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The sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is $\frac{n(n+1)(2n+1)}{6}$ Prove that the sum of the first $n$ squares $(1 + 4 + 9 + \cdots + n^2)$ is
$\frac{n(n+1)(2n+1)}{6}$.
Can someone pls help and provide a solution for this and if possible explain the question
| Let it be true for k so for $k^2=\frac{(k)(k+1)(2k+1)}{6}=$..(1). Now we will have to prove for $(k+1)^2$. So we have to prove $(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$. So we start ${(1^2+2^2...k^2+(k+1)^2)}=\frac{(k)(k+1)(2k+1)}{6}.(k^2+2k+1)=\frac{(k+1)(2k^2+k+6k+6)}{6}=\frac{(k+1)(k+2)(2k+3)}{6}$. Hope now you know factorizations and also hope its clear.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1544526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Indefinite integral - What am I doing wrong? We have the integral $\displaystyle \int \sqrt{4-x^2} dx$, and I want to evaluate it using integration by parts. So our first step gives us:
$$ \int \sqrt{4-x^2} dx = x \sqrt{4-x^2} + \int \dfrac{x^2}{\sqrt{4-x^2}} dx $$
Now we have to deal with $\displaystyle \int \dfrac{x^2}{\sqrt{4-x^2}} dx $. I do integration by parts again:
$$\displaystyle \int \dfrac{x^2}{\sqrt{4-x^2}} dx = x^2 \displaystyle \int \dfrac{1}{\sqrt{4-x^2}} dx -2 \displaystyle \int \dfrac{x}{\sqrt{4-x^2}} dx $$
We have $\displaystyle \int \dfrac{1}{\sqrt{4-x^2}} dx = \arcsin(\dfrac{x}{2})$ and $\displaystyle \int \dfrac{x}{\sqrt{4-x^2}} dx = -\sqrt{4-x^2}$, we have $$\displaystyle \int \sqrt{4-x^2} dx = x \sqrt{4-x^2} +x^2 \arcsin(\dfrac{x}{2}) + 2\sqrt{4-x^2} +C $$
However, this is incorrect. Can anyone tell me where I made a mistake?
| Usually for this form of integrand ($\sqrt{a^2-x^2}$) we use a trig sub with $x=a\sin\theta$. In our case $a=2$.
$$x=2\sin\theta, dx=2\cos\theta d\theta$$
Substituting we have
$$\int\sqrt{4-x^2}dx=\int2\cos\theta\sqrt{4-4\sin^2\theta}\,d\theta$$
$$=4\int\cos^2\theta\, d\theta$$
We can rewrite $\cos^2\theta$ as $\cos(2\theta)/2+1/2$.
$$=2\int\cos2\theta\,d\theta+2\theta$$
Let $u=2\theta$, $du=d\theta$, then
$$=\int\cos u \,du+u$$
$$=u+\sin u+C$$
$$=2\theta+\sin2\theta+C$$
Now we use the double angle formula.
$$=2\theta+2\sin\theta\cos\theta+C$$
$$=2\theta+2\sin\theta\sqrt{1-\sin^2\theta}+C$$
Solving for theta in our previous substitution we have $\theta=\arcsin x/2$. Therefore
$$\int \sqrt {4-x^2}=2\arcsin \left(\frac x 2\right) +\frac x 2 \sqrt{4-x^2}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1546594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Choosing numbers at random - expected value calculation From set $\{1,2,\ldots,49 \}$ we choose at random 6 numbers without replacing them. Let X denotes quantity of odd numbers chosen. Find $\mathbb{E}X$, how to find that? I have no idea whatsoever.
EDIT:: still looking for the sufficient explanation.
| We consider the $7$ cases and determine how many odd balls are present in each case.
The $7$ cases are:
*
*$6$ even : $\dbinom{24}{6}$
*$5$ even, $1$ odd : $\dbinom{24}{5}\cdot 25$
*$4$ even, $2$ odd : $\dbinom{24}{4}\cdot \dbinom{25}{2}$
*$3$ even, $3$ odd : $\dbinom{24}{3}\cdot \dbinom{25}{3}$
*$2$ even, $4$ odd : $\dbinom{24}{2}\cdot \dbinom{25}{4}$
*$1$ even, $5$ odd : $24\cdot \dbinom{25}{5}$
*$6$ odd : $\dbinom{25}{6}$
Calculate all of these, multiply each by the corresponding number of odd balls, for example, $\dbinom{24}{4}\cdot \dbinom{25}{2}\times 2$, add them all up and divide by the total number of cases, $\dbinom{49}{6}$
| {
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"url": "https://math.stackexchange.com/questions/1547386",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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How do i solve $\frac{dy}{dx}- \frac {dx}{dy}= \frac {x}{y}-\frac {y}{x}$? I came up with this question in my exam. But i didn't get it right. Can someone show me how to solve this differential equation
$$\frac{dy}{dx}- \frac {dx}{dy}= \frac {x}{y}-\frac {y}{x}$$
| Since $$\frac{dy}{dx}-\frac{dx}{dy}=\frac{dy}{dx}-\frac{1}{\frac{dy}{dx}}=\frac{x}{y}-\frac{y}{x}$$
Let $a=\frac{dy}{dx}$ and $b=\frac{x}{y}$. We are given that $$a-\frac{1}{a}=b-\frac{1}{b}$$
Rearranging gives $$(a-b)(1+\frac{1}{ab})=0$$
Which gives us $$a=b$$ or $$a=-\frac{1}{b}$$
Plugging $a,b$ back, we have $$\frac{dy}{dx}=\frac{x}{y}$$ or $$\frac{dy}{dx}=-\frac{y}{x}$$
If $\frac{dy}{dx}=\frac{x}{y}$, we have $y dy = x dx$, so integration gives $y^2=x^2+C$, or $y=\pm \sqrt{x^2+C}$.
If $\frac{dy}{dx}=-\frac{y}{x}$, we have $-\frac{dy}{y}=\frac{dx}{x}$, so integration gives $-\ln y = \ln x+C$, or $xy=C$, so $y=\frac{C}{x}$.
Therefore, the solution set is $y=\pm \sqrt{x^2+C}$ and $y=\frac{C}{x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1548569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that $\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}+\cdots =F_{n+1}$ where $F_{n+1}$ is the $n+1$ th Fibonacci number I have to proove this this identity which connects Fibonacci sequence and Pascal's triangle:
$$\begin{pmatrix}n\\0\end{pmatrix}+\begin{pmatrix}n-1\\1\end{pmatrix}+\dotsm+\begin{pmatrix}n-\lfloor\frac{n}{2}\rfloor\\\lfloor\frac{n}{2}\rfloor\end{pmatrix} = F_{(n+1)}$$
Example: ${6\choose0} + {5\choose1} + {4\choose2} + {3\choose3} = 13 = F_{7}$
| Let $G_n=\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}\cdots $.
Then, we have
$$G_1=1,G_2=2$$
$$G_{n}+G_{n+1}=\binom{n}{0}+\binom{n-1}{1}+\binom{n-2}{2}\cdots +\binom{n+1}{0}+\binom{n}{1}+\binom{n-1}{2}\cdots $$
$$=\binom{n+1}{0}+\left(\binom n0+\binom{n}{1}\right)+\left(\binom{n-1}{1}+\binom{n-1}{2}\right)+\left(\binom{n-2}{2}+\binom{n-2}{3}\right)+\cdots$$
$$=\binom{n+2}{0}+\binom{n+1}{1}+\binom{n}{2}+\cdots=G_{n+2}$$
So, we know that $G_{n}$ is $n+1$-th Fibonacci number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that the curve has two tangents I'm a little stuck on a math problem that reads as follows:
Show that the curve $x = 5\cos(t), y = 3\sin(t)\, \cos(t)$ has two tangents at $(0, 0)$ and find their equations
What I've Tried
*
*$ \frac{dx}{dt} = -5\sin(t) $
*$ \frac{dy}{dt} = 3\cos^2(t) - 3\sin^2(t) $ because of the product rule. We can simplify this to $ 3(\cos^2(t) - \sin^2(t)) \rightarrow 3\cos(2t) $
*In order to get the slope $ m $, we can write $$ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} $$
*Solving for $ \frac{dy}{dx} $ as follows:
*
*$ \frac{3\cos(2t)}{-5\sin(t)} $ can be rewritten as
*$ \frac{-3}{5}(\cos(2t)\csc(t)) = m $
*Plugging $ (0,0) $ back into the equations of $ x $ and $ y $ we have as follows:
*
*$ 5\cos(t) = 0 \rightarrow t = \frac{\pi}{2} $
*
*Note: I'm unsure what happens to the $ 5 $
*$ \frac{dx}{dt} $ evaluated at $ t = \frac{\pi}{2} $ gives us $ -5\sin(\frac{\pi}{2}) = -5 $
*$ \frac{dy}{dt} $ evaluated at $ t = \frac{\pi}{2} $ gives us $ 3\cos(\frac{2\pi}{2}) \rightarrow 3\cos(\pi) = -3 $
*$ \frac{dy}{dx} = \frac{3}{5} $
*Continuing on, if we add $ \pi $ to the value of $ t $ we get $ t = \frac{3\pi}{2} $. Plug the new value of $ t $ into the equations of $ x $ and $ y $
*
*$ \frac{dx}{dt} $ evaluated at $ t = \frac{3\pi}{2} $ gives us $ -5sin(\frac{3\pi}{2}) = 5 $
*$ \frac{dy}{dt} $ evaluated at $ t = \frac{3\pi}{2} $ gives us $ 3cos(\frac{6\pi}{2}) \rightarrow 3cos(3\pi) = -3 $
*$ \frac{dy}{dx} = -\frac{3}{5} $
*We now have our two slopes of the tangent lines:
*
*$ y = -\frac{5}{3}x $
*$ y = \frac{5}{3}x $
The issue is that webassign is claiming that the slopes are wrong as can be seen here:
Here is the solution in graph form that is correct:
p.s. My apologies if this is a repost. I've seen this response Show that the curve x = 5 cos t, y = 4 sin t cos t has two tangents at (0, 0) and find their equations. and followed it already with no avail.
| You may like to use a short method to prove this.
Eliminate t to get the Cartesian equation of curve as $f(x,y)=9x^4-225x^2+625y^2$
Now, equate the least degree terms to zero to get tangents at $(0,0)$ which gives $y=3x/5$ and $y=-3x/5$.
| {
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"url": "https://math.stackexchange.com/questions/1549432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many 4 digit numbers are divisible by 29 such that their digit sum is also 29? How many $4$ digit numbers are divisible by $29$ such that their digit sum is also $29$?
Well, answer is $5$ but what is the working and how did they get it?
| Given the $4$-digit number $\overline{abcd}$, the sum of digits being divisible by $29$ implies:
$$a+b+c+d=29,$$
because: $29<4\cdot 9<58$.
$\overline{abcd}$ is being divisible by $29$ implies:
$$1000a+100b+10c+d\equiv 14a+13b+10c+d\equiv 0\pmod{29}.$$
Hence:
$$\begin{cases} \ \ \ \ a+\ \ \ \ b+\ \ \ \ c+d=29 \\
14a+13b+10c+d=29n\end{cases}.$$
Multiply the first by $14$ and subtract the second:
$$b+4c+13d=29(14-n),9\le n<14,$$
because: $9+4\cdot 9+13\cdot 9=162<29\cdot 6=174$.
It is easy to check ($2\le b,c,d\le 9$):
$$\begin{align}&n=9 \Rightarrow b+4c+13d=145 \\
& \ \ \ d=9 \Rightarrow (b,c)=\color{red}{(4,6)}; (8,5);\\
&\ \ \ d=8 \Rightarrow (b,c)=(5,9); (9,8); \\
&\ \ \ (b,c,d)=(8,5,9), (5,9,8), (9,8,8);\\
&n=10 \Rightarrow b+4c+13d=116 \\
&\ \ \ d=8 \Rightarrow (b,c)=\color{red}{(4,2)};\\
&\ \ \ d=7 \Rightarrow (b,c)=\color{red}{(5,5)}; (9,4);\\
&\ \ \ d=6 \Rightarrow (b,c)=\color{red}{(2,9)}; (6,8);\\
&\ \ \ (b,c,d)=(9,4,7), (6,8,6);\\
&n=11 \Rightarrow b+4c+13d=87 \\
&\ \ \ d=5 \Rightarrow (b,c)=\color{red}{(2,5)}; \color{red}{(6,4)};\\
&\ \ \ d=4 \Rightarrow (b,c)=\color{red}{(3,8)}; \color{red}{(5,7)}; \color{red}{(9,6)}.
\end{align}$$
Hence:
$$\overline{abcd}=7859; 7598; 4988; 9947; 9686.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Divisibility of an expression Need some guidance
How to prove that $9\cdot n^9+7\cdot n^7+3\cdot n^3+n$ is divisible by $10$.
I've tried transforming the expression by adding $n^9$ and $-n^9$ in order to make a multiple of 10 but no use. I've even tried math. induction but got stuck, maybe this can not be proven. Are there any other methods?
Thanks in advance
| Let $P$ be a polynomial: $P(n)=9n^9+7n^7+3n^3+n$.
$P(n)$ is even for any $n$, because if $n$ is even then you're looking at a sum of $4$ even numbers, if it's odd at a sum of $4$ odd numbers.
Then you only have to prove that $5|P(n)$.
Notice that $n^5\equiv_5 n$ for any $n$. (Proved at the end)
Then $P(n)\equiv_5 -n^9-3n^7+3n^3+n \equiv_5 -n^4n^5-3n^2n^5+3n^3+n \equiv_5 -n^5 - 3n^3 + 3n^3 + n = \ \ n - n^5 \equiv_5 0$
Here's a nice proof of $5|n^5-n$:
$$n^5-n=n(n^4-1)=n(n^2-1)(n^2-1)=n(n^2-1)(n^2+5-4)=n(n^2-1)(n^2-4)+5n(n^2-1) = (n-2)(n-1)n(n+1)(n+2)+5(n-1)n(n+1)$$
Which is a sum of a product of five consecutive integers and something divisible by $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrate $\int \frac{\arctan\sqrt{\frac{x}{2}}dx}{\sqrt{x+2}}$ $$\int \frac{\arctan\sqrt{\frac{x}{2}} \, dx}{\sqrt{x+2}}$$
I've tried substituting $x=2\tan^2y$, and I've got:
$$\frac{1}{\sqrt2}\int\frac{y\sin y}{\cos^4 y} \, dy$$
But I'm not entirely sure this is a good thing as I've been unable to proceed any further from there.
| \begin{align}
u & = \arctan \sqrt{\frac x 2} \\[10pt]
du & = \frac{dx/2}{\left(1+ \dfrac x 2\right)2\sqrt{\dfrac x 2}} = \frac{dx}{(2+x)\sqrt{2x}} \\[10pt]
dv & = \frac{dx}{\sqrt{x+2}} \\[10pt]
v & = 2\sqrt{x+2}
\end{align}
\begin{align}
\int u\,dv & = uv - \int v\,du = 2\sqrt{x+2} \arctan \sqrt{\frac x 2} - \int \frac{\sqrt 2\,dx}{\sqrt{x+2}\sqrt x}
\end{align}
Then
$$
\sqrt{x^2+2x} = \sqrt{(x^2+2x + 1) - 1} = \sqrt{(x+1)^2 -1} = \sqrt{\sec^2\theta - 1},\quad\text{etc.}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1550405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
$\sum_{n=1} ^\infty \frac{8}{n15^n}$ without calculator I just can't figure out which function to use to get the sum. I tried with ln, but that gives me an alternating series.
| Hint
Consider $$A=\sum_{n=1} ^\infty \frac{8}{n15^n}={8}\sum_{n=1} ^\infty \frac{1}{n15^n}={8}\sum_{n=1} ^\infty \frac{x^n}{n}$$ where $x=\frac 1 {15}$.
You can recongnize that the summation is just Taylor expansion of $-\log(1-x)$ from which the problem becomes simple.
We then have $$A=-8\log(1-\frac 1 {15})=-8\log(\frac {14} {15})=8\log(\frac {15} {14})$$ Now, with no calculator, remember the fast convergent expansion $$\log \left(\frac{1+y}{1-y}\right)=2\left( y+\frac{ y^3}{3}+\frac{ y^5}{5}+\frac{ y^7}{7}+O\left(y^9\right)\right)$$ Setting $\frac{1+y}{1-y}=\frac {15} {14}$ gives $y=\frac {1} {29}$. So, $$8\log(\frac {15} {14})=16\left(\frac {1} {29}+\frac{1}{73167}+\frac{1}{102555745}+\frac{1}{120749134163}+\cdots\right)$$ Just using the first and second terms probably gives a sufficient approximation $$16\left(\frac {1} {29}+\frac{1}{73167}\right)=\frac{40384}{73167}\approx 0.5519428158$$ while the exact value would be $\approx 0.5519429719$.
Instead of Taylor series, you could use Padé approximants; a simple one would be $$\log(1+x)\approx \frac{x+\frac{x^2}{2}}{1+x+\frac{x^2}{6}}$$ which, for $x=\frac 1 {14}$, would lead to $$A\approx \frac{696}{1261}=0.5519429025$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Complex equation: $z^8 = (1+z^2)^4$ What's up with this complex equation?
$ z^8 = (1+z^2)^4 $
To start with, there seems to be a problem when we try to apply root of four to both sides of the equation:
$ z^8 = (1+z^2)^4 $
$ z^2 = 1 + z^2 $
which very clearly doesn't have any solutions, but we know there are solutions: the problem is from an exam, and, besides, wolphram alpha gladily gives them to us.
We've tried to solve it using the trigonomectric form, but the sum inside of the parenthesis is killing all of our attempts.
Any help? Ideas?
| $$z^8=(z^2+1)^4\Longleftrightarrow$$
$$z^4=(z^2+1)^2\Longleftrightarrow\space\space\vee\space\space z^4=-(z^2+1)^2\Longleftrightarrow$$
$$z^2=z^2+1\Longleftrightarrow\space\space\vee\space\space z^2=-1-z^2\Longleftrightarrow\space\space\vee\space\space z^4=-(z^2+1)^2\Longleftrightarrow$$
$$0\ne1\space\space\vee\space\space z^2=-1-z^2\space\space\vee\space\space\Longleftrightarrow z^4=-(z^2+1)^2\Longleftrightarrow$$
$$z^2=-1-z^2\Longleftrightarrow\space\space\vee\space\space z^4=-(z^2+1)^2\Longleftrightarrow$$
$$2z^2=-1\Longleftrightarrow\space\space\vee\space\space z^4=-(z^2+1)^2\Longleftrightarrow$$
$$z^2=-\frac{1}{2}\Longleftrightarrow\space\space\vee\space\space z^4=-(z^2+1)^2\Longleftrightarrow$$
$$z=\pm\frac{i}{\sqrt{2}}\Longleftrightarrow\space\space\vee\space\space z^4=-(z^2+1)^2\Longleftrightarrow$$
$$z=\pm\frac{i}{\sqrt{2}}\Longleftrightarrow\space\space\vee\space\space z^4=-z^4-2z^2-1\Longleftrightarrow$$
$$z=\pm\frac{i}{\sqrt{2}}\Longleftrightarrow\space\space\vee\space\space 2z^4+2z^2+1=0\Longleftrightarrow$$
Substitute $x=z^2$:
$$z=\pm\frac{i}{\sqrt{2}}\Longleftrightarrow\space\space\vee\space\space 2x^2+2x+1=0\Longleftrightarrow$$
$$z=\pm\frac{i}{\sqrt{2}}\Longleftrightarrow\space\space\vee\space\space x=\frac{-2\pm\sqrt{-4}}{4}\Longleftrightarrow$$
$$z=\pm\frac{i}{\sqrt{2}}\Longleftrightarrow\space\space\vee\space\space x=\frac{-2\pm 2i}{4}\Longleftrightarrow$$
$$z=\pm\frac{i}{\sqrt{2}}\Longleftrightarrow\space\space\vee\space\space z^2=\frac{-2\pm 2i}{4}\Longleftrightarrow$$
$$z=\pm\frac{i}{\sqrt{2}}\space\space\vee\space\space z=\pm\sqrt{\frac{-2\pm 2i}{4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $\int \frac{x+1}{(x^2-x+8)^3}\, dx$ Could you give me a hint on how to find $$\int \frac{x+1}{(x^2-x+8)^3}\, dx$$
It doesn't seem like partial fractions are the way to go with here and using the integration by parts method seems to be tedious.
I have also tried substituting $(x^2-x+8)$ but it gets even more complicated then.
Is there a way to solve this using only basic formulas?
| As a last resort, it amounts to being able to compute
$$I_n=\int\frac{\mathrm d\mkern 1mu x}{(x^2-x+8)^n}$$
This sort of integrals can be calculated recursively. I'll show how to obtain a relation between $I_2$ and $I_3$. $I_1$ is standard, and is an $\arctan$ after a change of variable.
Integrate $I_2$ by parts, setting
\begin{align*}u&=\frac1{(x^2-x+8)^2},&\mathrm d\mkern 1mu v&=\mathrm d\mkern 1mu x\\
\text{whence}\qquad \mathrm d\mkern 1mu u&=\frac{-2(2x-1)}{(x^2-x+8)^3}\mathrm d\mkern 1mu x,& v&=x
\end{align*}
This yields
\begin{align*}I_2&=\frac{x}{(x^2-x+8)^2}+2\int\frac{(2x^2-x)}{(x^2-x+8)^3}\mathrm d\mkern 1mu x\\
&=\frac{x}{(x^2-x+8)^2}+2\int\frac{(2x^2-2x+16)+(x-8)}{(x^2-x+8)^3}\mathrm d\mkern 1mu x\\
&=\frac{x}{(x^2-x+8)^2}+4I_2+2\int\frac{x-8}{(x^2-x+8)^3}\mathrm d\mkern 1mu x\\
\end{align*}
Now
$$\int\frac{x-8}{(x^2-x+8)^3}\mathrm d\mkern 1mu x=\int\frac{\frac12(2x-1)-\frac{15}2}{(x^2-x+8)^3}\mathrm d\mkern 1mu x=-\frac1{4(x^2-x+8)^2}-\frac{15}2I_3$$
from which we deduce the relation
$$15I_3-3I_2=\frac{2x-1}{(x^2-x+8)^2}.$$
Similarly, you can integrate $I_1$ by parts to obtain a relation between $I_2$ and $I_1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the (complex) solutions $z$ to $z^4 = w$
What are possible solutions to $$z^4 = w$$ where $z \in \mathbb{C} $ and $w \in \mathbb{R}$?
Here is my attempt:
Write both numbers in polar form: $r^4(\cos 4 \theta + i\sin 4 \theta) = w(\cos 0 + i \sin 0) $.
Working with the conventions of $r \geq 0$ and $0 \leq \theta < 2\pi$ I get the following:
$r^4 = w \iff r = w^\frac{1}{4}$
and
$4\theta = n2\pi$ where $n \in \mathbb{Z}$.
$\Rightarrow \theta = \frac{n}{2}\pi$
Thus we get four possible solutions:
$z = w^\frac{1}{4}(\cos 0 + i \sin 0) = w^\frac{1}{4}$
$z = w^\frac{1}{4}(\cos \frac{\pi}{2} + i \sin \frac{\pi}{2})= iw^\frac{1}{4}$
$z = w^\frac{1}{4}(\cos \pi + i \sin \pi) = -w^\frac{1}{4}$
$z = w^\frac{1}{4}(\cos \frac{3\pi}{2} + i \sin \frac{3\pi}{2}) = -iw^\frac{1}{4}$
Is this legit, something missing or very much not rigorous?
| If $w < 0$, then the equation $r = w^{1 / 4}$ is not even sensible, as any fourth root of $w$ is nonreal (and anyway in this case there is not a preferred choice among these) but $r$ is a real variable. The fix is to write any $w < 0$ in polar form as $$w = |w|e^{\pi i} = |w|(\cos \pi + i \sin \pi).$$ Then, regardless of the sign of $w$ we have $r = |w|^{1 / 4}$, and when $w < 0$ our equation in $\theta$ becomes $4 \theta = (2n + 1) \pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1554941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
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What is the integral of $\frac{x-1}{(x+3)(x^2+1)}$? I've worked with partial fractions to get the integral in the form $$\int\frac{A}{x+3} + \frac{Bx + C}{x^2+1}\,dx$$ Is there a quicker way?
| First of all: you have to be carefull with your notation because you mean:
$$\int\left(\frac{A}{x+3} + \frac{Bx + C}{x^2+1}\right)\space\text{d}x$$
And Partial fractions is the most easy way:
$$\int\frac{x-1}{(x+3)(x^2+1)}\space\text{d}x=$$
$$\int\left(\frac{2x-1}{5(x^2+1)}-\frac{2}{5(x+3)}\right)\space\text{d}x=$$
$$\frac{1}{5}\int\frac{2x-1}{x^2+1}\space\text{d}x-\frac{2}{5}\int\frac{1}{x+3}\space\text{d}x=$$
$$\frac{1}{5}\int\left(\frac{2x}{x^2+1}-\frac{1}{x^2+1}\right)\space\text{d}x-\frac{2}{5}\int\frac{1}{x+3}\space\text{d}x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find Least Squares Regression Line I have a problem where I need to find the least squares regression line. I have found $\beta_0$ and $\beta_1$ in the following equation
$$y = \beta_0 + \beta_1 \cdot x + \epsilon$$
So I have both the vectors $y$ and $x$.
I know that $\hat{y}$ the vector predictor of $y$ is $x \cdot \beta$ and that the residual vector is $\epsilon = y - \hat{y}$.
I know also that the least squares regression line looks something like this $$\hat{y} = a + b \cdot x$$
and that what I need to find is $a$ and $b$, but I don't know exactly how to do it. Currently I am using Matlab, and I need to do it in Matlab. Any idea how should I proceed, based on the fact that I am using Matlab?
Correct me if I did/said something wrong anyway.
| Sequence of $m$ measurements:
$$\left\{ x_{k}, y_{k} \right\}_{k=1}^{m}$$
Model:
$$
y(x) = \beta_{0} + \beta_{1} x
$$
Linear system:
$$
\begin{align}
%
\mathbf{A} \, \beta &= y \\
% A
\left[ \begin{array}{cc}
1 & x_{1} \\
1 & x_{2} \\
\vdots & \vdots \\
1 & x_{m}
\end{array} \right]
% beta
\left[ \begin{array}{cc}
\beta_{1} \\
\beta_{2}
\end{array} \right]
%
&=
\left[ \begin{array}{c}
y_{1} \\
y_{2} \\
\vdots \\
y_{m}
\end{array} \right]
%
\end{align}
$$
Least squares solution:
$$
\beta_{LS} = \left\{
\beta \in \mathbb{C}^{2} \colon
\lVert
\mathbf{A} \,x - y
\rVert_{2}^{2}
\text{ is minimized}
\right\}
$$
Solution type: we have full column rank. Solution is unique - a point.
Solution method 1: Normal equations
$$
\begin{align}
%
\mathbf{A}^{*} \,\mathbf{A} \, \beta &= \mathbf{A}^{*} \,y \\
% A
\left[ \begin{array}{cc}
\mathbf{1} \cdot \mathbf{1} & \mathbf{1} \cdot x \\
x \cdot \mathbf{1} & x \cdot x
\end{array} \right]
% beta
\left[ \begin{array}{cc}
\beta_{1} \\
\beta_{2}
\end{array} \right]
%
&=
\left[ \begin{array}{c}
\mathbf{1} \cdot y \\
x \cdot y
\end{array} \right]
%
\end{align}
$$
$$
\Downarrow
$$
$$
\begin{align}
%
\beta &= \left( \mathbf{A}^{*} \, \mathbf{A} \right)^{-1} \mathbf{A}^{*} \,y \\
% beta
\left[ \begin{array}{cc}
\beta_{1} \\
\beta_{2}
\end{array} \right]
%
&=
% inv
\left(
\left( \mathbf{1} \cdot \mathbf{1} \right)
\left( x \cdot x \right)
-
\left( \mathbf{1} \cdot x \right)^{2}
\right)^{-1}
\left[ \begin{array}{rr}
x \cdot x & -\mathbf{1} \cdot x \\
-\mathbf{1} \cdot x & \mathbf{1} \cdot \mathbf{1}
\end{array} \right]
%
\left[ \begin{array}{c}
\mathbf{1} \cdot y \\
x \cdot y
\end{array} \right]
%
\end{align}
$$
Solution method 2: Moore-Penrose pseudoinverse:
$$
\beta = \mathbf{A}^{+} y
$$
The MATLAB intrinsic mldivide is one option.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1556864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof of trignometric identity Could you help me prove this? I've gotten stuck, need some help..
$$\sin^2\Theta + \tan^2\Theta = \sec^2\Theta - \cos^2\Theta$$
Here's what I've done so far:
Left Side:
$$\sin^2\Theta + \frac{\sin^2\Theta}{\cos^2\Theta}=\frac{\sin^2\Theta\cos^2\Theta+\sin^2\Theta}{\cos^2\Theta}$$
Right Side:
$$\frac{1}{\cos^2\Theta} - \cos^2\Theta = \frac{1-\cos^2\Theta\cos^2\Theta}{\cos^2\Theta}$$
Thanks in advance.
Note: I have tried simplifying it even further but I'm not getting the results, so I've left it at the points that I'm sure of.
| Am I the only one here who was required to work on only one side when doing trig proofs?
The strategy for all such proofs is to work on both sides until you can get them to be equal, then retrace your steps and put them all on a single side.
To wit,
$$ \sin^2x + \tan^2x = \sec^2x - \cos^2x$$
Add $\cos^2 x$ to both sides
$$(\sin^2 x + \cos^2 x) + \tan^2 x = \sec^2 x$$
Using a trig identity gives
$$(1) + \tan^2 x = \sec^2 x$$
Which is itself an identity.
Now we achieved this after working on both sides, but this work reveals the "trick": let's use $\sin^2 x + \cos^2 x = 1$ on one side.
Working on the left hand side, we'll use the fundamental identity in this form: $\sin^2 x = 1 - \cos^2 x$.
$$ \sin^2x + \tan^2x = \sec^2x - \cos^2x$$
$$(1 - \cos^2 x) + \tan^2 x = \sec^2x - \cos^2x$$
$$(1 + \tan^2 x) - \cos^2 x = \sec^2x - \cos^2x$$
$$\sec^2 x - \cos^2 x = \sec^2x - \cos^2x$$
Working on the right hand side is similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1557632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An exercise concerning complex numbers Assume that $| z + 1 | > 2$. Show that $|z^3 + 1| > 1$.
My try was:
$$|z^3 + 1| = |z + 1| |z^2 - z + 1| > 2 |z^2 - z + 1| $$
but I'm stuck proving that $|z^2 - z + 1| > \frac 1 2$
| Write $z = -1+ w$, so $|z+1|>2$ means $|w| > 2$. Then
$z^3 + 1 = w^3 - 3 w^2 + 3 w = w (w^2 - 3 w + 3)$. The claim is that
if $|w| > 2$, $|w^2 - 3 w + 3| > 1/2$. If $w = r \exp(i\theta)$,
$$|w^2 - 3 w + 3|^2 = r^4 - 6 r^3 \cos(\theta) + (3 + 12 \cos(\theta)^2) r^2 - 18 r \cos(\theta)+ 9$$
Call that $F(r,\theta)$. We have
$$ \dfrac{\partial F}{\partial \theta} = 6 r \sin(\theta) (r^2 - 4 r \cos(\theta) + 3) $$
Thus for any $r > 2$ the minimum of $F(r,\theta)$ must occur at
one of the values $\theta = 0, \pi$, or $\pm \arccos((r^2+3)/(4r))$
(note that $0 < (r^2+3)/(4r) \le 1$ if $2 < r \le 3$).
For $r>2$ we have $$\eqalign{F(r,0) &= (r^2 - 3 r + 3)^2 > 1\cr
F(r,\pi) &= (r^2+3r+3)^2 > 169\cr F(r,\pm \arccos((r^2+3)/(4r))) &=
\dfrac{(r^2-3)^2}{4} > \dfrac{1}{4}\cr}$$
Thus the minimum is greater than $1/4$, which establishes the claim.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1558075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Sum of a Series With Denominators of the form $(2^i) (3^j)(5^k)$ Can anyone solve this?
Find the sum of the series $1 + \frac{1}{2} +\frac{1}{3} + \frac{1}{4} + \frac{1}{5}+ \frac{1}{6} + \frac{1}{8} + \frac{1}{9} + \cdots,$ where the denominators are of the form $(2^i) (3^j)(5^k)$?
The test came with the next answer choices:
a) $\frac{7}{2}$
b) $1$
c) $3$
d) $\frac{15}{4}$
e) $\frac{3}{2}$
I thought this would tend to infinity. Maybe the test answer choices were wrong..
| We have an Euler product:
$$ \prod_{p\leq 5}\left(1-\frac{1}{p}\right)^{-1} = \prod_{p\leq 5}\left(1+\frac{1}{p}+\frac{1}{p^2}+\ldots \right) = \sum_{n\in A}\frac{1}{n} $$
where $A$ is the set of positive integers whose prime divisors are $\leq 5$.
It follows that our series equals:
$$ 2\cdot \frac{3}{2}\cdot\frac{5}{4} = \color{red}{\frac{15}{4}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Linear Algebra - seemingly incorrect result when looking for a basis
For the following matrix $$ A = \begin{pmatrix}
3 & 1 & 0 & 0 \\
-2 & 0 & 0 & 0 \\
-2 & -2 & 1 & 0 \\
-9 & -9 & 0 & -3
\end{pmatrix} $$
Find a basis for the eigenspace $E_{\lambda}(A)$ of each eigen value.
First step is to find the characteristic polynomial to find the eigenvalues. $$\det(A- I\lambda) = 0 \implies \lambda ^4 - \lambda^3 - 7 \lambda^2 + 13 \lambda -6 = 0 $$ $$\lambda = 1 \lor \lambda = 2 $$ Now that we have the eigen values, to find a basis for $E_{\lambda}(A)$ for the eigen values $\lambda$, we find vectors $\mathbf{v}$ that satisfy $(A-\lambda I)\mathbf{v} = 0$:
$$(A - 2I)\mathbf{v} = \begin{pmatrix}
1 & 1 & 0 & 0 \\
-2 & -2 & 0 & 0 \\
-2 & -2 & -1 & 0 \\
-9 & -9 & 0 & -5
\end{pmatrix} \begin{pmatrix}
v_1 \\
v_2 \\
v_3 \\
v_4
\end{pmatrix} = \begin{pmatrix}
0 \\
0 \\
0 \\
0
\end{pmatrix} \implies
\begin{cases}
v_1 + v_2 = 0\\
-2v_1 -2v_2 = 0 \\
-2v_1 - 2v_2 -v_3 = 0 \\
-9v_1 -9v_2 - 5v_4 = 0
\end{cases}$$ $\implies v_2 = -v_1 \land v_3 = 0 \land v_4 = 0$. So $(v_1, v_2, v_3, v_4) = (v_1, -v_1, 0, 0) = v_1(1, -1, 0, 0)$ so $(1, -1, 0, 0)$ is the basis.
However, this is apparently incorrect. What have I done wrong here?
Also, for the eigenvalue $\lambda = 1$ I do get the correct answer, so only the part above gives me problems.
| If you calculate $\det(A-\lambda I)$ by the first row (and all the minors too) you get
$$
\det(A-\lambda I)=(3-\lambda)\lambda(1-\lambda)(-3-\lambda)+2(1-\lambda)(-3-\lambda)=(3+\lambda)(1-\lambda)(\lambda^2-3\lambda+2)
=(3+\lambda)(1-\lambda)(\lambda-1)(\lambda-2).
$$
So the eigenvalues are $1$ (with multiplicity $2$), $2$, and $-3$.
The eigenvectors (and so the eigenspace) for $2$ you found correctly. You still need to find the eigenspaces for $1$ and for $-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $a^2+b^2\geq \frac{c^2}{2}$ and friends if $a+b\geq c\geq0$ Sorry for my inequality spam, but I got to prepare for my exams today :( Here's another:
Problem:
Prove$$a^2+b^2\geq \frac{c^2}{2}$$$$a^4+b^4\geq \frac{c^4}{8}$$$$a^8+b^8\geq \frac{c^8}{128}$$ if $a+b\geq c\geq0$
Attempt:
Working backwards:
$$a^2+b^2\geq \frac{c^2}{2}$$
$$\implies 2a^2+2b^2\geq c^2$$
I am stuck on the first one, let alone the others. I know by AM-GM, $\frac{a^2+b^2}{2}\geq ab$ but how can I use it here?
| All of them follow from Holder's inequality: for all $a_{ij}>0$:
$$\prod_{i=1}^k\sum_{j=1}^m a_{ij}^k\ge \left(\sum_{j=1}^m\prod_{i=1}^k a_{ij}\right)^k$$
$k=2$ gives Cauchy-Schwarz inequality. In this case, let $k=8, m=2, a_{ij}=1$ for all $i\in\{1,2,\ldots,7\}, j\in\{1,2\}$ and $a_{81}=a, a_{82}=b$:
$$(1^8+1^8)^7(a^8+b^8)\ge (a+b)^8\ge c^8$$
Similarly for the other two. Though this is unneeded for your simple inequalities: $2\left(a^2+b^2\right)\ge a^2+b^2+2ab=(a+b)^2\ge c^2$ and then, like Macavity suggests, replace $a,b,c$ with $a^2,b^2, c^2/2$ or with $a^4,b^4,c^4/2^3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Series' convergence - making my ideas formal
Find the collection of all $x \in \mathbb{R}$ for which the series $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot x^n$ converges.
My first step was the use the ratio test:
$$ \lim_{n \to \infty} \dfrac{(3^{n+1}+n+1) \cdot |x|^{n+1}}{(3n+n) \cdot |x|^n} = \lim_{n \to \infty} \dfrac{3^{n+1}|x|+n|x|+|x| }{3^n+n} = \lim_{n \to \infty} ( \dfrac{3^{n+1}|x|}{3^n+n} + \dfrac{n|x|}{3^n+n} + \dfrac{|x|}{3^n+n}) $$
$$ = \lim_{n \to \infty} (\dfrac{3|x|}{1+\dfrac{n}{3^n}} + \dfrac{|x|}{\dfrac{3^n}{n} + 1} + \dfrac{|x|}{3^n + n}) = 3|x| + 0 + 0 = 3|x| $$
So we want $3|x| < 1$, i.e. $|x| < \dfrac{1}{3}$. But we have to also check the points $x=\dfrac{1}{3}, -\dfrac{1}{3}$, where the limit equals $1$.
For $x=\dfrac{1}{3}$ we have $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot (\dfrac{1}{3})^n = \displaystyle \sum_{n=1}^\infty (1 + n \cdot \dfrac{1}{3}^n) $ which obviously diverges.
For $x=-\dfrac{1}{3}$ we have $\displaystyle \sum_{n=1}^\infty (3^n + n)\cdot (-\dfrac{1}{3})^n = \displaystyle \sum_{n=1}^\infty ((-1)^{n} + n \cdot (-\dfrac{1}{3})^n)$. Now I know this diverges because $(-1)^n$ and $(-\dfrac{1}{3})^n$ have alternating coefficients. But is there a theorem that I can use here?
So we conclude that $|x| < \dfrac{1}{3}$, but is the above work enough to show it conclusively?
| For $x=-1/3$, we can separate the even $n$ and odd $n$ terms and obtaining:
$$A=\sum_{n = 1}^{\infty} \left((-1)^n + n \left(-\frac 1 3\right)^n\right)
=\sum_{k = 1}^{\infty} \left((-1)^{2k} + (2k) \left(-\frac 1 3\right)^{2k}\right)
+\sum_{k = 1}^{\infty} \left((-1)^{2k+1} + (2k+1) \left(-\frac 1 3\right)^{2k+1}\right)$$
$$A=\sum_{k = 1}^{\infty} \left(1+ (2k) \left(\frac 1 3\right)^{2k}-1 - (2k+1) \left(\frac 1 3\right)^{2k+1}\right)=\sum_{k = 1}^{\infty} \left((2k) \left(\frac 1 3\right)^{2k}- (2k+1) \left(\frac 1 3\right)^{2k+1}\right)=\sum_{k = 1}^{\infty} \left(\frac 1 3\right)^{2k+1}\left(4k-1\right)$$
You can now apply the ratio test to the last series.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral of $\sqrt{R^2-x^2}$ Is there any way to compute the integral
\begin{equation}
\int_{-a}^{a} \sqrt{R^2-x^2} \,\mathrm{d}x, \qquad 0<a<R
\end{equation}
without using trig substitution or integration by parts?
I'm thinking to relate this to area of circle, but I couldn't find the relationship.
Thank you.
| Without trigonometric substitution, integration by parts, or appeal to the geometric interpretation of the integral, we need to devise some quite fortuitous manipulations. We proceed, therefore, and write
$$\begin{align}
\sqrt{R^2-x^2}&=\frac12 \left(\sqrt{R^2-x^2}+\sqrt{R^2-x^2}\right)\\\\
&=\frac12 \left(\sqrt{R^2-x^2}+\sqrt{R^2-x^2}\right)+\frac{x^2}{\sqrt{R^2-x^2}}-\frac{x^2}{\sqrt{R^2-x^2}}\\\\
=&\frac12 \left(\sqrt{R^2-x^2}-\frac{x^2}{\sqrt{R^2-x^2}}\right)+\frac12\left(\sqrt{R^2-x^2}+\frac{x^2}{\sqrt{R^2-x^2}}\right)\\\\
&=\frac12 \frac{d\left(x\sqrt{R^2-x^2}\right)}{dx}+\frac12\left(R^2-x^2\right)\left(\frac{1}{\sqrt{R^2-x^2}}+\frac{x^2}{(R^2-x^2)^{3/2}}\right)\\\\
&=\frac12 \frac{d\left(x\sqrt{R^2-x^2}\right)}{dx}+\frac12\left(R^2-x^2\right)\frac{d\left(x/\sqrt{R^2-x^2}\right)}{dx}\\\\
&=\frac12 \frac{d\left(x\sqrt{R^2-x^2}\right)}{dx}+\frac12R^2\left(\frac{1}{1+\left(\frac{x}{\sqrt{R^2-x^2}}\right)^2}\right)\frac{d\left(x/\sqrt{R^2-x^2}\right)}{dx}\\\\
&=\frac{d}{dx}\left(\frac12 x\sqrt{R^2-x^2}+\frac12 R^2 \arctan\left(\frac{x}{\sqrt{R^2-x^2}}\right) \right)
\end{align}$$
Therefore, we arrive at
$$\begin{align}
\int_{-a}^a\sqrt{R^2-x^2}\,dx&=2\int_0^a \sqrt{R^2-x^2}\,dx\\\\
&=\left.\left( x\sqrt{R^2-x^2}+ R^2 \arctan\left(\frac{x}{\sqrt{R^2-x^2}}\right) \right)\right|_0^a\\\\
&= a\sqrt{R^2-a^2}+ R^2 \arctan\left(\frac{a}{\sqrt{R^2-a^2}}\right)
\end{align}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Fun Q6: Side length of the pentagon in a five sided star? Consider a regular pentagon of side length $a$. If you form a 5-sided star using the vertices of the pentagon, then you'll get a pentagon inside that star. What is the side length of that pentagon?
In general, for a n-sided star, what is the side length of the n-sided regular polygon in the star? Take the distance between two adjacent vertices be $a$
| Let $x$ be the length of small n-sided regular polygon in the star & $a$ be the distance between two adjacent vertices, then the angle of spike of regular star polygon is given as
$$\alpha=\frac{\pi}{\text{number of vertices (points) in the star}}=\frac{\pi}{n}$$
Now, draw a perpendicular from one vertex of star to the side of the small regular polygon to obtain a right triangle .
Using geometry of right triangle, the length of perpendicular drawn to the side of small regular polygon can be obtained
$$=\frac{a}{2}\csc\frac{\pi}{n}-\frac{x}{2}\cot\frac{\pi}{n}$$
Hence, in right triangle, one should have
$$\tan\frac{\pi}{2n}=\frac{\frac{x}{2}}{\frac{a}{2}\csc\frac{\pi}{n}-\frac{x}{2}\cot\frac{\pi}{n}}$$
$$x=\frac{a\tan\frac{\pi}{2n}\csc\frac{\pi}{n}}{1+\tan\frac{\pi}{2n}\cot\frac{\pi}{n}}$$
$$x=\frac{a\sin\frac{\pi}{2n}}{\sin\frac{\pi}{n}\cos\frac{\pi}{2n}+\cos\frac{\pi}{n}\sin\frac{\pi}{2n}}$$
$$x=\frac{a\sin\frac{\pi}{2n}}{\sin\left(\frac{\pi}{n}+\frac{\pi}{2n}\right)}$$
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{x=\frac{a\sin\frac{\pi}{2n}}{\sin\frac{3\pi}{2n}}}}$$
$$\forall \ \ n=2k+1\ \ (k\in N)$$
Hence for a regular pentagon in the star, setting $n=5$, the side of regular pentagon
$$x=\frac{a\sin\frac{\pi}{10}}{\sin\frac{3\pi}{10}}=a\frac{\sin18^\circ}{\cos36^\circ}=a\frac{\frac{\sqrt 5-1}{4}}{\frac{\sqrt 5+1}{4}}=\color{red}{\frac{a}{2}(3-\sqrt 5)}$$
Edited details: If the star regular polygon has $2n$ no. of vertices which is obtained by placing two congruent $n$-sided regular polygons one on the other in symmetrical staggered manner similar to a hexagram then a generalized formula for calculating side $x$ of $2n$-sided regular polygon in the star (having $2n$ no. of vertices & $a$ is the distance between two adjacent vertices) can be derived as follows (see figure below)
The angle of spike of regular star polygon is given as
$$\alpha=\text{interior angle of a n-sided regular polygon}=\frac{(n-2)\pi}{n}$$
Now, draw a perpendicular from one vertex of star to the side of small regular polygon to obtain a right triangle.
Using geometry of right triangle, the length of perpendicular drawn to the side of small regular polygon can be obtained
$$=\frac{a}{2}\csc\frac{\pi}{2n}-\frac{x}{2}\cot\frac{\pi}{2n}$$
Hence, in right triangle, one should have
$$\tan\frac{(n-2)\pi}{2n}=\frac{\frac{x}{2}}{\frac{a}{2}\csc\frac{\pi}{2n}-\frac{x}{2}\cot\frac{\pi}{2n}}$$
$$\cot\frac{\pi}{n}=\frac{x}{a\csc\frac{\pi}{2n}-x\cot\frac{\pi}{2n}}$$
$$x=\frac{a\csc\frac{\pi}{2n}\cot\frac{\pi}{n}}{1+\cot\frac{\pi}{n}\cot\frac{\pi}{2n}}$$
$$x=\frac{a\cos\frac{\pi}{2n}}{\cos\frac{\pi}{n}\cos\frac{\pi}{2n}+\sin\frac{\pi}{n}\sin\frac{\pi}{2n}}$$
$$x=\frac{a\cos\frac{\pi}{n}}{\cos\left(\frac{\pi}{n}-\frac{\pi}{2n}\right)}$$
$$\bbox[5px, border:2px solid #C0A000]{\color{blue}{x=\frac{a\cos\frac{\pi}{n}}{\cos\frac{\pi}{2n}}}}$$
$$\forall \ \ \ \ n\ge 3\ \ (n\in N)$$
Hence for a regular hexagon in the hexagram (see in the above diagram) , setting $2n=6$ or $n=3$, the side of regular hexagon
$$x=\frac{a\cos\frac{\pi}{3}}{\cos\frac{\pi}{6}}=a\frac{\frac{1}{2}}{\frac{\sqrt 3}{2}}=\color{red}{\frac{a}{\sqrt3}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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find the matrix representation of $T$ relative to $B,B'$ where $T(p(x)) = (x+1)p'(x^2)$ Hi I really need help with this linear algebra question.
Let $T:P_2\to P_3$ be defined by $T(p(x)) = (x+1)p'(x^2)$ and let $B = (1,x+1,x^2+x)$ and $B' = (x^3,x^3+x,x^2+x,x+1)$ be ordered basis for $P_2$ and $P_3$ respectively.
a.) find the matrix representation of $T$ relative to $B,B'$
b.) Use your answer for part a.) to compute $T(x^2+3x+5)$
So far I have evaluated:
$T(p(1)) = 0$
$T(p(x+1)) = 2x^2+4x+2$
$T(p(x^2+x)) = 4x^4+10x^3+8x^2+2x$
but this is where i get confused, the third polynomial has degree 4 which is not in $P_3$ right? did i make an error in my calculation somewhere or am i going about this completely wrong? any help is greatly appreciated thanks.
| So, first of all, the expression $T(p(1))$ doesn't make sense. The symbol $p(x)$ is supposed to represent a polynomial. So your computation would look like this:
If $p(x) = 1$, then $p'(x) = 0$, and so $p'(x^2) = 0$. So $T(1) = (x+1)(0) = 0$.
If $p(x) = x + 1$, then $p'(x) = 1$, and so $p'(x^2) = 1$. So $T(x + 1) = (x + 1)(1) = x + 1$.
If $p(x) = x^2 + x$, then $p'(x) = 2x + 1$, and so $p'(x^2) = 2x^2 + 1$. So $T(x^2 + x) = (x + 1)(2x^2 + 1) = 2x^3 + 2x^2 + x + 1$.
Now you have to find each of your answers in $B'$ coordinates.
$T(1) = 0$, which is $\begin{bmatrix} 0 \\ 0 \\ 0 \\ 0 \end{bmatrix}$ in $B'$ coordinates.
$T(x + 1) = x + 1$, which is the fourth element of $B'$. So $T(x + 1) = \begin{bmatrix} 0 \\ 0 \\ 0 \\ 1 \end{bmatrix}$.
$T(x^2 + x) = 2x^3 + 2x^2 + x + 1$. It isn't obvious how to write this as a linear combination of elements in $B'$, but we can figure this out using 'standard' coordinates on $P_3$. That is, using coordinates from the basis $\{x^3,x^2,x,1\}$. The resulting linear system is:
$$\begin{bmatrix} 1 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 0 & 1 \end{bmatrix}\mathbf{x} = \begin{bmatrix} 2 \\ 2 \\ 1 \\ 1 \end{bmatrix}$$
Where $\mathbf{x}$ is the desired $B'$-coordinate vector of $2x^3 + 2x^2 + x + 1$. Some row reduction gives:
$\mathbf{x} = \begin{bmatrix} 4 \\ -2 \\ 2 \\ 1 \end{bmatrix}$
And so the matrix for $T$ is:
$$\begin{bmatrix} 0 & 0 & 4 \\ 0 & 0 & -2 \\ 0 & 0 & 2 \\ 0 & 1 & 1 \end{bmatrix}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to take an integral using half angle trigonometric substitution. So i have this question which is asking to take the integral using a predefined trigonometric substitution which is $$u=\tan\frac{x}{2}$$
and the integral equation is $$\int\frac{\sin x\ dx}{(6\cos x-2)(3-2\sin x)}$$ How would i go on about this problem? Because to begin with i do not know how i would even use the given substitution method. Any help is appreciated thank you.
| By letting $u=\tan\frac{x}2$, we get
$$\int\frac{\sin x}{(6\cos x-2)(3-2\sin x)}d x=\int\frac{u d u}{(1-2u^2)(3u^2-4u+3)}\\
=\frac{8}{49}\int\frac{(9u+4)du}{1-2u^2}+\frac{12}{49}\int\frac{(9u-8)du}{3u^2-4u+3},$$
the later two are both integration of rational functions.
$$\int\frac{(9u+4)du}{1-2u^2}=9\int\frac{udu}{1-2u^2}+4\int\frac{du}{1-2u^2}\\
=-\frac94\int\frac{d(1-2u^2)}{1-2u^2}+\sqrt2\int\left(\frac{1}{1-\sqrt2u}+\frac{1}{1-\sqrt2u}\right)du\\
=-\frac94\ln|1-2u^2|+\sqrt2\ln\frac{1+\sqrt2u}{1-\sqrt2u}+C_1$$
$$\int\frac{(9u-8)du}{3u^2-4u+3}=\frac13\int\frac{9\left(u-\frac23\right)-2}
{\left(u-\frac23\right)^2+\frac59}du=3\int\frac{\left(u-\frac23\right)}
{\left(u-\frac23\right)^2+\frac59}du+\frac23\int\frac{du}
{\left(u-\frac23\right)^2+\frac59}\\
=\frac32\int\frac{d\left(u-\frac23\right)^2}{\left(u-\frac23\right)^2+\frac59}
+\frac2{\sqrt5}\int\frac{d\frac3{\sqrt5}\left(u-\frac23\right)}{\left[\frac3{\sqrt5}\left(u-\frac23\right)\right]^2+1}\\
=\frac32\ln\left(u^2-\frac43u+1\right)-\frac2{\sqrt5}\arctan\left[\frac3{\sqrt5}\left(u-\frac23\right)\right]+C_2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Triangle of numbers sum What is the total of all these numbers?
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
. . . . . .
1 2 ........N
The answer should be a single binomial coefficient. It looks pretty easy, but I cannot come to a solution.
| The total of the $1$'s is $N$. The total of the $2$'s is $2(N-1)$, the toal of the $3$'s is $3(N-2)$,..., the total of the $N$'s is $N(N-(N-1))=N$.
Throwing this into a sum we find:
$$\sum_{i=1}^N i \cdot (N-(i-1)) = \sum_{i=1}^N i \cdot N - \sum_{i=1}^N i^2 + \sum_{i=1}^N i$$
$$= N \frac{N(N+1)}{2} - \frac{N(N+1)(2N+1)}{6} + \frac{N(N+1)}{2}$$
$$= \frac{N(N+1)^2}{2} - \frac{N(N+1)(2N+1)}{6}.$$
We can factor this into $$N(N+1) \left( \frac{(N+1)}{2} - \frac{(2N+1)}{6} \right)$$
$$=N(N+1) \left( \frac{N+2}{6} \right) = \frac{N(N+1)(N+2)}{6} = \frac{(N+2)!}{(N-1)!3!} = { N+2 \choose 3 }$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Help on solving the equation $\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{a+x}}=\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{a-x}}$ Could you give me some help on finding the roots (if any) of the following equation:
$$
\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{a+x}}=\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{a-x}}
$$
I tried to apply some classic approaches, but I had no luck... Could you lend me a hand? Thanks in advance!
| $$\frac{\sqrt{a+x}}{\sqrt{a}+\sqrt{a+x}}=\frac{\sqrt{a-x}}{\sqrt{a}-\sqrt{a-x}}$$
$$\frac{\sqrt{1+\frac xa}}{\sqrt{1}+\sqrt{1+\frac xa}}=\frac{\sqrt{1-\frac xa}}{\sqrt{1}-\sqrt{1-\frac xa}}$$
$$\frac xa \mapsto y$$
$$\frac{\sqrt{1+y}}{1+\sqrt{1+y}}=\frac{\sqrt{1-y}}{1-\sqrt{1-y}}$$
$$\sqrt{1+y}-\sqrt{1-y^2}=\sqrt{1-y}+\sqrt{1-y^2}$$
$$\sqrt{1+y}-\sqrt{1-y}=2\sqrt{1-y^2}$$
$$1+y+1-y-2\sqrt{1-y^2}=4(1-y^2)$$
$$4y^2-2=2\sqrt{1-y^2}$$
$$2y^2-1=\sqrt{1-y^2}$$
$$4y^4-4y^2+1=1-y^2$$
$$4y^4-3y^2=0$$
$$y^2(4y^2-3)=0$$
$$y=0,\quad y=\pm\frac{\sqrt3}2$$
$$x=0,\quad x=\pm\frac{\sqrt3}2a$$
| {
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"url": "https://math.stackexchange.com/questions/1568526",
"timestamp": "2023-03-29T00:00:00",
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The probability that each child gets at least one chocolate is?
5 different chocolates are to be distributed among 4 children.The
probability that each child gets at least one chocolate is ?
Total number of ways is $4^5$.Got that.After that what to do?
| We count the number of "favourables." The numbers involved are very small, so we use a counting procedure that would be inefficient for larger numbers.
The lucky child can be chosen in $\binom{4}{1}$ ways. For each such way, the chocolates she gets can be chosen in $\binom{5}{2}$ ways. And for every way of doing this, there are $3!$ ways to distribute the rest of the chocolates, one to each remaining child.
Remark: The total number of ways to distribute the chocolates is $4^5$, not $5^4$. For each of chocolates $C_1$ to $C_5$, there are $4$ ways to choose the child it will be given to.
Added: We sketch the standard Inclusion/Exclusion approach to the problem. There are $4^5$ ways to distribute the chocolates with no restriction. We now count the bad ways, in which at least one sad little child gets nothing. Call the children A, B, C, D. There are $3^5$ ways to distribute the chocolates so that A gets nothing, for we are distributing $5$ chocolates among $3$ kids. The same is true for B getting nothing, and so on. So our first estimate of the number of bad distributions is $4\cdot 3^5$, which we write as $\binom{4}{1}3^5$.
However, we have double-counted the number of ways in which, for example, A and B both get nothing. There are $\binom{4}{2}$ ways to choose two children to get nothing, and for each way there are $2^5$ ways to distribute the chocolates among the other two kids.
So our second estimate for the number of bad distributions is $\binom{4}{1}3^5-\binom{4}{2}2^5$. However, we have taken away too much, for we have removed once too many times the $\binom{4}{3}1^5$ ways in which all the chocolates go to one kid. We conclude that the number of bad distributions is
$$\binom{4}{1}3^5-\binom{4}{2}2^5+\binom{4}{3}1^5.$$
Subtract this from the total $4^5$ to find the number of good distributions.
The above idea works in general.
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit exists theoritically,but graphically there is a vertical asymptote there.Why is it so. In finding the limit of $\lim_{x\to 0}\frac{2^x-1-x}{x^2}$
I used the substitution $x=2t$
$\lim_{x\to 0}\frac{2^x-1-x}{x^2}=\lim_{t\to 0}\frac{2^{2t}-1-2t}{4t^2}=\frac{1}{4}\lim_{t\to 0}\frac{2^{2t}-2\times2^t+2\times2^t+1-2-2t}{t^2}$
$=\frac{1}{4}\lim_{t\to 0}\frac{(2^{t}-1)^2+2\times2^t-2-2t}{t^2}$
$=\frac{1}{4}\lim_{t\to 0}\frac{(2^{t}-1)^2}{t^2}+\frac{1}{2}\lim_{t\to 0}\frac{2^t-1-t}{t^2}$
$\lim_{x\to 0}\frac{2^x-1-x}{x^2}=\frac{1}{2}(\ln2)^2$
But when i see the graph of the function,there is a vertical asymptote at $x=0$
And before $x=0$,function is approaching $\infty$ and after $x=0$,function is approaching $-\infty$.
That means limit should not exist,theoritically limit is coming but graphically limit does not exist.What is wrong?Why is it so?I do not understand.Please help me.
| $\lim_{x\to 0}\frac{2^x-1-x}{x^2}
$
For small $x$,
$2^x
=e^{x \ln 2}
\approx 1+x \ln 2+O(x^2)
$
so
$\frac{2^x-1-x}{x^2}
=\frac{1+x \ln 2 +O(x^2)-1-x}{x^2}
=\frac{x (\ln 2-1) +O(x^2)}{x^2}
=\frac{(\ln 2-1) +O(x)}{x}
\to \infty
$
as
$x \to 0$.
In general,
$\lim_{x \to 0}\frac{a^x-1-x\ln a}{x^2}
$
exists
and is
$\begin{array}\\
\lim_{x \to 0}\frac{e^{x \ln a}-1-x\ln a}{x^2}
&=\lim_{x \to 0}\frac{1+(x \ln a)+(x \ln a)^2/2+O(x^3)-1-x\ln a}{x^2}\\
&=\lim_{x \to 0}\frac{(x \ln a)^2/2+O(x^3)}{x^2}\\
&=\lim_{x \to 0}( \ln a)^2/2+O(x)\\
&=( \ln a)^2/2\\
\end{array}
$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve $x^2\equiv -3\pmod {\!91}$ by CRT lifting roots $\!\bmod 13\ \&\ 7$ Question 1) Solve $$x^2\equiv -3\pmod {13}$$
I see that $x^2+3=13n$. I don't really know what to do? Any hints?
The solution should be $$x\equiv \pm 6 \pmod {13}$$
Question 2) $\ $ [note $\bmod 7\!:\ x^2\equiv -3\equiv 4\iff x\equiv \pm 2.\,$ Here we lift to $\!\!\pmod{\!91}\ $ -Bill]
Given $x\equiv \pm 6 \pmod {13}$ and $x\equiv \pm 2 \pmod {7}$ find solutions $\pmod {91}$. I see that $91=13 \times 7$, does it mean I have to use chinese remainder theorem on 4 equations? If,so $x=6\times 13\times 7 \times 7\times (13\times 7 \times 7)^{-1}...$
| The numbers are small, and one could find the answers without the full CRT machinery. But we will go ahead and use a "general" procedure.
The idea is that to solve $x\equiv a\pmod{7}$, $x\equiv b\pmod{13}$, you use
$$x\equiv (C)(13)(a)+(D)(7)(b)\mod{91},$$
where $C$ is the inverse of $13$ modulo $7$ and $D$ is the inverse of $7$ modulo $13$. We can see I think easily that $C=-1$ and $D=2$ will work, so we get
$$x\equiv -13a+14b\pmod{91}.\tag{1}$$
Now let us for example take $a=-2$ and $b=6$, one of your $4$ possibilities. That gives $x\equiv 110\equiv 19\pmod{91}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Green's Theorem on Line Integral I am asked to find the line integral for the following field:
$$F = (e^{y^2}-2y)i + (2xye^{y^2}+\sin(y^2))j$$
On the line segment with points $(0,0),(1,2)$ and $(3,0)$. I have to do it with Greens theorem. This is the setup I have so far.
$$\int_C F \cdot dr = \iint_D \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\, dA$$
$$\frac{\partial Q}{\partial x} = ye^{y^2} \qquad \frac{\partial P}{\partial y}=2ye^{y^2}-2$$
Line from $(0,0)$ to $(1,2)$ is $y = 2x$. Line from $(1,2)$ to $(3,0)$ is $y = -x + 3$
$$\int_0^1\int_0^{2x}ye^{y^2}-2ye^{y^2}-2 + \int_1^3\int_0^{-x+3}ye^{y^2}-2ye^{y^2}-2 $$
$$\int_0^1\int_0^{2x}-ye^{y^2}-2 \, dy\,dx+ \int_1^3\int_0^{-x+3}-ye^{y^2}-2 \,dy\,dx$$
$$\int_0^1\left[\frac{e^{y^2}}{2} - 2y\right]_0^{2x} + \int_1^3\left[-\frac{e^{y^2}}{2} - 2y\right]_0^{-x+3}$$
$$\int_0^1\frac{e^{4x^2}}{2}-\frac{1}{2}-4x \,dx+ \int_1^3-\frac{e^{(-x+3)^2}}{2}-2(-x+3)+\frac{1}{2}\,dx$$
Let $u = -x + 3$ and $du = -1$.
$$\int_0^1\frac{e^{4x^2}}{2}-\frac{1}{2}-4x \,dx+ \int_1^3\frac{e^{(u)^2}}{2}+2(u)-\frac{1}{2}\,du$$
$$\left[\frac{e^{4x^2}}{8x}-\frac{x}{2}-2x^2 \right]_0^1+ \left[\frac{e^{(u)^2}}{4u}+u^2+\frac{u}{2}\right]_1^3$$
$$\left[\frac{e^{4x^2}}{8x}-\frac{x}{2}-2x^2 \right]_0^1+ \left[\frac{e^{(-x+3)^2}}{4(-x+3)}+(-x+3)^2+\frac{-x+3}{2}\right]_1^3$$
$$\frac{e^{4}}{8}-\frac{1}{2}-2 \,-\frac{e^{4}}{8}-4-1$$
$$\frac{1}{2}-2 \,-5$$
But the answer key says the answer is $-3$. Where have I gone wrong?
| The problem lies in two places: the first one in $$\frac{\partial Q}{\partial x} = 2ye^{y^2},$$ but this is only algebraic, even though it made your life much harder.
The second one is much more sensible. Look at Green's Theorem
$$\oint_C (P dx + Q dy) = \iint_D \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\,dA,$$ where D is CLOSED domain and $C = \partial D$.
Therefore, the integral you are suppose to have is
$$\int_C F \cdot dr = -\left[\iint_{\triangle} \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\, dA + \int_3^0 F(t,0) \cdot (1,0) \, dt\right],$$
where, $\triangle$ is the triangular domain of the question, $((0,0),(1,2),(3,0))$, and the the last term is the integral to bound you domain. Note that we integrate from $3 \to 0$ in order to close the domain, and we need the negative of RHS, because the chosen parametrization, that needs to respect the right-hand rule.
So, let
$$G(x,y) = \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = 2$$
We have
\begin{align}
\int_C F \cdot dr &= -\left[ \int_0^1\int_0^{2x} G(x,y) \, dy\,dx + \int_1^3\int_0^{-x+3}G(x,y) \,dy\,dx + \int_3^0 F(t,0) \cdot (1,0) \, dt,\right]\\
&= -\left[ \int_0^1\int_0^{2x} 2 \, dA + \int_1^3\int_0^{-x+3} 2 \,dA + \int_3^0 1 \, dt,\right]\\
&= -\left[ 2\int_0^1\int_0^{2x} \, dA + 2\int_1^3\int_0^{-x+3} \,dA - \int_0^3 \, dt,\right]\\
&=-\left[2\dfrac{1\times2}{2} + 2\dfrac{2 \times 2}{2} - 3 \right]\\
&=-3
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Differentiating $x^2=\frac{x+y}{x-y}$ Differentiate:
$$x^2=\frac{x+y}{x-y}$$
Preferring to avoid the quotient rule, I take away the fraction:
$$x^2=(x+y)(x-y)^{-1}$$
Then:
$$2x=(1+y')(x-y)^{-1}-(1-y')(x+y)(x-y)^{-2}$$
If I were to multiply the entire equation by $(x-y)^2$ then continue, I get the solution. However, if I continue the following, I don't. Likely some place I erred, but I cannot figure out where:
Expansion:
$$2x=(x-y)^{-1}+y'(x-y)^{-1}-(x+y)(x-y)^{-2}+y'(x+y)(x-y)^{-2}$$
Preparing to isolate for $y'$:
$$2x-(x-y)^{-1}+(x+y)(x-y)^{-2}=y'(x-y)^{-1}+y'(x+y)(x-y)^{-2}$$
$$2x-(x-y)^{-1}+(x+y)(x-y)^{-2}=y'[(x-y)^{-1}+(x+y)(x-y)^{-2}]$$
Isolating $y'$:
$$y'=\frac{2x-(x-y)^{-1}+(x+y)(x-y)^{-2}}{(x-y)^{-1}+(x+y)(x-y)^{-2}}$$
Multiple top and bottom by $(x-y)$:
$$y'=\frac{2x(x-y)-1+(x+y)(x-y)^{-1}}{1+(x+y)(x-y)^{-1}}$$
Then, inserting $x^2$ into $(x+y)(x-y)^{-1}$, I get:
$$y'=\frac{2x(x-y)-1+x^2}{1+x^2}$$
While the answer states:
$$y'=\frac{x(x-y)^2+y}{x}$$
Which I do get if I multiplied the entire equation by $(x-y)^2$ before. It does not seem to be another form of the answer, as putting $x=2$, the denominator cannot match each other. Where have I gone wrong?
| You are forgetting that there is a relation between $x$ and $y$. Your answer and the official "correct" answer are the same. One way to see this: make the substitution $x^2=\frac{x+y}{x-y}$ into your answer and:
$$\begin{align}
\frac{2x(x-y)-1+x^2}{1+x^2}&=\frac{2x(x-y)-1+\frac{x+y}{x-y}}{1+\frac{x+y}{x-y}}\\
&=\frac{2x(x-y)^2-(x-y)+(x+y)}{(x-y)+(x+y)}\\
&=\frac{2x(x-y)^2+2y}{2x}\\
&=\frac{x(x-y)^2+y}{x}\\
\end{align}$$
| {
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The value of double integral $\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dx\,dy$? Given double integral is :
$$\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dx\,dy$$
My attempt :
We can't solve since variable $x$ can't remove by limits, but if we change order of integration, then
$$\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dy\,dx$$
$$\implies\int _0^1\int _0^{\frac{1}{x}}\frac{x}{1+y^2}\:dy\,dx = \frac{1}{2}$$
Can you explain in formal way, please?
Edit : This question was from competitive exam GATE. The link is given below on comments by Alex M. and Martin Sleziak(Thanks).
| Right way is:
$$\int\limits_0^1x\:dx\int\limits_0^\dfrac1x \dfrac{dy}{1+y^2} = \int\limits_0^1\arctan y\:\Biggl.\Biggr|_0^{\dfrac1x} x\:dx =\int\limits_0^1x\arctan \dfrac1x\:dx = \int\limits_0^1\left(\dfrac{\pi}2-\arctan x\right)x\:dx =$$$$ \left.\dfrac{\pi}2\dfrac{x^2}2\right|_0^1 -\int\limits_0^1\arctan x\: d\dfrac{x^2}2 = \dfrac{\pi}4-\left.\dfrac{x^2}2\arctan x\right|_0^1 +{\dfrac12\int\limits_0^1\dfrac{x^2}{1+x^2}\:dx} =$$$$ \dfrac{\pi}4-\dfrac{\pi}8 + \dfrac12\int\limits_0^1\left(1 - \dfrac1{1+x^2}\right)dx = \dfrac{\pi}8+\left.\dfrac12(x-\arctan x)\right|_0^1=\dfrac12$$
To change order of integral, you build the region of integration in the chart, where you can see that the region of integration is made up of a square and curved trapezoid, so:
$$\int\limits_0^1x\:dx\int\limits_0^\dfrac1x \dfrac1{1+y^2}\:dy = \int\limits_0^1 \dfrac1{1+y^2}\:dy\int\limits_0^1 x\:dx + \int\limits_1^\infty \dfrac1{1+y^2}\:dy\int\limits_0^\dfrac1y x\:dx = $$$$ \arctan y\:\Biggl.\Biggr|_0^1\cdot\left.\dfrac{x^2}2\right|_0^1 + \int\limits_1^\infty \left.\dfrac{x^2}2\right|_0^\dfrac1y\dfrac1{1+y^2}dy = \dfrac{\pi}8 + \dfrac12\int\limits_1^\infty \dfrac1{1+y^2}\dfrac1{y^2}\:dy = $$$$\dfrac{\pi}8+\dfrac12\int\limits_1^\infty \left(\dfrac1{y^2}-\dfrac1{1+y^2}\right)\:dy = \dfrac{\pi}8+\dfrac12\left(-\dfrac1y-\arctan y\right)\Biggr.\Biggr|_1^\infty = \dfrac12$$
And this way of calculation does not seem more simple.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1579690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
why $ 1 - \cos^2x = \sin^2x $? I'm trying to prove this result $$\lim_{x\to 0} \frac{1 - \cos(x)}{x} = 0$$ In this process I have come across an identity $1-\cos^2x=\sin^2x$. Why should this hold ? Here are a few steps of my working:
\begin{array}\\
\lim_{x\to 0} \dfrac{1 - \cos(x)}{x}\\ = \lim_{x\to 0} \left[\dfrac{1 - \cos(x)}{x} \times \dfrac{1 + \cos(x)}{1 + \cos(x)}\right] \\
=\lim_{x\to 0} \left[\dfrac{1 - \cos^2(x)}{x(1+\cos(x))}\right] \\
=\lim_{x\to 0} \left[\dfrac{\sin^2(x)}{x(1+\cos(x))}\right]
\end{array}
| Let $F(x)=\sin ^2 x + \cos ^ 2 x$.
$$F'(x)=2 \sin x\cos x- 2 \cos x\sin x=0$$
Since $F(0)=1$ and $F$ is constant, we get
$$\sin ^2 x + \cos ^ 2 x=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
For give permutation $\sigma\in S_{13}$ solve equation $x^3=\sigma$ We have this permutation
$$\sigma =\left({\begin{array}{*{20}c}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13\\ 3 & 2 & 1 & 11 & 5 & 8 & 13 & 10 & 9 & 12 & 4 & 6 & 7\end{array}}\right)\ $$
Find $x^3=\sigma\in S_{13}$
The cycles are (1 3)(2)(4 11)(5)(6 8 10 12)(9)(13 7)
| The permutation with the cycles $(1\ 3)(2)(4\ 11)(5)(6\ 12\ 10\ 8)(9)(13\ 7)$ does the job.
You only have to invert the $4$-cycle. The other cycles remain. You can also take the cycle $(2\ 5\ 9)$ to get another solution.
| {
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"url": "https://math.stackexchange.com/questions/1585581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How find gcd polynomials? How to find gcd of polynomials $gcd(x^3+x^2-x-1,3x^2+2x-1)$ ??
I divide of polynomials. It worked like this $\frac 13 x - \frac19$,$ R\left( x\right) =-\frac89x -\frac89$
| You want the gcd of $f(x) = x^3+x^2-x-1$ and its derivative. Now
$$f(x) = x^2 (x+1) - (x+1) = (x^2-1)(x+1) = (x-1)(x+1)^2$$ so
$$\gcd(f(x), f'(x)) = x+1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove $1^2-2^2+3^2-4^2+......+(-1)^{k-1}k^2 = (-1)^{k-1}\cdot \frac{k(k+1)}{2}$ I'm trying to solve this problem from Skiena book, "Algorithm design manual".
I don't know the answer but it seems like the entity on the R.H.S is the summation for series $1+2+3+..$. However the sequence on left hand side is squared series and seems to me of the form:
$-3-7-11-15\ldots $
I feel like its of the closed form:
$\sum(-4i+1)$
So how do I prove that the equality is right?
| No one's pointing out my favorite result that $n^2$ is the sum of the first odd numbers?
$(n + 1)^2 - n^2 = (n^2 + 2n + 1) - n^2 = 2n + 1$.
and the result follows...
Case 1: $k=2m$ is even.
$1 - 2^2 + 3^2 - 4^2 + .... + (k-1)^2 - (k)^2 = -(2^2 -1) - (4^2 - 3^2) - ...- (k^2 - (k-1)^2)=$
$-(3) - (-7)-....-(2k + 1) = \sum_{j = 1}^m-(4j + 1)=$
$-4(\sum_{j=1}^m j) + m(\sum_{j=1}^m j)=$
$-4(\frac{(m(m+1)}{2}) + (m) = -2m(m+1) + m = m(-2(m+1) + 1) = m(-2m - 1) =$
$ -m(2m + 1)= -\frac{k(k + 1)}{2}$
Case 2: $k = 2m + 1$ is odd.
$[1 - 2^2 + 3^2 - 4^2 + .... + (k-2^2 - (k-1)^2 ] + k^2 =$
$ -\frac{(k-1)k}{2} + (k^2) = k[-\frac{k-1}{2} + k] =$
$k[\frac{-(k-1)+2k}{2} ]=\frac{k(k + 1)}{2}$
So
sum = $(-1)^{k-1}\frac{k(k + 1)}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Prove that $\frac{ab}{a^5+b^5+ab}+\frac{bc}{b^5+c^5+bc}+\frac{ca}{c^5+a^5+ca} \leq 1.$ The following problem was on the IMO 1996 shortlist :
Let $a,b,c$ be positive real numbers such that $abc = 1$. Prove that
$$\dfrac{ab}{a^5+b^5+ab}+\dfrac{bc}{b^5+c^5+bc}+\dfrac{ca}{c^5+a^5+ca} \leq 1.$$
I tried factoring out things but that didn't seem to work. I don't see how to factor the denominator so I get stuck.
| since
$$a^5+b^5\ge a^2b^3+a^3b^2$$
so
$$\sum\dfrac{ab}{a^5+b^5+ab}\le\sum\dfrac{1}{ab^2+a^2b+1}=\sum\dfrac{abc}{ab^2+a^2b+abc}=\sum\dfrac{c}{a+b+c}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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find the maximum possible area of $\triangle{ABC}$ Let $ABC$ be of triangle with $\angle BAC = 60^\circ$
. Let $P$ be a point in its interior so that $PA=1, PB=2$ and
$PC=3$. Find the maximum area of triangle $ABC$.
I took reflection of point $P$ about the three sides of triangle and joined them to vertices of triangle. Thus I got a hexagon having area double of triangle, having one angle $120$ and sides $1,1,2,2,3,3$. We have to maximize area of this hexagon. For that, I used some trigonometry but it went very complicated and I couldn't get the solution.
| Let $\mathcal{A}$ be the area of $\triangle ABC$.
Let $\theta$ and $\phi$ be the angles $\angle PAC$ and $\angle BAP$ respectively.
We have $\theta + \phi = \angle BAC = \frac{\pi}{3}$.
As functions of $\theta$ and $\phi$, the side lengths $b$, $c$ and area $\mathcal{A}$ are:
$$
\begin{cases}
c(\theta) &= \cos\theta + \sqrt{2^2-\sin^2\theta}\\
b(\phi) &= \cos\phi + \sqrt{3^2-\sin^2\phi}
\end{cases}
\quad\text{ and }\quad
\mathcal{A}(\theta) = \frac{\sqrt{3}}{4} c(\theta)b\left(\frac{\pi}{3}-\theta\right)
$$
In order for $\mathcal{A}(\theta)$ to achieve maximum has a particular $\theta$,
we need
$$\frac{d\mathcal{A}}{d\theta} = 0
\iff \frac{1}{\mathcal{A}}\frac{d\mathcal{A}}{d\theta} = 0
\iff \frac{1}{c}\frac{dc}{d\theta} - \frac{1}{b}\frac{db}{d\phi} = 0
\iff \frac{\sin\theta}{\sqrt{2^2-\sin^2\theta}} - \frac{\sin\phi}{\sqrt{3^2-\sin^2\phi}} = 0$$
This implies
$$\frac{\sin\theta}{2} = \frac{\sin\phi}{3}
= \frac13 \sin\left(\frac{\pi}{3} - \theta\right)
= \frac13 \left(\frac{\sqrt{3}}{2}\cos\theta - \frac12\sin\theta\right)
\iff 4\sin\theta = \sqrt{3}\cos\theta$$
and hence
$$\theta = \tan^{-1}\left(\frac{\sqrt{3}}{4}\right) \approx 0.4086378550975924
\;\;( \approx 23.41322444637054^\circ )$$
Furthermore, we have
$\displaystyle\;\frac{\sin\theta}{2} = \frac{\sin\phi}{3} = \frac{\sqrt{3}}{2\sqrt{19}}\;$.
Substitute this into the expression for side lengths and area, we get
$$
\begin{cases}
c &= \frac{4+\sqrt{73}}{\sqrt{19}}\\
b &= \frac{7+3\sqrt{73}}{2\sqrt{19}}
\end{cases}
\quad\implies\quad
\mathcal{A} = \frac{\sqrt{3}}{8}(13+\sqrt{73}) \approx 4.664413635668018
$$
Please note that the condition $\displaystyle\;\frac{\sin\theta}{2} = \frac{\sin\phi}{3}$ is equivalent to $\angle ABP = \angle ACP$. If one can figure out why these two angles equal to each other when $\mathcal{A}$ is maximized, one should be able to derive all the result here w/o using any calculus.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$ The other day I came across this problem:
Let $x$, $y$, $z$ be real
numbers. Prove that
$$(x^2+1)(y^2+1)(z^2+1) + 8 \geq 2(x+1)(y+1)(z+1)$$
The first thought was power mean inequality, more exactly : $AM \leq SM$ ( we noted $AM$ and $SM$ as arithmetic and square mean), but I haven't found anything helpful.
(To be more specific, my attempts looked like this :
$\frac{x+1}{2} \leq \sqrt{\frac{x^2+1}{2}}$)
I also take into consideration Cauchy-Buniakowsky-Scwartz or Bergström inequality, but none seems to help.
Some hints would be apreciated.
Thanks!
| For all the equations of symmetrical type, the extreme value, if exists, is achieved when variables are equal.
It is not difficult to prove it. If you have $f(x,y,z)$ symmetrical over $x,y,z$ then all their derivatives are the same, so whatever condition is needed for $x$ is the same for $y$ and the same for $z$. This means that the extreme value, if exists, is for $x=y=z$.
$$(x^2+1)(y^2+1)(z^2+1) + 8 - 2(x+1)(y+1)(z+1)=0$$
$$x=y=z$$
$$(x^2+1)^3 + 8 - 2(x+1)^3=0$$
$$x=y=z=1$$
For any other value, since we can calculate $x=y=z=2$ for example and find that the expression is greater than 0, it is
$$(x^2+1)(y^2+1)(z^2+1) + 8 - 2(x+1)(y+1)(z+1)>0$$
All equations of this type can be solved this way as long as you can calculate the extreme value.
(Usually this is not the intention regarding the solution, since this method is removing the flavor from the equations completely.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Proving a trigonometric inequality I've been having difficulty with the following,
Prove that,
$[\sin^{n+1}(x)]^{2}+[\cos^{n+1}(x)]^{2} \geq (\frac12)^{n}$ where $x$ is real and $n$ is a non-negative integer.
I've tried an inductive approach but have struggled with the inductive step.
| use Holder inequality
$$[(\cos^2{x})^{n+1}+(\sin^2{x})^{n+1}][1+1]^{n} \geq (\cos^2{x}+\sin^2{x})^{n+1}=1$$
or
Use AM-GM inequality we have
$$(\sin^2{x})^{n+1}+\dfrac{1}{2^{n+1}}+\dfrac{1}{2^{n+1}}+\cdots+\dfrac{1}{2^{n+1}}\ge (n+1)\cdot\sqrt[n+1]{\dfrac{1}{2^{n(n+1)}}(\sin^2{x})^{n+1}}=\dfrac{n+1}{2^{n}}\sin^2{x}$$
so
$$(\sin^2{x})^{n+1}+\dfrac{n}{2^{n+1}}\ge\dfrac{n+1}{2^{n+1}}\sin^2{x}\tag{1}$$
the same as
$$(\cos^2{x})^{n+1}+\dfrac{n}{2^{n+1}}\ge\dfrac{n+1}{2^{n+1}}\cos^2{x}\tag{2}$$
$(1)+(2)$
$$\sin^{2n+2}{x}+\cos^{2n+2}{x}\ge\dfrac{1}{2^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove for every odd integer $a$ that $(a^2 + 3)(a^2 + 7) = 32b$ for some integer $b$. I've gotten this far:
$a$ is odd, so $a = 2k + 1$ for some integer $k$.
Then $(a^2 + 3).(a^2 + 7) = [(2k + 1)^2 + 3] [(2k + 1)^2 + 7]$
$= (4k^2 + 4k + 4) (4k^2 + 4k + 8) $
$=16k^4 + 16k^3 + 32k^2 + 16k^3 + 16k^2 + 32k + 16k^2 + 16k + 32$
$=16k^4 + 32k^3 + 64k^2 + 48k + 32$
But this isn't a multiple of 32, at most I could say $(a^2 + 3)(a^2 + 7) = 16b$ for some integer $b$
| insert $a=2b+1$ and you get
$((2b+1)^2 + 3)((2b+1)^2 + 7) /32= 1/2 (b^2+b+1) (b^2+b+2)$
either $(b^2+b+1)$ or $(b^2+b+2)$ will be even so the RHS is an integer for all b.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Prove that if $a,b,$ and $c$ are positive real numbers, then $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \geq ab + bc + ca$.
Prove that if $a,b,$ and $c$ are positive real numbers, then $\dfrac{a^3}{b}+\dfrac{b^3}{c}+\dfrac{c^3}{a} \geq ab + bc + ca$.
I tried AM-GM and it doesn't look like AM-GM or Cauchy-Schwarz work here. The $ab+bc+ca$ reminds of a cyclic expression, so that may help by factoring the inequality and getting a true statement.
| For positive $x$, $y$, $z$, $a$, $b$, and $c$, note that that
\begin{align*}
(x+y+z)^2 &= \left(\frac{x}{\sqrt{a}}\sqrt{a} + \frac{y}{\sqrt{b}}\sqrt{b} +\frac{z}{\sqrt{c}}\sqrt{c}\right)^2\\
&\le \left(\frac{x^2}{a} + \frac{y^2}{b} +\frac{z^2}{c}\right)(a+b+c).
\end{align*}
That is,
\begin{align*}
\frac{x^2}{a} + \frac{y^2}{b} +\frac{z^2}{c} \ge \frac{(x+y+z)^2}{a+b+c}.
\end{align*}
Then
\begin{align*}
\frac{a^3}{b} + \frac{b^3}{c} +\frac{c^3}{a} &=\frac{a^4}{ab} + \frac{b^4}{bc} +\frac{c^4}{ac}\\
&\ge \frac{\left(a^2+b^2+c^2\right)^2}{ab+bc+ac}\\
&\ge \frac{\left(ab+bc+ac\right)^2}{ab+bc+ac}\\
&=ab+bc+ac.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 7,
"answer_id": 2
} |
Solve integral $\int{\frac{x^2 + 4}{x^2 + 6x +10}dx}$ Please help me with this integral:
$$\int{\frac{x^2 + 4}{x^2 + 6x +10}}\,dx .$$
I know I must solve it by substitution, but I don't know how exactly.
| \begin{align}
\int \frac{x^2+4}{x^2+6x+10}dx&=\int \frac{x^2+6x+10-(6x+18)+12}{x^2+6x+10}dx\\
&=\int1dx-3\int \frac{2x+6}{x^2+6x+10}dx+12\int \frac{1}{x^2+6x+10}dx\\
&=x-3\ln(x^2+6x+10)+12\int \frac{1}{(x+3)^2+1}dx\\
&=x-3\ln(x^2+6x+10)+12\int \frac{\sec^2u}{\tan^2u+1}du\\
&=x-3\ln(x^2+6x+10)+12\int 1du\\
&=x-3\ln(x^2+6x+10)+12\tan^{-1}(x+3)+C\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Evaluate the definite integral $\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}dx$ Problem :
Determine the value of $$\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}\ \text dx$$
My approach: using $\int^a_0f(x)\ \text dx = \int^a_0 f(a-x)\ \text dx$,
$$
\begin{align}
\frac{105}{19}\int^{\pi/2}_0 \frac{\sin 8x}{\sin x}\ \text dx &= \frac{105}{19}\int^{\pi/2}_0 \frac{\sin (4\pi -8x)}{\cos x}\ \text dx\\
&= \frac{105}{19}\int^{\pi/2}_0 -\frac{\sin 8x}{\cos x}\ \text dx
\end{align}
$$
But it seems it won't work please help thanks
| Using formulas $\sin2a = 2\sin a\cos a,\quad \cos a \cos b = \frac12(\cos(a+b)-\cos(a-b))$, have:
$$\dfrac{\sin 8x}{\sin x} = 8\cos 4x\cos 2x\cos x = 4\cos4x(\cos3x+\cos x) = 2(\cos7x+\cos 5x+\cos 3x +\cos x),$$so
\begin{align}
\dfrac {105}{19}\int_0^\limits\dfrac\pi2\dfrac{\sin 8x}{\sin x} &= \dfrac {210}{19}\int_0^\limits\dfrac\pi2(\cos7x+\cos 5x+\cos 3x +\cos x)\,dx =\\
&= \dfrac {210}{19}\left(\frac17\sin7x+\frac15\sin 5x+\frac13\sin 3x +\sin x\right)\biggr|_0^\frac\pi2 =
\\&= \frac{210}{19}\left(-\frac17+\frac15-\frac13+1\right) = \frac{210}{19}\cdot\frac{76}{105} = 8.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 1
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Prove that $\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$ Prove that $$\sum_{n=1}^{\infty}\left(\frac{1}{(8n-7)^3}-\frac{1}{(8n-1)^3}\right)=\left(\frac{1}{64}+\frac{3}{128\sqrt{2}}\right)\pi^3$$
I don't have an idea about how to start.
| For $b \in \{0,\ldots,7\}$, consider the function $G(b) := \sum_{n \geq 1} \frac{e^{-2\pi i nb/8}}{n^3}$. Observe that
\begin{equation*}
G(b) = \sum_{0 \leq a \leq 7} e^{-2\pi i ab/8} \sum_{n \geq 1} \frac{1}{(8n-a)^3}.
\end{equation*}
Now, we know that if $a,a' \in \{0,\ldots,7\}$ then
\begin{equation*}
\frac{1}{8}\sum_{0 \leq b \leq 7} e^{2\pi i(a-a')b/8} = \begin{cases} 1& \text{ if $a = a'$} \\ 0& \text{ otherwise}. \end{cases}
\end{equation*}
Therefore, we have
\begin{equation*}
\frac{1}{8} \sum_{0 \leq b \leq 7}e^{-2\pi i(7b)/8} G(b) = \sum_{n \geq 1} \frac{1}{(8n-7)^3},
\end{equation*}
and thus your sum in question is precisely
\begin{equation*}
\frac{1}{8}\sum_{0 \leq b \leq 7} \left(e^{-2\pi i(7b)/8} -e^{-2\pi ib/8}\right)G(b).
\end{equation*}
Now, observe that $G(b) = \sum_{n \geq 1} a_n e^{-2\pi int}$, where $t = \frac{b}{8}$ and $a_n = \frac{1}{n^3}$. This looks like a Fourier series. Can you determine which function it represents? If so then you would be nearly done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Find the summation $\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$ What is the value of the following sum?
$$\frac{1}{1!}+\frac{1+2}{2!}+\frac{1+2+3}{3!}+ \cdots$$
The possible answers are:
A. $e$
B. $\frac{e}{2}$
C. $\frac{3e}{2}$
D. $1 + \frac{e}{2}$
I tried to expand the options using the series representation of $e$ and putting in $x=1$, but I couldn't get back the original series. Any ideas?
| Starting with $\sum_{i=1}^ni=\frac{n(n+1)}{2}$, the sum simplifies to
\begin{align*}
z=\frac{1}{2}\sum_{i=1}^\infty \frac{i(i+1)}{i!}&=\frac{1}{2}\sum_{i=1}^\infty \frac{i+1}{(i-1)!}\\
&=\frac{1}{2}\sum_{i=0}^\infty\frac{i+2}{i!}\\
&=\frac{1}{2}\sum_{i=0}^\infty\left(\frac{i}{i!}+\frac{2}{i!}\right)\\
&=\frac{1}{2}\left(\sum_{i=1}^\infty\frac{1}{(i-1)!}+\sum_{i=0}^\infty\frac{2}{i!}\right)\\
&=\frac{1}{2}\left(\sum_{i=0}^\infty\frac{1}{i!}+2\sum_{i=0}^\infty\frac{1}{i!}\right)\\
&=\frac{1}{2}(e+2e)=\frac{3e}{2}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1597328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Definite integral $\int_0^1 \frac{\arctan x}{x\,\sqrt{1-x^2}}\,\text{d}x$ Wanting to calculate the integral $\int_0^1 \frac{\arctan x}{x\,\sqrt{1-x^2}}\,\text{d}x$ it will certainly already known to many of you that an interesting way to attack it is to refer to the method of integration and differentiation with respect to a parameter, getting $\frac{\pi}{2}\,\log\left(1+\sqrt{2}\right)$.
Instead, what it does not at all clear is how software such as Wolfram Mathematica can calculate that result in an exact manner and not only approximate. Can someone enlighten me? Thanks!
| I thought it might be instructive to present two approaches that begin with the Feyman "trick" for differentiating under the integral. We write
$$I(a)=\int_0^1\frac{\arctan (ax)}{x\sqrt{1-x^2}}\,dx$$
Then, we differentiate with $I(a)$ to find
$$I'(a)=\int_0^1\frac{1}{(1+x^2a^2)\sqrt{1-x^2}}\,dx$$
can be evaluated by first substituting $x=\sin u$ so that
$$\begin{align}
I'(a)&=\int_0^{\pi/2}\frac{1}{1+a^2\sin^2 u}\,du\\\\
&=\frac12 \int_{-\pi/2}^{\pi/2}\frac{1}{1+a^2\sin^2 u}\,du
\end{align}$$
Next, we use the trigonometric identity $\sin^2u=\frac{1-\cos(2u)}{2}$ so that
$$\begin{align}
I'(a)&=\int_0^{\pi/2}\frac{1}{1+\frac12a^2-\frac12a^2\cos (2u)}\,du \tag 1\\\\
&=\frac14 \int_{-\pi}^{\pi}\frac{1}{1+\frac12a^2-\frac12a^2\cos (u)}\,du \tag 2
\end{align}$$
We pursue evaluation of $(1)$ using the Weirestrass Substitution and evaluation of $(2)$ using contour integration.
First, we enforce the substitution $2u=\tan(x/2)$ in $(1)$. Then, we obtain
$$\begin{align}
I'(a)&=2\int_{0}^{\infty}\frac{1}{1+4(1+a^2)x^2}\,dx\\\\
&=\frac{2}{2\sqrt{1+a^2}}\left.\arctan\left(2\sqrt{1+a^2}\,x\right)\right|_{0}^{\infty}\\\\
&=\frac{\pi}{2\sqrt{1+a^2}}
\end{align}$$
Alternatively, we let $u=e^{iz}$ in $(2)$ and write
$$\begin{align}
I'(a)&=\frac{i}{2a^2}\oint_{|z|=1}\frac{1}{z^2-\left(1+\frac{2}{a^2}\right)z+1}\,dz\\\\
&=\frac{i}{a^2}(2\pi i) \,\,\text{Res}\left(\frac{1}{z^2-2\left(1+\frac{2}{a^2}\right)z+1},z=\left(1+\frac{2}{a^2}\right)-\frac{2\sqrt{1+a^2}}{a^2}\right)\\\\
&=\pi/2\sqrt{1+a^2}
\end{align}$$
Finally, following the approach used by @Manu we find from $I'(a)$, $I(1)$ is
$$I(1)=\pi \log(1+\sqrt{2})$$
And we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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For all $n >0$, $ \left(1+\frac{1}{n} \right)^n = 1+ \sum_{k=1}^n\bigl[ \frac{1}{k!} \prod_{r=0}^{k-1}(1-\frac{r}{n}) \bigr]$ I am working through some problems on induction and I have been stuck on this one for a while. If anyone has any hints. I can show it is true for the $n=1$ and $n=2$ case but I am having difficulty on the induction step.
This is what I have so far,
$ (1+\frac{1}{n} )^n = 1+ \sum\limits_{k=1}^n[ \frac{1}{k!} \prod\limits_{r=0}^{k-1}(1-\frac{r}{n}) ]$
$n =1 $ case
$ (1+\frac{1}{1} )^1 = 1+ \sum\limits_{k=1}^1[ \frac{1}{k!} \prod\limits_{r=0}^{k-1}(1-\frac{r}{1}) ]$
$ 2 = 1+ \frac{1}{1!} \prod\limits_{r=0}^{0}(1-\frac{r}{1}) ]= 1+(1-\frac{0}{1})=1+1=2$
So, 2=2. True
$n=2$ case
$ (1+\frac{1}{2} )^2 = 1+ \sum\limits_{k=1}^2[ \frac{1}{k!} \prod\limits_{r=0}^{k-1}(1-\frac{r}{2}) ]$
$ (1+\frac{1}{2} )^2 = 1+ [ \frac{1}{1!} \prod\limits_{r=0}^{0}(1-\frac{r}{2}) ]+\frac{1}{2!} \prod\limits_{r=0}^{1}(1-\frac{r}{2}) ]$
$ (1+\frac{1}{2} )^2 = 1+ [ \frac{1}{1!} (1-\frac{0}{2}) ]+[\frac{1}{2!} (1-\frac{1}{2}) ]$
$ (1+\frac{1}{2} )^2 = 1+ [ \frac{1}{1!} (1) ]+[\frac{1}{2!} (\frac{1}{2}) ]$
$ (1+\frac{1}{2} )^2 = 1+ 1+ \frac{1}{4} ]$
And finally, $\frac{9}{4} = \frac{9}{4}$
So the n=2 step is also true.
Induction Step
Show that the $(n+1)^{th}$ case is true when the $n^{th}$ case is true.
Now, I assume the $n^{th}$ case is true. So,
$ (1+\frac{1}{n} )^n = 1+ \sum\limits_{k=1}^n[ \frac{1}{k!} \prod\limits_{r=0}^{k-1}(1-\frac{r}{n}) ]$
then I multiply both sides by $ (1+\frac{1}{n} )^{n+1} $, but I haven't gotten far from there. Any hints would be great!
| While the following development is not an induction proof, I thought it would be instructive to present a direct proof of the equality of interest. To that end, we have from the binomial theorem
$$\begin{align}
\left(1+\frac1n\right)^n&=\sum_{k=0}^n\binom{n}{k}\frac1{n^k}\\\\
&=\sum_{k=0}^n\frac{n!}{k!(n-k)!n^k}\\\\
&=\sum_{k=0}^n\frac{1}{k!}\frac{\prod_{j=0}^{k-1}(n-j)}{n^k}\\\\
&=\sum_{k=0}^n\frac{1}{k!}\prod_{j=0}^{k-1}\left(1-\frac{j}{n}\right)
\end{align}$$
And we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Modular arithmatic $x^a \mod N= a\cdot b$ $x^3 \equiv 2 \pmod{15}$
How to solve it mod $3$ and mod $5$?
And how does the Chinese remainder theorem help?
I want a general method to follow in case the modulus is composite number.
Thanks in advance!
| $x^3\equiv 2\pmod{15}\Rightarrow \begin{cases}x^3\equiv 2\pmod{3}\\x^3\equiv 2\pmod{5}\end{cases}$
It is clear that if $x\equiv 0$ or $x\equiv 1\pmod{3}$ that $x^3\not\equiv 2\pmod{3}$ so we know that $x\equiv 2\pmod{3}$. Checking confirms this since $2^3=8\equiv 2\pmod{3}$
For the second implication, it should be clear that $x\equiv 0,1,4\pmod{5}$ will result in $x^3\equiv 0,1,4\pmod{5}$ respectively. We are left to check when $x\equiv 3$ and $2$.
$2^3=8\equiv 3\pmod{5}$
$3^3=27\equiv 2\pmod{5}$
Thus, we have the system of congruencies $\begin{cases}x\equiv 2\pmod{3}\\x\equiv 3\pmod{5}\end{cases}$
Applying chinese remainder theorem and solving the system yields $x\equiv 8\pmod{15}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim \frac{10+3^x}{20-3^x}$ as $x$ goes to $\infty$
Find $\lim \frac{10+3^x}{20-3^x}$ as $x$ goes to $\infty$
I tried $\displaystyle \lim_{x \to \infty} \frac{10+3^x}{20-3^x} =\lim_{x \to \infty} \frac{\frac{d}{dx}10+3^x}{\frac{d}{dx}20-3^x}= -\lim_{x \to \infty}\frac{3^x \log{3}}{3^x \log{3}} = -1.$
However, this is wrong.
EDIT: It's in fact correct. I do wish to see if there are other alternatives, though.
| Notice, $$\lim_{x\to \infty}\frac{10+3^x}{20+3^x}$$
$$=\lim_{x\to \infty}\frac{3^x\left(\frac{10}{3^x}+1\right)}{3^x\left(\frac{20}{3^x}-1\right)}$$
$$=\lim_{x\to \infty}\frac{\frac{10}{3^x}+1}{\frac{20}{3^x}-1}$$
$$=\frac{0+1}{0-1}=-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599591",
"timestamp": "2023-03-29T00:00:00",
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Let $0 \le a \le b \le c$ and $a+b+c=1$. Show that $a^2+3b^2+5c^2 \ge 1$
Let $0 \le a \le b \le c$ and $a+b+c=1$. Show that $a^2+3b^2+5c^2
\ge 1$.
My solution: since $a+b+c=1$ we have to show that $a^2+3b^2+5c^2\ge1=a+b+c$
Since $a,b,c \ge 0 $ the inequality is true given that every term on the left hand side of the inequality is greater or equal to the corresponding term on the right.
However I am not sure if I am reasoning correctly, as the hint from my book seems to depict the problem in a harder way than I am ,as it suggests to square the expression $a+b+c=1$ and so on...
So my question is wheter I am overlooking some detail in the problem which makes my solution inadequate.
| It is not true that $a^2\geq a$ for all $a\geq0$ (e.g. for $a=\frac12$), nor that $3b^2\geq b$ or $5c^2\geq c$.
Proof
Because $c\geq a$, it suffices to prove $3a^2+3b^2+3c^2=1$. This follows from AM-QM: $\frac{a^2+b^2+c^2}3\geq\left(\frac{a+b+c}3\right)^2$.
Or by using $3x^2\geq2x-\frac13$ (indeed, this is equivalent to $3(x-\frac13)^2\geq0$): $3a^2+3b^2+3c^2\geq2a+\frac13+2b+\frac13+2c+\frac13=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the common integer solutions to $a + b = c \cdot d$ and $a \cdot b = c + d$ I find nice that $$ 1+5=2 \cdot 3 \qquad 1 \cdot 5=2 + 3 .$$
Do you know if there are other integer solutions to
$$ a+b=c \cdot d \quad \text{ and } \quad a \cdot b=c+d$$
besides the trivial solutions $a=b=c=d=0$ and $a=b=c=d=2$?
| I found that, through solving the equations, we will have following (if a>b):
$$a=\frac{cd+\sqrt{c^2d^2-4(c+d)}}{2}$$
and
$$b=\frac{cd-\sqrt{c^2d^2-4(c+d)}}{2}$$
I think it will be useful for you to find all of the solutions.
Only you should find the solutions for:
$$c^2d^2-4(c+d)=m^2$$
where $m$ is any positive integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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If $a^2 + b^2 = 1$, show there is $t$ such that $a = \frac{1 - t^2}{1 + t^2}$ and $b = \frac{2t}{1 + t^2}$ My question is how we can prove the following:
If $a^2+b^2=1$, then there is $t$ such that $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$
| We show that except for the case $a=-1$, $b=0$, there always is a $t$ satisfying the conditions of the OP. The calculation is geometric. A similar calculation is of importance at the beginning of the theory of elliptic curves.
Let $P=(a,b)\ne(-1,0)$ be on the unit circle $x^2+y^2=1$. Suppose that the line joining $(-1,0)$ to $(a,b)$ has slope $t$. The line has equation $y=t(x+1)$. Substituting in $x^2+y^2=1$ we get after a little calculation that $(1+t^2)x^2+2tx+t^2-1=0$. One of the roots is $-1$. The product of the roots is $\frac{t^2-1}{1+t^2}$, so the other root $a$ is given by
$$a=\frac{1-t^2}{1+t^2}.$$
Now from $y=t(x+1)$ we get
$$b=t\left(\frac{1-t^2}{1+t^2}+1\right) =\frac{2t}{1+t^2}.$$
Remark: The "exceptional" point $(-1,0)$ can be dealt with by letting $t=\infty$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$\lim\frac{\cos{2x^2}-1}{x^2\sin{x^2}}$ as $x$ goes to $0$
Calculate $\displaystyle \lim_{x \to 0}\frac{\cos{2x^2}-1}{x^2\sin{x^2}}$
I tried L'hopital, but the denominator gets more and more complicated.
How does one calculate this limit?
| Recall that $\cos(2t) = 1-2\sin^2(t)$. Hence, we have
$$\cos(2x^2)-1 = -2\sin^2(x^2)$$
Hence, we have
$$\lim_{x \to 0} \dfrac{\cos(2x^2)-1}{x^2\sin(x^2)} = \lim_{x \to 0} \dfrac{-2\sin^2(x^2)}{x^2\sin(x^2)} = -2 \lim_{x \to 0} \dfrac{\sin(x^2)}{x^2} = -2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1600629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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limit as n goes to infinity of $\frac{\sqrt{n^3-3}-\sqrt{n^3+2n^2+3}}{\sqrt{n+2}}.$
How do you go about solving $$\lim_{n\to\infty}\frac{\sqrt{n^3-3}-\sqrt{n^3+2n^2+3}}{\sqrt{n+2}}.$$
I know that I have to fix the top so that it is not $(\infty - \infty$),
but if I multiple it by
$$\frac{\sqrt{n^3-3}+\sqrt{n^3+2n^2+3}}{\sqrt{n^3-3}+\sqrt{n^3+2n^2+3}},$$
the bottom part becomes very ugly and extremely hard to deal with.
| This is a straightforward calculation: Expanding the fraction by $\sqrt{n^3-3}+\sqrt{n^3+2n^2+3}$ gives
\begin{align*}
&\, \frac{\sqrt{n^3-3}-\sqrt{n^3+2n^2+3}}{\sqrt{n+2}} \\
=&\, \frac{(\sqrt{n^3-3}-\sqrt{n^3+2n^2+3)}(\sqrt{n^3-3}+\sqrt{n^3+2n^2+3})}{\sqrt{n+2} (\sqrt{n^3-3}+\sqrt{n^3+2n^2+3})} \\
=&\, \frac{(n^3-3)-(n^3+2n^2+3)}{\sqrt{(n+2)(n^3-3)} + \sqrt{(n+2)(n^3+2n^2+3)}} \\
=&\, \frac{-2n^2-6}{\sqrt{n^4+2n^3-3n-6} + \sqrt{n^4+4n^3+4n^2+3n+6}}.
\end{align*}
Dividing numerator and denumerator by $n^2$ gves
\begin{align*}
&\, \frac{-2n^2-6}{\sqrt{n^4+2n^3-3n-6} + \sqrt{n^4+4n^3+4n^2+3n+6}} \\
=&\, \frac{-2-\frac{6}{n^2}}{\sqrt{1+\frac{2}{n}-\frac{3}{n^3}-\frac{6}{n^4}} + \sqrt{1+\frac{4}{n}+\frac{4}{n^2}+\frac{3}{n^3}+\frac{6}{n^4}}}.
\end{align*}
Taking the limit $n \to \infty$ results in $-2/2 = -1$.
| {
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Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ Real numbers $x,y$ satisfies $x^2+y^2=1.$If the minimum and maximum value of the expression $z=\frac{4-y}{7-x}$ are $m$ and $M$ respectively,then find $2M+6m.$
Let $x=\cos\theta$ and $y=\sin\theta$,because $\sin^2\theta+\cos^2\theta=1$.
Then we need to find the minimum and maximum value of the expression $\frac{4-\sin\theta}{7-\cos\theta}$.
I differentiated it and equated it to zero to find the critical points or points of extrema.
They are $\theta_1=\arcsin(\frac{1}{\sqrt{65}})-\arctan(\frac{7}{4})$ and $\theta_2=\arccos(\frac{1}{\sqrt{65}})+\arctan(\frac{4}{7})$
I found $\frac{4-\sin\theta_1}{7-\cos\theta_1}$ and $\frac{4-\sin\theta_2}{7-\cos\theta_2}$.
$\frac{4-\sin\theta_1}{7-\cos\theta_1}=\frac{3}{4}$ and $\frac{4-\sin\theta_2}{7-\cos\theta_2}=\frac{5}{12}$
This method is full of lengthy calculations.I want to know is there an elegant solution possible for this problem which is short and easy.
| $$y-4=z(x-7)\tag{1}$$
is an equation of all the lines that pass $(7,4)$ with slope $z$. When $z$ is at its minimum or maximum then the line touches the unit circle. Equation of tangent lines with slope $z$ to a unit circle is
$$y=zx\pm\sqrt{z^2+1}\tag{2}$$
Comparing $(1)$ and $(2)$,
$$-7z+4=\pm\sqrt{z^2+1}$$
$$(7z-4)^2=z^2+1$$
$$48z^2-56z+15=0$$
$$(4z-3)(12z-5)=0$$
$$\therefore M=\frac34, m=\frac5{12}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to properly set up partial fractions for repeated denominator factors I was just trying to solve a problem that had the following item which I needed to split into separate generating functions:
$$\frac{x}{(1-2x)^2(1-5x)}$$
I had assumed I needed to split it into:
$$\frac{A}{1-2x} + \frac{B}{1-2x} + \frac{C}{1-5x}$$
But according to Wolfram Alpha it appears I had to split it into:
$$\frac{A}{1-2x} + \frac{B}{(1-2x)^2} + \frac{C}{1-5x}$$
Can anyone explain the intuition behind this? Is this a general rule that when you have a repeated factor in the denominator, you split it into all powers of that factor?
| Consider the simplest of cases. $$\frac{\xi}{x^3}$$ Assuming $\xi$ is some polynomial, one could carry on long division to determine the quotient and the remainder. Thus one could get $$\frac{\xi}{x^3}=q(x) + \frac{r(x)}{x^3}$$ Now what can we say for certain about $r(x)$ ? We can say that it is the remainder thus of smaller degree than $x^3$. The most general possible polynomial of degree 2 or smaller is $A+Bx+Cx^2$. Then $$\frac{r(x)}{x^3}=\frac{A+Bx+Cx^2}{x^3}$$ Further more, we can split the right side $$\frac{r(x)}{x^3}=\frac{A}{x^3}+\frac{Bx}{x^3}+\frac{Cx^2}{x^3}$$ which becomes
$$\frac{r(x)}{x^3}=\frac{A}{x^3}+\frac{B}{x^2}+\frac{C}{x}$$ Thus the above would be a sensible what to try to split any proper fraction $\frac{r(x)}{x^3}$ Moreover, the similar idea would hold for $\frac{r(x)}{(x+a)^3}$, meaning a sensible way to split it would be $$\frac{r(x)}{(x+a)^3}=\frac{A}{(x+a)^3} +\frac{B}{(x+a)^2}+\frac{C}{(x+a)} $$ hope that helps..
| {
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integrate $\int \cos^{4}x\sin^{4}xdx$
$$\int \cos^4x\sin^4xdx$$
How should I approach this?
I know that $\sin^2x={1-\cos2x\over 2}$ and $\cos^2x={1+\cos2x\over 2}$
| we now that
$$ \sin 2x=2\cos x\sin x$$
$$\cos x\sin x=\frac{\sin 2x}{2}$$
so...
$$\cos^4 x\sin^4 x=\frac{\sin^4 2x}{16}$$
$$\frac{\sin^4 2x}{16}=\frac{(\frac{1-\cos 4x}{2})^2}{16}=\frac{1-2\cos 4x+\cos^24x}{64}=\frac{1-2\cos 4x+\frac{1+\cos 8x}{2}}{64}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Limit of: $\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$ I want to find the limit of: $$\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$$
I tried expanding it by $$ \frac{(n^3+n^{1/2})^{1/3}+(n^3-1)^{1/3}}{(n^3+n^{1/2})^{1/3}+(n^3-1)^{1/3}} $$
but it didn't help much. Wolfram says the answer is $\frac{3^{1/2}}{3}$.
Any help would be greatly appreciated.
| Rewrite, $\lim_{n \to \infty}(\sqrt[3]{n^3+\sqrt{n}}-\sqrt[3]{n^3-1})\cdot \sqrt{(3n^3+1)}$
as $n\left(\left(1+\frac{1}{n^{\frac{5}{2}}} \right)^{\frac{1}{3}} -\left(1-\frac{1}{n^3} \right)^{\frac{1}{3}} \right)\times \sqrt{3}n^{\frac{3}{2}}\left(1+\frac{1}{3n^3}\right)^{\frac{1}{2}}$
Now use the Binomial theorem for fractional powers.
| {
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Prove the inequality $\tan{\frac{\pi\sin{x}}{4\sin{\alpha}}}+\tan{\frac{\pi\cos{x}}{4\cos{\alpha}}} > 1$
Prove the inequality $$\tan{\dfrac{\pi\sin{x}}{4\sin{\alpha}}}+\tan{\dfrac{\pi\cos{x}}{4\cos{\alpha}}} > 1$$
for any $x, \alpha$ with $0 \leq x \leq \dfrac{\pi}{2}$ and $\dfrac{\pi}{6} < \alpha < \dfrac{\pi}{3}$.
The best idea I had was to use the identity $\tan(A+B) = \dfrac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)}$. Thus we may say that $\tan \left({\dfrac{\pi \sin{x}}{4 \sin{\alpha}}}+\dfrac{\pi\cos{x}}{4\cos{\alpha}} \right) \left(1-\tan{\dfrac{\pi\sin{x}}{4\sin{\alpha}}} \tan{\dfrac{\pi\cos{x}}{4\cos{\alpha}}} \right) = \tan{\dfrac{\pi\sin{x}}{4\sin{\alpha}}}++\tan{\dfrac{\pi\cos{x}}{4\cos{\alpha}}}.$ Then I just have to show its greater than $1$ on these intervals.
| Since $\dfrac{\pi}{6} < \alpha < \dfrac{\pi}{3}$, we have $\sin \alpha > \dfrac{1}{2}$ and $\cos \alpha > \dfrac{1}{2}$.
Since $0 \le x \le \dfrac{\pi}{2}$, we have $0 \le \sin x \le 1$ and $0 \le \cos x \le 1$.
Hence, $0 \le \dfrac{\pi\sin x}{4\sin \alpha} < \dfrac{\pi}{2}$ and $0 \le \dfrac{\pi\cos x}{4\cos \alpha} < \dfrac{\pi}{2}$.
Therefore, $\tan \dfrac{\pi\sin x}{4\sin \alpha} \ge 0$ and $\tan \dfrac{\pi\cos x}{4\cos \alpha} \ge 0$ for all $0 \le x \le \dfrac{\pi}{2}$ and $\dfrac{\pi}{6} < \alpha < \dfrac{\pi}{3}$.
Now, we beak the proof into three cases:
Case I: If $x > \alpha$, then $\sin x > \sin \alpha$. Hence, $\tan \dfrac{\pi\sin x}{4\sin \alpha} > \tan\dfrac{\pi}{4} = 1$.
Therefore, $\tan \dfrac{\pi\sin x}{4\sin \alpha} + \tan \dfrac{\pi\cos x}{4\cos \alpha} > 1 + 0 = 1$.
Case II: If $x < \alpha$, then $\cos x > \cos \alpha$. Hence, $\tan \dfrac{\pi\cos x}{4\cos \alpha} > \tan\dfrac{\pi}{4} = 1$.
Therefore, $\tan \dfrac{\pi\sin x}{4\sin \alpha} + \tan \dfrac{\pi\cos x}{4\cos \alpha} > 0 + 1 = 1$.
Case III: If $x = \alpha$, then $\sin x = \sin \alpha$ and $\cos x = \cos \alpha$.
Therefore, $\tan \dfrac{\pi\sin x}{4\sin \alpha} + \tan \dfrac{\pi\cos x}{4\cos \alpha} = \tan \dfrac{\pi}{4} + \tan \dfrac{\pi}{4} = 1+1 = 2 > 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607661",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is $\sqrt {12} = 2 \sqrt 3$? Why $\sqrt {12} = 2 \sqrt 3$? It is obvious? If we considered the function $f(s) = s^2 $ it is injective on positive numbers so we obtain the conclusion. But in the same time it is an equality between irrational numbers. Suppose that we know just to compute the square roots.
| Let's assume (unless that's the part you want proven) that there is only 1 positive number $x$ with the property $x^2=12$. We'll denote that number by $\sqrt{12}$. Likewise let $\sqrt{3}$ be the positive $y$ such that $y^2=3$.
We know that $\Bbb R$ is a field and thus multiplication is commutative. We also know that because $2\gt 0$ and $\sqrt{3}\gt 0$ that $2\sqrt{3}\gt 0$. Then $$(2\sqrt{3})^2 = (2\sqrt{3})(2\sqrt{3}) = (2^2)(\sqrt{3})^2 = 4\cdot 3=12=\sqrt{12}^2 \\ \implies 2\sqrt{3}=\sqrt{12}$$
| {
"language": "en",
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"source": "stackexchange",
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Is there an easy way to compute the determinant of matrix with 1's on diagonal and a's on anti-diagonal? \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & a \\ 0 & 1 & 0 & 0 & a & 0 \\0 & 0 & 1 & a & 0 & a \\0 & 0 & a & 1 & 0 & a \\0 & a & 0 & 0 & 1 & 0 \\a & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}
Thanks
| We have the matrix
$$A= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & a \\ 0 & 1 & 0 & 0 & a & 0 \\0 & 0 & 1 & a & 0 & a \\0 & 0 & a & 1 & 0 & a \\0 & a & 0 & 0 & 1 & 0 \\a & 0 & 0 & 0 & 0 & 1 \\ \end{bmatrix}$$
Elininating the $a$s below the diagonal by adding multiples of the first, second and third line of $A$, we obtain the upper triangular matrix $A'$:
$$A'= \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & a \\ 0 & 1 & 0 & 0 & a & 0 \\0 & 0 & 1 & a & 0 & a \\0 & 0 & 0 & 1-a^2 & 0 & a-a^2 \\0 & 0 & 0 & 0 & 1-a^2 & 0 \\0 & 0 & 0 & 0 & 0 & 1-a^2 \\ \end{bmatrix}$$
Since adding a mulitple of a row to another row does not alter the determinant, we can say that $\det(A) = \det(A')$. Furthermore, the determinant of an upper triangular matrix is the product of its diagonal entries. Thus, we have:
$$\det(A) = \det(A') = (1-a^2)^3$$
Alternatively, eliminate the $a$s above the diagonal by adding multiples of the fourth, fifth and sixth row. The determinant of a lower triangular matrix can be obtained in the same way.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability that $2^a+3^b+5^c$ is divisible by 4
If $a,b,c\in{1,2,3,4,5}$, find the probability that $2^a+3^b+5^c$ is divisible by 4.
For a number to be divisible by $4$, the last two digits have to be divisible by $4$
$5^c= \_~\_25$ if $c>1$
$3^1=3,~3^2=9,~3^3=27,~3^4=81,~ 3^5=243$
$2^1=2,~2^2=4,~2^3=8,~2^4=16,~2^5=32$
Should I add all possibilities? Is there a simpler method?
| Let's consider the remainders when dividing by $4$. If $a=1$ then $2^a=2$ has the remainder $2$. Otherwise, $2^a$ has the remainder zero. If $b$ is odd then $3^b$ has remainder $3$, but if $b$ is even $3^b$ has remainder $1$. Whatever $c$ is, $5^c$ will have remainder $1$. (All of these statements are provable by induction, but the patterns are obvious.)
Then there are just two basic ways for the sum to have remainder zero when divided by $4$:
Case 1: $a=1$, $b$ is even, $c$ is anything. This gives remainders $2+1+1$ or zero.
Case2: $a>1$, $b$ is odd, $c$ is anything. This gives remainders $0+3+1$ or zero.
Now count each case and add the counts.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find prob. that only select red balls from $n$ (red+blue) balls There are 4 blue balls and 6 red balls(total 10 balls). $X$ is a random variable of the number of selected balls(without replacement), in which
$$P(X=1)=0.1$$
$$P(X=2)=0.5$$
$$P(X=3)=0.2$$
$$P(X=4)=0.1$$
$$P(X=10)=0.1$$
Then, what is probability of only selecting red balls?
This is what I have tried:
The (conditional) probability that all $r$ of the balls are selected from the red is just: ${6\choose r}\big/{10\choose r}$, for $0\leq r\leq 6$ , and $0$ elsewhere.
That is, let $N_R$ be the number of red balls selected, and $N_r$ the total number of balls selected, then:
$$\mathsf P(N_r=N_R\mid N_r=r) = \frac{6!/(6-r)!}{10!/(10-r)!} \mathbf 1_{r\in\{1\ldots 6\}}$$
As the number of balls selected is a random variable with the specified distribution, then the probability that all balls selected are red is:
$$\begin{align}
\mathsf P(N_R=N_r) & =\frac{1}{10}\frac{6!\,(10-1)!}{(6-1)!\,10!}+\frac 5{10}\frac{6!\,(10-2)!}{(6-2)!\,10!}+\frac{1}{5}\frac{6!\,(10-3)!}{(6-3)!\,10!}+...
\\[1ex] &
\end{align}$$
| Let $A = \{\text{Choose only red}\}.$ Let $X$ be the number of balls drawn, and let $R$ be the number of red balls drawn. Then
\begin{align*}
P(A) &=\sum_{k = 0}^{6} P(R = k|X = k)P(X = k)\tag 1\\
&=\sum_{k = 0}^{6} \frac{\binom{6}{k}}{\binom{10}{k}}p_k\\
&=\frac{\binom{6}{1}}{\binom{10}{1}}(.1)+\frac{\binom{6}{2}}{\binom{10}{2}}(.5)+\frac{\binom{6}{3}}{\binom{10}{3}}(.2)+\frac{\binom{6}{4}}{\binom{10}{4}}(.1)+\frac{\binom{6}{5}}{\binom{10}{5}}(0)+\frac{\binom{6}{6}}{\binom{10}{6}}(0)\tag2\\
&= \frac{187}{700},
\end{align*}
where in $(1)$ it has to be the case that the number the random number you draw has to equal the number of red balls you draw; in $(2)$, $p_5 = p_6 = 0$ since according to the distribution of $X$, you cannot draw 5 or 6 times.
So, our answers agree, almost. If you complete your calculation (since you didn't post it), our answers should agree.
| {
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Compute this integral (Is there a trick hidden to make it eassier?) I need some tips to compute this integral:
$$ \int\,\dfrac{\sqrt{x^2-1}}{x^5\sqrt{9x^2-1}}\,dx $$
What I did was express the denominator in the following form:
$$ \int\,\dfrac{\sqrt{x^2-1}}{x^5\sqrt{9x^2-1}}\,dx = \int\,\dfrac{\sqrt{x^2-1}}{x^5\sqrt{8x^2+x^2-1}}\,dx $$
Then, I made the change $x = \sec{\theta}$, then
$$ \int\,\dfrac{\sqrt{x^2-1}}{x^5\sqrt{8x^2+x^2-1}}\,dx = \int\,\dfrac{\sqrt{\sec^2{\theta}-1}}{\sec^5{\theta}\sqrt{8\sec^2{\theta}+\sec^2{\theta}-1}}\sec{\theta}\tan{\theta}\,d{\theta} $$
Trying to symplify this expression, I came to this:
$$ \dfrac{1}{4}\int\,\dfrac{\sin^2(2\theta)\cos{\theta}}{\sqrt{8\sin^2{\theta}+1}}\,d{\theta} $$
I feel this integral can be computed using some kind of "trick", but I can't see it. Thanks for your help and have a nice day!
| $$I=\int\,\dfrac{\sqrt{x^2-1}}{x^5\sqrt{9x^2-1}}\,dx$$
Let $u=\displaystyle\frac{\sqrt{x^2-1}}{\sqrt{9x^2-1}}$, which means that $\displaystyle\frac{du}{dx}=\frac{8x}{(x^2-1)^\frac{1}{2}(9x^2-1)^\frac{3}{2}}$
$$\Rightarrow I=\int\,\dfrac{u}{x^5}\,\frac{dx}{du}du$$
$$=\frac{1}{8}\int\,\dfrac{1}{x^6}\frac{\sqrt{x^2-1}}{\sqrt{9x^2-1}}\,\left((x^2-1)^\frac{1}{2}(9x^2-1)^\frac{3}{2}\right)du$$
$$=\frac{1}{8}\int\,\dfrac{(x^2-1)(9x^2-1)^2}{x^6}\,du$$
By some algebra you will find that $x=\displaystyle\frac{\sqrt{u^2-1}}{\sqrt{9u^2-1}}$, so
$$I=\frac{1}{8}\int\,\dfrac{\left(\frac{u^2-1}{9u^2-1}-1\right)\left(9\left(\frac{u^2-1}{9u^2-1}\right)-1\right)^2}{\left(\frac{u^2-1}{9u^2-1}\right)^3}\,du$$
which simplifies to
$$I=-64\int\,\dfrac{u^2}{(u^2-1)^3}\,du$$
Which can be found by the substitution $v=u^2-1$. This gives that
$$I=2\ln\left(\frac{1+u}{1-u}\right)-\frac{8(u^3+u)}{(u^2-1)^2}$$ You should get an answer by back-substitution. Perhaps a trigonometric substitution like you suggested would have made it simpler but this seemed to work fine.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying nested square roots ($\sqrt{6-4\sqrt{2}} + \sqrt{2}$) I guess I learned it many years ago at school, but I must have forgotten it. From a geometry puzzle I got to the solution
$\sqrt{6-4\sqrt{2}} + \sqrt{2}$
My calculator tells me that (within its precision) the result equals exactly 2, but I have no idea how to transform the calculation to symbolically get to that result.
(I can factor out one $\sqrt{2}$ from both terms, but that does not lead me anywhere, either)
| $$\sqrt{6-4\sqrt{2}}+\sqrt{2}=\sqrt{6-\sqrt{4^2}\sqrt{2}}+\sqrt{2}=$$
$$\sqrt{6-\sqrt{16}\sqrt{2}}+\sqrt{2}=\sqrt{6-\sqrt{16\cdot2}}+\sqrt{2}=$$
$$\sqrt{6-\sqrt{32}}+\sqrt{2}=\sqrt{\sqrt{6^2}-\sqrt{32}}+\sqrt{2}=$$
$$\sqrt{\sqrt{36}-\sqrt{32}}+\sqrt{2}=\sqrt{\sqrt{4\cdot9}-\sqrt{4\cdot8}}+\sqrt{2}=$$
$$\sqrt{2\sqrt{9}-2\sqrt{8}}+\sqrt{2}=\sqrt{2\left(\sqrt{9}-\sqrt{8}\right)}+\sqrt{2}=$$
$$\sqrt{2}\sqrt{\sqrt{9}-\sqrt{8}}+\sqrt{2}=\sqrt{2}\left(\sqrt{\sqrt{9}-\sqrt{8}}+1\right)=$$
$$\sqrt{2}\left(\sqrt{3-2\sqrt{2}}+1\right)=\sqrt{2}\left(\sqrt{2\left(\frac{3}{2}-\sqrt{2}\right)}+1\right)=$$
$$\sqrt{2}\left(\sqrt{2}\sqrt{\frac{3}{2}-\sqrt{2}}+1\right)=
\sqrt{2}\left(\sqrt{2}\left(1-\frac{1}{\sqrt{2}}\right)+1\right)=$$
$$\sqrt{2}\left(\sqrt{2}-\frac{\sqrt{2}}{\sqrt{2}}+1\right)=\sqrt{2}\left(\sqrt{2}-1+1\right)=$$
$$\sqrt{2}\left(\sqrt{2}\right)=\left(\sqrt{2}\right)^2=2$$
EDIT:
$$\sqrt{\frac{3}{2}-\sqrt{2}}=\sqrt{\frac{3}{2}-\frac{2\sqrt{2}}{2}}=$$
$$\sqrt{\frac{3-2\sqrt{2}}{2}}=\sqrt{\frac{\left(2-\sqrt{2}\right)^2}{4}}=$$
$$\frac{\sqrt{\left(2-\sqrt{2}\right)^2}}{\sqrt{4}}=\frac{2-\sqrt{2}}{2}=1-\frac{\sqrt{2}}{2}=1-\frac{1}{\sqrt{2}}$$
| {
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Prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+b}{a+c}+\frac{b+c}{b+a}+\frac{c+a}{c+b}.$
Prove that for all positive real numbers $a,b,$ and $c$, we have $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}.$$
What I tried is saying $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} = \dfrac{a^2c+b^2a+c^2b}{abc} \geq \dfrac{3abc}{abc} = 3$. Then how can I use this to prove that $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}$?
| Note that, $0=\ln\dfrac{a}{b}+\ln\dfrac{b}{c}+\ln\dfrac{c}{a}=\ln\dfrac{a+b}{a+c}+\ln\dfrac{b+c}{b+a}+\ln\dfrac{c+a}{c+b}$ and taking $\ln\dfrac{a}{b}\ge\ln\dfrac{b}{c}\ge\ln\dfrac{c}{a}$ WLOG gives us $ac\ge b^2$ ,$\;$ $a^2\ge bc$ $\;$ and $\;$ $ab\ge c^2$ also because $(a+c)^2=a^2+c^2+ac+ac\ge \dfrac{(a+c)^2}{2}+b^2+ac\ge(a+c)b+b^2+ac=(a+b)(b+c)$
$$\ln\dfrac{c+a}{c+b}\ge\ln\dfrac{a+b}{a+c}\ge\ln\dfrac{b+c}{b+a}$$ $\;$ or
$$\ln\dfrac{c+a}{c+b}\ge\ln\dfrac{b+c}{b+a}\ge\ln\dfrac{a+b}{a+c}$$ must be satisfied.
$\\$
Also it is easy to show $\dfrac{a}{b}\ge\dfrac{c+a}{c+b}$ because $ab+ac\ge ab+bc$ and $\dfrac{c}{a}\le\dfrac{b+c}{b+a}$ because $bc+ca\le ab+ac$ also $\dfrac{c}{a}\le\dfrac{a+b}{a+c}$ because $a^2\ge ac$ and $ab\ge c^2$.
$\\$
So; $$\left(\ln\dfrac{c+a}{c+b}, \ln\dfrac{a+b}{a+c}, \ln\dfrac{b+c}{b+a}\right) \prec \left(\ln\dfrac{a}{b}, \ln\dfrac{b}{c}, \ln\dfrac{c}{a}\right)$$
or $$\left(\ln\dfrac{c+a}{c+b}, \ln\dfrac{b+c}{b+a}, \ln\dfrac{a+b}{a+c}\right) \prec \left(\ln\dfrac{a}{b}, \ln\dfrac{b}{c}, \ln\dfrac{c}{a}\right)$$
and by Karamata's inequality with using
$\\$
$f(x)=e^x$ ($f''(x)=e^x>0$) $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solution to $p^3-p+1=a^2$ What are the solutions to $p^3-p+1=a^2$ where $p$ is prime and $a$ is natural?
I found the solutions:
$p=3$ and $a=5$
$p=5$ and $a=11$
and one solution when $p$ is not a prime:
$p=56$ and $a=419$, i think
I don't know what to do else. I tried to find limits for $a$ or limits for $p$, I tried to do something with factoring but I got nothing. Please help.
| If the equation is true, than $a^2 \equiv 1 \pmod p$. This implies that $a \equiv 1$ or $-1 \pmod p$, since $p$ is prime.
If $a \equiv 1 \pmod p$, then $a=pk+1$. This implies that $p^2-pk^2-(2k+1)=0$. This implies that $2k+1$ is divisible by $p$. $k=\frac{pb-1}{2}$ for some natural number $b$. Putting this into our equation, notice that $(\frac{pb-1}{2})^2+b=p$ This implies that $(pb-1)^2+4b=4p$. However, since $b \ge 1$,we can say that $(pb-1)^2+4b \ge (p-1)^2+4 > 4p$ if $p \ge 7$. This implies that $p=5,2,3$. However, since $2k+1$ is divisible by $p$, $p$ is not $2$. Manual computation gives us that $p=5$
In the same way, it $a \equiv -1 \pmod p$, $p$ is 3.
Therefore, $p$ is $3$ or $5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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To find the sum of the series $\,1+ \frac{1}{3\cdot4}+\frac{1}{5\cdot4^2}+\frac{1}{7\cdot4^3}+\ldots$ The answer given is $\log 3$.
Now looking at the series
\begin{align}
1+ \dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot4^2}+\dfrac{1}{7\cdot4^3}+\ldots &=
\sum\limits_{i=0}^\infty \dfrac{1}{\left(2n-1\right)\cdot4^n}
\\
\log 3 &=\sum\limits_{i=1}^\infty \dfrac{\left(-1\right)^{n+1}\,2^n}{n}
\end{align}
How do I relate these two series?
| Hint: a common series that is used for computing log of any real number is
$$
\log\left(\frac{1+x}{1-x}\right)=2\left(x+\frac{x^3}3+\frac{x^5}5+\frac{x^7}7+\dots\right)
$$
$u=\frac{1+x}{1-x}\iff x=\frac{u-1}{u+1}$
| {
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How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
| Using geometric series
Consider $S_k = 1 + 2\sum_{i=0}^{k-1}4^i$. It follows that $S_k$ is in $\mathbb{N}^{+}$. However,
$$S_k = 1 + 2\cdot\frac{4^k - 1}{4 - 1} = \frac{2^{2k+1}+1}{3}$$
QED
Using recurrence
For $k\in\mathbb{N}$, consider the recurrence $J_{k+1} = 4J_{k} - 1$, with $J_0 = 1$.
Since $J_0 = 1$, it follows that each of $J_k$ is in $\mathbb{N}^{+}$.
Solving the recurrence yields: $J_k = \frac{1}{3}(2^{2k + 1}+ 1)$.
The characteristic equation $r^2 - 5r + 4 = 0$ follows from $J_{k+2} = 5J_{k+1} - 4J_{k}$. Hence, $J_k = A\lambda_1^k + B\lambda_2^k$, with $\lambda_1 = 4$ and $\lambda_2 = 1$ as solutions to the characteristic equation. As $J_0 = 1 \implies J_1 = 3$, it follows that $A = \frac{2}{3}$ and $B = \frac{1}{3}$.
QED
| {
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To evaluate the given determinant Question: Evaluate the determinant
$\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|$
My answer:
$\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|= \left|
\begin{array}{cc} b^2c^2 & bc & c \\
c^2a^2 & ca & a \\
a^2b^2 & ab & b \\
\end{array}
\right| + \left|
\begin{array}{cc} b^2c^2 & bc & b \\
c^2a^2 & ca & c \\
a^2b^2 & ab & a \\
\end{array}
\right|= abc \left|
\begin{array}{cc} bc^2 & c & 1 \\
ca^2 & a & 1 \\
ab^2 & b & 1 \\
\end{array}
\right| +abc \left|
\begin{array}{cc} b^2c & b & 1 \\
c^2a & c & 1 \\
a^2b & a & 1 \\
\end{array}
\right|$
how do I proceed from here?
| Your given matrix is : $\left|
\begin{array}{cc} b^2c^2 & bc & b+c \\
c^2a^2 & ca & c+a \\
a^2b^2 & ab & a+b \\
\end{array}
\right|$
Determinant of the given matrix is :
$\implies b^2c^2[ca^2 + abc - abc + a^2b] - bc[a^3c^2 + a^2bc^2 - a^2b^2c - a^3b^2] + (b+c)[a^3bc^2 - a^3b^2c]$
$\implies b^2c^2[ca^2 + a^2b] - bc[a^3c^2 + a^2bc^2 - a^2b^2c - a^3b^2] + (b+c)[a^3bc^2 - a^3b^2c]$
$\implies a^2b^2c^3 - a^2b^3c^2 - a^3bc^3 - a^2b^2c^3 + a^2b^3c^2 + a^3b^3c + a^3b^2c^2 - a^3b^3c + a^3bc^3 - a^3b^2c^2$
$\implies a^2b^2c^3 - a^2b^2c^3 - a^2b^3c^2 + a^2b^3c^2 - a^3bc^3 + a^3bc^3 + a^3b^3c - a^3b^3c + a^3b^2c^2 - a^3b^2c^2$
$\implies 0$
Therefore, given matrix is singular.
| {
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How to calculate $\int_a^bx^2 dx$ using summation? So for this case, we divide it to $n$ partitions and so the width of each partition is $\frac{b-a}{n}$ and the height is $f(x)$.
\begin{align}
x_0&=a\\
x_1&=a+\frac{b-a}{n}\\
&\ldots\\
x_{i-1}&=a+(i-1)\frac{b-a}{n}\\
x_i&=a+i\frac{b-a}{n}
\end{align}
So I pick left point, which is $x_{i-1}$
I start with
\begin{align}
\sum\limits_{i=1}^{n} \frac{b-a}{n}f\left(a+\frac{(i-1)(b-a)}{n}\right)
&=\frac{b-a}{n}\sum\limits_{i=1}^{n} \left(a+\frac{(i-1)(b-a)}{n}\right)^2\\
&=\frac{b-a}{n}\left(na^2+ \frac{2a(b-a)}{n}\sum\limits_{i=1}^{n}(i-1)+\frac{(b-a)^2}{n^2} \sum\limits_{i=1}^{n}(i-1)^2\right)
\end{align}
Here I am stuck because I don't know what$ \sum\limits_{i=1}^{n}(i-1)^2$is (feel like it diverges). Could someone help?
| $$\sum_{i=1}^n (i-1)^2$$
$$=\sum_{i=1}^n (i^2-2i+1)$$
$$=\sum_{i=1}^n i^2-2\sum_{i=1}^ni+\sum_{i=1}^n1$$
$$=\frac{n(n+1)(2n+1)}{6}-2\cdot \frac{n(n+1)}{2}+n$$
$$=\frac{(n^2+n)(2n+1)}{6}-n(n+1)+n$$
$$=\frac{2n^3+3n^2+n}{6}-n^2$$
$$=\frac{2n^3-3n^2+n}{6}$$
$$=\frac{n(n-1)(2n-1)}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Over an integral arising from Kepler's problem [also: generally useful integral, NOT DUPLICATE!] This post might appear as a duplicate of the following:
Over an integral arising from Kepler's problem [also: generally useful integral]
So recalling quickly:
$$\Phi(\epsilon) = \frac{1}{2\pi}\int_0^{2\pi}\ \frac{\sin^2\theta\ \text{d}\theta}{(1 + \epsilon\cos\theta)^2}$$
Note: The integral arises in studying the Kepler's problem, which is why $\epsilon\in [0;\ 1]$.
Necessary Note I tried to solve it with complex analysis, as you can see in the above link, but in my book it's solved by series with lots of unknown passages, which I will write down here.
$$
\begin{align*}
\frac{1}{2\pi}\int_0^{2\pi}\ \frac{\sin^2\theta\ \text{d}\theta}{(1 + \epsilon\cos\theta)^2} & = \sum_{n = 1}^{+\infty}\ (-1)^n\epsilon^n(n+1)\cdot\frac{1}{2\pi}\int_0^{2\pi}\ (1 - \cos\theta)\cos^n\theta\ \text{d}\theta
\\\\
& = \sum_{n = 0}^{+\infty}\ \epsilon^{2n}(2n+1)\left[\frac{1}{2^{2n}} \binom{2n}{n} - \frac{1}{2^{2n+2}}\binom{2n+2}{n+1}\right]
\\\\
& = \sum_{n = 0}^{+\infty}\ \epsilon^{2n}\frac{2n+1}{2^{2n+2}} \frac{(2n)!}{(n!)^2} \left[4 - \frac{(2n+1)(2n+2)}{(n+1)^2}\right]
\\\\
& = \sum_{n = 0}^{+\infty}\ \epsilon^{2n} \frac{(2n+1)!}{2^{2n+1}(n!)^2 (n+1)}
\\\\
& = \frac{1}{\epsilon^2}\ \sum_{n = 0}^{+\infty} \epsilon^{2(n+1)}(-1)^{n+1}\binom{-1/2}{n+1}
\\\\
& = \frac{1}{\epsilon^2}[(1 - \epsilon^2)^{-1/2} - 1]
\end{align*}
$$
Can anybody help with that? It seems really messy..
P.s. I wrote down here exactly the passages of the book.
What I need
I am not familiar with the passage from the first row to the second one (namely: where the integral has been solved, in which we pass to the binomial coefficient). In addition to that, I didn't understand how the very last series has been summed.
| There is an elementary path:
$$I = \frac{1}{2\pi}\int_0^{2\pi}\frac{\sin^2\theta}{(1+\varepsilon\cos\theta)^2}\,d\theta =\frac{1}{2\pi\varepsilon}\int_0^{2\pi}\sin\theta\,d\left(\dfrac{1}{1+\varepsilon\cos\theta}\right)$$
By parts:
$$ I = \frac{1}{2\pi\varepsilon}\dfrac{\sin\theta}{1+\varepsilon\cos\theta}\,\biggr|_0^{2\pi} - \dfrac{1}{2\pi\varepsilon}\int_0^{2\pi}\frac{\cos\theta\,d\theta}{1+\varepsilon\cos\theta} = 0 - \frac{1}{2\pi\varepsilon^2}\int_0^{2\pi}\frac{\varepsilon\cos\theta+1-1}{1+\varepsilon\cos\theta}\,d\theta$$$$ = -\frac{1}{2\pi\varepsilon^2}\theta\,\biggr|_0^{2\pi} + \frac{1}{2\pi\varepsilon^2}\int_0^{2\pi}\frac{d\theta}{1+\varepsilon\cos\theta} = -\frac{1}{\varepsilon^2} + \frac{1}{\pi\varepsilon^2}\int_0^{\pi}\frac{d\theta}{1+\varepsilon\cos\theta}.$$
Applying universal trigonometric substitution,
$$\theta = 2\arctan u,\quad d\theta = \dfrac{2}{1+u^2}du,\quad\cos\theta = \dfrac{1-u^2}{1+u^2},$$
we get:
$$I=-\dfrac{1}{\varepsilon^2}+\dfrac{2}{\pi\varepsilon^2}\int_0^\infty\frac{d\theta}{1+\varepsilon+(1-\varepsilon)u^2},$$
$$I=-\dfrac{1}{\varepsilon^2}+\dfrac{2}{\pi\varepsilon^2(1-\varepsilon)}\sqrt{\dfrac{1-\varepsilon}{1+\varepsilon}}\arctan\sqrt{\dfrac{1+\varepsilon}{1-\varepsilon}}u\,\biggr|_0^\infty,$$
$$ \boxed {I = \dfrac{1}{\varepsilon^2}\left(\dfrac{1}{\sqrt{1-\varepsilon^2}}-1\right)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate this integral $\int \frac{x^2}{4x^4+25}dx$. I have to calculate this integral $\int \frac{x^2}{4x^4+25}dx$.
I dont have any idea about that.
I thought about parts integration.
Thanks.
| Hint:
$4x^4+25$ factors as the product of two irreducible quadratic polynomials:
$$4x^4+25=(2x^2+5)^2-20x^2=(2x^2-2\sqrt 5x+5)(2x^2+2\sqrt 5x+5,$$
whence the partial fractions decomposition:
$$\frac{x^2}{4x^4+25}=\frac{Ax+B}{2x^2-2\sqrt 5x+5}+\frac{Cx+D}{2x^2+2\sqrt 5x+5}.$$
As $\dfrac{x^2}{4x^4+25}$ is an even function, we have $C=-A,\enspace B=D$. Setting $x=0$ shows $B=D=0$, and reducing to the same denominator yields $A=\dfrac{\sqrt 5}{20}$. Hence we have to compute:
$$\frac{\sqrt5}{20}\int\frac{ x}{2x^2-2\sqrt 5x+5}\,\mathrm d x-\frac{\sqrt5}{20}\int\frac{x}{2x^2+2\sqrt 5x+5}\,\mathrm d x$$
Now split each integral:
\begin{align*}
\int\frac{ x}{2x^2-2\sqrt 5x+5}\,\mathrm d x&= \frac14\int\frac{4x-2\sqrt5}{2x^2-2\sqrt 5x+5}\,\mathrm d x+\frac12\int\frac{\mathrm d x}{4x^2-4\sqrt 5x+10} \\[1ex]
&=\frac14\ln(2x^2-2\sqrt 5x+5)+\frac14\int\frac{\mathrm d(2x-\sqrt5)}{(2x-\sqrt5)^2+5} \\[1ex]
&= \frac14\ln(2x^2-2\sqrt 5x+5)+\frac1{4\sqrt5}\,\arctan(2x-\sqrt5)
\end{align*}
Similarly,
$$\int\frac{ x}{2x^2+2\sqrt 5x+5}\,\mathrm d x=\frac14\ln(2x^2+2\sqrt 5x+5)+\frac1{4\sqrt5}\,\arctan(2x+\sqrt5)$$
Thus we obtain
$$\int\frac{x^2}{4x^4+25}\,\mathrm dx=\frac{\sqrt5}{80}\ln\biggl(\frac{2x^2-2\sqrt 5x+5}{2x^2+2\sqrt 5x+5}\biggr)+\frac{\sqrt5}{400}(\arctan(2x-\sqrt5)-\arctan(2x+\sqrt5)). $$
The last term may be further simplified using the formula $\;\arctan p-\arctan q \equiv\arctan\dfrac{p-q}{1+pq}\mod\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I evaluate this series? How do I evaluate this series:
\begin{equation}
\sum_{n=2}^\infty \frac{\prod_{k=1}^{n-1} (2k-1) }{2^nn!} = \frac{1}{8} + \frac{1}{16} + \frac{5}{128} + \frac{7}{256} +\ldots
\end{equation}
I wanted to use the Comparison test to show convergence, but I didn't know what to compare it to since my series has a product in the numerator...
I'm lost as to how to evaluate it.
| Here are hints to show that the series converges. (Actually computing the value will take more work.)
Hint1: The series is $\frac1{2\cdot4} + \frac{1\cdot3}{2\cdot4\cdot6} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10} + \cdots$.
Hint2: One attempt is to cancel the $1$ with the $2$, the $3$ with the $4$, etc. to get $\frac1{2\cdot4} + \frac{1\cdot3}{2\cdot4\cdot6} + \frac{1\cdot3\cdot5}{2\cdot4\cdot6\cdot8} + \frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10} + \cdots \le \frac14 + \frac16 + \frac18 + \frac1{10} +\cdots$. But the right hand side diverges, so this is no good, so you need to be more clever.
Hint3: $\frac{1\cdot3\cdot5\cdot7}{2\cdot4\cdot6\cdot8\cdot10} = \frac{(\sqrt1)^2\cdot(\sqrt3)^2\cdot(\sqrt5)^2\cdot(\sqrt7)^2}{2\cdot4\cdot6\cdot8\cdot10} = \sqrt1 \cdot \frac{\sqrt1\cdot\sqrt3}2 \cdot \frac{\sqrt3\cdot\sqrt5}4 \cdot \frac{\sqrt5\cdot\sqrt7}6 \cdot \frac{\sqrt7}{8\cdot10} \le \frac{\sqrt7}{8\cdot 10} $. So the entries grow on the order of $\frac{\sqrt n}{n^2}=\frac1{n^{1.5}}$, which converges by the $p$-series test.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627248",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Proving trig identity using De Moivre's Theorem
Question: Prove $$\cos(3x) = \cos^3(x) - 3\cos(x)\sin^2(x) $$ by using De'Moivres Theorem
So far (learning complex numbers at the moment) that De Moivre's theorem states that
if $z$ $=$ $r\text{cis}(\theta)$ then $z^n = r^n\text{cis}(n\theta)$
so with this question I was thinking if
$$ z = \cos(3\theta) + i\sin(3\theta) $$
then
$$ z = (\cos(\theta) + i\sin(\theta))^3 $$
and then expanding and comparing the real part? Is that the right way to go for this question?
| The solution can be completed in this manner:
We know, by De-Moivre's Theorem,
$$(\cos x + i \sin x)^3=\cos 3x + i \sin 3x$$
Therefore, we can write
$$\cos^3 x + 3i\cos^2x\sin x + 3i^2\cos x\sin^2 x + i^3 \sin^3 x=\cos 3x + i \sin 3x$$
or, $$\cos^3 x + 3i\cos^2x\sin x - 3\cos x\sin^2 x - i \sin^3 x=\cos 3x + i \sin 3x$$
or, $$(\cos^3 x- 3\cos x\sin^2 x) + i(3\cos^2x\sin x - \sin^3 x)=\cos 3x + i \sin 3x$$
As far as your problem is concerned, just compare the real parts of this equation.
Hope this helps you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solving the equation $\frac{x+\sqrt 3}{\sqrt x + \sqrt {x+\sqrt 3}} + \frac{x-\sqrt 3}{\sqrt x - \sqrt {x-\sqrt 3}} = \sqrt x$ for $x$ I have the following equation:
$$\frac{x+\sqrt 3}{\sqrt x + \sqrt {x+\sqrt 3}} + \frac{x-\sqrt 3}{\sqrt x - \sqrt {x-\sqrt 3}} = \sqrt x$$
I know that $x=2$, but I don't know steps to get there.
| Take lcm & simplify as follows $$\frac{x+\sqrt 3}{\sqrt x+\sqrt{x+\sqrt 3}}+\frac{x-\sqrt 3}{\sqrt x-\sqrt{x-\sqrt 3}}=\sqrt x$$
$$\frac{(x+\sqrt 3)(\sqrt x-\sqrt{x-\sqrt 3})+(x-\sqrt 3)(\sqrt x+\sqrt{x+\sqrt 3})}{(\sqrt x+\sqrt{x+\sqrt 3})(\sqrt x-\sqrt{x-\sqrt 3})}=\sqrt x$$
$$2x\sqrt x-x\sqrt{x-\sqrt 3}-\sqrt 3\sqrt{x-\sqrt 3}+x\sqrt{x+\sqrt 3}-\sqrt 3\sqrt{x+\sqrt 3}$$$$=x\sqrt x+x\sqrt{x+\sqrt 3}-x\sqrt x-x\sqrt{x-\sqrt 3}-\sqrt x\sqrt{x^2-3}$$
$$x\sqrt x+\sqrt x\sqrt{x^2-3}-\sqrt 3\left(\sqrt{x+\sqrt 3}+\sqrt{x-\sqrt 3}\right)=0$$
re-arranging & taking squares on both the sides,
$$\left(\sqrt x(x+\sqrt{x^2-3})\right)^2=\left(\sqrt 3\left(\sqrt{x+\sqrt 3}+\sqrt{x-\sqrt 3}\right)\right)^2$$
$$2x^3-9x=(2x^2-6)\sqrt{x^2-3}$$
$$2x^3-9x=2(x^2-3)^{3/2}$$
taking squares on both the sides & simplifying, one should get
$$27x^2-108=0$$
$$x^2=4\iff x=\pm 2$$
but $\sqrt x$ is undefined for $x<0$ hence, the correct value is $$\boxed{\color{red}{x=2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Calculate the limit $\left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right)$ I have to calculate the following limit:
$$\lim_{n\to\infty} \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right)$$
I believe the limit equals $1$, and I think I can prove it with the squeeze theorem, but I don't really know how.
Any help is appreciated, I'd like to receive some hints if possible.
Thanks!
| This is the final answer I got, thanks to all the help:
For every $n>0$,$\frac{n}{\sqrt{n^2+n}}\le \left(\frac{1}{\sqrt{n^2+1}}+\frac{1}{\sqrt{n^2+2}}+...+\frac{1}{\sqrt{n^2+n}}\right) \le \frac{1}{\sqrt{n^2+n}}$
Using the squeeze theorem, we calculate the middle expression's limit.
$\lim_\limits{n \to \infty} \frac{n}{\sqrt{n^2+n}}=\lim_\limits{n \to \infty} \frac{\sqrt{n^2}}{\sqrt{n^2+n}}= \lim_\limits{n \to \infty}\sqrt{\frac{ {n^2}}{{n^2+n}}}=\lim_\limits{n \to \infty}\sqrt{\frac{ \frac{n^2}{n^2}}{{\frac{n^2}{n^2}+\frac{n}{n^2}}}}=\lim_\limits{n \to \infty} \sqrt{\frac{1}{1+\frac{1}{n}}}=1$ from limits arithmetic.
Likewise, we can calculate the right hand side and reach to the conclusion that the original sequence approaches $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to evaluate $\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$? I have problems to solve this limit
$$\lim _{x\to \infty }\left(x\sqrt{x^2-x}-x^2\cos\left(\frac{1}{\sqrt{x}}\right)\right)$$
I tried with Taylor:
$$\lim _{x\to \infty }\left(x^2\sqrt{1-\frac{1}{x}}-x^2\left(1-\frac{1}{2x}+\frac{1}{24x^2}\right)\right)=\lim _{x\to \infty }\left(x^2-x^2+\frac{x}{2}-\frac{1}{24}\right)=\infty$$
that is the wrong result, in must be $\color{red}{-\frac{1}{6}}$
| Hint: You need to use the Taylor series expansion for the square root as well as the cosine. Expanding $\sqrt{1-u}$ around $u=0$ we have that $$\sqrt{1-u}=1-\frac{u}{2}-\frac{u^2}{8}+O(u^3)$$ and so as $x\rightarrow \infty$ $$x^2\sqrt{1-\frac{1}{x}}=x^2-\frac{x}{2}-\frac{1}{8}+O\left(\frac{1}{x}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$ I saw some resolutions here like $\sqrt{x+\sqrt{x+\sqrt{x}}}- \sqrt{x}$, but I couldn't get the point to find
$\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$.
I tried $\frac{1}{x}.(\sqrt{x+\sqrt{x+\sqrt{x}}})=\frac{\sqrt{x}}{x}\left(\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}} \right)=\frac{1}{\sqrt{x}}\left(\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}} \right)$ but now I got stuck. Could anyone help?
| $\dfrac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x} = \sqrt{\dfrac{x+\sqrt{x+\sqrt{x}}}{x^2}}\\
= \sqrt{\dfrac{1}{x} + \dfrac{\sqrt{x+\sqrt{x}}}{x^2}}
= \sqrt{\dfrac{1}{x} + \sqrt{\dfrac{1}{x^3}+\sqrt{\dfrac{1}{x^7}}}}$
then the limit when $x \rightarrow \infty$ is clearly $0$
| {
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"url": "https://math.stackexchange.com/questions/1629846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$ If $x^2+3x+5=0$ and $ax^2+bx+c=0$ have a common root and $a,b,c\in \mathbb{N}$, find the minimum value of $a+b+c$
Using the condition for common root, $$(3c-5b)(b-3a)=(c-5a)^2$$
$$3bc-9ac-5b^2+15ab=c^2+25a^2-10ac$$
$$25a^2+5b^2+c^2=15ab+3bc+ac$$
There is pattern in the equation. The coefficients of the middle terms and first terms of both sides are $(5\times 5,5)$ and $(5\times 3,3)$.
I tried to use Lagrange multipliers. But isnt there any simpler way to minimize $a+b+c$?
| See one root is complex so other os obviously its conjugate as $b^2-4ac<0$ so the equations have both roots in common and as $a,b,c$ have no common factor thus $min(a+b+c)=1+3+5=9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Working out expression values What is the value of this expression?
$$\frac{1}{\dfrac{1}{\frac{1}{2} + \frac{1}{3}} + \dfrac{1}{\frac{1}{4} + \frac{1}{5}}}$$
I thought I'd start by working out 1/2 + 1/3,which is 5/6, and then working out 1/4 + 1/5, which is 9/20. Then I added them together, which is 1 17/60. But how do I proceed?
| As you've noticed, $\frac{1}{2}+\frac{1}{3}=\frac{5}{6}$ and $\frac{1}{4}+\frac{1}{5}=\frac{9}{20}$. Therefore:
$$\begin{align}
\cfrac{1}{
\cfrac{1}{ \cfrac{1}{2}+\cfrac{1}{3} } + \cfrac{1}{ \cfrac{1}{4}+\cfrac{1}{5} }
}
& =\cfrac{1}{ \cfrac{1}{ \cfrac{5}{6} }+\cfrac{1}{ \cfrac{9}{20} } }
\\[2ex] & =\cfrac{1}{\cfrac{6}{5}+\cfrac{20}{9}}
\\[2ex] & =\cfrac{1}{\cfrac{6\cdot 9+20\cdot 5}{5\cdot 9}}
\\[2ex] & =\cfrac{5\cdot 9}{6\cdot 9+20\cdot 5}
\\[2ex] & = \ldots
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find $\alpha^3 + \beta^3$ which are roots of a quadratic equation. I have a question.
Given a quadratic polynomial, $ax^2 +bx+c$, and having roots $\alpha$ and $\beta$. Find $\alpha^3+\beta^3$. Also find $\frac1\alpha^3+\frac1\beta^3$
I don't know how to proceed. Any help would be appreciated.
| Just to be different.
If $\alpha$ is a solution of $ax^2 + bx + c = 0$ Then
$a\alpha^2 + b\alpha + c = 0$
So
$\quad \alpha^2 = -\dfrac{b\alpha + c}{a}$
$\quad \alpha^3 = -\dfrac{b\alpha^2 + c\alpha}{a}
= -\dfrac{b\left( -\dfrac{b\alpha + c}{a} \right)+ c\alpha}{a}
= -\dfrac{ -b^2\alpha - bc + ac\alpha}{a^2}
= -\dfrac{(ac -b^2)\alpha - bc}{a^2}$
Similarly,
$\quad \beta^3 = -\dfrac{(ac -b^2)\beta - bc}{a^2}$
So $\alpha^3 + \beta^3 =
-\dfrac{(ac -b^2)(\alpha + \beta) - 2bc}{a^2} =
\dfrac{(b^2 - ac)(\alpha + \beta) + 2bc}{a^2}$
Since $\alpha + \beta = -\dfrac ba$, we see that
$\alpha^3 + \beta^3 =
\dfrac{(b^2 - ac) \left(-\dfrac ba \right) + 2bc}{a^2} =
\dfrac{(-b^3 + abc) + 2abc}{a^3} =
\dfrac{3abc - b^3}{a^3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Can I use GAP to show block structures in a multiplication group clearer $\ $? The usual output of GAP for the multiplication table of the group $S3$ is
$$\pmatrix{1&2&3&4&5&6\\2&1&4&3&6&5\\3&6&5&2&1&4\\4&5&6&1&2&3\\5&4&1&6&3&2\\6&3&2&5&4&1}$$
The table
$$\pmatrix{1&2&3&4&5&6\\2&3&1&6&4&5\\3&1&2&5&6&4\\4&5&6&1&2&3\\5&6&4&3&1&2\\6&4&5&2&3&1}$$
would show the block-structure (two $3\times 3$ blocks with entries $1-3$ and two $3\times 3$-blocks with entries $4-6$) much clearer.
Can I display such a multiplication table with gap ?
| Let N be a normal subgroup of G. Then
MultiplicationTable(Flat(List(RightCosets(G, N),i->List(i))));
produces a table with block structure.
For example:
gap> G:=QuaternionGroup(8);
<pc group of size 8 with 3 generators>
gap> N:=NormalSubgroups(G)[5];
Group([ y2 ])
gap> Display(MultiplicationTable(Flat(List(RightCosets(G, N),i->List(i)))));
[ [ 1, 2, 3, 4, 5, 6, 7, 8 ],
[ 2, 1, 4, 3, 6, 5, 8, 7 ],
[ 3, 4, 2, 1, 8, 7, 5, 6 ],
[ 4, 3, 1, 2, 7, 8, 6, 5 ],
[ 5, 6, 7, 8, 2, 1, 4, 3 ],
[ 6, 5, 8, 7, 1, 2, 3, 4 ],
[ 7, 8, 6, 5, 3, 4, 2, 1 ],
[ 8, 7, 5, 6, 4, 3, 1, 2 ] ]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Can you find the maximum or minimum of an equation without calculus? Without using calculus is it possible to find provably and exactly the maximum value
or the minimum value of a quadratic equation
$$ y:=ax^2+bx+c $$
(and also without completing the square)?
I'd love to know the answer.
| This is almost the same as completing the square but .. for giggles.
If $a = 0$ we know $y = xb + c$ will get "extreme" and "extreme" positive and negative values of $x$ so no max or minimum is possible.
So if $ax^2 + bx + c = a(x^2 + x b/a)+c := a(x^2 + b'x) + c$ So finding the max/min is simply a matter of finding the max/min of $x^2 + b'x$ and multiplying by $a$ and adding $c$.
In other words .... wolog $a = 1$ and $c = 0$.
Now we know $x^2 + bx$ has only a min as $x^2$ is positive and as $|x|$ increases the $x^2$ term "overpowers" the $bx$ term. Or if $x > |b|/2$ then $(x+ h)^2 + b(x + h) = x^2 + bx +h(2x + b) + h^2 > 0$ so the expression has no max value.
So we want to find the minimum of $x^ + b'x = x(x + b)$. Let $y := x - b'/2$ then $x(x + b')=(y -b'/2)(y + b'/2)= y^2 - (b'^2/4)$. As $y^2 \ge 0$ the min will occur when $y = 0$ or in other words, $x= b'/2 = b/2a$
So the max/min of $ax^2 + bx + c$ occurs at $x = b/2a$ and the max/min value is $b^2/4 + b^2/2a + c$
Okay, that really was the same thing as completing the square but it didn't feel like it so what the @@@@.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find a thousand natural numbers such that their sum equals their product The question is to find a thousand natural numbers such that their sum equals their product. Here's my approach :
I worked on this question for lesser cases :
\begin{align}
&2 \times 2 = 2 + 2\\
&2 \times 3 \times 1 = 2 + 3 + 1\\
&3 \times 3 \times 1 \times 1 \times 1 = 3 + 3 + 1 + 1 + 1\\
&7 \times 7 \times 1 \times 1 \times \dots\times 1 \text{ (35 times) } = 7 + 7 + 1 + 1 .... \text{ (35 times) }
\end{align}
Using this logic, I seemed to have reduced the problem in the following way.
$a \times b \times 1 \times 1 \times 1 \times\dots\times 1 = a + b + 1 + 1 +...$
This equality is satisfied whenever $ ab = a + b + (1000-n)$
Or $ abc\cdots n = a + b + \dots + n + ... + (1000 - n)$
In other words, I need to search for n numbers such that their product is greater by $1000-n$ than their sum. This allows the remaining spots to be filled by $1$'s. I feel like I'm close to the answer.
Note : I have got the answer thanks to Henning's help. It's
$112 \times 10 \times 1 \times 1 \times 1 \times ...$ ($998$ times)$ = 10 + 112 + 1 + 1 + 1 + ...$ ($998$ times)
This is for the two variable case. Have any of you found answers to more than two variables ?
$abc...n = a + b + c + ... + n + (1000 - n) $
| Do we have an alternative to fill up with ones?
Let $N=1000$, then
\begin{align}
\sum_{i=1}^N n_i &= \prod_{i=1}^N n_i \iff \\
N n_a &= n_g^N
\end{align}
where $n_a$ is the arithmetic mean and $n_g$ the geometric mean of the numbers $n_i$.
Those numbers on the left and right hand side of the equation
drift apart very fast.
Already for $n_a = n_g = 2$ we would have $2000$ vs. $2^{1000} \approx 10^{301}$.
Assuming $n_a \approx n_g$ we estimate:
$$
N x = x^N \Rightarrow \\
N = x^{N-1} \Rightarrow \\
x = \sqrt[N-1]{N}
$$
For $N=1000$ this gives $x=\sqrt[999]{1000}=1.0069\dotso$.
Such an estimated mean gives not much room for numbers $n_i > 1$.
$$
1.007 = \frac{1000+a}{1000} = 1 + a/1000 \Rightarrow \\
a = 7
$$
Spreading an excess of $7$ over a couple of numbers is too pessimistic.
$$
1.007 = \sqrt[1000]{b} \Rightarrow \\
b \approx 1070
$$
That looks better, the product has to be equal with a sum something above $1000$.
Have any of you found answers to more than two variables ?
A solution with three numbers different from $1$ is
$x=67$, $y=z=4$. This gives $67+4+4+997=1072$.
Also $67\times 4 \times 4 \times 1^{997}=1072$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of $x$ which is correct I have one exercise which is $$(x+2013)(x+2014)(x+2015)(x+2016)+1=0$$
I tag $A=x+2013$ or other for many ways but still can not find the first $x$ value. please help.
| $a(a+1)(a+2)(a+3)+1=0$ thus $(a^2+3a)(a^2+3a+2)+1=0$ thus let $a^2+3a=x$ thus $b^2+2b+1=0 so b=-1,-1$ thus $a^2+3a+1=0$ thus $a=\frac{-3\pm(\sqrt{5})}{2}$ and accordingly $x$ will be approximated
| {
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"timestamp": "2023-03-29T00:00:00",
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Image of a family of circles under $w = 1/z$ Given the family of circles $x^{2}+y^{2} = ax$, where $a \in \mathbb{R}$, I need to find the image under the transformation $w = 1/z$. I was given the hint to rewrite the equation first in terms of $z$, $\overline{z}$, and then plug in $z = 1/w$.
However, I am having difficulty doing this.
I completed the square in $x^{2}+y^{2}=ax$ to obtain $\left(x - \frac{a}{2} \right)^{2} + y^{2} = \left(\frac{a}{2} \right)^{2}$.
Then, given that $\displaystyle x = Re(z) = \frac{z+\overline{z}}{2}$ and $\displaystyle y = Im(z)= \frac{-(z-\overline{z})}{2i}$, I made those substitutaions and my equation became $\left( \frac{z+\overline{z}-a}{2}\right)^{2} - \left(\frac{\overline{z}-z}{2} \right)^{2} = \left(\frac{a}{2} \right)^{2}$.
Then, sbustituting in $z = \frac{1}{w}$, this became $\displaystyle \frac{\left(\frac{1}{w} + \frac{1}{\overline{w}} - a \right)^{2}}{4} - \frac{\left(\frac{1}{\overline{w}} - \frac{1}{w} \right)^{2}}{4} = \frac{a^{2}}{4}$.
Beyond this, my algebra gets very wonky.
Could someone please tell me what my final result should be? Knowing that would allow me to work backwards and then apply these methods to other problems (of which I have many to do!).
Thanks.
| Hint: This transformation can be seen as successive transformations $$Z = \frac{z}{|z|^2} , \,\,\,w = \overline {Z}$$
since $z \cdot \overline{z} = |z|^2$.
Then $z = \frac{1}{w} = \frac{\overline w}{|w|^2}$ and $$x = \frac{u}{u^2 + v^2} \,\,\,\text{and} \,\,\,\,y = \frac{-v}{u^2+v^2}$$
where $w = u + iv$.
Edit: Now study $$A(x^2+y^2) + Bx + Cy + D = 0$$
complete the square and make the substitutions to see that is taken to
$$D(u^2 + v^2) + Bu - Cv + A = 0 $$
which is a circle or a line., depending on whether $D \neq 0$ or $D=0.$
See what happens when a cirle $A \neq 0$ is passing or not through the origin.
| {
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"timestamp": "2023-03-29T00:00:00",
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Showing $\int_{1}^{0}\frac{\ln(1-x)}{x}dx=\frac{\pi ^{2}}{6}$ Is there way to show $$\int_{1}^{0}\frac{\ln(1-x)}{x}dx=\frac{\pi ^{2}}{6}$$ without using the Riemann zeta function?
| Note: As pointed out by Hans, this answer uses the fact that $\zeta(2) = \dfrac{\pi ^2}{6}$ and therefore isn't what the OP asked for.
$$\frac{\ln(1+x)}{x} = 1 - \frac{x}{2} + \frac{x^2}{3}...$$
$$\int\frac{\ln(1+x)}{x}dx = x - \frac{x^2}{4} + \frac{x^3}{9}...+C$$
$$\int\limits^{0}_{1}\frac{\ln(1+x)}{x}dx = -1 + \frac{1}{4} - \frac{1}{9}...$$
After question edit:
We have $$\frac{\ln(1-x)}{x} = - 1 + \frac{x}{2} - \frac{x^2}{3}...$$
$$\int\frac{\ln(1-x)}{x}dx = -x + \frac{x^2}{4} - \frac{x^3}{9}...+C$$
Therefore $$\int\limits^{0}_{1}\frac{\ln(1-x)}{x}dx = 1 + \frac{1}{4} + \frac{1}{9}...$$
$$=\frac{\pi ^2}{6}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$27$ balls into $3$ cells Spreading $15$ white balls and $12$ black balls into $3$ cells, each of which can contain any number of balls.
$(A.)$ Find the probability that in each cell there will be exactly $5$ white balls.
$(B.)$ Find the probability that in each cell there will be exactly $5$ white balls and exactly $4$ black balls.
$(C.)$ Find the probability that in one cell there will be the whole the black balls and $3$ white balls, and there will be an even number of white balls in the two other cells.
My attempt:
$(A.)$ $\Omega$ is all the possible spredings $|\Omega|=3^{15}3^{12}=3^{27}$ because each ball "choose a cell"
Let A be the event that in each cell there will be exactly $5$ white balls,
$$|A|=\binom{15}{5}\binom{10}{5}\binom{5}{5}\\
\Longrightarrow P[A]=\frac{\binom{15}{5}\binom{10}{5}\binom{5}{5}}{3^{27}}=\frac{6250}{2541865828329}$$
$(B.)$ Let $B$ be the event that in each cell there will be exactly $4$ black balls. $$|B|=\binom{12}{4}\binom{8}{4}\binom{4}{4}\\
\Longrightarrow A \text{ and } B \text{ are independent I think so: }P[A\cap B]=P[A]P[B]=\frac{\binom{15}{5}\binom{10}{5}\binom{5}{5}}{3^{27}}\cdot \frac{\binom{12}{4}\binom{8}{4}\binom{4}{4}}{3^{27}}=\frac{6250}{2541865828329} \cdot \frac{3850}{847288609443}$$
$(C.)$ Let $C_1$ be the event that in one cell there will be the whole the black balls and $3$ white balls, $|C_1|=\binom{12}{12}\binom{15}{3}\times 3$, and let $C_2$ be the event that there will be an even number of white balls in the two other cells, $|C_2|=\binom{8}{4}\binom{4}{4}\times 2$.
$C_1$ and $C_2$ are independent so $$P[C_1\cap C_2]=P[C_1]P[C_2]\\
P[C_1]=\frac{\binom{12}{12}\binom{15}{3}\times 3}{3^{27}}\\
P[C_2]=\frac{\binom{8}{4}\binom{4}{4}\times 2}{3^{27}}\\
\Longrightarrow =\frac{910}{7625597484987}\cdot \frac{70}{2541865828329}$$
My attempt is correct?
| $(A)$ You distributed all of the balls in the denominator but only distributed the white balls in the numerator. $|A|$ should be
$$
\binom{15}{5}\binom{10}{5}\binom{5}{5}3^{12}
$$
$(B)$ There is a similar error.
$(C)$ If you take away $3$ white balls, $12$ remain not $8$.
| {
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What is the integral value of $\frac{\tan 20^\circ+\tan40^\circ+\tan80^\circ-\tan60^\circ}{\sin40^\circ}$? I have tried possibly all approaches.
I first expressed $80$ as $60+20$ and $40$ as $60-20$ and then used trig identities.I later used conditional identities expressing $\tan 20^\circ+\tan40^\circ+\tan120^\circ$ as $\tan 20^\circ \tan40^\circ \tan120^\circ$. But I really can't get to the end of it .
Please help.
| $$\tan20^\circ-\tan60^\circ=-\dfrac{\sin(60-20)^\circ}{\cos20^\circ\cdot\cos60^\circ}=-\dfrac{2\sin40^\circ}{\cos20^\circ}$$
$$\tan40^\circ+\tan80^\circ=\dfrac{\sin(40+80)^\circ}{\cos40^\circ\cos80^\circ}$$
Adding $(1),(2)$
$$\dfrac{\sin120^\circ}{\cos40^\circ\cos80^\circ}-\dfrac{2\sin40^\circ}{\cos20^\circ} =\dfrac{\sin120^\circ\cos20^\circ-2\sin40^\circ\cos40^\circ\cos80^\circ}
{\cos20^\circ\cos40^\circ\cos80^\circ}$$
Now $S=\sin120^\circ\cos20^\circ-2\sin40^\circ\cos40^\circ\cos80^\circ$
$2S=\sin(120+20)^\circ+\sin(120-20)^\circ-2\sin80^\circ\cos80^\circ$
$=\sin(180-40)^\circ+\sin100^\circ-\sin160^\circ$
$=\sin40^\circ+\sin80^\circ-\sin20^\circ$
$=\sin40^\circ+2\sin30^\circ\cos50^\circ$
$=2\sin40^\circ$
Formulas used :
*
*$\sin(180^\circ-A)=\sin A$
*Prosthaphaeresis Formula $:\sin C-\sin D$
*$\sin2y=2\sin y\cos y$
*$2\sin A\cos B=\sin(A+B)+\sin(A-B)$
Now use Upon multiplying $\cos(20^\circ)\cos(40^\circ)\cos(80^\circ)$ by the sine of a certain angle, it gets reduced. What is that angle? to find the answer to be $$\dfrac1{\dfrac18}=?$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that for each $n \geq 2$, $\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$ Need to show that for each $n \in \mathbb{N}$, with $n \geq 2$,
$$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{n^2}\right) = \frac{n + 1}{2n}$$
How to start the proof by induction? Is there any way to show this?
| Base case:
For $n=2$,
$$LHS=(1-\frac{1}{4})=\frac{3}{4}$$
$$RHS=\frac{(2+1)}{2\cdot 2}=\frac{3}{4}$$
Hence for $n=2$, the equality holds.
Induction hypothesis:
Let it hold for some $n=k$.
Then,
$$\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{k^2}\right) = \frac{k + 1}{2k}$$
For $n=k+1$,
$$LHS=\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{k^2}\right)\left(1 - \frac{1}{(k+1)^2}\right)$$
Inductive step:
Using the induction hypothesis, for $n=k+1$,
$$LHS=\left(1 - \frac{1}{4}\right)\left(1 - \frac{1}{9}\right) \cdots \left(1 - \frac{1}{k^2}\right)\left(1 - \frac{1}{(k+1)^2}\right)=\frac{k + 1}{2k}\left(1 - \frac{1}{(k+1)^2}\right)$$
$$=\frac{(k+1)^2-1}{2k(k+1)}=\frac{(k^2+2k)}{2k(k+1)}=\frac{k+2}{2(k+1)}=\frac{(k+1)+1}{2(k+1)}=RHS$$
Thus, whenever the equality holds for $n=k\ge2$, it also hold for $n=k+1$.
Using the principle of mathematical induction, it holds for all $n\ge2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$\lim_{n \to \infty}(\frac{a_n}{\sqrt{a_n^2+1}})=\frac{1}{2}$ - show that $a_n$ is convergent sequence Problem:
Show that $a_n$ is convergent sequence and find a limit of $a_n$.
$$\lim_{n \to \infty}(\frac{a_n}{\sqrt{a_n^2+1}})=\frac{1}{2}$$
I tried to look at this as normal limit problem so I wrote this:
$$\lim_{n \to \infty}(\frac{a_n}{\sqrt{a_n^2+1}})=\lim_{n \to \infty}(\frac{1}{\sqrt{1+1}})=\frac{1}{2}$$
But I didn't get anything which can help me to solve a problem.
| You can invert the function $y = {x \over \sqrt{x^2 + 1}}$ as follows.
$$y^2 = {x^2 \over x^2 + 1} = 1 - {1 \over x^2 + 1}$$
$$1 - y^2 = {1 \over x^2 + 1}$$
$${1 \over 1 - y^2} = x^2 + 1$$
$${1 \over 1 - y^2} - 1 = x^2$$
So we have
$$x^2 = {y^2 \over 1 - y^2}$$
Seeing that $x$ and $y$ must have the same sign, we have
$$x = {y \over \sqrt{1 - y^2}}$$
Hence if for your sequence $x_n$ you write $y_n = {x_n \over \sqrt{1 + x^2}}$, then you have
$$x_n = {y_n \over \sqrt{1 - y_n^2}}$$
Since $\lim_{n \rightarrow \infty} y_n = {1 \over 2}$, by the continuity of
${y \over \sqrt{1 - y^2}}$ you have
$$\lim_{n \rightarrow \infty} x_n = {{1 \over 2} \over \sqrt{1 - {1 \over 4}}}$$
$$= {1 \over \sqrt{3}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}$ I am trying to re-learn some basic math and I realize I have forgotten most of it.
Evaluate $$\sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1}$$
Call the terms $S_n$ and the total sum $S$.
$$S_n < \frac{1}{n^3} \Rightarrow \sum_{n=1}^{\infty} \frac{n}{n^4+n^2+1} = S < \infty$$
$$S_n = \frac{n}{n^4+n^2+1} = \frac{n}{(n^2+1)^2-1}$$
It has been more than a few years since I did these things.
I would like a hint about what method I should try to look for?
Thanks.
| Notice, use partial fractions as follows $$\sum_{n=1}^{\infty}\frac{n}{n^4+n^2+1}$$$$=\sum_{n=1}^{\infty}\frac{n}{(n^2-n+1)(n^2+n+1)}$$
$$=\frac 12\sum_{n=1}^{\infty}\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right)$$
$$=\frac 12\lim_{n\to \infty}\left(\left(\frac{1}{1}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)+\ldots +\left(\frac{1}{n^2-n+1}-\frac{1}{n^2+n+1}\right)\right)$$
$$=\frac 12\lim_{n\to \infty}\left(1-\frac{1}{n^2+n+1}\right)$$
$$=\frac 12\left(1-0\right)=\color{red}{\frac 12}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1645776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solving Trigonometric Equation.
Solve for $\theta$ $[0°<\theta<180°]$
$$\sin2\theta + \sin4\theta=\cos\theta + \cos3\theta.$$
My solution is here:
$$\sin2\theta + \sin4\theta=\cos\theta + \cos3\theta.$$
After using the transformation formula, I got
$$\sin3\theta\cos\theta=\cos3\theta\cos\theta.$$
I could not proceed from here.
| \begin{array}{rcl}
\sin 2\theta+\sin 4\theta &=& \cos \theta+\cos 3\theta \\
2\sin 3\theta \cos \theta &=& 2\cos 2\theta \cos \theta \\
\cos \theta \, (\sin 3\theta-\cos 2\theta) &=& 0 \\
\cos \theta \, (3\sin \theta-4\sin^{3} \theta+2\sin^{2} \theta-1)
&=& 0 \\
\cos \theta \, (1-\sin \theta)(4\sin^{2} \theta+2\sin \theta-1) &=& 0 \\
\end{array}
$\therefore \, \cos \theta=0$, $\, \sin \theta=1 \,$, or
$\displaystyle \, \sin \theta=\frac{-1\pm \sqrt{5}}{4}$
$\because \, 0 < \theta < 180^{\circ}$, reject $\, \sin \theta \leq 0$
$\therefore \, \theta = 18^{\circ}$, $90^{\circ}$ (twice) or $\, 162^{\circ}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Intergate $\int \frac{x}{(x^2-3x+17)^2}\ dx$
$$\int \frac{x}{(x^2-3x+17)^2}\ dx$$
My attempt:
$$\int \frac{x}{(x^2-3x+17)^2}\ dx=\int \frac{x}{\left((x-\frac{3}{2})^2+\frac{59}{4}\right)^2}\ dx$$
let $u=x-\frac{3}{2}$
$du=dx$
$$\int \frac{u+\frac{3}{2}}{\left((u)^2+\frac{59}{4}\right)^2}\ du$$
How can I continue from here?
| Separate the integral into two integrals:
$\displaystyle\int\frac{u}{u^2 + \left(\sqrt{\frac{59}{4}}\right)^{2}} + \displaystyle\int\frac{\frac{3}{2}}{u^2 + \left(\sqrt{\frac{59}{4}}\right)^{2}}$
Use another $u$-substitution on the first integral and then use trig substitution on the second one.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve without involving hyperbolic function. How to solve this integral without involving hyperbolic functions?
$$\int \frac{1}{4-5\sin^2 x}dx$$
The answer is $\frac{1}{4}(\ln (\sin x+2 \cos x)-\ln(2\cos x-\sin x))+c$
| \begin{align}I&= \int \frac{\sec^2 x}{4\sec^2 x -5\tan^2 x}\, \mathrm dx \\&= \int\frac{\mathrm du}{4 - u^2} \,\,\,\;\;\; [\textrm{substituting} \;\tan x = u\; \textrm{and using the differential}\; \mathrm du = \sec^2 x\,\mathrm dx \,.]\\& = \frac{1}{4} \ln \left|\frac{2+ u}{2-u}\right| +\rm C\,\,\,\;\;\; \left[\textrm{using the formula} \; \int \frac{1}{a^2 -x^2}\,\mathrm dx= \frac{1}{2a} \ln \left|\frac{a-x}{a+x}\right| \right]\;.\end{align}
Now replace the value of $u$ and the desired result will come out.
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding $a^5 + b^5 + c^5$
Suppose we have numbers $a,b,c$ which satisfy the equations
$$a+b+c=3,$$
$$a^2+b^2+c^2=5,$$
$$a^3+b^3+c^3=7.$$
How can I find $a^5 + b^5 + c^5$?
I assumed we are working in $\Bbb{C}[a,b,c]$. I found a reduced Gröbner basis $G$:
$$G = \langle a+b+c-3,b^2+bc+c^2-3b-3c+2,c^3-3c^2+2c+\frac{2}{3} \rangle$$
I solved the last equation for $c$ and got 3 complex values. When I plug into the 2nd equation $(b^2+bc+c^2-3b-3c+2)$ I get a lot of roots for $b$, and it is laborious to plug in all these values.
Is there a shortcut or trick to doing this? The hint in the book says to use remainders. I computed the remainder of $f = a^5 + b^5 + c^5$ reduced by $G$:
$$\overline{f}^G = \frac{29}{3}$$
How can this remainder be of use to me?
Thanks. (Note: I am using Macaualay2)
| Given $a+b+c=3$ and $a^2+b^2+c^2 =5$ and $a^3+b^3+c^3=7$
Using $$ab+bc+ca = \frac{1}{2}\left[(a+b+c)^2-(a^2+b^2+c^2)\right] = 2$$
and $$a^3+b^3+c^3-3abc=(a+b+c)\left[a^2+b^2+c^2-ab-bc-ca\right]=9$$
So $$7-3abc=9\Rightarrow abc=-\frac{2}{3}$$
Now Let $(t-a)\;,(t-b)\;,(t-c)$ be the root of cubic equation in terms of $t\;,$ Then
$$(t-a)(t-b)(t-c) =0\Rightarrow t^3-(a+b+c)t^2+(ab+bc+ca)t-abc=0$$
So $$t^3-3t^2+2t+\frac{2}{3}=0\Rightarrow t^4-3t^3+2t^2+\frac{2}{3}t=0......(1)$$
So $$\sum a^4-3\sum a^3+2\sum a^2+\frac{2}{3}\sum a=0$$, Where $\sum a^{n} = a^n+b^n+c^n\;,$ for $n=1,2,3,4,5$
So $$\sum a^4-3(7)+2\cdot 5+\frac{2}{3}\cdot 3=0\Rightarrow \sum a^4=9$$
Now $$t^5-3t^4+2t^3+\frac{2}{3}t^2=0$$ from equation $(1)$
So $$\sum a^5-3\sum a^4+2\sum a^3+\frac{2}{3}\sum a^2=0$$
So $$\sum a^5-3(9)+2(7)+\frac{2}{3}\cdot 5=0\Rightarrow \sum a^5 = \frac{29}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
If $a$ is a real root of $x^5 − x^3 + x − 2 = 0$, show that $\lfloor a^6 \rfloor = 3$.
Obviously since this is a 5th degree polynomial, solving it is not going to be possible (or may be hard). However I think that factoring it to get $x^5 − x^3 + x − 2 = (x^2-x+1)(x^3+x^2-x-2)$ will help. We know both roots of the quadratic are complex, so we need only focus on the cubic $x^3+x^2-x-2$. How can we use this to show that the real root $a$ of it has $\lfloor a^6 \rfloor = 3$?
| Let $f(x)=x^5-x^3+x-2$.
*
*First step. Differenciate the polynomial to obtain:
$$f'(x)=5x^4-3x^2+1$$
which has no zeros. Hence, $f$ has exactly one root.
*Second step. We have that $f(1)=-3<0$ and $f(2)=24>0$. Then the root lies in the interval $(1,2)$.
*Third step. Let $a$ be the root of $f$. We have
$$a^6=a\cdot a^5=a(a^3-a+2)$$
Now define $g(x)=x(x^3-x+2)$. Also differentating, we can see easily that $g$ is increasing in $[1,2]$, so now you have to try more values on $f$, to get finer bounds for the root. If we have that $l<a<u$ and $\lfloor g(l)\rfloor=\lfloor g(u)\rfloor$, then $\lfloor a^6 \rfloor$ is also the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652437",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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