the points where the two curves intersect. Hence,
set up the following equation:
$$ 3x^2 + 1 = 4$$.
From this equation, we find two solutions: $ x = \pm 1 $
\begin{align*}
du &= -dx \\
V &= \pi \int_{-1}^{1} (4 - (3x^2 + 1))^2 \,\,\, dx = \pi \int_{-1}^{1} (3 - 3x^2)^2 \,\,\, dx \\
V &= 9 \pi \int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \int_{-1}^{1} x^4 - 2x^2 + 1 \,\,\, dx \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \frac{x^5}{5} - \frac{2x^3}{3} + x \Big|_{-1}^{1} \\
%
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &=
\frac{^5}{5} - \frac{2(1)^3}{3} + 1 - \left( \frac{(-1)^5}{5} - \frac{2(-1)^3}{3} - 1 \right) \\
%
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \frac{1}{5} - \frac{2}{3} + 1 - \left( -\frac{1}{5} + \frac{2}{3} - 1 \right) \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \frac{1}{5} - \frac{2}{3} + 1 + \frac{1}{5} - \frac{2}{3} + 1 \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= 2 - \frac{4}{3} + \frac{2}{5} = 2 - \frac{20}{15} + \frac{6}{15} \\
\int_{-1}^{1} (x^2 - 1)^2 \,\,\, dx &= \frac{16}{15} \\
V &= 9 \pi \left( \frac{16}{15} \right) \\
V &= \frac{ 48 \pi }{5}
%
\end{align*}
The book's answer is:
$$ \frac{144 \pi}{15} $$
I claim my answer is right. That is, the book failed to simplify its answer. Am I missing something? Please comment.
| $$\frac{144 \pi}{15}$$
dividing the numerator and denominator by three:
$$\frac{48 \pi}{5}$$
In the future, if you don't feel like simplifying, you can just divide the two answers and see if it lines up
$$\frac{144 \pi}{15} \approx 30.1593$$
$$\frac{48 \pi}{5} \approx 30.1593$$
So, yes, you're right.
| {
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Computing characteristic polynomial of $4 \times 4$ matrix Given the following matrix
$$A = \begin{bmatrix} 1 & -3 & 0 & 3 \\ -2 & -6 & 0 & 13 \\ 0 & -3 & 1 & 3 \\ -1 & -4 & 0 & 8\end{bmatrix}$$
I need to calculate $|\lambda I - A|$.
I get to a very complicated determinant. Is there an easier way? The answer says it’s $(\lambda - 1)^4$.
| $$\lambda I- A=
\begin{bmatrix}
\lambda -1&3&0&-3 \\
2&\lambda + 6&0&-13 \\
0&3&\lambda -1&-3 \\
1&4&0&\lambda-8
\end{bmatrix}
$$
By developing the determinant using the third column, we'll have:
$$\det(\lambda I -A)= (\lambda -1) \cdot \begin{bmatrix}
\lambda -1&3&-3 \\
2&\lambda + 6&-13 \\
1&4&\lambda-8
\end{bmatrix} $$ $$= (\lambda -1)\left( (\lambda -1)\cdot \begin{bmatrix}\lambda + 6&-13 \\ 4& \lambda-8 \end{bmatrix} - 3 \cdot \begin{bmatrix}2&-13 \\ 1& \lambda-8 \end{bmatrix}
+3 \cdot \begin{bmatrix}2&\lambda +6 \\ 1& 4 \end{bmatrix} \right)$$
by solving this, you will get your answer. I don't think it's too complicated to handle. notice that $$\begin{bmatrix}\lambda + 6&-13 \\ 4& \lambda-8 \end{bmatrix}, \begin{bmatrix}2&-13 \\ 1& \lambda-8 \end{bmatrix}
, \begin{bmatrix}2&\lambda +6 \\ 1& 4 \end{bmatrix}$$ are three minors which you will have to find their determinant.
| {
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One number is removed from the set of integers from $1$ to $n.$ The average of the remaining numbers is $163/4.$ Which integer was removed?
One number is removed from the set of integers from $1$ to $n.$ The average of the remaining numbers is $\dfrac{163}4$. Which integer was removed?
Source. British Mathematical Olympiad 2010/11, Round 1, Problem 1
I was hoping if someone could spot the flaw in my working for this question.
Attempt.
I began by letting the integer that was removed be $x$.
Then: $$\frac{1 + 2 + \cdots + (x-1) + (x+1) +\cdots + n} {n-1} = \frac{163}{4}$$
There is two arithmetic sums in the denominator, the first from 1 to $x$ and the second from $x+1$ to $n$.
These are equal to $\frac{x(x-1)}{2}$ and $\frac{(n-x)(n+x+1)}{2}$, and subbing in to first equation this gives:
$$\frac{x(x-1) + (n-x)(n+x+1)}{2(n-1)} = \frac{163}{4}$$
which reduces to:
$$\frac{n^2 + n - 2x}{2(n-1)} = \frac {163}{4}$$
And then:
$$2(n^2 + n -2x) = 163(n-1)$$
At first I thought you could consider factors, as 163 was prime then:
$n-1 = 2$ giving $n = 3$ and $n^2 + n - 2x = 163$, which using $n=3$ gives $x= -75.5$ which isn't our positive integer.
I then tried considering a quadratic in $n$ and using the discriminant but again that just looked to give a negative value of $x.$ I would be grateful for any help
| This is not how you are supposed to solve it, but I feel like cheating. We have the equation $2(n^2+n-2x)=163(n-1)$, and $1\leq x\leq n$.
If you assume that $x=1$ then you solve for $n$ using the quadratic formula, and you obtain $79.5$.
If you assume that $x=n$ then you solve for $n$ you obtain $81.5$. Thus $n=80$ or $n=81$.
If $n=80$ then you can solve for $x$ and obtain $83=4x$, wrong. So $n=81$. Solving again yields $4n=244$, and ding ding, we have a winner.
As everyone else seems to have done this in the way I considered cheating, I should expand on the way I initially did it, which is completely different.
We have $2(n^2+n-2x)=163(n-1)$, and $1\leq x\leq n$. We see that $n$ is odd, so $n=2m+1$. Substituting in and cancelling the 2s yields
$$(2m+1)^2+(2m+1)-2x=163m$$
or
$$4m^2+2-2x=157m.$$
Write $y=x-1$ and also, the LHS is even, so $m=2a$ is even. More substituting and removing the $2$ from both sides yields
$$8a^2-y=157a.$$
The crucial point: we see that $a\mid y$. Since $n=2m+1=4a+1$, and $y$ is a multiple of $a$, $y=\alpha a$ for $\alpha$ between $1$ and $4$. Dividing through by $a$ yields
$$8a-\alpha=157.$$
Taking congruences modulo $8$ yields $\alpha\equiv 3\bmod 8$, so $\alpha=3$. Thus $y=3a$, so $x=3a+1$ and $n=4a+1$. We put this back into the top equation, $2(n^2+n-2x)=163(n-1)$, to obtain $a=20$, so $n=81$, $x=61$.
| {
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If $\lim_{x\to 0} \frac{x^2 \sin (bx)}{ax-\sin x}=1$, then find $a,b$ $$\lim_{x\to 0} \frac{x^2 \frac{\sin bx}{bx} (bx)}{x(\frac{ax}{x}-\frac{\sin x}{x})}$$
$$=\lim_{x\to 0} \frac{bx^2 \frac{\sin bx}{bx}}{a-\frac{\sin x}{x}}$$
$$=0$$
Which is obviously wrong. I think I am suffering from lack of conceptual clarity in limits, so what exactly is going wrong in this question?
| $$L=\lim_{x\to 0} \frac{x^2\sin bx}{ax-\sin x}=\lim_{x\to 0} \frac{x^2(bx-b^3x^3/6+...)}{ax-x+x^3/6}=\lim_{x\to 0} \frac{bx^3-b^5 x^5/6}{ax-x+x^3/6}$$
If $L=1$, then $a=1,b=1/6.$
| {
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Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. So here is the Question :-
Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. Find $f(-3)$ .
What I tried:- Since it's given that $f(x)$ is a polynomial of degree $3$ , I can assume $f(x) = ax^3 + bx^2 + cx + d$ for some integers $a,b,c,d$ and $a\neq 0$. Then we have :-
$$ax^3 + bx^2 + cx + d = (x^2 - 1)y + (2x - 5)$$
$$ax^3 + bx^2 + cx + d = (x^2 - 4)z + (-3x + 4)$$
This gives that $(x^2 - 1)y + (2x - 5) = (x^2 - 4)z + (-3x + 4)$ . But I am not sure how to proceed further since we have $3$ variables to deal with , and I am stuck here.
Any hints or explanations for this problem will be greatly appreciated !!
| Working $\,\bmod (x^2-1)\,$ we have that $\,x^2 \equiv 1\,$, then:
$$
f(x) = ax^3+bx^2+cx+d\equiv ax+b+cx+d=(a+c)x+(b+d)
$$
Comparing to the known remainder $\,2x-5\,$:
$$
\begin{cases}
a+c=2
\\ b+d = -5 \tag{1}
\end{cases}
$$
Repeating the steps $\,\bmod (x^2-4)\,$ and remainder $\,-3x + 4\,$:
$$
f(x) \equiv 4ax+4b+cx+d=(4a+c)x+(4b+d) \;\;\implies\;\;
\begin{cases}
4a+c=-3
\\ 4b+d = 4 \tag{2}
\end{cases}
$$
Solving the four equations $\,(1), (2)\,$ for $\,a,b,c,d\,$ gives the polynomial $\,f(x)\,$.
| {
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Wilson's theorem and fractions
What is the remainder of $m$ satisfying $$\frac{1}{1}+\frac{1}{2}+\frac{1}{3} +\dots+\frac{1}{33}=\frac{m}{33!}$$ upon division by $17$?
The factorial here hints me towards Wilson's theorem, but I cannot seem to get this in any way to the form such that I could use it. How should I go about this?
| Note that $$m=\frac{33!}{1}+\frac{33!}{2}+\cdots+\frac{33!}{33}$$
And we have to find the remainder when $m$ is divided by $17$.
Note that out of all the above terms, all terms except $\dfrac{33!}{17}$ are divsible by $17$. Thus $m\equiv \frac{33!}{17} \pmod{17}$.
But we have $$\begin{aligned}\frac{33!}{17}&=33\times 32\times \cdots \times 18 \times 16\times \cdots \times 1 \\&\equiv (-1)\times (-2)\times\cdots\times (-16)\times16\times\cdots \times1 \pmod{17} \\ &\equiv (16!)^2\pmod{17}\end{aligned}$$
Now from Wilson's theorem, we have $16!\equiv -1\pmod{17} \implies (16!)^2\equiv 1 \pmod{17}$
Thus $m\equiv \boxed{1} \pmod{17}$
| {
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How to obtain the sum of the series $\sum_{n=0}^{\infty}\frac{1}{2^{n}(3n+1)}$? How to prove what follows?
$$\sum_{n=0}^{\infty}\frac{1}{2^{n}(3n+1)}=\frac{2^{\frac{1}{3}}}{3}\ln\left(\frac{\sqrt{2^{\frac{2}{3}}+2^{\frac{1}{3}}+1}}{2^{\frac{1}{3}}-1}\right)+\frac{\sqrt[3]{2}}{3}\arctan\left(\frac{2^{\frac{2}{3}}+1}{\sqrt{3}}\right)-\frac{2^{\frac{1}{3}}\pi}{6\sqrt{3}}$$
My attempt:
$$\sum_{n=0}^{\infty}\frac{1}{2^n(3n+1)}=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{2^n(3n+1)}|_{x=1}$$
We put $$S(x)=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{2^n(3n+1)}\implies S^{'}(x)=\sum_{n=0}^{\infty}\frac{x^{3n}}{2^n(3n+1)}$$
$$S^{'}(x)=\sum_{n=0}^{\infty}\frac{(\frac{x^3}{2})^n}{3n+1}=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(1-\frac{3n}{3n+1})=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n-\sum_{n=0}^{\infty}(\frac{x^3}{2})^n\frac{3n}{3n+1}=\frac{1}{2-\frac{x^3}{2}}-\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(\frac{3n}{3n+1})=\alpha-\beta$$
Where
$$\beta=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(\frac{3n}{3n+1})$$
So $$\beta=?$$
Waiting for your help to find a beta or prove equal above.
| You can modify your attempt a little bit to get an easier way of solving this problem:
$$\sum_{n=0}^{\infty} \dfrac{1}{2^n(3n+1)} = \sum_{n=0}^{\infty} \dfrac{1}{(\sqrt[3]{2})^{3n}(3n+1)} = \sqrt[3]{2} \sum_{n=0}^{\infty} \dfrac{1}{(\sqrt[3]{2})^{3n+1}(3n+1)}$$
And now you can find the value of $$S(x) = \sum_{n=0}^{\infty} \dfrac{x^{3n+1}}{3n+1}$$ and replace $x = \dfrac{1}{\sqrt[3]{2}}$ back into $S(x)$
| {
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INMO : Prove that $\sqrt[3]{a}$ and $\sqrt[3] {b}$ themselves are rational numbers Let $a$ and $b$ be two positive rational numbers such that $\sqrt[3] {a} + \sqrt[3]{b}$ is also a rational number. Prove that $\sqrt[3]{a}$ and $\sqrt[3] {b}$ themselves are rational numbers.
My first response was, isn't is obvious? but then I tried..
if $a=b$ then we prove it easily . So suppose $a\ne b$.
Let $s=\sqrt[3]{a} + \sqrt[3]{b}$.
then $s^3 = a + b + 3 s \sqrt[3]{a} \sqrt[3]{b}$
then the product $p=\sqrt[3]{a} \sqrt[3]{b}$ is a rational number.
But what should I do next ?
| From your work, we can write $$\frac{p}{\sqrt[3]{a}} + \sqrt[3]{a} = s,$$ so $$\sqrt[3]{a} = \frac{s\pm \sqrt{s^2-4p}}{2}.$$
Cube both sides and solve for $\sqrt{s^2-4p}$ (noting $s^2-p \neq 0$) to show it's rational, which then shows $\sqrt[3]{a}$ is rational.
| {
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Question on determinant Given $x \in \Bbb R$ and
$$P = \begin {bmatrix}1&1&1\\0&2&2\\0&0&3\end {bmatrix}, \qquad Q=\begin {bmatrix}2&x&x\\0&4&0\\x&x&6\end {bmatrix}, \qquad R=PQP^{-1}$$
show that
$$\det R = \det \begin {bmatrix}2&x&x\\0&4&0\\x&x&5\end {bmatrix}+8$$
for all $x \in \Bbb R$.
My attempt: $|R|=\frac{|P||Q|}{|P|}=|Q|$
$$\left|\begin {array}&2&x&x\\0&4&0\\x&x&6\end {array}\right|=\left|\begin {array}&2&x&x\\0&4&0\\x&x&5\end {array}\right|+\left|\begin {array}&2&x&x\\0&4&0\\x&x&1\end {array}\right|=\left|\begin {array}&2&x&x\\0&4&0\\x&x&6\end {array}\right|+8-4x^2$$
What's my mistake?
| Hint.
$$
\det\left|\begin {array}&2&x&x\\0&4&0\\x&x&6\end {array}\right|=\det\left|\begin {array}&0&0&4\\2&x&x\\x&6&x\end {array}\right| = \det\left|\begin {array}&0&0&4\\2&x&x\\x&5&x\end {array}\right|+8
$$
| {
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Are there any elementary functions $\beta(x)$ that follows this integral $\int_{y-1}^{y} \beta(x) dx =\cos(y)$ Are there any simple functions $\beta(x)$ that follows this integral $$\int_{y-1}^{y} \beta(x) dx =\cos(y)$$
I think there is an infinite amount of solutions that are continuous everywhere but how can I find one that only uses elementary functions?
| Let $f(y) :=\beta(y) - (A\sin y + B \cos y) = \beta(y-1) - (A\sin (y-1) + B\cos (y-1))$
then $(A+1)\sin y + B\cos y =A\sin(y-1) +B \cos(y-1) $.
$=(A\cos 1 + B \sin 1)\sin y + (B \cos 1 - A\sin 1)\cos y$
$\therefore A+1 = (A\cos 1 + B \sin 1) \; , \; B = (B \cos 1 - A\sin 1) $
$A+1 = \frac{1}{A}((A^2+B^2)\cos 1 - B^2) $ ( $\because B(\cos 1 - 1) = A \sin 1 $)
$\Rightarrow (A^2 + B^2)(\cos 1 -1) = A , (A^2+B^2)\sin 1 = B$
Let $A^2 + B^2 = R$. then $R = R^2 ( 2 - 2\cos 1) $ , $R= \frac{1}{2-2\cos 1}$
so $ A = -1/2$ , $ B = \frac{\sin 1}{2(1-\cos 1)}$.
Now, $f(y)$ is periodic. ( period of $f(y)$ is $1$)
and
$ \int_{y-1}^{y} f(x)dx = \cos y - \int_{y-1}^{y} (A\sin x +B \cos x)dx \\ =\cos y + A(\cos y - \cos (y-1)) - B(\sin y - \sin (y-1))$.
because period of $f(y)$ is $1$, $\int_{y-1}^{y} f(x) dx = \int_{0}^{1} f(x) dx$
$= \cos1 + (A\cos 1 - A)-B(\sin 1) =\frac{1}{2} + \frac{1}{2} \cos 1 - \frac{\sin^2 1}{2(1-\cos 1)}$.
Finally, we get $ \beta(y) = -\frac{1}{2}\sin y + \frac{\sin 1}{2(1 - \cos 1)} \cos y + (\frac{1}{2} + \frac{1}{2} \cos 1 - \frac{\sin^2 1}{2(1-\cos 1)} ) $
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"language": "en",
"url": "https://math.stackexchange.com/questions/3778088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
How to prove distance from foci on an ellipse is equal to twice the semi-major axis (for specific ellipse) Prove that for any point (x,y) on the conic, the sum of the distances to the two foci is always twice the semi-major axis.
I know that this can be proven in general for all ellipses but the practice question specifically asks for this to be proven for $\frac{x^2}{9}$ + $\frac{y^2}{4}$ = 1. I feel like I'm really close but I've managed to math myself into a corner somehow.
Let the foci ($\sqrt{5}$, 0) and (-$\sqrt{5}$, 0) be denoted as F and F'. Let the point on the conic be denoted P(x,y). We are required to show PF + PF' = 2a. In this case, since a = 3, 2a = 6.
PF = $\sqrt{(x-\sqrt{5})^2 + y^2}$ and PF' = $\sqrt{(x+\sqrt{5})^2 + y^2}$
By rearranging the equation for the ellipse, we get y$^2$ = 4 - $\frac{4}{9}$x$^2$.
Substitute this into PF and PF' to get:
PF = $\sqrt{(x-\sqrt{5})^2 + 4 - \frac{4}{9}x^2}$ = $\sqrt{\frac{5}{9}x^2 - 2\sqrt{5}x + 9}$ = $\sqrt{(x - \frac{9\sqrt{5}}{5})^2}$ = x - $\frac{9\sqrt{5}}{5}$
PF' = $\sqrt{(x+\sqrt{5})^2 + 4 - \frac{4}{9}x^2}$ = $\sqrt{\frac{5}{9}x^2 + 2\sqrt{5}x + 9}$ = $\sqrt{(x + \frac{9\sqrt{5}}{5})^2}$ = x + $\frac{9\sqrt{5}}{5}$
Therefore PF + PF' = 2x
And then I got stuck
| An ellipse is a plane curve surrounding two foci, such that for all points on the curve, the sum of the two distances to the foci is a constant. One starts with $\sqrt{(x-c)^2+y^2}+\sqrt{(x+c)^2+y^2}=2a$ (your question) to arrive at $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $2a$ is any constant (which ends up being the length of the semi-major axis), $b^2=a^2-c^2$, and the foci are $(-c,0),(+c,0)$. Note that $a,b,c\in\mathbb R^+$.
Addendum
Finding the distance of either foci from a point $P(x,y)$ on the ellipse, $$PF=\sqrt{(x-c)^2+y^2}=\sqrt{(x-c)^2+b^2-\frac{b^2}{a^2}x^2}=\sqrt{\frac{c^2}{a^2}x^2-2cx+a^2}=\left\lvert\frac ca x-a\right\rvert=a-ex$$
$$PF^\prime=\sqrt{(x+c)^2+y^2}=\sqrt{(x+c)^2+b^2-\frac{b^2}{a^2}x^2}=\sqrt{\frac{c^2}{a^2}x^2+2cx+a^2}=\left\lvert\frac ca x+a\right\rvert=a+ex$$
since $x\in[-a,+a]$, where $e=\frac ca$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Evaluating the surface integral $\iint_S {({x^2} + {y^2})} \,dS$ using spherical coordinates For the integral $$\iint\limits_S {({x^2} + {y^2})} \,dS\quad,\,S:{x^2} + {y^2} + {z^2} = 2z$$The
correct answer is $${{8\pi } \over 3}$$ I used Spherical coordinate system,it turns to $$\int_0^{2\pi } {d\theta \int_0^{{\pi \over 2}} {({r^2}{{\sin }^2}\varphi } )({r^2}\sin \varphi )\,d\varphi } ,r = 2\cos \varphi $$Then use $r = 2\cos \varphi$,it turns to $$32\pi \int_0^{{\pi \over 2}} {{{\sin }^3}\varphi {{\cos }^4}\varphi \,d\varphi } = {{64} \over {35}}\pi $$Doesn't match the answer,I wonder where am I wrong.
| As an extra, there is a way of computing this integral without ever integrating by using spherical coordinates, which all hinges on symmetry. Notice that the sphere could also be written as
$$x^2+y^2+(z-1)^2=1$$
Since the integrand doesn't depend on the location in $z$ at all, we could safely translate the sphere down and be assured that
$$\iint\limits_{x^2+y^2+(z-1)^2=1}x^2+y^2\:dS = \iint\limits_{x^2+y^2+z^2=1}x^2+y^2\:dS$$
holds. Second, notice that by rotational symmetry (or just pairwise swapping the variables AKA reflectional symmetry) on the second surface we have that
$$\iint\limits_{x^2+y^2+z^2=1}x^2\:dS = \iint\limits_{x^2+y^2+z^2=1}y^2\:dS = \iint\limits_{x^2+y^2+z^2=1}z^2\:dS \equiv I$$
which is a quantity we will denote by $I$. Then the beauty of being on a sphere tells us that
$$3I = \iint\limits_{x^2+y^2+z^2=1}x^2+y^2+z^2\:dS = \iint\limits_{x^2+y^2+z^2=1}1\:dS = 4\pi$$
Thus the answer to the problem is $2I$, or
$$\frac{8\pi}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
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