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Find The Equation This is the question:
Find the equations of the tangent lines to the curve $y = x − \frac 1x + 1$ that are parallel to the line $x − 2y = 3$.
There are two answers: 1) smaller y-intercept; 2) larger y-intercept
The work:
The slope of the line is (1/2).
$y' = ((x + 1) - (x - 1))/(x + 1)^2 = 2/(x + 1)^2 = 1/2$
=> $(x + 1)^2 = 4$
=> $(x + 1) = 2$ and $ x + 1 = -2$
=> $x = 1$ and $x = -3$
At $x = 1$, $y = (x - 1)/(x + 1) = 0$
The equation of the tangent is $y/(x - 1) = (1/2)$
=> $2y = x - 1$
=> $x - 2y - 1 = 0$
At $x = -3$,$ y = (x - 1)/(x + 1) = -4/-2 = 2$
The equation of the tangent is $(y - 2)/(x + 3) = (1/2)$
=> $2y - 4 = x + 3$
=>$ x - 2y + 7 = 0$
| Two lines are parallel if they have the same slope, and the slope of a tangent line to a curve is the value of the derivative of the curve's function at that point. Therefore, follow this strategy.
1) Find the slope of the given line.
2) Find the derivative of the function defining the curve.
3) Set those expressions equal to each other and solve. This gives you the $x$-coordinate(s) of the intersection point(s) of the curve and the desired tangent line(s).
4) Find the $y$-coordinate for each $x$-coordinate you got in step 3.
5) Use the slope (from step 1) and the point(s) (from step 3) to find the equation(s) of the tangent line(s).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$, where numbers $a$ and $b$ are rational If $a$ and $b$ are rational numbers such that $\sqrt{4 -2 \sqrt{3}} = a + b\sqrt{3}$
Then what is the value of $a$? The answer is $-1$.
$$\sqrt{4 - 2\sqrt{3}} = a + b\sqrt{3}$$
$$4 - 2\sqrt{3} = 2^2 - 2\sqrt{3}$$ Let $u =2$ hence,
$$\sqrt{u^2 - \sqrt{3}u} = a + b\sqrt{3}$$
$$u^2 - \sqrt{3}u = u(u - \sqrt{3})$$
$$a + b\sqrt{3} = \sqrt{u}\sqrt{u - \sqrt{3}}$$
What should I do?
| We can start with $\sqrt{4-2\sqrt{3}} = a + b\sqrt{3}$ and square both sides. This gives us
$$ 4-2\sqrt{3} = a^2 + 2ab\sqrt{3} + 3b^2 $$
Because $a$ and $b$ are said to be rational, we know that the only term which will be a multiple of $\sqrt{3}$ is $2ab\sqrt{3}$ From this, we can split our system into two equations:
$$ 2ab = -2 $$
and
$$ a^2 + 3b^2 = 4$$
Let's use the first equation to sub $b = -1/a$. This gives us
$$ a^2 + 3/a^2 = 4 \Rightarrow a^4 -4a^2 + 3 = 0 $$
The quadratic formula then gives us $a = 1$ and $a = -1$, which would make $b = -1$ and $b = 1$, respectively. Because $\sqrt{3} > 1$ and the square root is always defined as positive, $\sqrt{4-2\sqrt{3}} = 1 - \sqrt{3}$ is an extraneous solution. Therefor, the only solution is $a = -1$ and $b = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Question in relation to completing the square In description of "completing the square" at http://www.purplemath.com/modules/sqrquad.htm the following is given :
I'm having difficulty understanding the third part of the transformation.
Where is
$ -\frac{1}{4}$ derived from $-\frac{1}{2}$ ?
Why is $ -\frac {1}{4}$ squared to obtain $ \frac {1}{16}$ ?
| In general, to complete the square of something like $x^2 \pm Bx = \pm C$, you take half the coefficient of $x$, which is $B$, and then square it.
For the third step, they first take half of the $x$ coefficient, which is $-\frac{1}{2}$. Half of $-\frac{1}{2}$ is $-\frac{1}{4}$ since
$$ \frac{-\frac{1}{2}}{2} = \frac{-\frac{1}{2}}{\frac{2}{1}}= -\frac{1}{2} \cdot \frac{1}{2} = -\frac{1}{4}.$$
Then, they square $-\frac{1}{4}$ which is $\frac{1}{16}$ since
$$ \left(-\frac{1}{4}\right)^2 = -\frac{1}{4} \cdot -\frac{1}{4} = \frac{1}{16}.$$
The reason why they add $\left (\frac{B}{2} \right)^2$ to both sides of the equation is to make a squared binomial:
$$x^2 + Bx + \left (\frac{B}{2} \right)^2 = \left(x + \frac{B}{2} \right)^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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What is wrong with this approach evaluating $\int \sec \theta \ d \theta$? When integrating a function like $\sin^m \theta \cdot \cos^n \theta$ where where $m,n$ are nonnegative integers and $n$ is odd, a common approach is to peel off one power of $\cos \theta$ and then rewrite the resulting even power of $\cos \theta$ in terms of $\sin \theta$ using $\cos^2 \theta = 1 - \sin^2 \theta$. What remains is a polynomial in $\sin \theta$, multiplied by $\cos \theta$, so the substitution $u = \sin \theta$ can be used. Since $\sec \theta =(\sin \theta)^0 (\cos \theta)^{-1}$ and $-1$ is odd, it seems like a good idea to try out the same approach here.
\begin{align*}
\int \sec \theta \ d \theta
&= \int \frac{\cos \theta}{\cos^2 \theta} \ d \theta && \text{Splitting off a power of $\cos \theta$}\\
&=\int \frac{ \cos \theta}{1-\sin^2 \theta} \ d \theta && \text{Eliminating the even power of $\cos \theta$} \\
&=\int \frac{1}{1-u^2}\ du && \text{Substituting $u=\sin \theta$}\\
&=\frac{1}{2} \int \left( \frac{1}{1+u} + \frac{1}{1-u} \right) \ du&& \text{Partial fractions}\\
&=\frac{1}{2} \left( \log |1+u| + \log |1-u| \right) && \\
&=\frac{1}{2} \log |1-u^2| && \\
&=\frac{1}{2} \log| \cos^2 \theta| && \text{Restoring $u=\sin \theta$} \\
&=\frac{1}{2} \log (\cos^2 \theta) && \text{Eliminating redundant absolute signs}\\
&= \log \sqrt{ \cos^2 \theta}&& \\
&= \log|\cos \theta|. \\
\end{align*}
I didn't bother adding the customary constant because the domain of $\sec \theta$ is disconnected and so you would actually need to add a piecewise constant function in order to specify the general antiderivative.
My questions are:
Question 1: Since $\frac{d}{d \theta} \log|\cos \theta| = - \tan \theta$, where does this calculation go wrong?
And also:
Question 2: Can this approach be salvaged?
| In your fifth line,
$$\log |1+u| + \log |1-u|$$
should be
$$\log |1+u| - \log |1-u|\ .$$
This then gives
$$\eqalign{
\frac12\log\Bigl|\frac{1+u}{1-u}\Bigr|
&=\frac12\log\Bigl(\frac{1+\sin\theta}{1-\sin\theta}\Bigr)\cr
&=\frac12\log\Bigl(\frac{1+\sin\theta}{1-\sin\theta}
\frac{1+\sin\theta}{1+\sin\theta}\Bigr)\cr
&=\log\frac{1+\sin\theta}{\cos\theta}\cr
&=\log(\sec\theta+\tan\theta)\ .\cr}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Let $a$, $b$ and $c$ be the three sides of a triangle. Show that $\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3$.
Let $a$, $b$ and $c$ be the three sides of a triangle.
Show that $$\frac{a}{b+c-a}+\frac{b}{c+a-b} + \frac{c}{a+b-c}\geqslant3\,.$$
A full expanding results in:
$$\sum_{cyc}a(a+b-c)(a+c-b)\geq3\prod_{cyc}(a+b-c),$$ or
$$\sum_{cyc}(a^3-ab^2-ac^2+2abc)\geq\sum_{cyc}(-3a^3+3a^2b+3a^2c-2abc),$$ but it becomes very ugly.
| Since $a,b,c$ are sides of a triangle, we can set $a = x+y$, $b = x+z$, $c = y+z$.
Plugging that in gives
\begin{align}
\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c} &= \frac{x+y}{2z} + \frac{x+z}{2y} + \frac{y+z}{2x} \\
&= \frac{xy(x+y) + xz(x+z) + yz(y+z)}{2xyz}\\
&= \frac{x^2y+ xy^2 + x^2z + xz^2 + y^2z + yz^2}{2xyz}\\
&\ge \frac{3}{xyz}\sqrt[6]{x^6y^6z^6}\\
&= 3
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1155955",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 6
} |
Integral $\int \limits _0 ^\pi |\sin x + \cos x|\; dx$ $$\int \limits _0 ^\pi |\sin x + \cos x|\; dx$$
If I divide integral in two parts $\int\limits_0^{\frac{\pi}{2}}{(\sin x + \cos x)\,dx}$ and $\int\limits_{\frac{\pi}{2}}^\pi{(\sin x - \cos x)\,dx}$...I am getting $4$...Am I right?
| Though answers were already given, I'd like to suggest a different approach:
We see that the absolute value of $\cos(x) + \sin(x)$ is required.
Also, we are dealing with sinusoidal functions, which can be easily dealt with the square function, so:
$$
\begin{align}
\int\limits_0^\pi \lvert \cos(x) + \sin(x) \rvert &= \\
&= \int\limits_0^\pi \sqrt{\left(\cos(x) + \sin(x) \right)^2} \\
&= \int\limits_0^\pi \sqrt{\cos^2(x) + 2 \sin(x) \cos(x) + \sin^2(x) } \\
&= \int\limits_0^\pi \sqrt{1 + 2 \sin(x) \cos(x) } \\
&= \begin{bmatrix} \text{Sinusoidal identity:} \\ 2 \sin(x) \cos(x) = \sin(2x) \end{bmatrix} \\
&= \int\limits_0^\pi \sqrt{1 + \sin(2x) } \\
\end{align}
$$
Now we should notice the function that we're dealing with (omitting the root, insignificant):
Picture was taken from WA for 1 + sin(2x).
So we're performing a sum operation over the rooted sine-wave, treating all values as positive.
This is exactly like performing the same sum operation over the cosine function, with different limits - from $-\frac{\pi}{2}$ to $\frac{\pi}{2}$ (due to phase shift symmetry):
$$
\begin{align}
\int\limits_0^\pi \sqrt{1 + \sin(2x) } &= \\
&= \int\limits_{-\frac{\pi}{2}}^\frac{\pi}{2} \sqrt{1 + \cos(2x) } \\
&= \begin{bmatrix} \text{Sinusoidal identity:} \\ cos(2x) = 2 \cos^2(x) - 1 \end{bmatrix} \\
&= \int\limits_{-\frac{\pi}{2}}^\frac{\pi}{2} \sqrt{1 + 2 \cos^2(x) - 1 } \\
&= \sqrt{2} \, \int\limits_{-\frac{\pi}{2}}^\frac{\pi}{2} \cos(x) \\
&= 2\sqrt{2}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Sum $\cot^2\frac{\pi}{2m+1}+\cot^2\frac{2\pi}{2m+1}+\cot^2\frac{3\pi}{2m+1}+\ldots+\cot^2\frac{m\pi}{2m+1}?$ How to find
$$\cot^2\frac{\pi}{2m+1}+\cot^2\frac{2\pi}{2m+1}+\cot^2\frac{3\pi}{2m+1}+\ldots+\cot^2\frac{m\pi}{2m+1}?$$
Of course, the number $m$ is assumed to be a positive integer.
| Here is a rather elementary and simple way to sum this series:
We know from De Moivre's Formula that
$$(\cos\theta+\iota\sin\theta)^n=\cos n\theta+\iota\sin n\theta$$
Expanding the LHS using Binomial Theorem,
$$\binom{n}{0}\cos^n\theta+\binom{n}{1}\cos^{n-1}\theta(\iota\sin\theta)+\cdots+\binom{n}{n}\iota^n\sin^n\theta=\cos n\theta+\iota\sin n\theta$$
Equating imaginary parts,
$$\binom{n}{1}\cos^{n-1}\theta(\sin\theta)-\binom{n}{3}\cos^{n-3}\theta(\sin\theta)^3+\binom{n}{5}\cos^{n-5}\theta(\sin\theta)^5-\cdots=\sin n\theta$$
Now let $n=2m+1$,then:
$$\binom{2m+1}{1}\cos^{2m}\theta(\sin\theta)-\binom{2m+1}{3}\cos^{2m-2}\theta(\sin\theta)^3+\cdots=\sin (2m+1)\theta$$
Dividing by $\sin^{2m+1}\theta$, we have a result:
$$\binom{2m+1}{1}(\cot^2\theta)^m-\binom{2m+1}{3}(\cot^2\theta)^{m-1}+\cdots={\sin (2m+1)\theta \over \sin^{2m+1}\theta}\tag{i}$$
Now consider the following equation of the $m^{th}$ degree:
$$\binom{2m+1}{1}x^m-\binom{2m+1}{3}x^{m-1}+\binom{2m+1}{5}x^{m-2}-\cdots=0$$
Making use of the preceding result (i), we can say that $\cot^2{\pi \over 2m+1}$,$\cot^2{2\pi \over 2m+1},\ldots,\cot^2{m\pi \over 2m+1}$ are the $m$ roots of this equation.
Using Vieta's formulas, the sum of roots of this equation (which is the required sum) must be this:
$${2m+1 \choose 3}\over{2m+1 \choose 1}$$
Which on simplification reduces to this:
$${2m^2-m}\over{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1157958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
How to calculate derivative of $f(x) = \frac{1}{1-2\cos^2x}$? $$f(x) = \frac{1}{1-2\cos^2x}$$
The result of $f'(x)$ should be equals
$$f'(x) = \frac{-4\cos x\sin x}{(1-2\cos^2x)^2}$$
I'm trying to do it in this way but my result is wrong.
$$f'(x) = \frac {1'(1-2\cos x)-1(1-2\cos^2x)'}{(1-2\cos^2x)^2} =
\frac {1-2\cos^2x-(1-(2\cos^2x)')}{(1-2\cos^2x)^2} = $$
$$=\frac {-2\cos^2x + 2(2\cos x(\cos x)')}{(1-2\cos^2x)^2} =
\frac {-2\cos^2x+2(-2\sin x\cos x)}{(1-2\cos^2x)^2} = $$
$$\frac {-2\cos^2x-4\sin x\cos x}{(1-2\cos^2x)^2}$$
| Avoid the quotient formula for functions of the form $1/g(x)$. Rather consider
$$
f(x)=\frac{1}{1-2\cos^2x}=\frac{1}{g(x)}
$$
where $g(x)=1-2\cos^2x$. Since $\Bigl(\dfrac{1}{x}\Bigr)'=-\dfrac{1}{x^2}$, you have, by the chain rule,
$$
f'(x)=-\frac{1}{g(x)^2}g'(x)
$$
and
$$
g'(x)=4\sin x\cos x,
$$
so
$$
f'(x)=-\frac{4\sin x\cos x}{(1-2\cos^2x)^2}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1158652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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The Sum of $\sum\limits_{n=1}^\infty \left(\sin\frac{10}{n} -\sin\frac{10}{n+1}\right)$ $$\sum\limits_{n=1}^\infty \left(\sin\frac{10}{n} - \sin\frac{10}{n+1}\right)$$
I see that as $n$ approaches $\infty$ that it'll be 0 thus convergent. However, I'm unclear of the manipulation that's implemented to get an actual result ($\sin(10) = -0.5442$). Can these $\sin$ elements be put together in such a way to be able to evaluate $\sin$ at some real number or is there a need for something else to be done?
| Just telescope it: look at the few initial terms: $$\sin 10 - \sin 5 + \sin 5 - \sin(10/3) + \sin(10/3)- \sin(10/4)+\cdots.$$ Meaning: $$\sum_{k=1}^n \left(\sin \frac{10}{k}-\sin\frac{10}{k+1}\right) = \sin 10 - \sin\frac{10}{n+1}.$$This way: $$\begin{align}\sum_{n=1}^{+\infty}\left(\sin\frac{10}{n} - \sin\frac{10}{n+1}\right) &= \lim_{n \to +\infty} \sum_{k=1}^n \left(\sin\frac{10}{k} - \sin\frac{10}{k+1}\right)\\ &= \lim_{n \to +\infty}\left(\sin 10 - \sin\frac{10}{n+1}\right) = \sin 10. \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1159411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all complex numbers satisfying the equation Find all complex numbers $z$ satisfying the equation $$\left|z+\frac{1}{z}\right|=2.$$
| Clearly, $z\ne0$
$$\implies|z^2+1|=2|z|$$
Let $z=re^{iy}$ where $r(\ge0),y$ are real
$$2r^2=(r^2\cos2y+1)^2+(r^2\sin2y)^2=r^4+2r^2\cos2y+1$$
$$(r^2)^2-4r^2\sin^2y+1=0$$
$$\implies r^2=\frac{4\sin^2y\pm\sqrt{(4\sin^2y)^2-4}}2=2\sin^2y\pm\sqrt{4\sin^4y-1}$$
As $4\sin^4y-1<4\sin^4y, r^2=2\sin^2y+\sqrt{4\sin^4y-1}$
As $r^2$ is real, we need $4\sin^4y-1\ge0\iff2\sin^2y\ge1\iff\cos2y=1-2\sin^2y\le0$
$\cos2y\le0\iff2m\pi+\dfrac\pi2\le2y\le2m\pi+\dfrac{3\pi}2$ where $m$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1159933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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contour integral piecewise Evaluate $$\int \limits_\gamma \frac1{z-1}dz$$
along the path:
$$\gamma(t) = \begin{cases}(1+i)t, & 0\leq t\leq 1 \\\\ t+i(2-t), & 1\leq t \leq2\end{cases}$$
I know how to do simple questions of these but I am unsure about this one. This is what I tried. Let $f(z)$ represent the equation given in the integral.
$$\gamma'(t) = \begin{cases}t+i, & 0\leq t\leq 1 \\\\ 1-i, & 1\leq t \leq2\end{cases}$$
and
$$f(\gamma(t)) = \begin{cases}\frac1{(1+i)t-1}, & 0\leq t\leq 1 \\\\ \frac1{t+i(2-t)-1}, & 1\leq t \leq2\end{cases}$$
Then what do you do?
| Actually $\gamma'(t) = 1 + i$ for $0 \le t < 1$ and $\gamma'(t) = 1 - i$ for $1 < t \le 2$. So your contour integral becomes
$$\int_0^1 f(\gamma(t))\gamma'(t)\, dt + \int_1^2 f(\gamma(t))\gamma'(t)\, dt = \int_0^1 \frac{1 + i}{(1 + i)t - 1}\, dt + \int_1^2 \frac{1 - i}{t - 1 + (2-t)i}\, dt.$$
To evaluate the last two integrals, express
$$\frac{1 + i}{(1 + i)t - 1} = \frac{(1 + i)[(1 - i)t - 1]}{|(1 + i)t - 1|^2} = \frac{2t - 1 - i}{(t - 1)^2 + t^2} = \frac{2t - 1 - i}{2t^2 - 2t + 1}$$
and
$$\frac{1 - i}{(t - 1) + (2 - t)i} = \frac{(1 - i)[(t - 1) - (2 - t)i}{|(t - 1) + (2 - t)i|^2} = \frac{2t - 3 - i}{(t - 1)^2 + (2-t)^2} = \frac{2t - 3 - i}{2t^2 - 6t + 5}.$$
We have
\begin{align}\int_0^1 \frac{2t - 1 - i}{2t^2 - 2t + 1} \, dt &= \frac{1}{2}\int_0^1 \frac{4t - 2}{2t^2 - 2t + 1}\, dt - \int_0^1 \frac{i}{2t^2 - 2t + 1}\, dt\\
&= \frac{1}{2}\log|2t^2 - 2t + 1|\bigg|_{t = 0}^1 - \frac{i}{2}\int_0^1 \frac{dt}{(t - \frac{1}{2})^2 + \frac{1}{4}}\\
&= -\frac{i}{2}\int_0^1 \frac{dt}{(t - \frac{1}{2})^2 + \frac{1}{4}}\\
&= -\frac{i}{2}\cdot 2\arctan\left[2\left(t - \frac{1}{2}\right)\right]\bigg|_{t = 0}^1\\
&= -\frac{\pi}{2}i.
\end{align}
A similar argument gives
$$\int_1^2 \frac{2t - 3 - i}{2t^2 - 6t + 5}\, dt = -\frac{\pi}{2}i.$$
Therefore
$$\int_\gamma \frac{dz}{z-1} = -\frac{\pi}{2}i - \frac{\pi}{2}i = -\pi i.$$
| {
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"url": "https://math.stackexchange.com/questions/1162126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability urn-envelope question Urn 1 contains 14 envelopes--6 with \$1 bills and 8 with \$5 bills. Urn 2 has 8 envelopes--3 with \$1 bills and 5 with \$5 bills. 3 bills are randomly transferred from urn 1 to urn 2. What is the probability of drawing a\$1 bill from urn 2?
Attempted answer:
Let $O_x$ be the event where $x$ \$1 bills are transferred from urn 1 to urn 2.
Let $A$ be the event that a \$1 bill is drawn from urn 2.
We either transfered 0 \$1 bills, 1 \$1 bills, 2 \$1 bills or 3 \$1 bills.
By law of total probability we have:
$$P(A) = P(O_3)P(A|O_3) + P(O_2)P(A|O_2) + P(O_1)P(A|O_1) + P(O_0)P(A|O_0)$$
$$P(A) = \left( \frac{6}{14}*\frac{5}{13}*\frac{4}{12} \right) \left( \frac{6}{11} \right) + \left( \frac{6}{14}*\frac{5}{13}*\frac{8}{12} \right) \left( \frac{5}{11} \right) + \left( \frac{6}{14}*\frac{8}{13}*\frac{7}{12} \right) \left( \frac{4}{11} \right) + \left( \frac{8}{14}*\frac{7}{13}*\frac{6}{12} \right) \left( \frac{3}{11} \right)$$
$$P(A) = \frac{178}{1001}$$
The answer is $\frac{30}{77}$
| Hint:
(1,1,1), (1,1,5), (5,5,1), (5,5,5) are the possible bills transferred
Similar to your approach:
P(USD1/Urn2) $$=\frac{{6\choose3}}{{14\choose3}}.\frac{6}{11}+\frac{{6\choose2}.{8\choose1}}{{14\choose3}}.\frac{5}{11}+\frac{{6\choose1}.{8\choose2}}{{14\choose3}}.\frac{4}{11}+\frac{{8\choose3}}{{14\choose3}}.\frac{3}{11}$$
$$ = \frac{720+3600+4032+1008}{24024} $$
$$= \frac{9360}{24024} $$
$$= \frac{30}{77}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1163245",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Assume that the integer $r$ is a primitive root of the prime $p$, where $p \equiv 1 \pmod{8}$. Assume that the integer $r$ is a primitive root of the prime $p$, where $p \equiv 1 \pmod{8}$.
Show that the solutions of the quadratic congruence $x^2 \equiv 2\pmod{p}$ are given by
$x \equiv \pm (r^{7(p-1)/8}+r^{(p-1)/8}) \pmod{p}$
Here is my attempt so far
$$r \text{ is a primitive root } \implies \exists k: 2\equiv r^k \pmod{p}$$
I am not sure why the question is assuming the given quadratic congruence always has a solution. But if it has a solution, from Euler's criterian we have $$2^{(p-1)/2}\equiv 1\pmod{p}$$
I feel stuck at this point.. Any help is appreciated thanks!
| Let $b=r^{(p-1)/8}$. Then $a=r^{7(p-1)/8}+r^{(p-1)/8}=b^7+b$ and so $a^2=b^{14}+2b^8+b^2$.
We want to prove that $a^2=2$.
Clearly, $b^8=1$. It remains to prove that $b^{14}+b^2=0$.
Now, $b^{14}+b^2=b^{6}+b^2=b^2(b^4+1)$.
But $b^4+1=0$ because $0=b^8-1=(b^4+1)(b^4-1)$, and $b^4-1\ne0$.
Thus, $x=\pm a$ are two solutions of $x^2=2$. That equation cannot have more than two solutions, since $\mathbb Z/p$ is a field. Hence, we have found them all.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1165869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Series Convergence $\sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{2n+2}{2n+4}\right)^n$ I have to show if this series converges or diverges. I tried using asymptotics, but it's not formally correct as they should work only when the arguments are extremely small. Any ideas?
$$
\sum_{n=1}^{\infty} \frac{1}{n} \left(\frac{2n+2}{2n+4}\right)^n
$$
| Recall that, as $x \to 0$, by the Taylor expansion, we have
$$
\begin{align}
e^x& =1+x+\mathcal{O}(x^2)\\
\ln (1+x)&=x-\frac {x^2}{2}+\mathcal{O}(x^3)
\end{align}
$$ giving, as $n \to \infty$,
$$
n\ln \left(1-\frac {2}{2n+4}\right)=n \left(-\frac {2}{2n+4}+\mathcal{O}(\frac {1}{n^2})\right)=-1+\mathcal{O}(\frac {1}{n})
$$ and
$$
\begin{align}
\frac{1}{n} \left(\frac{2n+2}{2n+4}\right)^n&=\frac{1}{n} \left(\frac{2n+4-2}{2n+4}\right)^n\\\\
&=\frac{1}{n} \left(1-\frac {2}{2n+4}\right)^n\\\\
&=\frac{1}{n}e^{n\ln (1-\frac {2}{2n+4})}\\\\
&=\frac{1}{n}e^{-1+\mathcal{O}(\frac {1}{n})}\\\\
&\sim \frac{e^{-1}}{n}
\end{align}
$$ thus your initial series is divergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1166192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Prove that $n^3+2$ is not divisible by $9$ for any integer $n$ How to prove that $n^3+2$ is not divisible by $9$?
| Consider the following $9$ cases:
*
*$n\equiv0\pmod9 \implies n^3\equiv 0\pmod9 \implies n^3+2\equiv 2\pmod9$
*$n\equiv1\pmod9 \implies n^3\equiv 1\pmod9 \implies n^3+2\equiv 3\pmod9$
*$n\equiv2\pmod9 \implies n^3\equiv 8\pmod9 \implies n^3+2\equiv10\equiv1\pmod9$
*$n\equiv3\pmod9 \implies n^3\equiv 27\equiv0\pmod9 \implies n^3+2\equiv 2\pmod9$
*$n\equiv4\pmod9 \implies n^3\equiv 64\equiv1\pmod9 \implies n^3+2\equiv 3\pmod9$
*$n\equiv5\pmod9 \implies n^3\equiv125\equiv8\pmod9 \implies n^3+2\equiv10\equiv1\pmod9$
*$n\equiv6\pmod9 \implies n^3\equiv216\equiv0\pmod9 \implies n^3+2\equiv 2\pmod9$
*$n\equiv7\pmod9 \implies n^3\equiv343\equiv1\pmod9 \implies n^3+2\equiv 3\pmod9$
*$n\equiv8\pmod9 \implies n^3\equiv512\equiv8\pmod9 \implies n^3+2\equiv10\equiv1\pmod9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
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Proof of an Limit Using the formal definition of convergence, Prove that $\lim\limits_{n \to \infty} \frac{3n^2+5n}{4n^2 +2} = \frac{3}{4}$.
Workings:
If $n$ is large enough, $3n^2 + 5n$ behaves like $3n^2$
If $n$ is large enough $4n^2 + 2$ behaves like $4n^2$
More formally we can find $a,b$ such that $\frac{3n^2+5n}{4n^2 +2} \leq \frac{a}{b} \frac{3n^2}{4n^2}$
For $n\geq 2$ we have $3n^2 + 5n /leq 3n^2.
For $n \geq 0$ we have $4n^2 + 2 \geq \frac{1}{2}4n^2$
So for $ n \geq \max\{0,2\} = 2$ we have:
$\frac{3n^2+5n}{4n^2 +2} \leq \frac{2 \dot 3n^2}{\frac{1}{2}4n^2} = \frac{3}{4}$
To make $\frac{3}{4}$ less than $\epsilon$:
$\frac{3}{4} < \epsilon$, $\frac{3}{\epsilon} < 4$
Take $N = \frac{3}{\epsilon}$
Proof:
Suppose that $\epsilon > 0$
Let $N = \max\{2,\frac{3}{\epsilon}\}$
For any $n \geq N$, we have that $n > \frac{3}{\epsilon}$ and $n>2$, therefore
$3n^2 + 5n^2 \leq 6n^2$ and $4n^2 + 2 \geq 2n^2$
Then for any $n \geq N$ we have
$|s_n - L| = \left|\frac{3n^2 + 5n}{4n^2 + 2} - \frac{3}{4}\right|$
$ = \frac{3n^2 + 5n}{4n^2 + 2} - \frac{3}{4}$
$ = \frac{10n-3}{8n^2+4}$
Now I'm not sure on what to do. Any help will be appreciated.
| It simple
$$ \lim_{x \to \infty} \dfrac{3 + \dfrac{5}{n}}{4 + \dfrac{2}{n^2}}$$
$$ = \dfrac{3}{4}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Polynomial division challenge Let $x,y,n \in \mathbb{Z} \geq 3$, Find $A,B$ such that $$x^{n-1}+x^{n-2}y+x^{n-3}y^2+\cdots+x^2y^{n-3}+xy^{n-2}+y^{n-1}= A(x^2+xy+y^2)+B$$ What is the best method to approach this?
| As we have homogeneous polynomials, we first solve the problem for the de-homogeneised polynomials, setting $y=1$, then homogeneise the results.
So we have to find $A, B$ such that
\begin{align*}
& x^{n-1}+x^{n-2}+x^{n-3}+\cdots+x^2+x+1= A(x^2+x+1)+B \\
\iff (&x-1)(x^{n-1}+x^{n-2}+x^{n-3}+\cdots+x^2+x+1)=A(x-1)(x^2+x+1)+B(x-1)\\
\iff\vphantom{(} & x^n-1=A(x^3-1)+B(x-1)
\end{align*}
Now let $r=n\bmod 3$. It is easy to check the quotient and the remainder of the division of $x^n-1$ by $\,x^3-1\,$ are:
$$ A=x^{n-3}+x^{n-6}++\cdots+x^r, \quad B = x^r -1. $$
Hence, for the original problem with homogeneous polynomial, we'll have:
$$A=x^{n-3}+x^{n-6}y^3++\cdots+x^ry^{n-3-r},\quad B=x^ry^{n-3-r}-y^{n-3}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1170331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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In an equilateral spherical triangle, show that $\sec A=1+\sec a$ Q. In an equilateral spherical triangle, show that $\sec A=1+\sec a$
So $A$ is the vertex or the angle of the triangle and $a$ is the side of the equilateral spherical triangle.
I started off the proof by using the law of cosines:
$$\cos(a)-\cos(a)\cos(a)=\sin(a)\sin(a)\cos(A)$$
and after simplifying it a bit, I obtained:
$$\cos(a)-\cos^2(a)=\sin^2(a)\cos(A)$$
I replaced $\sin^2(a)$ with $(1-\cos^2(a))$.
Then I obtained:
$$\cos(a)-\cos^2(a)=(1-\cos^2(a))\cos(A)$$
and I realized on the left side, I can pull a $\cos(a)$. So through factoring:
$$\cos(a)(1-\cos(a))=(1-\cos^2(a))\cos(A)$$
Either I'm not seeing it but I do not how to proceed after this. If anyone can help, I'd like that. Thanks!
| lambda, mu are lattitude and longitude. $R = 1$.
$O = (0,0,0)$
$A = (\cos \lambda_a \cos \mu_a, \cos \lambda_a \sin \mu_a, \sin \lambda_a)$
$B = (\cos \lambda_b \cos \mu_b, \cos \lambda_b \sin \mu_b, \sin \lambda_b)$
$C = (\cos \lambda_c \cos \mu_c, \cos \lambda_c \sin \mu_c, \sin \lambda_c)$
The planes $AOB$ and $AOC$ have the line $OA$ in common. Let $\alpha = \angle (AOB, AOC)$.
$OB \cdot OC = \cos \beta$.
You claim that $\frac{1}{\cos \alpha} = 1 + \frac{1}{\cos \beta}$ (*)
Angle between normals implies vectorial product: $\alpha = \angle (OA \times OB, OA \times OC)$.
$\cos \alpha = (OA \times OB) \cdot (OA \times OC)$
$(*) \Rightarrow \cos \beta = \cos \alpha \cos \beta + \cos \alpha \Rightarrow \cos \beta (1 - \cos \alpha) = \cos \alpha$
$\cos \alpha = b_x c_x + b_y c_y + b_z c_z$
$A \times B = (a_y b_z - a_z b_y, - a_z b_x + a_x b_z, a_x b_y - a_y b_x)$
$A \times C = (a_y c_z - a_z c_y, - a_z c_x + a_x c_z, a_x c_y - a_y c_x)$
$\cos \beta = (a_y b_z - a_z b_y)(a_y c_z - a_z c_y) + (- a_z b_x + a_x b_z)(- a_z c_x + a_x c_z) + (a_x b_y - a_y b_x)(a_x c_y - a_y c_x)$
$\cos \beta = (a_y^2 b_z c_z - a_y a_z b_z c_y - a_y a_z b_y c_z + a_z^2 b_y c_y) + (a_z^2 b_x c_x - a_x a_z b_x c_z - a_x a_z b_z c_x + a_x^2 b_z c_z) + (a_x^2 b_y c_y - a_x a_y b_y c_x - a_x a_y b_x c_y + a_y^2 b_x c_x)$
$\cos \beta = a_y^2 (\cos \alpha - b_y c_y) + a_z^2 (\cos \alpha - b_z c_z) + a_x^2 (\cos \alpha - b_x c_x) + (- a_y a_z b_z c_y - a_y a_z b_y c_z) + (- a_x a_z b_x c_z - a_x a_z b_z c_x) + (- a_x a_y b_y c_x - a_x a_y b_x c_y)$
$\cos \beta = \cos \alpha - a_x^2 b_x c_x - a_y^2 b_y c_y - a_z^2 b_z c_z + (- \cos \lambda_a \sin \mu_a \sin \lambda_a \sin \lambda_b \cos \lambda_c \sin \mu_c - \cos \lambda_a \sin \mu_a \sin \lambda_a \cos \lambda_b \sin \mu_b \sin \lambda_c) + (- \cos \lambda_a \cos \mu_a \sin \lambda_a \cos \lambda_b \cos \mu_b \sin \lambda_c - \cos \lambda_a \cos \mu_a \sin \lambda_a \sin \lambda_b \cos \lambda_c \cos \mu_c) + (- \cos \lambda_a \cos \mu_a \cos \lambda_a \sin \mu_a \cos \lambda_b \sin \mu_b \cos \lambda_c \cos \mu_c - \cos \lambda_a \cos \mu_a \cos \lambda_a \sin \mu_a \cos \lambda_b \cos \mu_b \cos \lambda_c \sin \mu_c)$
$\cos \beta = \cos \alpha - a_x^2 b_x c_x - a_y^2 b_y c_y - a_z^2 b_z c_z + \cos \lambda_a [(- \sin \mu_a \sin \lambda_a \sin \lambda_b \cos \lambda_c \sin \mu_c - \sin \mu_a \sin \lambda_a \cos \lambda_b \sin \mu_b \sin \lambda_c) + (- \cos \mu_a \sin \lambda_a \cos \lambda_b \cos \mu_b \sin \lambda_c - \cos \mu_a \sin \lambda_a \sin \lambda_b \cos \lambda_c \cos \mu_c) + (- \cos \mu_a \cos \lambda_a \sin \mu_a \cos \lambda_b \sin \mu_b \cos \lambda_c \cos \mu_c - \cos \mu_a \cos \lambda_a \sin \mu_a \cos \lambda_b \cos \mu_b \cos \lambda_c \sin \mu_c)]$
I may have not defined right.
| {
"language": "en",
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"source": "stackexchange",
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Find two linearly independent solutions of the differential equation $(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$ I want to find two linearly independent solutions of the differential equation
$$(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$$
Previously I have seen that the following holds for the differential equation $y''+ \frac{1}{x}y'-\frac{1}{x^2}y=0, x>0$:
*
*We are looking for solutions of the differential equation of the form $x^r$. Then the function $x^r$ is a solution of the differential equation at $(0,+\infty)$ if:
$$r(r-1)x^{r-2}+ \frac{1}{x} r x^{r-1}- \frac{1}{x^2}x^r=0 \forall x >0 \Rightarrow r=1 \text{ or } r=-1$$
*
*So, the functions $y_1(x)=x, y_2= \frac{1}{x}$ are solutions of the differential equation and it also holds that they are linearly indepedent since $W(y_1, y_2) \neq 0$
For this differential equation
$$(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3}$$
I thought the following:
$(3x-1)^2 y''+(9x-3)y'-9y=0 \text{ for } x> \frac{1}{3} \Rightarrow y''+ \frac{3}{3x-1}y'-\frac{9}{(3x-1)^2}y=0$
*
*We are looking for solutions of the differential equation of the form $\left( x- \frac{1}{3}\right)^r$.
Then the function $\left( x- \frac{1}{3}\right)^r$ is a solution of the differential equation at $( \frac{1}{3},+\infty)$ if:
$$r(r-1) \left( x- \frac{1}{3}\right)^{r-2}+ \frac{1}{ \frac{3x-1}{3}} r \left( \frac{3x-1}{3}\right)^{r-1}- \frac{9}{(3x-1)^2} \left( x- \frac{1}{3}\right)=0 \Rightarrow \dots \Rightarrow r= \pm 1$$
Therefore, the functions $z_1(x)=x- \frac{1}{3}, z_2(x)=\frac{1}{x- \frac{1}{3}}$ are solutions of the differential equation at $\left( \frac{1}{3}, +\infty\right)$.
$$z_1(x) z_2'(x)-z_1'(x) z_2(x)=\frac{-2}{x- \frac{1}{3}} \neq 0$$
So, $z_1, z_2$ are linearly independent solutions of the differential equation.
Thus, the general solution of $y''+ \frac{3}{3x-1}y'-\frac{9}{(3x-1)^2}y=0$ is of the form:
$$c_1 \left( x- \frac{1}{3} \right)+ c_2 \left( \frac{1}{x- \frac{1}{3}}\right) | c_1, c_2 \in \mathbb{R}, x> \frac{1}{3}$$
EDIT:
We set $t=x-\frac{1}{3}$ and we have:
$$\frac{dy}{dt}=\frac{dy}{dx} \frac{dx}{dt}=\frac{dy}{dx}$$
$$\frac{d^2y}{dt^2}=\frac{d}{dt} \left( \frac{dy}{dt}\right)=\frac{d}{dt} \left( \frac{dy}{dx} \right)=\frac{dx}{dt} \frac{d}{dx} \left( \frac{dy}{dx} \right)=\frac{d^2y}{dx^2}$$
$$y''(x)+ \frac{1}{x-\frac{1}{3}}y'(x)-\frac{1}{\left( x-\frac{1}{3}\right)^2}y(x)=0 \\ \Rightarrow y''(t)+\frac{1}{t}y'(t)-\frac{1}{t^2}y(t)$$
Two linearly independent solutions are $y_1(t)=t$ and $y_2(t)=\frac{1}{t}, y \in (0,+\infty)$.
Thus, two linearly independent solutions of $y''+\frac{1}{x-\frac{1}{3}}y'-\frac{1}{\left( x-\frac{1}{3} \right)^2}y=0, x> \frac{1}{3}$
are $y_1(x)=x-\frac{1}{3}, y_2(x)=\frac{1}{x-\frac{1}{3}}$
Is it right or have I done something wrong?
| Let $t = x - \frac{1}{3}$, then
$$\frac{dy}{dt} = y'$$
$$ \frac{d^2y}{dt^2} = y''$$
The equation becomes
$$ 9t^2 \frac{d^2y}{dt^2} + 9t \frac{dy}{dt} - 9y = 0 $$
This is of course the Cauchy-Euler equation. To solve, let $y = t^m$, so on and so forth.
| {
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"url": "https://math.stackexchange.com/questions/1172614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Quadratic formula question: Missing multiplying factor of A? I have a very simple problem which must have a simple answer and I was wondering if anyone can point out my error.
I have the following quadratic equation to factor:
$2x^2+5x+1$
Which is of the form:
$Ax^2+Bx+C$
All I want to now do is factor this into the form:
$(x+\alpha)(x+\beta)$
So using the quadratic formula
$x = \frac{-B\pm\sqrt{B^2-4AC}}{2A}$
I get
$x = \frac{-5 \pm \sqrt{17}}{4}$
So I would think then that
$(x - \frac{-5 + \sqrt{17}}{4})(x - \frac{-5 - \sqrt{17}}{4}) = 2x^2+5x+1 $
But it doesn't seem to work. Any x that I choose results in the answer being off by a factor of A (in this case A=2).
What silly thing am I missing?
| We can obtain the correct factorization by completing the square.
\begin{align*}
2x^2 + 5x + 1 & = 2\left(x^2 + \frac{5}{2}x\right) + 1\\
& = 2\left(x^2 + \frac{5}{2}x + \frac{25}{16}\right) - \frac{25}{8} + 1\\
& = 2\left(x + \frac{5}{4}\right)^2 - \frac{17}{8}\\
& = 2\left[\left(x + \frac{5}{4}\right)^2 - \frac{17}{16}\right]\\
& = 2\left[x + \frac{5}{4} + \frac{\sqrt{17}}{4}\right]\left[x + \frac{5}{4} - \frac{\sqrt{17}}{4}\right]\\
& = 2\left[x + \frac{5 + \sqrt{17}}{4}\right]\left[x - \frac{5 + \sqrt{17}}{4}\right]
\end{align*}
However, when we solve the equation $2x^2 + 5x + 1 = 0$ by completing the square, we first transform it into the monic equation
$$x^2 + \frac{5}{2}x + \frac{1}{2} = 0$$
whose roots are the same as those of the original equation. When we derive the Quadratic Formula, we complete the square on the equation
$$ax^2 + bx + c = 0$$
where $a \neq 0$. The first step in the derivation is to divide $ax^2 + bx + c = 0$ by $a$ to transform it into the monic equation
$$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$
whose roots are equivalent to those of the equation $ax^2 + bx + c = 0$.
If we factor the monic equation, we obtain
$$\left(x - \frac{-b + \sqrt{b^2 - 4ac}}{2a}\right)\left(x - \frac{-b + \sqrt{b^2 - 4ac}}{2a}\right) = 0$$
The sum of the roots is $-\dfrac{b}{a}$ and the product of the roots is $\dfrac{c}{a}$.
To obtain the original equation from the monic equation, we must multiply the monic equation by $a$. Thus, the factorization of the original equation is
\begin{align*}
ax^2 + bx + c & = a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right)\\
& = a\left(x - \frac{-b + \sqrt{b^2 - 4ac}}{2a}\right)\left(x - \frac{-b + \sqrt{b^2 - 4ac}}{2a}\right)
\end{align*}
What you found by using the sum and product of the roots was the factorization of the monic equation
$$x^2 + \frac{5}{2}x + \frac{1}{2} = 0$$
To obtain the original equation, you must multiply your factorization by $a = 2$.
| {
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"url": "https://math.stackexchange.com/questions/1175195",
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"source": "stackexchange",
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What does orthogonality mean in function space? The functions $x$ and $x^2 - {1\over2}$ are orthogonal with respect to their inner product on the interval [0, 1]. However, when you graph the two functions, they do not look orthogonal at all. So what does it truly mean for two functions to be orthogonal?
| Functions can be added together, scaled by constants, and taken in linear combination--just like the traditional Euclidean vectors.
$$\vec{u} = a_1\vec{v}_1 + a_2\vec{v}_2 + \cdots a_n\vec{v}_n$$
$$g(x) = a_1f_1(x) + a_2f_2(x) + \cdots + a_nf_n(x) $$
Just as $\left( \begin{array}{c} 5\\ 2\\ \end{array} \right) =
5\left( \begin{array}{c} 1\\ 0\\ \end{array} \right) +
2\left( \begin{array}{c} 0\\ 1\\ \end{array} \right) $, I can say $5x + 2x^2 - 1 = 5(x) + 2(x^2 - \frac{1}{2})$
"Orthogonality" is a measure of how much two vectors have in common. In an orthogonal basis, the vectors have nothing in common. If this is the case, I can get a given vector's components in this basis easily because the inner product with one basis vector makes all other basis vectors in the linear combination go to zero.
If I know $\left( \begin{array}{c} 1\\ 0\\ 0.5\\ \end{array} \right)$,
$\left( \begin{array}{c} 0\\ 1\\ 0\\ \end{array} \right)$ and
$\left( \begin{array}{c} 2\\ 0\\ -4\\ \end{array} \right)$ are orthogonal. I can quickly get the components of any vector in that basis:
$$\left( \begin{array}{c} 8\\ -2\\ 5\\\end{array} \right) =
a\left( \begin{array}{c} 1\\ 0\\ 0.5\\ \end{array} \right) +
b\left( \begin{array}{c} 0\\ 1\\ 0\\ \end{array} \right) +
c\left( \begin{array}{c} 2\\ 0\\ -4\\ \end{array} \right)$$
$$\left( \begin{array}{c} 1\\ 0\\ 0.5\\ \end{array} \right) \cdot \left( \begin{array}{c} 8\\ -2\\ 5\\\end{array} \right) =
\left( \begin{array}{c} 1\\ 0\\ 0.5\\ \end{array} \right) \cdot
\big[
a\left( \begin{array}{c} 1\\ 0\\ 0.5\\ \end{array} \right) +
b\left( \begin{array}{c} 0\\ 1\\ 0\\ \end{array} \right) +
c\left( \begin{array}{c} 2\\ 0\\ -4\\ \end{array} \right)
\big]$$
I know the $b$ and $c$ term disappear due to orthogonality, so I set them to zero and forget all about them.
$$8.25 = 1.25 a$$
I can also get function basis components easily this way. Take the fourier series on $[0,T]$ for example, which is just a (infinitely long) linear combination of vectors/functions:
$$
f(x) = a_0 + \sum_n^\infty a_ncos(\frac{2\pi n}{T}x) + \sum_n^\infty b_nsin(\frac{2\pi n}{T}x)
$$
I know that all the $cos$ and $sin$ basis functions are orthogonal, so I can take the inner product with $cos(\frac{2\pi 5}{T}x)$ and easily get a formula for the $a_5$ coefficient because all the other terms vanish when I do an inner product.
$$\int cos(\frac{2\pi 5}{T}x) f(x) dx = a_5 \int cos^2(\frac{2\pi 5}{T}x) dx$$
Of course I could do this in general with $cos(\frac{2\pi q}{T}x)$ to get any of the $a_q$ components.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1176941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 4
} |
Intersection between two planes and a line? What is the coordinates of the point where the planes: $3x-2y+z-6=0$ and $x+y-2z-8=0$ and the line: $(x, y, z) = (1, 1, -1) + t(5, 1, -1)$ intersects with eachother?
I've tried letting the line where the two planes intersect eachother be equal to the given line, this results in no solutions.
I have tried inserting the lines x, y and z values into the planes equations, this too, results in no solutions.
According to the answer sheet the correct solution is: $\frac{1}{2}(7,3,-3)$
| Isolate a variable in the planar equations and set the resulting expressions equal to each other (because they intersect):
$z= 6+2y-3x$, and $z=\frac{x+y-8}{2}$
so $6+2y-3x=\frac{x+y-8}{2}$.
Solving for x yields:
$\frac{20+3y}{7}=x$
and setting $y$ as the parameter ($y=s$), and substituting back into the original equation we have the equation of the line of intersection:
$x = \frac{20+3s}{7}$
$y=s$
$z=6+2s-\frac{60+9s}{7}$
Now, if the lines intersect the formulas for x and y must be equal so:
$s=1+t$
$\frac{20+3s}{7}=1+5t$
Solving yields $s=1.5$, which when plugged into our formulas for x,y, and z in terms of s yields $(3.5,1.5,-1.5)$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Prove $3a^4-4a^3b+b^4\ge0$ . $(a,b\in\mathbb{R})$ $$3a^4-4a^3b+b^4\ge0\ \ (a,b\in\mathbb{R})$$
We must factorize $3a^4-4a^3b+b^4\ge0$ and get an expression with an even power like $(x+y)^2$ and say an expression with an even power can not have a negative value in $\mathbb{R}$.
But I don't know how to factorize it since it is not in the shape of any standard formula.
| $$3a^4-4a^3b+b^4$$
$$=3a^4-3a^3b-a^3b+b^4$$
$$=3a^3(a-b)-b(a^3-b^3)$$
$$=(a-b)\Big(3a^3-b(a^2+b^2+ab)\Big)$$
$$=(a-b)(3a^3-a^2b-b^3-ab^2)$$
The cubic is zero if $a=b$, so try taking $(a-b)$ out again:
$$(a-b)(3a^3-3a^2b+2a^2b-2b^3+b^3-ab^2)$$
$$=(a-b)\Big(3a^2(a-b)+2b(a^2-b^2)-b^2(a-b)\Big)$$
$$=(a-b)^2(3a^2+b^2+2ab)$$
$$=(a-b)^2\Big((a+b)^2+2a^2\Big)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find the coordinate of third point of equilateral triangle. I have two points A and B whose coordinates are $(3,4)$ and $(-2,3)$ The third point is C. We need to calculate its coordinates.
I think there will be two possible answers, as the point C could be on the either side of line joining A and B.
Now I put AB = AC = BC.
We calculate AB by distance formula : $\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
= $ \sqrt{5^2 + 1^2} = \sqrt{26}$
I plug this value into the distance of AC and BC and finally equate to get :
$5x + y = 6$
Now what can I do? There are two variables, I am getting equation of a line! How can I solve this?
| Call the position of point $C$ by the coords $(a, b)$. The equations for $C$ are then
$$
\sqrt{(a-3)^2 + (b - 4)^2} = \sqrt{26} \\
\sqrt{(a+2)^2 + (b - 3)^2} = \sqrt{26}
$$
Squaring both, we get
$$
(a-3)^2 + (b - 4)^2 = 26 \\
(a+2)^2 + (b - 3)^2 = 26
$$
$$
a^2 - 6a + 9 + b^2 - 8b + 16= 26 \\
a^2 + 4a + 4 + b^2 - 6b + 9= 26
$$
Subtracting these two gives
$$
-10a + 5 - 2b + 7 = 0
$$
or
$$
6 = 5a + b
$$
which is a line both points must lie on. Writing this as
$$
b = 6 - 5a
$$
we can substitute in either equation. Let's got with the second:
$$
a^2 + 4a + 4 + b^2 - 6b + 9= 26
$$
becomes
$$
a^2 + 4a + 4 + (6-5a)^2 - 6(6-5a) + 9= 26
$$
which is a quadratic that can now be solved for the two possible values of $a$.
(Once you do so, you use $b = 6 - 5a$ to find the corresponding $b$-values.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Find all integer solutions of equality Find all integer solutions of equation $$x^3+(x+1)^3+...+(x+7)^3=y^3$$ I've solved it by opening brackets and consideration of signs but I think there is simpler way of solving it .
| Let
$$P(x)=x^3+(x+1)^3+(x+2)^3+(x+3)^3+(x+4)^3+(x+5)^3+(x+6)^3+(x+7)^3$$
so
$$P(-x-7)=-P(x)$$
on the other hand we can
$$P(x)=8x^3+84x^2+420x+784$$
if $x\ge 0$,we have
$$(2x+7)^3=8x^3+84x^2+294x+343<P(x)<8x^3+120x^2+600x+1000=(2x+10)^3$$
so we have
$$(2x+7)<y<2x+10$$
so
$$P(x)=(2x+8)^3\Longrightarrow -12x^2+36x+272=0$$
this equation no integer roots,
simaler we have
$$P(x)-(2x+9)^3=-24x^2-66x+55=0$$
also have no integer roots.
Now we have prove $x\ge 0 $ or $x\le -7$ this equation have no integer roots
But if $-6\le x\le 1$, we can easy to find when
$x=-2\Longrightarrow P(-2)=216=6^3$
$x=-3\Longrightarrow P(-3)=64=4^3$
$x=-4,P(-4)=-64=(-4)^3$
$x=-5,P(-5)=216^3=(-6)^3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
how to show $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ I know that $\frac{1+\cos x+\sin x}{1+\cos x-\sin x} = \frac{1+\sin x}{\cos x}$ is correct. But having some hard time proofing it using trig relations. Some of the relations I used are
$$
\sin x \cos x= \frac{1}{2} \sin(2 x)\\
\sin^2 x + \cos^2 x= 1\\
\sin^2 x = \frac{1}{2} - \frac{1}{2} \cos(2 x)\\
\cos(2 x) = 2 \cos^2(x)-1
$$
And others. I tried starting with multiplying numerator and denominator with either $\cos x$ or $\sin x$ and try to simplify things, but I seem to be going in circles.
Any hints how to proceed?
| Hint:
$$= \frac{(1 + \cos x + \sin x)(1 - \cos x + \sin x)}{(1 + \cos x - \sin x)(1 - \cos x + \sin x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to explain the formula for the sum of a geometric series without calculus? How to explain to a middle-school student the notion of a geometric series without any calculus (i.e. limits)? For example I want to convince my student that
$$1 + \frac{1}{4} + \frac{1}{4^2} + \ldots + \frac{1}{4^n} = \frac{1 - (\frac{1}{4})^{n+1} }{ 1 - \frac{1}{4}}$$
at $n \to \infty$ gives 4/3?
| $$
\frac{1}{3}-\frac{1}{4}=\frac{1}{12}=\frac{1}{4}\cdot\frac{1}{3}
$$
so
$$
\frac{1}{3}=\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{3}
$$
Now ask your 14 year old to plug in this expression for $1\over 3$ into itself, quite funny, bewildering and strange at first sight:
$$
\frac{1}{3}=\frac{1}{4}+\frac{1}{4}\cdot \left(\frac{1}{4}+\frac{1}{4}\cdot\frac{1}{3}\right)=
\frac{1}{4}+\frac{1}{4^2}+\frac{1}{4^2}\frac{1}{3}
$$
Repeat two or three times, then discuss the difference
$$
\frac{1}{3}-\left(\frac{1}{4}+\frac{1}{4^2}+\dots+\frac{1}{4^n}\right)
$$
edit
For example I want to convince my student that
this is impossible without talking about the notion a limit. How do you convince a student that $1,{1\over 2},{1 \over 3},\dots$ goes to zero? What does goes to even mean? In this situation
$$
1+\frac{1}{4}+\frac{1}{4^2}+\dots=\frac{4}{3}
$$
You cannot convince somebody that this is true without defining the meaning of these little dots on the left side.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1185259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 13,
"answer_id": 2
} |
Find all possible values of $ a^3 + b^3$ if $a^2+b^2=ab=4$. Find all possible values of $a^3 + b^3$ if $a^2+b^2=ab=4$.
From $a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)(4-4)=(a+b)0$. Then we know $a^3+b^3=0$.
If $a=b=0$, it is conflict with $a^2+b^2=ab=4$.
If $a\neq0$ and $b\neq0$, then $a$ and $b$ should be one positive and one negative. This contradict with $ab$=4.
I don't know the solution.
Any help please.
| As has been observed, one approach to this problem is to solve the equations over the complex numbers, and then check directly that $a^3+b^3=0$ in all cases. However, it's both useful and more general to observe that the conclusion can be reached without actually solving the equations: since $a^2+b^2=a b$, we have
$(*)\ \ a^3+b^3=(a+b)(a^2-a b + b^2)=0.$
The problem makes sense over any field -- in fact in any commutative ring with $1$ (this condition is only needed so that "$4$" makes sense.) In addition to being simple, an advantage of the solution $(*)$ is that it works regardless of the setting. This is relevant, since the problem did not specify a domain for $a$ and $b$, so a solution based on calculating the roots in a particular context involves an inherent assumption.
For example, in the field ${\Bbb F}_{37}$ (i.e., the integers mod $37$), the equations have the four solutions $(a,b)$=$(9,21)$, $(16,28)$, $(21,9)$, and $(28,16)$. You can check that $a^3+b^3=0$ in each case, but you're back to square one when faced with ${\Bbb F}_{61}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1186290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
Finding eigenvalues to classify the steady state of a system I have this system of differential equations which model chemical concentrations in a certain reactions:
$$\dot{x}=a-x-\frac{2xy}{1+x^2}\qquad \dot{y}=bx\left(1-\frac{y}{1+x^2}\right)$$
for $a,b >0$ and $x(t),y(t)\geq 0$,
and I want to find the steady state of it and then classify it.
I began by finding the only possible steady state, which occurs at $\left(\frac{a}{3},\frac{a^2}{9}+1\right)$, the intersection of $y=\frac{(x^2+1)(a-x)}{2x}$ and $y=1+x^2$.
I then calculated the Jacobian of the system to be
$$\left[\begin{matrix}
\frac{2(x^2-1)y}{(x^2+1)^2}-1 & \frac{-2x}{x^2+1} \\
\frac{b(x^2-1)y}{(x^2+1)^2}+b & \frac{-bx}{x^2+1}
\end{matrix}\right]$$
From here I substituted the values of $x$ and $y$ at the calculated steady state into the Jacobian:
$$\left[\begin{matrix}
\frac{2(\frac{a^2}{9}-1)}{(\frac{a^2}{9}+1)}-1 & \frac{-2a}{3(\frac{a^2}{9}+1)} \\
\frac{b(\frac{a^2}{9}-1)}{(\frac{a^2}{9}+1)}+b & \frac{-ba}{3(\frac{a^2}{9}+1)}
\end{matrix}\right]$$
Now to find the stability of the point $\left(\frac{a}{3},\frac{a^2}{9}+1\right)$, I need to find the eigenvalues of this matrix. I began by trying to calculate this by brute force and it is very messy. Is there an easier way? Have I missed some simplification? Have I made a mistake that is making this a much more complicated calculation than it should be?
I found the characteristic polynomial to be:
$$\frac{8a^6b^2}{243\left(\frac{a^8}{2187} + \frac{4a^6}{243} + \frac{2a^4}{9} + \frac{4a^2}{3} + 3\right)} -
\frac{8a^4b^2}{27\left(\frac{a^8}{2187} + \frac{4a^6}{243} + \frac{2a^4}{9} + \frac{4a^2}{3} + 3\right)} -
\frac{4a^4b^2}{27\left(\frac{a^6}{243} + \frac{a^4}{9} + a^2 + 3\right)} -
\frac{8a^3b\lambda}{27\left(\frac{a^6}{729} + \frac{a^4}{27} + \frac{a^2}{3} + 1\right)} +
\frac{8a^5b\lambda}{243\left(\frac{a^6}{729} + \frac{a^4}{27} + \frac{a^2}{3} + 1\right)} -
\frac{4a^4b^2\lambda}{27\left(\frac{a^6}{243} + \frac{a^4}{9} + a^2 + 3\right)} -
\frac{4a^3b\lambda^2}{27\left(\frac{a^4}{81} + \frac{2*a^2}{9} + 1\right)} -
\frac{4a^3b\lambda}{27\left(\frac{a^4}{81} + \frac{2*a^2}{9} + 1\right)}$$
and I have no idea how I would solve this for the eigenvalues $\lambda$. There must be a simpler way!
| i am going to set $a = 3$ see what happens. i will see if by scaling time if this can be done. i will make a change of variable $$x = 1 + u, y = 2 + v.$$ then
$$u' = 3 - (1+u) - \frac{2(1+u)(2 + v)}{1 + (1 + u)^2 } = \frac{(2-u)(2+2u+u^2)-2(2+2u+v+uv)}{2+2u+u^2} = \frac{(4-2u+4u+\cdots) -(4+4u+2v+\cdots)}{2+2u+u^2} = -u -v+\cdots$$ and
$$ v' = b(1+u)\left(1 - \frac{2+v}{2+2u+u^2}\right) = b(1+u)\frac{2u-v+u^2}{2+2u+u^2}= b(2u-v+\cdots)$$ the linear stability is determined by
$$u' = -u - v, \, v' = 2bu - bv. $$ the characteristic equation is $$\lambda^2 + (b+1) \lambda + 3b = 0, \lambda = \frac{-b-1 \pm \sqrt{b^2-10b + 1}}2 $$ so if $b > 5 + 2\sqrt 6,$ then the real part of $\lambda < 0$ implying a stable spiral. if $\lambda < 5 + 2\sqrt 6,$ then because roots are negative, we have a stable focus.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1186690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is equality $\sin(\frac{3\pi}{2}-x) = -\cos(x)$ always true? $\sin(\frac{3\pi}{2}-x) = -\cos(x)$
Is above equality always true even if $x \gt \frac{\pi}{2}$?
| why not $$\sin\left(\frac{3\pi} 2 - x\right) = \sin \left(\frac{3\pi} 2\right)\cos x -\cos\left(\frac{3\pi} 2\right) \sin x = -\cos x?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find the derivative using the chain rule and the quotient rule
$$f(x) = \left(\frac{x}{x+1}\right)^4$$ Find $f'(x)$.
Here is my work:
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4\left(x+1\right)^3x^4}{\left(x+1\right)^8}$$
$$f'(x) = \frac{4x^3\left(x+1\right)^4-4x^4\left(x+1\right)^3}{\left(x+1\right)^8}$$
I know the final simplified answer to be:
$${4x^3\over (x+1)^5}$$
How do I get to the final answer from my last step? Or have I done something wrong in my own work?
| I usually try to avoid the quotient rule if possible.
$$ f(x) = \left(\frac{x}{x+1}\right)^4 = \left(\frac{x+1-1}{x+1}\right)^4 = \left(\frac{x+1}{x+1} + \frac{-1}{x+1}\right)^4 = \left(1 - \frac{1}{x+1}\right)^4 $$
Applying the chain rule yields
$$ f'(x) = 4 \cdot \left(1-\frac{1}{x+1}\right)^3 \cdot \frac{1}{(x+1)^2} = 4 \cdot \left(\frac{x}{x+1}\right)^3 \cdot \frac{1}{(x+1)^2} = 4 \cdot \frac{x^3}{(x+1)^3} \cdot \frac{1}{(x+1)^2} = \frac{4x^3}{(x+1)^5} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
For $x, y \in \Bbb R$ such that $(x+y+1)^2+5x+7y+10+y^2=0$. Show that $-5 \le x+y \le -2.$ I have a problem:
For $x, y \in \Bbb R$ such that $(x+y+1)^2+5x+7y+10+y^2=0$. Show that
$$-5 \le x+y \le -2.$$
I have tried:
I write $(x+y+1)^2+5x+7y+10+y^2=(x+y)^2+7(x+y)+(y+1)^2+10=0.$
Now I'm stuck :(
Any help will be appreciated! Thanks!
| Set $x+y=c\iff y=c-x$
We have $(c+1)^2+5c+2y+10+y^2=0\iff y^2+2y+c^2+7c+11=0$
As $y$ is real, we need the discriminant $(2)^2-4\cdot1\cdot(c^2+7c+11)\ge0$
$\iff(c+2)(c+5)\le0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Integrate $\int_0^1 \frac{\ln(1+x^a)}{1+x}\, dx$ I have recently met with this integral:
$$\int_0^1 \frac{\ln(1+x^a)}{1+x}\, dx$$
I want to evaluate it in a closed form, if possible.
1st functional equation:
$\displaystyle f(a)=\ln^2 2-f\left ( \frac{1}{a} \right )$ since:
$$\begin{aligned}
f(a)=\int_{0}^{1}\frac{\ln (1+x^a)}{1+x}\,dx &= \int_{0}^{1}\ln \left ( 1+x^a \right )\left ( \ln (1+x) \right )'\,dx\\
&= \ln^2 2 - \int_{0}^{1}\frac{ax^{a-1}\ln (1+x)}{1+x^a}\,dx\\
&\overset{y=x^a}{=\! =\! =\!}\ln^2 2 - \int_{0}^{1}\frac{a y^{(a-1)/a}\ln \left ( 1+y^{1/a} \right )}{1+y}\frac{1}{a}y^{(1-a)/a}\,dy \\
&= \ln^2 2 -f\left ( \frac{1}{a} \right )
\end{aligned}$$
2nd functional equation:
$\displaystyle f(a)=-\frac{a\pi^2}{12}+f(-a)$ since:
$$\begin{aligned}
f(a)=\int_{0}^{1}\frac{\ln (1+x^a)}{1+x}\,dx &=\int_{0}^{1}\frac{\ln \left ( x^a\left ( 1+x^{-a} \right ) \right )}{1+x}\,dx \\
&= a\int_{0}^{1}\frac{\ln x}{1+x}\,dx+\int_{0}^{1}\frac{\ln (1+x^{-a})}{1+x}\,dx\\
&= a\int_{0}^{1}\frac{\ln x}{1+x}\,dx+f(-a)\\
&= a\int_{0}^{1}\ln x \sum_{n=0}^{\infty}(-1)^n x^n \,dx +f(-a)\\
&= a\sum_{n=0}^{\infty}(-1)^n \int_{0}^{1}\ln x \cdot x^n \,dx +f(-a)\\
&=\cdots\\
&=-\frac{a\pi^2}{12}+f(-a)
\end{aligned}$$
What I did try was:
1st way
$$\begin{aligned}
\int_{0}^{1}\frac{\log(1+x^a)}{1+x}\,dx &=\int_{0}^{1}\log(1+x^a)\sum_{n=0}^{\infty}(-1)^n x^n\,dx \\
&= \sum_{n=0}^{\infty}(-1)^n \int_{0}^{1}\log(1+x^a)x^n\,dx\\
&= \sum_{n=0}^{\infty}(-1)^n \{ \left[ \frac{x^{n+1}\log(1+x^a)}{n+1} \right]_0^1- \\
& \quad \quad \quad \quad \quad \frac{a}{n+1}\int_{0}^{1}\frac{x^{n+1}x^{a-1}}{1+x^a}\,dx \}
\end{aligned}$$
2nd way
I tried IBP but I get to an unpleasant integral of the form $\displaystyle \int_{0}^{1}\frac{\ln(1+x)x^{a-1}}{1+x^a}\,dx$.
3nd way
It was just an idea ... expand the nominator into Taylor Series , swip integration and summation and get into a digamma form . This way suggests that a closed form is far away ... since we are dealing with digammas here.
The result I got was:
$$\begin{aligned}
\int_{0}^{1}\frac{1}{1+x}\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}x^{ak} &=\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\int_{0}^{1}\frac{x^{ak}}{1+x}dx \\
&= 1/2\sum_{k=1}^{\infty}\frac{(-1)^{k-1}}{k}\left[\psi\left(\frac{ak}{2}+1\right)-\psi\left(\frac{ak}{2}+1/2\right)\right] \\
\end{aligned}$$
and I can't go on.
| Another series approach might offer a hint:
$$\int_0^1 x^pdp=\frac{1}{p+1}$$
$$\int_0^1 x^{ak} x^ldp=\frac{1}{ak+l+1}$$
$$\int_0^1 \frac{\ln(1+x^a)}{1+x}dx=\sum_{k=1}^{\infty} \sum_{l=0}^{\infty} \frac{(-1)^{k+l+1}}{k(ak+l+1)}=\sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k} \Phi(-1,1,ak+1)$$
Here $\Phi$ is Lerch Trancendent. I only consider $a>0$, otherwise this series might diverge. But really, we only need the case $a>1$ as you showed.
The double sum, despite being unwieldy, offers some insight into $f(a)$. For example, we can guess that for $a \to +\infty$ the asymptotic is $f(a) \to \frac{\pi^2}{12a}$:
$$\frac{1}{a} \sum_{k=1}^{\infty} \sum_{l=1}^{\infty} \frac{(-1)^{k+l}}{k^2+kl/a} \asymp^{a \to +\infty} \frac{1}{a} \sum_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^2}=\frac{\pi^2}{12a}$$
This is not a proof! But the asymptotic is right, as is shown in this answer for a related integral.
I really suggest you read the links provided in the comments to find out about possible closed forms, but just in case I aslo compile the simplest cases with closed forms here:
$$
\begin{matrix}
a & f(a) \\
-2 & \dfrac{7 \pi ^2}{48}+\dfrac{3 \ln ^2 2}{4} \\
-1 & \dfrac{ \pi ^2}{12}+\dfrac{\ln ^2 2}{2}\\
-\dfrac{1}{2} & \dfrac{\pi ^2}{16}+\dfrac{\ln ^2 2}{4}\\
0 & \ln ^2 2 \\
\dfrac{1}{2} & \dfrac{\pi ^2}{48}+\dfrac{\ln ^2 2}{4} \\
1 & \dfrac{\ln ^2 2}{2} \\
2 & -\dfrac{\pi ^2}{48}+\dfrac{3 \ln ^2 2}{4}\\
\end{matrix}
$$
See also this answer for a more comprehensive approach
| {
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"timestamp": "2023-03-29T00:00:00",
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prove increasing/decreasing sequence Are these two statements true? If so, how does one prove them?
1) For each integer k (positive or negative), the sequence
$a_n = (1 + k/n) ^ n$ (1)
is increasing (at least after a certain number n).
2) For each integer k (positive or negative) the sequence
$a_n = (1 + k/n) ^{n+1} $ (2)
is decreasing (at least after a certain number n).
Note that I have no problem proving that the two sequences are convergent and to find their limits but I have really hard time proving formally that these are monotonic (after certain n).
In fact for statement (1) seems one can prove it this way.
Note that:
$(1 + \frac{k+1}{n} )^n = \frac{(1 + \frac{k}{n+1})^{n+1}(1 + \frac{1}{n})^{n}}{1+\frac{k}{n+1}}$ (3)
Now apply induction on k. Based on (3), the (k+1)-sequence ${a_n}$ must be increasing because on the right hand side we have: the increasing k-sequence ${a_{n+1}}$ multiplied by the increasing 1-sequence ${a_n}$, and then divided by a decreasing sequence. So the left hand side must be increasing too.
Is this proof correct? Actually I think it only works for k>0 otherwise the sequence in the divisor (1 + k/(n+1)) is not decreasing but is increasing. Also, what is the proof for (2)? Should be something analogous, I guess, like the above proof for (1).
| Consider
\begin{align*}
\log(a_n) &= n\log\left(1 + \frac{k}{n}\right) \\
&= n\left(\frac{k}{n} - \frac{(\frac{k}{n})^2}{2} \mp \ldots \right)\\
& = \left(k - \frac{k^2}{2n} + \frac{k^3}{3n^2} \mp \ldots \right)
\end{align*}
where the series converges absolutely for $n > k$.
Then
\begin{align*}
\log(a_n) - \log(a_{n-1}) &= \left(k - \frac{k^2}{2n} + \frac{k^3}{3n^2} \mp \ldots \right) - \left(k - \frac{k^2}{2(n-1)} + \frac{k^3}{3(n-1)^2} \mp \ldots \right) \\
&= k^2\left(\frac{1}{2n - 2} - \frac{1}{2n}\right) + O(k(k/n)^2)
\end{align*}
This is negative for $n >> k$, yielding the result. Same technique works for (2).
| {
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"source": "stackexchange",
"question_score": "3",
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Show that $\binom{2n}{n}$ is an even number, for positive integers $n$. I would appreciate if somebody could help me with the following problem
Show by a combinatorial proof that
$$\dbinom{2n}{n}$$
is an even number, where $n$ is a positive integer.
I tried to solve this problem but I can't.
| You should be familiar with the recursive definition of the binomial coefficients, that $\binom{n}{r} = \binom{n-1}{r-1}+\binom{n-1}{r}$
\begin{matrix}1\\
1 & 1\\
1 & 2 & 1\\
1 & 3 & 3 & 1\\
1 & 4 & 6 & 4 & 1\\
\vdots & & \vdots & & \vdots&\ddots\\
& \cdots& \binom{n-1}{r-1} & \binom{n-1}{r} & \cdots\\
& \cdots& & \binom{n}{r} & \cdots\end{matrix}
Now, notice that $\binom{2n}{n} = \binom{2n-1}{n-1} + \binom{2n-1}{n}$
Notice further that $\binom{2n-1}{n-1} = \binom{2n-1}{(2n-1)-(n-1)} = \binom{2n-1}{n}$
So, $\binom{2n}{n} = 2\cdot \binom{2n-1}{n}$ and is therefore even.
The combinatorial proof of the recursive definition is that to choose $r$ objects out of $n$ with one special object, either the special object is in the choosing or it is not. Breaking it into cases, if it is, you still need to choose $r-1$ objects out of the $n-1$ remaining. If it is not, then you still need to choose $r$ objects out of the $n-1$ remaining.
Hence $\binom{n}{r} = \binom{n-1}{r-1}+\binom{n-1}{r}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Conditional Extremum, need help finding the extreme points in calculation. Find the conditonal extremums of the following $$u=xyz$$ if $$(1) x^2+y^2+z^2=1,x+y+z=0.$$
First i made the Lagrange function $\phi= xyz+ \lambda(x^2+y^2+z^2-1) + \mu (x+y+z) $, now making the derivatives in respect to x, y, z equal to zero i get the system :
$ \phi_x'=yz+2\lambda x+ \mu $
$\phi _y'=xz+2\lambda y+ \mu $
$\phi _z'=xy+2\lambda z+ \mu$
Using these systems and (1) , 6 points are found. I can't seem to calculate these points. Can anyone help with the calculation, i dont know how to come about the result..
| $x+y+z=0$ means $x,y,z$ have different signs,but for three ones ,there are two ones have same sign. WLOG, let $xy>0 \implies xy \le \dfrac{(x+y)^2}{4}$
we have $x^2+y^2+(x+y)^2=1 \ge \dfrac{(x+y)^2}{2}+(x+y)^2 \iff (x+y)^2 \le \dfrac{2}{3}$
when $x=y=\pm \dfrac{\sqrt{6}}{6} ,(x+y)^2$ get max and $xy$ get max also.
$u=-xy(x+y)$,
when $x>0,y>0 \implies x+y>0 \implies xy(x+y)\le \dfrac{(x+y)^3}{4}\le \dfrac{\sqrt{6}}{18} \implies u \ge - \dfrac{\sqrt{6}}{18} $
so the min points are $(\dfrac{\sqrt{6}}{6},\dfrac{\sqrt{6}}{6},-\dfrac{\sqrt{6}}{3})$
when $x<0,y<0 \implies x+y<0 \implies xy(x+y)\ge \dfrac{(x+y)^3}{4}\ge- \dfrac{\sqrt{6}}{18} \implies u \le \dfrac{\sqrt{6}}{18} $ or premium
so the max points are $(-\dfrac{\sqrt{6}}{6},-\dfrac{\sqrt{6}}{6},\dfrac{\sqrt{6}}{3})$
| {
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"source": "stackexchange",
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Show limit of function $\frac{xy(x+y)}{x^2-xy+y^2}$ for $(x,y)\to(0,0)$ Show $$\lim_{(x,y)\to (0,0)} xy \frac{(x+y)}{x^2-xy+y^2}=0$$
If I approach from $y=\pm x$, I get $0$. Is that sufficient?
| From $\bigl(|x|-|y|\bigr)^2\geq0$ we get $2|xy|\leq x^2+y^2$ and then
$$|x|+|y|\leq\sqrt{2(x^2+y^2)},\qquad x^2-xy+y^2\geq{1\over2}(x^2+y^2)\ .$$
This implies
$$\bigl|f(x,y)\bigr|=\left|{xy(x+y)\over x^2-xy+y^2}\right|\leq{{1\over2}(x^2+y^2)\sqrt{2(x^2+y^2)}\over{1\over2}(x^2+y^2)}=\sqrt{2(x^2+y^2)}\ .$$
Therefore, given an $\epsilon>0$, we have $\bigl|f(x,y)\bigr|<\epsilon$ as soon as $\sqrt{x^2+y^2}<\delta:={\displaystyle{\epsilon\over\sqrt{2}}}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$f(x) = (\cos x - \sin x) (17 \cos x -7 \sin x) $ $f(x) = (\cos x - \sin x) (17 \cos x -7 \sin x)$
Determine the greatest and least values of $\frac{39}{f(x)+14}$ and state a value of x at which greatest values occurs.
Do I just use a graphing calculator for this? Is there a way I could do this without a graphing calculator?
| you can do this without calculus. here is a way. i will use $t$ for $x.$ we have
$$\begin{align}y &= 14 + (\cos t - \sin t)(17 \cos t - 7 \sin t) \\&= 14 + 17 \cos^2 t-24 \sin t \cos t+7 \sin ^2 t\\
&=14 + \frac{17}{2}(1 + \cos 2t)-12 \sin 2t+\frac 72(1-\cos 2t)\\
&=26+5\cos 2t-12\sin 2t\\
&=26 + 13\cos(2t+\phi), \text{ where} \cos \phi = \frac 5{13}, \sin \phi = \frac{12}{13} \end{align}$$ so the maximum value of $y$ is $39$ and minimum value of $y$ is $13.$
therefore $$1 \le \frac {39}{ 14 + (\cos t - \sin t)(17 \cos t - 7 \sin t)} \le 3.$$
| {
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"url": "https://math.stackexchange.com/questions/1202320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability Question (Colored Socks)
In a drawer Sandy has 5 pairs of socks, each pair a different color. On Monday Sandy selects two individual socks at random from the 10 socks in the drawer. On Tuesday Sandy selects 2 of the remaining 8 socks at random and on Wednesday two of the remaining 6 socks at random. The probability that Wednesday is the first day Sandy selects matching socks is $\tfrac{m}{n}$, where $m$ and $n$ are relatively prime positive integers. Find $m+n$.
She has $5$ different colors: $a, b, c, d, e$.
The number of ways of selecting $2$ socks on Wednesday is:
$$\binom{6}{2}$$
But how do I go further?
| There are three cases to consider:
*
*Four socks, each of a different color, are selected on Monday and Tuesday, then a pair is selected on Wednesday from the one pair and four single socks that remain. The probability of this occurring is
$$1 \cdot \frac{8}{9} \cdot \frac{6}{8} \cdot \frac{4}{7} \cdot \frac{\binom{2}{2}}{\binom{6}{2}}$$
since the first sock is guaranteed to be of a distinct color, $8$ of the remaining $9$ socks are a different color than that of the first sock that is selected, $6$ of the remaining $8$ socks are of a different color than those of both of the first two socks that have been selected, $4$ of the remaining $7$ socks are of a different color than those of the first three socks that have been selected, and we must choose both socks of the fifth color from the six remaining socks.
*Four socks, of three different colors, are selected on Monday and Tuesday, then a pair is selected on Wednesday from the two remaining pairs and two remaining individual socks. The probability of this occurring is
$$1 \cdot \frac{8}{9} \cdot \left(\frac{2}{8} \cdot \frac{6}{7} + \frac{6}{8} \cdot \frac{2}{7}\right) \cdot \frac{4}{6} \cdot {1}{5}$$
On Monday, the probabilities are the same as in the first case. On Tuesday, there are two possibilities. One is that a sock of one of the $2$ colors already selected on Monday is selected from one of the $8$ remaining socks, then a sock of one of the other $3$ colors is selected from the $7$ remaining socks. The other possibility is that one of the $6$ socks of one the $3$ remaining colors is selected from the $8$ socks that remain, then one of the $2$ socks of a color already selected on Monday is selected from one of the $7$ remaining socks. On Wednesday, there are $6$ socks left. One of the $4$ socks from the two remaining pairs must be selected first, then the only sock of that color that remains must be selected from the $5$ remaining socks.
*Four socks, of two different colors, are selected on Monday and Tuesday, then a pair is selected on Wednesday from one of the three remaining pairs. The probability of this occurring is
$$1 \cdot \frac{8}{9} \cdot \frac{2}{8} \cdot \frac{1}{7} \cdot 1 \cdot \frac{1}{5}$$
On Monday, the probabilities are the same as in the preceding cases. On Tuesday, the first sock selected must be one of the $2$ socks of a color already selected on Monday, then the other sock that is selected must match the other color selected on Monday. On Wednesday, only the three colors that have not already been used are available, so the first sock is guaranteed to be of a different color than those selected on Monday and Tuesday. Finally, we must select the only remaining sock that matches the first selection on Wednesday from the five remaining socks.
Since these cases are mutually exclusive, you can add the probabilities to determine the probability that Wednesday is the first day that Sandy selects a matching pair of socks.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Using Laplace Transforms to Evaluate Integrals
I'm trying to solve $$\int_0^{\infty} \, \frac{e^{-2t}\cos(3t)-e^{-4t}\cos(2t)}{t}dt.$$
I'm sure that this involves Laplace transforms, I'm just not sure how.
I would start by separating and making two integrals, where the second would solve in the same procedure as the first. Now I would have
$\int_{0}^{\infty} e^{-2t}cos(3t)t^{-1}dt$, where from here I'm not sure what to do. Normally I would use something like $L{tf(t)}=-F'(s)$, but that doesn't apply here. Any help is appreciated, thanks
| *
*Method 1
By using the integral
\begin{align}
\int_{0}^{\infty} e^{-u t} \, du = \frac{1}{t}
\end{align}
the integral
\begin{align}
I = \int_0^{\infty} \, \frac{e^{-2t}\cos(3t)-e^{-4t}\cos(2t)}{t}dt
\end{align}
becomes
\begin{align}
I &= \int_0^{\infty} \int_{0}^{\infty} (e^{-2t}\cos(3t)-e^{-4t}\cos(2t) ) ds dt \\
&= \int_{0}^{\infty} \, ds \, \left[ \int_{0}^{\infty} e^{-(s+2)t} \cos(3t) \, dt - \int_{0}^{\infty} e^{-(s+4)t} \cos(2t) \, dt \right] \\
&= \int_{0}^{\infty} \left[ \frac{s+2}{(s+2)^{2} + 3^{3}} - \frac{s+4}{(s+4)^{2} + 4} \right] \, ds \\
&= \frac{1}{2} \left[ \ln\left( \frac{(s+2)^{2} + 9}{(s+4)^{2} + 4} \right) \right]_{0}^{\infty} \\
&= - \frac{1}{2} \, \ln\left( \frac{13}{20} \right).
\end{align}
*Method 2
Using the Laplace transform of a of a function divided by the variable is under the rule
\begin{align}
\mathcal{L}\left\{ \frac{f(t)}{t} \right\} = \int_{s}^{\infty} F(u) \, du
\end{align}
where $F(s)$ is the transformed function. Fron this rule it is seen that
\begin{align}
I &= \int_{2}^{\infty} \frac{u}{u^{2} + 3^{2}} \, du - \int_{4}^{\infty} \frac{u}{u^{2} + 2^{2}} \, du \\
&= \frac{1}{2} \left[ \ln(u^{2} + 9) \right]_{2}^{\infty} - \frac{1}{2} \left[ \ln(u^{2} + 4) \right]_{4}^{\infty} \\
&= \frac{1}{2} \, \lim_{u \rightarrow \infty} \left\{ \ln\left( \frac{u^{2}
+ 9}{u^{2} + 4} \right) \right\} - \frac{1}{2} \, \ln\left(\frac{2^{2} + 9}
{4^{2} + 4} \right) \\
&= \frac{1}{2} \, \lim_{u \rightarrow \infty} \left\{ \ln\left( \frac{1 + \frac{9}{u^{2}} }{ 1 + \frac{4}{u^{2}} } \right) \right\} - \frac{1}{2} \,
\ln\left( \frac{13}{20} \right) \\
&= - \frac{1}{2} \ln\left( \frac{13}{20} \right).
\end{align}
| {
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Sum $\pmod{1000}$
Let $$N= \sum_{k=1}^{1000}k(\lceil \log_{\sqrt{2}}k\rceil-\lfloor \log_{\sqrt{2}}k \rfloor).$$ Find $N \pmod{1000}$.
Let $\lceil x \rceil$ be represented by $(x)$ and $\lfloor x \rfloor$ be represented by $[x]$.
Consider $0 < x < 1$ then:
$$(x) - [x] = 1 - 0 = 1$$
Consider $x=0$ then:
$$(x) - [x] = 0$$
Consider $x=1$ then:
$$(x) - [x] = 1-1 = 0$$
So we must find $k$ such that:
$$\log_{\sqrt{2}}k \in \mathbb{Z}$$
$$\log_{\sqrt{2}}k = \frac{2\log k}{\log 2} = y \in \mathbb{Z}$$
$$\log k = \log(\sqrt{2}^y)\implies k = \sqrt{2}^y$$
So for a few integer $y$ values,
$$k = \sqrt{2}, 2^1, \sqrt{2}^3, 2^2,..., 2^3$$...
$k$ is an integer for powers of $2$.
$$\log_{\sqrt{2}} 2^x = x\log_{\sqrt{2}} 2 = x \frac{2\log_2 2}{\log_2 2} = 2x \in \mathbb{Z}$$
The sum is $1$ for $k=1$ but then $0$ for $k=2$.
$2^9 = 512 < 1000$ but $2^{10} = 1024 > 1000$.
but how can I compute the sum?
| You've correctly identified two statements:
$\lceil x \rceil - \lfloor x \rfloor = \begin{cases} 0 & \text{ if } x \in \mathbb{Z} \\ 1 & \text{ if } x \notin \mathbb{Z} \end{cases}$
For integers $k$, $\log_{\sqrt{2}} k \in \mathbb{Z} \iff k = 2^r$ for some integer $r$.
Combining these, we get
$\lceil \log_{\sqrt{2}} k \rceil - \lfloor \log_{\sqrt{2}} k \rfloor = \begin{cases} 0 & \text{ if } k = 2^r, \text{ some }r \\ 1 & \text{ otherwise } \end{cases}$
So
$$\sum_{k=1}^{1000}k(\lceil \log_{\sqrt{2}}k\rceil-\lfloor \log_{\sqrt{2}}k \rfloor) = \sum_{1 \leq k \leq 1000\text{, } k \neq 2^r\text{ for any } r \in \mathbb{Z}} k$$
$$ = \sum_{k=1}^{1000} k - \sum_{r=0}^92^r$$
which I'm pretty certain you can work out.
| {
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Limit of $\sqrt{25x^{2}+5x}-5x$ as $x\to\infty$ $\hspace{1cm} \displaystyle\lim_{x\to\infty} \left(\sqrt{25x^{2}+5x}-5x\right) $
The correct answer seems to be $\frac12$, whereas I get $0$.
Here's how I do this problem:
$$ \sqrt{25x^{2}+5x}-5x \cdot \frac{\sqrt{25x^{2}+5x}+5x}{\sqrt{25x^{2}+5x}+5x} = \frac{25x^2+5x - 25x^2}{\sqrt{25x^{2}+5x} +5x} = \frac {5x}{\sqrt{25x^{2}+5x}+5x} $$
$\sqrt{25x^{2}+5x}$ yields a bigger value than $5x$ as $x$ becomes a very big number. So the denominator is clearly bigger than numerator. So in this case, shouldn't the answer be $0$?
However, if I keep going and divide both numerator and denominator by $x$ I get:
$$ \frac{5}{ \frac{\sqrt{25x^2+5x}}{x} + 5 }$$
In the denominator, $\frac{\sqrt{25x^2+5x}}{x}$
yields a big number (because top is increasing faster than the bottom), in fact, it goes to infinity as $x$ goes to infinity.
In that case, it's just $5$ divided by something going to infinity, therefore, the answer should be $0$, but it's not, why?
| \begin{align}
\lim_{x\to\infty} \left(\sqrt{25x^{2}+5x}-5x\right)
&= \lim_{x\to\infty} x\left(\sqrt{25+5/x}-5\right) \\
&= \lim_{x\to\infty} \frac{\sqrt{25+5/x}-5}{1/x}
\end{align}
now we can apply L'Hospital's rule:
\begin{align}
= \lim_{x\to\infty}
\left.\frac{-5}{2x^2\sqrt{25+5/x} }
\right/
\frac{-1}{x^2}
&=
\lim_{x\to\infty}
\frac{5}{2\sqrt{25+5/x}}=\frac{1}{2}.
\end{align}
| {
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"source": "stackexchange",
"question_score": "1",
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Showing $1+2+\cdots+n=\frac{n(n+1)}{2}$ by induction (stuck on inductive step) This is from this website:
Use mathematical induction to prove that
$$1 + 2 + 3 +\cdots+ n = \frac{n (n + 1)}{2}$$
for all positive integers $n$.
Solution to Problem 1: Let the statement $P(n)$ be
$$1 + 2 + 3 + \cdots + n = \frac{n (n + 1)}{2}.$$
STEP 1: We first show that $P(1)$ is true.
Left Side $= 1$
Right Side $= \frac{1 (1 + 1)}{2} = 1$
Both sides of the statement are equal hence $P(1)$ is true.
STEP 2: We now assume that $P(k)$ is true
$$1 + 2 + 3 + \cdots + k = \frac{k (k + 1)}{2}$$
and show that $P(k + 1)$ is true by adding $k + 1$ to both sides of the above statement
$$
\begin{align}
1 + 2 + 3 + \cdots + k + (k + 1) &= \frac{k (k + 1)}{2} + (k + 1) \\
&= (k + 1)\left(\frac{k}{2} + 1\right) \\
&= \frac{(k + 1)(k + 2)}{2}
\end{align}
$$
The last statement may be written as
$$1 + 2 + 3 + \cdots + k + (k + 1) = \frac{(k + 1)(k + 2)}{2}$$
Which is the statement $P(k + 1)$.
My question is how in the very last line is the statement $P(k + 1)$ equal to $\frac{(k + 1)(k + 2)}{2}$. I don't get the last step.
| induction is basically saying that if it is true for this step, it is true for the next step. so assuming $1+2+3...+k=k(k+1)/2$, ie it is true for step k, we have to show that it must be true for step k+1, the next step. the final line shows how, by going through some algebra, adding all the numbers up to k+1 equals putting k+1 into the formula $n(n+1)/2$, written as $1+2+3...+k+[k+1]=[k+1]([k+1]+1)/2$. therefore, if it is true for step 1, it is for step 2, and 3...
| {
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Divergence of $\prod_{n=2}^\infty(1+(-1)^n/\sqrt n)$. Looking looking for a verification of my proof that the above product diverges.
$$\begin{align}
\prod_{n=2}^\infty\left(1+\frac{(-1)^n}{\sqrt n}\right) & =\prod_{n=1}^\infty\left(1+\frac1{\sqrt {2n}}\right)\left(1-\frac1{\sqrt{2n+1}}\right)\\
& =\prod_{n=1}^\infty\left(1-\frac1{\sqrt{n_1}}+\frac1{\sqrt {2n}}-\frac1{\sqrt{2n(2n+1)}}\right)\\
& =\prod_{n=1}{\sqrt{2n(2n+1)}-\sqrt{2n}+\sqrt{2n+1}-1\over\sqrt{2n(2n+1)}}\\
& \ge\prod_{n=1}^\infty{\sqrt{2n(2n+1)+2n+1}-\sqrt{2n}-1\over\sqrt{2n(2n+1)}},\quad\sqrt x+\sqrt y\ge\sqrt{x+y}\\
& = \prod_{n=1}^\infty{2n-\sqrt{2n}\over\sqrt{2n(2n+1)}}\\
& = \prod_{n=1}^\infty{2n-1\over\sqrt{2n+1}}
\end{align}$$
This last product diverges since
$$\lim_{n\to\infty}{2n-1\over\sqrt{2n+1}}=\infty.$$
I'm suspicious because
$$\lim_{n\to\infty}\left(1+{(-1)^n\over\sqrt n}\right)=1.$$
| Your calculation is wrong, but the conclusion is correct.
$$ \left( 1 + \dfrac{1}{\sqrt{2n}}\right) \left( 1 - \dfrac{1}{\sqrt{2n+1}}\right) = 1 - \dfrac{1}{2n} + O(n^{-3/2})$$
The infinite product diverges, but to $0$, not $+\infty$.
| {
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If $A,B>0$ and $A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$. If $A,B>0$ and $\displaystyle A+B = \frac{\pi}{3},$ Then find Maximum value of $\tan A\cdot \tan B$.
$\bf{My\; Try::}$ Given $$\displaystyle A+ B = \frac{\pi}{3}$$ and $A,B>0$.
So we can say $$\displaystyle 0< A,B<\frac{\pi}{3}$$. Now taking $\tan $ on both side, we get
$$\displaystyle \tan(A+B) = \tan \left(\frac{\pi}{3}\right).$$ So $\displaystyle \frac{\tan A+\tan B}{1-\tan A\tan B} = \sqrt{3}$.
Now Let $\displaystyle \tan A\cdot \tan B=y\;,$ Then $\displaystyle \tan B = \frac{y}{\tan A}.$
So $$\displaystyle \frac{\tan A+\frac{y}{\tan A}}{1-y}=\sqrt{3}\Rightarrow \tan^2 A+y=\sqrt{3}\tan A-y\sqrt{3}\tan A$$
So equation $$\tan^2 A+\sqrt{3}\left(y-1\right)\tan A+y=0$$
Now for real values of $y\;,$ Given equation has real roots. So its $\bf{Discrimnant>0}$
So $$\displaystyle 3\left(y-1\right)^2-4y\geq 0\Rightarrow 3y^2+3-6y-4y\geq 0$$
So we get $$3y^2-10y+3\geq 0\Rightarrow \displaystyle 3y^2-9y-y+3\geq 0$$
So we get $$\displaystyle y\leq \frac{1}{3}\cup y\geq 3$$, But above we get $\displaystyle 0<A,B<\frac{\pi}{3}$
So We Get $$\bf{\displaystyle y_{Max.} = \left(\tan A \cdot \tan B\right)_{Max} = \frac{1}{3}}.$$
My Question is can we solve above question using $\bf{A.M\geq G.M}$ Inequality or Power Mean equality.
Thanks
| By using $1-\tan A\tan B=\frac{\tan A+\tan B}{\tan(A+B)}$, it follows that for $A>0,B>0,A+B=\pi/3$,
$$\tan A\tan B=\frac{\tan A(\sqrt{3}-\tan A)}{1+\sqrt{3}\tan A}.$$
Hence
$$\begin{align}\max_{A>0,B>0,A+B=\pi/3} \tan A\tan B&=\max_{0<A<\pi/3} \frac{\tan A(\sqrt{3}-\tan A)}{1+\sqrt{3}\tan A}=\max_{1<s<4} \frac{(s-1)(4-s)}{3s}\\
&=\frac{1}{3}\max_{1<s<4}\left(5-\left(s+\frac{4}{s}\right)\right)=
\frac{1}{3}\left(5-\min_{1<s<4}\left(s+\frac{4}{s}\right)\right)
\\&=\frac{1}{3}(5-4)=\frac{1}{3}\end{align}$$
where $s=1+\sqrt{3}\tan A$ and by AGM-inequality
$$s+\frac{4}{s}\geq 2\sqrt{s\cdot \frac{4}{s}}=4$$
with equality for $s=2\in(1,4)$, that is $A=B=\pi/6$.
| {
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"url": "https://math.stackexchange.com/questions/1207197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the product, if we know 5 of them We have 4 positive,but not necessarily integers $a, b, c,$ and $d$. So we have 6 options how multiply two of them. And we know 5 of 6 products $2, 3, 4, 5$ and $6$. Find the last product.
| One way is to find out which two pairs of the $5$ known numbers give the same product, and then use this product to divide the remaining known number. In your example, by inspection, $2 \times 6$ = $3 \times 4$ = $12$, then $12 / 5 = 2.4$, so the last product is $2.4$.
The reason is: We assume the one pair of product is $(a \times b) \times (c \times d)$, the other pair of product is $(a \times c) \times (b \times d)$, and $(a \times b) \times (c \times d) = (a \times c) \times (b \times d) = (a \times d) \times (b \times c)$, so if we find two pairs of the $5$ known numbers can give the same product, it will be the product of $abcd$. Dividing it by the remaining number $(a \times d)$ gives the remaining product $(b \times c)$.
Edit: This method may sometimes give more than one solution. Consider the case when the given $5$ products are $8, 12, 16, 24$ and $32$, you may pair up $8 \times 24 = 12 \times 16 = 192$, and then divide by $32$ to get $6$ (the $4$ original numbers are $2, 3, 4, 8$ or $\sqrt{3}, \sqrt{12}, \sqrt{\frac{64}{3}}, \sqrt{48}$). On the other hand, you may also pair up $12 \times 32 = 16 \times 24 = 384$, and then divide by $8$ to get $48$ (the $4$ original numbers are then $2, 4, 6, 8$ or $\sqrt{6}, \sqrt{\frac{32}{3}}, \sqrt{24}, \sqrt{96}$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit as $x$ tend to zero of: $x/[\ln (x^2+2x+4) - \ln(x+4)]$ Without making use of LHôpital's Rule solve:
$$\lim_{x\to 0} {x\over \ln (x^2+2x+4) - \ln(x+4)}$$
$ x^2+2x+4=0$ has no real roots which seems to be the gist of the issue.
I have attempted several variable changes but none seemed to work.
| Rewrite it as
$$\lim\limits_{x\rightarrow0}\frac{x}{\ln \left( \frac{x^2+2x+4}{x+4}\right)}$$
Apply L'Hopital's rule:
$$\lim\limits_{x\rightarrow0}\frac{1}{\frac{x^2+8x+4}{(x+4)(x^2+2x+4)}}$$
Then simply evaluate the limit:
$$\frac{1}{\frac{0^2+8\times0+4}{(0+4)(0^2+2\times0+4)}} = \frac{1}{\frac{4}{16}} = 4$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Generating function for sequence $a_n = \lceil \sqrt{n} \rceil $ In one of books for discrete mathematics i came across sum to calculate $$\sum_{k=0}^n \lceil \sqrt{n} \rceil$$ which was fairly easy, but this sum intrigued me what is generating function for $$\sum_{n=0}^\infty \lceil \sqrt{n} \rceil x^n $$ so from that i can easily calculate $$\sum_{n=0}^\infty x^n \sum_{k=0}^n \lceil \sqrt{k} \rceil $$
My attempt:
Our sequence looks like this $0, 1, 2, 2, 2, 3, 3, 3, 3, 3, 4 \dots$
We see, difference between adjancent positions is 0 or 1. Let's write it down: $0, 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, \dots$
First zero seems quite uncomfortable, so let's ignore it (but we will have to multiply final generating function by $x$).
So, now we deal with $1, 1, 0, 0, 1, 0, 0, 0, 0, 1, \dots$
Between adjancent ones there is: 0 zeros, then 2 zeros, then 4 zeros, 6 zeros and so on...
I can't really see some pattern here (how to pack zeros and ones in one sequence), i would really appreciate some hints or solutions to this, because it appears as really interesting problem. Cheers
| As far as I can tell, the answer can only be written in terms of the Jacobi theta functions (or the like). Specifically, we have
$$
\begin{align}
\sum_{n=0}^\infty \lceil \sqrt{n} \rceil x^n
& = x + 2x^2 + 2x^3 + 2x^4 + 3x^5 + 3x^6 + 3x^7 + 3x^8 + 3x^9 + \cdots \\
& = (x + x^2 + x^3 + \cdots) + (x^2 + x^3 + x^4 + \cdots)
+ (x^5 + x^6 + x^7 + \cdots) + \cdots \\
& = \frac{x}{1-x} + \frac{x^2}{1-x} + \frac{x^5}{1-x} + \cdots \\
& = \frac{x}{1-x} (1 + x + x^4 + x^9 + x^{16} + \cdots) \\
& = \frac{x}{1-x} \sum_{k=0}^\infty x^{k^2} \\
& = \Bigl(\frac{x}{1-x}\Bigr) \Bigl(\frac{\vartheta_3(0, x) - 1}{2}\Bigr)
\end{align}
$$
Don't know if this helps you at all.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that equation $x^6+x^5-x^4-x^3+x^2+x-1=0$ has two real roots Prove that equation
$$x^6+x^5-x^4-x^3+x^2+x-1=0$$ has two real roots
and $$x^6-x^5+x^4+x^3-x^2-x+1=0$$
has two real roots
I think that:
$$x^{4k+2}+x^{4k+1}-x^{4k}-x^{4k-1}+x^{4k-2}+x^{4k-3}-..+x^2+x-1=0$$
and
$$x^{4k+2}-x^{4k+1}+x^{4k}+x^{4k-1}-x^{4k-2}-x^{4k-3}-..+x^2+x-1=0$$
has two real roots but i don't have solution
| since $ f(\pm \infty) = \infty, f(0) = -1$ shows that $f$ has one positive root and one negative root. so $f$ has at least two real roots.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How to differentiate $y=\sqrt{\frac{1+x}{1-x}}$? I'm trying to solve this problem but I think I'm missing something.
Here's what I've done so far:
$$g(x) = \frac{1+x}{1-x}$$
$$u = 1+x$$
$$u' = 1$$
$$v = 1-x$$
$$v' = -1$$
$$g'(x) = \frac{(1-x) -(-1)(1+x)}{(1-x)^2}$$
$$g'(x) = \frac{1-x+1+x}{(1-x)^2}$$
$$g'(x) = \frac{2}{(1-x)^2}$$
$$y' = \frac{1}{2}(\frac{1+x}{1-x})^{-\frac{1}{2}}(\frac{2}{(1-x)^2})
$$
| Write:
$$ y^2 = \frac{ 1 + x}{1 -x } \iff y^2 = 1 + \frac{2x}{1-x}$$
so taking derivative with repect to $x$ gives
$$ 2y y' = \frac{2(1-x) + 2x}{(1-x)^2} =\frac{2}{(1-x)^2} $$
so
$$ y' = \sqrt{ \frac{1-x}{1+x} } (1-x)^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1223727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Proof of sum results I was going through some of my notes when I found both these sums with their results
$$
x^0+x^1+x^2+x^3+... = \frac{1}{1-x}, |x|<1
$$
$$
0+1+2x+3x^2+4x^3+... = \frac{1}{(1-x)^2}
$$
I tried but I was unable to prove or confirm that these results are actually correct, could anyone please help me confirm whether these work or not?
| Hints:
$$(x^0+x^1+x^2+x^3+...x^n)(1-x)=x^0-x^1+x^1-x^2+x^2-x^3+x^3\cdots+x^{n}-x^{n+1}\\
=1-x^{n+1}$$
and
$$(x^0+2x^1+3x^2+4x^3+...(n+1)x^n)(1-2x+x^2)\\
=x^0-2x^1+x^2+2x^1-4x^2+2x^3+3x^2-6x^3+3x^4\cdots\\+(n+1)x^n-2(n+1)x^{n+1}+(n+1)x^{n+2}\\
=1-(n+2)x^{n+1}+(n+1)x^{n+2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If $\frac{a}{b}=\frac{x}{y}$, is $\frac{x-a}{y-b}=\frac{x}{y}$? Does this hold? $b,y \neq 0$, $b \neq y$.
| \begin{align}
\frac{a}{b}&=\frac{x}{y}\\
\frac{x}{a}&=\frac{y}{b}\\
\frac{x}{a}-1&=\frac{y}{b}-1\\
\frac{x-a}{a}&=\frac{y-b}{b}\\
\frac{x-a}{y-b}&=\frac{a}{b}\\
\frac{x-a}{y-b}&=\frac{x}{y}
\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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To prove $\prod\limits_{n=1}^\infty\cos\frac{x}{2^n}=\frac{\sin x}{x},x\neq0$ Prove $$\prod_{n=1}^\infty\cos\frac{x}{2^n}=\frac{\sin x}{x},x\neq0$$
This equation may be famous, but I have no idea how to start. I suppose it is related to another eqution:
(Euler)And how can I prove the $follwing$ eqution?
$$\sin x=x(1-\frac{x^2}{\pi^2})(1-\frac{x^2}{2^2\pi^2})\cdots=x\prod_{n=1}^\infty (1-\frac{n^2}{2^2\pi^2})$$
I can't find the relation of the two. Maybe I am stuck in a wrong way,thanks for your help.
| By double angle formula we have $$\sin\left(x\right)=2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)=4\sin\left(\frac{x}{4}\right)\cos\left(\frac{x}{4}\right)\cos\left(\frac{x}{2}\right)=\dots=2^{n}\sin\left(\frac{x}{2^{n}}\right)\prod_{k\leq n}\cos\left(\frac{x}{2^{k}}\right)$$ now remains to note that $$\lim_{n\rightarrow\infty}2^{n}\sin\left(\frac{x}{2^{n}}\right)=x.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Solve $\frac{|x|}{|x-1|}+|x|=\frac{x^2}{|x-1|}$ Solve $\frac{|x|}{|x-1|}+|x|=\frac{x^2}{|x-1|}$.What will be the easiest techique to solve this sum ?
Just wanted to share a special type of equation and the fastest way to solve it.I am not asking for an answer and i have solved it in my answer given below.Thank You for viewing.
| To me the easiest and most systematic way to solve it is to explicitly write out the absolute value which means breaking the equation into regions:
$$
\frac{|x|}{|x - 1|} + |x| = \frac{x^2}{|x - 1|} \rightarrow \begin{cases}
\left(\frac{1}{x - 1} + 1\right)x = \frac{x^2}{x - 1} & 1 < x < \infty \\
\left(\frac{1}{1 - x} + 1\right)x = \frac{x^2}{1 - x} & 0 \leq x < 1 \\
-\left(\frac{1}{1 - x} + 1\right)x = \frac{x^2}{1 - x} & -\infty < x < 0
\end{cases}
$$
$\frac{1}{x - 1} + 1 = \frac{1 + x - 1}{x - 1} = \frac{x}{x - 1}$ and $\frac{1}{1 - x} + 1 = \frac{1 + 1 - x}{1 - x} = \frac{2 - x}{1 - x}$ which gives:
$$
\frac{|x|}{|x - 1|} + |x| = \frac{x^2}{|x - 1|} \rightarrow \begin{cases}
\frac{x^2 - x^2}{x - 1} = 0 & 1 < x < \infty \\
\frac{2x - x^2 - x^2}{1 - x} = 0 & 0 \leq x < 1 \\
\frac{-2x + x^2 - x^2}{1 - x} = 0 & -\infty < x < 0
\end{cases}
$$
Which gives:
$$
\frac{|x|}{|x - 1|} + |x| = \frac{x^2}{|x - 1|} \rightarrow \begin{cases}
0 = 0 & 1 < x < \infty \\
2x\frac{1 - x}{1 - x} = 0 & 0 \leq x < 1 \\
-\frac{2x}{1 - x} = 0 & -\infty < x < 0
\end{cases}
$$
Which means that this equation is true for all (real) values $x > 1$ and $x = 0$--there are no negative values of $x$ which satisfy this equation since the third case only has a solution at $x = 0$.
| {
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Demand $z=x+y$ and $x^2/4 + y^2/5 + z^2/25 = 1$. What is the maximum value of $f(x,y,z) = x^2+y^2+z^2$?
Demand $z=x+y$ and $x^2/4 + y^2/5 + z^2/25 = 1$. What is the maximum value of $f(x,y,z) = x^2+y^2+z^2$?
I've been attempting this with Lagrange multipliers in a few different ways. However, the resulting system of equations with two lagrangians has so many variables that it becomes very complicated. Can someone show how this is to be done manually?
I also attempted turning it into two unknowns by replacing $z$ with $x+y$. However, this also led nowhere.
| Beginning with the equation for the Lagrange Multipliers, it is a matter of (tedious) algebraic manipulation with a goal to systematically eliminate parameters. We have $x+y-z=0$ and $\frac{x^2}{4}+\frac{y^2}{5}+\frac{z^2}{25}=1$, $f(x,y,z)=x^2+y^2+z^2$
$$\begin{pmatrix} 2x \\ 2y \\ 2z \end{pmatrix} = \lambda \begin{pmatrix} 1 \\ 1 \\ -1 \end{pmatrix} + \gamma \begin{pmatrix} x/2 \\ 2y/5 \\ 2z/25 \end{pmatrix} \implies \begin{cases} 2x= \lambda+\frac{1}{2}\gamma x \\ 2y= \lambda+\frac{2}{5}\gamma y \\ 2z= -\lambda+\frac{2}{25}\gamma z \end{cases}$$
$$\begin{cases} \lambda = 2x-\frac{1}{2}\gamma x \\ \lambda = 2y-\frac{2}{5}\gamma y \\ \lambda = -2z + \frac{2}{25}\gamma z \end{cases} \implies \begin{cases} \frac{1}{2}\gamma x-2x = \frac{2}{5}\gamma y-2y \\ \frac{1}{2}\gamma x-2x = -\frac{2}{25}\gamma z +2z\end{cases} \implies \begin{cases} \gamma=\frac{2x-2y}{x/2-2y/5} \\ \gamma = \frac{2x+2z}{x/2+2z/25} \end{cases}$$
It then remains to solve the system of equations:
$$\frac{2x-2y}{x/2-2y/5} = \frac{2x+2z}{x/2+2z/25}\ \ \ \ \wedge \ \ \ \ x+y-z=0 \ \ \ \ \wedge \ \ \ \ \frac{x^2}{4}+\frac{y^2}{5}+\frac{z^2}{25}=1$$
The solutions to this system may be any of the following: maxima, minima, or saddle points of f(x,y,z) under the given constraints.
edit: One of the resulting quadratics is factorable.
$$\frac{2x-2y}{x/2-2y/5} = \frac{2x+2z}{x/2+2z/25} \implies (x-y)(25x+4z)=(5x+5z)(5x-4y)$$ After expanding and eliminating the z variable (z=x+y), the quadratic reduces to: $$21x^2+10xy-16y^2=0$$ $$(3x-2y)(7x+8y)=0$$ Thus, we have $y=(3/2)x$ or $y=(-7/8)x$, hence, an explicit solution from the quadratic formula using the ellipsoid constraint. Plugging these back into f, we see a maximum at $f(2\sqrt{\frac{5}{19}},3\sqrt{\frac{5}{19}},5\sqrt{\frac{5}{19}})=10$.
| {
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Calculate $\int_0^{2\pi} \frac{\sin(t) + 4}{\cos(t) + \frac{5}{3}}dt$ I have to calculate $\int_0^{2\pi} \frac{\sin t + 4}{\cos t + \frac{5}{3}}dt$ using complex analysis.
I was thinking of setting $z(t) = re^{it} $ but I'm not sure what $r$ to pick or can I just pick any and is this even useful? Do I have to worry about the numerator of the integral? Before this I only had to calculate integral around some curve and then look at the singular values and use the residue theorem. Now it seems I have to do it the other way around?
| HINT: split the integral into two summands:
$$\int_0^{2\pi} \frac{\sin t + 4}{\cos t + \frac{5}{3}} dt = \int_0^{2\pi} \frac{\sin t}{\cos t + \frac{5}{3}} dt + \int_0^{2\pi} \frac{dt}{\cos t + \frac{5}{3}} =$$
$$=\left. -\log \left( \cos t + \frac{5}{3} \right) \right|_0^{2\pi} + 4\int_{|z|=1} \frac{1}{\frac{z+z^{-1}}{2} + \frac{5}{3}} \frac{dz}{iz}$$
Where you substitute $z=e^{it}$, so that $dz=ie^{it} dt= iz dt$ and $\cos t = \frac{e^{it}+e^{-it}}{2}=\frac{z+z^{-1}}{2}$.
Continuing, you get
$$0+ \frac{24}{i} \int_{|z|=1} \frac{dz}{(z+3)(3z+1)} = \frac{24}{i} \left(2\pi i \operatorname{Res} \left( \frac{1}{(z+3)(3z+1)} , -\frac{1}{3} \right)\right) = 48 \pi \frac{1}{8} = 6 \pi$$
| {
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How to find value of $x$ in this formula I have this formula:
$$1-\frac 1x=y$$
How do I invert this so that, if I have value of $y$, I want to find value of $x$.
I know, but I am pretty dense in math :(
I dont even know what category to put this under! Any help is appreciated!
| \begin{align}
1 - \frac{1}{x} &= y \\
1 &= \frac{1}{x} + y \\
1 - y &= \frac{1}{x} \\
x &= \frac{1}{1-y}
\end{align}
If $y = 1$, the above solution is undefined because there is no $x$ such that $1 - \frac{1}{x} = y$ because we would have
\begin{align}
1 - \frac{1}{x} &= 1 \\
\frac{1}{x} &= 0 \\
\end{align}
which is undefined (looking at a graph of $f(x) = \frac{1}{x}$ is another way to see this).
| {
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How can you derive $\sin(x) = \sin(x+2\pi)$ from the Taylor series for $\sin(x)$? \begin{eqnarray*}
\sin(x) & = & x - \frac{x^3}{3!} + \frac{x^5}{5!} - \ldots\\
\sin(x+2π) & = & x + 2\pi - \frac{(x+2π)^3}{3!} + \frac{(x+2π)^5}{5!} - \ldots \\
\end{eqnarray*}
Those two series must be equal, but how can you show that by only manipulating the series?
| Note that $\sin$ is the unique solution of $y'' = -y$ subject to $y(0) = 0, y'(0) = 1$. Note that $x \mapsto \sin (x+2 \pi)$ also satisfies the same
differential equation, so the question boils down to showing that
$\sin (2 \pi) = 0, \cos(2 \pi) = 1$ (where $\cos = \sin'$).
Consider $\eta(x) = \sin^2 x + \cos^2 x$, we see $\eta(0) = 1$ and $\eta(x)'= 0$, hence $\sin^2 x + \cos^2 x =1$ for all $x$, and so we only need to establish that $\cos (2 \pi) = 1$.
It is straightforward to establish that there is a smallest $T>0$ such that $
\cos T = 0$. Rudin has a nice method in "Real & Complex Analysis". Alternatively, suppose $\cos x >0$ for all $x \ge 0$. We have
$\cos 0 = 1$, and so we have some $x_0>0$ such that $\sin x_0 >0$. By
presumption, $\sin $ is strictly increasing on $x \ge 0$, and since $\cos' x = - \sin x$, we have $\cos' x \le - \sin x_0$ for all $x \ge x_0$,
which contradicts $\cos x >0$ for all $x \ge 0$. Since $\cos$ is continuous,
there is a smallest positive number $T$ such that $\cos T = 0$, and we can see that $\sin T = 1$, since $\sin$ is increasing on $[0,T)$.
We define $\pi = 2 T$.
Let $\zeta(x) = \cos ( {\pi \over 2}-x)$. We note that $\zeta(0) = 0, \zeta'(0) = 1$ and
$\zeta'' = - \zeta$, and so by uniqueness we have $\sin x= \cos ({\pi \over 2}-x) = \cos (x-{\pi \over 2})$ for all $x$ (since $\cos $ is even), and
differentiating gives
$\cos x = -\sin (x-{\pi \over 2})$ (since $\cos' = -\sin$). It is
straightforward to establish that $\cos (2 \pi) = 1$ from this.
Addendum: To show that the solution of $y''+y = 0$ is unique we could use the Picard Lindelöf theorem. Alternatively, suppose $y_1,y_2$ are two solutions with
the same initial conditions and let $\delta = y_1-y_2$. Let $V = \delta^2 + (\delta')^2$ and note that $V' = 0$. Since $V(0) = 0$, we have $V(x) = 0$ for all $x$, that is, $y_1 = y_2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233961",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "24",
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How prove this geometry inequality $R_1^4+R_2^4+R_3^4+R_4^4+R_5^4\geq {4\over 5\sin^2 108^\circ}S^2$ Zhautykov Olympiad 2015 problem 6 This links discusses the olympiad problem which none of students could solve , meaning it is very hard.
Question:
The area of a convex pentagon $ABCDE$ is $S$, and the circumradii of the triangles $ABC$, $BCD$, $CDE$, $DEA$, $EAB$ are $R_1$, $R_2$, $R_3$, $R_4$, $R_5$. Prove the inequality
$$R_1^4+R_2^4+R_3^4+R_4^4+R_5^4\geq {4\over 5\sin^2 108^\circ}S^2$$
A month ago,I tried to solve this problem,because I know the following Möbius theorem(1880):
Theorem:
Let $a,b,c,d,e$ are area of $\Delta ABC,\Delta BCD,\Delta CDE,\Delta DEA,\Delta EAB$,then we have
$$S^2-S(a+b+c+d+e)+ab+bc+cd+de+ea=0$$
This is a result by Gauss 1880,you can see this paper (for proof):(Gauss Carl Fridrich Gauss Werke Vol 4.2ter Abdruck,1880:406-407)
Using this Theorem ,in 2002 ,Chen Ji and Xiongbin proved this result:
$$a^2+b^2+c^2+d^2+e^2\ge\dfrac{20S^2}{(5+\sqrt{5})^2}$$
I use the well known result: $$a\le\dfrac{3\sqrt{3}}{4}R^2_{1},b\le\dfrac{3\sqrt{3}}{4}R^2_{2},c\le\dfrac{3\sqrt{3}}{4}R^2_{3},d\le\dfrac{3\sqrt{3}}{4}R^2_{4},e\le\dfrac{3\sqrt{3}}{4}R^2_{5}$$
so we have
$$R_1^4+R_2^4+R_3^4+R_4^4+R_5^4\ge\dfrac{16}{27}(a^2+b^2+c^2+d^2+e^2)
\ge\dfrac{16\cdot 20}{27(5+\sqrt{5})^2}S^2$$
But I found
$$\dfrac{16\cdot 20}{27(5+\sqrt{5})^2}<\dfrac{32}{5(5+\sqrt{5})}={4\over 5\sin^2 108^\circ}$$
so my work can't solve this contest problem. If we can solve following question,then the Olympiad problem can be solved:
Question 2:
$$a^2+b^2+c^2+d^2+e^2\ge\dfrac{54}{5(5+\sqrt{5})}S^2$$
| Dan Schwarz (one of the major problem proposers for EGMO, RMM, Balkan...) has posted a solution at here. I'll briefly sketch it here.
First, one takes the midpoints $M_1$, ..., $M_5$ of $A_1A_2$, ..., $A_5A_1$. Then at each angle $A_i$, one takes the circumcircle of triangle $M_{i-1}A_iM_{i+1}$ (which has radius $\frac 12 R_i$) and the point diametrically opposite $A_i$ on it, say $X_i$. This gives a quadrilateral $A_iA_{i-1}X_iA_{i+1}$, with area at most $(\frac 12 R_i)^2 \cdot 2\sin A_i$. Consider all five of these quadrilaterals, as shown.
It's not too hard to show that these five quadrilaterals cover the entire pentagon (look at perpendicular bisectors). So summing gives
$$ S \le \frac{1}{2} \sum_i R_i^2 \sin A_i. $$
Then, by the Cauchy-Schwarz Inequality, we have
$$ 4S^2 \le \left( \sum_i R_i^2 \sin A_i \right)^2 \le \left( \sum_i R_i^4 \right) \left( \sum_i \sin^2 A_i \right). $$
So it remains to show $$\sum_i \sin^2 A_i \le 5\sin^2 (108^{\circ})$$ reducing this to a purely algebraic problem (with $0^{\circ} \le A_i \le 180^{\circ}$ and $\sum A_i = 540^{\circ}$).
It's tempting to try and apply, say, Jensen's Inequality, but the function $\sin^2 x$ has a few inflection points, so one has to proceed more delicately using a fudging argument (along the lines of the $n-1$ equal-value trick). Indeed, we do this by showing that if $A_1 \le A_2 \le A_3 \le A_4 \le A_5$ and $A_1 < 108^{\circ}$, then we can replace $A_1$ with $108^{\circ}$ and $A_5$ by $A_1+A_5-108^{\circ}$ while increasing $\sum_i \sin^2 A_i$. Repeating this process until all $A_i$ are equal completes the proof.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
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Euclidean distance and dot product I've been reading that the Euclidean distance between two points, and the dot product of the two points, are related. Specifically, the Euclidean distance is equal to the square root of the dot product. But this doesn't work for me in practice.
For example, let's say the points are $(3, 5)$ and $(6, 9)$. The Euclidean distance is $\sqrt{(3 - 6)^2 + (5 - 9)^2}$, which is equal to $\sqrt{9 + 16}$, or $5$. However, the dot product is $(3 * 6 + 5 * 9)$, which is $63$, and the square root of this is not $5$.
What am I getting wrong?
| In $\mathbb{R}^{2}$, the distance between $\displaystyle X = \begin{pmatrix} x_{1} \\ x_{2} \end{pmatrix}$ and $\displaystyle Y = \begin{pmatrix} y_{1} \\ y_{2} \end{pmatrix}$ is defined as :
$$ d(X,Y) = \Vert Y - X \Vert = \sqrt{\left\langle Y-X,Y-X \right\rangle}. $$
with $\left\langle Y-X,Y-X \right\rangle = (y_{1}-x_{1})^{2} + (y_{2}-x_{2})^{2}.$
So, if $\displaystyle X=\begin{pmatrix} 3 \\ 5 \end{pmatrix}$ and $\displaystyle Y = \begin{pmatrix} 6 \\ 9 \end{pmatrix}$,
$$\begin{align*}
d(X,Y) &= {} \sqrt{\left\langle \begin{pmatrix} 3 \\ 4 \end{pmatrix},\begin{pmatrix} 3 \\ 4 \end{pmatrix}\right\rangle} \\[2mm]
&= \sqrt{9 + 16} \\[2mm]
&= \sqrt{25} \\[2mm]
&= 5.
\end{align*} $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
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Solving integral $\int \frac{3x-1}{\left(x^2+16\right)^3}$ I need to solve this one integral.
$$\int \frac{3x-1}{\left(x^2+16\right)^3}$$
You need to use the method of undetermined coefficients. That's what I get:
$$(3x-1) = (Ax + B)(x^{2}+16)^{2} + (Cx + D)(x^{2}+16) + (Ex + F)$$
$$1: 256B + 16D + F = -1$$
$$x: 256A + 16C + E = 3$$
$$x2: 32B + D = 0$$
$$x3: 32A + C = 0$$
$$x4: B = 0$$
$$x5: A = 0$$
$$A = 0;B = 0;C = 0;D = 0;E = 3;F = -1;$$
It turns out that I'm back to the same integral. What is wrong I do?
| As your integrand is already decomposed in partial fractions, start with the next step. The derivative of $x^2 + 16$ is $2x$, hence we have
$$ \int \frac{3x-1}{(x^2+ 16)^3}\, dx = \frac 32 \int\frac{2x}{(x^2 +16)^3}\,dx -\int\frac{1}{(x^2+16)^3}\, dx $$
In the first term, let $u = x^2+16$, giving
$$ \int \frac{2x}{(x^2+16)^3}\, dx = \int u^{-3}\, du = -\frac 12 u^{-2}
= -\frac 1{2(x^2 + 16)^2} $$
In the second term, we let $v = \frac x4$, giving
$$ \int \frac{1}{(x^2 + 16)^3}\, dx = 4\int \frac{1}{(16v^2 + 16)^3}\, dv
= \frac 1{1024}\int \frac{dv}{(v^2 + 1)^3} $$
Now we have by partial integration that
$$ \int \frac{dv}{(v^2 + 1)^\alpha} = \frac{v}{2(\alpha-1)(v^2+ 1)^{\alpha - 1}} + \frac{2\alpha -3}{2\alpha - 2}\int \frac{dv}{(v^2 + 1)^{\alpha -1}} $$
Hence
\begin{align*}
\int \frac{dv}{(v^2 + 1)^3} &= \frac v{4(v^2 + 1)^2} + \frac{3}4 \int \frac {dv}{(v^2 + 1)^2}\\
&= \frac v{4(v^2 + 1)^2} + \frac 34 \cdot \frac{v}{2(v^2 + 1)} + \frac 34 \cdot \frac 12 \int \frac{dv}{v^2 + 1} \\
&= \frac v{4(v^2 + 1)^2} +\frac{3v}{8(v^2 + 1)} + \frac 38 \arctan v
\end{align*}
Let $v = \frac x4$ again and you are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove the set $\mathcal{O}_d := \left\{\frac{a + b\sqrt{d}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\}$ is a ring.
Prove that if $d$ is a non-square integer with $d \equiv 1 \mod 4$ then the set $$\mathcal{O}_d := \left\{\frac{a + b\sqrt{d}}{2}:a,b \in \mathbb{Z}, a \equiv b \mod 2 \right\}$$ is a ring, and in particular a integral domain.
Little bit stuck here.
For $x,y \in \mathcal{O}_d,$ we define $+: \mathcal{O}_d \times \mathcal{O}_d \to \mathcal{O}_d$ by $(x + y) = \dfrac{a + b \sqrt{d}}{2} + \dfrac{a' + b'\sqrt{d}}{2} = \dfrac{(a+a') + (b+b')\sqrt{d}}{2} = \dfrac{a'+b'\sqrt{d}}{2} + \dfrac{a + b\sqrt{d}}{2} = (y + x) \in \mathcal{O}_d.$
We have that $0 = \dfrac{0 + 0\sqrt{d}}{2} \in \mathcal{O}_d$ and $\dfrac{a + b \sqrt{d}}{2} + \dfrac{-a + -b\sqrt{d}}{2} = \dfrac{(a-a) + (b-b)\sqrt{d}}{2} = 0 .$ Hence $-a \in \mathcal{O}_d.$ Also $$\begin{align} (x + y) + z &= \left(\dfrac{a + b\sqrt{d}}{2} + \dfrac{a' + b'\sqrt{d}}{2}\right) + \dfrac{a'' + b''\sqrt{d}}{2} \\&= \dfrac{(a + a') + (b + b')\sqrt{d}}{2} + \dfrac{a'' + b''\sqrt{d}}{2} \\&= \dfrac{(a + a' + a'') + (b + b' + b'')\sqrt{d}}{2} \\&= \dfrac{a + b\sqrt{d}}{2} + \dfrac{(a'+a'')+(b+b'')\sqrt{d}}{2} \\&= \dfrac{a + b\sqrt{d}}{2} + \left(\dfrac{a'+b'\sqrt{d}}{2} + \dfrac{a''+b''\sqrt{d}}{2}\right) \\&= x + (y+z) \end{align}$$
for all $x,y,z \in \mathcal{O}_d.$
Hence $\mathcal{O}_d$ is a group under addition.
I've had difficulty showing the multiplicative operation $\circ: \mathcal{O}_d \times \mathcal{O}_d \to \mathcal{O}_d$ stays in the set. How do we define the operation so that $x \circ y \in \mathcal{O}_d?$ When I multiply two elements from $\mathcal{O}_d$ I get $x \circ y = \left(\dfrac{a + b \sqrt{d}}{2} \right)\left(\dfrac{a' + b' \sqrt{d}}{2} \right) = \dfrac{(a + b \sqrt{d})(a' + b'\sqrt{d})}{4} = \dfrac{aa' + ab'\sqrt{d} +ba'\sqrt{d} + bb'd}{4}= \dfrac{aa' + bb'd + (ab' +ba')\sqrt{d}}{4}$ which doesn't seem to be an element of $\mathcal{O}_d.$
| $aa'+bb'd\equiv aa'+bb'\equiv aa'+aa'\equiv 0\bmod 2$, so $u=\dfrac{aa'+bb'd}{2}\in\mathbb Z$.
$ab'+ba'\equiv aa'+aa'\equiv 0\bmod 2$, so $v=\dfrac{ab'+ba'}{2}\in\mathbb Z$.
Now note that $x\circ y$ writes $\dfrac{u+v\sqrt d}{2}$ with $u,v\in\mathbb Z$.
In the following $a-b=2k$, $a'-b'=2k'$ and $d=1+4l$.
We have $u-v=\dfrac{aa'+bb'd}{2}-\dfrac{ab'+ba'}{2}=\dfrac{a'(a-b)+b'(bd-a)}{2}=\dfrac{2ka'+b'(-2k+4bl)}{2}=$ $ka'+b'(-k+2bl)=k(a'-b')+2bb'l=2kk'+2bb'l\equiv 0\bmod 2$, so $x\circ y\in\mathcal O_d$.
| {
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"timestamp": "2023-03-29T00:00:00",
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$a^2b + abc + a^2c + ac^2 + b^2a + b^2c + abc +bc^2$ factorisation
I came across this from a university mathematics resource page but they do not provide answer to this.
What I did was this:
$(a^2+b^2+c^2)(a+b+c) - (a^3 + b^3 + c^3) + 2abc$
But I don't think this is the correct solution.
How should I spot how to factorise expression here?
I wish to learn more of this seemingly complex and uncommon algebra factorisation.
Can you recommend me a book or a website for this?
There seem to be only common factorisations when I google.
Many thanks in advance,
Chris
| $$a^2b+abc+a^2c+ac^2+b^2a+b^2c+abc+bc^2$$
$$=ab(a+b+c)+ac(a+b+c)+b^2c+bc^2$$
$$=(a+b+c)(ab+ac)+bc(b+c)$$
$$=a(a+b+c)(b+c)+bc(b+c)$$
$$=(b+c)[a(a+b+c)+bc]$$
$$=(b+c)[a^2+ab+ab+bc]$$
$$=(b+c)[a(a+b)+c(a+b)]$$
$$=(b+c)(a+b)(a+c)$$
This is the simplest method I could think of. Hope it helps.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $6^n - 1$ is always divisible by $5$ by induction I'm trying to prove the following, but can't seem to understand it. Can somebody help?
Prove $6^n - 1$ is always divisible by $5$ for $n \geq 1$.
What I've done:
Base Case:
$n = 1$: $6^1 - 1 = 5$, which is divisible by $5$ so TRUE.
Assume true for $n = k$, where $k \geq 1$:
$6^k - 1 = 5P$.
Should be true for $n = k + 1$
$6^{k + 1} - 1 = 5Q$
$= 6 \cdot 6^k - 1$
However, I am unsure on where to go from here.
| This is the inductive step written out:
$$
6 \cdot 6^k - 1 = 5 \cdot Q |+1; \cdot \frac{1}{6};-1 \Leftrightarrow 6^k - 1 = \frac{5\cdot Q-5}{6}\underset{P}{\rightarrow}5\cdot P = \frac{5\cdot Q - 5 }{6} | \cdot \frac{1}{5}; \cdot 6\Leftrightarrow Q=6\cdot P + 1
$$
$$
6^k - 1 = \frac{5\cdot Q-5}{6} \overset{Q}{\rightarrow}\ (6^k-1 = \frac{5\cdot (6\cdot P + 1)-5}{6}\Leftrightarrow 6^k-1 = 5\cdot P)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
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Logic behind finding a $ (2 \times 2) $-matrix $ A $ such that $ A^{2} = - \mathsf{I} $. I know the following matrix "$A$" results in the negative identity matrix when you take $A*A$ (same for $B*B$, where $B=-A$):
$$A=\begin{pmatrix}0 & -1\cr 1 & 0\end{pmatrix}$$
However, I am not certain how one would go about finding this matrix without guessing and checking. Is there some systematic way of doing so? I have tried by assuming the following:
$$\begin{pmatrix}a & b\cr c & d\end{pmatrix}^2 = \begin{pmatrix}-1 & 0\cr 0 & -1\end{pmatrix}$$
...and then get the following equations:
$$a*a+b*c=-1$$
$$a*b+b*d=0$$
$$a*c+c*d=0$$
$$b*c+d*d=-1$$
This only tells me that $a=-d$, and then as best I can tell leaves both $c$ and $d$ without a solution, so perhaps this isn't the best method, but its the only approach that's coming to mind.
Any suggestions?
| If you assume that $A$ is invertible, then you can write:
$$
AA = -\mathbb{I}_2 \rightarrow A^{-1}AA = -A^{-1}\mathbb{I}_2 \rightarrow A = -A^{-1}\mathbb{I}_2
$$
But we know the inverse for a $2\times2$ matrix--and it's going to create a set of four linear equations for four variables:
$$
\begin{pmatrix}
a& b \\
c & d
\end{pmatrix} = \begin{pmatrix}
-d & b \\
c & -a
\end{pmatrix}
$$
This creates four equations:
\begin{align}
a = -d \rightarrow a + d = 0 \\
b = b \\
c = c \\
d = -a \rightarrow a + d = 0
\end{align}
This suggests that this is trues for all values of $a$ and $d$ such that $a + d = 0$ and any values of $b$ and $c$. Let's check:
$$
\begin{pmatrix}
a & x \\
y & -a
\end{pmatrix}
\times \begin{pmatrix}
a & x \\
y & -a
\end{pmatrix}
=
\begin{pmatrix}
a^2 + xy & ax - ax \\
ay - ay & xy + a^2
\end{pmatrix}
$$
Then it's a matter of solving the equation $a^2 + xy = -1$. For instance, we could choose $a = 2$ and $xy = -5$:
$$
\begin{pmatrix}
2 & -5 \\
1 & -2
\end{pmatrix}
\text{ and its inverse }
\begin{pmatrix}
-2 & -5 \\
1 & 2
\end{pmatrix}
\text{ both work}
$$
...obviously there are many more you could choose. A general way would be:
$$
\begin{pmatrix}
a & b \\
-\frac{a^2 + 1}{b} & -a
\end{pmatrix}
\text{ or }
\begin{pmatrix}
a & -\frac{a^2 + 1}{b} \\
b & -a
\end{pmatrix}
$$
It may be worth noting that if $b = 0$ then $a^2 + 1 = 0$ must also be true so that the other value, $\frac{a^2 + 1}{b}$, can take on any value. In this case we get: $a^2 = -1 \rightarrow a = \pm i$ and gives the equation $a^2 + 2xy = -1 \rightarrow -1 + 2*x * 0 = -1 \rightarrow -1 = -1$ (which is true for any value of $x$--the non-zero off diagonal). So this gives the other answer as well:
$$
\begin{pmatrix}
\pm i & 0 \\
x & \mp i
\end{pmatrix}
\text{ or }
\begin{pmatrix}
\pm i & x \\
0 & \mp i
\end{pmatrix}
$$
You can have arbitrarily complex components. For instance choose $a = 5i$. This gives: $a^2 + xy = -1 \rightarrow -25 + xy = -1 \rightarrow xy = 24$. So we could choose 6 and 4, 3 and 8, 2 and 12...or any two real values (or any two complex values--although they would have to be conjugates). Another complex example:
$$
\begin{pmatrix}
5i & 3 \\
8 & -5i
\end{pmatrix}
$$
or perhaps:
$$
\begin{pmatrix}
3i+2 & -4 \\
3i - 1 & -3i - 2
\end{pmatrix}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the number of values for $x$ and $y$? I have come across numerous questions where I am asked for example, if $x$ and $y$ are non-negative integers and $3x + 4y = 96$, how many pairs of $(x,y)$ are there?
Usually, I just use trial and error and look at the multiples of $3$ and $4$. However, I was wondering whether there is a more efficient way to solve such problems.
Help would be appreciated.
Thank you :)
| To solve your particular problem, you can see that what you need to find is all values of $x$ for which $96 - 3x$ is a positive multiple of $4$.
That means you want $96-3x$ to be divisible by $4$, and since $96$ is divisible by $4$, that means that $3x$ must be as well. Now, since $3$ and $4$ are coprime, $3x$ is divisible by $4$ if and only if $x$ is divisible by $4$, meaning that the possible values for $x$ are $0,4,8,12,\dots, 32$ (since, if $x>32$, then $3x>96$ meaning that $y<0$).
Now, for every $x$, simply find $y$ as $y=\frac{96-3x}{4}$ so that you get the pairs
$$x=0, y=24\\
x=4, y=21\\
x=8, y=18\\
x=12, y=15\\
x=16, y=12\\
x=20, y=9\\
x=24, y=6\\
x=28, y=3\\
x=32, y=0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to Find the Function of a Given Power Series? (Please see edit below; I originally asked how to find a power series expansion of a given function, but I now wanted to know how to do the reverse case.)
Can someone please explain how to find the power series expansion of the function in the title? Also, how would you do it in the reverse case? That is, given the power series expansion, how can you deduce the function $\frac{x}{1-x-x^3}$?
I've also rewritten the function as $$\frac{x}{1-x-x^3} = x \cdot \frac{1}{1-x(1-x^2)}$$ which is of the form of the Maclaurin series $\frac{1}{1-x} = 1+x+x^2+x^3+\ldots$
So the series expansion would then be $$x \cdot (1+(x-x^3)+(x-x^3)^2+(x-x^3)^3+\ldots)$$
However, expanding this to find the simplified power series expansion becomes complicated. According to WolframAlpha (link: http://www.wolframalpha.com/input/?i=power+series+of+x%2F%281-x-x%5E3%29) it eventually works out to $$x+x^2+x^3+2x^4+3x^5+4x^6+6x^7+9x^8+13x^9+19x^{10} +\ldots$$ which is a function defined recursively by $f_n = f_{n-1} + f_{n-3}$ for all $n\gt 3$ with the initial condition that $f_1=f_2=f_3=1$
Appreciate any and all help!
Thanks for reading,
A
Edit: I actually want to find the reverse of the original question. Given the recursively defined function (see above) $$x+x^2+x^3+2x^4+3x^5+4x^6+6x^7+9x^8+13x^9+19x^{10} +\ldots ,$$ how can I show that the function of this power series expansion is $\frac{x}{1-x-x^3}$?
| $f_{n+3}=f_{n+2}+f_{n}; n\geq 1$ then $\sum^{\infty}_{n=1}f_{n+3}x^n = \sum^{\infty}_{n=1}f_{n+2}x^n+\sum^{\infty}_{n=1}f_{n}x^n$ with $f(x)=\sum^{\infty}_{n=1}f_{n}x^n$ then $\frac{f(x)-f_1x-f_2x^2-f_3x^3}{x^3}= \frac{f(x)-f_1x-f_2x^2}{x^2}+f(x)$ then
$$\frac{f(x)-x-x^2-x^3}{x^3}= \frac{f(x)-x-x^2}{x^2}+f(x)$$ then
$$f(x)= \frac{x}{1-x-x^3}$$
| {
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Inconsistent Matrices I'm teaching myself Linear Algebra and am not sure how to approach this problem:
Let A be a 4×4 matrix, and let b and c be two vectors in R4. We are told that the system Ax = b is inconsistent. What can you say about the number of solutions of the system Ax = c?
Bretscher, Otto (2013-02-21). Linear Algebra with Applications (2-Download) (5th Edition) (Page 35). Pearson HE, Inc.. Kindle Edition.
I start with a generic 4x4 matrix, a vector x, and another generic vector b:
$$
\left[
\begin{array}{cccc}
a & b & c & d \\
e & f & g & h \\
i & j & k & l \\
m & n & o & p
\end{array}
\right]
\left[
\begin{array}{c}
x_{0} \\
x_{1} \\
x_{2} \\
x_{3}
\end{array}
\right]
=
\left[
\begin{array}{c}
b_{0} \\
b_{1} \\
b_{2} \\
b_{3}
\end{array}
\right]
$$
And the definition of inconsistent in terms of systems of equations. In the case of A, A would be inconsistent if one of its rows looked like this:
$$
\left[
\begin{array}{cccc}
0 & 0 & 0 & q
\end{array}
\right]
$$
Where q does not equal zero.
My first hangup is--what does it even mean for this equation to be inconsistent? Given the above definition of inconsistent, only matrices with more than one column can be inconsistent. If this is the case, then A must be inconsistent, b must be undefined, and therefore c must be undefined as well.
But the solutions at the end of the book say Ax = c has infinitely many solutions or no solutions.
What am I missing?
| Perhaps things will go faster with a simpler example. Consider the inconsistent equations
$$
\begin{align}
x + y &= 1 \\
x + y &= 0
\end{align}
$$
The linear system is
$$
\begin{align}
%
\mathbf{A} x &= b\\
%
\left[
\begin{array}{cc}
1 & 1 \\
1 & 1 \\
\end{array}
\right]
\left[
\begin{array}{c}
x \\
y \\
\end{array}
\right]
&=
\left[
\begin{array}{r}
1 \\
0 \\
\end{array}
\right]
%
\end{align}
$$
There are no exact solutions for this problem.
But you are asking about
$$
\begin{align}
%
\mathbf{A} x &= c\\
%
\left[
\begin{array}{cc}
1 & 1 \\
1 & 1 \\
\end{array}
\right]
\left[
\begin{array}{c}
x \\
y \\
\end{array}
\right]
&=
\left[
\begin{array}{r}
c_{1} \\
c_{2} \\
\end{array}
\right]
%
\end{align}
$$
A solution exists when
$$
\left[
\begin{array}{r}
c_{1} \\
c_{2} \\
\end{array}
\right]
=
\alpha
\left[
\begin{array}{r}
1 \\
1 \\
\end{array}
\right]
$$
Obivous solutions are
$$
\left[
\begin{array}{c}
x \\
y \\
\end{array}
\right]
=
\alpha
\left[
\begin{array}{c}
1 \\
0 \\
\end{array}
\right]
%
\qquad
%
\left[
\begin{array}{c}
x \\
y \\
\end{array}
\right]
=
\alpha
\left[
\begin{array}{c}
0 \\
1 \\
\end{array}
\right]
%
\qquad
%
\left[
\begin{array}{c}
x \\
y \\
\end{array}
\right]
=
\frac{\alpha}{2}
\left[
\begin{array}{c}
1 \\
1 \\
\end{array}
\right]
%
$$
Linear systems either have no solution (no existence), a unique solution (existence and uniqueness), or an infinitude of solutions (existence, no uniqueness). We must have a infinite number of solutions. They are
$$
\left[
\begin{array}{c}
x \\
y \\
\end{array}
\right]
=
\left[
\begin{array}{c}
x \\
\alpha - x \\
\end{array}
\right]
$$
This is the set
$$
(x,y)_{soln} = \left\{
(x,y) \in \mathbb{C}^{2} \colon y = \alpha - x
\right\}
$$
Summary: The inconsistent system will either present no solutions or an infinite number of solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find $\max _{z: \ |z|=1} \ f \left( z \right)$ for $f \left( z \right) = |z^3 - z +2|$ Let $f : C \mapsto R $, $f \left( z \right) = |z^3 - z +2|$. How find $\max _{z: \ |z|=1} \ f \left( z \right)$ ?
| $\bf{My\; Solution::}$ Let $z=x+iy\;,$ Then $|x+iy| = 1\Rightarrow x^2+y^2 =1 $
Now We have To Maximize $$f(z) = \left|z^3-z+2\right| = \left|(x+iy)^3-(x+iy)+2\right|$$
We Get
$$f(x,y) = \left|x^3-iy^3+3ix^2y-3xy^2-x-iy+2\right|$$
$$f(x,y)=\left|x(x^2-3y^2)+iy(3x^2-y^2)-x-iy+2\right|\;,$$ Using $x^2+y^2 = 1$
$$f(x,y) = \left|x(4x^2-3)+iy(4x^2-1)-x-iy+2\right|=\sqrt{[(4x^3-4x)-2]^2+(4x^2y)^2}$$
Now Let
$$g(x,y) = (4x^3-4x-2)^2+4x^4y^2 = (4x^3-4x-2)^2+4x^4(1-x^2)\;,$$ Using $x^2+y^2 = 1$
So Here we have to Maximize $$g(x) = (4x^3-4x-2)^2+4x^4-4x^6$$
Using Derivative Test, You Can Maximize It.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Area of an equilateral triangle
Prove that if triangle $\triangle RST$ is equilateral, then the area of $\triangle RST$ is $\sqrt{\frac34}$ times the square of the length of a side.
My thoughts:
Let $s$ be the length of $RT$. Then $\frac s2$ is half the length of $\overline{RT}$. Construct the altitude from the $S$ to side $\overline{RT}$. Call the intersection point $P$. Now, you have a right triangle whose sides are $|\overline{RP}| = \frac s2$ and $|\overline{RS}| = s$. By the Pythagorean Theorem, $|\overline{SP}| = \sqrt{s^2 - \frac14 s^2} = \sqrt{ \frac34 s^2} = \frac{\sqrt{3}}{2} s$. The area of the triangle is $$\frac12 \left(|\overline{SP}|\right)\left(|\overline{RT}|\right) = \left(\frac12 s\right) \left(\frac{\sqrt{3}}{2} s\right) = \frac{\sqrt{3}}{4}s^2,$$ as suggested.
| Let the lenght of the side be $s.$
In an equilateral triangle, the lengths of the sides are equal.
So, $RS=ST=TR=s.$
All angles are $\frac{π}{3}$ $radians.$
Area of a triangle as we know, is $$\frac{(RS)(ST)\sin(S)}{2}.$$
$$=\frac{s^2\sin(\frac{π}{3})}{2}$$
The area of the $ΔRST$ is thus, $$\frac{s^2\sqrt3}{4}$$
$Quod $ $erat $ $demonstrandum$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integration with Limits Find $$\displaystyle \lim_{n \to \infty} \int^{1}_{0}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx$$
Now, the solution was hinted like this:
using the property $$f(x)=f(2a-x)$$ the limit becomes half
$$2 \displaystyle \int^{\frac{1}{2}}_{0} (x^{n}+(1-x)^{n})^{\frac{1}{n}} dx$$.....$$(1)$$
plz see carefully.
for $$x\in (0,\frac{1}{2})$$
$$x$$ has smaller value than $$(1-x)$$.
taking common $$(1-x)^{n}$$ in $$(1)$$
also $$\displaystyle \lim_{n\to \infty} (\frac{x}{1-x})^{n} \rightarrow 0$$
the integral reduces to
$$2 \displaystyle \int^{\frac{1}{2}}_{0} (1-x) \times (1^{n})^{\frac{1}{n}}.dx$$
$$2 \displaystyle \int^{\frac{1}{2}}_{0} (1-x).dx$$
$$\frac{3}{4}$$
| $$y=(x^{n}+(1-x)^{n})^{\frac{1}{n}}=(1-x)\left(1+\left(\frac{x}{1-x}\right)^{n}\right)^{\frac{1}{n}}$$
Now, in $0\leq x \leq \frac{1}{2}$
$$0\leq \frac{x}{1-x} \leq 1$$
$$0\leq \left(\frac{x}{1-x}\right)^{n} \leq 1$$
$$1\leq \left(1+\left(\frac{x}{1-x}\right)^{n}\right)^{\frac{1}{n}} \leq 2^{\frac{1}{n}}$$
$$(1-x)\leq y \leq 2^{\frac{1}{n}}(1-x)$$
$$\int_0^{\frac{1}{2}}(1-x)dx \leq \int_0^{\frac{1}{2}} y\:dx \leq 2^{\frac{1}{n}}\int_0^{\frac{1}{2}}(1-x)dx$$
$$\frac{3}{8} \leq \int_0^{\frac{1}{2}} y\:dx \leq 2^{\frac{1}{n}} \frac{3}{8}$$
for $x$ tending to infinity $2^{\frac{1}{n}}$ tends to $1$. So, the limit is :
$$\int_0^{\frac{1}{2}} y\:dx =\frac{3}{8}$$
$$ \lim_{n \to \infty} \int^{1}_{0}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx=2\lim_{n \to \infty} \int^{\frac{1}{2}}_{0}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx=\frac{3}{4}$$
In addition. A detailed answer to the question raised in several remarks : How to prove the last equation above ?
$\int_{\frac{1}{2}}^{1}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx$, with change of variable $x=(1-X)$ or $X=(1-x)$ , then $dx=-dX$. The lower bounday $x=\frac{1}{2}$ becomes $X=1-\frac{1}{2}=\frac{1}{2}$, the upper boundary $x=1$ becomes $X=1-1=0$.
$$
\int_{\frac{1}{2}}^{1}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx= \int_{\frac{1}{2}}^{0}((1-X)^{n}+(1-(1-X))^{n})^{\frac{1}{n}}(-dX)$$
$$
\int_{\frac{1}{2}}^{1}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx= -\int_{\frac{1}{2}}^{0}((1-X)^{n}+X^{n})^{\frac{1}{n}}dX= \int^{\frac{1}{2}}_{0}((1-X)^{n}+X^{n})^{\frac{1}{n}}dX
$$
A basic property of the defined integrals : one can change the symbol of the dummy variable without changing the value of the integral. So, one is allowed to remplace $X$ by $x$ or any other symbol.
$$
\int_{\frac{1}{2}}^{1}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx= \int^{\frac{1}{2}}_{0}((1-x)^{n}+x^{n})^{\frac{1}{n}}dx=\frac{3}{8}
$$
$$
\int^{1}_{0}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx= \int^{\frac{1}{2}}_{0}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx+\int_{\frac{1}{2}}^{1}(x^{n}+(1-x)^{n})^{\frac{1}{n}}dx=\frac{3}{8}+\frac{3}{8}=\frac{3}{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How can $n^5+4$ be a perfect square? How can one find all $n \in \mathbb{N}$ such that $n^5+4$ is a perfect square?
I see that $n^5=(x+2)(x-2)$ here im suck can someone help ?
| Partial solution:
$d=\gcd(x+2,x-2)$ is $1$, $2$ or $4$.
If $d=1$ then $x+2$ and $x-2$ are perfect fifth powers. That is clearly impossible.
If $d=2$ then $x+2=2a$ and $x-2=2^4b=16b$ where $a$ is odd and $b$ is an integer (or vice versa). Either way, we have that $x=a+8b$, which is odd; a contradiction.
If $d=4$ then $x+2=4a$ and $x-2=4a-4$, where $a$ is some integer, that is, $x=4a-2$. Then $n$ is even, that is, $n=2m$ and
$$2m^5=a^2-a$$
There is at least one example, for $a=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1256466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $f=x^4-4x^2+16\in\mathbb{Q}[x]$ is irreducible
Prove that $f=x^4-4x^2+16\in\mathbb{Q}[x]$ is irreducible.
I am trying to prove it with Eisenstein's criterion but without success: for p=2, it divides -4 and the constant coefficient 16, don't divide the leading coeficient 1, but its square 4 divides the constant coefficient 16, so doesn't work. Therefore I tried to find $f(x\pm c)$ which is irreducible:
$f(x+1)=x^4+4x^3+2x^2-4x+13$, but 13 has the divisors: 1 and 13, so don't exist a prime number p such that to apply the first condition: $p|a_i, i\ne n$; the same problem for $f(x-1)=x^4+...+13$
For $f(x+2)=x^4+8x^3+20x^2+16x+16$ is the same problem from where we go, if we set p=2, that means $2|8, 2|20, 2|16$, not divide the leading coefficient 1, but its square 4 divide the constant coefficient 16; again, doesn't work.. is same problem for x-2
Now I'll verify for $f(x\pm3)$, but I think it will be fall... I think if I verify all constant $f(x\pm c)$ it doesn't work with this method... so have any idea how we can prove that $f$ is irreducible?
| Below is an explicit proof. Note that $x^4-4x^2+16 = (x^2-2)^2 + 12$, which clearly has no real root. Hence, the only possible way to reduce $x^4-4x^2+16$ over $\mathbb{Q}$ is $(x^2+ax+b)(x^2+cx+d)$. However, the roots of $x^4-4x^2+16$ are $x = \pm \sqrt{2 \pm i\sqrt{12}}$, which are all complex numbers. Since complex roots occur in conjugate pairs, $\sqrt{2 \pm i\sqrt{12}}$ must be the roots of one of the factored quadratic. Hence, the factored quadratic must be
$$x^2-(\sqrt{2+i\sqrt{12}}+\sqrt{2-i\sqrt{12}})x + \sqrt{2+ i\sqrt{12}}\sqrt{2- i\sqrt{12}} = x^2-2\sqrt3 x+4$$
The other factored quadratic must be
$$x^2+(\sqrt{2+i\sqrt{12}}+\sqrt{2-i\sqrt{12}})x + \sqrt{2+ i\sqrt{12}}\sqrt{2- i\sqrt{12}} = x^2+2\sqrt3 x+4$$
Hence, $x^4-4x^2+16$ is irreducible over $\mathbb{Q}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1260722",
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"source": "stackexchange",
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Area enclosed by an equipotential curve for an electric dipole on the plane I am currently teaching Physics in an Italian junior high school. Today, while talking about the electric dipole generated by two equal charges in the plane, I was wondering about the following problem:
Assume that two equal charges are placed in $(-1,0)$ and $(1,0)$.
There is an equipotential curve through the origin, whose equation is
given by:
$$\frac{1}{\sqrt{(x-1)^2+y^2}}+\frac{1}{\sqrt{(x+1)^2+y^2}}=2 $$ and
whose shape is very lemniscate-like:
Is there a fast&tricky way to compute the area enclosed by such a curve?
Numerically, it is $\approx 3.09404630427286$.
| At the first step, I will introduce a proper curve linear coordinates for this problem. This will help to construct the integral for area. We can write the equation of these equi-potential curves as
$$\frac{1}{r_1}+\frac{1}{r_2}=C \tag{1}$$
where $C$ is some real constant and $r_1$ and $r_2$ are defined as
$$r_1=\sqrt{(x-a)^2+y^2} \\
r_2=\sqrt{(x+a)^2+y^2} \tag{2}$$
where $2a$ is the distance between the two like charges on the $x$-axis placed at $x=a$ and $x=-a$. Equation $(2)$ is introducing a new curve-linear coordinates $(r_1,r_2)$ which is called the two-center bipolar coordinates. The geometric interpretation is easy as it just describes the coordinates of a point in $xy$ plane via the distance of that point through two other points which are called the centers. You can take look at this link in WIKI or this post on MSE. However, they doesn't contain that much information.
Then we find $x$ and $y$ from equations in $(2)$ in terms of $r_1$ and $r_2$. For this purpose, subtract and add the equations in $(4)$ to get
$$\begin{align}
r_2^2-r_1^2&=4ax \\
r_2^2+r_1^2&=2(x^2+y^2+a^2)
\end{align}
\tag{3}$$
after some simplifications, $x$ and $y$ in terms of $r_1$ an $r_2$ will be
$$
\begin{align}
x &= \frac{1}{4a} (r_{2}^{2}-r_{1}^{2})\\
y &= \pm \frac{1}{4a} \left( \sqrt{16 a^2 r_{2}^{2}-(r_{2}^{2}-r_{1}^{2}+4a^2)^2} \right)
\end{align}
\tag{4}$$
For a given $(r_1,r_2)$ we will find two pairs $(x,y)$ and $(x,-y)$. It is evident from the above picture that why this happens.
The next step will be the construction of the integral for area. We set $C=2a=2$ and find the intersection of the $\infty$ shaped curve with the $x$-axis
$$y=0 \qquad \to \qquad \frac{1}{\sqrt{(x-1)^2}}+\frac{1}{\sqrt{(x+1)^2}}=2 \qquad \to \qquad x=-\phi,0,\phi \tag{5}$$
where $\phi=\frac{1+\sqrt{5}}{2}$ is the golden ratio number. So according to $(1)$, $(4)$, and $(5)$ the parametric equations of the curve in the second quadrant of $xy$ plane will be
$$
\begin{align}
x &= \frac{1}{4} \left(r_{2}^{2}-\left(\frac{r_2}{2r_2-1}\right)^{2}\right)\\
y &= \frac{1}{4} \sqrt{16 r_{2}^{2}-\left(r_{2}^{2}-\left(\frac{r_2}{2r_2-1}\right)^{2}+4\right)^2}
\end{align}
\qquad \qquad \phi-1 \le r_2 \le 1
\tag{6}$$
Finally, the integral to be evaluated for the area will be
$$\begin{align}
\text{Area} &=4 \int_{-\phi}^{0} y dx \\
&=4 \int_{\phi-1}^{1}y \frac{dx}{dr_2}dr_2 \\
&= \int_{\phi-1}^{1} \sqrt{16 r_{2}^{2}-\left(r_{2}^{2}-\left(\frac{r_2}{2r_2-1}\right)^{2}+4\right)^2} \left(\frac{r_2}{2}-\frac{r_2}{2(2r_2-1)^2}+\frac{r_{2}^{2}}{(2r_2-1)^3}\right) dr_2
\end{align}$$
The numerical value of area up to fifty digits is
$$\text{Area}=3.0940463058814386237217800770286020796565427678113$$
as stated in the question. However, finding a tricky way to evaluate the definite integral of the area in a closed form should be investigated. I did the computations in this MAPLE file which may be useful for anyone who reads this post.
| {
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"timestamp": "2023-03-29T00:00:00",
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The residue of $9^{56}\pmod{100}$ How can I complete the following problem using modular arithmetic?
Find the last two digits of $9^{56}$.
I get to the point where I have $729^{18} \times 9^2 \pmod{100}$. What should I do from here?
| $\bmod 4\!:\ 9^{56}\equiv 1^{56}\equiv 1\,\Rightarrow\, 9^{56}=4m+1$.
$$\bmod 25\!:\ 9^{56}\stackrel{(1)}\equiv 9^{56\pmod{\phi(25)}}\equiv 9^{56\pmod {20}}\equiv 9^{-4}\equiv \left(\frac{1}{81}\right)^2$$
$$\equiv \left(\frac{1+5\cdot 25}{6+3\cdot 25}\right)^2\equiv \left(\frac{126}{6}\right)^2\equiv 21^2\equiv (-4)^2\equiv 16\equiv 4m+1$$
$$\iff 4m\equiv 15\equiv 40\iff m\equiv 10\iff m=25l+10$$
$9^{56}=4(25l+10)+1=100l+41$. In $(1)$ I used Euler's theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1262409",
"timestamp": "2023-03-29T00:00:00",
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What is the connection between the discriminant of a quadratic and the distance formula? The $x$-coordinate of the center of a parabola $ax^2 + bx + c$ is $$-\frac{b}{2a}$$
If we look at the quadratic formula
$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
we can see that it specifies two points at a certain offset from the center
$$-\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$
This means that $\frac{\sqrt{b^2 - 4ac}}{2a}$ is the (horizontal) distance from the vertex to the roots. If I squint, the two squared-ish quantities being subtracted under a square root sign reminds me of the Euclidean distance formula
$$\sqrt{(x_0 - x_1)^2 + (y_0 - y_1)^2}$$
Is there a connection? If not, is there any intuitive or geometric reason why $\frac{\sqrt{b^2 - 4ac}}{2a}$ should be the horizontal distance from the vertex to the roots?
| We have $$f(x)=ax^2+bx+c=a(x+\frac{b}{2a})^2+\frac{4ac-b^2}{4a}$$ with $b^2-4ac\geq 0$ and $a\neq 0$. The vertex of the parabola is $$V\left(-\frac{b}{2a},\frac{4ac-b^2}{4a}\right)$$ and the root(s) are at $$R_1\left(-\frac{b}{2a}+\sqrt{\frac{b^2-4ac}{4a^2}},0\right)~\text{and}~R_2\left(-\frac{b}{2a}-\sqrt{\frac{b^2-4ac}{4a^2}},0\right).$$
*
*If $b^2-4ac=0$, then we only have one root at $x=-\frac{b}{a}$. Thus we get the point $V\left(-\frac{b}{2a},0\right)$, which simultaneously is the vertex of the parabola, meaning that the horizontal distance from the vertex to the root is $0$.
*If $b^2-4ac>0$, then we have two different roots and the vertex of the parabola does not lie on the x-axis. But as we are only interested in the horizontal distance we again look at $V_x\left(-\frac{b}{a},0\right)$. Now we can apply the formula for the eucledian distance to $V_x$ and $R_1$:
$$d(V_x,R_1)=\sqrt{\left(-\frac{b}{2a}+\sqrt{\frac{b^2-4ac}{4a^2}}-\left(-\frac{b}{2a}\right)\right)^2+\left(0-0\right)^2}=\sqrt{\frac{b^2-4ac}{4a^2}}.$$
For $V_x$ and $R_2$ we get
$$d(V_x,R_2)=\sqrt{\left(-\frac{b}{2a}-\sqrt{\frac{b^2-4ac}{4a^2}}-\left(-\frac{b}{2a}\right)\right)^2+\left(0-0\right)^2}=\sqrt{\left(-\frac{b^2-4ac}{4a^2}\right)^2}=\sqrt{\frac{b^2-4ac}{4a^2}}.$$
So $$d=\sqrt{\frac{b^2-4ac}{4a^2}}$$ is indeed the horizontal distance between the vertex and the roots.
This is also expressed in your formula, as $$-\frac{b}{2a}\pm \sqrt{\frac{b^2-4ac}{4a^2}}$$ only gives you the $x$-coordinate of the roots. As $-\frac{b}{2a}$ is the $x$-coordinate of the vertex, for the roots you only add/subtract $\sqrt{\frac{b^2-4ac}{4a^2}}$, meaning you go "left/right" starting at $V_x$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "19",
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Help me understand this solution $$2x^4yy'+y^4=4x^6$$
The way my teacher did it is:
First, he made a substitution:
$y=z^m$
$y'=mz^{m-1}z'$
$$2x^4 z^m mz^{m-1} z'+z^{4m}=4x^6$$
$$z'=\frac{4x^6-z^{4m}}{2mx^4z^{2m-1}}=f\left(\frac{z}{x}\right)$$
Then he wrote the part I don't understand (how he got $m$):
$2m-1=2$
$2m+1=4$
$m=\frac{3}{2}$ - for this $m$, the equation becomes homogeneous.
So we make a substitution: $y=z^{\frac{3}{2}}$
$$y'=\frac{3}{2}z^{\frac{1}{2}} z'$$
$$3x^4z^2z'+z^6=4x^6$$
Substitution: $z(x)=u(x) x$
$$\frac{u^2}{u-u^6-3u^3}du=\frac{dx}{3x}$$
We multiply by $-3$:
$$\frac{3u^2}{u^6+3u^3-u}du=-\frac{dx}{x}$$
Substitution: $u^3=t$, $3u^2\,du=dt$
$$\frac{x^3-y^2}{4x^3-y^2}=\frac{c1}{x^5}$$
$c1=c$ or $c1=-c$
So, all in all, the part I don't understand is how he got m.
| here ia nother way to do this problem.
a change of variable $y = u^k$ in $$2x^4 yy'+y^4 = 4x^6 \to 2kx^4u^{2k-1}u'+u^{4k} = 4x^6$$ now we will choose $k$ so that $2k-1 = 4k \to k = -1/2.$ and we have $$-x^4u'+u= 4x^6 \to e^{x^3/3}u' - \frac{e^{x^3/3}}{x^4}u = -4x^6e^{x^3/3} $$ on integration, we get $$u =-4e^{-x^3/3}\int 4x^6e^{x^3/3}\, dx, \quad y = \frac1{u^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the inverse of a number under a certain modulus How does one get the inverse of 7 modulo 11?
I know the answer is supposed to be 8, but have no idea how to reach or calculate that figure.
Likewise, I have the same problem finding the inverse of 3 modulo 13, which is 9.
| Here's an illustration of finding the multiplicative inverse of $37 \bmod 100$ using the extended Euclidean algorithm. (I used bigger numbers for this example so that the relationships are a little clearer).
On each line, $n=100s+37t$. We start the table with two lines giving $n=100$ and $n=37$ in the obvious way. Then $q$ gives the rounded-down ratio of the current and previous value of $n$. Using this, the next line is calculated by subtracting $q$ copies of the current line entries from the previous line entries.
$$\begin{array}{c|c|c|c}
n & s & t & q \\ \hline
100 & 1 & 0 & \\
37 & 0 & 1 & 2 \\
26 & 1 & -2 & 1 \\
11 & -1 & 3 & 2 \\
4 & 3 & -8 & 2 \\
3 & -7 & 19 & 1 \\
1 & 10 & \color{red}{-27} & 3 \\
\end{array}$$
On the last line, $1 = 10\times 100 + (-27)\times 37$, so $\color{red}{-27}\times 37 \equiv 1 \bmod 100$
Bringing the result positive, $-27 \equiv \color{red}{73} \bmod 100$. And $37 \times 73 = 2701 \equiv 1 \bmod 100 \quad \checkmark$.
As you can perhaps see, you don't actually need to calculate the $s$ values at all to get the modular inverse. I left them in to help understand the table.
The corresponding tables for your particular questions:
$$\begin{array}{c|c|c|c}
n & s & t & q \\ \hline
11 & 1 & 0 & \\
7 & 0 & 1 & 1 \\
4 & 1 & -1 & 1 \\
3 & -1 & 2 & 1 \\
1 & 2 & \color{red}{-3} & 3 \\
\end{array}$$
and $7^{-1} \equiv -3 \equiv \color{red}8 \bmod 11$. $\quad 7\times 8 = 56 = 55+1\quad \checkmark$
$$\begin{array}{c|c|c|c}
n & s & t & q \\ \hline
13 & 1 & 0 & \\
3 & 0 & 1 & 4 \\
1 & 1 & \color{red}{-4} & 3 \\
\end{array}$$
and $3^{-1} \equiv -4 \equiv \color{red}9 \bmod 13$. $\quad 3\times 9 = 27 = 26+1\quad \checkmark$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1266282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
For $n>2, n\in\mathbb{Z}$, why is this true: $\left\lfloor 1/\left(\frac{1}{n^2}+\frac{1}{(n+1)^2}+\cdots+\frac{1}{(n+n)^2}\right)\right\rfloor=2n-3$ Let $n>2$ be a positive integer, prove that
$$\left\lfloor \dfrac{1}{\dfrac{1}{n^2}+\dfrac{1}{(n+1)^2}+\cdots+\dfrac{1}{(n+n)^2}}\right\rfloor=2n-3?$$
before I use hand Calculation $n=2,3,4$,maybe I calculation some wrong,I can't use computer it!
so maybe this result is right. But how to prove the statement?
| Asymptotically,
$$ \sum_{i=n}^{2n} \dfrac{1}{i^2} = \dfrac{1}{2n} + \dfrac{5}{8n^2} + O(1/n^3)$$
so
$$ \dfrac{1}{\displaystyle\sum_{i=n}^{2n} \dfrac{1}{i^2}} = 2n - \dfrac{5}{2} + O(1/n) $$
Thus your equation will be true for sufficiently large $n$. With sufficiently good explicit bounds on the $O(1/n^3)$ term, you should be able to prove that it is true for all $n \ge 5$.
EDIT: For an explicit upper bound, since $1/x^2$ is convex,
$$\dfrac{1}{i^2} \le \int_{i-1/2}^{i+1/2} \dfrac{dt}{t^2}$$
so
$$ \sum_{i=n}^{2n} \dfrac{1}{i^2} \le \int_{n-1/2}^{2n+1/2} \dfrac{dt}{t^2} =
\dfrac{4(n+1)}{(2n-1)(4n+1)} $$
Call this upper bound $U(n)$. We have
$$ \dfrac{1}{\sum_{i=n}^{2n} \dfrac{1}{i^2} } \ge \dfrac{1}{U(n)} = 2 n - \dfrac{5}{2} + \dfrac{9}{4(n+1)} > 2 n - \dfrac{5}{2}$$
For a lower bound, consider
$$ g(i) = \int_{i-1/2}^{i+1/2} \left( \dfrac{1}{t^2} - \dfrac{1}{4t^4}\right)\; dt
= \dfrac{4 (4 i^2 - 16 i - 1)}{(4 i^2 - 1)^2} $$
Now
$$ \dfrac{1}{i^2} - g(i) = \dfrac{64 i^3 - 4 i^2 + 1}{(4 i^2-1)^2 i^2} > 0$$
for $i \ge 1$. Thus
$$ \sum_{i=n}^{2n} \dfrac{1}{i^2} \ge \int_{n-1/2}^{2n+1/2} \left(\dfrac{1}{t^2} - \dfrac{1}{4t^4}\right)\; dt = {\frac {8 \left( n+1 \right) \left( 96\,{n}^{4}-48\,{n}^{3}-32\,{
n}^{2}+5\,n+1 \right) }{3 \left( 2\,n - 1 \right) ^{3} \left( 4\,n+1
\right) ^{3}}}
$$
which I will call $L(n)$. Now
$$\dfrac{1}{L(n)} - (2n - 2) = -{\frac {384\,{n}^{5}-1760\,{n}^{4}+584\,{n}^{3}+492\,{n}^{2}-62
\,n-13}{ 8 \left( n+1 \right) \left( 96\,{n}^{4}-48\,{n}^{3}-32\,{n}^{2
}+5\,n+1 \right) }}
$$
which is negative for $n \ge 5$ (you can verify using Sturm's theorem that the numerator and denominator have no zeros for $n \ge 5$).
Thus $$\dfrac{1}{\displaystyle \sum_{i=n}^{2n} \dfrac{1}{i^2}} < 2n-2 \ \text{for}\ n \ge 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268447",
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"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to solve this nonstandard system of equations? How to solve this system of equations
$$\begin{cases}
2x^2+y^2=1,\\
x^2 + y \sqrt{1-x^2}=1+(1-y)\sqrt{x}.
\end{cases}$$
I see $(0,1)$ is a root.
| Solution.
First way
From the first equation, we have
$$\begin{cases}
2x^2\leqslant 1,\\
y^2 \leqslant 1
\end{cases}
\Leftrightarrow
\begin{cases}
-\dfrac{1}{\sqrt{2}} \leqslant x \leqslant \dfrac{1}{\sqrt{2}},\\
- 1 \leqslant y \leqslant 1.
\end{cases}$$
Then, the conditions of $x$ and $y$ are
$$\begin{cases}
0 \leqslant x \leqslant \dfrac{1}{\sqrt{2}},\\
- 1 \leqslant y \leqslant 1.
\end{cases}$$
We have $x^2 + y^2 = 1-x^2$. Therefore $x^2 + y^2 \leqslant 1$.
Another way,
$$1-x^2 = y \sqrt{1-x^2} -(1-y)\sqrt{x} \leqslant y \sqrt{1-x^2}.$$
Because
$$ y \sqrt{1-x^2} \leqslant \dfrac{y^2 + 1 - x^2}{2}.$$
Implies
$$1-x^2 \leqslant \dfrac{y^2 + 1 - x^2}{2} \Leftrightarrow x^2 + y^2 \geqslant 1 .$$
From $x^2 + y^2 \leqslant 1$ and $x^2 + y^2 \geqslant 1$, we have $x^2 + y^2 = 1.$
Solve
$$\begin{cases}
x^2 + y^2 = 1,\\
2x^2 + y^2 = 1,\\
0 \leqslant x \leqslant 1,\\
- 1 \leqslant y \leqslant \dfrac{1}{\sqrt{2}}
\end{cases} \Leftrightarrow \begin{cases}
x = 0,\\
y = 1.\end{cases}$$
Second way.
We have $2x^2 + y^2 = 1$, therefore $y=\sqrt{1 - 2x^2}$ or $y=-\sqrt{1 - 2x^2}.$
First case, $y=\sqrt{1 - 2x^2}$, subtitution the second equation, we get
$$x^2 + \sqrt{1 - 2x^2}\cdot\sqrt{1-x^2}=1+(1-\sqrt{1 - 2x^2})\sqrt{x}.$$
equavalent to
$$1 - x^2 - \sqrt{1 - 2x^2}\cdot\sqrt{1-x^2} + (1-\sqrt{1 - 2x^2})\sqrt{x}=0$$
or
$$\sqrt{1-x^2} (\sqrt{1-x^2} - \sqrt{1 - 2x^2}) + (1-\sqrt{1 - 2x^2})\sqrt{x}=0.$$
This is equavalent to
$$\dfrac{\sqrt{1-x^2}\cdot(1-x^2-1+2x^2)}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}}+\dfrac{(1-1+2x^2)\sqrt{x}}{1+\sqrt{1 - 2x^2}} = 0$$
or
$$x^2\left (\dfrac{\sqrt{1-x^2}}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}} + \dfrac{2\sqrt{x}}{1+\sqrt{1 - 2x^2}}\right )=0.$$
It is easy to see that
$$\dfrac{\sqrt{1-x^2}}{\sqrt{1-x^2}+\sqrt{1 - 2x^2}} + \dfrac{2\sqrt{x}}{1+\sqrt{1 - 2x^2}}> 0$$
Thus $x = 0$.
Second cases. $y=-\sqrt{1 - 2x^2}.$
We can check that
$$x^2 + y \sqrt{1-x^2} \leqslant \dfrac{1}{2}$$
and
$$1+(1-y)\sqrt{x} >1.$$
In this case, the given system of equations has no solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
For which values of $a,b,c$ is the matrix $A$ invertible? $A=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}$
$$\Rightarrow\det(A)=\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix}-\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix}+\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix}\\=ab^2-a^2b-ac^2+a^2c+bc^2-b^2c\\=a^2(c-b)+b^2(a-c)+c^2(b-a).$$
Clearly, $$\left\{\det(A)\neq0\left|\begin{matrix}c\neq b\\a\neq c\\b\neq a\\a,b,c\neq 0\end{matrix}\right.\right\}\\$$
Is it sufficient to say that the matrix is invertible provided that all 4 constraints are met? Would Cramer's rule yield more explicit results for $a,b,c$ such that $\det(A)\neq0$?
| The matrix is the special case of so-called Vandermonde Matrix (See Induction for Vandermonde Matrix).
det($A$)$=(a-b)(a-c)(b-c)$. So if $a\neq b,a\neq c, b\neq c, \space$ det$(A)\neq 0$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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If $\frac{a}{b}=\frac{b}{c}=\frac{c}{d}$, prove that $\frac{a}{d}=\sqrt{\frac{a^5+b^2c^2+a^3c^2}{b^4c+d^4+b^2cd^2}}$ What I've done so far;
$$\frac{a}{b}=\frac{b}{c}=\frac{c}{d}=k\\
a=bk, b=ck, c=dk\\
a=ck^2, b=dk^2\\
a=dk^3$$
I tried substituting above values in the right hand side of the equation to get $\frac{a}{d}$ but couldn't.
| You did fine. Repalicng your values in the term inside the radical, you have
$$a^5+b^2c^2+a^3c^2=d^5 k^{15}+d^5 k^{11}+d^4 k^6$$ $$b^4c+d^4+b^2cd^2=d^5 k^9+d^5 k^5+d^4$$ thet is to say that the ratio is $k^6$, its square root is $k^3$ which is $\frac ad$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
$\lim\limits_{h\to0} \frac{(x+h)^3 -x^3}{h}$ Compute the following limit I can not seem to figure out the direction to go with this problem. I'm not quite sure how to break it up.
| Others have pointed out here that $(x+h)^3 = x^3+3x^2h+3xh^2+h^3$.
Here's different approach. Recall that $a^3-b^3=(a-b)(a^2+ab+b^2)$. Consequently
\begin{align}
(x+h)^3-x^3 & = \Big((x+h)-x\Big)\Big( (x+h)^2 +(x+h)x + x^2 \Big) \\[12pt]
& = h\Big( (x+h)^2 +(x+h)x + x^2 \Big).
\end{align}
Now we might be tempted to do the routine simplifications of the expression in the large parentheses, but it's quicker to do this:
$$
\frac{(x+h)^3-x^3} h = \frac{h\Big( (x+h)^2 +(x+h)x + x^2 \Big)} h = (x+h)^2 +(x+h)x + x^2.
$$
We want the limit of that as $h\to0$, and now we can just plug in $h=0$ and get $x^2+x^2+x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1271900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$
Prove that number of zeros at the right end of the integer $(5^{25}-1)!$ is $\frac{5^{25}-101}{4}.$
Attempt:
I want to use the following theorem:
The largest exponent of $e$ of a prime $p$ such that $p^e$ is a divisor of $n!$ is given by $$e=[\frac{n}{p}]+[\frac{n}{p^2}]+[\frac{n}{p^3}]+\cdots$$.
The number of times the prime divisor $5$ is repeated in $(5^{25}-1)!$ equals the greatest exponent of $5$ contained in $(5^{25}-1)!$, which is $e_1=[\frac{(5^{25}-1)}{5}]+[\frac{(5^{25}-1)}{5^2}]+[\frac{(5^{25}-1)}{5^3}]+[\frac{(5^{25}-1)}{5^4}]+\cdots$
The number of times the prime divisor $2$ is repeated in $(5^{25}-1)!$ equals the greatest exponent of $2$ contained in $(5^{25}-1)!$, which is $e_2=[\frac{(5^{25}-1)}{2}]+[\frac{(5^{25}-1)}{2^2}]+[\frac{(5^{25}-1)}{2^3}+[\frac{(5^{25}-1)}{2^4}]+\cdots$
Therefore, the number of zeros at the right end equals the greatest exponent of $10$ contained in $(5^{25}-1)!$$=\min{(e_1, e_2)}$.
I am unable to simplify $e_1, e_2$. Please help me.
| Actually it would be:
$$\small e_1=\sum_{k=1}^{\infty} \left\lfloor\frac{5^{25}-1}{5^k}\right\rfloor=\sum_{k=1}^{\infty} \left\lfloor5^{25-k}-\frac1{5^k}\right\rfloor=\sum_{k=1}^{25} (5^{25-k}-1)\tag{$\because \left\lfloor-\frac1{5^k}\right\rfloor=-1$}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1272660",
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"source": "stackexchange",
"question_score": "3",
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Diagonalize tri-diagonal symmetric matrix How to diagonalize the following matrix?
\begin{pmatrix}
2 & -1 & 0 & 0 & 0 & \cdots \\
-1 & 2 & -1 & 0 & 0 & \cdots \\
0 & -1 & 2 & -1 & 0 & \cdots \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
\cdots & 0 & 0 & -1 & 2 & -1 \\
\cdots & 0 & 0 & 0 & -1 & 2 \\
\end{pmatrix}
It seems like we need to first compute eigenvalues and eigenvectors for this matrix, like the following:
\begin{vmatrix}
2 - \lambda & -1 & 0 & 0 & 0 & \cdots \\
-1 & 2 - \lambda & -1 & 0 & 0 & \cdots \\
0 & -1 & 2 - \lambda & -1 & 0 & \cdots \\
\cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\
\cdots & 0 & 0 & -1 & 2 - \lambda & -1 \\
\cdots & 0 & 0 & 0 & -1 & 2 - \lambda \\
\end{vmatrix}
But I just don't know what to do next.
| Let $\mathbf{x}$ be the eigenvector of $A$ with eigenvalue $\lambda$.
$$
A \mathbf{x} = \lambda \mathbf{x}
$$
Note that $\mathbf{x}$ is a solution to the following difference equation:
$$
-x_{k+1} + 2x_k - x_{k-1} = \lambda x_k,\quad k = 1, \dots, n\\
x_0 = 0,\quad x_{n+1} = 0.
$$
The characteristic equation for that linear reccurence is
$$
q^2 - (2 -\lambda) q + 1 = 0
$$
Observe that $q_1 q_2 = 1$ due to Viète theorem. The discriminant of the quadratics is $\mathscr D = (2 - \lambda)^2 - 4 = \lambda^2 - 4 \lambda$. Let's consider different cases for $\mathscr D = 0$, $\mathscr D > 0$ and $\mathscr D < 0$.
Case 1a. $\lambda = 0$. The $q_{1,2} = 1$, so the general solution is
$$
x_k = C_1 k + C_2.
$$
From the boundary conditions we conclude that $C_1 = C_2 = 0$. The solution is trivial, so $\lambda = 0$ is not an eigenvalue.
Case 1b. $\lambda = 4$. The $q_{1,2} = -1$ and the solution is
$$
x_k = (-1)^k (C_1 k + C_2).
$$
Again, $C_1 = C_2 = 0$ and $\lambda = 4$ is not an eigenvalue.
Case 2a. $\lambda < 0$ or $\lambda > 4$. $q_{1,2} \in \mathbb{R}, |q_1| > 1, |q_2| < 1$. Generic solution is
$$
x_k = C_1 q_1^k + C_2 q_2^k
$$
From left condition $C_1 = C_2$. Let $C_1 = 1$
$$
x_{n+1} = q_1^{n+1} + q_2^{n+1} = 0 \implies q_1^{n+1} = -q_2^{n+1}.
$$
If $n$ is even then $q_1 = -q_2$. But then $|q_1| = |q_2|$ and we know that $|q_1| > 1 > |q_2|$. Contradiction, no solution is possible.
If $n$ is odd then sides of $q_1^{n+1} = -q_2^{n+1}$ have different sign (left is positive, right is negative or zero). Also no solutions here.
Case 3. $0 < \lambda < 4$. The $q_{1,2} = \frac{2-\lambda \pm \sqrt{(2-\lambda)^2 - 4}}{2} = e^{\pm i\phi}, \phi \in \mathbb{R}$. General solution in this case will be
$$
x_k = C_1 \cos \phi k + C_2 \sin \phi k.
$$
From $x_0 = 0$ we obtain $C_1 = 0$. Let $C_2 = 1$.
$$
x_{n+1} = \sin \phi (n+1) = 0
$$
This equation has exactly $n$ different solutions which give different $x_k$ vectors.
$$
\phi_m = \frac{\pi m}{n+1},\quad m = 1, \dots, n
$$
So
$$
x^{(m)}_k = \sin \frac{\pi m k}{n+1}\\
\lambda^{(m)} = 4 \sin^2 \frac{\pi m}{2(n+1)}
$$
will be the $n$ different eigenpairs.
Edit So I actually didn't show how to diagonalize $A$. Here it is.
Since the matrix $A$ is symmertic, all its eigenvalues are orthogonal.
$$
(x^{m}_k, x^{\ell}_k) = \sum_{k=1}^{n} \sin \frac{\pi m k}{n+1} \sin \frac{\pi \ell k}{n+1} = \begin{cases} \frac{n+1}{2} & m = \ell\\0 & m \neq \ell\end{cases}
$$
It is the same functions that form discrete Fourier basis (discrete sine transform to be precise).
Let's make eigenvectors orthonormal (they are only orthogonal for now). Divide them by $\sqrt{\frac{n+1}{2}}$:
$$
\tilde{x}^{(m)}_k =\sqrt{\frac{2}{n+1}} \sin \frac{\pi m k}{n+1}, \qquad (\tilde{x}^{(m)}_k, \tilde{x}^{(\ell)}_k) = \delta_{m,\ell}
$$
Putting the $\tilde{x}^{(m)}_k$ vectors in matrix $U$
$$
U = \begin{pmatrix}
\tilde{x}^{(1)}_1 & \tilde{x}^{(2)}_1 & \dots & \tilde{x}^{(n)}_1 \\
\tilde{x}^{(1)}_2 & \tilde{x}^{(2)}_2 & \dots & \tilde{x}^{(n)}_2 \\
\vdots & \vdots & \ddots & \vdots \\
\tilde{x}^{(1)}_n & \tilde{x}^{(1)}_n & \dots & \tilde{x}^{(n)}_n \\
\end{pmatrix}, \qquad U_{ij} = \sqrt{\frac{2}{n+1}} \sin \frac{\pi i j}{(n + 1)}
$$
Matrix $U$ is orthogonal, so $U^\mathsf{T} = U^{-1}$. Also $U_{ij} = U_{ji}$, so $U^\mathsf{T} = U$. Writing the $A \mathbf{x} = \lambda \mathbf{x}$ in matrix form we have
$$
A U = U \Lambda
$$
where $$\Lambda = \operatorname{diag} \begin{pmatrix}
4\sin^2 \frac{\pi}{2(n+1)} &
4\sin^2 \frac{2\pi}{2(n+1)} & \dots &
4\sin^2 \frac{n\pi}{2(n+1)}
\end{pmatrix}.$$
Hence,
$$
A = U \Lambda U^T = U \Lambda U\\
\Lambda = U A U
$$
| {
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"url": "https://math.stackexchange.com/questions/1274739",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How does $-\frac{1}{x-2} + \frac{1}{x-3}$ become $\frac{1}{2-x} - \frac{1}{3-x}$ I'm following a solution that is using a partial fraction decomposition, and I get stuck at the point where $-\frac{1}{x-2} + \frac{1}{x-3}$ becomes $\frac{1}{2-x} - \frac{1}{3-x}$
The equations are obviously equal, but some algebraic manipulation is done between the first step and the second step, and I can't figure out what this manipulation could be.
The full breakdown comes from this solution
$$
\small\begin{align}
\frac1{x^2-5x+6}
&=\frac1{(x-2)(x-3)}
=\frac1{-3-(-2)}\left(\frac1{x-2}-\frac1{x-3}\right)
=\bbox[4px,border:4px solid #F00000]{-\frac1{x-2}+\frac1{x-3}}\\
&=\bbox[4px,border:4px solid #F00000]{\frac1{2-x}-\frac1{3-x}}
=\sum_{n=0}^\infty\frac1{2^{n+1}}x^n-\sum_{n=0}^\infty\frac1{3^{n+1}}x^n
=\bbox[4px,border:1px solid #000000]{\sum_{n=0}^\infty\left(\frac1{2^{n+1}}-\frac1{3^{n+1}}\right)x^n}
\end{align}
$$
Original image
| This problem all boils down to the following relationship $$-1 = \frac{-1}{1}=\frac{1}{-1}$$
Part one is easy if you just express the division as a multiplication
$$x=\frac{-1}{1}\implies -1=1\cdot x\implies -1=x$$
For part two,
$$x=\frac{1}{-1}\implies1=-1\cdot x$$
$$1+(-1)+x=-1\cdot x+(-1)+x$$
$$x+0=-1\cdot x + 1\cdot x + (-1)$$
$$x=x((-1)+1)+(-1)$$
$$x=x((-1)+1)+(-1)$$
$$x=0x+(-1)$$
$$x=0+(-1)$$
$$x=-1$$
This assumes that $0x=0$
$$0x=(0+0)x$$
$$0x=0x+0x$$
$$0x+(-0x)=0x+0x+(-0x)$$
$$0=0x+0$$
$$0=0x$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275071",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Determine values of k for a matrix to have a unique solution I have the following system and need to find for what values of $k$ does the system have
i) a unique solution
ii) no solution
iii) an infinite number of solutions
$(k^3+3k)x + (k+5)y + (k+3)z = k^5+(k+3)^2
ky + z = 3
(k^3+k^2-6k)z = k(k^2-9)$
Putting this into a matrix I get
$M =\left[
\begin{array}{cc}
k^3 + 3k & k+5 & k+3 & k^5+(k+3)^2 \\
0 & k & 1 & 3 \\
0 & 0 & k^3+k^2-6k & k(k^2-9)\\
\end{array}\right]$
I understand for ii) we need the bottom row to read $0 \ 0 \ 0 \ k(k^2-9)$ which it does for values $k=2$.
For iii) we need the bottom row to read $0 \ 0 \ 0 \ 0$ which we do when $k=0$ and $k=-3$.
Because the first column only contains a value for the top row, I can't use elementary row operations to chance the value in $a_{11}$, so I don't know how I can make it into the identity matrix?
| Coefficient matrix is given by
$A = \begin{pmatrix}
k^3+3k & k+5 & k+3\\
0 & k & 1 \\
0 & 0 & k^3+k^2-6k
\end{pmatrix}$
And, augmented matrix is given by
$[A:b] = \begin{pmatrix}
k^3+3k & k+5 & k+3 & k^5+(k+3)^2\\
0 & k & 1 & 3\\
0 & 0 & k^3+k^2-6k & k(k^2-9)
\end{pmatrix}$
(1) UNIQUE SOLUTION, if $rank(A) = rank([A:b]) =$ number of variables (here, number of variables is $3$).
Find out the values of $k$ for which $det(A) \neq 0$.
Values of $k$ are = $\mathbb{C}-\{0, \pm\sqrt{3} i, 2, -3\}$, where $\mathbb{C}$ denotes the set of complex numbers.
(2) NO SOLUTION, if $rank(A) \neq rank([A:b])$.
Check that, for $k = \pm\sqrt{3} i, 2$, $rank(A) = 2$ but $rank([A:b]) = 3$.
(3) INFINITELY MANY SOLUTION, if $rank(A) = rank ([A:b]) <$ number of variables.
Check that, for $k= 0, -3$, $rank(A) = rank ([A:b]) <$ number of variables.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1275774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Primitive element theorem, simple extension
Let $X$, $Y$ be indeterminates over $F_2$, the finite field with 2 elements.
Let $L = F_2(X, Y )$ and $K = F_2(u, v)$, where $u = X + X^2$, $v = Y + Y^2$.
Explain why $L$ is a simple extension of $K$. Find an element $\gamma \in L$ such
that $L = K(\gamma)$. [Hint: First show that $X, Y$ , and $X + Y$ are all
algebraic of degree 2 over K.]
I have shown that it is a simple extension but I have trouble finding the primitive element.
(I'm preparing for my prelims and I have been stuck at this for almost a week now. Please help me with a hint to solve this)
| Note that sending $X$ to $X+1$ is a $K$-automorphism of $L$ since this is always an automorphism of $K$ since $X$ is transcendental over $F_2$, and it fixes $u$ since $X+1+(X+1)^2=X+1+X^2+1=X+X^2$. Similarly sending $Y$ to $Y+1$ is a $K$-automorphism. Note that these automorphisms necessarily generate the Galois group of the extension since the extension has degree at most 4 over $K$ since $X$ and $Y$ each have degree 2 over the $K$. Note that $XY$ has four distinct images under the automorphism group, $XY$, $X(Y+1)=XY+1$, $(X+1)Y=XY+Y$, and $(X+1)(Y+1)=XY+X+Y+1$. Therefore $XY$ has minimal polynomial $(t-XY)(t-(XY+X))(t-(XY+Y))(t-(XY+X+Y+1))$ over $K$, and $XY$ is a primitive element for the extension.
To see this more directly note that if we let $w=XY$, then $$\frac{uv-w-w^2}{w}=\frac{XY+XY^2+X^2Y+X^2Y^2-XY-X^2Y^2}{XY}=X+Y,$$ so $X+Y\in K[w]$. In particular $X^2+X+X+Y=X^2+Y\in K[w]$. Multiplying by $X+Y$ gives $X^3+X^2Y+XY+Y^2\in K[w]$ subtracting $XY$ and $Y^2+X^2=(X+Y)^2$ gives $X^3+X^2Y+X^2=X(X^2+XY+X)\in K[w]$. Since $X^2+XY+X\in K[w]$, this implies $X\in K[w]$. Together with $X+Y\in K[w]$, we have $Y\in K[w]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1277574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Fermats Little Theorem: How $(3^7)^{17}$ will leave the same remainder when divided by $17$ as $3^7$? As it is used in this explanation, how and why $(3^7)^{17}$ will leave the same remainder when divided by $17$ as $3^7$?
Thanks!
| The question referred to is
Find the remainder when $54^{124}$ is divided by $17$
Step 1: Reduce the base according to the modulus
We can see that $54 \equiv 3 \bmod 17$, so $54^{124} \equiv 3^{124} \bmod 17$.
Step 2: Reduce the exponent according to the order of the base
We know that, since $17$ is prime, $a^{17} \equiv a$. Specifically, when $a$ is coprime to $17$, $a^{16} \equiv 1 \bmod 17$. So we can cast out $16$s from the exponent and get $54^{124} \equiv 3^{12} \bmod 17$.
Step 3: Calculate the result
The final step of calculating $ 3^{12} \bmod 17$ could (for example) be completed by considering
$$3^4 = 81 \equiv -4 \bmod 17, \\
\text{ then }\: 54^{124} \equiv 3^{12} \equiv (-4)^3 \equiv 16\cdot -4 \equiv-1\cdot -4 \equiv 4 \bmod 17
$$
Although a straightforward one-by-one exponentiation is easy enough here:
$$\begin{array}{c|c}
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 \\ \hline
3^n \bmod 17 & 3 & 9 & 10 & 13 & 5 & 15 & 11 & 16 & 14 & 8 & 7 & 4\\
\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1277858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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} |
Show that $f:= \frac{1 - 2x}{x^2 - 1}$ is monotonically increasing in the open interval $(-1, 1)$ Show that $f:= \frac{1 - 2x}{x^2 - 1}$ is monotonically increasing in the open interval $(-1, 1)$
To show a function is monotonically increasing, I started by saying that:
A function $f$ is monotonically increasing in an interval $(a, b)$ if for all $x, y \in (a, b)$, with $x$ < $y$, we have that $f(x) \leq f(y)$.
So I started by tryin to show:
Let $x$, $y$ be arbitrary fixed numbers in $(-1, 1)$, with $x < y$, we want to show that $f(x) \leq f(y)$:
$$f(x) = \frac{1 - 2x}{x^2 - 1} \leq \frac{1 - 2y}{y^2 - 1} = f(y)$$
We multiply the left side by $\frac{y^2 - 1}{y^2 - 1}$, which is positive because $y^2$ is always less than $1$, so we don't need to change the sign:
$$\frac{1 - 2x}{x^2 - 1}\cdot \frac{y^2 - 1}{y^2 - 1} \leq \frac{1 - 2y}{y^2 - 1}$$
We know multiply by $\frac{x^2 - 1}{x^2 - 1}$ the right side:
$$\frac{1 - 2x}{x^2 - 1}\cdot \frac{y^2 - 1}{y^2 - 1} \leq \frac{1 - 2y}{y^2 - 1} \cdot \frac{x^2 - 1}{x^2 - 1}$$
So we can simplify:
$$(1 - 2x)\cdot (y^2 - 1) \leq (1 - 2y) \cdot (x^2 - 1)$$
Now, how do we proceed? Is this correct?
| Without derivatives:
Note that we can write for $|x|<1$
$$f(x)=\frac{1-2x}{x^2-1}= \frac{-1/2}{x-1}+ \frac{-3/2}{x+1}$$
If we take $y>x$, then we have
$$f(y)-f(x) =\frac{y-x}{2(x-1)(y-1)}+\frac{3(y-x)}{2(x+1)(y+1)}>0$$
for all $-1<x<y<1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
$\lim_{x\rightarrow0}\frac{1-\left(\cos x\right)^{\ln\left(x+1\right)}}{x^{4}}$ Could you please check if I derive the limit correctly?
$$\lim_{x\rightarrow0}\frac{1-\left(\cos x\right)^{\ln\left(x+1\right)}}{x^{4}}=\lim_{x\rightarrow0}\frac{1-\left(O\left(1\right)\right)^{\left(O\left(1\right)\right)}}{x^{4}}=\lim_{x\rightarrow0}\frac{0}{x^{4}}=0$$
| $$1-(\cos{x})^{\log{(1+x)}} = 1-e^{\log{\cos{x}} \log{(1+x)}} $$
$$\begin{align}\log{\cos{x}} &= \log{\left ( 1-\frac{x^2}{2!} + \frac{x^4}{4!}+\cdots \right )}\\ &= \left (-\frac{x^2}{2!} + \frac{x^4}{4!}+\cdots \right ) - \frac12\left (-\frac{x^2}{2!} + \frac{x^4}{4!}+\cdots \right )^2+\cdots \\ &= -\frac{x^2}{2}-\frac{x^4}{12} + \cdots \end{align}$$
$$\begin{align}\log{\cos{x}} \log{(1+x)} &= \left ( -\frac{x^2}{2}-\frac{x^4}{12} + \cdots \right ) \left (x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4}+\cdots \right )\\ &= -\frac{x^3}{2} \left ( 1+\frac{x^2}{6}+\cdots \right ) \left (1 - \frac{x}{2} + \frac{x^2}{3} +\cdots \right ) \\ &= -\frac{x^3}{2} + \frac{x^4}{4} + \cdots \end{align}$$
$$\begin{align}1-e^{\log{\cos{x}} \log{(1+x)}} &= -\left (-\frac{x^3}{2} + \frac{x^4}{4} + \cdots \right ) - \frac1{2!} \left (-\frac{x^3}{2} + \frac{x^4}{4} + \cdots \right )^2+\cdots \\ &= \frac{x^3}{2} - \frac{x^4}{4} + \cdots\end{align}$$
Thus,
$$\lim_{x\to 0} \frac{1-(\cos{x})^{\log{(1+x)}}}{x^4} = \lim_{x\to 0} \left (\frac{1}{2 x} - \frac14 + \cdots \right) = \infty $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1279868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$ \lim_{x\to o} \frac{(1+x)^{\frac1x}-e+\frac{ex}{2}}{ex^2} $
$$ \lim_{x\to o} \frac{(1+x)^{\frac1x}-e+\frac{ex}{2}}{ex^2} $$
(can this be duplicate? I think not)
I tried it using many methods
$1.$ Solve this conventionally taking $1^\infty$ form in no luck
$2.$ Did this, expand $ {(1+x)^{\frac1x}}$ using binomial theorem got $\frac13$ then grouped coefficients of $x^0$ and it cancelled with $e$ then took coefficient of $x$ cancelled with $\frac{ex}{2}$ and so on very messy right ? at last I got $\frac13$ but that's not the expected answer!
I must have went wrong somewhere can anyone help me with this.
| let $y = \left(1 + \frac1n\right)^n.$ taking logarithm, we get
$$\begin{align} \ln y &= n\ln\left(1 + \frac 1n \right)\\
&= n \left(\frac 1n - \frac1{2n^2} +\frac1{3n^3}+\cdots \right) \\
& = 1 - \frac1{2n} + \frac{1}{3n^2}+ \cdots \end{align}$$
therefore $$\begin{align} y &= ee^{-\frac1{2n}} e^{\frac1{3n^2}}\\
&=e\left(1-\frac{1}{2n}+ \frac1{8n^2}+\cdots\right)
\left(1+\frac{1}{3n^2}+\cdots\right) \\
&=e\left(1- \frac1{2n} +\frac{11}{24n^2} + \cdots\right)
\end{align}$$
from this we get $$n^2\left(y - e + \frac e{2n}\right) = \frac{11e}{24} + \cdots $$ and $$\lim_{x \to 0}\frac{(1+ x)^{\frac1x}- e + \frac{ex}2}{ex} =\frac{11}{24}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
How to find cotangent? Need to find a $3\cot(x+y)$ if $\tan(x)$ and $\tan(y)$ are the solutions of $x^2-3\sqrt{5}\,x +2 = 0$.
I tried to solve this and got $3\sqrt{5}\cdot1/2$, but the answer is $-\sqrt{5}/5$
| From quadratic equation given,
$$\tan{x}+\tan{y}=3\sqrt5$$
$$\tan{x}\tan{y}=2$$
So by the Compound Angle Formula,
$$\tan{(x+y)}=\frac{\tan{x}+\tan{y}}{1-\tan{x}\tan{y}}=\frac{3\sqrt5}{-1}=-3\sqrt5$$
So we have
$$3\cot{(x+y)}=3\cdot{\frac1{\tan{(x+y)}}}=3\cdot{\frac1{-3\sqrt5}}=\frac{-\sqrt5}{5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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Proving simple trigonometric identity I need help with this one
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha = \sin \alpha
$$
I tried moving sin a on the other side of the eqation
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha - \sin \alpha = 0
$$
This are the operations I was able to do
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \mathrm{tan}^2\alpha} - \cos\alpha - \sin \alpha = 0
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \frac{\sin^2\alpha}{\cos^2\alpha}} - \cos\alpha - \sin \alpha = 0
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{\frac{\cos^2\alpha- \sin^2\alpha}{\cos^2\alpha} } - \cos\alpha - \sin \alpha = 0
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha*\cos^2\alpha + \cos^3 \alpha}{\cos^2\alpha- \sin^2\alpha} - \cos\alpha - \sin \alpha = 0
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\cos^3 \alpha}{\sin\alpha} - \cos\alpha - \sin \alpha = 0
$$
I don't see what else I can do with this, so I tried to solve the left part of the equation.
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \tan^2\alpha} - \cos\alpha = \sin \alpha
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{1- \frac{\sin^2\alpha}{\cos^2\alpha}} - \cos\alpha = \sin \alpha
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\sin\alpha + \cos \alpha}{\frac{\cos^2\alpha- \sin^2\alpha}{\cos^2\alpha} } - \cos\alpha = \sin \alpha
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - cos\alpha} + \frac{\sin\alpha\cdot\cos^2\alpha + \cos^3 \alpha}{\cos^2\alpha- \sin^2\alpha} - \cos\alpha = \sin \alpha
$$
$$
\frac{\sin^2 \alpha}{\sin\alpha - \cos\alpha} + \frac{\cos^3 \alpha}{\sin\alpha} - \cos\alpha = \sin \alpha
$$
And I get to nowhere again. I have no other ideas, I didn't see some formula or something. Any help is appreciated.
| Hint: Put the fractions over the same denominator $\left(\sin\left(\alpha\right)-\cos\left(\alpha\right)\right)\left(1-\tan^{2}\left(\alpha\right)\right)$, then eliminate the denominator on the left side. Now you have just to expand the left and right side, and finally use the identity $\tan\left(\alpha\right)=\frac{\sin\left(\alpha\right)}{\cos\left(\alpha\right)}.$ You will get the same expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Integration of complex functions with trig functions: $\int_0^{2 \pi} \frac{ d\theta}{5-\cos( \theta )}$ $\int_0^{2 \pi} \frac{ d\theta}{5-\cos( \theta )}$
How should I integrate this? Using the exponantial identities of trig? Any hints will be great...
Thank you!
| Here's an answer using Fourier series.
First fact: if $\beta\in(-1,1)$,
$$\sum_{n=0}^{+\infty}\beta^n\cos(n\theta)=\Re\left(\frac1{1-\beta\mathrm{e}^{i\theta}}\right)=\frac{1-\beta\cos\theta}{1+\beta^2-2\beta\cos\theta}=\frac12\frac{2-2\beta\cos\theta}{1+\beta^2-2\beta\cos\theta}=\frac12\frac{\bigl(1+\beta^2-2\beta\cos\theta\bigr)+1-\beta^2}{1+\beta^2-2\beta\cos\theta}=\frac12+\frac12\frac{1-\beta^2}{1+\beta^2-2\beta\cos\theta}.$$
Now we find $\beta\in(-1,1)$ such that $1+\beta^2=10\beta$ so that the denominator is a multiple of $5-\cos\theta$:
$$\beta=5-2\sqrt6$$
fulfills this requirement. Observe that $1-\beta^2=2-10\beta$, we'll use it below.
With this value of $\beta$ we have:
$$\sum_{n=0}^{+\infty}\beta^n\cos(n\theta)=\frac12+\frac1{2\beta}\frac{1-5\beta}{5-\cos\theta}.$$
Now, integrating on a period, we clearly have:
$$1=\frac1{2\pi}\int_0^{2\pi}\left(\frac12+\frac1{2\beta}\frac{1-5\beta}{5-\cos\theta}\right)\,\mathrm{d}\theta,$$
i.e.,
$$1=\frac12+\frac{1-5\beta}{4\beta\pi}\int_0^{2\pi}\frac1{5-\cos\theta}\,\mathrm{d}\theta,$$
hence
$$\int_0^{2\pi}\frac1{5-\cos\theta}\,\mathrm{d}\theta=\frac{2\beta\pi}{1-5\beta}=\frac{\pi\sqrt6}6.$$
As a by-product, we also obtain a more general formula: for all $n\in\mathbb{N}^*$,
$$\beta^n=\frac1\pi\int_0^{2\pi}\left(\frac12+\frac1{2\beta}\frac{1-5\beta}{5-\cos\theta}\right)\cos(n\theta)\,\mathrm{d}\theta,$$
hence
$$\beta^n=\frac1{2\beta\pi}\int_0^{2\pi}\frac{1-5\beta}{5-\cos\theta}\cos(n\theta)\,\mathrm{d}\theta,$$
hence
$$\frac{2\beta^{n+1}\pi}{1-5\beta}=\int_0^{2\pi}\frac{\cos(n\theta)}{5-\cos\theta}\,\mathrm{d}\theta,$$
(and this equality also happens to be correct for $n=0$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Another combined limit I've tried to get rid of those logarithms, but still, no result has came.
$$\lim_{x\to 0 x \gt 0} \frac{\ln(x+ \sqrt{x^2+1})}{\ln{(\cos{x})}}$$
Please help
| HINT:
To get rid of the $ln$ in the denominator, remember that $\ln \cos x= \frac{1}{2} \ln(1-\sin^2(x))$.
As for the $ln$ in the nominator, you'll have to calculate the limit: $\lim_{x\to 0^+} {\ln(x+ \sqrt{x^2+1})^{1/x^2}}$ wich is easier (it's zero).
SOLUTION:
\begin{align}
\lim_{x\to 0^+} \frac{\ln(x+ \sqrt{x^2+1})}{\ln{(\cos{x})}}
&=2\lim_{x\to 0^+} \frac{\ln(x+ \sqrt{x^2+1})}{\ln(1-\sin^2(x))}\tag{1}\\
&=2\lim_{x\to 0^+} \frac{\sin^2(x)}{\ln(1-\sin^2(x))} \cdot \frac{x^2}{\sin^2(x)} \cdot \frac{\ln(x+ \sqrt{x^2+1})}{x^2}\tag{2}\\
&=2\lim_{x\to 0^+} \frac{\sin^2(x)}{\ln(1-\sin^2(x))} \cdot \lim_{x\to 0^+} \frac{x^2}{\sin^2(x)} \cdot \lim_{x\to 0^+} \frac{\ln(x+ \sqrt{x^2+1})}{x^2}\tag{3}\\
&=2\cdot (-1) \cdot 1 \cdot \infty \tag{4}\\
&=-\infty \tag{5}\\
\end{align}
Explanation:
$(1)$: $\frac{1}{2} \ln(1-\sin^2(x))=\frac{1}{2} \ln(\cos^2(x))=\frac{1}{2} \cdot 2\ln(\cos x)=\ln \cos x$
$(4)$: $\lim_{x\to 0^+} \frac{\sin^2(x)}{\ln(1-\sin^2(x))} =\lim_{x\to 0^+} \frac{1}{\ln(1-\sin^2(x))^{1/\sin^2(x)}}=\lim_{y\to 0^+} \frac{1}{\ln(1-y)^{1/y}}=\frac{1}{\ln(\frac{1}{e})}=-1 $
and
$\lim_{x\to 0^+} \frac{\ln(x+ \sqrt{x^2+1})}{x^2}=\lim_{x\to 0^+} {\ln(x+ \sqrt{x^2+1})^{1/x^2}}=\ln(\infty)=\infty$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the integer x: $x \equiv 8^{38} \pmod {210}$ Find the integer x: $x \equiv 8^{38} \pmod {210}$
I broke the top into prime mods:
$$x \equiv 8^{38} \pmod 3$$
$$x \equiv 8^{38} \pmod {70}$$
But $x \equiv 8^{38} \pmod {70}$ can be broken up more:
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod {10}$$
But $x \equiv 8^{38} \pmod {10}$ can be broken up more:
$$x \equiv 8^{38} \pmod 5$$
$$x \equiv 8^{38} \pmod 2$$
In the end,I am left with:
$$x \equiv 8^{38} \pmod 5$$
$$x \equiv 8^{38} \pmod 2$$
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod 3$$
Solving each using fermat's theorem:
*
*$x \equiv 8^{38}\equiv8^{4(9)}8^2\equiv64 \equiv 4 \pmod 5$
*$x \equiv 8^{38} \equiv 8^{1(38)}\equiv 1 \pmod 2$
*$x \equiv 8^{38} \equiv 8^{6(6)}8^2\equiv 64 \equiv 1 \pmod 7$
*$x \equiv 8^{38} \equiv 8^{2(19)}\equiv 1 \pmod 3$
So now, I have four congruences. How can i solve them?
| The quick solution to the Chinese Remainder Theorem exercise is to find integers $w,x,y,z$ such that $105w + 70x+42y + 30z=1$. Then:
$$x\equiv 0\cdot 105w + 1\cdot 70x + 4\cdot 42y + 1\cdot 30z\pmod {210}$$
| {
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"url": "https://math.stackexchange.com/questions/1286716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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show that $\frac{1}{F_{1}}+\frac{2}{F_{2}}+\cdots+\frac{n}{F_{n}}<13$ Let $F_{n}$ is Fibonacci number,ie.($F_{n}=F_{n-1}+F_{n-2},F_{1}=F_{2}=1$)
show that
$$\dfrac{1}{F_{1}}+\dfrac{2}{F_{2}}+\cdots+\dfrac{n}{F_{n}}<13$$
if we use
Closed-form expression
$$F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right)$$
$$\dfrac{n}{F_{n}}=\dfrac{\sqrt{5}n}{\left(\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right)}$$
Well and now I'm stuck and don't know how to proceed
| Let $\phi=\frac{1+\sqrt{5}}{2}$. Then $\phi^2=\phi+1$.
By induction, we have $F_n > \phi^{n-2}$ for all $n\ge 1$, and so
$$
\sum_{n=1}^{N} \frac{n}{F_n}
<
\sum_{n=1}^{\infty} \frac{n}{F_n}
< \sum_{n=1}^{\infty} \frac{n}{\phi^{n-2}}
= \sum_{n=1}^{\infty} \frac{n\phi^2}{\phi^{n}}
= \phi^2 \sum_{n=0}^{\infty} \frac{n}{\phi^{n}}
= \frac{\phi^3}{(1-\phi)^2}
= \frac{1+2\phi}{2-\phi}
< 13
$$
because $\phi< \dfrac53$.
I've used that $\phi^2=\phi+1$ and so $\phi^3=\phi\phi^2=\phi^2+\phi=2\phi+1$.
(The value of the last fraction above is $\approx 11.09$. It is the same as Simon's bound.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 0
} |
Trigonometric Substitution in $\int _0^{\pi/2}{\frac{ x\cos x}{ 1+\sin^2 x} dx }$ Evaluate $$ \int _{ 0 }^{ \pi /2 }{ \frac { x\cos { (x) } }{ 1+\sin ^{ 2 }{ x } } \ \mathrm{d}x } $$
$$$$
The solution was suggested like this:$$$$
SOLUTION:
First of all its, quite obvious to have substitution $ \sin(x) \rightarrow x $
$$ I = \int_{0}^{1} \frac{\arcsin(x)}{1+x^2} \ \mathrm{d}x$$
Now using integration by parts,
$$ I = \frac{\pi^2}{8} - \int_{0}^{1} \frac{\arctan(x)}{\sqrt{1-x^2}} \ \mathrm{d}x$$
Could someone please explain these two steps to me? For example, how do we get $\arcsin(x)$ in the numerator?
Thanks a lot!
| You may obtain
$$
\int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx=\frac12 \log^2 (1+\sqrt{2}) . \tag1
$$
Proof. First observe that
$$\begin{align}
\int_0^\pi \frac{x\cos x}{1+\sin^2 x}\:dx &= \int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx + \int_{\pi/2}^{\pi} \frac{x\cos x}{1+\sin^2 x}\:dx \\
& = \int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx - \int_0^{\pi/2} \frac{(x+\pi/2)\sin x}{1+\cos^2 x}\:dx \\
&= 2\int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx - \pi\int_0^{\pi/2} \frac{\cos x}{1+\sin^2 x}\:dx \\
&= 2\int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx - \pi \left[\arctan u\right]_0^1\\
&= 2\int_0^{\pi/2} \frac{x\cos x}{1+\sin^2 x}\:dx - \frac{\pi^2}{4}.
\end{align}
$$
Then evaluate
$$
\int_0^{\pi}\frac{x \cos{x}}{1+\sin^2{x}} {\rm d}x
$$
using for example this approach.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1289387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Matrix Problem of form Ax=B The matrix $A$ is given by
$$\left(\begin{array}{ccc}
1 & 2 & 3 & 4\\
3 & 8 & 11 & 8\\
1 & 3 & 4 & \lambda\\
\lambda & 5 & 7 & 6\end{array} \right)$$
Given that $\lambda$=$2$, $B$=$\left(\begin{array}{ccc}
2 \\
4 \\
\mu \\
3 \end{array} \right)$ and $X$=$\left(\begin{array}{ccc}
x \\
y \\
z \\
t \end{array} \right)$
Find the value of $\mu$ for which the equations defined by $AX=B$ are consistent and solve the equations in this case. State the rank of A.
So I began by reducing matrix $A$ to reduced row echelon form (kind of like taking the null space, except I'm dealing with $Ax=B$ instead of $Ax=0$) but since I have 5 variables and only 4 equations, I'm not sure how to continue onward.
| Gauss-Jordan elimination is one of standard methods for solving linear systems.
$
\left(\begin{array}{cccc|c}
1 & 2 & 3 & 4 & 2 \\
3 & 8 &11 & 8 & 4 \\
1 & 3 & 4 & 2 & \mu \\
2 & 5 & 7 & 6 & 3
\end{array}\right)\sim
\left(\begin{array}{cccc|c}
1 & 2 & 3 & 4 & 2 \\
2 & 5 & 7 & 6 & 3 \\
3 & 8 &11 & 8 & 4 \\
1 & 3 & 4 & 2 & \mu
\end{array}\right)\sim
\left(\begin{array}{cccc|c}
1 & 2 & 3 & 4 & 2 \\
0 & 1 & 1 &-2 &-1 \\
0 & 2 & 2 &-4 & 0 \\
1 & 3 & 4 & 2 & \mu
\end{array}\right)\sim
\left(\begin{array}{cccc|c}
1 & 0 & 1 & 8 & 4 \\
0 & 1 & 1 &-2 &-1 \\
0 & 0 & 0 & 0 & 0 \\
1 & 3 & 4 & 2 & \mu
\end{array}\right)\sim
\left(\begin{array}{cccc|c}
1 & 0 & 1 & 8 & 4 \\
0 & 1 & 1 &-2 &-1 \\
0 & 3 & 3 &-6 & \mu-4 \\
0 & 0 & 0 & 0 & 0
\end{array}\right)\sim
\left(\begin{array}{cccc|c}
1 & 0 & 1 & 8 & 4 \\
0 & 1 & 1 &-2 &-1 \\
0 & 0 & 0 & 0 & \mu-1 \\
0 & 0 & 0 & 0 & 0
\end{array}\right)
$
In this case you should discuss what happens depending on the value of $\mu$. The third row corresponds to the equation $0x+0y+0z+0t=\mu-1$, or - to put it simpler - $0=\mu-1$. What can you say about solutions of this system if $\mu-1=0$? What can you say if $\mu-1\ne0$?
You can also check your work in WA like this or this
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Subsets and Splits