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Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x(3+x)}{\sqrt{3-x^2}(3+x)^2}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x}{\sqrt{3-x^2}(3+x)}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
At this point, I want to transform this derivative into the form of $\dfrac{3(x+1)}{(3+x)^2\sqrt{3-x^2}}$
How do I do this? This form is given by Wolfram:
https://www.wolframalpha.com/input/?i=derivative+%283-x%5E2%29%5E%281%2F2%29%2F%283%2Bx%29
| $f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$ is false !
Correct is:
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(-x^2)-\sqrt{3-x^2}}{(3+x)^2}$.
Now proceed !
| {
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"url": "https://math.stackexchange.com/questions/3525571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Probability of winning chocolates in throwing game In urn there are 12 balls from which 7 are green.John draws at random 4 balls.Then he hits a target with the green balls he drew(if there are any).His probability of hitting successfully is $\frac{1}{4}$ in every shot and if he succeeds he wins a chocolate for every successful hit.
a)What is the probability John to win exactly 2 chocolates?
b)if a) happened,what is the probability he drew exactly 3 green balls from the urn?
How I solved it I think : Let $A$ be the event John wins 2 chocolates, and let $H_{i}$ for $i$ in ${[0,1,2,3,4]}$ be the event - John drew exactly $i$ green balls.
So for a) we seek $P(A|H_{1} \cup H_{2} \cup H_{3} \cup H_{4}) =P(A|H_{0})P(H_{0})+(A|H_{1})P(H_{1})+P(A|H_{2})P(H_{2})+P(A|H_{3})P(H_{3})+P(A|H_{4})P(H_{4})$.
The cases where $i=0$ and $i=1$ , the probability is $0$ so we need to find
$P(A|H_{2})P(H_{2})+P(A|H_{3})P(H_{3})+P(A|H_{4})P(H_{4})$.
$P(H_{2}) = \frac{ \binom{7}{2}\binom{5}{2}}{\binom{12}{4}}$ ,
$P(H_{3}) = \frac{ \binom{7}{3}\binom{5}{1}}{\binom{12}{4}}$ ,$P(H_{4}) = \frac{ \binom{7}{4}\binom{5}{0}}{\binom{12}{4}}$ ,
$P(A|H_{2})=\frac{1}{4}\frac{1}{4}$ - since we want him to succeed two times
$P(A|H_{3})=3(\frac{1}{4}\frac{1}{4}\frac{3}{4})$ - since we want two success and 1 failure multiplied by 3 for the permutations
$P(A|H_{4})=6(\frac{1}{4}\frac{1}{4}\frac{3}{4}\frac{3}{4})$
Now we just plug them in above and find the answer
for b) I think we seek $P(H_{3}|A)=?$ which can be found applying Bayes's theorem and plugging in the values from a)
Is my solution correct?
| Note that John can only win 2 chocolates if he drew at least 2 green balls from the urn. So the first step is to find the probs of drawing 2, 3 and 4 greens from the urn.
We have $$p(2G)= 6\frac{7}{12}\frac{6}{11}\frac{5}{10}\frac{4}{9}=\frac{42}{99}$$ Similarly $$p(3G)=\frac{35}{99},p(4G)=\frac{7}{99}$$
We also need to work out $$p(2c|2G)=\frac{1}{4}\frac{1}{4}=\frac{8}{128},\quad p(2c|3G)=3\frac{1}{4}\frac{1}{4}\frac{3}{4}=\frac{18}{128},\quad p(2c|4G)=6\frac{9}{256}=\frac{3}{128}$$
So now we can apply Bayes to get that $$p(3G|2c)=\frac{p(2c|3G)p(3G)}{p(2c|2G)p(2G)+p(2c|3G)p(3G)+p(2c|4G)p(4G)}$$ $$=\frac{18\cdot35}{8\cdot42+18\cdot35+3\cdot7}=\frac{630}{987}\approx0.64$$
| {
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"url": "https://math.stackexchange.com/questions/3526086",
"timestamp": "2023-03-29T00:00:00",
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ordered pair of $(p,q)$ in $20!$
A rational number given in the form
$\displaystyle \frac{p}{q},\;\;p,q\in \mathbb{Z}^{+}\;,\frac{p}{q}\in(0,1)$ and $p,q$ are coprime to each other.If $pq=20!.$ Then number of ordered pair of $p,q$ are
what i try
$20!=2^{18}\cdot 3^{8}\cdot 5^4\cdot 7^2\cdot 11\cdot 13\cdot 17\cdot 19$
Let $p=2^{a_{1}}\cdot 3^{a_{2}}\cdot 5^{a_{3}}\cdot 7^{a_{4}}\cdot 11^{a_{5}}\cdot 13^{a_{6}}\cdot 17^{a_{7}}\cdot 19^{a_{8}}$
$q=2^{b_{1}}\cdot 3^{b_{2}}\cdot 5^{b_{3}}\cdot 7^{b_{4}}\cdot 11^{b_{5}}\cdot 13^{b_{6}}\cdot 17^{b_{7}}\cdot 19^{b_{8}}$
$0\leq a_{1}\leq 18,0\leq a_{2}\leq 8,0\leq a_{3}\leq 4,0\leq a_{4}\leq 2,0\leq a_{5}\leq 1$
$0\leq a_{6}\leq 1,0\leq a_{7}\leq 1,0\leq a_{8}\leq 1$
and $0\leq b_{1}\leq 18,0\leq b_{2}\leq 8,0\leq b_{3}\leq 4,0\leq b_{4}\leq 2,0\leq b_{5}\leq 1$
$0\leq b_{6}\leq 1,0\leq b_{7}\leq 1,0\leq b_{8}\leq 1$
How do i solve it Help me please
| Since $p.q=20!$ hence I can treat this like there are eight distinct objects i.e $2^{18},3^8,5^4,7^2,11,13,17,19$ need to be distributed into two different groups $p , q$ where
$$ $$
Total number of ways are $2^8$ since each object can go in any one of the group. And since $20!$ is not a perfect square hence $p\neq q$ .
$$ $$
For $\frac{p}{q} \in (0,1)$ we need to consider unordered pairs of solutions i.e $\frac{2^8}{2}=2^7=128$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve a system of equation: $\cos(2x) + \cos(y) = 1$, $\sin(2x) + \sin(y) = 1$ Solve a system of equation:
$$\cos(2x) + \cos(y) = 1$$
$$\sin(2x) + \sin(y) = 1$$
My idea:
Let's see what is product of this two equations.
$$\cos(2x)\sin(2x) + \cos(2x)\sin(y) + \cos(y)\sin(2x) + \cos(y)\sin(y) = 1$$
$$\cos(2x)\sin(2x) + \sin(2x+y) + \cos(y)\sin(y) = 1$$ But this idea didn't give me anything. Also if I sum I have problem... but this is high school problem so it must have some easy solution.
| HINT.-We have from the two given equations $$2\cos\left(\dfrac{2x+y}{2}\right)\cos\left(\dfrac{2x-y}{2}\right)=1\\2\sin\left(\dfrac{2x+y}{2}\right)\cos\left(\dfrac{2x-y}{2}\right)=1$$ so, by division $$\tan\left(\dfrac{2x+y}{2}\right)=1\Rightarrow\dfrac{2x+y}{2}=\dfrac{(2n+1)\pi}{4}$$ Each of the two given equations is an "oval" closed curve over some bounded region of the plan and this is repeated periodically because of the periodicity of functions considered.
Now taking $n=0$ the line $y=-2x+\dfrac{\pi}{2}$ cuts the axes in $(0,\dfrac{\pi}{2})$ and
$(\dfrac{\pi}{4},0)$ which give points of intersection satisfying the proposed equations. Thus the general solution, taked from paralelle lines to the first one is $$(x,y)=\left(\dfrac{\pi}{4},\dfrac{\pi}{2}\right),\left(\dfrac{5\pi}{4},\dfrac{\pi}{2}\right),\left(\dfrac{9\pi}{4},\dfrac{\pi}{2}\right),\cdots,\left(\dfrac{(4n+1)\pi}{4},\dfrac{\pi}{2}\right),\cdots$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Complicated question, please help me out I'm not able to get a proper defined function and not able to determine it's nature, pls help
Let the function $f:[0,1] \longrightarrow \Bbb R$ be defined by
$$f(x) = \max\left\{\frac{|x-y|}{x+y+1}\, :\, 0\leqslant y \leqslant 1\right\}.$$
Then what are the intervals on which the function is strictly increasing and strictly decreasing?
| For each $x\in [0,1]$, the $y\in [0,1]$ which maximizes $\frac{|x-y|}{x+y+1}$ also maximizes ${\left(\frac{|x-y|}{x+y+1}\right)}^2$.
Because $|x-y|^2 = (x-y)^2$, we may apply calculus techniques to the second one.
We could also not square, and consider when $y\leqslant x$ and when $y>x$, but squaring does away with cases.
Hence the maximum occurs when $y=0$, when $y=1$ or when $y$ is a root of the derivative
$$\frac \partial{\partial y}\left(\frac{{(x-y)}^2}{{(x+y+1)}^2}\right) = \frac{-2(2x+1)(x-y)}{(x+y+1)^3}.$$
This is $0$ only when $y=x$, but we see that in this situation the original functions attains its lowest value $(0)$, and hence the maximum occurs when $y=0$ or when $y=1$.
So the maximum is either $x/(x+1)$ or $(1-x)/(x+2)$.
Now, we have
$$\begin{align}
\frac x{x+1} > \frac {1-x}{x+2}
&\iff x(x+2) > (1+x)(1-x)
\\&\iff x^2 + 2x > 1-x^2
\\&\iff 2x^2+2x - 1 > 0 \iff x<\frac{-1-\sqrt3}2 \,\,\text{ or }\,\, x>\frac{-1+\sqrt3}2
\end{align}$$
It follows that
$$f(x)=\begin{cases}
\frac{1-x}{x+2}&;&x\in\left[0, \frac{-1+\sqrt3}2\right]\\
\frac{x}{x+1}&;&x\in\left(\frac{-1+\sqrt3}2, 1\right]\\
\end{cases}$$
Do you think you can take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$ $$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$$
I tried to used $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ but it did not worked out so I tried to use the squeeze theorem.
$$0=\sqrt[3]{x^3}-\sqrt{x^2}\leq \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}\leq\sqrt[3]{8x^3}-\sqrt{4x^2}=2-2=0$$
But on the right hand I have inrcrased $\sqrt{x^2-2x}$ rather then deceased
Another attempt:
$$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}=\lim_{x\to \infty} \sqrt[3]{x^3(1+\frac{2}{x^2})}-\sqrt{x^2(1-\frac{2}{x})}=\\=\lim_{x\to \infty} x\sqrt[3]{(1+\frac{2}{x^2})}-x\sqrt{(1-\frac{2}{x})}=\lim_{x\to \infty} x[\sqrt[3]{(1+\frac{2}{x^2})}-\sqrt{(1-\frac{2}{x})}]$$
| Let $\dfrac1x=h$ where $h>0$
$$\sqrt[3]{x^3+2x}= \dfrac{\sqrt[3]{1+2h^2}}h$$ and $$\sqrt{x^2-2x}=\dfrac{\sqrt{1-2h}}h$$
Now for $F=\lim_{y\to0}\dfrac{\sqrt[n]{1+my}-1}y,$ set $\sqrt[n]{1+my}-1=z,my=(1+z)^n-1$
$F=\lim_{z\to0}\dfrac{mz}{nz+O(z^2)}=\dfrac mn$
$$\implies\lim_{y\to0}\sqrt[n]{1+my}=\dfrac{my}n+1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Is the function $f$ continuous at $(0,0)$
Is the function $f$ continuous at $(0,0)$?
$f(x, y)$ := $\frac{xy}{|x|+|y|}$, if $(x, y)$ $\not= (0, 0)$
and $f(0, 0) := 0$
My attempt:
$f(x, y)$ = $\frac{xy}{x+y}$ if $(x,y) > (0,0)$ and
$f(x, y)$ = $\frac{xy}{-(x+y)}$ if $(x,y) < (0,0)$
So using polar coordinates:
$f(x, y)$ = $\frac{xy}{x+y}$ = $\frac{rcos\theta sin\theta}{cos\theta + sin\theta}$, thus $|f(x,y)| <= r = {(x^2 + y^2)}^{0.5} $ and thus $f$ is continuous.
$f(x, y)$ = $\frac{xy}{-(x+y)}$ = $\frac{-rcos\theta sin\theta}{cos\theta + sin\theta}$, thus $|f(x,y)| <= - r = {-(x^2 + y^2)}^{0.5} $ and thus $f$ is not continuous.
Is my answer correct?
| If $x=r\cos\theta$ and $y=r\sin\theta$, then\begin{align}\left\lvert\frac{xy}{\lvert x\rvert+\lvert y\rvert}\right\rvert&=\frac{r^2\left\lvert\cos(\theta)\sin(\theta)\right\rvert}{r\bigl(\lvert\cos\theta\rvert+\lvert\sin\theta\rvert\bigr)}\\&\leqslant\frac r2,\end{align}since $\left\lvert\cos(\theta)\sin(\theta)\right\rvert\leqslant\frac12$ and $\cos\theta\rvert+\lvert\sin\theta\rvert\geqslant1$. So, $f$ is continuous at $(0,0)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\sum\limits_{n=0}^\infty\binom{2n}n\frac n{4^n(n+1)^2}=\ln(16)-2$
Prove that
$$\sum\limits_{n=0}^\infty\binom{2n}n\frac n{4^n(n+1)^2}=\ln(16)-2$$
According to Wolfram the above holds. Could someone show me the steps for this?
| The $n$-th Catalan number is defined as
$$ C_n := \frac 1 {n+1}\binom{2n}n.$$
By inspecting the family of definite integrals
$$ J_n(\alpha) := \frac 1 \pi \int_0^\alpha \xi^{2n} \sqrt{\alpha^2-\xi^2} d\xi, \qquad \alpha > 0,\ n\geq 0, $$
it can be shown, via standard integration techniques, that $C_n = J_n(2)$:
equivalently, through the substitution $x=\xi^2$, $$ C_n = \frac{1}{2\pi} \int_0^4 x^n \sqrt{\frac{4-x}{x}} dx.$$
Then your sum $S$ becomes
$$\begin{split}
S &=\sum_{n=0}^\infty \frac{ n C_n}{(n+1) 4^n } = \sum_{n=0}^\infty \frac{(n+1-1) C_n}{(n+1)4^n} = \sum_{n=0}^\infty \frac{C_n}{4^n} - \sum_{n=0}^\infty \frac{C_n}{4^n(n+1)} \\
&= \frac{1}{2\pi} \int_0^4 \sqrt{\frac{4-x}{x}} \left[\sum_{n=0}^\infty \left( \frac{x}{4}\right)^n -\sum_{n=0}^\infty \frac{(x/4)^n}{n+1} \right] dx \\
&= \frac{1}{2\pi} \int_0^4 \sqrt{\frac{4-x}{x}} \left[ \frac{1}{1-x/4} + \frac{\log(1-x/4)}{x/4} \right]dx \\
&= \frac{1}{2\pi} \int_0^4 \sqrt{\frac{4-x}{x}} \left[ \frac{4}{4-x}+\frac 4 x \log\left(\frac{4-x}{4} \right) \right] dx \\
&=\frac{2}{\pi} \int_0^4 \frac{dx}{\sqrt{4-(x-2)^2}} + \frac 2 \pi \int_0^4 \frac 1 x \log\left(\frac{4-x}{4} \right) dx \\
&= \frac 2\pi \int_0^\pi dt + \frac{2}{\pi} \int_0^1 \frac{\log s}{1-s} \sqrt{\frac{s}{1-s}}ds \\
&=: 2+\frac 2\pi I,
\end{split}$$
where we have interchanged sum and integral by total convergence of power series, and employed the formula for geometric series and its integral, and parallel substitutions $x = 2-2\cos t$, $s = 1 - x/4$. The definite integral $I$ can be resolved as follows:
$$\begin{split}
I &= \int_0^1 \frac{\log s}{1-s} \sqrt{\frac{s}{1-s}}ds=\int_0^\infty \log \left( \frac y {y+1}\right) \frac{\sqrt y}{y+1} dy = 2\int_0^\infty \log \left( \frac {x^2} {x^2+1}\right) \frac{x^2}{x^2+1} dx \\
&= \underbrace{\left[(x-\arctan x)\log \left( \frac {x^2} {x^2+1}\right) \right]_{-\infty}^{+\infty}}_{=0} -2 \int_{-\infty}^\infty \frac{x-\arctan x}{x(1+x^2)}dx \\
&= -2\pi +2 \int_{-\pi/2}^{\pi/2} z \cot(z) dz = -2\pi + 4 \left(\underbrace{[z \log(\sin z)]_0^{\pi/2}}_{=0} - \int_0^{\pi/2} \log(\sin z) dz \right),
\end{split}$$
where we have employed substitutions $y = s/(1-s)$, $x = \sqrt y$, and then $z = \arctan x$. The remaining integral has been evaluated many times over on MSE (see this answer): in total, we have $$I = 2\pi(\log 2 - 1), \quad \implies \quad S = 2 + 4 (\log 2 -1) = \boxed{\log(16)-2} $$
as required.
| {
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integral ordered pair $(x,y)$ in $x^2+y^2=2x+2y+xy$
Find all possible integral ordered pair $(x,y)$ in $x^2+y^2=2x+2y+xy$
what i try $y^2-(x+2)y+x^2-2x=0$
$$y=\frac{x+2\pm\sqrt{(x+2)^2-4(x^2-2x)}}{2}$$
$$y=\frac{x+2\pm \sqrt{-3x^2+12x+4}}{2}$$
How do i solve it Help me please
| The equation is equivalent to $(x-y)^{2}+(x-2)^{2}+(y-2)^{2}=8$. However due to symmetry we can assume $x-y\geq 0$ but then there are only three possible choices $x-y=0,1,2$. If $x-y=0$ $(x,y)=(0,0),(4,4)$ if $x-y=1$ then $7=(y-1)^{2}+(y-2)^{2}=0+7=1+6=2+5=3+4$ so no solution.
If $x-y=2$ then $y^{2}+(y-2)^{2}=4=0+4=3+1=2+2$ thus only possible $y=0$ or $y=2$ thus $(x,y)=(2,0),(4,2)$.
Thus all possible solutions are $(x,y)=(2,0),(0,2),(4,2),(2,4),(0,0),(4,4)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Definite integral with irrational function I wonder if it's possible solve this definite integral in terms of elemental or special functions:
$$\int_0^1\sqrt{\sqrt{x^4+a}-x^2}\,dx$$
with $a>0$. I've tried with Wolfram Mathematica but doesn't give me anything.
| Let's rewrite the integral as
$$\int_0^1 \sqrt{\sqrt{x^4+b^2}-x^2}\:dx = \int_0^1 \frac{1}{2}\sqrt{\sqrt{1+\frac{b^2}{y^2}}-1}\:dy$$
for $a = b^2$ and letting $x^2 = y$. Now use the substitution $y = b\operatorname{csch}t$ to get
$$\int_{\sinh^{-1}(b)}^\infty \frac{b\cosh t}{2\sinh^2 t}\sqrt{\cosh t - 1}\:dt$$ $$ =
\frac{b}{4\sqrt{2}}\int_{\sinh^{-1}(b)}^\infty \left[\frac{1}{\cosh\left(\frac{t}{2}\right)-1} - \frac{1}{\cosh\left(\frac{t}{2}\right)+1} + \frac{2}{\cosh^2\left(\frac{t}{2}\right)}\right]\frac{1}{2}\sinh\left(\frac{t}{2}\right)\:dt$$
$$ = \frac{b}{4\sqrt{2}}\left[\log\left(\frac{\cosh\left(\frac{t}{2}\right)-1}{\cosh\left(\frac{t}{2}\right)+1}\right) - \frac{2}{\cosh\left(\frac{t}{2}\right)} \right]_{\sinh^{-1}(b)}^\infty$$
where we went from the first line to the second with the hyperbolic double angle identities: $$\cosh t = \cosh^2\left(\frac{t}{2}\right) + \sinh^2\left(\frac{t}{2}\right) = 1+2\sinh^2\left(\frac{t}{2}\right)$$
$$\sinh t = 2\sinh\left(\frac{t}{2}\right)\cosh\left(\frac{t}{2}\right)$$
At infinity both terms vanish. To evaluate the lower bound, we'll have to use the double angle formulas in reverse.
$$\frac{b}{4\sqrt{2}}\left[\frac{2}{\cosh\left(\frac{t}{2}\right)} - \log\left(\frac{\cosh\left(\frac{t}{2}\right)-1}{\cosh\left(\frac{t}{2}\right)+1}\right) \right]$$ $$ = \frac{b}{4\sqrt{2}}\left[\frac{2}{\sqrt{\cosh^2\left(\frac{t}{2}\right)}} - \log\left(\frac{\sqrt{\cosh^2\left(\frac{t}{2}\right)}-1}{\sqrt{\cosh^2\left(\frac{t}{2}\right)}+1}\right) \right]$$
$$= \frac{b}{4\sqrt{2}}\left[\frac{2\sqrt{2}}{\sqrt{1+\cosh t}} - \log\left(\frac{\sqrt{1+\cosh t}-\sqrt{2}}{\sqrt{1+\cosh t}+\sqrt{2}}\right) \right]$$ $$ = \frac{b}{4\sqrt{2}}\left[\frac{2\sqrt{2}}{\sqrt{1+\sqrt{1+\sinh^2 t}}} - \log\left(\frac{\sqrt{1+\sqrt{1+\sinh^2 t}}-\sqrt{2}}{\sqrt{1+\sqrt{1+\sinh^2 t}}+\sqrt{2}}\right) \right]$$
Then plugging in our point we get the answer
$$\boxed{\frac{\sqrt{a}}{2\sqrt{1+\sqrt{1+a}}}+\frac{\sqrt{a}}{4\sqrt{2}}\log\left(\frac{\sqrt{1+\sqrt{1+a}}+\sqrt{2}}{\sqrt{1+\sqrt{1+a}}-\sqrt{2}}\right)}$$ $$ \longrightarrow \frac{1}{2\sqrt{1+\sqrt{2}}} + \frac{1}{2\sqrt{2}}\log\left(\sqrt{1+\sqrt{2}}+\sqrt{2}\right)\approx 0.8622$$
which agrees with the value of the integral when $a=1$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$$
$$h(x)=\dfrac{x^2+x^2-2+2}{x-\dfrac{1}{x}}$$
$$h(x)=x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}$$
Case $1$: $x-\dfrac{1}{x}>0$
$$\dfrac{x^2-1}{x}>0$$
$$x\in(-1,0) \cup (1,\infty)$$
$$AM\ge GM$$
$$\dfrac{x-\dfrac{1}{x}+\dfrac{2}{x-\dfrac{1}{x}}}{2}>\sqrt{2}$$
$$h(x)\ge 2\sqrt{2}$$
$$x-\dfrac{1}{x}=\dfrac{2}{x-\dfrac{1}{x}}$$
$$x^2+\dfrac{1}{x^2}-2=2$$
$$x^2+\dfrac{1}{x^2}=4$$
$$x^4-4x^2+1=0$$
$$x^2=2\pm\sqrt{3}$$
Only $x=\sqrt{2+\sqrt{3}},-\sqrt{2-\sqrt{3}}$ are the valid solutions.
Case $2$: $x-\dfrac{1}{x}<0$
$$x\in(-\infty,-1) \cup (0,1)$$
$$h(x)=-\left(\dfrac{1}{x}-x+\dfrac{2}{\dfrac{1}{x}-x}\right)$$
By $AM\ge GM$, $h(x)\ge-2\sqrt{2}$
We will get this minimum value at $-\sqrt{2+\sqrt{3}},\sqrt{2-\sqrt{3}}$
So answer should have been $-2\sqrt{2}$ but actual answer is $2\sqrt{2}$. What am I missing here.
|
Your calculation is fine, except the interpretation. Note that,
$$h(-\sqrt{2-\sqrt3}) = h(\sqrt{2+\sqrt3}) = 2\sqrt2$$
As seen from the plot, $2\sqrt2 $ at $-\sqrt{2-\sqrt3}$ and $\sqrt{2+\sqrt3}$ are the two local minima, while $-2\sqrt2 $ are the local maxima.
| {
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recurrence initial conditions I'm working on a homework assignment involving recursion and I'm having trouble finding an easy way to determine the initial conditions. Heres the problem:
We want to tile ann×1 strip with tiles of three types: 1×1 tiles that are dark-blue, light-blue,and red; 2×1 green tiles, and 3×1 sky-blue tiles. Now give a formula with initial conditions for the number of tilings, considering that blue tiles cannot be next to each other.
I can understand that the recurrence equation is:
$B_n = B_{n-1}+3B_{n-2}+2B_{n-3}+B_{n-4}+B_{n-5}$
And I've found initial conditions for
$B_0=1$
$B_1 = 3$
$B_2 = 6$
$B_3 = 17$
However, I found these but actually writing down all the possible combinations of tiles, but $B_4$ is a huge possible list. Is there some method of combinatorics or permutations I can use to find the initial conditions for $B_4$?
| Let $a_n$ and $b_n$ the number of $n$-tilings that start with red or green, and blue, respectively. Also, let $c_n$ (your $B_n$) be the total number of $n$-tilings. Then $a_0=a_1=b_0=c_0=1$, $b_1=b_2=2$, $c_1=3$, and, by conditioning on the next tile, we see that
\begin{align}
a_n &= c_{n-1} + c_{n-2} &&\text{for $n \ge 2$}\\
b_n &= 2 a_{n-1} + a_{n-3} &&\text{for $n \ge 3$}\\
c_n &= a_n+b_n-[n=0] &&\text{for $n \ge 0$}
\end{align}
Hence
\begin{align}
c_n &= c_{n-1} + c_{n-2}+2 a_{n-1} + a_{n-3}\\
&=c_{n-1} + c_{n-2}+2 (c_{n-2} + c_{n-3}) + (c_{n-4} + c_{n-5})\\
&=c_{n-1} + 3c_{n-2}+2 c_{n-3} + c_{n-4} + c_{n-5},
\end{align}
as you had claimed.
We can also obtain generating functions as follows.
Let $A(z)=\sum_{n=0}^\infty a_n z^n$, $B(z)=\sum_{n=0}^\infty b_n z^n$, and $C(z)=\sum_{n=0}^\infty c_n z^n$. Then the recurrence relations imply
\begin{align}
A(z)-1 - z &=z (C(z)-1) + z^2 C(z) \\
B(z)-1 -2 z-2 z^2 &= 2z (A(z)-1-z) + z^3 A(z) \\
C(z) &= A(z)+B(z)-1
\end{align}
Solving for $A(z)$, $B(z)$, and $C(z)$ yields
\begin{align}
A(z) &= \frac{1}{1 - z - 3 z^2 - 2 z^3 - z^4 - z^5}\\
B(z) &= \frac{1 + z - 3 z^2 - z^3 - z^4 - z^5}{1 - z - 3 z^2 - 2 z^3 - z^4 - z^5}\\
C(z) &= \frac{1 + 2 z + z^3}{1 - z - 3 z^2 - 2 z^3 - z^4 - z^5}
\end{align}
Notice that the (common) denominator implies that each sequence satisfies the order-5 recurrence
$$f_n - f_{n-1} - 3 f_{n-2} - 2 f_{n-3} - f_{n-4} - f_{n-5} = 0,$$
as before.
Expanding the series for $C(z)$ yields
$$1 + 3 z + 6 z^2 + 18 z^3 + 43 z^4 + 113 z^5 + 287 z^6 + 736 z^7 + 1884 z^8 + 4822 z^9 + 12346 z^{10} + \dots .$$
In particular, $c_3 = 18$, which you could also have obtained directly from the recurrence relations as follows:
\begin{align}
a_2 &= c_1 + c_0 = 3 + 1 = 4\\
c_2 &= a_2 + b_2 = 4 + 2 = 6\\
a_3 &= c_2 + c_1 = 6 + 3 = 9\\
b_3 &= 2a_2 + a_0 = 2\cdot 4 + 1 = 9\\
c_3 &= a_3+b_3=9+9 = 18
\end{align}
| {
"language": "en",
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Show that $\int_{0}^{1}\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)} dx = 2\pi\frac{\sqrt{3}}{27}$ The given question is from Complex Variables written by Levinson and Redheffer, Problem 7 in Chapter 4. We want to show that $\int_{0}^{1}\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)}\,dx = 2\pi\frac{\sqrt{3}}{27}$. Below is my first attempt.
Let $C$ be the a contour that consists of:
a. A counter-clockwise oriented closed circle $C_1$ centered at the point $z = 1$ with radius $\epsilon$
b. A counter-clockwise oriented closed circle $C_0$ centered at the point $z = 0$ with radius $\epsilon$
c. A directed path $P_1$that starts from point $z = \epsilon$, ends at point $z =1 - \epsilon$ and joins $C_0$ and $C_1$ and lies in the upper half of $x$ axis.
d. A directed path $P_2$ that starts from point $z = 1 - \epsilon$, ends at point $z =\epsilon$ and joins $C_0$ and $C_1$ and lies in the lower half of $x$ axis.
Namely $C = C_1 \rightarrow P_2 \rightarrow C_0 \rightarrow P_1$. Also we know that the angle of $\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)}$ will change by $\frac{-2\,\pi\,i}{3}$ and hence on $P_2$ we have $\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)}\,e^{\large\frac{-2\,\pi\,i}{3}}$ while on $P_1$ we have just$\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)}$. I do not know if we need to find the residue at infinity but for the given function it does not exist. So I try $x^2 = (\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)} + x)\,[\sqrt[\leftroot{-3}\uproot{3}3]{x^4(1 - x)^2} - x\,\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)} + x^2]$ and try to find the contour integral of $\large\frac{x}{\sqrt[\leftroot{-3}\uproot{3}3]{x^4(1 - x)^2} + x\,\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)}] + x^2}$ around $C$. However, then I do not know how to work out the result when the denominator is $x^2$.
| Let $f(z)$ be the branch of $\sqrt[\leftroot{-3}\uproot{3}3]{z^2(1 - z)}$ that is real positive at $z$ on $P_1$. Then the value of $f(x)$ for $x>1$ is $\sqrt[\leftroot{-3}\uproot{3}3]{x^2(x-1)}\cdot e^{-\frac{\pi i}{3}}$ and $f(x)=\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1-x)}\cdot e^{-\frac{2\pi i}{3}}$ on $P_2$. Therefore, letting $ \epsilon\to 0$, we have$$
\int_0^1 \sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)} dx+e^{-\frac{2\pi i}{3}}\int_1^0
\sqrt[\leftroot{-3}\uproot{3}3]{x^2(1-x)}dx=2\pi i\operatorname{Res}(\infty).\tag{1}
$$
Evaluation of $\operatorname{Res}(\infty)$:
Since $\sqrt[\leftroot{-3}\uproot{3}3]{1-t}$ has its expansion
$$
\sqrt[\leftroot{-3}\uproot{3}3]{1-t}=1-\frac{t}{3}-\frac{t^2}{9}-\frac{5t^3}{81}-\cdots \quad(|t|<1),
$$
the expansion of $f(x)$ for $1<x<\infty$
is
\begin{align*}
f(x)&=\sqrt[\leftroot{-3}\uproot{3}3]{x^2(x-1)}\cdot e^{-\frac{\pi i}{3}}\\
&=e^{-\frac{\pi i}{3}}\cdot x\left(1-\frac{1}{3x}-\frac{1}{9x^2}-\frac{5}{81x^3}-\cdots \right).
\end{align*}
Thus we have $$\operatorname{Res}(\infty)=\frac{1}{9}e^{-\frac{\pi i}{3}}.$$
Recall that if $f(z)=\sum_{n=-\infty}^\infty \frac{c_n}{z^n} \,(r<|z|<\infty)$, then $\operatorname{Res}(\infty)=-c_1.$
From $(1)$ we have \begin{align*}
&\left(1-e^{-\frac{2\pi i}{3}}\right)\int_0^1 \sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)} dx
=\frac{2\pi i}{9}e^{-\frac{\pi i}{3}},\\
&\int_0^1 \sqrt[\leftroot{-3}\uproot{3}3]{x^2(1 - x)} dx=2\pi\frac{\sqrt{3}}{27}.
\end{align*}
| {
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Describe the set of points $z$ in the complex plane that satisfies the following equation. $|2z−i|=4$
I have tried solving it multiple times but I cannot. Here are my steps. I hope somebody can catch my mistakes.
$|2z−i|=4$
$|2(a+bi)−i|=4$
$|(2a+2bi)−i|=4$
$|(2a+2bi)^2+(0\cdot-i)^2|=4$ (because every number is a complex number)
$|2a^2-2b^2+4\cdot a\cdot bi+1|=4$
$\sqrt{2a^2-2b^2+4\cdot a\cdot bi+1}=4$
I can’t solve this square root. Where did i go wrong . According to the book the answer is
circle, center = $i/2$, radius= $2$
I have solved other problems of it sorts but this one i couldnt.
Thanks for reading and hopefully for answering.
| At the end of step 3) you have
$|2a + 2bi - i|=4$. Try to combine the $i$ terms:
$|2a + (2b-1)i| = 4$. What you have here is the complex number $w = 2z -i = 2a + (2b-1)i$ so that $Re(w) = 2a (= 2Re(z))$ and $Im(w) = 2b -1(= 2Re(z) - 1)$.
Now you can do the formula that $|a+bi| = \sqrt{a^2 + b^2}$ or in this case
$|2a + (2b-1)i| = \sqrt{(2a)^2 + (2b-1)^2} = 4$.
So $4a^2 + 4b^2 - 4b +1 = 16$ which I suppose you could solve
$(2b-1)^2 = 16-4a^2 = 4(4-a^2)$
$2b - 1 = \pm 2\sqrt{2-a^2}$
$b = \pm \sqrt{2-a^2}+\frac 12$ so $z$ is any complex number $j + ki$ where $k = \pm \sqrt{2-j^2} + \frac 12$.
But what does that mean?
Well, bear in mind a complex number can be viewed as a point in a plane.
So if $w = 2z-i$ then $w = x+yi$ is a point in the plane. We are given that $|w| = 4$ so that means that $w$, if viewed as the point $(x,y)$ is a distance of $4$ from the origin or that $(x,y)$ in in a circle of radius $4$ around the origin (and $x^2 + y^2 = 4^2$.
But if all possible $w$ form a circle of radius $4$ around the origin. Then $2z = w+i$ are the points that form of the circle around the point $(0,1) = 0 + 1i = i$ with radius $4$.
So $2z$ are all the points $x+ yi$ where $x^2 + (y-1)^2 = 4^2$
And so $z$ is all the points where the $w+ u i$ where $w,u$ are half the values of those points in the circle. So $z = w+ui$ where $(2w)^2 + (2u-1)^2 = 4^2$ or
$w^2 + (u-\frac 12)^2 = 4 = 2^2$ or
in other words $z$ is all the point in a circle of radius $2$ around the point $\frac 12i = 0 + \frac 12 i= (0, \frac 12)$.
It is all $z = w+ui$ where $w^2 +(u-\frac 12)i = 2^2$
| {
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Does the following series (inspired by harmonic things) converge? $$\left[e - \left(1+ \left(\frac{1}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4}} \right) \right) ^ {\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} \right)} \right] + $$
$$\left[e - \left(1+ \left(\frac{1}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5}} \right) \right) ^ {\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \right)} \right] +$$
$$\left[e - \left(1+ \left(\frac{1}{\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} +\frac{1}{6}} \right) \right) ^ {\left(\frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6}\right)} \right] + ...$$
| This series diverges, but it's somewhat instructive in why - this is an instance where we can just "peel back" all the layers of the expression by applying linear approximations and eventually reduce it to something trivial.
In particular, I assume that you know
$$\lim_{x\rightarrow 0}(1+x)^{1/x} = e$$
and thought to look at the rate of convergence of this, essentially, by looking at what happens if we approach $x=0$ fairly slowly and sum up the error terms. The trouble is that the behavior of the function $f(x)=(1+x)^{1/x}$ actually isn't all that interesting: it's some smooth function with non-zero derivative at $0$. Although it's not that important, the derivative turns out to be $-e/2$ at $0$ which means that if $\alpha < -e/2 < \beta$ then, for all small enough $x$ we have that $f(x)-e$ is between $\alpha x$ and $\beta x$ - so is essentially linear.
In particular, this suffices to say that if $x_n$ is a sequence with $0$ as first limit, then $\sum f(x_n)$ converges if and only if $\sum x_n$ does. Okay, so now we've unravelled the outermost detail just by noticing that "differentiable" is basically as good as "linear" here, which is not interesting for the matter of convergence.
So, let's define
$$x_n=\frac{1}{\frac{1}2 + \frac{1}3 + \ldots + \frac{1}n}$$
This, again, is easy enough to handle: that denominator is known to be asymptotic to $\log(n)$ - meaning, again, that it is always within ratio of $\log(n)$ for all large enough $n$. This can be derived via comparison with integrals of $\frac{1}x$. So, again, $\sum x_n$ is going to converge if and only if $\sum \frac{1}{\log(n)}$ converges.
At this point we can stop: $\frac{1}{\log(n)}$ is bigger than $\frac{1}n$, and we know that the harmonic series diverges (indeed we just used an asymptotic for it!), so clearly $\sum \frac{1}{\log(n)}$ diverges. Since the partial sums of this turn out to always be within a constant factor of those of the original, the original sum diverges.
| {
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"source": "stackexchange",
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If $f(x)$ is a polynomial of degree three leaves remainder $1$ when divided by $(x−1)^2$ and leaves remainder $–1$ when divided by $(x+1)^2$
$f(x)$ is a polynomial of degree three which leaves remainder $1$ when divided by $(x−1)^2$ and leaves remainder $–1$ when divided by $(x+1)^2$.
If $f(x)=0$ has roots $\alpha,\beta,\gamma$, then $$(\alpha\beta+\beta\gamma+\gamma\alpha)=\text?$$
The answer to the question is $-3$. How do I approach this problem?
| Hint
Let $f(x)=1+(x-1)^2(ax+b)=-1+(x+1)^2(cx+d)$
$ax^3+x^2(b-2a)+x(a-2b)+b-1=cx^3+x^2(d+2c)+x(c+2d)+d+1$
Now compare the coefficients of the different exponent of $x$
$\implies c=a$
$b-1=d+1\iff d=b-2$
$b-2a=d+2c\iff b=d+2(c+a)=d+4a$
$a-2b=c+2d\implies b=-d=2-b$
Hope you can take it from here?
| {
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Help with induction using divisibility. Need some help with this question for induction using divisibility: $8*19^n-2*3^{2n+2}$ is divisible by 10 for any positive integer.
What I have so far is proving the first value and $P_k$ and $P_{k+1}$
$P_1$ = $8*19^1-2*3^{2+2}$=-10; $P_1$ is true.
$P_k$ = $8*19^k-2*3^{2k+2}$= 10A
$P_{k+1}$ = $8*19^{k+1}-2*3^{2k+4}$ = 10B
I've just found myself stuck on the proof starting with:
$8*19^{k+1}-2*3^{2k+4}$=
Thanks for any help. If I could get an answer in the format of $P_k$ and $P_{k+1}$ that would greatly appreciated.
| Let $P(n)$ be the statement that $10 \mid 8 \cdot 19^n - 2 \cdot 3^{2n + 2}$.
If we wish to establish that the statement holds for all nonnegative integers, we have to verify that $P(0)$ holds since that will be our base case.
Since $8 \cdot 19^0 - 2 \cdot 3^2 = 8 - 18 = -10 = 10 \cdot (-1)$, $P(0)$ holds.
If we wish to establish that the statement holds for all positive integers, we have to verify that $P(1)$ holds since that will be our base case.
Since $8 \cdot 19^1 - 2 \cdot 3^4 = 8 \cdot 19 - 2 \cdot 81 = 152 - 162 = -10 = 10 \cdot (-1)$, $P(1)$ holds.
Assume $P(k)$ holds for some positive integer $k$. Then $10 \mid 8 \cdot 19^k - 2 \cdot 3^{2k + 2}$, so there exists $m \in \mathbb{Z}$ such that $8 \cdot 19^k - 2 \cdot 3^{2k + 2} = 10m$.
Let $n = k + 1$. Then
\begin{align*}
8 \cdot 19^n - 2 \cdot 3^{2n + 2} & = 8 \cdot 19^{k + 1} - 2 \cdot 3^{2(k + 1) + 2}\\
& = 19 \cdot 8 \cdot 19^k - 2 \cdot 3^{2k + 4}\\
& = 19 \cdot 8 \cdot 19^k - 9 \cdot 2 \cdot 3^{2k + 2}\\
& = (10 + 9) \cdot 8 \cdot 19^k - 9 \cdot 2 \cdot 3^{2k + 2}\\
& = 10 \cdot 8 \cdot 19^k + 9(8 \cdot 19^k - 2 \cdot 3^{2k + 2})\\
& = 10 \cdot 8 \cdot 19^k + 9 \cdot 10m && \text{by the induction hypothesis}\\
& = 10(8 \cdot 19^k + 9m)
\end{align*}
Thus, if $10 \mid 8 \cdot 19^k - 2 \cdot 3^{2k + 2}$, $10 \mid 8 \cdot 19^{k + 1} - 2 \cdot 3^{2(k + 1) + 2}$. Hence, $P(k) \implies P(k + 1)$.
Since $P(1)$ holds and $P(k) \implies P(k + 1)$ for each $k \geq 1$, $P(n)$ holds for each positive integer $n$. The same steps that establish $P(k) \implies P(k + 1)$ also apply when $k$ is a nonnegative integer.
In fact, since $P(0)$ holds, $P(1)$ holds, and $P(k) \implies P(k + 1)$ for each $k \geq 1$, $P(n)$ holds for each nonnegative integer $n$.
Had we wished to prove that $P(n)$ holds for each nonnegative integer $n$, we could have skipped the step of establishing that $P(1)$ holds and instead assumed that $P(k)$ holds for some nonnegative integer $k$.
| {
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On conjectured continued fractions and $e$ Playing around with numbers, I conjectured three incredibly interesting things:
$$9+\cfrac{1}{18+0\times 12\cfrac{1}{18+1\times 12+\cfrac{1}{18+2\times 12+\cfrac{1}{18+3\times 12+\ddots}}}}=\frac{4e^{1/3}-2}{e^{1/3}-1}$$
$$6+\cfrac{1}{9+0\times 6+\cfrac{1}{9+1\times 6+\cfrac{1}{9+2\times 6+\cfrac{1}{9+3\times 6+\ddots}}}}=\frac{4e^{2/3}-2}{e^{2/3}-1}$$
$$5+\cfrac{1}{6+0\times 4+\cfrac{1}{6+1\times 4+\cfrac{1}{6+2\times 4+\cfrac{1}{6+3\times 4+\ddots}}}}=\frac{4e-2}{e-1}$$
So, what in God's name is going on behind the scenes? Why does this seem to be true, why does it involve $e$, so many questions! All I did was play around on a calculator with some continued fractions, went to Wolfram Alpha and asked for the result to be written in terms of $e$ and then I noticed some patterns. But what is really going on? Well, apart from a bit of luck, I have no idea.
Any ideas? Thanks.
Edit:
This question may be of help, as it reveals general continued fractions regarding the hyperbolic tangent, which to those who don't know, is a function with respect to some value $\alpha$ defined as $\tanh(\alpha):=\frac{e^{2\alpha} -1}{e^{2\alpha}+1}.$
| First of all, the terms on the right are of the form
$$
\frac{4e^z-2}{e^z-1} = 4 + \frac2{e^z-1}
$$
Let's examine the last of your cases which is for $z=1$. It is known that
$$
e = 1 + \frac2{[1;6\;10\;14\;18\cdots]}
$$
hence
$$
4+\frac2{e-1} = 4+[1;6\;10\;14\;18\cdots] = [5;6\;10\;14\;18\cdots]
$$
which is your 3rd fraction. The Wikipedia page linked above has also fractions for $e^{x/y}$, you can use it to see your other equations just as easy:
$$e^{x/y} =
1+
\cfrac{2x}{2y-x+}\;
\cfrac{x^2}{6y+}\;
\cfrac{x^2}{10y+}\;
\cfrac{x^2}{14y+}\;
\cdots$$
thus
$$
\frac{2x}{e^{x/y}-1}
= 2y-x+\;
\cfrac{x^2}{6y+}\;
\cfrac{x^2}{10y+}\;
\cfrac{x^2}{14y+}\;
\cdots
$$
and with $x=1$ and $y=3$ we have:
$$\begin{align}
4+\frac2{e^{1/3}-1}
&= 4+2\cdot3-1+\;
\cfrac{1}{6\cdot3\,+}\;
\cfrac{1}{10\cdot3\,+}\;
\cfrac{1}{14\cdot3\,+}\;
\cdots\\
&= [9;18\;30\;42\;\cdots]\\
\end{align}$$
which is your 1st fraction. And finally, for the 2nd case, we rewrite the above to
$$\begin{align}
\frac{2x}{e^{x/y}-1}
&= 2y-x+\;
\cfrac{x}{\frac6xy+}\;
\cfrac{1}{\frac{10}x y+}\;
\cfrac{1}{\frac{14}x y+}\;
\cdots\\
&\stackrel{x=2}=
2y-2+
\cfrac{2}{3y+}\;
\cfrac{1}{5y+}\;
\cfrac{1}{7y+}\;
\cfrac{1}{9y+}\;
\cdots\\
\end{align}$$
dividing by $x=2$:
$$\begin{align}
\frac{2}{e^{2/y}-1}
&= y-1+\;
\cfrac{1}{3y+}\;
\cfrac{1}{5y+}\;
\cfrac{1}{7y+}\;
\cdots\\
&\stackrel{y=3}=
2+
\cfrac{1}{9+}\;
\cfrac{1}{15+}\;
\cfrac{1}{21+}\;
\cfrac{1}{27+}\;
\cdots\\
\end{align}$$
Adding 4 we arrive at the 2nd equation.
p.s.: If we didn't evaluate for $x$ and just continued, we'd get
$$\begin{align}
\frac{2}{e^{x/y}-1}
&= 2\frac yx - 1+\;
\cfrac{1}{\frac6xy+}\;
\cfrac{1}{\frac{10}x y+}\;
\cfrac{1}{\frac{14}x y+}\;
\cdots\\
\end{align}$$
which is after dividing the equation by $x$ and moving the $x$'s to the denominators in all fractions. This can be rewritten as, now with $z=2y/x$:
$$\begin{align}
4+\frac{2}{e^{2/z}-1}
&= 4 + z - 1+\;
\cfrac{1}{3 z+}\;
\cfrac{1}{5 z+}\;
\cfrac{1}{7 z+}\;
\cdots\\
&= [z+3;3z\;5z\;7z\;\cdots]
\end{align}$$
We then get your equations for $z=6$, $z=3$ and $z=2$, respectively.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3558182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Revisiting Ahmed Integral in $(0,\infty)$ A recent post in MSE:
Evaluate $\int_0^{\infty } \frac{\tan ^{-1}\left(a^2+x^2\right)}{\left(x^2+1\right)\sqrt{a^2+x^2}} \, dx$
re-emphasizes that when Ahmed Integral is converted to a two-dimensional integral in $x,y$, then the sameness of domain the $(0,1)$ of $x$ and $y$ makes it do-able. Further, it asks for the evaluation of a slightly different Integral, which is challenging.
In this light, now the question here is: How to find $$\int_{0}^{\infty} \frac{\tan^{-1}\sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}} dx?$$
Mind the upper limit that is $\infty$, there is square root sign in the argument of $\tan^{-1}$, and yes the integrand is the same as that of Ahmed integral.
| \begin{align}
&\int_{0}^{\infty} \frac{\tan^{-1}\sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}}~ dx\\
=& \int_{0}^{\infty}\int_0^1 \frac{1}{(1+x^2)(1+y^2(2+x^2))}dy~dx\\
= & \int_{0}^{1}\int_0^\infty \frac{1}{1+y^2}
\bigg( \frac{1}{1+x^2}-\frac{y^2}{1+y^2(2+x^2) }\bigg) dx~dy\\
= & \ \frac\pi2\int_{0}^1 \frac{1}{1+y^2}
\bigg( 1-\frac{y}{\sqrt{1+2y^2} }\bigg)dy\\
=& \ \frac\pi2 \left(\tan^{-1}y-\tan^{-1}\sqrt{1+2y^2}\right)\bigg|_0^1
=\frac\pi2\cdot \frac\pi{6}=\frac{\pi^2}{12}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
Find the open intervals where $x(x-6)^3$ is concave upwards and concave downwards. Find the open intervals where $f(x)=x(x-6)^3$ is concave upwards and concave downwards.
Solution: We want to find the concavity of $f(x)=x(x-6)^3$, so we have to take the second derivative, find the critical points (where it equals zero and DNE), then split up the number line by these points and test for the sign of $f''(x)$ in each of them. First lets get to the second derivative by using the product rule and the chain rule where appropriate:
$f'(x)=(x)'(x-6)^3+x((x-6)^3)'$
$f'(x)=(1)(x-6)^3+x(3(x-6)^2)(x-6)'$
$f'(x)=(x-6)^3+x(3(x-6)^2)(1)$
$f'(x)=(x-6)^3+3x(x-6)^2$
Okay, now let's do it again
$f''(x)=((x-6)^3)'+(3x)'(x-6)^2+(3x)((x-6)^2)'$
$f''(x)=3(x-6)^2(x-6)'+3(x-6)^2+(3x)(2(x-6))(x-6)'$
$f''(x)=3(x-6)^2(1)+3(x-6)^2+6x(x-6))(1)$
$f''(x)=3(x-6)^2+3(x-6)^2+6x(x-6))$
$f''(x)=6(x-6)^2+6x(x-6))$
Foiling out and collecting like terms yields:
$f''(x)=12x^2-108x+216$
$f''(x)=12(x^2-9x+18)$
$f''(x)=12(x-6)(x-3)$
We want to know when this equals zero...
$f''(x)=12(x-6)(x-3)=0$
$\rightarrow x=6,x=3$
So let's split up our number line by $x=3,6$ and test each region:
$f''(0)=(12)(-6)(-3)=12*18>0$
$f''(5)=12(-1)(2)<0$
$f''(7)=12(1)(4)>0$
So our function is concave upward on $(-\infty,0)$ and $(6,\infty)$, and concave downwards on $(3,6)$
| Set $z=x-6$;
This amounts to a translation of the coordinate system and does not change the characteristics of the function.
Then $Y=(z+6)z^3=z^4+6z^3;$
$Y'=4z^3+18z^2$;
$Y''=12z^2+36z=12z(z+3)$;
This is a parabola.
$Y'' \gt 0$: Convex;
$12z(z+3)\gt 0$;
1) $z>0$; 2) $z <-3$;
$Y'' <0$: Concave: $-3 <z <0$.
Reset to $x: x=z+6$;
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3563272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $n \ln \left(1+\frac{1}{n}\right) \geq \frac{2 n}{2 n+1}$
Hence or otherwise show that for all positive integers $n$
$$
n \ln \left(1+\frac{1}{n}\right) \geq \frac{2 n}{2 n+1}
$$
This is related to another part of this question where I proved $\ln \left(\frac{4-t}{t}\right) \geq 2-t$ for $0<t \leq 2$ by integrating $\frac{1}{s}+\frac{1}{4-s} \geq 1$(which is true for $0<s<4$) over $[t, 2] .$ However, I am not able to use this to prove that
$$n \ln \left(1+\frac{1}{n}\right) \geq \frac{2 n}{2 n+1}$$ for all positive integers. Any help would be appreciated.
| Integrating
$\dfrac1{1+t}
=\sum_{k=0}^{m-1}(-1)^k t^k+\dfrac{(-1)^mt^m}{1+t}
$,
we get
$\ln(1+x)
=\sum_{k=0}^{m-1}(-1)^k \dfrac{x^{k+1}}{k+1}+\int_0^x \dfrac{(-1)^mt^m}{1+t}dt
$
for any $m \ge 1$ and $x \ge 0$.
Therefore
$n\ln(1+1/n)
=\sum_{k=0}^{m-1}(-1)^k \dfrac1{(k+1)n^k}+n\int_0^{1/n} \dfrac{(-1)^mt^m}{1+t}dt
$.
Putting
$m = 2$, this is
$n\ln(1+1/n)
=1-\dfrac1{2n}+n\int_0^{1/n} \dfrac{t^2}{1+t}dt
=\dfrac{2n-1}{2n}+n\int_0^{1/n} \dfrac{t^2}{1+t}dt
$
so we want
$\dfrac{2n}{2n+1}
\le \dfrac{2n-1}{2n}+n\int_0^{1/n} \dfrac{t^2}{1+t}dt
$
or
$\dfrac{1}{2n^2(2n+1)}
\le \int_0^{1/n} \dfrac{t^2}{1+t}dt
$.
$\int_0^{1/n} \dfrac{t^2}{1+t}dt
\ge \int_0^{1/n} \dfrac{t^2}{1+1/n}dt
=\dfrac{(1/n)^3}{3(1+1/n)}
=\dfrac1{3n^2(n+1)}
$
so we want
$3n^2(n+1)
\le 2n^2(2n+1)
$
or
$3(n+1)
\le 2(2n+1)
$
or
$3n+3 \le 4n+2
$
or
$n \ge 1$
which is true.
Also,
$\int_0^{1/n} \dfrac{t^2}{1+t}dt
\lt \int_0^{1/n} t^2dt
=\dfrac1{3n^3}
$
so
$n\ln(1+1/n)
=1-\dfrac1{2n}+n\int_0^{1/n} \dfrac{t^2}{1+t}dt
\lt 1-\dfrac1{2n}+\dfrac1{3n^2}
\lt 1-\dfrac{3n-2}{6n^2}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Evaluate the indefinite integral $\int \frac{\sin^3\frac\theta 2}{\cos\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}} d\theta$ $$\int \frac{\sin^3\frac\theta 2}{\cos\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}} d\theta$$
My Attempt:
$$I = \int \frac{\sin^2\frac\theta2\cdot\sin\frac\theta2\cos\frac\theta2}{\cos^2\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}}
d\theta \\ = \frac12 \int \frac{(1-\cos\theta)\sin\theta}{(1+\cos\theta)\sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}}
d\theta$$
$$\text{let }\cos\theta = t \implies -\sin\theta d\theta = dt$$
$$I = \frac12 \int \frac{t-1}{(t+1) \sqrt{ t^3+t^2+t } } dt $$
I am not sure how to proceed further from here. Any hints/solutions on how to resolve the cubic expression under the square root?
| Use $s = \frac{t-1}{t+1} \implies t = \frac{1+s}{1-s}$
$$I = \frac{1}{2}\int \frac{2s}{\sqrt{3-2s^2-s^4}}\:ds = \frac{1}{2}\int\frac{2s}{\sqrt{4-(s^2+1)^2}}\:ds = \frac{1}{2}\sin^{-1}\left(\frac{s^2+1}{2}\right)$$
Then keep reversing the substitutions
$$I = \frac{1}{2}\sin^{-1}\left(\frac{\left(\frac{t-1}{t+1}\right)^2+1}{2}\right) = \frac{1}{2}\sin^{-1}\left(\frac{t^2+1}{(t+1)^2}\right)$$
$$\implies I = \frac{1}{2}\sin^{-1}\left[\frac{1}{4}\sec^4\left(\frac{\theta}{2}\right)-\tan^2\left(\frac{\theta}{2}\right)\right]+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3567411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 3,
"answer_id": 1
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Why does $3x^4 + 16x^3 + 20x^2 - 9x - 18$ = $(x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6})3$? $$3x^4 + 16x^3 + 20x^2 - 9x - 18 $$
When simplified I arrive to:
$$ (x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6}) $$
But the math book wrote:
$$ (x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6})3 $$
with that extra 3 at the end. The graph calculator seem to agree with that extra 3 as well. what did I do wrong?
| Let $ a\ne 0$.
If the equation $$ax^2+bx+c=0$$ has two roots $ x_1 $ and $ x_2 $, then
$$ax^2+bx+c=a(x-x_1)(x-x_2)$$
do not forget the coefficient $ a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Modular arithmetic and inverse functions The question is shown below:
Suppose $S=\{0,1,2,3,4,5,6,7,8,9,10\}$ and that the function $f:S\rightarrow S$ is given by:
$f(x)=6x^2+3x+8$ (mod 11)
Let $T=\{0,5\}$.
Find $f^{-1}\left(T\right)$.
Alright,
so my initial approach with this question was to find the inverse function,
$f^{-1}(x)=\dfrac{\pm\sqrt{24x-183}-3}{12}$, but $f^{-1}x\in S\geq0$
thus,
$f^{-1}(x)=\dfrac{\sqrt{24x-183}-3}{12}$
and then just churn out congruent values of $x=0,5$ in modulo 11 until I find a suitable value for $f^{-1}(x)$ within $S$, but it seems a bit tedious. Anyways, I repeated this churning process until I received the values of $\{2,3,6,10\}$.
Can anyone suggest a faster approach and please advise me if my approach is logically sound.
Many thanks :)
| Considering the quadratic $f(x)=6x^2 + 3x + 8$ we have a discriminant (in $\Bbb F_{11}$) of
$$D=3^2 - 4 \cdot 6 \cdot 8 \equiv 4 \pmod{11}$$
which is a square (whether it is $\ge 0$ or not is irrelevant, and we don't use $\sqrt{D}$ in finite fields. We just have the two roots $2, -2 \equiv 9$. So $f(x)$ has two roots $$\frac{-3 + 2}{12} = -1 \equiv 10 \text{ and } \frac{-3 -2}{12} = -5 \equiv 6$$
where we have to remember that $12 \equiv 1 \pmod{11}$ which is helpful.
So modulo $11$ we have that $f(x)=(x+1)(x+5)$ and we've already found two members of $f^{-1}[\{0,5\}]$ namely the roots $10$ and $6$.
Next solve $f(x)=10$ (see below for the correct value) or equivalently $$6x^2 + 3x -2 = 0$$ which has discriminant $$3^2 - 4\cdot 6 \cdot (-2) = 2 \pmod{11}$$ and $2$ is not a square modulo $11$ (the squares in $\Bbb F_{11}$ are $0^2= 0,1^2=1, 2^2=4, 3^2= 9, 4^2=5, 5^2=3, 6^2=3, 7^2=5, 8^2=9,9^2=4,10^2=1$ or use quadratic reciprocity theory, if you know it. So $f$ does not assume the value $10$.
Solve $f(x)=5$ or $6x^2 + 3x+ 3= 0$, discriminant $3^2 - 4\cdot 6 \cdot 3 \equiv 3 \pmod{11}$ which is $5^2$ and $(-5)^2 = 6^2$ so roots are
$$\frac{-3 + 5}{12}=2, \frac{-3+6}{12}=3 \pmod{11}$$
so $$f^{-1}[\{0,5\}] = \{2,3,6,10\}$$ without tedious brute force.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the MacLaurin series of $\frac{x+3}{2-x}$ I did:
$$\frac{x+3}{2-x} = \frac{x}{2-x}+\frac{3}{2-x} = x(\frac{1}{2-x})+3(\frac{1}{2-x}) = \\
= \frac{x}{2}(\frac{1}{1-\frac{x}{2}})+\frac{3}{2}(\frac{1}{1-\frac{x}{2}}) = \\
= \frac{x+3}{2}\sum(\frac{x}{2})^n = ...?$$
The answer my professor got was $\frac{(3 + x)}{(2 - x)} = \frac{3}{2} + \sum_{n=1}^∞ \frac{x^n5}{ 2^{n+1}}$
Unfortunately I forgot to write down how he did it. How do I get that solution and is mine necessarily incorrect/incomplete?
| $\begin{align}\frac{3 + x}{2 - x} &=\frac{-x-3}{x-2}\end{align}$
Use long division to find the quotient and remainder of $\frac{-x-3}{x-2}$ and then rewrite it as the quotient plus the remainder over the denominator:
$\begin{align}\frac{-x-3}{x-2}
&= -1-\frac{5}{x-2}=-1+\frac{5}{2-x}\\
&=-1-\frac{\frac52}{1-\frac x2}\\
&=-1+\frac{5}{2}.\frac{1}{1-\frac{x}{2}} \end{align}$
Since,
$\frac{1}{1-t} = 1+t+t^2+t^3+\>...$
$\begin{align}-1+\frac{5}{2}.\frac{1}{1-\frac{x}{2}}
&=-1+\frac52\left(1+ \frac x2 + \frac {x^2}{2^2} + \frac {x^3}{2^3}+\> …\right)\\
&=-1+\frac52+\frac52\left(\frac x2 + \frac {x^2}{2^2} + \frac {x^3}{2^3}+\> …\right)\\
&=\frac32+\frac52 \sum_{n=1}^∞ \frac{x^n}{ 2^{n}}\\
&=\frac32+\sum_{n=1}^∞ \frac{5x^n}{ 2^{n+1}}\end{align}$
Here is the answer that same with your professor. Have a nice day :D
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $\frac{(a+b+c)^2}{(a^2+b^2+c^2)}=\frac{\cot(A/2)+\cot(B/2)+\cot(C/2)}{\cot(A)+\cot(B)+\cot(C)}$ $\frac{(a+b+c)^2}{(a^2+b^2+c^2)}=\frac{\cot(A/2)+\cot(B/2)+\cot(C/2)}{\cot(A)+\cot(B)+\cot(C)}$
I need to solve this trigonometric identity for a trianlge.
I'm not allowed to use the formula $a+b+c=s$ where 's' is perimeter.
My Try
Using sine rule I was able to simplify,
\begin{align}\text{L. H. S.}& =\frac{(\sin A+\sin B+ \sin C)^2}{\sin^2A+\sin^2B+\sin^2C} \\
&=\frac{4\cos^2\dfrac C2\Bigl(\dfrac{\cos(A-B)}{2}+\sin\frac C2\Bigr)^2}{\sin^2A+\sin^2B+\sin^2C}
\end{align}
I'm unable to simplify further, Please give me hint. Thanks in advance.
| Start from the RHS and use the following formulas:
Step 1:
$\cot \frac{A}{2} = \frac{\Delta}{(s-b)(s-c)}$ etc. in the numerator
$\cot A = \frac{R}{abc} (b^2 + c^2 - a^2)$ etc. in the denominator
Step 2:
$R = \frac{abc}{4 \Delta}$
Step 3:
$\Delta^2 = s(s-a)(s-b)(s-c)$
You are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Divisibility by $33^{33}$ Let $P_n=(19+92)(19^2+92^2)\cdots(19^n+92^n)$ for each positive integer $n$. Determine , with proof the least positive integer $n$, if it exists , for which $P_n$ is divisible by $33^{33}$.
I have made no progress concrete enough to show .
| Modulo $3$, we have
$$19^k+92^k\equiv 1^k+(-1)^k\pmod 3 $$
and this is $0$ iff $k$ is odd. So for each odd $k$, the factor $19^k+92^k$ adds (at least) one factor $3$; and for even $k$, it doesn't.
This alone gives us
$3^m\mid P_{2m-1}$, or equivalently
$$3^{\lfloor\frac{n+1}2\rfloor}\mid P_n. $$
There may be higher powers of $3$ added by such a factor. Indeed,
$$19^k+92^k\equiv 1+2^k\equiv 0\pmod 9\iff k\equiv 3\pmod 6.$$
This improves our estimate to
$$3^{\lfloor\frac{n+1}2\rfloor+\lfloor \frac{n+3}6\rfloor}\mid P_n. $$
Hence for $3^{33}$, $n=49$ would be sufficient, but this is still not optimal (though we may suspect that we should rather do more work for the second part - divisibility by $11^{33}$).
So let's look at $11$:
We have $19\equiv -3\pmod{11}$ and $92\equiv 4\equiv (-3)\cdot 6\pmod{11}$, hence
$$19^k+92^k\equiv(-2)^k(1+6^k)\equiv 0\pmod{11} \iff k\equiv 5\pmod {10}.$$
This gives us
$$ 11^{\lfloor\frac{n+5}{10}\rfloor}\mid P_n.$$
Incidentally, we verify that $11^2\|19^5+92^5$ so that
$$ 11^{2\lfloor\frac{n+5}{10}\rfloor}\mid P_n.$$
From this, we find that $n=165$ will certainly be sufficient.
For the exact result (which is readily found numerically: $n=155$), you will need to investigate the factors modulo $1331$ (and even $14641$), I am afraid.
| {
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"url": "https://math.stackexchange.com/questions/3574903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding xy+yz+zx such that the given determinant = 0
$x≠y≠z$
$\begin{vmatrix}x&x^3&x^4-1\\y&y^3&y^4-1\\z&z^3&z^4-1\end{vmatrix} = 0$
Then xy+yz+zx = | A. x+y+z | B. $xyz$ | C. $xyz\over(x+y+z)$ | D. $(x+y+z)\over xyz$ |
Given Ans - D
What I did first was R1->R1-R3 & R2->R2-R3 and throwing (x-z) and (y-z) to the 0.....but this way the opened determinant is still too complex
What I did second was putting values of x and y but with that I was only able to eliminate option A & B
I need help with the correct approach (the correct row transformation) or any other method I can try.
| Hint:
$$\begin{vmatrix}x&x^3&x^4-1\\y&y^3&y^4-1\\z&z^3&z^4-1\end{vmatrix}=\begin{vmatrix}x&x^3&x^4\\y&y^3&y^4\\z&z^3&z^4\end{vmatrix}-\begin{vmatrix}x&x^3&1\\y&y^3&1\\z&z^3&1\end{vmatrix}=xyz\begin{vmatrix}1&x^2&x^3\\1&y^2&y^3\\1&z^2&z^3\end{vmatrix}-\begin{vmatrix}1&x&x^3\\1&y&y^3\\1&z&z^3\end{vmatrix}$$
Can you end it from here?
$$\begin{vmatrix}1&x^2&x^3\\1&y^2&y^3\\1&z^2&z^3\end{vmatrix}=(x-y)(y-z)(z-x)(xy+yz+zx)$$
$$\begin{vmatrix}1&x&x^3\\1&y&y^3\\1&z&z^3\end{vmatrix}=(x-y)(y-z)(z-x)(x+y+z)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3580468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Multinomial expansion sum Given,
$$ \frac{x^2+x+1}{1-x}= a_0+a_1x +a_2x^2+\cdots $$
then, find the sum:
$$ \sum^{50}_{r=1}a_r $$
I knew using multinomial expansion, that
$$ (1-x)^{-1} = 1+ x+x^2+\cdots$$
Hence, $$ \frac{x^2+x+1}{1-x}=1\times(1+x+x^2+\cdots) +x\times(1+x+x^2+\cdots)+x^2\times(1+x+x^2+\cdots) $$
Hence, sum
$ = 3(a_0+ a_1+a_2+\cdots+a_{48}) + 2a_{49}+a_{50} $
So, sum should be $147+3 = 150$. But, I was wrong. Any hints?
The correct answer was
149
Edit: from @John's answer, I got to know that I used symbol $a_r$ for multinomial coeff. and sum coeff. both. The latter $a_r$ was for multinomial actually.
| Let's start from here:
$$1\times(1+x+x^2+\cdots) +x\times(1+x+x^2+\cdots)+x^2\times(1+x+x^2+\cdots)$$
Let us rewrite this in a somewhat more friendly way as well by writing out what the sum would be, in effect:
$$\begin{array}{rrrrr}
1 & +x & +x^2 & +x^3 & \cdots \\
& x & +x^2 & +x^3 & \cdots \\
& & x^2 &+x^3 & \cdots \\
\hline
1 & +2x &+ 3x^2 &+ 3x^3 & \cdots
\end{array}$$
From the $x^2$ term onwards, there's always an $x^k$ ($k \ge 2$) expression in each of the infinite sums once you distribute the outside term, so there's three of them.
You want the sum of the coefficients for the $x,x^2,x^3,\cdots,x^{50}$ terms as well. Thus you want $2 + 3\cdot49$ as your sum, which is $149$.
Your error is likely that you just forgot that you're not meant to count $a_0$ (your constant term): the sum is from $r=1$ to $50$ of $a_r$. This would explain why you're precisely $1$ over the correct answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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How many subsets are there with exactly 8 elements in a set of 16 elements? Below is a problem I did which I believe I did correctly. I would like somebody to confirm that I did, or tell me where I went wrong.
Problem:
Consider a set with $16$ elements in it. How many subsets does it have with exactly $8$ elements?
Answer:
Let $c$ be the number of subsets with exactly $8$ elements.
\begin{align*}
c &= \frac{16(15)(14)(13)(12)(11)(10)(9)}{8!} \\
c &= \frac{2(15)(14)(13)(12)(11)(10)(9)}{7!} \\
c &= \frac{2(15)(2)(13)(12)(11)(10)(9)}{6!} \\
c &= \frac{2(15)(2)(13)(2)(11)(10)(9)}{5!} \\
c &= \frac{2(15)(2)(13)(2)(11)(10)(9)}{5(4)(3)(2)} \\
c &= \frac{2(3)(2)(13)(2)(11)(10)(9)}{4(3)(2)} \\
c &= 13(11)(10)(9) \\
c &= 12870
\end{align*}
| The number of subsets with exactly k elements ${ n \choose k}$
Your answer:
${ 16 \choose 8}$
R code
> choose(16,8)
[1] 12870
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving binomial summation $\sum_{k=0}^{\lfloor{n/2}\rfloor} \binom{n-k}{k} 2^{n-k}$ How can we solve the sum
$$\sum_{k=0}^{\lfloor{n/2}\rfloor} \binom{n-k}{k} 2^{n-k}$$
The problem arose from a counting question, but I am unable to solve this sum.
Edit:
The counting problem was similar to what @Phicar has written, ie, I looked up and the question is equivalent to fibonacci tiling in two colours.
| Here's the "snake-oil" approach that uses generating functions:
\begin{align}
\sum_{n=0}^\infty \sum_{k=0}^{\lfloor{n/2}\rfloor} \binom{n-k}{k} 2^{n-k} z^n
&= \sum_{k=0}^\infty \sum_{n=2k}^\infty \binom{n-k}{k} 2^{n-k} z^n\\
&= \sum_{k=0}^\infty 2^{-k} \sum_{n=2k}^\infty \binom{n-k}{k} (2z)^n\\
&= \sum_{k=0}^\infty 2^{-k} \sum_{n=0}^\infty \binom{n+k}{k} (2z)^{n+2k}\\
&= \sum_{k=0}^\infty 2^{-k} \frac{(2z)^{2k}}{(1-2z)^{k+1}}\\
&= \frac{1}{1-2z} \sum_{k=0}^\infty \left(\frac{2z^2}{1-2z}\right)^k\\
&= \frac{1}{1-2z} \cdot \frac{1}{1-\frac{2z^2}{1-2z}}\\
&= \frac{1}{1-2z-2z^2}\\
&= \frac{1+\sqrt 3}{2\sqrt 3}\cdot\frac{1}{1-(1+\sqrt 3)z} - \frac{1-\sqrt 3}{2\sqrt 3}\cdot\frac{1}{1-(1-\sqrt 3)z}\\
&= \frac{1+\sqrt 3}{2\sqrt 3}\sum_{n=0}^\infty ((1+\sqrt 3)z)^n - \frac{1-\sqrt 3}{2\sqrt 3}\sum_{n=0}^\infty((1-\sqrt 3)z)^n,
\end{align}
which implies that $$\sum_{k=0}^{\lfloor{n/2}\rfloor} \binom{n-k}{k} 2^{n-k}=\frac{(1+\sqrt 3)^{n+1} - (1-\sqrt 3)^{n+1}}{2\sqrt 3}.$$
Note that the recurrence $A_n=2(A_{n-1}+A_{n-2})$ mentioned by @Phicar is implied by the denominator $1-2z-2z^2$ of the generating function, without prior knowledge of what the sequence counts.
| {
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"answer_count": 2,
"answer_id": 0
} |
Comparing $2$ infinite continued fractions
$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} \\
B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}}$
Given the two infinite continued fractions $A$ and $B$ above, which is larger, $2A$ or $B?$
I used the golden ratio on the $2$ and came up with:
$A = 1 + \dfrac{1}{A} \\
B = 2 + \dfrac{1}{B}$
Converting to quadratic equations:
$A^2 - A - 1 = 0 \\
B^2 -2B - 1 = 0$
Resulting to:
$2A = 1 + \sqrt{5} > 1 + \sqrt{2} = B$
My Question is:
Are there any more ways to solve this type of problem?
|
Are there any more ways to solve this type of problem?
We can get some loose bounds by truncating the fractions.
$A = 1 +\dfrac{1}{1 + \left[\frac{1}{1 + \frac{1}{\ddots}}\right]}$
The quantity in the square brackets is clearly smaller than 11 (we have 11 divided by something larger than 11); it follows that $A>1+\frac{1}{1+1}=\frac{3}{2}$
Similarly, we have
$B = 2 +\dfrac{1}{2 + \left[\frac{1}{2 + \frac{1}{\ddots}}\right]}$
Here, we use the fact that the bracketed quantity is greater than zero to get $B<2+\frac{1}{2} = \frac{5}{2}$
Hence $\boxed{2A>B}$
| {
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} |
Evaluate: $S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$ Evaluate of this sum:
$$S=\sum_{j=1}^{\infty}\prod_{k=1}^{j}\frac{2k}{j+k+1}$$
Expand out the sum:
$$S=\prod_{k=1}^{1}\frac{2k}{k+2}+\prod_{k=1}^{2}\frac{2k}{k+3}+\prod_{k=1}^{3}\frac{2k}{k+4}+\cdots$$
$$S=\frac{2}{3}+\frac{2}{4}\cdot\frac{4}{5}+\frac{2}{5}\cdot\frac{4}{6}\cdot\frac{6}{7}+
\frac{2}{6}\cdot\frac{4}{7}\cdot\frac{6}{8}\cdot\frac{8}{9}+\cdots+\frac{2^nn!}{(2n)!\div (n+1)!}$$
I don't know what to do next...
| Amazing is to recognize some series.
Consider
$$S=\sum_{j=1}^{\infty} \frac{j!\,(j+1)!\, 2^j}{(2j+1)!}x^{2j}$$ Let $x=y \sqrt 2$ to make
$$S=\sum_{j=1}^{\infty}\frac{4^j\, j!\, (j+1)! }{(2 j+1)!}y^{2 j}=-\frac{y^2}{y^2-1}+\frac{1}{2 \left(y^2-1\right)}-\frac{\sin ^{-1}(y)}{2
\sqrt{1-y^2} \left(y^2-1\right) y}$$
Making $y=\frac 1 {\sqrt 2}$ gives the result.
If we expand the rhs as a Taylor series built around $y=\frac 1 {\sqrt 2}$ we have
$$S=\frac{\pi }{2}+\sqrt{2} (4+\pi )
\left(y-\frac{1}{\sqrt{2}}\right)+O\left(\left(y-\frac{1}{\sqrt{2}}\right)^2
\right)$$
$$S=\frac{\pi }{2}+(4+\pi ) (x-1)+O\left((x-1)^2\right)$$
| {
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"answer_count": 4,
"answer_id": 3
} |
Evaluate: $S=\sum_{n=1}^{\infty}\frac{(2n)!}{(2n+1)!!^2}$ How to evaluate this sum?
$$S=\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j(j-1/2)}{(j+1/2)^2}$$
$$\frac{j(j-1/2)}{(j+1/2)^2}=\frac{2j(2j-1)}{(2j+1)^2}$$
$$S=\prod_{j=1}^{1}\frac{2j(2j-1)}{(2j+1)^2}+\prod_{j=1}^{2}\frac{2j(2j-1)}{(2j+1)^2}+\prod_{j=1}^{3}\frac{2j(2j-1)}{(2j+1)^2}+\cdots$$
$$S=\frac{1\cdot2}{3^2}+\frac{1\cdot2}{3^2}\cdot\frac{3\cdot4}{5^2}+\frac{1\cdot2}{3^2}\cdot\frac{3\cdot4}{5^2}\cdot\frac{5\cdot6}{7^2}+\cdots+\frac{(2n)!}{(2n+1)!!^2}$$
$$S=\sum_{n=1}^{\infty}\frac{(2n)!}{(2n+1)!!^2}$$
| I suppose that there are several ways to arrive to the result.
I am not very happy with the following
$$S(x)=\sum_{n=1}^{\infty}\frac{(2n)!}{\big[(2n+1)!!\big]^2}x^{n-1}=\frac{2}{9} \, _3F_2\left(1,\frac{3}{2},2;\frac{5}{2},\frac{5}{2};x\right)$$
$$S(1)=2 C-1$$ where $C$ is Catalan constant.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Exact expression of a trigonometric integral Let $a>2$ be a real number and consider the following integral
$$
I(a)=\int_0^\pi\int_0^\pi \frac{\sin^2(x)\sin^2(y)}{a+\cos(x)+\cos(y)} \mathrm{d}x\,\mathrm{d}y
$$
My question. Does there exist a closed-form expression of $I(a)$?
Some comments. Since $a-2<a+\cos(x)+\cos(y)<a+2$ and $\int_0^\pi \int_0^\pi \sin^2(x)\sin^2(y)\ \mathrm{d}x\, \mathrm{d}y=\frac{\pi^2}{4}$, we have the following bounds
$$
\frac{\pi^2}{4(a+2)} < I(a) < \frac{\pi^2}{4(a-2)},
$$
however I didn't manage to find an exact expression for $I(a)$. Any help is welcome!
| Partial answer:
As the integrand is an even function, write $$I(a)=\frac14\int_{-\pi}^\pi\int_{-\pi}^\pi\frac{\sin^2x\sin^2y}{a+\cos x+\cos y}\,dx\,dy$$ and let $w=e^{ix}$. Then for $C_1=\{w:|w|=1\}$, we have \begin{align}\int_{-\pi}^\pi\frac{\sin^2x}{a+\cos x+\cos y}\,dx&=\oint_{C_1}\frac{\frac{(w-1/w)^2}{-4}}{b+\frac{w+1/w}2}\frac{dw}{iw}=\frac i2\oint_{C_1}\frac{(w^2-1)^2}{w^2(w^2+2bw+1)}\,dw\end{align} where $b:=a+\cos y$. Inside $C_1$, there is a double pole at $0$, so letting $f(w)$ be the integrand, \begin{align}\operatorname{Res}(f,0)=\lim_{w\to0}\frac d{dw}(w^2f(w))=-2b.\end{align} Solving the quadratic in the denominator of $f$ gives $w=-b\pm\sqrt{b^2-1}$ but only the positive root lies in $C_1$. As this is a simple pole, \begin{align}\operatorname{Res}(f,-b+\sqrt{b^2-1})=\lim_{w\to-b+\sqrt{b^2-1}}(w+b-\sqrt{b^2-1})f(w)=2\sqrt{b^2-1}.\end{align} Hence \begin{align}\int_{-\pi}^\pi\frac{\sin^2x}{a+\cos x+\cos y}\,dx=\frac i2\cdot2\pi i(2\sqrt{b^2-1}-2b)=-2\pi(\sqrt{b^2-1}-b)\end{align} so that \begin{align}I(a)&=-\frac\pi2\int_{-\pi}^\pi\sin^2y\left(\sqrt{(a+\cos y)^2-1}-a-\cos y\right)\,dy\\&=-\frac\pi2\int_{-\pi}^\pi\sin^2y\sqrt{(a+\cos y)^2-1}\,dy+\frac{a\pi^2}2.\end{align} Now let $z=e^{iy}$. Then for $C_2=\{z:|z|=1\}$, we have \begin{align}\int_{-\pi}^\pi\sin^2y\sqrt{(a+\cos y)^2-1}\,dy&=\oint_{C_2}\left(\frac{z-1/z}{2i}\right)^2\sqrt{\left(a+\frac{z+1/z}2\right)^2-1}\cdot\frac{dz}{iz}\\&=\frac i8\oint_{C_2}\frac{(z^2-1)^2}{z^4}\sqrt{(z^2+2az+1)^2-4z^2}\,dz.\end{align} Evidently, there is a pole of order $4$ at $0$ which lies in $C_2$. However, due to the presence of $\sqrt\cdot$, branch points also need to be considered, and solving the equation $z^2+2az+1=\pm2z$ reveals that only $z=\pm1-a+\sqrt{a^2\mp2a}$ are the two branch points in $C_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590145",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Integration through partial fractions with complex roots In integrating the following: $\frac{1}{(x^2+2x+3)^2}$ I am trying to use partial fraction decomposition as follows:
$\frac{1}{(x^2+2x+3)^2} = \frac{Ax + B}{x^2+2x+3} + \frac{Cx + D}{(x^2+2x+3)^2}$
Which gives me:
$1 = (Ax + B)(x^2 +2x +3) + (Bx + C)$
And that gets me back to where I started.
$\frac{1}{(x^2+2x+3)^2}$
What am I doing wrong?
| You don't need he partial fractions decomposition: as already observed, the substitution $u=x+1$ results in having to compute the integral $\;\int\frac{\mathrm du}{(u^2+2)^2}$.
Now this integral pertains to a well-known type:
$$I_n=\int\frac{\mathrm d x}{(x^2+a^2)^n},$$
which is easily computed though a recursion formula. Indeed
*
*$I_1=\dfrac1a\,\arctan \dfrac xa$.
*To establish the recursion formula, use integration by parts for $I_n$:
Setting $u=\dfrac1{(x^2+a^2)^n},\;\mathrm dv=\mathrm dx$, whence $\;\mathrm du=\dfrac{-2n x}{(x^2+a^2)^{n+1}}\,\mathrm dx,\;v= x$, so
$$I_n=\frac x{(x^2+a^2)^n}+2n\int\frac {x^2}{(x^2+a^2)^{n+1}}=\frac x{(x^2+a^2)^n}+2n(I_n-a^2I_{n+1}),$$
whence the relation
$$I_{n+1}=\frac1{2na^2}\frac x{(x^2+a^2)^n}+\frac{2n-1}{2na^2}I_n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$...
Let $a,b,c\in R$. If $f(x)=ax^2+bx+c$ is such that $a+b+c=3$ and $f(x+y)=f(x)+f(y)+xy$ for all $x,y\in R$. Then $\sum_{n=1}^{10} f(n)$ is
From the first equation
$$a(x^2+y^2+2xy)+b(x+y)+c=ax^2+bx+c+ay^2+by+c+xy$$
$$2axy=c+xy$$
Also, the summation will be
$$a(1^2+2^2+3^2...10^2)+b(1+2+3+4...+10)+10c$$
$$=385a+55b+10c$$
$$375a+45b+30$$
I don’t know what to do with the x and y terms. How do I proceed?
| You have that
$$f(x+y)=f(x)+f(y)+xy \tag{1}\label{eq1A}$$
for all $x,y\in R$. Your simplification of \eqref{eq1A} gives
$$2axy=c+xy \implies (2a-1)xy = c \tag{2}\label{eq2A}$$
Since this is true for every real $x$ and $y$, this means that $2a - 1 = 0 \implies a = \frac{1}{2}$ and $c = 0$. This thus gives that $b = 3 - a - c = \frac{5}{2}$. Can you finish the rest?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592681",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$ for all positive integers
I am trying to prove
$$(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$$
for all positive integers.
My attempts so far have been to Taylor expand the left hand side:
$$(n+1)^{2/3} -n^{2/3}\\
=n^{2/3}\big((1+1/n)^{2/3} -1\big)\\
=n^{2/3}\left(\sum_{\alpha=0}^\infty\frac{\frac{2}{3}(\frac{2}{3}-1)\cdots(\frac{2}{3}-\alpha+1)}{\alpha!} n^\alpha-1\right)\\
=n^{2/3}\sum_{\alpha=1}^\infty\frac{\frac{2}{3}(\frac{2}{3}-1)\cdots(\frac{2}{3}-\alpha+1)}{\alpha!} n^\alpha$$
I also tried proof by induction. Assume that it's true for n=k, so that
$$(n+1)^{2/3} -n^{2/3} < \frac{2}{3}n^{-1/3}\\
n^{2/3}\big((1+1/n)^{2/3} -1\big)<\frac{2}{3} n^{-1/3}\\
(1+1/n)^{2/3} -1<\frac{2}{3} n^{-1}$$
Then I want to prove that $(n+2)^{2/3} -(n+1)^{2/3} < \frac{2}{3}(n+1)^{-1/3}$. The left hand side is:
$$(n+2)^{2/3} -(n+1)^{2/3}\\
=(n+1)^{2/3}\left[\left(1+\frac{1}{n+1}\right)^{2/3}-1^{2/3}\right]\\
<(n+1)^{2/3}\cdot \frac{2}{3} n^{-1}\\
=\frac{2}{3}\frac{(n+1)^{2/3}}{n^{-1}}$$
But this is bigger than $\frac{2}{3}(n+1)^{-1/3}$, so I am stumped!
| An elementary way. Let $x=n^{1/3}\geq 1$ then it suffices to show that
$$(x^3+1)^{2/3} -x^2 <\frac{2}{3x}$$
that is
$$(x^3+1)^2<\left(x^2+\frac{2}{3x}\right)^3$$
or
$$x^6+2x^3+1<x^6+2x^3+\frac{4}{3}+\frac{8}{27x^3}$$
which trivially holds.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Question on sum of digits of squares Let $D$ be the function define as $D(b,n)$ be the sum of the base-$b$ digits of $n$.
Example: $D(2,7)=3$ means $7=(111)_2\implies D(2,7)=1+1+1=3$
Define $S_m(a)=1^m+2^m+3^m+...+a^m$ where $a,m\in\mathbb{Z}_+$
Can it be shown that
(1)$$D(a,S_2(a))\le 2(a-1)?$$
(2) $$D(a,S_2(a))< a\iff a\equiv5\mod6?$$
Note: For $a,m>1$
● $a^m<S_m(a)<a^{m+1}$
● $1\le D(a,S_m(a))\le(a-1)(m+1)$
● $D(a,S_m(a))=1+D(a,S_m(a-1))$proof
Edit
● $a\mid S_2(a)$ then $D(a+1,S_2(a+1))=a+1$
Proof:
let $b=a+1$.
Identically, we have
$$
S_2(n)
=
\sum_{i=1}^n i^2
=
\frac{n(n+1)(2n+1)}{6}
$$
hence
\begin{align*}
&a{\,|\,}S_2(a)\\[4pt]
\implies\;&a{\;|}\left(
\frac{a(a+1)(2a+1)}{6}
\right)\\[4pt]
\implies\;&6{\;|}\left((a+1)(2a+1)\right)\\[4pt]
\implies\;&6{\;|}\left(b(2b-1)\right)\\[4pt]
\implies\;&6{\,|\,}b\;\;\text{or}\;\;\Bigl(2{\,|\,}b\;\;\text{and}\;\;3{\;|\,}(2b-1)\Bigr)\\[4pt]
\end{align*}
If $6{\,|\,}b$, then
\begin{align*}
S_2(b)&=\frac{b(b+1)(2b+1)}{6}\\[4pt]
&=\frac{b^3}{3}+\frac{b^2}{2}+\frac{b}{6}\\[4pt]
&=
\left({\small{\frac{b}{3}}}\right)\!{\cdot}\,b^2
+
\left({\small{\frac{b}{2}}}\right)\!{\cdot}\,b^1
+
\left({\small{\frac{b}{6}}}\right)\!{\cdot}\,b^0
\end{align*}
hence
$$
D(b,S_2(b))
=
\left({\small{\frac{b}{3}}}\right)
+
\left({\small{\frac{b}{2}}}\right)
+
\left({\small{\frac{b}{6}}}\right)
=
b
$$
If $2{\,|\,}b\;\;$and$\;\;3{\;|\,}(2b-1)$, then $b\equiv 2\;(\text{mod}\;3)$, so
\begin{align*}
S_2(b)&=\frac{b(b+1)(2b+1)}{6}\\[4pt]
&=\frac{b^3}{3}+\frac{b^2}{2}+\frac{b}{6}\\[4pt]
&=
\left({\small{\frac{b+1}{3}}}\right)\!{\cdot}\,b^2
+
\left({\small{\frac{b-2}{6}}}\right)\!{\cdot}\,b^1
+
\left({\small{\frac{b}{2}}}\right)\!{\cdot}\,b^0
\end{align*}
hence
$$
D(b,S_2(b))
=
\left({\small{\frac{b+1}{3}}}\right)
+
\left({\small{\frac{b-2}{6}}}\right)
+
\left({\small{\frac{b}{6}}}\right)
=
b
$$
Thus, for all cases, we have $D(b,S_2(b))=b$.
| We can continue your casework on $ n \pmod{6}$.
Here's a sketch of it. Fill in the rest of the details yourself.
If $ n \equiv 0 \pmod{6}$, then $ \frac{ 2n^3 + 3n^2 + n } { 6} = \frac{2n}{6} \times n^2 + \frac{3n}{6} \times n + \frac{n}{6} \times 1 $,
so $D_2 = \frac{2n}{6} + \frac{3n}{6} + \frac{n}{6} = n $.
If $ n \equiv 1 \pmod{6}$, then $ \frac{ 2n^3 + 3n^2 + n }{6} = \frac{ 2n-2}{6} \times n^2 + \frac{ 5n - 5 } { 6} \times n + \frac{ n + 5 } { 6}\times 1 $,
so $D_2 = \frac{ 2n-2}{6} + \frac{ 5n-5}{6} + \frac{n+5}{6} = \frac{ 8n-2}{6}$.
If $ n\equiv 2 \pmod{6}$, then $ \frac{ 2n^3 + 3n^2 + n }{6} = \frac{2n - 4}{6} \times n^2 + \ldots$
so $D_2 = \ldots $
$\vdots$
If $ n \equiv 5 \pmod{6}$, then $ \frac{ 2n^3 + 3n^2 + n }{6} = \frac{2n+2}{6} \times n^2 + \frac{n+1}{6} \times n + 0 \times 1 $,
so $D_2 = \frac{2n+2}{6} + \frac{n+1}{6} = \frac{3n+3}{6} < n$.
I have not done the rest as yet, but I believe it will all work out (assuming the statement is true). I'd be happy to review your algebra if you are unable to conclude that $D_2 \leq 2 (n-1)$ and $ D_2 < n \iff n\equiv 5 \pmod{6}$.
| {
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Roots of the equation $(x – 1)(x – 2)(x – 3) = 24$ The equation $(x – 1)(x – 2)(x – 3) = 24$ has the real root equal to 'a' and the complex roots 'b' and 'c'. Then find the value of $\frac{bc}{a}$
My approach is as follow $y=f(x)=(x – 1)(x – 2)(x – 3) - 24=0$
$y'=3x^2-12x+11=0$
Solving we get $x=2\pm\sqrt{\frac{1}{3}}$
$f(2+\sqrt{\frac{1}{3}})<0$ & $ f(2-\sqrt{\frac{1}{3}})<0$
It is Local Minimum at $2+\sqrt{\frac{1}{3}}$ and Local Maximum at $2-\sqrt{\frac{1}{3}}$
By hit and trial I got $f(5)=0$ viz a=5
Given abc=30, therefore bc=6.
Hence the answer is $\frac{6}{5}$ which is correct.
My only concern is to find the real value without using any HIT and TRIAL.
| Noticing that $24=2\cdot3\cdot4$, $5$ must be a root. Then after long division by $x-5$, $x^2-x+6=0$.
Using Vieta,
$$\frac{bc}a=\frac65.$$
Now for a "general" solution, you first deplete the cubic by setting $z:=x-2$ and the equation is
$$(z+1)z(z-1)=z^3-z=24.$$
Then with $z:=\dfrac2{\sqrt 3}\cosh u$,
$$4\cosh^3u-3\cosh u=\cosh3u=36\sqrt3,$$
finally giving
$$x=\frac2{\sqrt3}\cosh\frac{\text{arcosh }36\sqrt3}3+2.$$
Needless to say, this is $5$.
| {
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Proving $\def\n#1{\left(\frac12+\sum\limits_{k=1}^n{#1}^{k^2}\right)}\n{a}\n{b}\ge{\n{(ab)}}^2$
Let $n$ be an even postive integer, and $a,b\in (-1,1)$, $a+b\ge0$. Show that
$$\left(\frac12+\sum_{k=1}^na^{k^2}\right)\left(\frac12+\sum_{k=1}^nb^{k^2}\right)\ge\left(\frac12+\sum_{k=1}^n(ab)^{k^2}\right)^2\tag{1}$$
It seems promising to use Cauchy-Schwarz inequality to prove it or other inequality: If $a,b>0$, then
$$\left(\frac{1}{2}+\sum_{k=1}^{n}a^{k^2}\right)\left(\frac{1}{2}+\sum_{k=1}^{n}b^{k^2}\right)\ge\left(\frac{1}{2}+\sqrt{\sum_{k=1}^{n}(a)^{k^2}\sum_{k=1}^{n}(b)^{k^2}}\right)^2,$$
but this is different from the RHS of $(1)$. Thanks.
| Partial answer
WLOG, assume that $a \ge b$. The inequality is written as
$$\frac{1}{2}(\sum a^{k^2} + \sum b^{k^2}) + \sum a^{k^2} \sum b^{k^2}
\ge \sum (ab)^{k^2} + (\sum (ab)^{k^2})^2. \tag{1}$$
If $b > 0$, the proof is easy. Indeed, since $0\le ab \le 1$, we have (by AM-GM)
$$\sum a^{k^2} + \sum b^{k^2} \ge \sum 2(\sqrt{ab})^{k^2} \ge \sum 2(ab)^{k^2}$$
and (by CBS)
$$\sum a^{k^2} \sum b^{k^2} \ge (\sum (\sqrt{ab})^{k^2})^2 \ge (\sum (ab)^{k^2})^2.
$$
So, the desired inequality in (1) is true.
It remains to prove the inequality in (1) under the condition $0\le -b \le a \le 1$.
*
*If $-b = 0$, clearly the inequality in (1) is true.
*If $a = 1$, it suffices to prove that
$$\frac{1}{2}(n + \sum b^{k^2}) + n \sum b^{k^2}
\ge \sum b^{k^2} + (\sum b^{k^2})^2.$$
It suffices to prove that $\sum b^{k^2} \ge -\frac{1}{2}$.
If $-b = 1$, clearly $\sum b^{k^2} = 0 \ge -\frac{1}{2}$.
If $0 < -b < 1$, clearly $f(2m) \triangleq \sum_{k=1}^{2m} b^{k^2}$ is non-increasing.
Also, $\sum_{k=1}^\infty b^{k^2} = \frac{1}{2}\vartheta_3(0, b) - \frac{1}{2} \ge -\frac{1}{2}$
where $\vartheta_3(z, q) = 1 + 2 \sum_{k=1}^\infty q^{k^2}\cos (2k z)$ is the Jacobi theta function
(by using the property $\vartheta_3(0, q)= \prod_{n=1}^\infty (1-q^{2n})(1+q^{2n-1})^2$).
Thus, we have $\sum b^{k^2} \ge -\frac{1}{2}$.
*If $-b = a$, to be continued.
*If $0 < -b < a < 1$, to be continued.
| {
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"url": "https://math.stackexchange.com/questions/3608799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If the equation $(x^2-4)^3(x^3+1)^n(x^2-5x+6)^m=0$ has 18 roots, find m+n. If the equation $(x^2-4)^3(x^3+1)^n(x^2-5x+6)^m=0$ has 18 roots, find $m+n$.
I did and I got
$$(x+2)^3(x-2)^{3+m}(x+1)^n(x^2-x+1)^n(x-3)^m=0$$, so I find $3+3+m+n+2n+m=18\implies 2m+3n=12$, the answer is m+n=5. What I have to do now?
| $2×3+3×n+2×m = 18$
Find $m+n$
$3*n+2*m = 12$
( Diophantine equation )
Offcourse $m,n € Z$ were $Z → (0,1,2,3,4,5,............)$
If we construct a table,
Put $n=0$, $m=6$
Put $n=1$, $m$ ≠ $Z$
Put $n=2$, $m=3$
Put $n=3$, $m$ ≠ $Z$
Put $n=4$, $m=0$
So there are only there values that works here $n=0$, $m=6$ and $n=2$, $m=3$ and $n=4$, $m=0$
Therefore $m+n$ is either $6$ or $5$ or $4$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that there do not exist any distinct natural numbers a,b,c,d such that
Show that there do not exist any distinct natural numbers a,b,c,d such that $a^3+b^3=c^3+d^3$ and $a+b=c+d$.
| Suppose $a+b=c+d$ and $a^3+b^3=c^3+d^3$.
$$a+b=c+d$$
$$(a+b)^3=(c+d)^3$$
$$a^3+b^3+3ab(a+b)=c^3+d^3+3cd(c+d)$$
$$3ab(a+b)=3cd(c+d)$$
$$ab=cd$$
Let $a+b=c+d=m$ and $ab=cd=n$
a and b are the roots of the quadratic equation
$$x^2-mx+n=0$$
by Vieta's relations because a+b=m and ab=n.
But c and d are also roots of the equation for similar reasons. But a quadratic equation can have at most two distinct roots.
Hence, a=c or a=d, so a,b,c,d are not distinct.
| {
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Is the integral $ \int_{1}^{2}\frac{dx}{\sqrt{x^2-x+1} - 1} $ converges or diverges. Determine if the following integral diverges/converges, if it converges, is it absolutely or conditionally converges.
$$
\int_{1}^{2}\frac{dx}{\sqrt{x^2-x+1} - 1}
$$
What i tried:
In that interval we know that:
$$
0 < x
$$
Therefore we can write:
$$
\frac{1}{\sqrt{x^2-x+1} - 1} < \frac{1}{\sqrt{x^2-2x+1} - 1} = \frac{1}{\sqrt{(x-1)^2} - 1}
$$
$$
= \frac{1}{x - 2}
$$
Now lets try to calculate the integral, using limits, we get:
$$
\lim_{t \to 2^-}\int_{1}^{t}\frac{1}{x-2} = ln(x-2)|_{1}^{t} = ln(t-2) - ln(1-2)
$$
I cant calculate this integral, namely i dont get any limit that is a number.
I thought that maybe i will get to a form like this:
$$
\int_{a}^{b}\frac{1}{(b-x)^\alpha}, \int_{a}^{b}\frac{1}{(x-a)^\alpha}
$$
But also, if i will, and conclude that $ \int \frac{1}{x - 2}$ diverges in that interval, i couldnt use the comparison test to conclude about the original function.
Can i have a hint?
Thank you.
| It diverges.
Multiplying by the conjugate:
$\frac{1}{\sqrt{x^2-x+1}-1} = \frac{\sqrt{x^2-x+1}+1}{x(x-1)}$.
A partial fraction decomposition yields:
$\frac{1}{x(x-1)}=\frac{1}{x}-\frac{1}{x-1}$ and therefore
$\frac{\sqrt{x^2-x+1}+1}{x(x-1)}=\frac{\sqrt{x^2-x+1}+1}{x}-\frac{\sqrt{x^2-x+1}+1}{x-1}$.
The right-hand side equals:
$\frac{\sqrt{x^2-x+1}}{x} - \frac{\sqrt{x^2-x+1}}{x-1} + \frac{1}{x} - \frac{1}{x-1}$
Also, remark that $\displaystyle \int_{1}^{2} \frac{\sqrt{x^2-x+1}}{x}dx < \infty$ and $\displaystyle \int_{1}^{2} \frac{dx}{x} < \infty$, hence it suffices to show that $\displaystyle \int_{1}^{2} \frac{\sqrt{x^2-x+1}}{x-1} dx$ diverges. Note that $\displaystyle \int_{1}^{2} \frac{x^2-x+1}{x-1} dx \geq \displaystyle \int_{1}^{2} \frac{dx}{x-1}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding number of integral solutions to an equation.
Find the number of integral solutions to:
$$x^2+y^2-6x-8y=0.$$
My attempt:
The equation can be rewritten as:
$$x^2+y^2-6x-8y+9+16=25,$$
basically adding 25 to both sides, or equivalently,
$$(x-3)^2+(y-4)^2=25.$$
This is a Pythagorean triplet. The only triplet of this form would be $3$, $4$, $5$, so the possibilities for $x - 3$ and $y - 4$ are:
1.) $3$, $4$
2.) $-3$, $-4$
3.) $-3$, $4$
4.) $3$, $-4$
For each of these pairs, there are two different pairs $(x,y)$. For instance, in the first pair we could have $x - 3 = 3$ and $y - 4 = 4$, or $x - 3 = 4$ and $y - 4 = 3$. So there are $8$ solutions.
But, shouldn't we also consider the pair $(0,5)$? Because
$$0^2+5^2=25.$$
Then we have two new pairs:
1.) $0$, $5$
2.) $0$, $-5$
This would give $4$ new solutions, so the total number of solutions should be $8+4=12$.
How is the answer just $8$ then? What am I missing?
| You are correct; there are solutions corresponding to $0^2+(\pm5)^2=25$, and there are $12$ integral solutions in total. Another way to see this, is to note that
$$(x-3)^2+(y-4)^2=25,$$
forces $|y-4|\leq5$, leaving $11$ possible values for $y$ and yielding $6$ different quadratics in $x$. A quick check then shows that only $4$ of them yield integral solutions for $x$.
| {
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Establish a lower bound for the generalized alternating harmonic series I'm given the following series:$$\sum_{n=1}^{\infty}{(-1)^{n-1}\over n^p}$$
I need to show that the sum is greater than $1/2$ for every $p > 0$. For $p \ge1$ this is obvious, as it follows by grouping terms $2$ by $2$, and the first sum is greater than $1/2$ while the other terms are all positive. I'm stuck with $0<p<1$.
| let $S_p(N)=\sum_{k=1}^{N}{k^{-p}}$; for notational simplicity we fix for now $0<p<1$ and let $S_p(N)=S(N)$.
We note that $\eta(p)=\sum_{n=1}^{\infty}{(-1)^{n-1}\over n^p} > S(2N)-2^{1-p}S(N)$ for any $N \ge 1$ as the remainder is an alternating sum with decreasing terms that starts at the positive $\frac{1}{(2N+1)^p}$
Using that $f(x)=x^{-p}$ is convex as $f''(x)>0, x>0$, we get $f(k) \le \int_{k-\frac{1}{2}}^{k+\frac{1}{2}}f(x)dx$ since $f(k+\alpha)+f(k-\alpha) \ge 2f(k), k \ge 1, 0 \le \alpha \le \frac{1}{2}$. Hence $S(N) \le \int_{\frac{1}{2}}^{N+\frac{1}{2}}f(x)dx=\frac{(N+\frac{1}{2})^{1-p}-\frac{1}{2}^{1-p}}{1-p}$, or
$-2^{1-p}S(N) \ge -\frac{(2N+1)^{1-p}}{1-p}+\frac{1}{1-p}$
Using the trapezoidal rule for $f$ which is convex, so the error term is negative, we get $\int_2^{2N}f(x)dx \le f(2)+..f(2N)-\frac{1}{2}f(2)+O(N^{-p})$, or
$S(2N) \ge \frac{(2N)^{1-p}}{1-p}+1+\frac{1}{2^{p+1}}-\frac{2^{1-p}}{1-p}+O(N^{-p})$
Hence $\eta(p) > \frac{(2N)^{1-p}}{1-p}+1+\frac{1}{2^{p+1}}-\frac{2^{1-p}}{1-p}+O(N^{-p})-\frac{(2N+1)^{1-p}}{1-p}+\frac{1}{1-p}$. Since the sum of terms in $N$ obviously goes to zero when $0<p<1$ fixed, it is enough to prove that
$1+\frac{1}{2^{p+1}}-\frac{2^{1-p}}{1-p}+\frac{1}{1-p}>\frac{1}{2}$ as then we pick $N$ large enough st the terms in $N$ are less than half $1+\frac{1}{2^{p+1}}-\frac{2^{1-p}}{1-p}+\frac{1}{1-p}-\frac{1}{2}>0$ and we get $\eta(p) > \frac{1}{2}$ as required
By bringing to the same denominator and allowing $0 \le p \le 1$ the required inequality is equivalent to:
edit - as pointed out the inequality is slightly more complicated but still elementary as follows:
$(3-p)2^p \ge 3+p$ with equality only for $p=0,1$
Here we let $q=2^p, p\log 2 =\log q, 1 \le q \le 2$, so we consider $g(q)=(3\log 2-\log q)q-3\log 2-\log q$ and we need to show that $g(q) >0, 1<q<2$
But $g''(q)=\frac{1}{q^2}-\frac{1}{q} \le 0$ as $q \ge 1$ so $g'$ decreasing and since $g'(1)=3\log 2-2>0, g'(2)=2\log 2-1.5<0$, it follows that $g$ strictly increases until some $1<q_0<2$ and then strictly decreases to $g(2)$, while $g(1)=g(2)=0$ ensuring $g(q)>0, 1<q<2$, so done!
Note that if we only require $\eta(p) \ge \frac{1}{2}$ we can directly apply the trapezoidal rule from $1$ to $2N$ rather than $2$ to $2N$ and then we get precisely $S(2N)-2^{1-p}S(N) \ge \frac{1}{2}-c_N$ where $|c_N| \to 0$ when $N \to \infty$, hence the result follows. To get a strict inequality (which is true as we saw) we need the extra work and that way we can get better estimates for $p$ away from $0$ or $1$
| {
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How to prove that the series $\sum\limits_{n=1}^{\infty} \log x_n$ is convergent?
Consider the sequence $\{x_n \}$ defined by $$x_n = e \left (\frac {n} {n+1} \right )^{n + \frac 1 2},\ n \geq 1.$$
Prove that the series $\sum\limits_{n=1}^{\infty} \log x_n$ is convergent.
My attempt: I find that \begin{align*} \sum\limits_{n=1}^{\infty} \log x_n & = \sum\limits_{n=1}^{\infty} \left ( 1 - \left (n + \frac 1 2 \right ) \log \left (1 + \frac 1 n \right ) \right ) \\ & = - \sum\limits_{n = 1}^{\infty} \sum\limits_{m=2}^{\infty} (-1)^m \left ( \frac {1} {m+1} - \frac {1} {2m} \right ) \frac {1} {n^m} \\ & = -\sum\limits_{n = 1}^{\infty} \sum\limits_{m=2}^{\infty} (-1)^m \frac {m-1} {2m(m+1)} \frac {1} {n^m} \end{align*}
From here how do I proceed further? Please help me in this regard.
| Notice, that:
$$\sum_{i=1}^n \log x_i=
\log \prod_{i=1}^n x_i =
\log \left( e\left(\frac{1}{2}\right)^{\frac{3}{2}} e\left(\frac{2}{3}\right)^{\frac{5}{2}} \dots e\left(\frac{n}{n+1}\right)^{\frac{2n+1}{2}} \right)=
\log \left( e^n \frac{n!}{(n+1)^{n+\frac{1}{2}}}\right)$$
Using Stirling's approximation:
$$\lim_{n \to \infty} \log \left( e^n \frac{n!}{(n+1)^{n+\frac{1}{2}}}\right) =
\lim_{n \to \infty} \log \left( e^{n + 1} \frac{(n+1)!}{(n+1)^{n + 1} \sqrt{n+1}} \frac{1}{e}\right) =
\log \frac{\sqrt{2 \pi}}{e}$$
Note, that this is the exact value of the sum, not only the proof of convergence.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why doesnt this trignometric substituition work? $$\frac 1 5\int\frac{x+1+5}{(x+1)^2+5}\,\mathrm dx$$
where i substitute $x+1 = \sqrt{5}\tan(\theta)$
After Substituition :
$$\frac15\int\frac{\sqrt5\tan(\theta)+5}{\tan^2(\theta)+1}\sqrt5\sec^2(\theta)\,\mathrm d\theta$$
but if i do $u = x +1$
and then further substitute and integrate i get :
$\int \frac u {(u^2 +5)} + \int \frac 5 {(u^2 +5)} $
which then gives :
$\frac {ln(x^2 + 2x +6)} 2 + \sqrt5 arctan(\frac {x+1}{\sqrt5}))$
not :
$ln(\frac{(x^2 + 2x + 6)} {\sqrt5}) + \sqrt5arctan(\frac {x + 1} {\sqrt5})$
| Where is your trig substitution failing?
$\int \frac{(x+1)+5}{(x+1)^2 + 5} \ dx$
$x+1 = \sqrt 5 \tan \theta\\
dx = \sqrt 5 \sec^2 \theta$
$\int \frac{(\sqrt 5 \tan \theta +5)(\sqrt 5 \sec^2 \theta)}{5\sec^2 \theta} \ d\theta\\
\int \frac{\tan \theta}{\sqrt 5} + \sqrt 5 \ d\theta$
That looks relatively simple from here.
$\frac {\ln |\sec\theta|}{\sqrt 5} + \sqrt 5 \theta + C$
$\theta = \arctan \frac {x+1}{\sqrt 5}\\
\sec\theta = \sqrt {\frac {(x+1)^2}{5} + 1} = \sec\theta = \sqrt {\frac {x^2 + 2x + 6}{5}}\\
\ln\sqrt {\frac {x^2 + 2x + 6}{5}} = \frac 12 \ln (x^2 + 2x + 6) - \frac 12 \ln 5 $
The $\frac 12 \ln 5$ is a constant and can be added to $C.$
$\frac {\ln {x^2 + 2x + 6}}{2\sqrt 5} + \sqrt 5 \arctan \frac {x+1}{\sqrt 5} + C$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving modulus equations with multiple unknowns I have the following equations:
$$8\equiv-(3a+7b)\mod11$$
and
$$9\equiv-(5a+4b)\mod11$$
How do I solve them?
I've tried to subtract one equation from the other, but I end up with the wrong answer.
It is supposed to be:
$a=2$ and $b=9$.
| By rearranging the first congruence equation $8 \equiv -3a -7b \pmod{11}$, we have $4b \equiv 8 + 3a \pmod{11}$. Since $\gcd(4, 11) = 1$, there is an inverse of $4$, which is $3$. We thus have
\begin{equation}
b \equiv 24 + 9a \equiv 2 - 2a \pmod{11}
\end{equation}
Plugging this into the other congruence equation $9 \equiv -5a - 4b \pmod{11}$. We have
\begin{equation}
9 \equiv -5a - 4(2 - 2a) \equiv 3a - 8 \pmod{11}
\end{equation}
which implies that $3a \equiv 17 \equiv 6 \pmod {11}$. Since $\gcd(3, 11) = 1$, we can divide by $3$. So, we have $a \equiv 2 \pmod{11}$. Using the same way, you can solve for $b$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Find the equation of a plane that contains a given line and is 2 units away from a given point. The given line has a vector of (2,-1,4) and contains the points A(-1,1,-1) and B(1,0,3). The given point is P(2,3,4). The problem asks you to find a plane such that its distance to P is 2 units and it contains line AB, and to justify the number of solutions it has.
I've tried making the plane's orthogonal vector (A,B,C) unitary and multiplying it by 2 to make its modulus 2, but I don't know how to find A, B and C from there.
Another approach is using the distance formula, but I get a single equation with 4 variables, some of which are squared.
| Let $T$ be a point of the sphere $s$ centered at $P$ withe radius $2$ such that the distance from the plane $p$ defined by $T$, $A$ and $B$ to $P$ is equal to $2$. Then $p$ is tangent to $s$ and therefore the vectors $TA$ and $TB$ are orthogonal to $TP$. So, you can find the point $T$ by solving the system$$\left\{\begin{array}{l}(x-2)^2+(y-3)^2+(z-4)^2=4\\(x-2)(x+1)+(y-3)(y-1)+(z-4)(z+1)=0\\(x-2)(x-1)+(y-3)y+(z-4)(z-3)=0,\end{array}\right.$$which is equivalent to$$\left\{\begin{array}{l}x^2-4 x+y^2-6 y+z^2-8 z+25=0\\x^2-x+y^2-4 y+z^2-3 z-3=0\\x^2-3 x+y^2-3 y+z^2-7 z+14=0.\end{array}\right.$$Now, if you subtract the first equation from each of the other two, you get the system$$\left\{\begin{array}{l}x^2-4 x+y^2-6 y+z^2-8 z+25=0\\3 x+2 y+5 z=28\\x+3 y+z=11.\end{array}\right.$$Solving the last two equations with respect to $x$ and $y$, you get that$$\left\{\begin{array}{l}x=\frac{62-13z}7\\y=\frac{5+2z}7.\end{array}\right.$$Finally, you put these values of $x$ and $y$ in the first equation in order to get a quadratic equation in $z$, whose roots are$$z=\frac{426\pm7\sqrt{138}}{111}.$$So, the solutions are$$\frac1{111}\left(192\mp13\sqrt{138},201\pm2\sqrt{138},426\pm7\sqrt{138}\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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When can a sum of 3 identical squares be written as a sum of 3 non-identical squares? Consider $3n^2$ for some positive integer $n$. When do there exist positive integers $n_1,n_2,n_3$, not all equal to $n$, such that
$$
3n^2 = n_1^2 + n_2^2 + n_3^2
$$
An example is $27$:
$$
27 = 3^2 + 3^2 + 3^2 = 1^2 + 1^2 + 5^2
$$
| I have one hint:
$(n-\alpha)^2 + (n- \alpha)^2 + (n+\alpha)^2 = n^2 -2n\alpha + \alpha^2 + n^2 -2n\alpha + \alpha^2 + n^2 +2n\alpha + \alpha^2 = \\ n^2 + n^2 + n^2 + 3\alpha^2 - 2n\alpha.$
So, if we want for the sum to have $3n^2$, then it is mandatory that we have $3\alpha^2-2n\alpha =0 \iff 3\alpha = 2n$, i.e. $\alpha = \frac{2n}{3}$.
Finally, the hint will work in a case when $n$ is divisible by $3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Question about $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$ $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$
Let $x=t^6$, then it becomes $\lim\frac{t^2 - (\sqrt[6]{2})^2}{t^3 - (\sqrt[6]{2})^3}$=$\lim\frac{(t+\sqrt[6]{2})(t-\sqrt[6]{2})}{(t-\sqrt[6]{2})(t^2+t\sqrt[6]{2}+(\sqrt[6]{2})^2}$=$\lim\frac{(t+\sqrt[6]{2})}{(t^2+t\sqrt[6]{2}+(\sqrt[6]{2})^2}$
But when $x\to 2$, t can go to $\sqrt[6]{2}$ or -$\sqrt[6]{2}$, which gives two different limits, where I was wrong? Thanks!
| You can only use $t$ going to $\sqrt[6]{2}$. This is because you have $t^3 = \sqrt{x} \gt 0$ for $x \to 2$, but $t = -\sqrt[6]{2}$ means $t^3 \lt 0$.
| {
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Find $f^{(80)}(27)$ where $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$
Suppose that $f(x)=(x+3)^{\frac{1}{3}}\cdot (x-27)$. Use a Taylor series expansion to find $f^{(80)}(27)$.
I tried the following:
\begin{align}
f'(x) &= (x+3)^{\frac{1}{3}}\cdot 1+(x-27)\cdot \frac{1}{3}(x+3)^{\frac{-2}{3}}\\
%
f''(x)
&= \frac{1}{3}(x+3)^{-\frac{2}{3}}+\frac{1}{3}(x-27)\cdot -\frac{2}{3}(x+3)^{-\frac{5}{3}}+\frac{1}{3}(x+3)^{-\frac{2}{3}} \\
&= \frac{2}{3}(x+3)^{-\frac{2}{3}}-\frac{1\cdot 2}{3^2}(x+3)^{-\frac{5}{3}}(x-27)\\
%
f'''(x) &= -\frac{2^2}{3^3}(x+3)^{-\frac{5}{3}}-\frac{1\cdot 2}{3^2}(x+3)^{-\frac{5}{3}}-\frac{1\cdot 2\cdot 5}{3^3}(x+3)^{-\frac{8}{3}}
\end{align}
This is very nasty, please help me to solve in some easy way.
| We first expand $f(x)$ at $x=27$. Because it already has a factor $(x-27)$,we just need to expand $(x+3)^{1/3}:=g(x)$.
Do the Taylor expansion:
$$
g(x) = g(27)+g'(27)(x-27)+\cdots+\frac{1}{79!}g^{(79)}(27)(x-27)^{79}+o((x-27)^{79})
$$
Then the Taylor expansion of $f(x)$ should be (according to the uniqueness of Taylor series):
$$
f(x) = g(27)(x-27)+g'(27)(x-27)^2+\cdots+\frac{1}{79!}g^{(79)}(27)(x-27)^{80}+o((x-27)^{80})
$$
So $f^{(80)}(27)= 80!\times\frac{1}{79!}g^{(79)}(27)=80g^{(79)}(27)$
And
$$
g^{(79)}(27) = \frac{1}{3}(\frac{1}{3}-1)\cdots(\frac{1}{3}-78)(27+3)^{\frac{1}{3}-79}
$$
Then it's solved. You can calculate it.
| {
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Calculating determinant of a symmetric matrix where the $k$th row is given by $[a_{k-1},a_k,...,a_0,a_1,...,a_{n-(k-1)}]$
For $j = 0,...,n$ set $a_{j} = a_{0} + jd$, where $a_{0}, d$ are fixed real numbers. Calculate the determinant of the $(n+1)\times (n+1)$ matrix
$$A = \begin{pmatrix}
a_{0} & a_{1} & a_{2} & \dots & a_{n}\\
a_{1} & a_{0} & a_{1} & \dots & a_{n-1}\\
a_{2} & a_{1} & a_{0} & \dots & a_{n-2}\\
\ldots & \ldots & \ldots & \ldots & \ldots\\
a_{n} & a_{n-1} & a_{n-2} & \dots & a_{0}
\end{pmatrix}.$$
How to calculate that? I haven't found any property of determinant of symmetric matrix which could help.
I've tried to use Gaussian elimination (subtracting each row from the row above it), but it didn't work
Gaussian elimination(subtracting each row from the row above it) brings to the matrix:
$$\begin{pmatrix}
-d & d & d & ... & d\\
-d & -d & d & ... & d\\
-d & -d & -d & .... & d\\
\ldots & \ldots & \ldots & \ldots & \ldots\\
a_{n} & a_{n-1} & a_{n-2} & ... & a_{0}
\end{pmatrix} = d^{n-1} \cdot \begin{pmatrix}
-1 & 1 & 1 & ... & 1\\
-1 & -1 & 1 & ... & 1\\
-1 & -1 & -1 & .... & 1\\
\ldots & \ldots & \ldots & \ldots & \ldots\\
a_{n} & a_{n-1} & a_{n-2} & ... & a_{0}
\end{pmatrix}$$
| Subtracting from each row the one above it, we shall obtain
$$ \begin{pmatrix}
a_{0} & a_{1} & a_{2} & ... & a_{n}\\
d & -d & -d & ... & -d\\
d & d & -d & .... & -d\\
\ldots & \ldots & \ldots & \ldots & \ldots\\
d & d & d & ... & -d
\end{pmatrix}$$
Now, subtracting from each column the one before it, we shall obtain
$$ \begin{pmatrix}
a_{0} & d & d & ... & d\\
d & -2d & 0 & ... & 0\\
d & 0 & -2d & .... & 0\\
\ldots & \ldots & \ldots & \ldots & \ldots\\
d & 0 & 0 & ... & -2d
\end{pmatrix}$$
Next, multiplying columns 2,3,$\ldots$,n+1 by $\frac{1}{2}$ and adding to the first we obtain:
$$ \begin{pmatrix}
a_{0}+\frac{nd}{2} & d & d & ... & d\\
0 & -2d & 0 & ... & 0\\
0 & 0 & -2d & .... & 0\\
\ldots & \ldots & \ldots & \ldots & \ldots\\
0 & 0 & 0 & ... & -2d
\end{pmatrix}$$
Finally, these row and column operations don't change the determinat of your matrix. Therefore $\det(A)=(a_o+\frac{nd}{2})(-2d)^n$.
| {
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} |
Using that $1 + z + z^{2} + ... + z^{n} = \frac{1-z^{n+1}}{1-z}$ and taking the real parts, prove that: $$ 1 + \cos \theta + \cos2\theta + ... + \cos n\theta = \frac12+\frac{\sin[(n + \frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})} $$
for $0 < \theta < 2\pi$.
Alright. What I have done is this, using the De Moivre's Formula:
$$ 1 + \cos \theta + \cos2\theta + ... + \cos n\theta = \operatorname{Re}(1 + (\cos\theta + i\sin\theta) + (\cos2\theta + i\sin2\theta) + ... + (\cos n\theta + i \sin n \theta))$$
That is equivalent to
$$ \operatorname{Re}(1 + e^{i\theta} + e^{2i\theta} + ... e^{ni\theta}) = \operatorname{Re} \biggl(\frac{1 - e^{(n+1)i\theta}}{1 - e^{i\theta}}\biggr)$$
I've reached to this point, but now I don't know what to do. Any hint or idea?
| $$ 1 + \cos \theta + \cos2\theta + ... + \cos n\theta = \frac{\sin[(n + \frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})} $$
Using De moivre's theorem you arrived that
$$ \operatorname{Re}(1 + e^{i\theta} + e^{2i\theta} + ... e^{ni\theta}) = \operatorname{Re} \biggl(\frac{1 - e^{(n+1)i\theta}}{1 - e^{i\theta}}\biggr)$$
Now will reverse-engineer the process back to trigonomety using
$$e^{ix} = \cos(x)+i\sin(x)$$
$$\operatorname{Re} \biggl( \frac{ 1 -( \cos{(n+1)\theta}+i\sin{(n+1)\theta} )}{ 1 - ( \cos{\theta}+i\sin{\theta} ) }\biggr)$$
$$\frac{ 1 - \cos{(n+1)\theta}-i\sin{(n+1)\theta}}{ 1 - \cos{\theta}- i\sin{\theta} }$$
We would now multiply the numerator and the denominator by the conjugate of the denominator $1 - \cos{\theta} + i\sin{\theta}$
$$\frac{ 1 - \cos{(n+1)\theta}-i\sin{(n+1)\theta} }{ 1 - \cos{\theta}- i\sin{\theta} }\cdot\frac{1 - \cos{\theta} + i\sin{\theta}}{1 - \cos{\theta} + i\sin{\theta}}$$
$$\frac{ (1 - \cos{(n+1)\theta}-i\sin{(n+1)\theta})\cdot(1 - \cos{\theta} + i\sin{\theta} )}{ (1 - \cos{\theta}- i\sin{\theta})\cdot(1 - \cos{\theta} + i\sin{\theta}) }$$
$$\frac{ (1 - \cos{(n+1)\theta}-i\sin{(n+1)\theta})\cdot(1 - \cos{\theta} + i\sin{\theta} )}{ (1 - \cos{\theta} )^2 - (i\sin{\theta})^2 }$$
Now since the denominator is off imaginary number, it's easy to equate $\mathbb{R}$
$$\frac{ (1 - \cos{(n+1)\theta}-i\sin{(n+1)\theta})\cdot(1 - \cos{\theta} + i\sin{\theta} )}{ (1 - \cos{\theta} )^2 + (\sin{\theta})^2 }$$
$$\frac{ (\sin{\theta}*\sin{(n+1)\theta}+\cos{\theta}*\cos{(n+1)\theta}-\cos{(n+1)\theta}-\cos{\theta}+1) + i\cdot(\cos{\theta}*\sin{(n+1)\theta}-\sin{(n+1)\theta}-\sin{\theta}*\cos{(n+1)\theta}+\sin{\theta}) }{ (1 - \cos{\theta} )^2 + (\sin{\theta})^2 }$$
so when we take away imaginary the expression becomes
$$\frac{ (\sin{\theta}*\sin{(n+1)\theta}+\cos{\theta}*\cos{(n+1)\theta}-\cos{(n+1)\theta}-\cos{\theta}+1) }{ (1 - \cos{\theta} )^2 + (\sin{\theta})^2 }$$
let's simplify further
$$\frac{-1\cos{(n+1)\theta}+\cos{n\theta}-\cos{\theta}+1 }{ 1-2\cos{\theta}+(\cos{\theta} )^2 + (\sin{\theta})^2 }$$
$$\frac{ -1\cos{(n+1)\theta}+\cos{n\theta}-\cos{\theta}+1 }{ 2 -2\cos{\theta} }$$
We still simplify further to reduce this
it turns out that $- H = \frac{1}{2} + \frac{\sin{(n+\frac{1}{2})\theta} }{2\sin{\frac{\theta}{2}}}$, and there was no mistake in my calculations
$$H = \frac{ -1\cos{(n+1)\theta}+\cos{n\theta}-\cos{\theta}+1 }{ 2 -2\cos{\theta} }$$
Proof of error, remember $$ 1 + \cos \theta + \cos2\theta + ... + \cos n\theta = \frac{\sin[(n + \frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})} $$, say $n=1$
But $$ 1+\cos{\theta} ≠ \frac{\sin[(n + \frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})}$$, for every $\theta$ it doesn't turn out to be equal because it is with a displacement of $\frac{1}{2}$
So that $$ 1 + \cos \theta + \cos2\theta + ... + \cos n\theta = \frac{1}{2} + \frac{\sin[(n + \frac{1}{2})\theta]}{2\sin(\frac{\theta}{2})} $$
Check $\theta$ and $n$ therein
| {
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} |
(Different) derivatives of $f(x) = \arcsin\left(\left(5 x + 12 \sqrt{1-x^2}\right)/13\right)$ via two different substitutions? We have been given a function to differentiate:
$$f(x) = \arcsin \left(\frac{5x + 12\sqrt{1-x^2}}{13}\right)$$
My teacher told me the method to substitute $ x= \sin\vartheta$ which would simplify the argument of $\arcsin$ to $$\frac{5}{13}\sin \vartheta + \frac{12}{13}\cos \vartheta $$ and further into $\sin(\vartheta + \alpha)$ where $ \alpha = \arctan\left(\frac{12}{5}\right)$ therefore the function gets reduced into $f(x) = \arctan\left(\frac{12}{5}\right) + \arcsin(x)$ giving $$f'(x) = \frac{1}{\sqrt{1-x^2}}$$. However when I substitute $x = \cos\vartheta $, the argument reduces to : $$\frac{5}{13}\cos\vartheta + \frac{12}{13}\sin\vartheta$$ and further into $\sin(\alpha + \vartheta)$ where $\alpha = \arctan\left(\frac{5}{12}\right)$ but this time the function gets reduced to $$f(x) = \arctan\left(\frac{5}{12}\right) + \arccos(x)$$
thus $$f'(x) = \frac{-1}{\sqrt{1-x^2}}$$ Hence we obtain two different derivatives for the same function and I cannot understand why. I tried plotting the argument and different simplification and got that they are not always equal but cannot figure out the reason.
| HINT.-$(5,12,13)$ is a Pythagorean triple so what is the relation between $\arctan\left(\frac{12}{5}\right)$ and $\arctan\left(\frac{5}{12}\right)$?
| {
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Proof existence of $\int_0^\infty \frac{1}{\sqrt{x}} \frac{1}{1+x^2} \,\mathrm dx$ - is this proof correct? I have to show that the integral
$$ \int_0^\infty \frac{1}{\sqrt{x}} \frac{1}{1+x^2} \,\mathrm dx $$
exists.
My approach: Let $b > 1$, then the integral
$$ \int_1^b \frac{1}{\sqrt x} \frac{1}{1+x^2} \,\mathrm dx $$
exists because
$$
0 < \int_1^b \frac{1}{\sqrt x} \underbrace{\frac{1}{1+x^2}}_{\leq \frac{1}{x^2}} \,\mathrm dx \leq \int_1^b \frac{1}{x^{5/2}} \,\mathrm dx = \left.-\frac{2}{3} x^{-3/2} \right|_1^b = \underbrace{-\frac{2}{3} b^{-3/2}}_{\xrightarrow{b \to \infty}{0}} + \frac{2}{3} \xrightarrow{b \to \infty} \frac{2}{3}.
$$
Let $0 < \epsilon < 1$, then
$$ \int_\epsilon^1 \frac{1}{\sqrt x} \underbrace{\frac{1}{1+x^2}}_{<1} \,\mathrm dx < \int_\epsilon^1 \frac{1}{\sqrt x} \,\mathrm dx = \left.2 \sqrt x \right|_\epsilon^1 = 2(1 - \sqrt \epsilon) \xrightarrow{\epsilon \searrow 0} 2 $$
and in total
$$ 0 < \int_0^\infty \frac{1}{\sqrt x} \frac{1}{1+x^2} \,\mathrm dx = \underbrace{\int_0^1 \frac{1}{\sqrt x} \frac{1}{1+x^2} \,\mathrm dx}_{\leq 2} + \underbrace{\int_1^\infty \frac{1}{\sqrt{x}} \frac{1}{1+x^2} \,\mathrm dx}_{\leq \frac{2}{3}} \leq 2 + \frac{2}{3}. $$
Is this a proper proof?
| Looks fine, here is another approach :
\begin{aligned}&\bullet \ f:x\mapsto\frac{1}{\sqrt{x}\left(1+x\right)}\textrm{ is continuous on }\left(0,+\infty\right)\cdot\\ &\bullet \ \lim_{x\to 0}{\frac{x^{\frac{3}{4}}}{\sqrt{x}\left(1+x\right)}}=0\textrm{, and thus }f\left(x\right)=\underset{\overset{x\to 0}{}}{\mathcal{o}}\left(x^{-\frac{3}{4}}\right)\textrm{, and }x\mapsto x^{-\frac{3}{4}}\textrm{ is integrable on }\left(0,a\right]\textrm{ for}\\ &\textrm{any }a\in\left(0,+\infty\right) \cdot\\ &\bullet \ \lim_{x\to +\infty}{\frac{x^{\frac{3}{2}}}{\sqrt{x}\left(1+x\right)}}=0\textrm{, thus }f\left(x\right)=\underset{\overset{x\to +\infty}{}}{\mathcal{o}}\left(x^{-\frac{3}{2}}\right)\textrm{, and }x\mapsto x^{-\frac{3}{2}}\textrm{ is integrable on }\left[a,+\infty\right)\\ &\textrm{for any }a\in\left(0,+\infty\right) \cdot \end{aligned}
Thus, our $ f $ is integrable on $ \left(0,+\infty\right)\cdot $
| {
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Prove that $\lim_{n\to\infty}a_n=\frac{\sum_{i=1}^k2ia_i}{k(k+1)}$
Given that a sequence $(a_n)$ satisfies $a_{n+k} = \dfrac{a_n + a_{n+1} + \cdots + a_{n+k-1}}{k}$ for $n\geq 1,$ where $k\in\mathbb{N},$ prove that $\lim\limits_{n\to\infty} a_n = \dfrac{2a_1}{k(k+1)}+\dfrac{4a_2}{k(k+1)}+\cdots +\dfrac{(2k)a_k}{k(k+1)}.$
I am not sure how to go about doing this. However, I understand why this might be the case. Informally, the $j$th term, where $1\leq j\leq k$ eventually occurs $j$ times more often than the first term. Also, when the first $k$ terms are $1,$ all the terms in the sequence are one, as evidenced by the proof below.
Proof: We proceed by strong induction. Suppose the first $k$ terms are $1.$ Then the $(k+1)$th term is also $1$ (since it is the average of the first $k$ terms). Assume that the first $k+m$ terms are $1$ for some $m\in\mathbb{N}.$ Then we have $a_{m+1} = \cdots = a_{k+m} = 1$ and $a_{k+m+1} = \dfrac{a_{m+1} + \cdots + a_{k+m}}k = 1,$ so the first $k+m+1$ terms are $1.$ Hence by strong induction, all the terms of the sequence are $1$ if the first $k$ terms are $1.$
Since the $j$th term, $1\leq j\leq k$ occurs $j$ times as often as the first term, we have that $\dfrac{1}a + \dfrac{2}a + \cdots + \dfrac{k}a = 1,$ where $\dfrac{1}a$ is the coefficient of $a_1$ in the limit of the sequence. Hence $\dfrac{k(k+1)}{2a} = 1\Rightarrow a = \dfrac{k(k+1)}2\Rightarrow \lim\limits_{n\to\infty} a_n = \dfrac{a_1}a + \dfrac{2a_2}a +\cdots + \dfrac{ka_k}a = \dfrac{2a_1}{k(k+1)} + \cdots + \dfrac{(2k)a_k}{k(k+1)}.$
Obviously the above reasoning is too informal to be considered a proof, so I am not sure how to prove this. If I could prove that the $j$th term eventually occurs $j$ times as often as the first term in the limit, it would be enough to prove this.
| Based on the comment by Paramanand Singh, we see that we have the relation
$kx_{n+k} = x_{n+k-1} + x_{n+k-2} + \cdots + x_n.$
Hence $kx_{n+k}+(k-1)x_{n+k-1} +\cdots + 2x_{n+2} +x_{n+1} = kx_{n+k-1}+(k-1)x_{n+k-2}+\cdots + 2x_{n+1}+x_n.$
Repeating this process $n-1$ more times, we see that $kx_{n+k}+(k-1)x_{n+k-1}+\cdots + 2x_{n+2}+x_{n+1}=kx_k+(k-1)x_{k-1}+\cdots + 2x_2+x_1\,\forall n\geq 0\tag{1}$ Hence we have that \begin{align*}\lim\limits_{n\to\infty} kx_{n+k} + (k-1)x_{n+k-1} + \cdots + x_n &= (k+(k-1)+\cdots + 1)\lim\limits_{n\to\infty} x_n \\
&= \dfrac{k(k+1)}2 \lim\limits_{n\to\infty} x_n\tag{2}\end{align*}
as $\lim\limits_{n\to\infty} x_{n+j} = \lim\limits_{n\to\infty} x_n$ for $0\leq j\leq k$ (observe that since the limit exists, any subsequence converges to that limit, so it doesn't matter whether $k$ is finite).
However, we also have from $(1)$ that $\lim\limits_{n\to\infty} kx_{n+k} + (k-1)x_{n+k-1} + \cdots + x_n = kx_k + (k-1)x_{k-1}+\cdots + x_1,$ so $(2)$ gives that $\lim\limits_{n\to\infty} x_n = \dfrac{\sum_{j=1}^k 2jx_j}{k(k+1)},$ as required.
| {
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What is $\frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$ for $x \rightarrow 0$? I am trying to find the limit
$\lim_{x \rightarrow0} \frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$
for $a,b \in \rm{I\!R}_{+}$. Applying L'Hospital's rule leads to
$\lim_{x \rightarrow0}\frac{\cos x \cdot \sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}{-2 \sin x \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x}) + \frac{\sin x \cdot (a^2 + b^2 + 2 ab \cos x )}{\sqrt{a^2+b^2+ 2 ab \cos x}} }$.
However, this remains with both, cosine and sine. Maybe one could use a trigonometric identity, which I cannot find.
| Let us denote by $\pi - x$ the angle between the sides $a$ and $b$ from a triangle with third side $c(x)$, the application of the cosine's law results into
\begin{align*}
c^{2}(x) = a^{2} + b^{2} - 2ab\cos(\pi - x)
\end{align*}
Thus the given expression can be rewrriten as
\begin{align*}
f(x) = \frac{ab\sin(x)}{2\sqrt{c^{2}(x)(a + b - \sqrt{c^{2}(x)})}} = \frac{ab\sin(x)}{2c(x)\sqrt{a+b-c(x)}}
\end{align*}
Moreover, the expression $ab\sin(x)/2$ is the area of such triangle.
According to the Heron's formula, we have that
\begin{align*}
A = \frac{1}{4}\sqrt{(a+b+c)(a + b - c)(a - b + c)(-a + b + c)}
\end{align*}
Consequently, one has that
\begin{align*}
f(x) & = \frac{ab\sin(x)\sqrt{(a+b+c(x))(a - b + c(x))(-a + b + c(x))}}{2c(x)\sqrt{(a+b+c(x))(a + b - c(x))(a - b + c(x))(-a + b + c(x))}}\\\\
& = \frac{\sqrt{(a+b+c(x))(a - b + c(x))(-a + b + c(x))}}{4c(x)}
\end{align*}
Therefore we have that
\begin{align*}
\lim_{x\rightarrow 0}f(x) & = \frac{\sqrt{(a+b+c(0))(a-b+c(0))(-a+b+c(0))}}{4c(0)} = \sqrt{\frac{ab}{2(a+b)}}
\end{align*}
and we are done.
| {
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Determine the probability that the chosen coin is coin $2.$
Coin $1$ is a fair coin and coin $2$ is an unfair coin such that the probability of getting heads is $0.6.$ One of the coins is chosen at random and flipped repeatedly until the first head is obtained. Suppose that the first head is observed in the fifth flip. What is the probability that coin $2$ was taken?
Suppose $A$ is the event that indicates coin $1$ was taken and $B$ is the event that indicates coin $2$ was taken. Let $X$ be the random variable that indicates number of flips required to get the first head. Since one of the coins was taken at random so $P(A) = P(B) = \frac 1 2.$ Also by that given condition $P(X=5 \mid B) = (0.4)^4 (0.6).$ So $P(X=5 \mid B^c) = P(X=5 \mid A) = (0.5)^5.$ We need to find out $P(B \mid X=5).$ Now by Bayes' theorem \begin{align*} P(B \mid X=5) & = \frac {P(X=5 \mid B) P(B)} {P(X=5 \mid B) P(B) + P(X=5 \mid B^c) P(B^c)} \\ & = \frac {(0.4)^4 \times (0.6) \times (0.5)} {(0.4)^4 \times (0.6) \times (0.5) + (0.5)^5 \times (0.5)} \\ & = \frac {(0.4)^4 \times (0.6)} {(0.4)^4 \times (0.6) + (0.5)^5} \\ & \simeq 0.33. \end{align*}
Am I correct? Please check my solution.
Thanks in advance.
| Yes, your approach and your result are correct.
| {
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Prove that if $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ then $ \frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1 $ Question -
Let $a, b, c$ be positive real numbers such that $\sqrt{a}+\sqrt{b}+\sqrt{c}=3$ . Prove that
$$
\frac{a^{2}}{a+2 b^{2}}+\frac{b^{2}}{b+2 c^{2}}+\frac{c^{2}}{c+2 a^{2}} \geq 1
$$
My try
$$
\frac{a^{2}}{a+2 b^{2}}=a-\frac{2 a b^{2}}{a+2 b^{2}} \geq a-\frac{2 a b^{2}}{3 \sqrt[3]{a b^{4}}}=a-\frac{2(a b)^{2 / 3}}{3}
$$
which implies that
$$
\sum_{c y c} \frac{a^{2}}{a+2 b^{2}} \geq \sum_{c y c} a-\frac{2}{3} \sum_{c y c}(a b)^{\frac{2}{3}}
$$
It suffices to prove that
$$
(a b)^{2 / 3}+( b c)^{2 / 3}+\left (c a)^{2 / 3} \leq 3\right.
$$
beacuse we can easily get that $\sum a \ge 3$
but i am not able to prove it..
note that we have to prove this using only am-gm or weighted am-gm or power mean or any kind of means inequality because author did not introduce any advance inequality yet...
any hints ???
thankyou
| It is enough to show that $\sqrt a + \sqrt b + \sqrt c = 3 \implies (ab)^{2/3} + (bc)^{2/3} + (ca)^{2/3} \leqslant 3$. For ease, let us replace $a, b, c$ with $x^6, y^6, z^6$, so we need to show for positives, $x^3+y^3+z^3=3 \implies (xy)^4+(yz)^4+(zx)^4\leqslant 3$. This one is in fact a known old chestnut.
We note by AM-GM, $xy \leqslant \frac13(x^3+y^3+1) = \frac13(4-z^3)$, hence $(xy)^4 \leqslant \frac13(4x^3y^3-x^3y^3z^3)$. Cyclically summing three such inequalities, we get
$$\sum (xy)^4 \leqslant \frac43\sum (xy)^3-(xyz)^3$$
Now with $X=x^3, Y = y^3, Z = z^3$, it is enough to show with $X+Y+Z=3$
$$4(XY+YZ+ZX) -3XYZ\leqslant 9$$
which is the well known Schur's inequality:
$$4(X+Y+Z)(XY+YZ+ZX) \leqslant (X+Y+Z)^3+9XYZ$$
| {
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"source": "stackexchange",
"question_score": "3",
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Proof $\frac{-\sqrt{2 - \sqrt{2 - \sqrt{2}}}}{\sqrt2} + \frac{\sqrt{2 + \sqrt2}}{\sqrt2 \sqrt{2 - \sqrt{2 - \sqrt2}}} =\sqrt{2-\sqrt{2+\sqrt2}}$ I want to prove the following equation
$$\frac{-\sqrt{2 - \sqrt{2 - \sqrt{2}}}}{\sqrt{2}} + \frac{\sqrt{2 + \sqrt{2}}}{\sqrt{2} \sqrt{2 - \sqrt{2 - \sqrt{2}}}}$$
I know this must be equal to $\sqrt{2-\sqrt{2+\sqrt{2}}}$, but I find it pretty hard to prove it manually, without requiring a calculator nor estimates (since the expression should be equal to one of the roots of the polinomyal $f(x)=x^8-8x^6+20x^4-16x^2+2$, and all of them are real and distinct).
Thanks in advance for your help.
| Note that the second term of the LHS can be expressed as
$$\begin{align}
\frac{\sqrt{2 + \sqrt{2} }}{\sqrt{2} \sqrt{2 - \sqrt{2 - \sqrt{2}}}}
=\frac{\sqrt{2 + \sqrt{2 - \sqrt{2}}}}
{\sqrt{2}}
\end{align}$$
which can be verified by cross multiply. Then,
$$LHS = \sqrt{ \frac{2 + \sqrt{2 - \sqrt{2}}} {2} }
- \sqrt{ \frac{2 - \sqrt{2 - \sqrt{2}}} {2} }=\sqrt{2-\sqrt{2+\sqrt{2}}}=RHS $$
where the denest formula below is used
$$\sqrt{a-\sqrt c} =\sqrt{\frac{a+\sqrt {a^2-c}}{2}} -\sqrt{\frac{a-\sqrt {a^2-c}}{2}} $$
with $a=2$ and $c=2+\sqrt2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root in terms of $\alpha$ Consider the polynomial $x^3+2x^2-5x+1$ with roots $\alpha$ and $\alpha^2+2\alpha-4$. Find the third root $\beta$ in terms of $\alpha$
I have that $\alpha^3+2\alpha^2-5\alpha+1 = 0$, so $\alpha^3 = -2\alpha^2+5\alpha -1$.
And, $(\alpha^2+2\alpha-4)^3+2(\alpha^2+2\alpha-4)^2-5(\alpha^2+2\alpha-4)+1=0$ gives $\alpha^6+6\alpha^5+2\alpha^4-13\alpha^2+54\alpha-11=0$
Additionally, $\alpha^6 = (-2\alpha^2+5\alpha -1)^2 = 4\alpha^4-20\alpha^3+29\alpha^2-10\alpha+1$
I do not know how to move forward from here. I tried setting $f(x)$ equal to the product of the roots and expanding that out to a 14-term polynomial with $\alpha$ and $\beta$ coefficients but that seems unproductive.
| Given three roots $\alpha, \beta, \gamma$ of a polynomial, it can generally be written as
$$(x-\alpha)(x-\beta)(x-\gamma)=0$$
Note how each of the roots contribute in making the equality work. On expanding the braces,
$$(x-\alpha)(x-\beta)(x-\gamma)=0$$
$$[(x^2-x(\beta)-x(\alpha)+(\alpha\beta)](x-\gamma)=0$$
$$(x^3-x^2(\gamma)-x^2(\beta)+x(\beta\gamma)-x^2(\alpha)+x(\alpha\gamma)+x(\alpha\beta)-(\alpha\beta\gamma)=0$$
$$x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\alpha\gamma)x -(\alpha\beta\gamma)=0$$
Note that if any given equation is scaled down in such a way that the coefficient of $x^3$ is $1$ then the coefficient of $x^2$ gives the negative of sum of roots. More precisely, sum of roots of cubic equation = $-$(coefficient of $x^2$)/(coefficient of $ x^3$)
I hope you can take over from here. As pointed out already sum of roots is $-2$
Bonus: It can also be seen that product of roots of any given cubic equation is equal to the negative of the constant term divided by coefficient of $x^3$ and the sum of product of roots taken two at a time is, well, (coefficient of $x$)/(coefficient of $x^3$)
These are enough conditions on the given roots to get you started.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that for all positives $a, b$ and $c$, $(\sum_{cyc}\frac{c + a}{b})^2 \ge 4(\sum_{cyc}ca)(\sum_{cyc}\frac{1}{b^2})$.
Prove that for all positives $a, b$ and $c$, $$\left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)^2 \ge 4(bc + ca + ab) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
Let $ca + ab = m$, $ab + bc = n$ and $bc + ca = p$, we have that $$\left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2}\right)^2 \ge 2(m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
$$\iff \left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2} - 1\right)^2 \ge 2 \cdot \sum_{cyc((m, n, p), (a, b, c))}\left[n \cdot \left(\frac{1}{c^2} + \frac{1}{a^2}\right)\right] + 1$$
Expanding $\displaystyle \sum_{cyc((m, n, p), (a, b, c))}\left[n \cdot \left(\frac{1}{c^2} + \frac{1}{a^2}\right)\right]$ gives $$2 \cdot \sum_{cyc}\frac{ca}{b^2} + \left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)$$
Let $\dfrac{b + c}{a} = m'$, $\dfrac{c + a}{b} = n'$ and $\dfrac{a + b}{c} = p'$, we have that $$(m' + n' + p' - 1)^2 \ge 2 \cdot \left[2 \cdot \sum_{cyc}\frac{ca}{b^2} + (m' + n' + p')\right] + 1$$
Moreover, $$(m')^2 + (n')^2 + (p')^2 = \sum_{cyc}\left[\left(\frac{c + a}{b}\right)^2\right] \ge 2 \cdot \sum_{cyc}\frac{ca}{b^2}$$
$$\implies (m' + n' + p' - 1)^2 \ge 2 \cdot \left[(m')^2 + (n')^2 + (p')^2 + m' + n' + p'\right] + 1$$
$$\iff -[(m')^2 + (n')^2 + (p')^2] + 2(m'n' + n'p' + p'm') - 4(m' + n' + p') \ge 0$$, which is definitely not correct.
Another attempt, let $(0 <) \ a \le b \le c \implies ab \le ca \le bc \iff ca + ab \le ab + bc \le bc + ca$
$\iff m \le n \le p$ and $a^2 \le b^2 \le c^2 \iff \dfrac{1}{a^2} \ge \dfrac{1}{b^2} \ge \dfrac{1}{c^2}$.
By the Chebyshev inequality, we have that $$3 \cdot \left(\frac{m}{a^2} + \frac{n}{b^2} + \frac{p}{c^2}\right) \le (m + n + p) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\right)$$
Any help would be appreciated.
| Let $a+b+c=x(a^2+b^2+c^2).$
Thus, $$\sum_{cyc}(a-xa^2)^2\geq0=\left(\sum_{cyc}(a-xa^2)\right)^2$$ or
$$\sum_{cyc}(a^2-2xa^3+x^2a^4)\geq\sum_{cyc}(a^2+2ab)-2x\sum_{cyc}(a^3+a^2b+a^2c)+x^2\sum_{cyc}(a^4+2a^2b^2)$$ or
$$\sum_{cyc}ab-x\sum_{cyc}(a^2b+a^2c)+x^2\sum_{cyc}a^2b^2\leq0.$$
Now, let $a^2b^2+a^2c^2+b^2c^2=A$, $\sum\limits_{cyc}ab(a+b)=B$ and $ab+ac+bc=C$.
We proved that there is $x$, id est, $x=\frac{a+b+c}{a^2+b^2+c^2}$ for which $A>0$ and $$Ax^2-Bx+C\leq0,$$ which gives $$\Delta\geq0$$ or
$$B^2-4AC\geq0$$ or
$$\left(\sum_{cyc}(a^2b+a^2c)\right)^2-4\sum_{cyc}ab\sum_{cyc}a^2b^2\geq0,$$ which is your inequality exactly.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Proving two binomial identities I would like to show that
\begin{align}
&\sum_{j=n-k}^n\binom nj(1-x)^{n-j-1}x^{j-1}(j-nx)\\
&\qquad=\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\sum_{k=0}^{n-1}\frac{(-1)^k}n\binom{n-1}k\binom n{n-k}(n-k)(1-x)^kx^{n-k-1}\\
&\qquad=(-1)^{n-1}\sum_{k=0}^{n-1}\binom{n-1}k\binom{n+k-1}k(-x)^k
\end{align}
I feel I exhausted all identities/properties of binomials without success. Mathematica says it is true, but how to show it?
| The first Identity follows by induction. When $k=0$
\begin{eqnarray*}
\binom{n}{n} \frac{x^{n-1}}{1-x} (n-nx) = \binom{n}{n} nx^{n-1}.
\end{eqnarray*}
Now assume the sum for $k$ and add the $(k+1)^{th}$ term ($j=n-k-1$)
\begin{eqnarray*}
\binom{n}{n-k}(n-k)(1-x)^kx^{n-k-1}+\binom{n}{n-k-1}(1-x)^{k}x^{n-k-2}(n-k-1-nx) =
\binom{n}{n-k-1}x^{n-k-2}(1-x)^k \left( (k+1)x +(n-k-1)-nx \right) =\binom{n}{n-k-1}(n-k-1)(1-x)^{k+1}x^{n-k-2}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3650275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
n-th power of a matrix using the division of polynomials. Consider the matrix
$$
A=\begin{pmatrix}
0 & 0 & 0\\
-2 & 1 & -1\\
2 & 0 & 2
\end{pmatrix}
$$
*
*Calculate $A^3-3A^2+2A$.
*What is the remainder of the division of the polynomial $X^n$ by the polynomial $X^3-3X^2+2X$.
*Calculate $A^n$ for every natural number $n$.
I was solving the following problem and I was stuck in it. For part 1) the answer was the zero matrix. In part 2) I use the usual division and i get the following
$$
X^n=X^{n-3}(X^3-3X^2+2X)+3X^{n-1}-2X^{n-2}.
$$
When I pass to part 3) and using part 1) and 2), we obtain
$$
A^n=3A^{n-1}-2A^{n-2}.
$$
Using the fact that $A^3-3A^2+2A=O_{3\times 3}$. but if I use this answer for calculating $A^2$ the answer is not correct, so I think $A^n$ obtained is not correct. Now, one can use the diagonalization of the matrix $A$ and obtain
$$
A^n=\begin{pmatrix}
0 & 0 & 0\\
-2^n & 1 & 1-2^n\\
2^n & 0 & 2^n
\end{pmatrix}
$$
Can you help me in proving part 2 (if not correct) and part 3 without using the diagonalization method.
| You could use induction.
For $n=1$, your statement is true.
Assuming
$$
A^n=\begin{pmatrix}
0 & 0 & 0\\
-2^n & 1 & 1-2^n\\
2^n & 0 & 2^n
\end{pmatrix}
$$
Then
$$
A^{n+1} = AA^n =
\begin{pmatrix}
0 & 0 & 0\\
-2 & 1 & -1\\
2 & 0 & 2
\end{pmatrix}
\begin{pmatrix}
0 & 0 & 0\\
-2^n & 1 & 1-2^n\\
2^n & 0 & 2^n
\end{pmatrix} =
\begin{pmatrix}
0 & 0 & 0\\
-2^n - 2^n & 1 & 1-2^n - 2^n\\
2\cdot2^n & 0 & 2\cdot2^n
\end{pmatrix} =
\begin{pmatrix}
0 & 0 & 0\\
-2^{n+1} & 1 & 1-2^{n+1}\\
2^{n+1} & 0 & 2^{n+1}
\end{pmatrix}
$$
So
$$
A^n = \begin{pmatrix}
0 & 0 & 0\\
-2^n & 1 & 1-2^n\\
2^n & 0 & 2^n
\end{pmatrix}
$$
Is true for every natural number $n$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Cauchy-Schwarz inequality problem with four variables Let a,b,c,d, are positive numbers. How can I prove
$a/(b+c) + b/(c+d) + c/(d+a) + d/(a+b) \ge 2$
I am unable to find a transformation which will lead me to the result. Any help would be very much appreciated.
I was proceeding as follows:
$E = a/(b+c) + b/(c+d) + c/(d+a) + d/(a+b) = a^2/(ab+ac) + b^2/(bc+bd) + c^2/(cd+ca) + d^2/(ad+bd)$
then I was using CS as:
$ E > (a+b+c+d)^2/( ab+ + bc + cd + ad + 2ac + 2bd)$
After this step I am getting lost because of the non-standard form which I have at the denominator.
I think I got it, now we can write:
$ E > (a+b+c+d)^2/( (a+ c)(b+d) + 2ac + 2bd) =E2$
since $(a+ c)^2 / 2 > 2ac $ and $(b+ d)^2 / 2 > 2bd $
We can write :
$E2 > (a+b+c+d)^2/( (a+ c)(b+d) + (a+ c)^2 / 2 + (b+ d)^2 / 2) =E3$
replacing $x=a+c$ and $y=b+d$ we get
$ E3 = 2 (x+y)^2/( 2xy + x^2+ y^2) = 2$
Thus $E > 2$
| Applying Titu's lemma, it suffices to show that (fill in the slight gap)
$$ (a+b+c+d)^2 \geq 2(ab+ac+bc+bd+cd+ac+da+bd ).$$
This is obviously true by expansion, since it becomes
$$ ( a -c)^2 + (b-d) ^2 \geq 0 .$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving an ode of first order I want to solve this first order ode:
$$ x'(1+t^2) \sin(x)-2t \cos(x)=0 $$
$x(1)= \frac{ \pi}{3} $
I want to use the separation method, so :
$ x' (1+t^2) \sin(x)= 2t \cos(x) $
$\leftrightarrow x'= \frac{2t \cos (x)}{(1+t^2) \sin(x) }$
$ \frac{dx}{dy} = \frac{2t \cos (x)}{(1+t^2) \sin(x) } $
$\leftrightarrow \frac{ \sin(x) }{ \cos(x) } dx= 2t \frac{1}{1+t^2} dt $
so i have to solve:
$ \int \frac{ \sin(x) }{ \cos(x) } dx= \int 2t \frac{1}{1+t^2} dt $
$\leftrightarrow- \log( |cos(x)|) + c_1 = \log(t^2+1)+ c_2 $
from here I am not sure: If I use the exponential functin on both sides, I get:
$ - \cos(x) = t^2+1 $
then what? Can I take $ \arccos $ ? but what is then $ \arccos(cos(x)) $?
appreciate any help solving this ode :-)
| From your last line you get $$\log\left(|\cos x |\cdot(t^2+1)\right)= \log c \hspace{1 cm} \text{(for some c)} \\ \implies \cos x\cdot(t^2+1)= \pm c =C \\ \implies \cos x = \frac{C}{t^2+1}$$
Using $x(1)=\frac{\pi}{3}$, $$\cos x =\frac{1}{t^2+1} \\ x= \cos^{-1}\left( \frac{1}{t^2+1} \right) + 2n\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3662146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability problem on umbrellas
Two absent-minded roommates, mathematicians, forget their umbrellas in some way or another. $A$ always takes his umbrella when he goes out, while $B$ forgets to take his umbrella with probability $1/2$. Each of them forgets his umbrella at a shop with probability $1/4$. After visiting three shops, they return home. Find the probability that they have only one umbrella.
My Attempt
$A_i/\bar{A}_i$ : $A$ remembers or forgets umbrella at shop $i$
$B_i/\bar{B}_i$ : $B$ remembers or forgets umbrella at shop $i$
$B_o/\bar{B}_o$ : $B$ remembers or forgets umbrella at shop home
$A_i,,B_i=3/4;\bar{A}_i,\bar{B}_i=1/4$ and $B_o,\bar{B}_o=1/2$
$$
\text{Req. Prob.}=A_1A_2A_3\bar{B}_o+A_1A_2A_3B_o\bar{B}_1+A_1A_2A_3B_oB_1\bar{B}_2+A_1A_2A_3B_oB_1B_2\bar{B}_3+\bar{A}_1B_oB_1B_2B_3+A_1\bar{A}_2B_oB_1B_2B_3+A_1A_2\bar{A}_3B_oB_1B_2B_3\\
=\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{1}{4}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)\Big(\frac{1}{4}\Big)+\Big(\frac{3}{4}\Big)^3\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^2\Big(\frac{1}{4}\Big)+\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3\\
+\Big(\frac{3}{4}\Big)\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3+\Big(\frac{3}{4}\Big)^2\Big(\frac{1}{4}\Big)\Big(\frac{1}{2}\Big)\Big(\frac{3}{4}\Big)^3\\
=\frac{27}{128}+\frac{27}{128*4}+\frac{81}{128*16}+\frac{243}{128*64}+\frac{27}{128*4}+\frac{81}{128*16}+\frac{243}{128*64}\\
=\frac{3726}{128*64}=\frac{3726}{8192}
$$
But my reference gives the solution $\frac{7278}{8192}$, so what am I missing in my attempt ?
Note: Please check $b$ part of link for a similar attempt to solve the problem to obtain the solution as in y reference.
| You want the probability that only one of the two remembers their umbrella at every point.
At the end of the trip, $A$ will have an umbrella if it is remembered at each of the three stores. This has a probability of $(3/4)^3$
At the end of the trip, $B$ will have an umbrella if it is remembered at home and each of the three stores. This has a probability of $(1/2)(3/4)^3$
So the probability that exactly one has an umbrella at the end is, indeed:$$\begin{align}&\dfrac {3^3}{4^3}\left(1-\dfrac{3^3}{2\cdot4^3}\right)+\left(1-\dfrac{3^3}{4^3}\right)\dfrac{3^3}{2\cdot4^3}\\[1ex]=&\dfrac{3^4\left(2^5-3^2\right)}{2^{12}}\\[1ex]=&\dfrac{1863}{4096}\\[1ex]=&\dfrac{3726}{8192}\end{align}$$
But my reference gives the solution $7278/8192$, so what am I missing in my attempt ?
The probability that at most one has an umbrella by the end is:$$1-\dfrac{3^3}{2\cdot 4^3}\cdot\dfrac{3^3}{4^3}=\dfrac{7278}{8192}$$
The difference being: the probability that they both forget the umbrella.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove this algebraic version of the sine law? How to solve the following problem from Hall and Knight's Higher Algebra?
Suppose that
\begin{align}
a&=zb+yc,\tag{1}\\
b&=xc+za,\tag{2}\\
c&=ya+xb.\tag{3}
\end{align}
Prove that
$$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}.\tag{4}$$
(I suppose that $x,y,z$ are real numbers whose moduli are not equal to $1$.)
I discovered this problem from chapter 3 of Prelude to Mathematics by W. W. Sawyer. Sawyer thought that this problem arose from the sine law: let $a,b,c$ be respectively the lengths of the edges opposite to three vertices $A,B,C$ of a triangle. Define $x=\cos A$ and define $y,z$ analogously. Now equalities $(1)-(3)$ simply relate $a,b$ and $c$ to each other by the cosines of the angles and $(4)$ is just a rewrite of the sine law
$$
\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.
$$
However, the algebraic version $(4)$ looks more general. For example, it does not state that $a,b,c$ must be positive or that they must satisfy the triangle inequality.
Sawyer wrote that this isn't a hard problem, but he didn't provide any solution. I can prove $(4)$ using linear algebra. Suppose that $(a,b,c)\ne(0,0,0)$ (otherwise $(4)$ is obvious). Rewrite $(1)-(3)$ in the form of $M\mathbf a=0$:
$$\begin{bmatrix}-1&z&y\\ z&-1&x\\ y&x&-1\end{bmatrix}\begin{bmatrix}a\\ b\\ c\end{bmatrix}=0.$$
Since $x^2,y^2,z^2\ne1$, $M$ has rank $2$ and $D=\operatorname{adj}(M)$ has rank $1$. Hence all columns of $D$ are parallel to $(a,b,c)^T$ and $\frac{d_{11}}{d_{21}}=\frac{d_{12}}{d_{22}}=\frac{a}{b}$. Since $M$ is symmetric, $D$ is symmetric too. Therefore $\frac{1-x^2}{1-y^2}=\frac{d_{11}}{d_{22}}=\frac{d_{11}d_{12}}{d_{21}d_{22}}=\frac{a^2}{b^2}$, i.e. $\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}$.
As this problem comes from Hall and Knight's book, I think there should be a more elementary solution. Any ideas?
| Let $a=0$.
Thus, $$xc=b$$ and $$xb=c,$$ which gives $$x^2bc=bc$$ or $$(x^2-1)bc=0$$ and since $x^2\neq1,$ we obtain $bc=0$ and from here $$a=b=c=0,$$ which gives that our statement is true.
Let $abc\neq0$.
Thus, $$\frac{zb}{a}+\frac{yc}{a}=1$$ and $$\frac{xc}{b}+\frac{za}{b}=1,$$ which gives
$$z^2+\frac{xyc^2}{ab}+\frac{xzc}{a}+\frac{yzc}{b}=1$$ or
$$\frac{1-z^2}{c^2}=\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ca},$$ which gives
$$\frac{a^2}{1-x^2}=\frac{b^2}{1-y^2}=\frac{c^2}{1-z^2}=\frac{1}{\frac{xy}{ab}+\frac{yz}{bc}+\frac{zx}{ca}}.$$
| {
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"answer_id": 3
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Show that the derivative of a series $\sum_{n=1}^\infty \frac{-2x}{(2+x^2)^{n+1}}$ converges uniformly Consider the derivative series
$$
D = \sum_{n=1}^\infty \frac{-2x}{(2+x^2)^{n+1}}
$$
of
$$
S = \sum_{n=1}^\infty \frac{1}{n} \frac{1}{(2+x^2)^n}
$$
I have shown that $S$ converges by the use of Weiterstrass' M-test by saying that
as $x \in \mathbb{R}$ we must have that
$$
\frac{1}{(2+x^2)^n} \leq \frac{1}{n}
$$
for all $n \in \mathbb{N}$. This means that
$$
\frac{1}{n} \frac{1}{(2+x^2)^n} \leq \frac{1}{n^2}
$$
and from analysis we know that
$$
\sum_{n=1}^\infty \frac{1}{n^2}
$$
converges. But I am struggling to prove that $D$ converges uniformly. Do I have to use the same approach? I found that for $x \in \mathbb{R}$ and for all $n \in \mathbb{N}$ that
$$
\frac{-2x}{(2+x^2)^{n+1}} \leq \frac{1}{2^n}
$$
which is a geometric series that converges. Can I use this? Thanks in advance.
| If $f(x)=\frac{-2x}{(2+x^2)^{n+1}}$, then$$f'(x)=2 \left(x^2+2\right)^{-n-2} \left((2 n+1) x^2-2\right).$$So, the maximum of $|f|$ is attained when $x=\pm\sqrt{\frac2{2n+1}}$ and$$\left|f\left(\pm\sqrt{\frac2{2n+1}}\right)\right|=2 \sqrt{2} \sqrt{\frac{1}{2 n+1}} \left(\frac{2}{2 n+1}+2\right)^{-n-1}.$$Since$$\lim_{n\to\infty}2 \sqrt{2} \sqrt{\frac{1}{2 n+1}} \left(\frac{2}{2 n+1}+2\right)^{-n-1}=0,$$the convergence is uniform by the Weierstrass $M$-test.
| {
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"url": "https://math.stackexchange.com/questions/3676001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving $(a+b+c)^2\prod_{cyc}(a+b)-4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab\geqq 0$ From Mr. Michael Rozenberg solution:
For $a,b,c>0$$,$ prove that$:$
$$(a+b+c)^2\prod_{cyc}(a+b)\geq4\sum_{cyc}(a^2b+a^2c)\sum_{cyc}ab,$$
I found two SOS proof:
1) $$\text{LHS-RHS}={\frac { \left( a-b \right) ^{2}\cdot \text{M}+ab \left( {a}^{2
}-2\,ab+ca+{b}^{2}+bc-2\,{c}^{2} \right) ^{2}}{a+b}}$$
Where $$\text{M}=\left( 2\,ab-ca-bc+{c}^{2} \right) ^
{2}+c \left( -c+a+b \right) ^{2} \left( a+b \right)$$
2) $$\text{LHS-RHS}=c \left( a-b \right) ^{2} \left( a+b-c \right) ^{2}+a \left( b-c \right) ^{2} \left( b+c-a \right) ^{2}+b \left( c-a \right) ^{2} \left( c+a-b \right) ^{2}\geqq 0$$
| I think, the shortest way here it's $uvw$. See here: https://artofproblemsolving.com/community/c6h278791
Indeed, this inequality is a linear inequality of $w^3$ and for the proof it's enough to consider two cases:
*
*$c=0$;
*$b=c$, which very easy to make.
Due to the Nguyenhuyen_AG's post your inequality we can rewrite in the following form.
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0.$ Prove that:
$$\frac{a^2+b^2+c^2}{ab+bc+ca} + \frac{8abc}{(a+b)(b+c)(c+a)} \geqslant 2.$$
There is the following stronger inequality.
Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0.$ Prove that:
$$\frac{a^2+b^2+c^2}{ab+bc+ca} + \frac{8abc}{(a+b)(b+c)(c+a)} \geqslant 2+\frac{16(a-b)^2(a-c)^2(b-c)^2}{(ab+ac+bc)(a+b+c)^4}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Sum of finite series using partial fraction I'm quite stuck with the following problem. I have seen on this forum that there is already an answer for the infinite sum to the problem but I can't seem to find how to find the sum for a finite value.
The first part of the questions asks to transform the given series using partial fractions, which I did as follows:
$$ \frac{1}{k(k + 2)} $$
Which becomes:
$$ \frac{1}{2} \left(\frac{1}{k} - \frac{1}{k + 2}\right) $$
The question now asks to evaluate the finite sum:
$$ \sum_{k=1}^{n} \frac{1}{2} \left(\frac{1}{k} - \frac{1}{k + 2}\right) $$
I have tried expanding the summation and I have been able to cancel out some terms, but I cannot seem to find a correct solution in the end. Has anyone any idea or method on how to evaluate these sums after rewriting them using partial fractions?
Thanks in advance!
| For instance, with $n = 10$ we have
$$
\sum_{k=1}^n \frac1k - \frac 1{k+2} = \\
\left(1 - \frac 13 \right) +
\left(\frac12 - \frac 14 \right) +
\left(\frac13 - \frac 15 \right) + \cdots +
\left(\frac 18 - \frac 1{10} \right) +
\left(\frac19 - \frac 1{11} \right) +
\left(\frac1{10} - \frac 1{12} \right) =\\
1 + \frac 12 + \left(\frac13 - \frac 1{3} \right) + \cdots +
\left(\frac1{10} - \frac 1{10} \right) - \frac 1{11} - \frac 1{12}.
$$
For a more formal approach, note that
$$
\sum_{k=1}^n \frac1k - \frac 1{k+2} =
\sum_{k=1}^n \frac1k - \sum_{k=1}^n \frac 1{k+2} =
\sum_{k=1}^n \frac1k - \sum_{k=3}^{n+2} \frac 1{k}\\
= \left(1 + \frac 12 + \sum_{k=3}^n \frac1k\right) -
\left(\frac1{n+1} + \frac 1{n+2} + \sum_{k=3}^n \frac1k\right).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cubic Discriminant Uses The discriminant for the cubic equation $ax^3+bx^2+cx+d=0$ is
$Δ\:=b^2c^2−4ac^3−4b^3d−27a^2d^2+18abcd$
And I am aware that you can determine the number of roots a cubic has using method shown below -
$Δ\:>0$: the equation has three distinct real roots
$Δ\:=0$: the equation has a repeated root and all its roots are real
$Δ\:<0$: the equation has one real root and two non-real complex conjugate roots
But I was wondering if one could determine whether a cubic has rational or integer roots, as you can do with the discriminant for quadratics, and if so what the method would be.
I have noticed that with the cubics I have checked: if the discriminant is a perfect square there are 3 integer solutions, although I have not checked many and I am not sure of the reasoning behind it.
Any help would be greatly appreciated.
| When a monic cubic has square discriminant but no rational roots, what we expect is real roots that can be written as (doubled) cosines, or sums of them.
$$ x^3 + x^2 - 2x - 1 $$
has $$ 2 \cos \frac{2 \pi}{7} \; , \; \; 2 \cos \frac{4 \pi}{7} \; , \; \; 2 \cos \frac{8 \pi}{7} \; , \; \; $$
more in a minute
$$ x^3 - 3x + 1 $$
has $$ 2 \cos \frac{2 \pi}{9} \; , \; \; 2 \cos \frac{4 \pi}{9} \; , \; \; 2 \cos \frac{8 \pi}{9} \; , \; \; $$
$$ $$
$$ x^3 + x^2 - 4x + 1 $$
has $$ 2 \cos \frac{2 \pi}{13} + 2 \cos \frac{10 \pi}{13}\; , \; \; 2 \cos \frac{4 \pi}{13} + 2 \cos \frac{6 \pi}{13} \; , \; \; 2 \cos \frac{8 \pi}{13} +2 \cos \frac{12 \pi}{13} \; , \; \; $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding volume of solid in one quadrant - divide total volume by 4? 8? 2? I want to find the volume of the solid produced by revolving the region enclosed by $y=4x$ and $y=x^3$ in the first quadrant. The wording about the first quadrant confuses me but here's my work so far:
I know the volume unrestrained by quadrant is:
$$V = \int_a^b \pi(f(x)^2 - g(x)^2) dx$$
Where $f(x) = 4x$ and $g(x) = x^3$. To find $a$ and $b$, I look for the largest and smallest intersection points between the two functions:
$$\begin{align}
f(x) &= g(x) \\
4x &= x^3 \\
0 &= x^3 - 4x \\
&= x(x-2)(x+2) \\
x &\in \{-2, 0, 2\}
\end{align}$$
Plugging all of these into the volume equation above:
$$\begin{align}
V &= \int_a^b \pi (f(x)^2 - g(x)^2)dx \\
&= \int_{-2}^2 \pi ((4x)^2 - (x^3)^2)dx \\
&= \pi \int_{-2}^2 (16x^2 - x^6)dx \\
&= \pi \left( \int_{-2}^2 16x^2 dx - \int_{-2}^2 x^6 dx \right) \\
&= \pi \left( \left[ \frac{16x^3}{3} \right]_{-2}^2 - \left[ \frac{x^7}{7} \right]_{-2}^2 \right) \\
&= \pi \left( \frac{16(2)^3}{3} - \frac{16(-2)^3}{3} - \frac{(2)^7}{7} - \frac{(-2)^7}{7} \right) \\
&= \pi \left( \frac{128}{3} + \frac{128}{3} - \frac{128}{7} + \frac{128}{7} \right) \\
&= \pi \left( \frac{256}{3} - \frac{256}{7} \right) \\
&= \pi \left( \frac{1024}{21} \right) \\
&= \frac{1024\pi}{21}
\end{align}$$
This is the volume for the entire function. I make an assumption that since I only want one quadrant and the function is symmetric about both the x- and y-axes, I simply divide it by four.
$$\begin{align}
V_{\text{whole}} &= \frac{1024\pi}{21} \\
V_{\text{one quadrant}} &= \frac{1024\pi}{21} \times \frac{1}{4} \\
&= \frac{256\pi}{21}
\end{align}$$
I have no way of verifying my results. Can my assumption be made, or there's a differing method I should be using here?
If I'm now working in 3D space, would I instead divide it by eight? But if I'm revolving around $x=0$, wouldn't the solid of revolution take four quadrants in 3D space, thus I should divide the total volume by 2?
| The first quadrant is defined by $\{(x,y)\in \mathbb R^2 | x,y\ge 0\}$. The $\ge$ instead of $>$ is debatable, but is of no consequence for this problem.
Thus your intersection points are $(0,0)$ and $(2,2)$. So, the volume integral would be $\int_0^2$.
To answer your other question, it would be twice the volume of the correct volume, since you are integrating an even function ($f^2-g^2$) from $-2$ to $2$. Intuitively you are doubling the volume by calculating the volume of the identical rotated ($\pi$ radians) region between $-2$ and $0$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Olympiad Inequality Question I am new to the Olympiad-style questions and I hope someone could correct my proof for this question as I do not have the answer for it. Please leave some constructive criticism if possible so I could improve. Thanks in advance!
Let $a,b,c$ be positive real numbers. Prove that:
$$a^3 +b^3 +c^3\geq a^2b+b^2c+c^2a$$
My attempt:
By AM-GM Inequality,
$$\frac{a^2+b^2}{2}\ge\sqrt{a^2 b^2}=ab$$
$$\frac{a^2+c^2}{2}\ge\sqrt{a^2 c^2} =ac$$
$$\frac{b^2+c^2}{2}\ge\sqrt{b^2 c^2} =bc$$
Next, multiply $a,b,$ or $c$ to get RHS of the inequality wanted above:
$\dfrac{a(a^2+b^2)}{2} \ge a^2b$, $\dfrac{b(b^2+c^2)}{2} \ge b^2c$, $\dfrac{c(a^2+c^2)}{2} \ge ac^2$
Adding up the inequalities gives us:
$$\dfrac{a^3+ab^2+b^3+bc^2+a^2c+c^3}{2} \ge a^2b+b^2c+ac^2$$
and rearranging the inequality gives us:
$$a^3+b^3+c^3 \ge 2(a^2b+b^2c+ac^2)-ab^2-bc^2-a^2c$$ which is generally true.
| Hint: $a^3+a^3+b^3 \ge 3a^2b$ by AM-GM. Do it $2$ more times with the pairs $(b,c)$ and $(c,a)$. Then add up the $3$ inequalities, and divide both sides by $3$ to complete the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Double integral over $x^2+y^2 \le 1$ I am trying to calculate the double integral
$\displaystyle \iint_{x^2+y^2\leq 1} (\sin x+y+3)\,dA$
Here are my attempts so far:
1) I used polar coordinates
$ x= r \sin(\theta)$
$y= r \cos (\theta)$
where $\theta \in [0,2 \pi]$ and $r \in [0,1]$ which gives
$\displaystyle \int_0^{2\pi} \int_0^1 \bigg(\sin\big(r \cos(\theta)\big)+ r \sin(\theta)+3 \bigg)r\, dr\, d\theta$
and stuck with fiding antiderivative of the function with respect to r
$ r \sin\big(r \cos(\theta)\big)$
2) I tried to divide the region into parts that
$A \cup B = \{ (x,y) : x^2+y^2 \leq 1\}$ where $A=\{ (x,y) : x^2+y^2 \leq 1$ and $x\geq 0 \}$ and $B=\{ (x,y) : x^2+y^2 \leq 1$ and $x <0 \}$ which gives me
$ \quad\displaystyle \int_{-1}^1 \int _0^{\sqrt{1-x^2}} \big( \sin x +y+3 \big)\,dy \,dx + \int_{-1}^1 \int _0^{\sqrt{1-x^2}} \big( \sin x +y+3 \big)\,dy\,dx$
$\displaystyle =\int_{-1}^1 \big( \sqrt{1-x^2} \sin x + \frac{1-x^2}{2}+3 \sqrt{1-x^2} \big)\,dx + \dotsm $
and stuck with finding antiderivative of the function
$\sqrt{1-x^2} \sin x$.
I will be glad to hear any tips. Thanks in advance.
| The integral of a sum is equal to the sum of the integrals of the summands (assuming all exist, and finitely many summands).
$$\iint_{x^2+y^2=1}\sin(x)+y+3\,dA = \iint_{x^2+y^2=1}\sin(x)\,dA + \iint_{x^2+y^2=1}y\,dA + \iint_{x^2+y^2=1}3\,dA$$
Hint: For any odd function $f$ (a function such that $f(-x) = -f(x)$ for all $x$), what is $\int_{-c}^cf(x)dx$?
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this problem with $P(Q(n))\equiv n\pmod p$ for all integers $n$, the degrees of $P$ and $Q$ are equal. $p$ is a prime. Let $K_p$ be the set of all polynomials with coefficients from the set $\{0,1,\dots ,p-1\}$ and degree less than $p$. Assume that for all pairs of polynomials $P,Q\in K_p$ such that $P(Q(n))\equiv n\pmod p$ for all integers $n$, the degrees of $P$ and $Q$ are equal. Determine all primes $p$ with this condition
I try:Suppose that $a\not\equiv b\pmod{p - 1}$ satisfied $ab\equiv 1\pmod{p - 1}$. Then $P(x) = x^a, Q(x) = x^b$ would have $P(Q(n)) = n^{ab}\equiv n\pmod{p}$ for all integers $n$ and the degrees of $P, Q$ are unequal and less than $p$ when considering $a, b$ as least residues. then I can't,Thanks
| To complete Peter's answer, which shows that $P=x^a$ and $Q=x^b$ works for some $(a,b)$ unless $p\in\{2,3,5,7,13\}$, here is a full explication in these cases:
If $p=2$ then, since $P$ and $Q$ can clearly not be constant, they must be linear, and so $p=2$ works.
If $p=3$, then $\deg P(Q(x))=(\deg P)(\deg Q)$ must be either $1$ or at least $3$. If it's $1$ then $\deg P=\deg Q=1$. Otherwise, neither can be $1$ (since both are $\leq 3$), so they must both be $2$, and so $p=3$ works.
For larger primes, we'll need the following result:
Lemma. If $P$ permutes the integers $\bmod p$, then $\deg P$ cannot divide $p-1$ unless $P$ is linear.
Proof. We know (for example via primitive roots or Newton sums) that
$$\sum_{x\in S}x^k\equiv \begin{cases}0&\text{if }1\leq k<p-1 \\ -1&\text{if }k=p-1,\end{cases}$$
for any complete residue system $S$ modulo $p$. if $P$ permutes such an $S$ and is of degree $d|p-1$, then
$$\sum_{x\in S}P(x)^{\frac{p-1}{d}}=0;$$
however, if you expand out $P(x)^{\frac{p-1}{d}}$ termwise, there is only one term of degree $p-1$ (which does not vanish) and all other terms are less, a contradiction. $\square$
Now, this shows that $p=5$ works, since the only nonlinear permutation polynomials mod $5$ can be of degree $3$. This can also be made to show that $p=7$ works; the only allowable polynomials are of degree $4$ and $5$. Now, consider a permutation polynomial $P$ of degree $4$. By replacing $P(x)$ with $aP(x+b)+c$ for some $b,c$ and nonzero $a$ (accompanied by corresponding changes in $Q$), we may assume that $P$ is monic and has no unit or $x^3$ coefficient. By replacing this $P(x)$ with $a^{-4}P(ax)$ for some $a\neq 0$ (and performing corresponding changes in $Q$) we may assume that the $x^2$ coefficient is either $0$, $1$, or $-1$.
Consider $P(x)=x^4+x^2+ax$. If $a=0$ this is clearly not a permutation polynomial as $P(x)=P(-x)$; otherwise, if $-a$ is in the image of $x^3+x$ mod $7$, this is not a permutation polynomial as we can pick $x\neq 0$ so that $P(x)=0$. This image is the set
$$\{2,3,4,5\},$$
so we only have to deal with the case of $a=\pm 1$. These are isomorphic by flipping $x$ and $-x$; and if $P(x)=x^4+x^2+x$ then
$$P(1)\equiv 3\equiv P(4).$$
So, there are no permutation polynomials in this case.
Consider $P(x)=x^4-x^2+ax$. We can deal with $a=0$ as before; now, if $-a$ is in the image of $x^3-x$ (and is nonzero) we are done by similar reasoning as the above. This happens for the set
$$a\in\{1,3,4,6\},$$
so we only care about $a=(\pm)2$. For this $a$, $P(1)\equiv P(2)\equiv 2$.
Now, consider $P(x)=x^4+ax$. By flipping the sign of the input of $P$, we need only consider $a\in\{0,1,2,3\}$. It is clear that $a=0$ does not work; also, $a=1$ fails since $P(-1)\equiv P(0)\equiv 0$, and $a=2$ fails since $P(1)\equiv P(3)\equiv 3$. However, $a=3$ works. So, we only need to consider the polynomial $x^4+3x$.
Now, we notice that $Q$ has to exactly invert $P$; from this, we know that the values of $Q$ modulo $p$ are exactly determined, and so we can find $Q$ directly using Lagrange interpolation; in other words, given $P$, only one unique $Q$ exists. Now observe the miraculous identity
\begin{align*}
(-x^4+3x)^4+3(-x^4+3x)
&\equiv(x^4-3x)^4-3x^4+2x\\
&\equiv x^{16}-12x^{13}+54x^{10}-108x^7+81x^4-3x^4+2x\\
&\equiv x^{16}+2x^{13}+5x^{10}+4x^7+x^4+2x\\
&\equiv x^4+2x+5x^4+4x+x^4+2x\\
&\equiv x\\
\end{align*}
where we have used Fermat's Little Theorem in the form $x^7\equiv x$. So, for this $P$, $Q$ is also of degree $4$, and thus $p=7$ works too.
Now, for $p=13$, we see that $P(x)=x^9+4x^7+12x^5+4x^3+10x$ and $Q(x)=x^5+x^3+8x$ are inverses, so $p=13$ fails. Thus, our answer is $\boxed{p\in\{2,3,5,7\}}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Determine the convergence of the series $\sum_{n=1} ^{\infty} \frac{5^{n}-2^{n}}{7^{n}-6^{n}}$ Does
$$\sum_{n=1} ^{\infty} \frac{5^{n}-2^{n}}{7^{n}-6^{n}}$$
converge?
I tried the ratio test but I failed.
| Ratio test:
$$
\frac{\displaystyle\frac{5^{n+1}-2^{n+1}}{7^{n+1}-6^{n+1}}}{\displaystyle\frac{5^{n}-2^{n}}{7^{n}-6^{n}}}
=\frac{7^{n}-6^{n}}{{7^{n+1}-6^{n+1}}}\,\frac{5^{n+1}-2^{n+1}}{5^{n}-2^{n}}
=\frac{1-(6/7)^n}{7-(6/n)^{n+1}}\,\frac{5-(2/5)^{n+1}}{1-(2/5)^n}\to\frac57<1,
$$
so the series converges.
| {
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"timestamp": "2023-03-29T00:00:00",
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Ahmed integral revisited $\int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} \, dx$ How to prove
$$\small \int_0^1 \frac{\tan ^{-1}\left(\sqrt{x^2+4}\right)}{\left(x^2+2\right) \sqrt{x^2+4}} dx=-\frac{\pi \:\arctan \left(\frac{1}{\sqrt{2}}\right)}{8}+\frac{\arctan \left(\frac{1}{\sqrt{2}}\right)\arctan \left(\sqrt{2}\right)}{4}+\frac{\pi }{4}\arctan \left(\frac{1}{\sqrt{5}}\right)\;?$$
I came across this Ahmed integral on the site "Art of problem solving", and have found no proof so far. (These two problems seems to be related though). Any help will be appreciated!
| Let
$$ I(a)=\int_0^1 \frac{\arctan(a\sqrt{x^2+4})}{(x^2+2)\sqrt{x^2+4}}dx,I(0)=0,I=I(1)$$
\begin{align}
I'(a)&=\int_0^1 \frac{1}{(x^2+4)[1+a^2(x^2+4)]}dx \\
&=\int_0^1 \frac{1}{x^2+4}dx-a^2\int_0^1 \frac{1}{1+a^2(x^2+4)}dx \\
&=\frac{1}{2} \arctan \frac{1}{2}-\frac{a}{\sqrt{1+4a^2}}\arctan \frac{a}{\sqrt{1+4a^2}} \\
\end{align}
Note
$$ \int \frac{a}{\sqrt{1+4a^2}}\arctan \frac{a}{\sqrt{1+4a^2}} da $$
$$ =\frac{1}{4} \int \arctan \frac{a}{\sqrt{1+4a^2}} d(\sqrt{1+4a^2}) $$
Using integration by parts,then I believe you can finish it
| {
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Finding the Domain? The set ${{x:\left|x+\frac{1}{x}\right|>6}}$ equals the set
My Approach
We know,IF $|f(x)| \geqslant a, \quad ;a>0$
then $f(x) \geqslant a \quad \cup \quad f(x) \leqslant-a$
$\frac{x^{2}+1}{x}> 6$ & $\frac{x^{2}+1}{x}<-6$
$\frac{x^{2}-6 x+1}{x}>0..........(1)$
$\frac{x^{2}+6 x+1}{x}<0$............(2)
$Using$ $Wavy$ $curve$ $method,$
From (1) I can find,
$x \in(0,3-2 \sqrt{2}) \cup(3+2 \sqrt{2}, \alpha)$
From (2) ,I can find,
$x \in(-\alpha,-3-2 \sqrt{2}) \cup(-3+2 \sqrt{2},0)$
Next I am unable to find any common between them.
| We have $x\neq0$ and $$x^2+1>6|x|$$ or
$$|x|^2-6|x|+1>0,$$ which is
$$|x|>3+2\sqrt2$$ or
$$|x|<3-2\sqrt2,$$ which gives the answer:
$$(-\infty,-3-2\sqrt2)\cup(-3+2\sqrt2,0)\cup(0,3-2\sqrt2)\cup(3+2\sqrt2,+\infty)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3696447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Where does the equation of asymptotes of a hyperbola come from? It's known that the asymptotes of a hyperbola $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ is given by $y=\pm\frac{b}{a}x$ if $a>b$.
I tried to find a proof of the fact that why the equations of these asymptotes are like that,however the only reference (Thomas calculus book) that I found explained
that the two asymptotes are derived by letting $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=0$.
It would be highly appreciated if someone prove why the equation of the asymptotes have such form.
| Edited to do it properly -- see below
Original post:
We have
$$y=b\sqrt{\frac{x^2}{a^2}-1}=\frac{b}{a}x\sqrt{1-\frac{a^2}{x^2}}$$
And as $x\to\pm\infty$, $\sqrt{1-\frac{a^2}{x^2}}\to 1$.
End of original post
But as mentioned in the comments, it is not enough to show that $\frac{y}{bx/a}\to 1$. We have to show that $y-\frac{b}{a} x\to 0$:
$$y-\frac{b}{a}x=\frac{b}{a}x\left(\sqrt{1-\frac{a^2}{x^2}}-1\right)$$
But
$$1-\frac{a^2}{x^2}\le\sqrt{1-\frac{a^2}{x^2}}<1$$
So
$$\left|\sqrt{1-\frac{a^2}{x^2}}-1\right|\le\frac{a^2}{x^2}$$
Therefore
$$\left|y-\frac{b}{a}x\right|\le\frac{b}{a}|x|\cdot\frac{a^2}{x^2}=\frac{ba}{|x|}$$
which tends to $0$ as $x\to\pm\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3698030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 1
} |
How to solve equation with multiple trigonometric functions? Solve for $x$:
$\arccos( \cos(x) y + z) = \arcsin( \sin(x) a+b)$.
| $$\cos^{-1}(\cos(x)y+z)=\sin^{-1}(\sin(x)a+b)$$
Taking Sine of both the sides,
$$\sin(\cos^{-1}(\cos(x)y+z))=\sin(\sin^{-1}(\sin(x)a+b))$$
$$\sin\left(\sin^{-1}\sqrt{1-(\cos(x)y+z)^2}\right)=\sin(x)a+b$$
$$\sqrt{1-(\cos(x)y+z)^2}=a\sqrt{1-\cos^2(x)}+b$$
$$1-(\cos(x)y+z)^2=\left(a\sqrt{1-\cos^2(x)}+b\right)^2$$
$$\left(1-(\cos(x)y+z)^2+a^2\cos^2(x)-a^2-b^2\right)^2=\left(2ab\sqrt{1-\cos^2(x)}\right)^2$$
$$\left((a^2-y^2)\cos^2(x)-2yz\cos(x)+1-z^2-a^2-b^2\right)^2=4a^2b^2-4a^2b^2\cos^2(x)$$
Thus, you get a quartic equation in terms of $\cos(x)$. Solve it to get value of $\cos x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3699991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $a, b, c\in\mathbb R^+, $ then prove that $a^3b+b^3c+c^3a\ge abc(a+b+c) .$ While trying to prove it, I proved the following two inequalities:
$a^4+b^4+c^4\ge abc(a+b+c)$ and
$(a^2b+b^2c+c^2a)(ab+bc+ca)\ge abc(a+b+c)^2.$
The later one, on some simplification gives
$a^3b+b^3c+c^3a\ge abc(ab+bc+ca).$
But we can't claim that $ab+bc+ca\ge a+b+c$ for all positive $a, b, c.$ So this doesn't help. So am not quite sure how to approach the inequality in question. Please suggest.. Thanks in advance. (BTW can we use Cauchy-Schwarz's inequality? I tried but couldn't think of a proper choice for two triplets.)
| Suppose $c=\min\{a,b,c\}.$ We have
$$\begin{aligned}a^3b+b^3c+c^3a-abc(a+b+c)&=c(a^3+b^3-a^2b-ab^2)+a(c^3+a^2b-bc^2-ca^2)\\&=c(a+b)(a-b)^2+a(c+a)(a-c)(b-c) \geqslant 0.\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Simplify $4^3\sin^4(20^\circ)\sin^2(70^\circ)-4\sqrt3\sin^3(20^\circ)\sin(70^\circ)+3$ I was trying to solve a question where the two sides of a triangle were $$\frac{a\sin(20^\circ)}{\sin(70^\circ)}$$and $$\frac{a\sin(60^\circ)\sin(30^\circ)}{\sin(70^\circ)\sin(40^\circ)}$$ and the ange between them was $70^\circ$ I used the law of cosines to try find the third side which I call $c$ and I soon found that $$\frac{a^2[4^3\sin^4(20^\circ)\sin^2(70^\circ)-4\sqrt3\sin^3(20^\circ)\sin(70^\circ)+3]}{4^3\sin^4(70^\circ)\sin^2(20^\circ)}=c^2$$ But after this I’m not able to simplify further, please help, there still is a possibility that I might’ve done something wrong that I reached this step, please assist.
| We have the following theorem:
$$
\prod_{0<k<n}2\sin\frac{k\pi}n=n.
$$
Particularly:
$$\begin{align}
\prod_{0<k<9}{\sin\frac{k\pi}9}=[2^4\sin(20^\circ)\sin(40^\circ)\sin(60^\circ)\sin(80^\circ)]^2=9\\
\implies \sin(20^\circ)\sin(40^\circ)\sin(60^\circ)\sin(80^\circ)=\frac{3}{16}.\tag1
\end{align}
$$
Let us apply this to your triangle:
$$AB=\frac{a\sin(20^\circ)}{\sin(70^\circ)},\quad AC=\frac{a\sin(60^\circ)\sin(30^\circ)}{\sin(70^\circ)\sin(40^\circ)},\\ \alpha=\measuredangle CAB=70^\circ,\quad\beta=\measuredangle ABC,\quad \quad\gamma=\measuredangle BCA.$$
Then we have by law of sines:
$$
\frac{\sin\beta}{\sin\gamma}=\frac{AC}{AB}=\frac{\sin(60^\circ)\sin(30^\circ)}{\sin(20^\circ)\sin(40^\circ)}\stackrel{(1)}
=\frac{16\sin(60^\circ)\sin(30^\circ)\sin(60^\circ)\sin(80^\circ)}3
=\frac{\sin(80^\circ)}{\sin(30^\circ)}.
$$
In view of $\beta+\gamma=110^\circ$ one concludes
$$\beta=80^\circ,\quad \gamma=30^\circ.$$
Finally:
$$
\frac{BC}{AB}=\frac{\sin\alpha}{\sin\gamma}\implies BC=\frac{a\sin(20^\circ)}{\sin(70^\circ)}\frac{\sin(70^\circ)}{\sin(30^\circ)}=2a\sin(20^\circ).
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Let $a, b, c>0$. Prove that $\sum \limits_{cyc}{\frac{a}{b+c}\left(\frac{b}{c+a}+\frac{c}{a+b}\right)}\le \frac{(a+b+c)^2}{2(ab+bc+ca)}$ Reducing this whole expression i finally came to this
$$\sum \limits_{cyc}\left(ab^4+a^4b+a^2b^2c\right)\geq \sum \limits_{cyc}\left(a^3b^2+a^2b^3+a^3bc\right)$$
Here I am stuck. I can't prove this.
So I thought maybe I should try in another way.
Let $3u=a+b+c$, $3v^2=ab+bc+ca$ and $w^3=abc$
Hence the whole expression comes to this inequality
\begin{align*} (a+b+c)^2&(a+b)(b+c)(c+a) \\
& \geq 4(ab+bc+ca)(ab^2+a^2b+bc^2+b^2c+ca^2+c^2a)\\
\implies (a+b+c)^2&\left((a+b+c)(ab+bc+ca)-abc\right)\\
& \geq 4(ab+bc+ca)\left((a+b+c)(ab+bc+ca)-3abc\right)\\
\implies (3u)^2\left(3u\times3v^2-w^3\right)&\geq 4\times3v^2\left(3u\times3v^2-3w^3\right)\\
\implies 9u^2(9uv^2-w^3)&\geq 12v^2(9uv^2-3w^3)\\
\implies 9u^3v^2-u^2w^3&\geq 12uv^4-4v^2w^3\end{align*}
Here again, I am stuck. How can I prove this inequality?
| Another way.
Afte using your $uvw$'s substitution we see that our inequality is a linear inequality of $w^3$,
which by $uvw$ says that it's enough to prove our inequality in the following cases.
*
*$w^3\rightarrow0^+$.
Let $c\rightarrow0^+$ and $b=1$.
We obtain:
$$a(a+1)(a-1)^2\geq0;$$
2. Two variables are equal.
Let $b=c=1$.
We obtain:
$$a^2(a-1)^2\geq0.$$
About $uvw$ see here: https://artofproblemsolving.com/community/c6h278791
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How do you find appropriate trig substitution for $\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$? I want to solve the below:
$$\int \frac{\sqrt{16x^2 - 9}}{x} \, dx$$
I know for trig substitution, if I have something in the form of $\sqrt{x^2-a^2}$, I can use $x = a\sec{u}$; it just so happens my integral has a numerator in this form: $\sqrt{16x^2 - 3^2}$ so I know to use $x = 3\sec u$:
$$
\begin{align}
& \int \frac{\sqrt{16x^2 - 9}}{x} \, dx \\
= {} & \int \frac{\sqrt{16x^2 - 3^2}}{x} \, dx \\
= {} & \int \frac{\sqrt{16(3\sec u)^2 - 3^2}}{3\sec u} 3\sec u\tan u \, du \\
= {} & \int \frac{(\sqrt{16(3\sec u)^2 - 3^2)}(3\sec u\tan u)}{3\sec u} \, du \\
= {} & \int \sqrt{(16(3\sec u)^2 - 3^2)}(\tan u) \, du
\end{align}
$$
This doesn't seem to make it easy. However, using a calculator online, it suggests I instead use $x = \dfrac{3}{4}\sec{u}$ which simplifies the integral to a crisp $\int 3\tan^2 u \, du$.
My question is, how did the calculator get $a = \dfrac{3}{4}$ and is there a way to determine an ideal trig substitution for a given function?
| Given $\int \frac{\sqrt{16x^2}-9}{x}dx$ and that $\sqrt{x^2-a^2} \Rightarrow x=a \sec \theta \wedge a \sec \theta \tan \theta d\theta =dx$
Then,
$$\int \frac{\sqrt{16x^2}-9}{x}dx \Rightarrow \int \frac{\sqrt{16(3 \sec\theta)^2}-3^2}{3 \sec \theta} 3 \sec \theta \tan \theta d\theta $$
$$ = 12\int \tan \theta \sqrt{\sec^2 \theta}-9 $$
$$ = 12\sqrt {\sec^2\theta} -81\theta +C$$
by factoring out constants and integrating the sum term by term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 6
} |
If $d \equiv 3 \pmod 4 $ then $x^2 − dy^2 = −1$ has no solutions in positive integers $x, y$. In this case, I have come to the conclusion that $x^2 \equiv 0,1 \pmod 4$, $y^2 \equiv 0,1 \pmod 4$ but I am not sure what this means for $dy^2$ and how this won't have any solutions for the above equation. Can someone clarify this?
| $d\equiv3\equiv-1\pmod4$, so $x^2-dy^2\equiv x^2+y^2\pmod 4$.
Since $x^2\equiv0$ or $1$ and $y^2\equiv 0$ or $1$, $x^2+y^2\equiv 0, 1, $ or $2$, but not $3$ ($\equiv-1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3706277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Deriving the expansion of $\sin (\alpha - \beta)$ using $\sin x = \sqrt{1-\cos^2 x}$ I was deriving the expansion of the expansion of $\sin (\alpha - \beta)$ given that $\cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta$
Now, my textbook has done it in a different manner but I thought of doing it using the simple trigonometric identity $\sin^2 x + \cos^2 x = 1 \implies \sin x = \sqrt{1-\cos^2 x}$. I thought that it would be pretty easy (it probably is), until I got stuck in the final part which included the modulus function.
Here's how I did it :
$$\sin (\alpha - \beta) = \sqrt {1 - \cos^2 (\alpha - \beta)} = \sqrt{1-(\cos \alpha \cos \beta + \sin \alpha \sin \beta)^2}$$
By substituting $1$ as $\sin^2\alpha + \cos^2\alpha$ and expanding $(\cos \alpha \cos \beta + \sin \alpha \sin \beta)^2$, we get :
$$\therefore \sin (\alpha - \beta) = \sqrt{\sin^2\alpha + \cos^2\alpha - \cos^2\alpha\cos^2\beta-
\sin^2\alpha\sin^2\beta-2\sin\alpha\sin\beta\cos\alpha\cos\beta}$$
$$\therefore \sin(\alpha-\beta) = \sqrt{\sin^2\alpha (1-\sin^2\beta)+\cos^2\alpha(1-\cos^2\beta)-2\sin\alpha\sin\beta\cos\alpha\cos\beta}$$
$$\therefore \sin(\alpha - \beta) = \sqrt{\sin^2\alpha\cos^2\beta+\cos^2\alpha\sin^2\beta-2\sin\alpha\sin\beta\cos\alpha\cos\beta}$$
$$\therefore \sin(\alpha - \beta) = \sqrt{(\sin\alpha\cos\beta - \cos\alpha\sin\beta)^2} = |\sin\alpha\cos\beta-\cos\alpha\sin\beta|$$
Now, how do I get rid of the modulus sign? I do know that I must decide whether the expression inside the modulus functions in positive or negative, but I can't seem to decide how.
Thanks!
| What is your question?
At start when you accepted
$$\sin x =\pm \sqrt{1-\cos^2 x}, \tag1$$
at the end why don't you accept
$$\sin (\alpha - \beta) = \pm \sqrt{1-(\cos \alpha \cos \beta + \sin \alpha \sin \beta)^2}\tag2$$
in that very same sense? What is extra in (2) ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3707342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$\int \frac{x^2\,dx}{(a-bx^2)^2}$ How do I integrate $\int \frac{x^2\,dx}{(a-bx^2)^2}$
I've tried substitution and partial fraction decomposition, but I'm not getting anywhere.
| Almost without any substitution.
Let $x=\frac{\sqrt{a} }{\sqrt{b}}y$
$$I=\int \frac{x^2}{(a-b\,x^2)^2}\,dx=\frac{1}{\sqrt{a}\,\, b^{3/2}}\int \frac{y^2}{\left(1-y^2\right)^2}\,dy$$ Now, using partial fraction decomposition
$$\frac{y^2}{\left(1-y^2\right)^2}=\frac{1}{4 (y-1)}-\frac{1}{4 (y+1)}+\frac{1}{4 (y+1)^2}+\frac{1}{4 (y-1)^2}$$ does not seem to make any problem
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3709833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Eigenvectors of $\begin{pmatrix}2&-2\\ -4&-2\end{pmatrix}$ - What am I doing wrong? Eigenvalues are $2\sqrt{3}$ and $-2\sqrt{3}$, I'll calculate the eigenvector for $2\sqrt{3}$ here
We've got:
$\begin{pmatrix}2-2\sqrt{3}&-2\\ -4&-2-2\sqrt{3}\end{pmatrix}\begin{pmatrix}y_1\\ y_2\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}$
I multiply the first row by negative 1, I get:
$\begin{pmatrix}-2+2\sqrt{3}&2\\ -4&-2-2\sqrt{3}\end{pmatrix}\begin{pmatrix}y_1\\ y_2\end{pmatrix}=\begin{pmatrix}0\\ 0\end{pmatrix}$
I add the first and second row together, I get:
$\left(-6+2\sqrt{3}\right)y_1-2\sqrt{3}y_2=0$
So the solution would be
$\left(-\sqrt{3}+1\right)y_1=y_2$
The eigenvector is $\begin{pmatrix}y\\ \left(1-\sqrt{3}\right)y\end{pmatrix}$
But that's apparently wrong according to this calculator
What's wrong here?
| You did nothing wrong. What you got were that the eigenvectors corresponding to the eigenvalue $2\sqrt3$ are those of the form $\left(1,1-\sqrt3\right)^Ty$, with $y\ne0$. That calculator got the vectors of the form $\left(-\frac{\sqrt3+1}2,1\right)^Ty$ with $y\ne0$. But$$\left(-\frac{\sqrt3+1}2\right)\begin{pmatrix}1\\1-\sqrt3\end{pmatrix}=\begin{pmatrix}-\frac{\sqrt3+1}2\\1\end{pmatrix}.$$So, it's the same answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3711355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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$f(x - f(y))= f(f(y)) - 2xf(y) + f(x)$
Let $x,y \in \mathbb R$. Determine all $f : \mathbb R \to \mathbb R$ that satisfies $$f(x - f(y)) = f(f(y)) - 2xf(y) + f(x)$$
My solution: $f(x) = 0$ is obvious. So, I’ll consider other functions.
$x = f(y)$ $\to$ $f(0) = f(x) - 2x^2 + f(x)$
Which implies $f(x) = 2x^2 + \frac{f(0)}{2}$.
Do we have to find $f(0)$ or I can just put it back to the original statement, then derive $f(0)$ and conclude?
| Here is a solution: Suppose that $f$ solves the functional equation, and write $g(x) = f(x) - x^2$.
Claim. $ g(x-f(y)) = g(x) + \frac{f(0)}{2} $ holds for all $x, y \in \mathbb{R}$.
Proof. Using the functional equation, we get
\begin{align*}
g(x-f(y))
&= \bigl( f(f(y)) - 2xf(y) + f(x) \bigr) - \bigl( x - f(y) \bigr)^2 \\
&= g(f(y)) + g(x)
\end{align*}
Plugging $x = f(y)$, we get $2g(f(y)) = g(0) = f(0)$, hence we get $g(f(y)) = \frac{f(0)}{2}$ for any $y$. Then plugging this to the above identity completes the proof. $\square$
Since $f \equiv 0$ is a trivial solution, it suffices to consider the case where $f$ is not identically zero, and we do so. Let $a$ satisfy $f(a) \neq 0$. Then plugging $y = t - f(a)$,
\begin{align*}
g(x) + \tfrac{f(0)}{2}
&= g\bigl(x - f(t - f(a))\bigr) \\
&= g\bigl(x - (t - f(a))^2 - g(t - f(a))\bigr) \\
&= g\bigl(x - t^2 + 2f(a)t - f(a)^2 - g(t) - \tfrac{f(0)}{2}\bigr) \\
&= g\bigl(x + 2f(a)t - f(a)^2 - \tfrac{f(0)}{2} - f(t)\bigr) \\
&= g\bigl(x + 2f(a)t - f(a)^2 - \tfrac{f(0)}{2}\bigr) + \tfrac{f(0)}{2}.
\end{align*}
Then plugging $x = f(a)^2 + \frac{f(0)}{2}$ shows that $g(2f(a)t)$ does not depend on the value of $t$, hence $g$ is constant. Then the above claim shows that $f(0) = 0$, and therefore $f(x) = x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3713529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How to prove $ \int_{1}^{ \infty} \frac{1}{ (1+x^3)^3 } dx$ is convergent? I am trying to determine whether the following improper integral is convergent or not.
$$ \int_{1}^{ \infty} \frac{1}{ (1+x^3)^3 } dx$$
I tried the following:
$l = \lim_{x \to a} ((x-a)^k)f(x)$, if $l \in [0, \infty)$ and $k < 1$, then the integral is convergent. But I can't use it well.
Can someone help me?
Thank you.
| If $x \in [1,\infty)$, then $\frac{1}{(1+x^3)^3} < \frac{1}{x^6}<\frac{1}{x^2}$. So for every $n \in \Bbb N$, we have that $\int_{1}^n \frac{1}{(1+x^3)^3}<\int_{1}^n\frac{1}{x^2}$. But $\int_{1}^n\frac{1}{x^2} = 1 - \frac{1}{n}$. Therefore
\begin{equation}
\int_{1}^n \frac{1}{(1+x^3)^3}<1-\frac{1}{n}
\end{equation}
Now taking limits we get
\begin{equation}
\int_{1}^{\infty} \frac{1}{(1+x^3)^3}= \lim_{n \to \infty}\int_{1}^n \frac{1}{(1+x^3)^3} < \lim_{n \to \infty}\left( 1-\frac{1}{n} \right) = 1
\end{equation}
This shows the improper integral converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3715071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to computer $f(\frac{1}{2})$ given $f(f(x)) = x^2 + \frac{1}{4}$? I have observed that $f(f(\frac{1}{2})) = \frac{1}{2}$ and $f(f(f(x))) = f(x^2 + \frac{1}{4})$, and when $x = \frac{1}{2}$, we have $f(\frac{1}{2}^2 + \frac{1}{4}) = f(\frac{1}{2})$. But I don't know how to proceed, or if any of these observations are helpful. How can I from this compute $f(\frac{1}{2})$ and what is the intuition or reasoning that would guide you towards the answer?
| Denote
$
c = f\left(\frac{1}{2}\right).
$
We have
$$
f\left(f\left( \frac{1}{2}\right)\right) = f(c) = \frac{1}{2}.
$$
$$
f\left(f\left( c\right)\right) = f\left(\frac{1}{2}\right) = c^2 + \frac{1}{4} = c.
$$
So
$$
\left(c - \frac{1}{2} \right)^2 = 0.
$$
This means that the only possible value for $c$ (if such function exists, of course) is $\frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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How to find eigenvalues of a Pauli matrix resolved direct product matrix? For a $2 \times 2$ hermitian matrix one can resolve the matrix
in terms of Pauli matrices like this
\begin{align}
H &= \begin{pmatrix}
a & b \\
b & -a
\end{pmatrix} \\
&= a \sigma_z + b \sigma_x \\
\end{align}
Here, I've assumed $ a,b \in \mathbb{R} $, Using Pauli matrix identities one
can find the eigenvalues of $H$ are $\pm \sqrt{a^2+b^2}$
I was thinking if this process can be generalized for higher dimensional matrices.
Lets say, we have a $(2n\times2n)$ matrix and this matrix can be written in terms of
Pauli matrices
\begin{align}
H = \sigma_{x}\otimes \mathbf{A} + \sigma_{y}\otimes \mathbf{B}
\end{align}
For example we set $n=2$
\begin{align}
\mathbf A = \begin{pmatrix}
e_1 & 0 \\
0 & e_2
\end{pmatrix} \\
\mathbf B = \begin{pmatrix}
b_1 & b_2 \\
b_2 & b_4
\end{pmatrix}
\end{align}
Can I find the eigenvalues using Pauli matrix identities as the $2\times2$ matrix, here also?
In $2\times2$ matrix one would do
\begin{align}
H^2 = \left(
\begin{array}{cc}
a^2+b^2 & 0 \\
0 & a^2+b^2 \\
\end{array}
\right) \\
E^2 =\left(
\begin{array}{cc}
a^2+b^2 & 0 \\
0 & a^2+b^2 \\
\end{array}
\right)
\end{align}
Thus one obtains the eigenvalue of H as $E=\pm \sqrt{a^2+b^2}$.
When I'm doing the same thing here
\begin{align}
H^2 &= (\sigma_{x}\otimes \mathbf{A} + \sigma_{y}\otimes \mathbf{B})^2 \\
\implies E^2 \mathbb{I}_{2n} &= \sigma_x\sigma_x \otimes \mathbf{A}^2 + \sigma_y\sigma_y\otimes \mathbf{B}^2
-\sigma_x\sigma_y \otimes \mathbf{A}\mathbf{B} - \sigma_y\sigma_x\otimes \mathbf{B}\mathbf{A} \\
\implies E^2 \mathbb{I}_{2n}&= \mathbb{I}_2 \otimes \mathbf{A}^2 + \mathbb{I}_2\otimes \mathbf{B}^2
+i\sigma_z \otimes \mathbf{A}\mathbf{B} - i\sigma_z\otimes \mathbf{B}\mathbf{A} \\
\implies E^2 \mathbb{I}_{2n}&= \mathbb{I}_2 \otimes (\mathbf{A}^2 + \mathbf{B}^2) + i \sigma_z \otimes [\mathbf{A},\mathbf{B}]
\end{align}
Here $[\mathbf{A},\mathbf{B}] = \mathbf{A}\mathbf{B} - \mathbf{B}\mathbf{A}$.
Now, when I take trace of the last equation $(n =2)$
\begin{align}
4 E^2 = Tr \mathbf{A}^2 + Tr \mathbf{B}^2
\end{align}
There should be no other terms as Pauli matrices are traceless,
so
\begin{align}
E = \pm \frac{1}{2}\sqrt{4 + b_1^2 + 2 b_2^2 + b_4^2 + e_1^2 + e_2^2}
\end{align}
Both of the solutions are 2 fold degenerate,
But the solutions are wrong, Mathematica gives some complicated and totally different output
\begin{align}
\sqrt{\frac{1}{2} \left(\lambda-\sqrt{\lambda^2-4 \delta}\right)} \\
\sqrt{\frac{1}{2} \left(\lambda-\sqrt{\lambda^2+4 \delta}\right)} \\
- \sqrt{\frac{1}{2} \left(\lambda-\sqrt{\lambda^2-4 \delta}\right)} \\
- \sqrt{\frac{1}{2} \left(\lambda-\sqrt{\lambda^2-4 \delta}\right)} \\
\end{align}
Here $\lambda = Tr \mathbf{A}^2 + Tr \mathbf{B}^2$, and $\delta = \text{Det}\: H$.
What I'm doing wrong here and how to get the correct expressions?
| The problem is your assumption that $H^2 = E^2 I_{2n}$.
In the $2 \times 2$ case, we use the equation $H^2 = E^2 I_2$ in order to exploit the fact that any trace-zero $2 \times 2$ matrix $H$ will be such that $H^2$ is a multiple of the identity. However, this trick no longer works in the generalized setting since it is no longer true that a trace-zero Hermitian matrix $H$ is such that $H^2$ is a multiple of the identity.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3721622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find integers $1+\sqrt2+\sqrt3+\sqrt6=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$ Root numbers Problem (Math Quiz Facebook):
Consider the following equation:
$$1+\sqrt2+\sqrt3+\sqrt6=\sqrt{a+\sqrt{b+\sqrt{c+\sqrt{d}}}}$$
Where $a,\,b,\,c,\,d$ are integers. Find $a+b+c+d$
I've tried it like this:
Let $w=\sqrt6,\, x=\sqrt3, \, y=\sqrt2, z=1$
$$\begin{align}
(y+z)^2 &= (y^2 + z^2) + 2yz\\
y+z &= \sqrt{(y^2 + z^2) + 2yz}\\
y+z &= \sqrt{3 + \sqrt{8}}
\end{align}$$
Let $y+z=f$
$$\begin{align}
(x+f)^2 &= (x^2 + f^2) + 2xf\\
x+f &= \sqrt{(x^2 + f^2) + 2xf}\\
x+f &= \sqrt{(9+\sqrt8) + 2\sqrt{9+3\sqrt8}}
\end{align}$$
And I don't think this going to work since there's still a root term on the bracket that is $9+\sqrt8$. I need another way to make it as an integer.
| First an answer $$1+\sqrt2+\sqrt3+\sqrt6=\sqrt{21+\sqrt{413+\sqrt{4656+ \sqrt{16588800}}}}.$$
Then an explanation.
Everything takes place inside the field $L=\Bbb{Q}(\sqrt2,\sqrt3)$. By elementary Galois theory the quadratic subfields of $L$ are $\Bbb{Q}(\sqrt2)$, $\Bbb{Q}(\sqrt3)$ and $\Bbb{Q}(\sqrt6)$. The number $c+\sqrt d$ must be an element of $L$, so we can conclude that $d=\ell^2 e$ with some integer $\ell$ and $e\in \{2,3,6\}$ with the choice of $e$ depending on a circumstance we don't know yet.
The key question is the following:
Which elements of $L$ have squares in the subfield $\Bbb{Q}(\sqrt e)$?
The answer is, again by elementary Galois theory, that for example the square of a number $z=(a+b\sqrt2+c\sqrt3+d\sqrt6)$ is in $\Bbb{Q}(\sqrt6)$ if and only if either $a=d=0$ or $b=c=0$. Similarly with the other intermediate fields. This comes from the relevant automorphism of $L$ needing to have $z$ as an eigenvector belonging to one of the eigenvalues $+1$ or $-1$.
Let $\alpha=1+\sqrt2+\sqrt3+\sqrt6$. Then
$$
\alpha^2=12+8\sqrt2+6\sqrt3+4\sqrt6.
$$
In view of the previous observation we need to find integers $m,n$ such that $(\alpha^2-m)^2-n$ only contains terms with two of the alternative square roots.
Expanding gives
$$
(\alpha^2-m)^2=m^2-8 \sqrt{6} m-12 \sqrt{3} m-16 \sqrt{2} m-24 m+192 \sqrt{6}+272 \sqrt{3}+336 \sqrt{2}+476.$$
We need one of the square roots to disappear from this by careful choice of $m$.
Because $12\nmid 272$ we cannot make $\sqrt3$ disappear. The choice $m=24$ would make $\sqrt6$ disappear, but then we need to choose $n=476$ to kill the coefficient of $1$. The catch is that then $(\alpha^2-24)^2-476<0$ which is killjoy. It would lead to the answer
$$\alpha=\sqrt{24+\sqrt{476-\sqrt{5376+1536 \sqrt{6}}}},$$
but the negative square root is disallowed, I think.
Therefore we must kill the $\sqrt2$-terms from $(\alpha^2-m)^2$. This forces the choice $m=21$, when
$$
(\alpha^2-21)^2=413+20\sqrt3+24\sqrt6.
$$
This, in turn, forces $n=413$. As the last step we calculate
$$
(20\sqrt3+24\sqrt6)^2=4656+2880\sqrt2=4656+\sqrt{16588800}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
How to show $\sum_{k=0}^{\infty} \frac{1}{k!} \left( \int_{1}^{x} \frac{1}{t} \ dt \right)^k =x$? There are a lot of ways to show that $e^x$ and $\ln(x)$ are inverse functions of each other depending on how you define them. I am trying to show that given the definitions
$$ e^x:= \sum_{k=0}^{\infty} \frac{x^k}{k!} \qquad \text{and} \qquad \ln(x) := \int_{1}^{x} \frac{1}{t} \ dt$$
then $$ e^{\ln(x)}=\sum_{k=0}^{\infty} \frac{1}{k!} \left( \int_{1}^{x} \frac{1}{t} \ dt \right)^k =x$$
My attempt:
My idea was to show that $\frac{d^2}{dx^2} e^{\ln(x)} =0$, and then use the initial conditions I can get by evaluating the definitions at specific values to figure out that the constant of integration must be $0$.
Doing this I get
$$
\frac{d^2}{dx^2}\sum_{k=0}^{\infty} \frac{1}{k!} \left( \int_{1}^{x} \frac{1}{t} \ dt \right)^k = \sum_{k=0}^{\infty} \frac{1}{k!} \frac{d^2}{dx^2} \left( \int_{1}^{x} \frac{1}{t} \ dt \right)^k
$$
From here I use the fact that $\frac{d^2}{dx^2} f(g(x)) = g'(x)^2 f''(g(x)) + f'(g(x))g''(x)$, which applied to this gives me
\begin{align*}
=& \sum_{k=0}^{\infty} \frac{1}{k!} \left[\left(\frac{1}{x}\right)^2 k(k-1)\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-2} + k\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-1} \left(-\frac{1}{x^2}\right) \right]\\
=&\frac{1}{x^2}\left[ \sum_{k=0}^{\infty} \frac{1}{(k-2)!}\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-2} - \sum_{k=0}^{\infty} \frac{1}{(k-1)!}\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-1} \right]= \frac{1}{x^2} \left(\frac{1}{(-2)!}\right) \left[\ln(x) \right]^{-2}
\end{align*}
which is the point where I noticed I may have made several mistakes in the process, since this last result didn't make much sense to me.
Could anyone tell me where my mistakes are? And also, does anyone know another way to rigorously show this result from the definitions in the beginning? Thank you!
| By taking into account the comment by @Pythagoras, my attempt can be corrected to get the following:
\begin{align}
\frac{d^2}{dx^2} e^{\ln(x)} & = \frac{d^2}{dx^2}\left(\underbrace{1}_{k=\color{blue}{0}} + \underbrace{\int_{1}^{x} \frac{1}{t} \ dt}_{k=\color{blue}{1}} +\sum_{k={\color{blue}{2}}}^{\infty} \frac{1}{k!} \left( \int_{1}^{x} \frac{1}{t} \ dt \right)^k \right)\\
& = -\frac{1}{x^2}+\sum_{k=2}^{\infty}\frac{1}{k!}\left[\left(\frac{1}{x}\right)^2 k(k-1)\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-2} + k\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-1} \left(-\frac{1}{x^2}\right) \right] \\
& \overset{\color{purple}{j =k-1}}{=} -\frac{1}{x^2}+\left[\frac{1}{x^2}\sum_{\color{purple}{j }=1}^{\infty}\frac{1}{(\color{purple}{j }-1)!}\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{\color{purple}{j }-1} - \frac{1}{x^2}\sum_{k=2}^{\infty}\frac{1}{(k-1)!}\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-1} \right] \\
& = -\frac{1}{x^2}+\left[\underbrace{\frac{1}{x^2}(1)}_{j=\color{green}{1}} + \frac{1}{x^2}\sum_{j=\color{green}{2}}^{\infty}\frac{1}{(j-1)!}\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{j-1} - \frac{1}{x^2}\sum_{k=2}^{\infty}\frac{1}{(k-1)!}\left( \int_{1}^{x} \frac{1}{t} \ dt \right)^{k-1} \right] \\
& = -\frac{1}{x^2}+\frac{1}{x^2}\\
& = 0
\end{align}
where we apply $\frac{d^2}{dx^2} f(g(x)) = g'(x)^2 f''(g(x)) + f'(g(x))g''(x)$ to $f(x) = x^k$ and $g(x) = \int_{1}^{x} \frac{1}{t} \ dt $. The previous result can thus be combined with
$$
e^{\ln(\color{blue}{1})} = 1 +\sum_{k=1}^{\infty} \frac{1}{k!} \left( \underbrace{\int_{1}^{\color{blue}{1}} \frac{1}{t} \ dt}_{0} \right)^k =1
$$
And
$$
\frac{d}{dx}e^{\ln(x)}\Bigg\vert_{x=\color{blue}{1}} = \frac{d}{dx}\left(1 +\int_{1}^{x} \frac{1}{t} \ dt +\sum_{k=2}^{\infty} \frac{1}{k!} \left(\int_{1}^{x} \frac{1}{t} \ dt \right)^k\right)\Bigg\vert_{x=\color{blue}{1}} = \frac{1}{\color{blue}{1}}+\sum_{k=2}^{\infty} \frac{k}{k!} \left( \underbrace{\int_{1}^{\color{blue}{1}} \frac{1}{t} \ dt}_{0} \right)^{k-1}\left(\frac{1}{\color{blue}{1}}\right) =1
$$
to conclude that
$$
\boxed{e^{\ln(x)}=x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Finding all real $a$ such that $16x^4-(a)x^3+(2a+17)x^2−(a)x+16=0$ has four distinct roots in geometric progression
Determine all real values of the parameter, $a$, for which the equation $$16x^4−(a)x^3+(2a+17)x^2−(a)x+16=0$$ has exactly four distinct real roots that form a geometric progression?
I noticed that the coefficients are symmetric: namely, the first coefficient is the same as the fifth one, the second is the same as the fourth, and the third is the same as the third.
I don’t know how to proceed using Vieta's formula.
| Let's do like derivation of Vieta's formulas rather than the formulas themselves.
Let $b,\,bq,\,bq^2,\,bq^3$ be the roots of $$p(x)=(x-b)(x-bq)(x-bq^2)(x-bq^3)$$ $$\hbox{and }f(x)=x^4−\frac{a}{16}x^3+\frac{2a+17}{16}x^2−\frac{a}{16}x+1.$$
It can be verified that $$p(x)=b^4 q^6 - b^3 (q^3 + q^2 + q + 1) q^3 x + \\
b^2 (q^4 + q^3 + 2 q^2 + q + 1) q x^2 - b (q^3 + q^2 + q + 1) x^3 + x^4$$
But $p(x)$ and $f(x)$ are same degree polynomials with highest degree coefficient $1$ and the same roots $\Rightarrow$ they have the same coefficients.
$$\begin{cases}
b^4 q^6=1\\
16 b^3 (q^3 + q^2 + q + 1) q^3=a\\
16 b^2 (q^4 + q^3 + 2 q^2 + q + 1) q = 2a+17\\
16 b (q^3 + q^2 + q + 1) = a
\end{cases}$$
Hence $a=170$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving polynomials using De Moivre's theorem Given $\cos 4\theta=8\cos^4\theta−8\cos^2\theta+1$, solve $16x^4-16x^2+1=0$.
The textbook's answers are $x=\pm\cos\dfrac{\pi}{12},\pm\cos\dfrac{5\pi}{12}$. I managed to get two of the four answers and i can not figure out what i did wrong.
My Attempt
Let $x=\cos\theta$
$16\cos^4\theta-16\cos^2\theta+1=0$
$2(8\cos^4\theta-8\cos^2\theta+1)=1$
$2\cos4\theta=1$
$\cos4\theta=\dfrac{1}{2}$
$\cos4\theta=\cos\dfrac{\pi}{3}$
$\therefore4\theta=\dfrac{\pi}{3}+k\pi$, for any integer k
I chose $0\le k \le 3$ as the other solutions would repeat its values for x.
$\theta=\dfrac{\pi}{12},\dfrac{\pi}{3},\dfrac{7\pi}{12},\dfrac{5\pi}{6}$
$\therefore x=\cos\dfrac{\pi}{12},\cos\dfrac{\pi}{3},\cos\dfrac{7\pi}{12},\cos\dfrac{5\pi}{6}$
$x=\cos\dfrac{\pi}{12},\cos\dfrac{\pi}{3},-\cos\dfrac{5\pi}{12},\cos\dfrac{5\pi}{6}$
| You wrote:
$$\cos4θ=\cfrac{1}{2}$$
$$\cos4θ=cos\cfrac{\pi}{3}$$
$$\therefore4θ=\cfrac{\pi}{3}+k\pi$$
This contains some errors
The $\cos$ and the $\sin$ function take each value to times in the interval $[0,2\pi]$ except $1$ and $-1$.
So the following holds:
$$\cos \phi=\cos (2\pi-\phi)$$
The period of these trigonometric functions is $2\pi$, and not $\pi$
Therefore you final line is
$$\therefore4θ=\pm\cfrac{\pi}{3}+2k\pi$$
After removing the duplicates four solutions remain.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3729917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Divisibility Number Theory problem, explanation needed I can't understand the solution of the following problem:
$x$,$y$,$z$ are pairwise distinct natural numbers show that $(x-y)^5$ + $(y-z)^5$ + $(z-x)^5$ is divisible by $5(x-y)(y-z)(z-x)$. No need to explain the div. by 5.
The sol. says:
$(x-y)^5$ + $(y-z)^5$ + $(z-x)^5$ is $zero$ for $x=y$, $y=z$, $z=x$. So the terms $(x-y)$, $(y-z)$, $(z-x)$ can be factored out.
This is the 106th problem chap. 6 form "Problem solving strategies" by A. Engel
If you have alternative solution pls feel free to post it.
| Let $x-y=a$ and $y-z=b$.
Thus, $z-x=-(a+b)$ and $$\sum_{cyc}(x-y)^5=a^5+b^5-(a+b)^5=-5a^4b-10a^3b^2-10a^2b^3-5ab^4=$$
$$=-5ab(a^3+2a^2b+2ab^2+b^3)=-5ab(a+b)(a^2-ab+b^2+2ab)$$ and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3731462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.