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Proving bounds of an integral $\displaystyle\frac{1}{\sqrt{3}} \leq \int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{x^3 +4}} \leq 1$
I found $f'(x)=\displaystyle\frac{-3x^2}{2(x^3 + 4)^{\frac{3}{2}}}.$
Set $f'(x) = 0$. Got $x = 0$ as a local max.
$f(0) = 1/\sqrt{5}$
$f(2) = 1/\sqrt{12}$
So $1/\sqrt{3} \leq 1/\sqrt{12} \leq$ the integral $\leq 1/\sqrt{5} \leq 1$.
Correct?
Unsure how to progress from here
| Hint:
$$\int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{(2)^3 +4}} \leq \int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{x^3 +4}} \leq \int_{0}^{2} \frac{\mathrm{d}x}{\sqrt{(0)^3 +4}} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2924252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the area bounded by $y = 2 {x} - {x}^2 $ and straight line $ y = - {x}$ $$
y =\ 2\ {x} - {x}^2
$$
$$
y =\ -{x}
$$
According to me , the area
$$
\int_{0}^{2}{2x\ -\ { x} ^2}\, dx \ + \int_{2}^{3}{\ {x} ^2\ -\ 2{x} }\, dx \\
$$
Which gives the area $ \frac{8}{3}$
But the answer is $ \frac{9}{2}$
| To find the points of intersection between the curve $y = 2x - x^2$ and $y = -x$, equate the two expressions.
\begin{align*}
2x - x^2 & = -x\\
3x - x^2 & = 0\\
x(3 - x) & = 0
\end{align*}
which has solutions $x = 0$ and $x = 3$. These are the limits of integration.
Since
\begin{align*}
y & = 2x - x^2\\
& = -x^2 + 2x\\
& = -(x^2 - 2x)\\
& = -(x^2 - 2x + 1) + 1\\
& = -(x - 1)^2 + 1
\end{align*}
the graph of $y = 2x - x^2$ is a parabola with vertex $(1, 1)$ that opens downwards. Hence, it is above the line $y = -x$ in the interval $(0, 3)$, as shown in the figure below.
We want to find the area of the shaded region. Since it lies below the parabola and above the line, its area is
$$\int_{0}^{3} [2x - x^2 - (-x)]~\textrm{d}x = \int_{0}^{3} (3x - x^2)~\textrm{d}x = \frac{9}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2926222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why isn't this approach in solving $x^2+x+1=0$ valid? There is this question in which the real roots of the quadratic equation have to be found:
$x^2 + x + 1 = 0$
To approach this problem, one can see that $x \neq 0$ because:
$(0)^2 + (0) + 1 = 0$
$1 \neq 0$
Therefore, it is legal to divide each term by $x$:
$x + 1 + \frac{1}{x} = 0$
$x = -1 - \frac{1}{x}$
Now, substitute $x$ into the original equation and solve:
$x^2 + (-1-\frac{1}{x}) + 1 = 0$
$x^2-\frac{1}{x} = 0$
$x^3 = 1$
$x = 1$
to get $x = 1$. Clearly this isn't the right answer. But why? Thanks.
| Your error lies in your last line, where you go from $x^3 = 1$ to $x = 1$. There are actually three solutions to $x^3 = 1$. They are as follows
$$\begin{align}
x_1 &= 1, & \text{or} \\
x_2 &= -\frac{1}{2} + \frac{\sqrt 3}{2} i, & \text{or} \\
x_3 &= -\frac{1}{2} - \frac{\sqrt 3}{2} i
\end{align}$$
Only solutions $x_2$ and $x_3$ solve the original problem, so solution $x_1$ can be omitted. Potentially introducing extraneous solutions is a risk you take when you perform a substitution like this.
This arises from the fact that your substitution is changing the degree of your equation from degree 2 to degree 3, so an additional solution must be introduced (assuming no repeated solutions).
| {
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"timestamp": "2023-03-29T00:00:00",
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$\ \lim_{x\to0} \frac{x}{\cos(\frac{\pi}{2}-x)} $ without L'Hospital's rule
How do I find the following?$$\ \lim_{x\to0} \frac{x}{\cos(\frac{\pi}{2}-x)} $$
I have tried to use trig identities :
$$ \frac{x}{\cos(\frac{\pi}{2}-x)} = \frac{x}{\cos\frac{\pi}{2}\cos x+\sin\frac{\pi}{2}\sin x} = $$
Can't really see anything out of that?
edit- so based on Math lover hint :
$$\ \frac{x}{\cos \frac{\pi}{2}\cos x + \sin \frac{\pi}{2}\sin x} = \frac{x}{0 \cdot \cos x + 1\cdot \sin x } = \frac{x}{\sin x} = \frac{0}{0} = 0 $$
| As an alternative by $f(x)=\cos\left(\frac{\pi}{2}-x\right)$ from the definition of derivative
$$\lim_{x\to0} \frac{x}{\cos\left(\frac{\pi}{2}-x\right)}=\lim_{x\to0} \frac{x-0}{\cos\left(\frac{\pi}{2}-x\right)-\cos\left(\frac{\pi}{2}\right)}=\frac1{f'(0)}=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2930703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve $\sin^{3}x+\cos^{3}x=1$
Solve for $x\\ \sin^{3}x+\cos^{3}x=1$
$\sin^{3}x+\cos^{3}x=1\\(\sin x+\cos x)(\sin^{2}x-\sin x\cdot\cos x+\cos^{2}x)=1\\(\sin x+\cos x)(1-\sin x\cdot\cos x)=1$
What should I do next?
| Let use by $t= \tan (x/2)$
*
*$\sin x=\frac{2t}{1+t^2}$
*$\cos x=\frac{1-t^2}{1+t^2}$
to obtain
$$2t^6-8t^3+6t^2=0$$
$$\iff t^2(t^4-4t+3)=t^2(t-1)^2(t^2+2t+3)=0$$
and since $t^2+2t+3>0$ the solutions are
*
*$t=0 \implies \frac x 2=k\pi\implies x=2k\pi$
*$t=1 \implies \frac x 2=\frac{\pi}4+k\pi \implies x=\frac{\pi}2+2k\pi $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2931449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 2
} |
Solve $7^x+x^4+47=y^2$
Solve $$7^x+x^4+47=y^2$$ where $x, y \in \mathbb{N}^*$
If $x$ is odd then the left term is congruent with $3$ mod $4$ so it couldn't be a perfect square, so we deduce that $x=2a$ and the relation becomes $$49^a+16a^4+47=y^2$$ and it is easy to see that the left term is divisible by $16$ so we obtain that $y=4b$, so we have to find $a$ and $b$ such that $$49^a+16a^4+47=16b^2$$From this point I was completely stuck. I think that there are no solutions but how can I prove it?
| If $x$ is even and let $x=2a$, then $(7^a)^2<7^x+x^4+47<(7^a+1)^2=7^x+2\times7^a+1$ if $(2a)^4+47<2\times7^a+1$, which is true for $a \ge 4$. Therefore, it is enough to consider only $x=2, 4$ and $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2931537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 0
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Is discrete uniform distribution fully characterized by first two moments? Let $U$ be the discrete uniform on $\{0,1,\dots,n\}$. If a discrete random variable $X$ satisfies:
$$
\begin{align}
\mathbb{E}[X] &= \mathbb{E}[U] = n/2 \\
\mathrm{Var}(X) &= \mathrm{Var}(U) = ((n+1)^2-1)/12,
\end{align}
$$
is this sufficient to conclude that $X \equiv U$? For the case of $n=1$ and $n=2$ it can be shown to be true.
Addition: $X$ is known to have full support over $\{0,1,\dots,n\}$.
| No. For $n=3$: consider the distribution of $X$ given by $p=(1/5,2/5,1/10,3/10)$. (There are many other possible choices.)
You have
$$
\mathbb{E}[X] = \sum_{k=0}^3 k p_k = 0\cdot \frac{1}{5} + 1\cdot\frac{2}{5}+2\cdot\frac{1}{10}+3\cdot\frac{3}{10} = \frac{3}{2} = \mathbb{E}[U_3]
$$
and
$$
\mathbb{E}[X^2] = \sum_{k=0}^3 k^2 p_k = 0\cdot \frac{1}{5} + 1\cdot\frac{2}{5}+4\cdot\frac{1}{10}+9\cdot\frac{3}{10} = \frac{7}{2} = \mathbb{E}[U_3^2]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Solving integer equation $a^3+b^3=a^2+72ab+b^2$
Find all pairs of positive integers $(a;b)$ that satisfy
$$a^3+b^3=a^2+72ab+b^2.$$
I have already solved this by letting $S=a+b$, $P=ab$. Then I have $S^3-S^2=(3S+70)P$, which will result in $3S+70$ dividing $5110S$, or $357700$ is divisible by $3S+70$. The result is $(a;b)\in\left\{(1;9),(9;1),(37;37)\right\}$.
However, this strategy is extremely complicated and time-consuming, by listing all divisors of $357700$. I do not know any other way to access this problem. Many thanks.
| Let $a=x+y, b=x-y$. Then:
$$a^3+b^3=a^2+72ab+b^2 \Rightarrow \\
(x+y)^3+(x-y)^3=(x+y)^2+72(x+y)(x-y)+(x-y)^2 \Rightarrow \\
2x^3+6xy^2=2x^2+72x^2 -72y^2+2y^2 \Rightarrow \\
y^2=\frac{(37-x)x^2}{35+3x}\ge 0.$$
Note that:
$$x=\frac{a+b}{2}>0 \ \ \text{and} \ \ 37-x\ge0 \Rightarrow 0<x\le 37.$$
One can quickly check (if not further analysis) all numbers to find: $(x,y)=(5,4),(37,0)$.
Hence:
$$(a,b)=(9,1),(37,37) \ \ \text{and} \ \ (1,9) \ \text{(due to symmetry)}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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Calculate limit of $\frac{1}{n}\cdot (1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n})$ How to prove that
$$\lim_{n\to\infty}(\frac{1}{n}\cdot(1+\sqrt{2}+\sqrt{3}+\cdots+\sqrt{n})) = +\infty$$
using only basic limit operations and theorems?
| You have
$$\begin{align}
\frac{1}{n}\sum_{k=1}^n \sqrt{k}
&\geq \frac{1}{n}\sum_{k=\lfloor n/2\rfloor}^n \sqrt{k}
\geq \frac{1}{n}\sum_{k=\lfloor n/2\rfloor}^n \sqrt{\frac{n}{2}} \\
&= \frac{n-\lfloor n/2\rfloor+1}{n}\cdot \sqrt{\frac{n}{2}}
\geq \frac{n}{2n}\cdot \sqrt{\frac{n}{2}} \\
&= \frac{\sqrt{n}}{2\sqrt{2}} \xrightarrow[n\to\infty]{} \infty
\end{align}$$
where the only "trick" lies in the first inequality: dropping some (positive) terms from the sum actually helps to get an easy lower bound on the sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Radius and angular derivatives expressed in Cartesian coordinates Given a point with Cartesian coordinates $(x,y)$ and with Cartesian velocity $(\dot{x},\dot{y})$, I would like to express its radius $r$, its angle $\theta$, its radius velocity $\dot{r}$, and its angular velocity $\dot{\theta}$.
First, I know that $x = r \cos\theta$ and $y = r \sin\theta$, such that $r = \sqrt{x^2+y^2}$ and $\theta = \pm \arccos (x/r)$.
Second, I know that:
\begin{align}
\dot{x} = - r \dot\theta \sin\theta + \dot{r} \cos\theta & \quad (1)\\
\dot{y} = r \dot\theta \cos\theta + \dot{r} \sin\theta & \quad (2)
\end{align}
I should be able do derive an expression of $\dot{r}$ and $\dot\theta$ from these two equations.
Here is my first attempt, using the fact that $\cos^2\theta + \sin^2\theta = 1$:
\begin{align}
\dot{r}^2 = \dot{x}^2 + \dot{y}^2 + (r \dot\theta)^2 + 2 r \dot\theta (\dot{x} \sin\theta - \dot{y} \cos\theta) & \quad (3)\\
(r \dot\theta)^2 = \dot{x}^2 + \dot{y}^2 + \dot{r}^2 - 2 \dot{r} (\dot{x} \cos\theta + \dot{y} \sin\theta) & \quad (4)
\end{align}
Now, substituting (4) in (3), I get an equation of $\dot{r}$ where $\dot\theta$ does not appear, but which is quite intricate:
$$\dot{x}^2 + \dot{y}^2 + \sqrt{\dot{x}^2 + \dot{y}^2 + \dot{r}^2 - 2 \dot{r} (\dot{x} \cos\theta + \dot{y} \sin\theta)} (\dot{x} \sin\theta - \dot{y} \cos\theta) - \dot{r} (\dot{x} \cos\theta + \dot{y} \sin\theta) = 0$$
I have the feeling that I am doing the right thing, but that I am missing some intuitive shortcut to properly isolate $\dot{r}$.
| Following the method provided by Rodney Dunning in another answer:
*
*Taking the time-derivative of $r^2 = x^2 + y^2$, we get
$$2 r \dot{r} = 2 x \dot{x} + 2 y \dot{y} \text{,}$$
and we may write
$$\boxed{\dot{r} = \frac{x \dot{x} + y \dot{y}}{r}} \text{.}$$
*Taking the time-derivative of $\tan\theta = \frac{y}{x}$, we get
$$\frac{\dot\theta}{\cos^2\theta} = \frac{x \dot{y} - y \dot{x}}{x^2} \text{,}$$
and we may write
$$\boxed{\dot\theta = \frac{x \dot{y} - y \dot{x}}{x^2} \cos^2\theta} \text{.}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the range of $ (\arcsin x)^2 + (\arccos x)^2 $ without using derivatives? I am expected to find the range of the following function:
$$
(\arcsin x)^2 + (\arccos x)^2
$$
I individually added up the range of $(\arcsin x)^2$ $(\arccos x)^2$
But that gave wrong answer ?
What to do with this problem?
| Let me write $f(x) = \arcsin^2 x + \arccos^2 x$. The natural domain of $f$ is the intersection of natural domains of $\arcsin$ and $\arccos$: $$\mathcal D(f) = \mathcal D(\arcsin) \cap \mathcal D(\arccos) = [-1,1]\cap [-1,1] = [-1,1].$$
From the identity $\sin x = \cos(\frac\pi 2 - x)$, it follows that $\arcsin x + \arccos x = \frac\pi 2$. Thus, we have $$ f(x) = \arcsin^2 x + \arccos^2 x = \arcsin^2 x + (\frac\pi 2 - \arcsin x)^2 = 2\arcsin^2 x -\pi \arcsin x + \frac{\pi^2}4.$$
Let us define $g(x) = 2x^2 - \pi x + \frac{\pi^2}4$. We can see that $f(x) = g(\arcsin x)$. The range of $f$ is thus:
$$f([-1,1]) = g(\arcsin ([-1,1])) = g([-\frac\pi 2,\frac\pi 2]).$$
The graph of $g$ is a parabola. To determine $g([-\frac\pi 2,\frac\pi 2])$, let us first write $g(x) = 2(x-\frac\pi 4)^2+\frac{\pi^2}8$.
The minimum of $g$ on $\mathbb R$ occurs at $x_0 = \frac\pi 4$ and is equal to $\frac{\pi^2}8$. Since $x_0\in [-\frac\pi 2,\frac\pi 2]$, it is also the minimum of $g$ on the segment $[-\frac\pi 2,\frac\pi 2]$.
The maximum of $g$ then occurs at one of the boundary points $\{\pm\frac\pi 2\}$. Calculate $g(\frac\pi 2) = \frac{\pi^2}4$ and $g(-\frac\pi 2) = \frac{5\pi^2}4$. Since $g(-\frac\pi 2)$ is greater, that is the maximum of $g$ on $[-\frac\pi 2,\frac\pi 2]$.
We conclude that $g([-\frac\pi 2,\frac\pi 2]) = [\frac{\pi^2}8,\frac{5\pi^2}4]$ and that is the range of $f$.
Visual aid. The red line is the range of $\arcsin$ and the blue line is the range of $f$:
| {
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Prove that the quantity is an integer I want to prove that $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$.
I have thought to use induction.
Base Case: For $n=1$, $\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6}=\frac{1}{3}-\frac{1}{2}+\frac{1}{6}=0 \in \mathbb{Z}$.
Induction hypothesis: We suppose that it holds for $n=k$, i.e. that $\frac{k^3}{3}-\frac{k^2}{2}+\frac{k}{6} \in \mathbb{Z}$.
Induction step: We want to show that it holds for $n=k+1$.
$$\frac{(k+1)^3}{3}-\frac{(k+1)^2}{2}+\frac{k+1}{6}=\frac{k^3}{3}+\frac{k^2}{2}+\frac{k}{6}$$
Is everything right? If so, then we cannot use at the induction step the induction hypothesis, can we?
Or can we not get the desired result using induction?
| To prove $$\frac{n^3}{3}-\frac{n^2}{2}+\frac{n}{6} \in \mathbb{Z}, \forall n \geq 1$$
We take common denominator and prove the numerator is a multiple of $6.$
The numerator factors as $$n(2n-1)(n-1)$$
One of $n$ or $n-1$ is even so the product is multiple of $2$
The remainder of n in dividing by $3$ is either $0$ or $1$ or $2$
In either of these cases the product $$n(2n-1)(n-1)$$ is divisible by $3$
Thus the numerator is always a multiple of $6$ which makes the fraction an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2946269",
"timestamp": "2023-03-29T00:00:00",
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$\int_{-\infty}^{\infty}\frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}}dx=\zeta(2)$ Without Feynman Integration How do I find $$\int_{-\infty}^{\infty}\frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}}dx=\zeta(2)$$
without Feynman integration? I saw this video, which gives $$\int_{0}^{1}\frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}}dx=\frac{5\pi^2}{96}$$
Via Feynman integration, but I would like to know another method.
| Notice that $\frac{\arctan x}{x} = \int_{0}^{1} \frac{dy}{1+x^2y^2}$. So
\begin{align*}
\int_{-\infty}^{\infty} \frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}} \, dx
&= \int_{-\infty}^{\infty} \int_{0}^{1} \frac{1}{(x^2+1)(1+(x^2+2)y^2)} \, dydx.
\end{align*}
Notice that
$$
\frac{1}{(x^2+1)(1+(x^2+2)y^2)}
= \frac{1}{(x^2+1)(y^2+1)} - \frac{y^2}{(y^2+1)(1+(x^2+2)y^2)}. $$
So interchanging the order of integration,
\begin{align*}
\int_{-\infty}^{\infty} \frac{\arctan\sqrt{x^2+2}}{(x^2+1)\sqrt{x^2+2}} \, dx
&= \int_{0}^{1} \left( \frac{\pi}{y^2+1} - \frac{\pi y}{(y^2+1)\sqrt{2y^2+1}} \right) \, dy \\
&= \pi \left[ \arctan(y) - \arctan\sqrt{2y^2+1} \right]_{0}^{1} \\
&= \frac{\pi^2}{6}.
\end{align*}
| {
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Prove that $\frac{3}{5} + \frac{4}{5}i$ is not a root of unity I want to prove that $z=\frac{3}{5} + \frac{4}{5}i$ is not a root of unity, although its absolute value is 1.
When transformed to the geometric representation: $$z=\cos{\left(\arctan{\frac{4}{3}}\right)} + i\sin{\left(\arctan{\frac{4}{3}}\right)}$$ According to De Moivre's theorem, we get:
$$z^n= \cos{\left(n\arctan{\frac{4}{3}}\right)} + i\sin{\left(n\arctan{\frac{4}{3}}\right)}$$
Now, if for $n\in \mathbb{N}: z^n=1$, then the imaginary part of the expression above must be zero, therefore:
$$\sin{\left(n\arctan{\frac{4}{3}}\right)}=0 \iff n\arctan{\frac{4}{3}} = k\pi, \ \ \ k \in \mathbb{N}$$
And we get that for $z$ to be a root of unity for some natural number $n$, $n$ must be in the form:
$$n = \frac{k\pi}{\arctan{\frac{4}{3}}}, \ \ \ k \in \mathbb{N}$$
On the other hand, for $z^n=1$ it must be that:
$$\cos{\left(n\arctan{\frac{4}{3}}\right)} = 1 \iff n\arctan{\frac{4}{3}} = 2l\pi, \ \ \ l \in \mathbb{N}$$
and thus
$$n = \frac{2l\pi}{\arctan{\frac{4}{3}}}, \ \ \ l \in \mathbb{N}$$
By comparing those two forms of $n$, it must be the case that $k=2l$ and for $n$ to satisfy $z^n = 1$. What follows is that $n$ should be in the form
$$n = \frac{2l\pi}{\arctan{\frac{4}{3}}}, \ \ \ l \in \mathbb{N}$$
But, at the same time, $n$ must be a natural number. Should I prove now that such $n$ cannot even be a rational number, let alone a natural one? Or how should I approach finishing this proof?
| From here, we know that
If $x$ is a rational multiple of $\pi$, then $2\cos(x)$ is an algebraic integer.
So if $x=\arctan(4/3)$ is a rational multiple of $\pi$, then $2\cos(x)=6/5$ is an algebraic integer, which is also a rational number, and hence an integer. This is a contradiction. Therefore $\frac{3+4i}5$ is not a root of unity.
Hope this helps.
| {
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Prove $\lim\limits_{x \to \pm\infty}\dfrac{x^3+1}{x^2+1}=\infty$ by the definition. Problem
Prove $\lim\limits_{x \to \pm\infty}\dfrac{x^3+1}{x^2+1}=\infty$ by the definition.
Note:
The problem asks us to prove that, no matter $x \to +\infty$ or $x \to -\infty$, the limit is $\infty$,which may be $+\infty$ or $-\infty.$
Proof
$\forall M>0$,$\exists X=\max(1,M+1)>0, \forall|x|>X$οΌ
\begin{align*}
\left|\frac{x^3+1}{x^2+1}\right|&=\left|x-\frac{x-1}{x^2+1}\right|\\&\geq |x|-\left|\frac{x-1}{x^2+1}\right|\\&\geq |x|-\frac{|x|+1}{x^2+1}\\&\geq |x|-\frac{x^2+1}{x^2+1}\\&=|x|-1\\&>X-1\\&\geq M.
\end{align*}
Please verify the proof above.
| Your proof is correct.
You may consider shorter proofs by using some simplifications.
Since $x\to \infty $ we assume $x>1$
$$ \frac {x^3+1}{x^2+1} \ge\frac {x^3}{2x^2}=x/2$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Limit of $\frac{y^4\sin(x)}{x^2+y^4}$ when $(x,y) \to (0,0)$ I have to find the following limit when $x,y$ tend towards $0$. I think the limit doesn't exist (thanks to Wolfram Alpha), but all I can find no matter what path I use (I've tried $y=xΒ², x=y^2, x=0, y=0$) is that the limit equals $0$ (because of the $sin(0)$). Any tips?
$$\frac{y^4\sin(x)}{x^2+y^4}$$
| Let $|x|,|y| <1$, $x,y$ real.
$0\le \left | \dfrac{y^4 \sin x }{x^2+y^4} \right | \le $
$\dfrac{y^4|x|}{x^4 + y^4} \le$
$\dfrac{(y^4+x^4)|x|}{x^4+ y^4}= |x| \le \sqrt{x^2+y^2}.$
Choose $\delta =\epsilon.$
Used : $|\sin x| \le |x|$
| {
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"source": "stackexchange",
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Find the zeros of $f(z)=z^3-\sin^3z$ I want to find the zeros of $f(z)$, $$f(z)=z^3-\sin^3z$$
My attempt
$f(z)=0$
$z^3-(z-z^3/3!+z^5/5!-\dots)^3=0$
$z^3-z^3(1-z^2/3!+z^4/5!-\dots)^3=0$
$z^3[1-(1-z^2/3!+z^4/5!-\dots)^3]=0$
So $z=0$ is a zero of order $3$.
I don't feel good about this answer. Please give me some hints if I am incorrect.
Edit: What is the order of root $z=0$?
| Hint: The root at zero is more than fourth order. Also notice that
$z^3 - \sin^3 z \approx 0$
when $z= 5.3341 - 2.4614 i$. That should lead you to some more roots.
You can compute the lower order terms of the Taylor Series from the algebra in your "My Attempt" section.
$$z^3 - \sin^3z = z^3 - (z-z^3/3!+ z^5/5!-z^7/7!+z^9/9!-O\left(z^{11}\right))^3$$
$$= z^3 - \left(z^3-\frac{z^5}{2}+\frac{13 z^7}{120}-\frac{41 z^9}{3024}+O\left(z^{11}\right)\right)$$
$$=\frac{z^5}{2}-\frac{13 z^7}{120}+\frac{41 z^9}{3024}-O\left(z^{11}\right)$$
So, fifth order at $z=0$.
| {
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Find the volume of the solid obtained by rotating the region bounded by: Q: "Find the volume of the solid obtained by rotating the region bounded by:
$ $ $ $ $ $ $ $ $ $ $ $ $ $ $ $$x + y = 4$, $ $ $ $$x = 5 - (y-1)^2$
$ $ $ $ $ $ $ $ $ $ $ $ $ $ About the $x-axis$."
I have attempted this problem for a while now and keep messing up somewhere. It would much be appreciated if you could point out my mistake and lead me in the right direction.
Attempt: $ $ $ y = 4 - x$, $ $ $y = \sqrt{5-x}+1$,$ $ a= 1 $ $ b= 4
$\pi\int_1^4(\sqrt{5-x}+1)^2-(4-x)^2dx$ = $\pi\int_1^4-x^2+7x+2\sqrt{5-x}-10 dx$
$\pi(\frac{-x^3}3 +\frac{7x^2}2 +\frac{4/(5-x)^{3/2}}3-10x) |^4_1$ = $\pi(\frac{-64}3+\frac{112}2+\frac{4}3-40+\frac{1}3-\frac{7}2-\frac{32}3+10)$
= -24.60914245312
| I think shells will be easier than washer on this one.
Sketch your region. We have a line intersecting a parabola, but the parabola is sideways to the normal orientation.
shells $V = 2\pi \int_0^3 y((5-(y-1)^2)-(4-y))\ dy$
$2\pi \int_0^3 3y^2-y^3\ dy\\
2\pi (y^3 - \frac 14 y^4|_0^3) = 2\pi(27 -\frac {81}{4}) = \frac {27}{2}\pi$
by washers.
If you graph your region, you will see a bulge to the right of the points of intersection (4,0) that you must account for.
$V = \pi \int_1^4 (\sqrt{5-x} + 1)^2 - (4-x)^2 \ dx+\int_4^5 (\sqrt{5-x} + 1)^2 - (-\sqrt{5-x} + 1)^2 \ dx$
| {
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"source": "stackexchange",
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Prove that $7^n+2$ is divisible by $3$ for all $n β \mathbb{N}$
Use mathematical induction to prove that $7^{n} +2$
is divisible by $3$ for all $n β \mathbb{N}$.
I've tried to do it as follow.
If $n = 1$ then $9/3 = 3$.
Assume it is true when $n = p$. Therefore $7^{p} +2= 3k $ where $k β \mathbb{N} $. Consider now $n=p+1$. Then
\begin{align}
&7^{p+1} +2=\\
&7^p\cdot7+ 2=\\
\end{align}
I reached a dead end from here. If someone could help me in the direction of the next step it would be really helpful. Thanks in advance.
| Show that $7^n\equiv 1 (\text{mod} 3)$
$(6 + 1)(6 + 1) = 36 + 12 + 1$
$(6 + 1)(6 + 1)(6 + 1) = 216 + 72 + 6 + 36 + 12 + 1$
So all $(6 + 1)^n$ involve the sum of a series of terms which are multiples of $6$ (divisible by $3$) except for the last term which is $1$
Hence $7^n\equiv 1 (\text{mod} 3)$
and $7^n + 2$ is always divisible by $3$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Induction: How to prove that $ab^n+cn+d$ is divisible by $m$.
If $a+d$, $(b-1)c$, $ab-a+c$ are divisible by $m$, prove that
$ab^n+cn+d$ is also divisible by $m$.
I want to prove this by induction. For proving $ab^{k+1}+c(k+1)+d$ is divisible by $m$, i want to prove that $ab^k(b-1)+c$ is divisible by $m$ and then add it to $ab^{k}+ck+d$. Any idea how to prove $ab^k(b-1)+c$ is divisible by $m$? Or is there a better way to solve the problem?
Thanks in advance.
| Without induction. Let's do some tagging 1st
$$m \mid a+d \tag{1}$$
$$m \mid (b-1)c \tag{2}$$
$$m \mid ab-a+c \tag{3}$$
then
$$m \mid ab^n+cn+d \iff m \mid a\left(b^n-1\right)+cn+\color{red}{a+d} \overset{(1)}{\iff}\\
m \mid a\left(b^n-1\right)+cn \iff \\
m \mid a(b-1)\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)+cn \iff \\
m \mid (ab-a+c-c)\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)+cn \iff \\
m \mid \color{red}{(ab-a+c)\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)}-c\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)+cn \overset{(3)}{\iff}\\
m \mid c\left(b^{n-1}+b^{n-2}+...+b^2+b+1\right)-cn \iff \\
m \mid c\left(b^{n-1}-1\right)+c\left(b^{n-2}-1\right)+...+c\left(b^2-1\right)+\color{red}{c\left(b-1\right)} \overset{(2)}{\iff}\\
m \mid c\left(b^{n-1}-1\right)+c\left(b^{n-2}-1\right)+...+c\left(b^2-1\right)$$
which is true because for $\forall k\geq 2$ we have
$$c\left(b^k-1\right)=\color{red}{c\left(b-1\right)}\left(b^{k-1}+b^{k-2}+...+b^{2}+b+1\right)$$
and from $(2)$
$$m \mid c\left(b^k-1\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How big does $r$ need to be, to ensure that $x^4+y^4 >r^2$ for all $x^2+y^2 =r^2$? Consider the circle $x^2+y^2 = r^2$ for some fixed $r>0$. How big does $r$ need to be, to ensure that $$x^4+y^4 >r^4$$ for all $x^2+y^2 =r^2$? I know that $x^2<x^4$ whenever $|x| >1$, but I'm not sure how to use that here.
| $x^4+y^4=r^4-2r^4\cos^2{\theta}\sin^2{\theta}=r^4(1-\frac{1}{2}\sin^2{2\theta})$
The minimum value of the term in parentheses is $\frac{1}{2}$
So $s^4<\frac{1}{2}r^4$
| {
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Maximise $(x+1)\sqrt{1-x^2}$ without calculus Problem
Maximise $f:[-1,1]\rightarrow \mathbb{R}$, with $f(x)=(1+x)\sqrt{1-x^2}$
With calculus, this problem would be easily solved by setting $f'(x)=0$ and obtaining $x=\frac{1}{2}$, then checking that $f''(\frac{1}{2})<0$ to obtain the final answer of $f(\frac{1}{2})=\frac{3\sqrt{3}}{4}$
The motivation behind this function comes from maximising the area of an inscribed triangle in the unit circle, for anyone that is curious.
My Attempt
$$f(x)=(1+x)\sqrt{1-x^2}=\sqrt{(1-x^2)(1+x)^2}=\sqrt 3 \sqrt{(1-x^2)\frac{(1+x)^2}{3}}$$
By the AM-GM Inequality, $\sqrt{ab}\leq \frac{a+b}{2}$, with equality iff $a=b$
This means that
$$\sqrt 3 \sqrt{ab} \leq \frac{\sqrt 3}{2}(a+b)$$
Substituting $a=1-x^2, b=\frac{(1+x)^2}{3}$,
$$f(x)=\sqrt 3 \sqrt{(1-x^2)\frac{(1+x)^2}{3}} \leq \frac{\sqrt 3}{2} \left((1-x^2)+\frac{(1+x)^2}{3}\right)$$
$$=\frac{\sqrt 3}{2} \left(\frac{4}{3} -\frac{2}{3} x^2 + \frac{2}{3} x\right)$$
$$=-\frac{\sqrt 3}{2}\frac{2}{3}(x^2-x-2)$$
$$=-\frac{\sqrt 3}{3}\left(\left(x-\frac{1}{2}\right)^2-\frac{9}{4}\right)$$
$$\leq -\frac{\sqrt 3}{3}\left(-\frac{9}{4}\right)=\frac{3\sqrt 3}{4}$$
Both inequalities have equality when $x=\frac{1}{2}$
Hence, $f(x)$ is maximum at $\frac{3\sqrt 3}{4}$ when $x=\frac{1}{2}$
However, this solution is (rather obviously I think) heavily reverse-engineered, with the two inequalities carefully manipulated to give identical equality conditions of $x=\frac{1}{2}$. Is there some better or more "natural" way to find the minimum point, perhaps with better uses of AM-GM or other inequalities like Jensen's inequality?
| By AM-GM
$$(1+x)\sqrt{1-x^2}=\frac{1}{\sqrt3}\sqrt{(1+x)^3(3-3x)}\leq$$
$$\leq\frac{1}{\sqrt3}\sqrt{\left(\frac{3(1+x)+3-3x}{4}\right)^4}=\frac{3\sqrt3}{4}.$$
The equality occurs for $1+x=3-3x,$ which says that we got a maximal value.
| {
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Verify proof that $x_n = \sum_{k=1}^n {k \over 2^k}$ is bounded and find its supremum and infinum The problem I'm solving states:
Let $n\in \mathbb N$ and $x_n$ be a sequence:
$$
x_n = \sum_{k=1}^n {k \over 2^k}
$$
Prove $x_n$ is bounded and find $\sup\{x_n\}$ and $\inf\{x_n\}$
Let $S_n$ denote the sum, then:
$$
S_n = \frac{1}{2^1} + \frac{2}{2^2} + \frac{3}{2^3} + \dots + \frac{n}{2^n}
$$
Multiply $S_n$ by $1 \over 2$:
$$
{1\over 2} S_n = \frac{1}{2^2} + \frac{2}{2^3} + \frac{3}{2^4} + \dots + \frac{n}{2^{n+1}}
$$
Subtract the sums:
$$
\begin{align}
S_n - { 1 \over 2 } S_n &= \sum_{k=1}^n {1\over 2^k} - {n\over 2^{n+1}} = \\
&= 1 - {1 \over 2^n } - {n \over 2^{n+1}} = \\
&= 2 - {n \over 2^n} - {2 \over 2^n} < 2
\end{align}
$$
Therefore:
$$
\inf\{x_n\} = {1\over 2} \le x_n < 2 = \sup\{x_n\} \\
$$
Have I done it the right way?
| It's may be easier to do it this way: since $(x_n)_n$ is strictly increasing, $\inf x_n=x_1=1/2$ and $\sup x_n=\lim x_n$, which you can calculate by noting that $f(x)=\sum_{n=0}x^n=\frac{1}{1-x};\ |x|<1.$
Now, $f$ may be differentiated inside its interval of convergence, so $x\sum_{n=1}nx^{n-1}=\frac{x}{(1-x)^{2}}.$ To finish, evaluate this expression at $x=1/2.$
| {
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The rth term in $(1+x)^{1/x} = e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ Let
$$(1+x)^{\frac {1}{x}} = e.G(x)$$
Taking logarithm on both sides,
$$\frac {1}{x} \log {(1+x)} = 1 + \log {G(x)}$$
Putting in the Taylor expansion for $\log {(1+x)}$ we have,
$$\frac {1}{x}(x - \frac {1}{2}x^2 + \frac {1}{3}x^3 - ....) = 1 + \log {G(x)}$$
Solving for $G(x)$ we have,
$$G(x) = e^{-\frac {1}{2}x + \frac {1}{3}x^2 - ...}$$
The difficulty starts here, for in order to get the desired Taylor expansion $e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ I have to plug in the entire expansion $-\frac {1}{2}x + \frac {1}{3}x^2 - ...$ for the variable in the Taylor expansion for $e$. Is there any alternative way? My main question is : what is the rth term of the desired expansion $e[1 - \frac {1}{2}x + \frac {11}{24}x^2 - \frac {7}{16}x^3 + ....]$ in its closed form?
| \begin{align}
\dfrac1e(1+x)^{1/x}
&= \exp\left(\dfrac{\ln(1+x)-x}{x}\right) \\
&= \exp\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right) \\
&= 1+\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right)+\dfrac12\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right)^2 \\
& +\dfrac16\left(-\dfrac{x}{2}+\dfrac{x^2}{3}-\dfrac{x^3}{4}+\cdots\right)^3+\cdots\\
&= 1-\frac{1}{2}x+\left(\dfrac13+\dfrac18\right)x^2+\left(-\dfrac14+\dfrac122\dfrac{-1}{2}\dfrac13+\dfrac16\dfrac{-1}{8}\right)x^3+\cdots
\end{align}
| {
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Proof verification of $x_n = \sqrt[3]{n^3 + 1} - \sqrt{n^2 - 1}$ is bounded
Let $n \in \mathbb N$ and:
$$
x_n = \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1}
$$
Prove $x_n$ is bounded sequence.
Start with $x_n$:
$$
\begin{align}
x_n &= \sqrt[^3]{n^3 + 1} - \sqrt{n^2 - 1} = \\
&= n \left(\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}}\right)
\end{align}
$$
From here:
$$
\sqrt[^3]{1 + {1\over n^3}} \gt 1 \\
\sqrt{1 - {1\over n^2}} \lt 1
$$
Therefore:
$$
\sqrt[^3]{1 + {1\over n^3}} - \sqrt{1 - {1\over n^2}} \gt 0
$$
Which means $x_n \gt 0$.
Consider the following inequality:
$$
\sqrt[^3]{n^3 + 1} \le \sqrt{n^2 + 1} \implies \\
\implies x_n < \sqrt{n^2 + 1} - \sqrt{n^2 - 1}
$$
Or:
$$
x_n < \frac{(\sqrt{n^2 + 1} - \sqrt{n^2 - 1})(\sqrt{n^2 + 1} + \sqrt{n^2 - 1})}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} = \\
= \frac{2}{\sqrt{n^2 + 1} + \sqrt{n^2 - 1}} <2
$$
Also $x_n \gt0$ so finally:
$$
0 < x_n <2
$$
Have i done it the right way?
| This may be weaker than the other ways (that are nice, including OP's), but seemed like an easy way to see it; note that $n \le \sqrt[3]{n^3+1} \le n+1$, and $n-1 \le \sqrt{n^2-1} \le n$ [can be verified by cubing & squaring, respectively]
So $\sqrt[3]{n^3+1} - \sqrt{n^2-1} \ge n - n = 0$ and $\sqrt[3]{n^3+1} - \sqrt{n^2-1} \le (n+1) - (n-1) = 2$.
| {
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Evaluate $\lim\limits_{n \to \infty}\frac{n}{\sqrt[n]{n!}}$. Solution
Notice that
$$(\forall x \in \mathbb{R})~~e^x=1+x+\frac{x^2}{2!}+\cdots+\frac{x^n}{n!}+\cdots.$$
Let $x=n$ where $n\in \mathbb{N_+}$. Then we obtain
$$e^n=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\cdots>\frac{n^n}{n!}.$$
Thus, we obtain
$$e>\frac{n}{\sqrt[n]{n!}}.\tag1$$
Moreover, we can find that, for $k=0,1,\cdots,n-1.$ $$\frac{n^k}{k!}< \frac{n^n}{n!}.$$
Thus
\begin{align*}
e^n&=1+n+\frac{n^2}{2!}+\cdots+\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)(n+2)}+\cdots\right]\\
&< (n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot\left[\frac{n}{n+1}+\frac{n^2}{(n+1)^2}+\cdots\right]\\
&=(n+1)\cdot\frac{n^n}{n!}+\frac{n^n}{n!}\cdot n\\
&=(2n+1)\cdot \frac{n^n}{n!},
\end{align*}
which shows that
$$\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}.\tag2$$
Combining $(1)$ and $(2)$, we have
$$e>\frac{n}{\sqrt[n]{n!}}>\frac{e}{\sqrt[n]{2n+1}}\to e(n \to \infty).$$
By the squeeze theorem,
$$\lim_{n \to \infty}\frac{n}{\sqrt[n]{n!}}=e.$$
Please correct me if I'm wrong. Many thanks.
| For the denominator, rewrite it as $ e^{\frac{\sum_{k=1}^{n} \log k}{n}}$ then approximate the sum of logs with the integral ($n \log n -n) $ to get the result.
| {
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How can partial fractions be used for deductions?
Find partial fractions of the expression,$\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}$
. Hence deduce that; $\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)}+\frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)}+\frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)}=a+b+c+d-p-q-r-s$
My Working
I was able to calculate partial fractions as follows,
$\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}=1+\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)(x-a)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)(x-b)}+..$
But I cannot proceed to deduction part. Highly appreciated if someone can give me a hint to work this out. Thank you!
| You already got the correct partial fraction decomposition
$$
\frac{(x-p)(x-q)(x-r)(x-s)}{(x-a)(x-b)(x-c)(x-d)}
=1+\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)} \cdot \frac{1}{x-a} \\
+ \frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)} \cdot \frac{1}{x-b} \\
+ \frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)} \cdot \frac{1}{x-c} \\
+ \frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)} \cdot \frac{1}{x-d} \, .
$$
With the substitution $x = \frac 1y$ we get
$$ \tag{*}
\frac{(1-py)(1-qy)(1-ry)(1-sy)}{(1-ay)(1-by)(1-cy)(1-dy)}
=1+\frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)} \cdot \frac{y}{1-ay} \\
+ \frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)} \cdot \frac{y}{1-by} \\
+ \frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)} \cdot \frac{y}{1-cy} \\
+ \frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)} \cdot \frac{y}{1-dy} \, .
$$
For small $y$ the left-hand side has the development
$$
\frac{1 - (p+q+r+s)y + O(y^2)}{1-(a+b+c+d)y + O(y^2)}
= (1 - (p+q+r+s)y + O(y^2))(1+(a+b+c+d)y + O(y^2) \\
= 1 + (a+b+c+d-p-q-r-s)y + O(y^2)
$$
for $y \to 0$. On the right-hand side we have
$$
\frac{y}{1-ay} = y(1+ay + O(y^2)) = y + O(y^2)
$$
and similarly for the other fractions $\frac{y}{1-by}$, $\frac{y}{1-cy}$
and $\frac{y}{1-dy}$.
Therefore a comparison of the $y$-terms (i.e. the derivates with respect
to $y$ at $y = 0$) in equation $(**)$ gives the desired identity
$$
a+b+c+d-p-q-r-s = \frac{(a-p)(a-q)(a-r)(a-s)}{(a-b)(a-c)(a-d)}+\frac{(b-p)(b-q)(b-r)(b-s)}{(b-a)(b-c)(b-d)} \\
+\frac{(c-p)(c-q)(c-r)(c-s)}{(c-a)(c-b)(c-d)} \\
+\frac{(d-p)(d-q)(d-r)(d-s)}{(d-a)(d-b)(d-c)} \, .
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2969555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$
Find the remainder $R$ of $(1^2+1)(2^2+1)...(p^2+1)$ divided by $p$, with $p$ being a prime number greater than $3$.
For $p \equiv 1 \mod 4$, there exists an integer $j$ such that $p\mid j^2+1$ (since $-1$ is a quadratic residue of $p$), therefore $R=0$.
For $p \equiv 3 \mod 4$, how can we find $R$? Does $R$ depend on the value of $p$?
| Let $p$ be a prime, with $p\equiv 3\;(\text{mod}\;4)$.
Claim:
$$\prod_{k=1}^p (1+k^2)\equiv 4\;(\text{mod}\;p)$$
Proof:
\begin{align*}
\prod_{k=1}^p (1+k^2)\;(\text{mod}\;p)
&\equiv
\prod_{k=1}^{p-1} (1+k^2)\;(\text{mod}\;p)
\\[4pt]
&\equiv
\left(\prod_{k=1}^{\frac{p-1}{2}}(1+k^2)\right)^{\!\!2}\;(\text{mod}\;p)
\\[4pt]
\end{align*}
Thus, to prove the claim, it suffices to show
$$
\prod_{k=1}^{\frac{p-1}{2}}(1+k^2)\equiv 2\;(\text{mod}\;p)
$$
Note that
$$1+1^2,1+2^2,...,1+\left({\small{\frac{p-1}{2}}}\right)^{\!2}$$
are distinct mod $p$, and are the roots of the congruence $f(x)\equiv 0\;(\text{mod}\;p)$, where
$$f(x)=(x-1)^{\frac{p-1}{2}}-1$$
Since $p\equiv 3\;(\text{mod}\;4)$, it follows that $\frac{p-1}{2}$ is odd, hence the constant term of $f$ is $-2$.$\\[4pt]$
Note that $f$ is monic, hence by Vieta's formula, since $\frac{p-1}{2}$ is odd, and the constant term of $f$ is $-2$, the product in $Z_p$ of the roots of $f$ is $2$, so we get
$$
\prod_{k=1}^{\frac{p-1}{2}}(1+k^2)\equiv 2\;(\text{mod}\;p)
$$
which proves the claim.
Thus, if $p$ is prime, with $p\equiv 3\;(\text{mod}\;4)$, we get $R\equiv 2^2\;(\text{mod}\;p)$, hence
$$
\begin{cases}
R=4,\;\text{if}\;p > 3\\[4pt]
R=1,\;\text{if}\;p = 3\\
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2972213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Proving the given inequalities Q: Prove the given inequalities for positive a,b,c:$(i) \left[\frac{bc+ca+ab}{a+b+c}\right]^{a+b+c}>\sqrt{(bc)^a.(ca)^b.(ab)^c}$$(ii) \left(\frac{a+b+c}{3} \right)^{a+b+c}<a^ab^bc^c<\left(\frac{a^2+b^2+c^2}{a+b+c}\right)^{a+b+c} $I know that G.M$\le$A.M and somehow i guess it must be used.But i really struggling to prove this kind of inequality.Any hints or solution will be appreciated.And i do apologize if this question is very basic.Thanks in advance.
| $a)$We have by the weighted AM-GM inequality: $\sqrt{(ab)^{\frac{c}{a+b+c}}\cdot (bc)^{\frac{a}{a+b+c}}\cdot (ca)^{\frac{b}{a+b+c}}}\le \sqrt{\dfrac{3abc}{a+b+c}}\le \dfrac{ab+bc+ca}{a+b+c}\iff(ab+bc+ca)^2 \ge 3abc(a+b+c)\iff (ab)^2+(bc)^2+(ca)^2 \ge abc(a+b+c) $ which is true.
$b)$ Consider $f(x) = x\ln x \implies f''(x) = \dfrac{1}{x} >0 $. Thus $f$ is convex and using $f(a)+f(b)+f(c) \ge 3f\left(\frac{a+b+c}{3}\right)$, the answer follows for the left inequality. For the right one, apply again the weighted AM-GM inequality $a^{\frac{a}{a+b+c}}\cdot b^{\frac{b}{a+b+c}}\cdot c^{\frac{c}{a+b+c}}\le \dfrac{a^2+b^2+c^2}{a+b+c}$ . QED.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Shortest distance between a point and a parabola. Find the shortest distance from the point $(1, 0)$ to the parabola $y^2 = 4x$.
As-
Let $D^2=(x-1)^2 + y^2$. Now reduce this to one variable by putting $y^2 = 4x$ to get $D^2 = (x-1)^2+ 4x$. We are working with $D^2$ to make the calculations easier. Clearly $$\frac{d}{dx}D^2 = 2(x-1)+4 = 0 \implies x = -1$$ At $x = -1$, we have $\frac{d^2}{dx^2}D^2 = 2 < 0$ which shows that $D^2$ is minimum. At this $x = -1$.
But this is not a point on parabola. Am i doing some mistake or we have to follow some other method to get the minima?
| Let $\left(\frac{y^2}{4},y\right)$ be a point on our parabola.
Thus, we need to find a minimal value of the following expression.
$$\sqrt{\left(\frac{y^2}{4}-1\right)^2+y^2},$$ which is equal to $1$ for $y=0$.
We'll prove that it's a minimal value.
Indeed, we need to prove that:
$$\left(\frac{y^2}{4}-1\right)^2+y^2\geq1$$ or
$$\frac{y^2}{2}+\frac{y^4}{16}\geq0,$$ which is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Sum the first $n$ terms of the series $1 \cdot 3 \cdot 2^2 + 2 \cdot 4 \cdot 3^2 + 3 \cdot 5 \cdot 4^2 + \cdots$ The question
Sum the first $n$ terms of the series:
$$ 1 \cdot 3 \cdot 2^2 + 2 \cdot 4 \cdot 3^2 + 3 \cdot 5 \cdot 4^2 + \cdots. $$
This was asked under the heading using method of difference and the answer given was
$$ S_n = \frac{1}{10}n(n+1)(n+2)(n+3)(2n+3). $$
My approach
First, I get $$ U_n=n(n+2)(n+1)^2. $$
Then I tried to make $U_n = V_n - V_{n-1}$
in order to get $S_n = V_n - V_0$.
But I really don't know how can I figure this out.
| Rewrite the formula as
$$\sum_2^{n+1}k^2(k+1)(k-1)=\sum_2^{n+1}k^4-\sum_2^{n+1}k^2.$$
It is a well known formula that
$$1+2^2+3^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}.$$
Due to Fermat, we also have
$$1+2^4+3^4+\cdots+n^4=\frac15n^5+\frac12n^4+\frac13n^3-\frac1{30}n.$$
Just substitute in your values to get the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2973979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the smallest integer greater than 1 such that $\frac12$ of it is a perfect square and $\frac15$ of it is a perfect fifth power?
What is the smallest integer greater than 1 such that $\frac12$ of it is a perfect square and $\frac15$ of it is a perfect fifth power?
I have tried multiplying every perfect square (up to 400 by two and checking if it is a perfect 5th power, but still nothing. I don't know what to do at this point.
| Here's a very unsophisticated approach: Let $n$ be the smallest such integer. Then there exist integers $a$ and $b$ such that $n=5a^5$ and $n=2b^2$. It follows that $a$ is a multiple of $2$, say $a=2a_1$, and $b$ is a multiple of $5$, say $b=5b_1$. Then
$$n=2^5\cdot5\cdot a_1^5\qquad\text{ and }\qquad n=2\cdot5^2\cdot b_1^2.$$
This in turn shows that $a_1$ is a multiple of $5$, say $a_1=5a_2$, and $b_1$ is a multiple of $2$, say $b_1=2b_2$. Then
$$n=2^5\cdot5^6\cdot a_2^5\qquad\text{ and }\qquad n=2^3\cdot5^2\cdot b_2^2.$$
This in turn shows that $b_2$ is a multiple of both $2$ and $5^2$, say $b_2=2\cdot5^2\cdot b_3$. Then
$$n=2^5\cdot5^6\cdot a_2^5\qquad\text{ and }\qquad n=2^5\cdot5^6\cdot b_3^2.$$
This shows that $n\geq2^5\cdot5^6$, and as you might expect a quick check shows that $n=2^5\cdot5^6$ does indeed work, so $n=2^5\cdot5^6=500000$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2976181",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
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Derivation of the tangent half angle identity
I'm having trouble proceeding from
$$\frac{\sin(\theta)}{1+\cos(\theta)}$$
to
$$\tan\left(\frac{\theta}{2}\right)$$
Context:
Consider the function $f$ defined for all $(x,y)$ such that $y \neq 0$, with the rule $$f(x,y) = \frac{y}{\sqrt{x^2+y^2}+x}$$
Show that $$f(r\cos(\theta),r\sin(\theta)) = \tan\left(\frac{\theta}{2}\right)$$
So far I've done:
$$f(r\cos(\theta),r\sin(\theta)) = \frac{r\sin(\theta)}{\sqrt{r^2\cos^2(\theta) + r^2\sin^2(\theta)}+r\cos(\theta)} = \frac{r\sin(\theta)}{\sqrt{r^2}+r\cos(\theta)}\\=\frac{\sin(\theta)}{1+\cos(\theta)}$$
Using $$\cos(\theta) = 2\cos^2\left(\frac{\theta}{2}\right)-1 \implies 2\cos^2\left(\frac{\theta}{2}\right)=\cos(\theta)+1$$
We get $$f(r\cos(\theta),r\sin(\theta)) = \frac{\sin(\theta)}{2\cos^2\left(\frac{\theta}{2}\right)}$$
But I can't see how to proceed from here to the required result. Thanks in advance for any help!
| Let $L=\frac{\sin \theta}{1+\cos \theta}$ and $R=\tan(\theta/2)$.
Left
\begin{align}
\frac{\textrm{d}L}{\textrm{d}\theta} &= \frac{\cos\theta(1+\cos\theta)-\sin\theta (-\sin\theta)}{(1+\cos \theta)^2}\\
&=\frac{1}{1+\cos\theta}
\end{align}
Again
\begin{align}
\frac{\textrm{d}^2L}{\textrm{d}\theta^2} &= -(1+\cos\theta)^{-2}(-\sin\theta)\\
&=\frac{\sin\theta}{1+\cos\theta}\frac{1}{1+\cos\theta}\\
&=L\frac{\textrm{d}L}{\textrm{d}\theta}
\end{align}
Right
\begin{align}
\frac{\textrm{d}R}{\textrm{d}\theta} &= \frac{\sec^2(\theta/2)}{2}\\
&= \frac{1+\tan^2(\theta/2)}{2}\\
2\frac{\textrm{d}R}{\textrm{d}\theta} &= 1+R^2
\end{align}
Again
\begin{align}
2\frac{\textrm{d}^2R}{\textrm{d}\theta^2} &= 2R\frac{\textrm{d}R}{\textrm{d}\theta}\\
\frac{\textrm{d}^2R}{\textrm{d}\theta^2} &= R\frac{\textrm{d}R}{\textrm{d}\theta}
\end{align}
Conclusion
\begin{align}
L&=R\\
\frac{\sin \theta}{1+\cos \theta}&=\tan(\theta/2)
\end{align}
Edit
Assume that the questioner and his friend were asked to solve a differential equation $y''=yy'$ with $y(0)=0$ and $y'(0)=1/2$.
Making the story short, he and his friend found the solution, $y=\frac{\sin\theta}{1+\cos\theta}$ and $y=\tan(\theta/2)$, respectively.
He now can proceed from $\frac{\sin\theta}{1+\cos\theta}$ to $\tan(\theta/2)$
without doubt.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How many integers $a,b,c$, both positive and negative, such that $P=a^b b^c c^a$ is a prime number?
How many integers $a,b,c$, both positive and negative, such that $P=a^b b^c c^a$ is a prime number ?
If $a,b,c$ are positive, then two of $a,b,c$ equal to $1$. Assume that $b=c=1$, then $a$ is any prime number.
WLOG, if $c<0$ and $a,b>0$, then $a$ must be even, so $a=2$, thus $b=1$ and $c=-1$
If $a,b,c<0$, then $P$ is smaller than $1$, so $P$ won't be a prime.
However if only one of $a,b,c$ is positive, how can we find $a,b,c$ ?
| WLOG assume $b,c<0$, $a>0$. Then $a^b$ and $b^c$ will be fractional unless $a=1$, $b=-1$. So we need to find $1\cdot (-1)^c \cdot c=p$ for $c<0$. Since $c$ is negative, it must also be odd so that $(-1)^c$ will be negative and the entire product will be positive. Hence, $c$ can be the negative of any odd prime.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2977702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Find $\cos(\alpha+\beta)$ if $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$
If $\alpha$, $\beta$ are the roots of the equation $a\cos x+b\sin x=c$, then prove that $\cos(\alpha+\beta)=\dfrac{a^2-b^2}{a^2+b^2}$
My Attempt
$$
b\sin x=c-a\cos x\implies b^2(1-\cos^2x)=c^2+a^2\cos^2x-2ac\cos x\\
(a^2+b^2)\cos^2x-2ac\cos x+(c^2-b^2)=0\\
\implies\cos^2x-\frac{2ac}{a^2+b^2}\cos x+\frac{c^2-b^2}{a^2+b^2}=0
$$
$$
a\cos\alpha+b\sin\alpha=c\implies a\cos^2\alpha\cos\beta+b\sin\alpha\cos\alpha\cos\beta=c\cos\alpha\cos\beta\\
a\cos\beta+b\sin\beta=c\implies a\sin\alpha\sin\beta\cos\beta+b\sin\alpha\sin^2\beta=c\sin\alpha\sin\beta\\
c\cos(\alpha+\beta)=a\cos\beta+a\sin\alpha\cos\beta.(\sin\beta-\sin\alpha)+b\sin\alpha+b\sin\alpha\cos\beta(\cos\alpha-\cos\beta)\\
$$
I think its getting complicated to solve now. What is the simplest way to solve this kind of problems?
| To begin with, notice that
\begin{align*}
& a\cos(x) + b\sin(x) = c \Longleftrightarrow \frac{a}{\sqrt{a^{2}+b^{2}}}\cos(x) + \frac{b}{\sqrt{a^{2}+b^{2}}}\sin(x) = \frac{c}{\sqrt{a^{2}+b^{2}}}\Longleftrightarrow\\
& \sin(\theta + x) = \frac{c}{\sqrt{a^{2}+b^{2}}}\quad\text{where}\quad \sin(\theta) = \frac{a}{\sqrt{a^{2}+b^{2}}}\,\,\text{and}\,\,\cos(\theta) = \frac{b}{\sqrt{a^{2}+b^{2}}}\\\\
&\therefore \alpha = \arcsin\left(\frac{c}{\sqrt{a^{2}+b^{2}}}\right) - \theta\quad\text{and}\quad\beta = \pi - \arcsin\left(\frac{c}{\sqrt{a^{2}+b^{2}}}\right) - \theta
\end{align*}
Finally, we get
\begin{align*}
\cos(\alpha+\beta) = \cos(\pi-2\theta) = -\cos(2\theta) = 2\sin^{2}(\theta)-1 = \frac{2a^{2}}{a^{2}+b^{2}} - 1 = \frac{a^{2}-b^{2}}{a^{2}+b^{2}}
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Determinant 4x4 This looks very simple but I guess I made a mistake somwhere because the result should be 17
$$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 1 & 2 & 0 \\ 1 & -2 & 0 & 1 \\ 2 & -1 & -1 & 1 \end{vmatrix}$$
First I added negative 1st row to 3rd
$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 1 & 2 & 0 \\ 0 & -4 & -4 & 3 \\ 2 & -1 & -1 & 1 \end{vmatrix}$
Then I multiplicate 2nd row times $4$, so the whole determinant will be multiplied with $0.25$.
Then I added 3rd row to the second
$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 0 & 4 & 0 \\ 0 & -4 & -4 & 3 \\ 2 & -1 & -1 & 1 \end{vmatrix}$
Now I cross 3rd column and 2nd row which would be $4((-1)^5)$ plus the rest which would be $-2(-5)$ = $10$ not $17$
Thanks for help
| You start with
$$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 1 & 2 & 0 \\ 1 & -2 & 0 & 1 \\ 2 & -1 & -1 & 1 \end{vmatrix}$$
and then, to the third row, you subtract the first, giving
$$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 1 & 2 & 0 \\ 0 & -4 & -4 & 3 \\ 2 & -1 & -1 & 1 \end{vmatrix}$$
then duplicate 2nd row to get
$$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & 2 & 4 & 0 \\ 0 & -4 & -4 & 3 \\ 2 & -1 & -1 & 1 \end{vmatrix}$$
and then, to the 2nd row, add the 3rd, to get
$$\begin{vmatrix} 1 & 2 & 4 & -2 \\ 0 & -2 & 0 & 3 \\ 0 & -4 & -4 & 3 \\ 2 & -1 & -1 & 1 \end{vmatrix}$$
notice how the last element of the 2nd row now is a $3$.
Using Laplace formula over the 1st column we get
$$1\times \begin{vmatrix}-2 & 0 & 3 \\ -4 & -4 & 3 \\ -1 & -1 & 1 \end{vmatrix} - 2\times \begin{vmatrix} 2 & 4 & -2 \\ -2 & 0 & 3 \\ -4 & -4 & 3 \end{vmatrix}$$
Because
$$\begin{vmatrix}-2 & 0 & 3 \\ -4 & -4 & 3 \\ -1 & -1 & 1 \end{vmatrix} = 2, \begin{vmatrix} 2 & 4 & -2 \\ -2 & 0 & 3 \\ -4 & -4 & 3 \end{vmatrix} = -16$$
we get
$$0.5\times\left(1\times 2 - 2\times (-16) \right) = 17$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2981600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Determining what powers come out after polynomial multiplication Is there a quick method to determine what powers come out after polynomial multiplication? Specifically, I'm working with raising a polynomial by an integral power, so the binomial/multinomial theorem would be useful (though I have no idea how to use it).
For example, expanding out $(x+x^2+x^5)^3$ gives me:
$$=x^{15}+3x^{12}+3x^{11}+3x^9+6x^8+3x^7+x^6+3x^5+3x^4+x^3$$
Is there a quick way to know what powers come out (i.e. $15,12,11,\cdots$) just using the given powers of the unexpanded polynomial.
| So for this case, the powers you can get are all the sums you can make using three terms (due to the power outside of the parenthesis) and the numbers 1,2, and 5 (the powers of the X's).
So since we allow for a given number to be repeated, there should be 10 different powers in the expansion (choose 3 terms from 3 choices with replacement):
$1+1+1=3$
$1+1+2=4$
$1+2+2=5$
$2+2+2=6$
$5+1+1=7$
$5+2+1=8$
$5+2+2=9$
$5+5+1=11$
$5+5+2=12$
$5+5+5=15$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding an integrating factor and solving: $(2x \sin(x + y) + \cos(x + y))dx + \cos(x + y)dy = 0$ I am trying to find an integrating factor and solve the following differential equation:
$$(2x \sin(x + y) + \cos(x + y))dx + \cos(x + y)dy = 0$$
These are my steps:
$$(2x \sin(x + y) + \cos(x + y)) + \cos(x + y)dy/dx = 0$$
I check if the equation is exact:
\begin{equation}
\partial U_{xy} = \partial U_{xy}
\end{equation}
\begin{equation}
2x\cos \left(x+y\right)-\sin \left(x+y\right) \neq -\sin \left(x+y\right),
\end{equation}
Its not so I need to find an integrating factor such that
$$\frac{d}{dy} \left( ΞΌ(x)2x\cos \left(x+y\right)-\sin \left(x+y\right) \right)= \frac{d}{dx} \left( ΞΌ(x)-\sin \left(x+y\right) \right)$$
And at this point I simply get stuck. Any help or advice would be appreciated.
| $$(2x \sin(x + y) + \cos(x + y))dx + \cos(x + y)dy = 0$$
$$2x \sin(x + y)dx +( \cos(x + y))(dx +dy) = 0$$
Substitute $v=x+y$
$$2x \sin(v)dx + \cos(v)dv = 0$$
It's not exact. Multiply by $\mu=e^{x^2}$ as integrating factor
$$2xe^{x^2} \sin(v)dx + e^{x^2}\cos(v)dv = 0$$
The diffrential is exact..
$$\boxed{e^{x^2} \sin(x+y)=K}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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length of tangent to a curve passing through another point Let $A$ be a point on the curve
$\mathcal{C} : x^2+y^2-2x-4=0$
If the tangent line to $\mathcal{C}$ at $A$ passes through $P(4,3)$, then what is the length of AP?
Please, include a general method of approaching similar kind of questions.
| Welcome to MSE.Hint: first you have to find the center (o) of the curve which is a circle with equation:
$x^2-2x+y^2-4=(x-1)^2+(y-0)^2-5=0r^2$ or$(x-1)^2+(y-0)^2=5=r^2$
Where r is the radius of circle and O(-1, 0) is its center. In right triangle OAP we can write:
$AP^2=OP^2-r^2$
$OP^2=(4-1)^2+(3-0)^2=3^2+3^2=18$
$AP^2=18-5=13$ β $AP=\sqrt {13}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Inequality in difference of square roots I think this inequality is true and I'm trying to prove it:
For real, non-negative $a$ and $b$
$$\lvert \sqrt{a} - \sqrt{b}\rvert \le \lvert \sqrt{a - b} \rvert$$
Closest thing I've found is equation
$$\sqrt{a} + \sqrt{b} = \sqrt{a + b + \sqrt{4ab}}$$
But this breaks if I set $b$ to $-b$ because then you're taking the root of a negative.
Does anyone have a proof of this, or a more general result that implies this?
| This is not true if $a < b$. If $a < b $ then $a -b < 0$ and $\sqrt{a-b}$ does not exist and $|\sqrt{a-b}|$ does not exist. And so it is not true that $|\sqrt{a} - \sqrt{b}| \le |\sqrt{a-b}|$.
I think what you meant was to prove $|\sqrt{a} -\sqrt{b}| \le \sqrt{|a - b|}$ which is true.
Notice: $\min (a,b) \le \sqrt{ab}\le \max(a,b)$. Why? If $c = \min (a,b)$ and $d = \max(a,b)$ then $c =\sqrt{c\cdot c}\le \sqrt{c\cdot d}(=\sqrt{ab})\le \sqrt{d\cdot d} = d$.
So. Letting $c = \min(a,b)$ and $d=\max(a,b)$ we have
$|\sqrt a-\sqrt b|^2 = (\sqrt{d} - \sqrt{x})^2 = d -2\sqrt{cd} + c\le d - 2c+c = d-c= |a-b|$.
And as we are dealing with non-negative values it follows
$\sqrt {|\sqrt{a} - \sqrt{b}|^2}\le \sqrt{|a-b|}$ or
$|\sqrt{a} - \sqrt{b}|^2 \le |a-b|$
.....
Although an easier way to do it is if we don't get bogged down in absolute values; they are only there to assure to take the positive values anyway.
For any $0 \le b\le a$ then
$\sqrt a - \sqrt b \le \sqrt{a-b} \iff$
$(\sqrt a - \sqrt b)^2 \le (\sqrt{a-b})^2 \iff$
$a -2\sqrt{ab} + b \le a -b\iff$
$b \le \sqrt{ab}$ and as $b < a$ it follow $b = \sqrt{b^2}\le \sqrt{ab}$.
....
Or directly:
$\sqrt{a - b} = \sqrt{a -2b + b}=\sqrt{a -2\sqrt{b}\sqrt{b} + b} \ge \sqrt{a - 2\sqrt{a}\sqrt{b} + b}=\sqrt{(\sqrt a - \sqrt b)^2} = \sqrt a - \sqrt b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Evaluating $\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}$ without L'Hopital I have the following limit question, where different indices of roots appear in the numerator and the denominator
$$\lim_{x\to3}\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1}.$$
As we not allowed to use L'Hopital, I want to learn how we can proceed algebraically.
| A possible purely algebraic way can use
$$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$$
Using this formula note that
*
*$\sqrt[4]{x-2} - 1 = \frac{\left(\sqrt[4]{x-2}\right)^4 - 1^4 }{\left(\sqrt[4]{x-2}\right)^3 + \left(\sqrt[4]{x-2}\right)^2+ \sqrt[4]{x-2} + 1} = \frac{x-3 }{\left(\sqrt[4]{x-2}\right)^3 + \left(\sqrt[4]{x-2}\right)^2+ \sqrt[4]{x-2} + 1}$
*$\sqrt[3]{x+5} - 2 = \frac{\left(\sqrt[3]{x+5}\right)^3- 2^3 }{\left(\sqrt[3]{x+5}\right)^2+ \sqrt[3]{x+5}\cdot 2 + 2^2} = \frac{x-3 }{\left(\sqrt[3]{x+5}\right)^2+ \sqrt[3]{x+5}\cdot 2 + 2^2}$
So, you get
$$\frac{\sqrt[3]{x+5}-2}{\sqrt[4]{x-2}-1} = \frac{\left(\sqrt[4]{x-2}\right)^3 + \left(\sqrt[4]{x-2}\right)^2+ \sqrt[4]{x-2} + 1}{\left(\sqrt[3]{x+5}\right)^2+ \sqrt[3]{x+5}\cdot 2 + 2^2} \stackrel{x \to 3}{\longrightarrow}\frac{4}{12} = \frac{1}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2986763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Show that $2 < (1+\frac{1}{n})^{n}< 3$ without using log or binommial coefficient $2 <(1+\frac{1}{n})^{n}< 3$
Is it possible to show the inequality without using binommial coefficients thus only by induction? The leftinequality can be shown using bernoulli inequality.
| I have found the answer to my problem in a different forum:
We show that $b_n > b_{n+1}$ with $b_n:= (1+\frac{1}{n})^{n+1}$
Proof:
To show: $\frac{b_n}{b_{n+1}}\geq 1$
$\frac{b_n}{b_{n+1}} = \frac{\left(1+\frac{1}{n}\right)^{n+1}}{\left(1+\frac{1}{n+1}\right)^{n+2}} = \frac{1}{1+\frac{1}{n}} \cdot \left(\frac{1+\frac{1}{n}}{1+\frac{1}{n+1}}\right)^{n+2} = \frac{1}{1+\frac{1}{n}} \cdot \left(1+\frac{1}{n(n+2)}\right)^{n+2}$
Exploiting the bernoulli inequality
$\frac{1}{1+\frac{1}{n}} \cdot \left(1+\frac{1}{n(n+2)}\right)^{n+2} \geq 1$
$b_5 < 3$ and $(1+\frac{1}{n})^n < b_n $
Thus $\forall n\geq 5: (1+\frac{1}{n})^n < 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\sum_{(a,b,c)\in T}\frac{2^a}{3^b 5^c}$, for $T$ the set of all positive integer triples $(a,b,c)$ forming a triangle
Let $T$ be the set of all triples $(a,b,c)$ of positive integersΒ for which there exist triangles with side lengths $a$, $b$, $c$. Express
$$\sum\limits_{(a,b,c)\,\in\,T}\;\frac{2^a}{3^b 5^c}$$
as a rational number in lowest terms
I really don't know to start with this question. My solution: I have written equation of solution of triangle in form of $a$, $b$, and $c$; i.e, $s/(s-a)$, where $s = (a+b+c)/2$.
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
\newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
\newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
\newcommand{\dd}{\mathrm{d}}
\newcommand{\ds}[1]{\displaystyle{#1}}
\newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
\newcommand{\pars}[1]{\left(\,{#1}\,\right)}
\newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
\newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
\newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
\newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
$\ds{\sum_{\pars{a,b,c}\ \in\ T}
{2^{a} \over 3^{b}5^{c}}:\ {\LARGE ?}\qquad
\\[2mm] T \equiv
\braces{\pars{a,b,c}\ \mid\ a,b,c \in \mathbb{N}_{\geq 1}\ \mbox{and}\ \verts{b - c} < a < b + c}}$
I'll evaluate
\begin{align}
&\bbox[10px,#ffd]{\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}
\sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}}
\bracks{\vphantom{\large A}\verts{b - c} < a < b + c}}
\\[5mm] = &\
\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}
\sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}}
\bracks{b < c}\bracks{\vphantom{\large A}c - b < a < c + b}
\\[2mm] + &\
\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}
\sum_{c = 1}^{\infty}{x^{a} \over \pars{yz}^{b}}
\bracks{b = c}\bracks{\vphantom{\large A}a < 2b}
\\[2mm] + &\
\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}
\sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}}
\bracks{b > c}\bracks{\vphantom{\large A}b - c < a < b + c}
\\[5mm] = &\
\overbrace{\sum_{b = 1}^{\infty}{1 \over y^{b}}
\sum_{c = b + 1}^{\infty}{1 \over z^{c}}
\sum_{a = c - b + 1}^{c + b - 1}x^{a}}
^{\mbox{See expression}\ {\large\eqref{2}}}\ +\
\overbrace{\sum_{b = 1}^{\infty}{1 \over \pars{yz}^{b}}
\sum_{a = 1}^{2b - 1}x^{a}}
^{\mbox{See expression}\ {\large\eqref{3}}}
\\[2mm] + &\
\underbrace{\quad\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}
\sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}}
\bracks{b > c}\bracks{\vphantom{\large A}b - c < a < b + c}\quad}
_{\ds{\mbox{It's equal to the}\ first\ term\
\mbox{under the exchange}\ y \leftrightarrow z}}
\label{1}\tag{1}
\end{align}
The RHS first term becomes:
\begin{align}
&\bbox[#ffd,10px]{\sum_{b = 1}^{\infty}{1 \over y^{b}}
\sum_{c = b + 1}^{\infty}{1 \over z^{c}}
\sum_{a = c - b + 1}^{c + b - 1}x^{a}} =
\sum_{b = 1}^{\infty}{1 \over y^{b}}
\sum_{c = b + 1}^{\infty}{1 \over z^{c}}\,
x^{c - b + 1}\,{x^{2b - 1} - 1 \over x - 1}
\\[5mm] = &\
\sum_{b = 1}^{\infty}{1 \over y^{b}}
\,{x^{b} - x^{-b + 1} \over x - 1}
\sum_{c = b + 1}^{\infty}\pars{x \over z}^{c}
\\[5mm] = &\
{1 \over x - 1}\sum_{b = 1}^{\infty}
\,\bracks{\pars{x \over y}^{b} - {x \over \pars{xy}^{b}}}
{\pars{x/z}^{b + 1} \over 1 - x/z}
\\[5mm] = &\
{z \over \pars{x - 1}\pars{z - x}}\sum_{b = 1}^{\infty}
\,\bracks{{x \over z}\pars{x^{2} \over yz}^{b} - {x^{2} \over z} \pars{1 \over yz}^{b}}
\\[5mm] = &\
{z \over \pars{x - 1}\pars{z - x}}
\,\bracks{{x \over z}{x^{2}/\pars{yz} \over 1 - x^{2}/\pars{yz}} - {x^{2} \over z}{1/\pars{yz} \over 1 - 1/\pars{yz}}}
\\[5mm] = &\
\bbx{{x^{3} \over \pars{x - 1}\pars{z - x}\pars{yz - x^{2}}} -
{x^{2} \over \pars{x - 1}\pars{z - x}\pars{yz - 1}}}
\label{2}\tag{2}
\end{align}
The RHS second term in \eqref{1} is given by
\begin{align}
&\bbox[10px,#ffd]{\sum_{b = 1}^{\infty}{1 \over \pars{yz}^{b}}\,
\sum_{a = 1}^{2b - 1}x^{a}} =
\sum_{b = 1}^{\infty}\pars{1 \over yz}^{b}\,
x\,{x^{2b - 1} - 1 \over x - 1}
\\[5mm] = &\
{1 \over x - 1}\bracks{%
\sum_{b = 1}^{\infty}\pars{x^{2} \over yz}^{b} -
x\sum_{b = 1}^{\infty}\pars{1 \over yz}^{b}}
\\[5mm] = &\
{1 \over x - 1}\bracks{%
{x^{2}/\pars{yz} \over 1 - x^{2}/\pars{yz}} -
x\,{1/\pars{yz} \over 1 - 1/\pars{yz}}}
\\[5mm] = &\
\bbx{{1 \over x - 1}\pars{%
{x^{2} \over yz - x^{2}} -{x \over yz - 1}}}
\label{3}\tag{3}
\end{align}
The last term in \eqref{1} RHS is found, as pointed out above, with the exchange
$\ds{y \leftrightarrow z}$ in \eqref{2}.
The final result is given by
$$
\bbx{\bbox[10px,#ffd]{\sum_{a = 1}^{\infty}\sum_{b = 1}^{\infty}
\sum_{c = 1}^{\infty}{x^{a} \over y^{b}z^{c}}
\bracks{\vphantom{\large A}\verts{b - c} < a < b + c}} =
{x\pars{x + yz} \over \pars{x - y}\pars{x - z}\pars{yz - 1}}}
$$
When $\ds{x = 2\,,\ y = 3}$ and $\ds{z = 5}$, the above expression is reduced to:
$$
\bbx{\sum_{\pars{a,b,c}\ \in\ T}
{2^{a} \over 3^{b}5^{c}} = {17 \over 21}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2990350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
If $\sin x+\sin^2x+\sin^3x=1$, then find $\cos^6x-4\cos^4x+8\cos^2x$
If $\sin x+\sin^2x+\sin^3x=1$, then find $$\cos^6x-4\cos^4x+8\cos^2x$$
My Attempt
\begin{align}
\cos^2x&=\sin x+\sin^3x=\sin x\cdot\big(1+\sin^2x\big)\\
\text{ANS}&=\sin^3x\cdot\big(1+\sin^2x\big)^3-4\sin^2x\cdot\big(1+\sin^2x\big)^2+8\sin x\cdot\big(1+\sin^2x\big)\\
&=\sin x\cdot(1+\sin^2x)\bigg[\sin^2x\cdot(1+\sin^2x)^2-4\sin x\cdot(1+\sin^2x)+8\bigg]\\
&=
\end{align}
I don't think its getting anywhere with my attempt, so how do I solve it ?
Or is it possible to get the $x$ value that satisfies the given condition $\sin x+\sin^2x+\sin^3x=1$ ?
Note: The solution given in my reference is $4$.
| Given $$\sin x+\sin^2x+\sin^3x=1$$$$\sin x+\sin^3x=1-\sin^2x$$$$(\sin x+\sin^3x)^2=(1-\sin^2x)^2$$
$$\sin^2x+\sin^6x+2\sin^4x=\cos^4x$$
$$1-\cos^2x+(1-\cos^2x)^3+2(1-\cos^2x)^2=\cos^4x$$
$$1-\cos^2x+1-3\cos^2x+3\cos^4x-\cos^6x+1-4\cos^2x+2\cos^4x=\cos^4x$$$$\cos^6x-4\cos^4x+8\cos^2x=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
General way of solve $ax^2+by+c=0$ For example ,the diophantine equation
$$x^2+1=25y$$
we can solve this by finding particular solution $(x,y)=(7,2)$
and using this , we can get general solution.
My question is
"To solve $ax^2+by+c=0$ $(a, b, c \in \mathbb{Z})$ we must find particular solution or
not?"
| Since x and y have a difference like $d$,we can have a single variable quadratic equation by transformation.
If $x=y Β±d$ we have:
$a(y Β±d)^2+by+c=0$
$ay^2 +(b Β±2ad)y Β±ad^2+c=0$
$\Delta=(b Β±2ad)^2-4a(c Β±ad^2)>=0$
Limitation of value of $\Delta$ gives us a short range of numbers to try for solution. For example we solve $x^2+1=25y$:
We consider two cases:
$|x|>|y|$; so we can transform equation as follows:
Let $x= y+d$ , we have:
$(y+d)^2-25y+1=0$
$y^2+(2d-25)y+d^2+1=0$
$\Delta=-100 d +621$
For a solution in set of real numbers we must have:
$\Delta=-100 a +621>=0$ β $100a<=621$ β$a<6$ .
Now we have a small range to try, for example :
$a=5$ β$\Delta=121$ β $y=2$ and $y=13$
$y=2 β x=7$
$y=13 β x=18$.
Now suppose $|x|<|y|$, we have:
$(y-d)^2+1-25y=0$
$y^2+(-2a-25)y+a^2+1=0$
Which gives $y=2, x=7$ and $y=41, x=32$
For a general formula for x and y, the homogeneous equation $x\times x-25y=0$ gives $x=25$ and $y=1$ and we have:
$x=7$ as a single solution so the equation for x is $x=25t +7$.
Where t β Z. For y we put x in equation:
$(25t+7)^2+1=25y$ which gives $y=25t^2+14t+2$.
Hence there are infinitely many solutions for this equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2994635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Recurrence Relation - # of binary strings with given property Let $a_n$ be the number of binary strings of length $n$ with the property that each entry is adjacent to at least one entry of the same type.
ex: $11000111$ is a valid string but $11011000$ is not valid
$\textbf{(a) Find $a_1,a_2,a_3,a_4,a_5,a_6,a_7$}$
If someone can check that my attempt is correct, I would really appreciate it.
$a_1=0$ since we cannot have just $0$ or just $1$ as there will be no adjacent of the same type
$a_2=2$: either $00$ or $11$
$a_3=2$: either $000$ or $111$
$a_4=4$:
Reasoning:
$\textbf{If we start with a $0$}$: For the second entry we have $1$ choice as we are forced to put a $0$ since we started with a $0$. For the third entry we have $2$ choices, and similarly for the fourth entry we have $1$ choice. So there are $2$ such strings.
$\textbf{If we start with a $1$}$: For the second entry we are forced to put a $1$. For the third entry we have $2$ choices, and for the fourth entry we have $1$ choice. So there are $2$ such strings.
So $a_4=2+2=4$ strings.
Following the same method for the remaining:
$a_5=4$
$a_6=8$
$a_7=8$
$\textbf{(b) Find the recurrence relation for $a_n$}$
$$a_n=
\begin{cases}
2a_{n-2}&n \text{ even},\\
a_{n-1}&n \text{ odd}
\end{cases}$$
| You can take this as a sequence of stretches of at least two equal symbols. In ordinary generating function terms a sequence of two or more would be represented by:
$\begin{align*}
z^2 + z^3 + \dotsb
&= \frac{z^2}{1 - z}
\end{align*}$
A sequence of the above in turn would be:
$\begin{align*}
1 + \frac{z^2}{1 - z} + \left(\frac{z^2}{1 - z}\right)^2 + \dotsb
&= \frac{1}{1 - \frac{z^2}{1 - z}} \\
&= \frac{1 - z}{1 - z - z^2}
\end{align*}$
Consider the Fibonacci sequence, defined by $F_{n + 2} = F_{n + 1} + F_n$ with $F_0 = 0, F_1 = 1$. It's generating function is:
$\begin{align*}
F(z)
&= \sum_{n \ge 0} F_n z^n
= \frac{z}{1 - z - z^2}
\end{align*}$
But also:
$\begin{align*}
\sum_{n \ge 0} F_{n + 1} z^n
&= \frac{F(z) - F_0}{z}
= \frac{1}{1 - z - z^2}
\end{align*}$
We see that our generating function is:
$\begin{align*}
\frac{1 - z}{1 - z - z^2}
&= \sum_{n \ge 0} (F_{n + 1} - F_n) z^n \\
&= \sum_{n \ge 0} F_{n - 1} z^n
\end{align*}$
So the number of sequences of length $n$ is $F_{n - 1}$, where we extend to $F_{-1} = 1$ (as it should be to get one sequence of length 0).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2996975",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Given big rectangle of size $x, y$, count sum of areas of smaller rectangles. Let's say we have two integers $x$ and $y$ that describe one rectangle, if this rectangle is splitten in exactly $x\cdot y$ squares, each of size $1\cdot 1$, count the sum of areas of all rectangles that can be formed from those squares.
For example if $x = 2, y = 2$, we have $4$ rectangles of size $(1,1)$, $2$ rectangles of size $(1, 2)$, $2$ rectangles of size $(2, 1)$ and $1$ rectangle of size $(4,4)$. So the total answer is $4\cdot 1+ 2\cdot2 + 2\cdot2+1\cdot4 = 16$
Is there easy way to solve this for different $x$ and $y$?
| I assume, in your example, you mean there is one rectangle of size $(2, 2)$, because there are none $(4, 4)$.
In general, we have $(x - a + 1) \times (y - b + 1)$ rectangles of size $(a, b)$: we need to choose how many empty columns we leave before the rectangle $(0, 1, 2, \dots \text{ or } x-a)$, and how many rows we leave above it $(0, 1, 2, \dots \text{ or } y-b)$
So, in total, we have this area
$$
\sum_{a=1}^x \sum_{b=1}^y (x - a + 1) \times (y - b + 1) \times ab = \\
\sum_{a=1}^x \Big( a (x - a + 1) \sum_{b=1}^y b (y - b + 1)\Big) = \\
\big(\sum_{a=1}^x a (x - a + 1)\big) \big( \sum_{b=1}^y b (y - b + 1)\big)
$$
We can simplify this further by examining the sequence $A_n = \sum_{a=1}^n a (n - a + 1)$.
If we examine the first few elements of this sequence $(1, 4, 10, 20, 35, 56)$, we can notice that the sequence of differences $(1, 3, 6, 10, 15, 21)$ is exactly the sequence of triangular numbers. If that holds true in general, $A_n$ is the sum of the first $n$ triangular numbers, i.e. $A_n = \frac{n(n+1)(n+2)}{6}$ as seen here.
Here's a proof that this indeed holds in general:
$$
\begin{align}
A_n - A_{n-1} = 1 \times n &+ 2 \times (n-1) + 3 \times (n-2) + \dots + n \times 1 \\
&- 1 \times (n-1) - 2 \times (n-2) - 3 \times (n-3) - \dots = (n-1) \times 1 = \\
&= n + (2 - 1) \times (n - 1) + (3 - 2) \times (n - 2) + \dots + (n - 2 - n + 1) \times 1 \\
&=\sum_{i=1}^n i = \frac{n(n+1)}{2},
\end{align}
$$
which is indeed the $n^{\text{th}}$ triangle number.
Thus, the answer to the original question is
$$\frac{x(x+1)(x+2)}{6} \times \frac{y(y+1)(y+2)}{6}$$
and, in the case $x=y=2$ this is indeed 4.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Eigenvalue of a given operator If $u_0$ is a positive radial symmetric nontrival solution of
$$
-\frac{1}{2}\frac{d^2u}{dx^2}+\lambda u -u^3=0
$$
Then how to show $-3\lambda$ is a eigenvalue of
$$
Lu=-\frac{1}{2}\frac{d^2u}{dx^2}+\lambda u -3u_0^2 u
$$
and the corresponding eigenfunction is $u_0^2$ ?
What I try:
\begin{align}
Lu_0^2
&=-\frac{1}{2}\frac{d^2u_0^2}{dx^2}+\lambda u_0^2 -3u_0^4 \\
&=-(\frac{du_0}{dx})^2 -(u_0 \frac{d^2 u_0}{dx^2}) + \lambda u_0^2 -3 u_0^4 \\
&=-(\frac{du_0}{dx})^2-\frac{1}{2}(u_0 \frac{d^2 u_0}{dx^2})-2u_0^4
+ [-\frac{1}{2}(u_0 \frac{d^2 u_0}{dx^2})+ \lambda u_0^2 - u_0^4] \\
&=-(\frac{du_0}{dx})^2-\frac{1}{2}(u_0 \frac{d^2 u_0}{dx^2})-2u_0^4
\end{align}
then I don't how to deal the $-(\frac{du_0}{dx})^2$, what should I do ?
| Here is my try. Let $v_0=\frac{du_0}{dx}$. From
$$-\frac12\frac{d^2u_0}{dx^2}+\lambda u_0-u_0^3=0\implies \frac{d^2u_0}{dx^2}=2\lambda u_0-2u_0^4,$$
we have as you computed
$$Lu_0^2=-v_0^2-u_0\frac{d^2u_0}{dx^2}+\lambda u_0^2-3u_0^4=-v_0^2-u_0(2\lambda u_0-2u_0^4)+\lambda u_0^2-3u_0^4.$$
So $Lu_0^2=-v_0^2-\lambda u_0^2+u_0^4$. Also from
$$-\frac12\frac{d^2u_0}{dx^2}+\lambda u_0-u_0^3=0\implies-\frac{1}{2}v_0\frac{dv_0}{dx}+\lambda u_0\frac{du_0}{dx}-u_0^3\frac{du_0}{dx}=0,$$
we get by integrating wrt $x$ that
$$-\frac{v_0^2}{4}+\frac{\lambda u_0^2}{2}-\frac{u_0^4}{4}=c$$
for some constant $c$. So $-v_0^2+2\lambda u_0^2-u_0^4=4c$. That is,
$$Lu_0^2=-v_0^2-\lambda u_0^2+u_0^4=(-v_0^2+2\lambda u_0^2-u_0^4)-3\lambda u_0^2=4c-3\lambda u_0^2.$$
Presumably, you have some conditions to make $c=0$ (is there anything missing?), because if $c=0$, $Lu_0^2=-3\lambda u_0^2$ as required.
| {
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Can area of rectangle be greater than the square of its diagonal?
Q: A wall, rectangular in shape, has a perimeter of 72 m. If the length of its diagonal is 18 m, what is the area of the wall ?
The answer given to me is area of 486 m2. This is the explanation given to me
Is it possible to have a rectangle of diagonal 18 m and area greater than the area of a square of side 18 m ?
| No, use the Pythagorean Theorem.
$$c^2 = a^2+b^2$$
$c$ is the length of the diagonal. It divides the rectangle into two congruent right triangles with hypotenuse $c$. $a$ and $b$ are the pairs of sides of the rectangle (and the other two sides of each congruent right triangle).
Recall for any real number, its square must be non-negative.
$$(a-b)^2 \geq 0 \implies a^2-2ab+b^2 \geq 0 \implies \color{blue}{a^2+b^2 \geq 2ab}$$
The area of the rectangle is $ab$, but $c^2 \geq 2ab$, so the square of the diagonal is at least twice the area of the rectangle.
Now, to find the area itself.
For the diagonal:
$$c^2 = a^2+b^2$$
$$\implies 18^2 = a^2+b^2$$
$$\color{blue}{324 = a^2+b^2} \tag{1}$$
For the perimeter:
$$2(a+b) = 72$$
$$a+b = 36$$
Now, define one variable in terms of the other.
$$\color{purple}{a = 36-b} \tag{2}$$
Combine $(1)$ and $(2)$.
$$324 = a^2+b^2 \implies 324 = (36-b)^2+b^2$$
$$324 = 36^2-2(36)b+b^2+b^2 \implies 324 = 1296-72b+2b^2 \implies 2b^2-72b+972 = 0$$
But $$\Delta = b^2-4ac$$
$$\Delta = 72^2-4(2)(972) = -2592$$
$$\implies \Delta < 0$$
Thus, there is no solution. (No such rectangle exists.)
| {
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Proof of a Well-Known Fibonacci Identity Involving Cubes of Fibonacci Numbers The following is due to Lucas in 1876:
$$F_{n + 1}^3 + F_n^3 - F_{n - 1}^3 = F_{3n}$$
I am unable to locate an elementary proof of this identity, and am unable to reproduce it myself. Would anyone mind sharing a proof or a source?
| This is probably not the most elegant way, but you can use the explicit formula for the Fibonacci numbers.
$$
F_n = \frac{1}{\sqrt{5}} \tau^n + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^n
$$
Here $\tau = \frac{1}{2}+\frac{\sqrt{5}}{2}, -1/\tau = \frac{1}{2}-\frac{\sqrt{5}}{2}$. This formula for $F_n$ can now be plugged into the left-hand side of the desired identity.
$$
\begin{align*} F_{n+1}^3 + F_n^3 - F_{n-1}^3
= & \left(\frac{1}{\sqrt{5}} \tau^{n+1} + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^{n+1}\right)^3\\
& + \left(\frac{1}{\sqrt{5}} \tau^n + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^n\right)^3\\
& - \left(\frac{1}{\sqrt{5}} \tau^{n-1} + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^{n-1}\right)^3
\end{align*}$$
Expanding out the terms takes a little work but will eventually lead to the desired expression
$\frac{1}{\sqrt{5}} \tau^{3n} + \left(-\frac{1}{\sqrt{5}}\right)\left(-\frac{1}{\tau}\right)^{3n}$ for $F_{3n}$.
As an example, considering the leading terms of the three cubes yields
$$
\frac{1}{5\sqrt{5}}\tau^{3n+3}+\frac{1}{5\sqrt{5}}\tau^{3n} - \frac{1}{5\sqrt{5}}\tau^{3n-3}
= \frac{1}{5\sqrt{5}}\left(\tau^3 + 1 - \frac{1}{\tau^3}\right)\tau^{3n}
$$
Since $\tau^3 = 2+ \sqrt{5}$ and $-1/\tau^3 = 2-\sqrt{5}$, we see that this simplifies to $(1/\sqrt{5})\tau^{3n}$, which is the first term of the expression for $F_{3n}$.
I leave the consideration of the other terms to the reader.
| {
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Attempt at sequence proof $\frac{n+3}{n^2 -3}$ converges to $0$
Prove convergence of the following sequence: $$\frac{n+3}{n^2 -3} \rightarrow 0$$
Proof discussion:
Notice that since whenever $n>3$, we have $n^2 -3 >0$, we also know that $n+3 >0$, so $\frac{n+3}{n^2 -3}>0$. This means we can drop the absolute value signs in:
$$ \left|\frac{n+3}{n^2 -3}-0\right|=\frac{n+3}{n^2 -3} $$
We now notice that for $n>3$ also $n^2 -9>0$ and $n^2 -3 > n^2 -9$ so $\frac{1}{n^2 -3}< \frac{1}{n^2 -9}$ we can thus write:
$$\frac{n+3}{n^2 -3}<\frac{n+3}{n^2 -9}=\frac{(n+3)}{(n+3)(n-3)}=\frac{1}{n-3} $$
To be able to complete this proof we want that $\frac{1}{n-3}<\epsilon$, we write $n-3>\frac{1}{\epsilon}$ or $n> \frac{1}{\epsilon} +3$. If we pick $n_0 =\lceil\frac{1}{\epsilon} +3\rceil$, it will also be automatically larger than $3$. We can now write our proof:
Proof:
For all $\epsilon>0$, we let $n_0=\lceil{\frac{1}{\epsilon}+3 }\rceil$ then for all $n>n_0$, we know that:
$$|a_n-0|=\left|\frac{n+3}{n^2-3} \right|<\frac{n+3}{n^2-9}=\frac{1}{n-3}< \frac{1}{\frac{1}{\epsilon}+3-3}=\epsilon$$
And hence our sequence converges to $0$ $\square$.
Is my proof okay?
| Alternatively, decompose
$$\frac{n+3}{n^2-3}=\frac1{n-\sqrt3}+(3-\sqrt3)\frac1{n^2-3}$$
and the two terms are of the form $\dfrac1\infty$.
| {
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What is $\lim_{x \to 3} (3^{x-2}-3)/(x-3)(x+5)$ without l'HΓ΄pital's rule? I'm trying to solve the limit $\lim_{x \to 3} \frac{3^{x-2}-3}{(x-3)(x+5)}$
but I don't know how to proceed: $\lim_{x \to 3} \frac{1}{x+5}$ $\lim_{x \to 3} \frac{3^{x-2}-3}{x-3}$ = $1\over8$ $\lim_{x \to 3} \frac{\frac{1}{9}(3^{x}-27)}{x-3}$
Any hints? Thanks in advance.
| We have that with $y=x-3 \to 0$
$$\lim_{x \to 3} \frac{3^{x-2}-3}{(x-3)(x+5)}=\lim_{y \to 0} \frac{3\cdot 3^{y}-3}{y(y+8)}=\lim_{y \to 0} \frac{3}{y+8}\frac{3^{y}-1}{y}$$
then recall that by standard limits
$$\frac{3^{y}-1}{y}=\frac{e^{y\log 3}-1}{y\log 3}\cdot \log 3 \to \log 3$$
| {
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Show that $P(x,y)=0$ is an ellipse if $b^2-4ac<0$. I tried the following:
I write the polynomial $P(x, y) = ax^2+bxy+cy^2+dx+ey+h$ in the form $P(x, y) = Ax^2 + Bx + C$ where $A$, $B$, and $C$ are polynomial functions of $y$. This $P(x, y) = Q(x)$ has the discriminant $\Delta_x(y) = (b^2 β 4ac)y^2 + (2bd β 4ae)y + (d^2 β 4ah)$. The discriminant of the $\Delta_x(y) = (b^2 β 4ac)y^2 + (2bd β 4ae)y + (d^2 β 4ah)$ is $\Delta_0 = 16(a^2e^2-aebd+acd^2+ah(b^2-4ac))$.
My questions:
i. The book says that if $b^2-4ac<0$, one of the following occurs: $K_1={\{y \ | \ \Delta_x(y) \ge 0}\} = \emptyset$, $K_2={\{y \ | \ \Delta_x(y) \ge 0}\} = {\{y_0}\}$, or there exist real numbers $\alpha$ and $\beta$ such that $K_3={\{y \ | \ \Delta_x(y) \ge 0}\} = {\{\alpha \le y \le \beta}\}$. How to show that only one of this three cases happens?
ii. For cases $K_1$ and $K_2$ it is easy to know that $P(x,y)=0$ is empty or a single point, respectively. But how case $K_3$ implies $P(x,y)=0$ to be ellipse? (for example it can be a part of a hyperbola too [the normal one $\pi/2$ rotated] since the domain is an interval and for any point on domain there are two points $y$ fitting in the equation - also IF for the $P(x, y)$ there are only possible shapes: circle/ellipse, parabola and hyperbola then the only choice remains is circle/ellipse).
| For i.
Let $Y=\Delta_x(y)$. Then, note that $Y=\Delta_x(y)$ is the equation of a parabola. Moreover, the coefficient of $y^2$ is $b^2-4ac$ which is negative, so the parabola opens down. So, only one of the three cases happens.
For ii.
for example it can be a part of a hyperbola too [the normal one $\pi/2$ rotated] since the domain is an interval and for any point on domain there are two points $y$ fitting in the equation
It is impossible that $P(x,y)$ is a part of a hyperbola. Note that, in the first place, $P(x,y)=0$ has no restrictions on $x,y$. So, it is impossible that $P(x,y)=0$ is a part of something. In the case $K_3$ where there exist real numbers $\alpha,\beta$ such that $\alpha\le y\le \beta$, the whole curve $P(x,y)=0$ is included in that interval. This immediately implies that $P(x,y)=0$ is neither a hyperbola nor a parabola.
also IF for the $P(x, y)$ there are only possible shapes: circle/ellipse, parabola and hyperbola then the only choice remains is circle/ellipse).
I think that in this context, it may be supposed that circles are a special case of ellipses.
Anyway, in the case $K_3$, $P(x,y)=0$ can be a circle.
If $a=1,b=0,c=1,d=-2,e=-2$ and $h=0$, then we have
$$b^2-4ac=-4\lt 0,\qquad \Delta_x(y) \ge 0\iff 1-\sqrt 2\le y\le 1+\sqrt 2$$
and
$$P(x,y)=0\iff (x-1)^2+(y-1)^2=2$$
which is the equation of a circle.
Added :
Suppose that $ax^2+bxy+cy^2$ becomes $AX^2+BXY+CY^2$ by the rotation of $\theta$ :
$$x=X\cos\theta+Y\sin\theta,\qquad y=-X\sin\theta+Y\cos\theta$$
Then, we get
$$\begin{align}A&=a\cos^2\theta-b\sin\theta\cos\theta+c\sin^2\theta
\\\\B&=(a-c)\sin(2\theta)+b\cos(2\theta)
\\\\C&=a\sin^2\theta+b\sin\theta\cos\theta+c\sin^2\theta\end{align}$$
Since
$$A+C=a+c,\qquad A-C=(a-c)\cos(2\theta)-b\sin(2\theta)$$
we get
$$B^2-4AC=B^2-(A+C)^2+(A-C)^2=(a-c)^2+b^2-(a+c)^2=b^2-4ac$$
So, when $B^2=0$, we get $$b^2-4ac=-4AC$$
It follows that if $b^2-4ac\lt 0$, i.e. $AC\gt 0$, i.e. either $A\gt 0,C\gt 0$ or $A\lt 0,C\lt 0$, then the equation $$AX^2+0XY+CY^2+\cdots =0$$ represents an ellipse.
| {
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Evaluate and Simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$ I am trying to evaluate and simplify $\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$.
I am getting $\frac{11}{10}$ but the answer is $\frac{3-4\sqrt{3}}{10}$
My Process:
$\cos[\cos^{-1}(\frac{3}{5}) + \frac{\pi}{3}]$
$\cos[\cos^{-1}(\frac{3}{5})] + \cos(\frac{\pi}{3})$
$(\frac{2}{2}) \cdot \frac{3}{5} + \frac{1}{2} \cdot (\frac{5}{5})$
$\frac{6}{10} + \frac{5}{10}$
$\frac{11}{10}$
| By compound-angle formula,
\begin{eqnarray*}
\cos(\cos^{-1}(\frac{3}{5})+\frac{\pi}{3}) & = & \cos\left(\cos^{-1}(\frac{3}{5})\right)\cos\frac{\pi}{3}-\sin\left(\cos^{-1}(\frac{3}{5})\right)\sin(\frac{\pi}{3})\\
& = & \frac{3}{5}\cdot\frac{1}{2}-\sin\left(\cos^{-1}(\frac{3}{5})\right)\cdot\frac{\sqrt{3}}{2}.
\end{eqnarray*}
To evaluate $\sin\left(\cos^{-1}(\frac{3}{5})\right)$, we let $\theta=\cos^{-1}(\frac{3}{5})$.
Then $\cos\theta=\frac{3}{5}$. Recall that $\sin^{2}\theta+\cos^{2}\theta=1$
and observe that $0<\theta<\frac{\pi}{2}$. Therefore $\sin\theta>0$ and
$\sin\theta$ is given by $\sin\theta=\sqrt{1-\cos^{2}\theta}=\frac{4}{5}$.
It follows that
\begin{eqnarray*}
\cos(\cos^{-1}(\frac{3}{5})+\frac{\pi}{3}) & = & \frac{3}{5}\cdot\frac{1}{2}-\frac{4}{5}\cdot\frac{\sqrt{3}}{2}\\
& = & \frac{3}{10}-\frac{2\sqrt{3}}{5}
\end{eqnarray*}
| {
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Classifying singularities of $f(z) = \frac{1}{(z-2)^2}+e^{\frac{1}{3-z}}$
Classify all the singularities of $f(z) = \frac{1}{(z-2)^2}+e^{\frac{1}{3-z}}$
I know that the singularities of $f$ are exactly $z=2,z=3$. I just want to check that if my solution for the classification is correct:
$z=2$: since $1/(3-z)$ is analytic at $z=2$, then the composition $e^{\frac{1}{3-z}}$ is also analytic at this point. Analyitic functions admit Taylor expansion, hence the expansion of $e^{\frac{1}{3-z}}$ will only give "positive coeffcients" (coefficients with positive powers), so it suffices to only look for the Laurent expansion of $\frac{1}{(z-2)^2}$, which is already in this form. Thus, $z=2$ is a pole of order 2.
$z=3$: now $\frac{1}{(z-2)^2}$ is analytic at this point, so by the argument above we don't have to care about its series expansion. Now, since $e^z$ is entire, this series admits expansion in the region $3<|z|<\infty$, so we can write:
$$e^{\frac{1}{3-z}} = \sum_{n=0}^{\infty}\frac{1}{n!(3-z)^n}.$$
The uniqueness of the laurent series guarantee us that this series above is the Laurent expansion of $e^{\frac{1}{3-z}}.$ Therefore, $z=3$ is an essential singularity.
Is it correct? Or I do have to find the series expansion in each case?
| Your reasoning is correct.
$e^{-\frac{1}{z-3}}$ is analytic at $z=2$, so it can be expanded in a Taylor series about that point in the region $0<|z-2|<1$:
\begin{align}e^{-\frac{1}{z-3}}=e+e(z-2)+\frac{3e}{2}(z-2)^2+\cdots\end{align}
Therefore, the Laurent series of $f(z)$ about $z=2$ in the region $0<|z-2|<1$ is given by:
\begin{align}f(z)=&\frac{1}{(z-2)^2}+e+e(z-2)+\frac{3e}{2}(z-2)^2\\&+\cdots\end{align}
Similarly, $\frac{1}{(z-2)^2}$ is analytic at $z=3$, so it can be expanded in a Taylor series about that point in the region $0<|z-3|<1$:
\begin{align}\frac{1}{(z-2)^2}=&1-2(z-3)+3(z-3)^2+\cdots\end{align}
Therefore, the Laurent series of $f(z)$ about $z=3$ in the region $0<|z-3|<1$ is given by:
\begin{align}f(z)=&\cdots+\frac{1}{2}\frac{1}{(z-3)^2}-\frac{1}{z-3}+2\\&-2(z-3)+3(z-3)^2+\cdots\,\end{align}
and we arrive at your conclusions.
| {
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Prove the limit lim nββ $ {(2n^7 + 3n^5 + 4n) \over (4n^7 β 7n^2 + 5)} $ = 1/2 Im trying to use the formal definition of a limit to prove
$\lim \limits_{x \to β} $$ {(2n^7 + 3n^5 + 4n)\over(4n^7 β 7n^2 + 5)}$ = 1/2
I understand this problem is done backwards so I set up the equation
$ {(2n^7 + 3n^5 + 4n)\over(4n^7 β 7n^2 + 5)}$ - ${1\over 2}$
$\left|{(2n^7 + 3n^5 + 4n)\over(4n^7 β 7n^2 + 5)} - {1\over 2} \right|$ < Ξ΅
Both the denominator and numerator are always positive so we can drop the absolute value and simplify and rewrite
$ {(6n^3-7)\over(8n^7-14n^2+10)}$ < $ {(6n^5-7n^2+8n+5)\over(8n^7-14n^2+10)}$ < Ξ΅
$ {(1)\over(4n^4)}$ = $ {(n^3)\over(4n^7)}$ < $ {(3n^3)\over(4n^7)}$ = $ {(6n^3)\over(8n^7)}$ < $ {(6n^3)\over(8n^7-14n^2+10)}$ < Ξ΅+7
$ {(1)\over(Ξ΅+7)}$ < $4n^4$
$ {1\over4Ξ΅+28^{1\over4}}$ < n = N
Now I have my N so I can writ my proof:
Let Ξ΅, N > 0 and N < n. Suppose N = $ {1\over4Ξ΅+28^{1\over4}}$ . Then it must follow that:
$ {1\over4Ξ΅+28^{1\over4}}$ < n
$ {1\over4Ξ΅+28}$ < $n^4$
$ {1\overΞ΅+7}$ < $4n^4$
$ {1\over 4n^4}$ < $Ξ΅+7$
from there on I dont know how to use my inequalities to get back to what I started with.
| Note that\begin{align}\left\lvert\frac{2n^7+3n^5+4n}{4n^7-7n^2+5}-\frac12\right\rvert&=\frac{\lvert6n^5+7n^2+8n-5\rvert}{\lvert8n^7-14n^2+10\rvert}\\&\leqslant\frac{26n^5}{8n^7-14n^2-10}\\&=\frac{26}{8n^2-14n^{-3}-10n^{-5}}\\&\leqslant\frac{26}{8n^2-24},\end{align}if $n>1$. Now, use the fact that\begin{align}\frac{26}{8n^2-24}<\varepsilon&\iff8n^2-24>\frac{26}\varepsilon\\&\iff n>\sqrt{\frac{13}{4\varepsilon}+3}.\end{align}
| {
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A Fibonacci convolution
A Fibonacci convolution. Recall that $$F(x)=\sum_{n=0}^\infty F_n x^n =\frac{x}{1-x-x^2} =\frac{1}{\sqrt{5}} \left(\frac{1}{1-\Phi x} -\frac{1}{1-\bar{\Phi}x}\right).$$
(a) Prove that $\displaystyle \sum_{n=0}^\infty F_{n+1} x^n =\frac{1}{1-x-x^2}$.
(b) Prove that $\displaystyle \sum_{n=0}^\infty (2F_{n+1} -F_n)x^n =\frac{2-x}{1-x-x^2} =\sum_{n=0}^\infty (\Phi^n +\bar{\Phi}^n) x^n$.
(c) Prove that $\displaystyle 5F(x)^2 =\sum_{n=0}^\infty \binom{n+1}{1} \Phi^n x^n -2\sum_{n=0}^\infty F_{n+1} x^n +\sum_{n=0}^\infty \binom{n+1}{1} \bar{\Phi}^n x^n$.
(d) Prove that $\boldsymbol{\displaystyle \sum_{k=0}^n F_k F_{n-k} =\frac{2n F_{n+1} -(n+1)F_n}{5}}$.
I've gotten through (a)-(c) but I don't know how to start (d).
| Since you have already mastered a.) to c.) we can conveniently use the results to prove d.).
We obtain
\begin{align*}
\color{blue}{5F(x)^2}&=\sum_{n=0}^{\infty}(n+1)\left(\Phi^n+\bar{\Phi}^n\right)x^n-2\sum_{n=0}^\infty F_{n+1}x^n\tag{1}\\
&=\sum_{n=0}^\infty(n+1)\left(2F_{n+1}-F_n\right)x^n-2\sum_{n=0}^\infty F_{n+1}x^n\tag{2}\\
&\,\,\color{blue}{=\sum_{n=0}^\infty\left(2nF_{n+1}-(n+1)F_n\right)x^n}
\end{align*}
and d.) follows.
Comment:
*
*In (1) we apply c.) by using $\binom{n+1}{1}=n+1$ and collecting the first and the last sum.
*In (2) we use from b.) the identity $2F_{n+1}-F_n=\Phi^n+\bar{\Phi}^n$.
| {
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limit of $\sqrt{x^6}$ as $x$ approaches $-\infty$ I need to solve this limit:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}}$$
The answer is $-3$, but I got 3 instead. This is my process:
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}} =
\lim_{x \to - \infty}{\frac {\sqrt{x^6(9-\frac {5}{x^2})}}{x^3(1-\frac {2}{x}+\frac{1}{x^3})}} =
\lim_{x \to - \infty}{\frac {\sqrt{x^6}\sqrt{(9-\frac {5}{x^2})}}{x^3(1-\frac {2}{x}+\frac{1}{x^3})}} =
\lim_{x \to - \infty}{\frac {\require{cancel} \cancel{x^3} \sqrt{(9-\frac {5}{x^2})}}{\require{cancel} \cancel{x^3}(1-\frac {2}{x}+\frac{1}{x^3})}} =
\frac {3}{1} = +3$$
I've been told that in the third step the $\sqrt{x^6}$ should be equal $\textbf{-}\sqrt{x^3}$, but I didn't understand why.
I'll be glad to get your help!
Thank you.
| To check or avoid confusion with sign let $y=-x \to \infty$ then
$$\lim_{x \to - \infty}{\frac {\sqrt{9x^6-5x}}{x^3-2x^2+1}}=\lim_{y \to \infty}{\frac {\sqrt{9y^6+5y}}{-y^3-2y^2+1}}=-3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3017300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
} |
Does $z^4+2z^2+z=0$ have complex roots? Does $z^4+2z^2+z=0$ have complex roots? How to find them? Besides $z=0$, I got the equation $re^{3i\theta}+2re^{i\theta}=e^{i(\pi+2k\pi)}$, $k\in \mathbb Z$. How to find the complex roots?
| Enough to check the discriminant for $ax^3+bx^2+cx+d=0$:
$$
\Delta_3=-27 a^2 d^2+18 a b c d-4 a c^3-4 b^3 d+b^2 c^2
$$
$$
\Delta_3=\begin{cases} >0 & \text{3 distinct real roots}\\
<0 & \text{1 real, 2 conjugate complex roots}\\
=0 & \text{3 real roots with duplicates}\\
\end{cases}
$$
In your case, it's $a=1$, $b=0$, $c=2$ and $d=1$, so $\Delta_3=-59$, hence the equation has one real and two complex conjugate roots, with all 3 distinct. As for finding the roots themselves, the Cardano formula helps check this out. The final result is:
$$
x_1=\frac{\sqrt[3]{\frac{1}{2} \left(\sqrt{177}-9\right)}}{3^{2/3}}-2 \sqrt[3]{\frac{2}{3
\left(\sqrt{177}-9\right)}}
$$
$$
x_2=\left(1+i \sqrt{3}\right)
\sqrt[3]{\frac{2}{3 \left(\sqrt{177}-9\right)}}-\frac{\left(1-i \sqrt{3}\right)
\sqrt[3]{\frac{1}{2} \left(\sqrt{177}-9\right)}}{2\ 3^{2/3}}
$$
$$
x_3=\left(1-i
\sqrt{3}\right) \sqrt[3]{\frac{2}{3 \left(\sqrt{177}-9\right)}}-\frac{\left(1+i
\sqrt{3}\right) \sqrt[3]{\frac{1}{2} \left(\sqrt{177}-9\right)}}{2\
3^{2/3}}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3018093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Values of $x$ satisfying $\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$
For what values of $x$ between $0$ and $\pi$ does the inequality $\sin x\cdot\cos^3 x>\sin^3x\cdot\cos x$ hold?
My Attempt
$$
\sin x\cos x\cdot(\cos^2x-\sin^2x)=\frac{1}{2}\cdot\sin2x\cdot\cos2x=\frac{1}{4}\cdot\sin4x>0\implies\sin4x>0\\
x\in(0,\pi)\implies4x\in(0,4\pi)\\
4x\in(0,\pi)\cup(2\pi,3\pi)\implies x\in\Big(0,\frac{\pi}{4}\Big)\cup\Big(\frac{\pi}{2},\frac{3\pi}{4}\Big)
$$
But, my reference gives the solution, $x\in\Big(0,\dfrac{\pi}{4}\Big)\cup\Big(\dfrac{3\pi}{4},\pi\Big)$, where am I going wrong with my attempt?
| The given solution is wrong; you are correct. At $x=\frac{7\pi}8\in\left(\frac{3\pi}4,\pi\right)$, we have that $$\frac14\sin4x=\frac14\sin\frac{7\pi}2=-\frac14<0$$ which is a contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3021679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Express a parametrization of a curve $C$ as a function of $X$ and $Y$ Express the curve $C$ with parametrization $\{(\cos 3t, \sin 2t) : t \in \mathbb{R} \}$ as a function of $X$ and $Y$.
Here's what I have (I don't know if this is helpful)
I let $X= \cos 3t$ and $Y= \sin 2t$. So I have $X= \cos^3 t -3\sin^2 t \cos t$ and $Y=2 \cos t \sin t$.
I don't know how to go on from here. Please help. Thanks!
| Answer:
$$
X^4 - X^2 + 4 Y^6 - 6 Y^4 + \frac{9}{4} Y^2 = 0
$$
How I got it:
We can replace the trigonometric parametrization $(\cos(u), \sin(u))$ of the unit circle with a rational one: $\left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right)$. So we equate $\cos(u) = x = \frac{1-t^2}{1+t^2}$ and $\sin(u) = y =\frac{1-t^2}{1+t^2}$. Your expressions for $X$ and $Y$ then become
\begin{align*}
X &= \cos^3(u) - 3 \cos(u) \sin^2(u) = x^3 - 3 xy^2 = \frac{-t^{6} + 15 t^{4} - 15 t^{2} + 1}{(1+t^2)^3}\\
Y &= 2 \cos(u) \sin(u) = 2 x y = \frac{-4 t^{3} + 4 t}{(1+t^2)^2}
\end{align*}
In the polynomial ring $\mathbb{Q}[t,z,X,Y]$ we form the ideal
$$
I = ((1+t^2)^3 X -(-t^{6} + 15 t^{4} - 15 t^{2} + 1), (1+t^2)^2 Y - (-4 t^{3} + 4 t), (1+t^2)z - 1)
$$
and compute a GrΓΆbner basis with respect to the elimination ordering $t > z > X > Y$. The last entry of this GrΓΆbner basis is the desired polynomial in $X$ and $Y$.
The proof is in the pudding, though: here is a SageMathCell comparing the two plots.
Actually, a slightly better way is to instead compute a GrΓΆbner basis for the ideal $(X - (x^3 - 3xy^2), Y - (2xy), x^2 + y^2 - 1)$ in $\mathbb{Q}[x,y,X,Y]$ with respect to the elimination ordering $x > y > X > Y$. This yields the same answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
square root system of equations
I have a system of equations as follows that I need to solve for $x$:
$$
\sqrt{1 - x^2} + \sqrt{4 - x^2} = z\\
\sqrt{4 - y^2} + \sqrt{9 - y^2} = z
$$
Originally, I was trying to put $x$ in terms of $z,$ but Iβm at a loss how I would solve this system. Trying to solve it just makes it more complicated, no matter if I multiply by the reciprocal or square both sides.
| Both the $x$ and $y$ equations have the form
$$\sqrt{a^2-h^2} + \sqrt{b^2 - h^2} = c$$
For everything to be real, we need $\sqrt{|a^2-b^2|} \le c \le a + b$.
Assume this is satisfied. For positive solution of $h$, this equation has a geometric interpretation. It is the equation for the height $h$ of a triangle of side $a,b,c$ with respect to side $c$. We can use
Heron's formula to figure out the area $\Delta$ of the triangle and then
$$\begin{align}h = \frac{2\Delta}{c}
&= \frac{1}{2c}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}\\
&= \frac{1}{2c}\sqrt{4a^2b^2 - (c^2-a^2-b^2)^2}\end{align}$$
Apply this to the equations at hand. For positive $x$ and $y$ solutions, we obtain
$$\begin{align}
\sqrt{1 - x^2} + \sqrt{4-x^2} = z &\implies
x = \frac{1}{2z}\sqrt{16 - (z^2-5)^2}\\
\sqrt{4 - y^2} + \sqrt{9-y^2} = z &\implies
y = \frac{1}{2z}\sqrt{144 - (z^2 - 13)^2}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3025753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Number of solutions of the equation $\cos(\pi\sqrt{x-4})\cos(\pi\sqrt{x})=1$
Find the number of solutions of the equation $\cos(\pi\sqrt{x-4})\cos(\pi\sqrt{x})=1$
\begin{align}
2\cos(\pi\sqrt{x-4})&.\cos(\pi\sqrt{x})=2\\\implies\cos\Big[\pi(\sqrt{x-4}+\sqrt{x})\Big]&+\cos\Big[\pi(\sqrt{x-4}-\sqrt{x})\Big]=2\\
\implies\cos\Big[\pi(\sqrt{x-4}+\sqrt{x})\Big]=1\quad&\&\quad\cos\Big[\pi(\sqrt{x-4}-\sqrt{x})\Big]=1\\
\pi(\sqrt{x-4}+\sqrt{x})=2n\pi\quad&\&\quad\pi(\sqrt{x-4}-\sqrt{x})=2m\pi\\
\sqrt{x-4}+\sqrt{x}=2n\quad&\&\quad\sqrt{x-4}-\sqrt{x}=2m\\
2\sqrt{x}=2(n-m)\quad&\&\quad2\sqrt{x-4}=2(n+m)\\
\sqrt{x}=n-m\quad&\&\quad\sqrt{x-4}=n+m\quad\&\quad x\geq4
\end{align}
How do I properly find the solutions ?
Or can I simply say
$$
x=(n-m)^2=(n+m)^2-4nm=x-4-4nm\implies nm=-1\\
\implies x=\bigg[n+\frac{1}{n}\bigg]^2\in\mathbb{Z}\implies n,\frac{1}{n}\in\mathbb{Z}\\
\implies n\neq0\implies n=1,x=4\text{ is the only solution}
$$
| More simply we need as a necessary condition
*
*$\pi\sqrt{x-4}=k_1\pi$
*$\pi\sqrt{x}=k_2\pi$
that is
*
*$\sqrt{x-4}=k_1 \implies x=k_1^2+4$
*$\sqrt{x}=k_2\implies x=k_2^2$
that is
$$k_1^2+4=k_2^2 \iff k_2^2-k_1^2=4$$
and the only possible solutions is $k_1=0 \implies x=4$ which works by inspection.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Upper bound for a ratio of two least common multiples Let $\text{lcm}(x)$ be the least common multiple of $\{1,2,3,\dots,x\}$
Let $x\#$ be the the primorial for $x$.
It occurs to me that for $x \ge 10$:
$$\frac{\text{lcm}(x^2+x)}{\text{lcm}(x^2)} < 2^x\frac{(x^2+x)\#}{(x^2)\#}$$
Am I right?
Here is my thinking:
(1) $\dfrac{\text{lcm}(x+1)}{\text{lcm}(x)} > 1$ if and only if $x+1$ is a power of a prime.
(2) If a prime $x^2 \ge p > x$, then $p^2 \ge (x+1)^2 > x^2 + x$ so it will divide out in the ratio.
(3) If a prime $x \ge p \ge (\sqrt{x}+1)$, then $p^2 \le x^2$ and $p^4 \ge (\sqrt{x}+1)^4 \ge {4\choose0}x^2 + {4\choose1}x^{3/2} + {4\choose2}x + {4\choose3}x^{1/2} + 1 > x^2+x$
(4) But there can only be one such $x \ge p \ge \sqrt{x} + 1$ such that $p^3 | \frac{(x^2+x)!}{(x^2)!}$ since $(p+2)^3 - p^3 \ge {3\choose1}2p^2 + {3\choose2}4p + 1 > 2x$
(5) Using Hanson's result that $\text{lcm}(x) < 3^x$, we have:
$$\frac{\text{lcm}(x^2 + x)}{\text{lcm}(x^2)} < 3^{\sqrt{x}+1}x\frac{(x^2+x)\#}{(x^2)\#}$$
(6) $2^x > 3^{\sqrt{x}+1}x$ since $2^{10} = 1024 > 969>3^{\sqrt{10}+1}\times10$ and $2^{\sqrt{x}} > 3(3x)^{1/\sqrt{x}}$
Edit: Forgot the case where a prime might be between $\sqrt{x}$ and $\sqrt{x}+1$
So, I've updated my argument.
| HINT $1$
Note that
$$r(x)={\large \dfrac {\mathrm{lcm}(x^2+x)}{\mathrm{lcm}(x^2)}\cdot{\dfrac {(x^2)\#}{(x^2+x)\#}} = \prod_{p_k<x+1} p_k^{[\log_{p_k}(x^2+x)]-[\log_{p_k}(x^2)]}}.$$
Calculations using Wolfram Alpha of function
$$\bar r(x) = \left(\dfrac45\right)^x\prod_{p_k<x+1} p_k^{[\log_{p_k}(x^2+x)]-[\log_{p_k}(x^2)]}$$
gives the plot of
which shows that $\bar r(x) \leq 1,$ and that leads to the empiric estimation
$$r(x)\leq \left(\dfrac54\right)^x,\quad x\in\mathbb N.$$
The reason is that all primes greater than $x,$ contains both in the ratio for lcm and the ratio for primorial. Therefore, correction of $r(m)$ happens at least for the squares of primes.
Let $x=11,$ then the interval $[x^2+1,x^2+x] = [122,132]$ contains new degrees $2^7=128$ and $5^3=125.$ So $r(11)=2\cdot5=10.$
This means that complex consideration leads to the better estimation.
HINT $2$
For $x>>1$
$$\dfrac{(x^2+x)\#}{x^2\#}\approx \left(x^2+\frac12x\right)^d,$$
where
$$d=\pi(x^2+x)-\pi(x^2)\approx \int\limits_{x^2}^{x^2+x}\dfrac{\mathrm dt}{\log t} = x^2\int\limits_{x^2}^{x^2+x}\dfrac1{\log t}\mathrm d\dfrac t{x^2} = x^2\int\limits_{1}^{1+1/x}\dfrac{\mathrm du}{\log(x^2u)}$$
$$ = \dfrac{x^2}{2\log x}\int\limits_{1}^{1+1/x}\dfrac{\mathrm du}{1+\dfrac{\log u}{2\log x}}
\approx \dfrac{x^2}{2\log x}\int\limits_{1}^{1+1/x}\left(1-\dfrac{\log u}{2\log x}\right)\,\mathrm du$$
$$ = \dfrac{x^2}{2\log x}\left(u - \dfrac{u\log u -1}{2\log x}\right)\bigg|_1^{1+\frac1x} = \dfrac{x^2}{2\log x}\left(\dfrac1x-\dfrac{(1+\frac1x)\log (1+\frac1x)}{2\log x}\right),$$
$$d\approx\dfrac x{2\log x} - \dfrac{x+1}{4\log^2x}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.
Prove that if $a, b, c \in \mathbb{Z^+}$ and $a^2+b^2=c^2$ then ${1\over2}(c-a)(c-b)$ is a perfect square.
I have tried to solve this question and did pretty well until I reached the end, so I was wondering if I could get help on that part. Here is what I did.
$$a^2+b^2=c^2$$ $$b^2=(c-a)(c+a)$$ Since $a, b, c > 0 \therefore (c+a) \ne 0$ $$\therefore c-a={b^2\over c+a}$$ Similarly we get, $$c-b={a^2\over c+b}$$ $$\therefore {1\over2}(c-a)(c-b)={1\over2}({b^2\over c+a})({a^2\over c+b})$$ $$={(ab)^2\over 2c^2+2ab+2bc+2ca}$$ $$={(ab^2)\over a^2+b^2+c^2+2ab+2bc+2ca}$$ $$={(ab)^2\over (a+b+c)^2}$$ $$=({ab\over a+b+c})^2$$ However, I was unable to prove that ${ab\over a+b+c} \in \mathbb{Z}$ Is there a way to prove it? Thank you
| $$(a+b-c)^2 = 2(c-a)(c-b)$$ is an equivalent statement of the Pythagorean formula. It is a well known fact that either a or b must be even. So from there, identify that $b^2=(c-a)(c+a) \text{ and } a^2=(c-b)(c+b)$. So it is unavoidable that either $(c-a)$ or $(c-b)$ must be even. This covers off the right side of the equation. On the left side, you have 2 odd numbers and an even number being added/subtracted together, resulting in an guaranteed even number. The final step is to simply divide both sides by 4!
$$\bigg(\frac{a+b-c}{2}\bigg)^2 = \frac{(c-a)(c-b)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3029557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
Number Theory: Prove that $\gcd(a,b) \le \sqrt{a+b}$
For positive integers $a$ and $b$, we know $\dfrac{a+1}{b} + \dfrac{b+1}{a}$ is also
a positive integer. Prove that $\gcd(a,b) \le \sqrt{a+b}$.
Using BΓ©zout's lemma, we know that $\gcd(a, b) = sa + tb$. I want to prove that $(sa+tb)^2 \le a+b$. We know $ab\,|\,a(a+1) + b(b+1)$.
Therefore, $(sa + tb)^2 \le (sa)^2 + (tb)^2 + 2st(a(a+1)+b(b+1))$.
I'm not sure how can continue from here. Any ideas to continue, or for a better way to prove the statement?
Thanks in advance.
| Hagen von Eitzen does not seem to want to explain his answer further, so I will attempt to do so here:
Suppose that $d|a$ and $d|b$. Then we have the following:
\begin{align*}
\frac{a+b}{d^2} &= \frac{a^2+a}{d^2}+\frac{b^2+b}{d^2}-\frac{a^2}{d^2}-\frac{b^2}{d^2} \\ &=\left(\frac{a+1}{b}+\frac{b+1}{a}\right)\cdot\frac{a}{d}\cdot\frac{b}{d}-\frac{a^2}{d^2}-\frac{b^2}{d^2}
\end{align*}
The RHS is an integer by the hypothesis of the problem. The LHS is positive by the hypothesis of the problem. Thus we can safely conclude that $\frac{a+b}{d^2}\geq 1$ for all common divisors of $a$ and $b$, which is what we wanted to show.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3030484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The series $\frac{1}{2}+\frac{2}{5}+\frac{3}{11}+\frac{4}{23}+...$ Consider the expression $\frac{1}{2}+\frac{2}{5}+\frac{3}{11}+\frac{4}{23}+...$
Denote the numerator and the denominator of the $j^\text{th}$ term by $N_{j}$ and $D_{j}$, respectively. Then, $N_1=1$, $D_1=2$, and, for every $j>1$, $$
N_j=
N_{j-1}+1\qquad D_j=
2D_{j-1}+1$$
What is the $50^\text{th}$ term?
Must we evaluate that term-by-term until we reach the $50^\text{th}$ term?
What is the sum of the first $25$ term?
Must we add them one-by-one?
What is the exact value of the sum to $\infty$?
| If $d_{j+1}=2d_j+1$ then $d_{j+1}+1=2(d_j+1)\Rightarrow d_n+1=(d_1+1)2^{n-1}=3\cdot2^{n-1}$.
So 50th term is $\frac{50}{3\cdot2^{49}-1}$
| {
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"url": "https://math.stackexchange.com/questions/3032189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Seeking Methods to solve $F\left(\alpha\right) = \int_{0}^{1} x^\alpha \arcsin(x)\:dx$ I'm looking for different methods to solve the following integral.
$$ F\left(\alpha\right) = \int_{0}^{1} x^\alpha \arcsin(x)\:dx$$
For $\alpha > 0$
Here the method I took was to employ integration by parts and then call to special functions, but can this equally be achieved with say a Feynman Trick? or another form integral transform?
My approach in detail:
Employ integration by parts:
\begin{align}
v'(x) &= x^\alpha & u(x) &= \arcsin(x) \\
v(x) &= \frac{x^{\alpha + 1}}{\alpha + 1} & u'(x) &= \frac{1}{\sqrt{1 - x^2}}
\end{align}
Thus,
\begin{align}
F\left(\alpha\right) &= \left[\frac{x^{\alpha + 1}}{\alpha + 1}\cdot\arcsin(x)\right]_0^1 - \int_0^1 \frac{x^{\alpha + 1}}{\alpha + 1} \cdot \frac{1}{\sqrt{1 - x^2}} \:dx \\
&= \frac{\pi}{2\left(\alpha + 1\right)} - \frac{1}{\alpha + 1}\int_0^1 x^{\alpha + 1}\left(1 - x^2\right)^{-\frac{1}{2}} \:dx
\end{align}
Here make the substitution $u = x^2$ to obtain
\begin{align}
F\left(\alpha\right) &= \frac{\pi}{2\left(\alpha + 1\right)} - \frac{1}{\alpha + 1}\int_0^1 \left(\sqrt{u}\right)^{\alpha + 1}\left(1 - u\right)^{-\frac{1}{2}} \frac{\:du}{2\sqrt{u}} \\
&= \frac{\pi}{2\left(\alpha + 1\right)} - \frac{1}{2\left(\alpha + 1\right)}\int_0^1 u^{\frac{\alpha}{2}}\left(1 - u\right) ^{-\frac{1}{2}} \:du \\
&= \frac{1}{2\left(\alpha + 1\right)} \left[ \pi - B\left(\frac{\alpha + 2}{2}, \frac{1}{2} \right) \right]
\end{align}
\begin{align}
F\left(\alpha\right) &=\frac{1}{2\left(\alpha + 1\right)} \left[ \pi - \frac{\Gamma\left(\frac{\alpha + 2}{2}\right)\Gamma\left(\frac{1}{2}\right)}{\Gamma\left(\frac{\alpha + 2}{2} + \frac{1}{2}\right)} \right] \\
&= \frac{1}{2\left(\alpha + 1\right)} \left[ \pi - \frac{\Gamma\left(\frac{\alpha + 2}{2}\right)\sqrt{\pi}}{\Gamma\left(\frac{\alpha + 3}{2}\right) } \right] \\
&= \frac{\sqrt{\pi}}{2\left(\alpha + 1\right)} \left[ \sqrt{\pi} - \frac{\Gamma\left(\frac{\alpha + 2}{2}\right)}{\Gamma\left(\frac{\alpha + 3}{2}\right) } \right]
\end{align}
Edits:
Correction of original limit observation (now removed)
Correction of not stating region of convergence for $\alpha$.
Correction of 1/sqrt to sqrt in final line.
Thanks to those commentators for pointing out.
| Answer 2.0:
We know that for $|x|<1$,
$$\arcsin x=\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+1)}x^{2k+1}$$
Where $$(1/2)_k=\frac{\Gamma(1/2+k)}{\Gamma(1/2)}$$
Hence we may begin with
$$F(a)=\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+1)}\int_0^1x^{2k+a+1}\mathrm dx=\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+1)(2k+2+a)}$$
Then we play with the fractions a little to get
$$F(a)=\frac1{a+1}\sum_{k\geq0}\frac{(1/2)_k}{k!}\bigg[\frac1{2k+1}-\frac1{2k+2+a}\bigg]$$
$$F(a)=\frac1{a+1}\sum_{n\geq0}\frac{(1/2)_n}{n!}\frac1{2n+1}-\frac1{a+1}\sum_{k\geq0}\frac{(1/2)_k}{k!}\frac1{2k+2+a}$$
$$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\sum_{k\geq0}\frac{(1/2)_k}{k!(2k+2+a)}$$
$$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\sum_{k\geq0}\frac{(1/2)_k}{k!}\int_0^1x^{2k+1+a}\mathrm dx$$
$$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\int_0^1\sum_{k\geq0}\frac{(1/2)_k}{k!}x^{2k+1+a}\mathrm dx$$
$$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\int_0^1\frac{x^{a+1}}{\Gamma(1/2)}\sum_{k\geq0}\frac{\Gamma(1/2+k)}{k!}x^{2k}\mathrm dx$$
$$F(a)=\frac\pi{2(a+1)}-\frac1{a+1}\int_0^1\frac{x^{a+1}}{\sqrt{1-x^2}}\mathrm dx$$
$u=x^2$:
$$F(a)=\frac\pi{2(a+1)}-\frac1{2(a+1)}\int_0^1u^{a/2}(1-u)^{-1/2}\mathrm du$$
$$F(a)=\frac\pi{2(a+1)}-\frac{\Gamma(\frac{a+2}2)\Gamma(\frac12)}{2(a+1)\Gamma(\frac{a+3}2)}$$
$$F(a)=\frac{\sqrt{\pi}}{2(a+1)}\bigg[\sqrt{\pi}-\frac{\Gamma(\frac{a+2}2)}{\Gamma(\frac{a+3}2)}\bigg]$$
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"language": "en",
"url": "https://math.stackexchange.com/questions/3033251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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Find $\lim_{n\to\infty} \cos(\frac{\pi}{4}) \cos(\frac{\pi}{8})\ldots \cos(\frac{\pi}{2^n}) $ I already know that $$ a_n = \cos\left(\frac{\pi}{2^{n+1}}\right) = \overbrace{\frac{\sqrt{2+\sqrt{2+\ldots + \sqrt{2}}}}{2}}^{n\text{ roots}}$$
Also I know that $$\lim_{n\to\infty} 2\cos\left(\frac{\pi}{2^n}\right) = 2
\text{ and if } a_n \xrightarrow {n\to\infty} a \text{ then } \sqrt[n]{a_1 a_2 \ldots a_n} \xrightarrow{n\to\infty} a $$
With that method I only got indeterminate form
$$ \lim_{n\to\infty} \cos\left(\frac{\pi}{4}\right) \cos\left(\frac{\pi}{8}\right)\ldots \cos\left(\frac{\pi}{2^n}\right) = \Big(\frac{\sqrt[n]{a_1 a_2 \ldots a_n}}{2}\Big)^n = 1^\infty $$
Anyone knows a working solution?
| If $x_n=\cos(\frac{\pi}{4}) \cos(\frac{\pi}{8})\ldots \cos(\frac{\pi}{2^n}) $ then $\ $
$$x_n\sin (\frac{\pi}{2^n})= \cos(\frac{\pi}{4}) \cos(\frac{\pi}{8}) \ldots
\cos(\frac{\pi}{2^n}) \sin (\frac{\pi}{2^n}) $$
$$=\frac{1}{2^1} \cos(\frac{\pi}{4}) \cos(\frac{\pi}{8}) \ldots
\cos(\frac{\pi}{2^{n-1}}) \sin (\frac{\pi}{2^{n-1}}) $$
$$ =\ldots= \frac{1}{2^{n-1}} $$
So $$x_n=\frac{1}{2^{n-1}\sin (\frac{\pi}{2^n})} $$
So $\lim_{n\to \infty }x_n=\frac{2}{\pi} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3036917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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} |
$\lim_{n \to \infty}(1+\frac{1}{n^2})(1+\frac{2}{n^2})...(1+\frac{n}{n^2})=e^{\frac{1}{2}}$. Here is the beginning of a proof:
Suppose $0<k \leq n$,
$1+\frac{1}{n}<(1+\frac{k}{n^2})(1+\frac{n+1-k}{n^2})=1+\frac{n+1}{n^2}+\frac{k(n+1-k)}{n^4}\leq 1+\frac{1}{n}+\frac{1}{n^2}+\frac{(n+1)^2}{4n^4}$.
I'm confused by the second inequality above.
| As an alternative, we have that
$$\left(1+\frac{1}{n^2}\right)\left(1+\frac{2}{n^2}\right)\ldots\left(1+\frac{n}{n^2}\right)=\prod_{k=1}^{n}\left(1+\frac{k}{n^2}\right)=e^{\sum_{k=1}^{n} \log\left(1+\frac{k}{n^2}\right) }$$
and
$$\sum_{k=1}^{n} \log\left(1+\frac{k}{n^2}\right)=\sum_{k=1}^{n}\left(\frac{k}{n^2}+k^2O\left(\frac{1}{n^4}\right)\right)=\frac1{n^2}\sum_{k=1}^{n}k+O\left(\frac{1}{n^4}\right)\sum_{k=1}^{n}k^2\to \frac12$$
indeed
*
*$\sum_{k=1}^{n}k=\frac{n(n+1)}{2}\implies \frac1{n^2}\sum_{k=1}^{n}k=\frac{n(n+1)}{2n^2}\to \frac12$
*$\sum_{k=1}^{n}k^2=\frac{n(n+1)(2n+1)}{6}\implies O\left(\frac{1}{n^4}\right)\sum_{k=1}^{n}k^2=O\left(\frac{1}{n}\right)\to 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3039690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Arranging 3 girls and 9 boys
Three girls A, B and C, and nine boys are lined up in a row. In how many ways this can be done if B must lie between A and C, and A, B must be separated by exactly 4 boys.
I have used the following approach. First select 4 boys ${9 \choose 4}$ and then combine them in between A and B. After that, partition the remaining 5 boys into 3 (similar to r balls into n boxes, here n is $(3-1)$) with ${ 5 + 2 -1\choose 2 - 1} $. Hence, the answer should be: ${9 \choose 4}{ 5 + 2 -1\choose 2 - 1} 4! 5! $. Can anybody tell me if this is correct or not?
| As pointed out in the comments by Barry Cipra, you made two errors. They are:
*
*$A$ can be to the left or right of $B$, so you have to double your answer.
*The number of ways to distribute $5$ identical objects to $3$ compartments is $\binom{5 + 3 - 1}{3 - 1}$.
With these corrections, your answer becomes
$$2\binom{9}{4}\binom{5 + 3 - 1}{3 - 1}4!5!$$
which is correct.
Method 1: We modify your approach.
Choose whether $A$ or $C$ is to the left of $B$. Suppose it is $A$. Let $x_1$ be the number of positions to the left of $A$ to be filled with boys; let $x_2$ be the number of positions to the right of $B$ and left of $C$ to be filled with boys; let $x_3$ be the number of positions to the right of $C$ to be filled with boys. Since $4$ boys must be between $A$ and $B$, we obtain
\begin{align*}
x_1 + 4 + x_2 + x_3 & = 9\\
x_1 + x_2 + x_3 & = 5
\end{align*}
which is an equation in the nonnegative integers. A particular solution of this equation corresponds to the placement of $3 - 1 = 2$ addition signs in a row of $5$ ones. For instance,
$$1 1 + 1 1 1 +$$
corresponds to the solution $x_1 = 2$, $x_3 = 3$, $x_3 = 0$. The number of such solutions is
$$\binom{5 + 3 - 1}{3 - 1} = \binom{7}{2}$$
since we must choose which two of the seven positions required for five ones and two addition signs will be filled with addition signs. Once this choice is made, there are $9!$ ways to arrange the boys in the positions chosen for them. Hence, the number of admissible arrangements is
$$\binom{2}{1}\binom{7}{2}9!$$
Method 2: We have seven objects to arrange, a block of length $6$ consisting of girls $A$ and $B$ and four positions for boys who will be placed between them, girl $C$, and five positions for the other boys. We choose a position for the block in seven ways and a position for girl $C$ in six ways. If girl $C$ is to the left of the block, girl $A$ must occupy the right end of the block and girl $B$ must occupy its left end. If girl $C$ is to the right of the block, girl $A$ must occupy the left end of the block and girl $B$ must occupy its right end. The boys can be arranged in the nine open positions in $9!$ ways. Hence, the number of admissible arrangements is
$$7 \cdot 6 \cdot 9!$$
| {
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"url": "https://math.stackexchange.com/questions/3041775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to evaluate $\int \frac{x^3}{\sqrt {x^2+1}}dx$ Evaluate $$\int \frac{x^3}{\sqrt {x^2+1}}dx$$
Let $u={x^2}+1$. Then $x=\sqrt {u-1}$ and $dx=\frac{1}{2\sqrt{u-1}}du$.
Therefore, $$\int \frac{(u-1)\sqrt{u-1}}{\sqrt u}\frac{1}{2\sqrt {u-1}}du$$
$$=\frac{1}{2}\int {\sqrt u}-{\frac {1}{\sqrt u}du}$$
$$={\frac{2}{3}}{{(x^2+1})}^{3/2}-{\frac{1}{2}}{\sqrt {x^2+1}}+C$$for some constant $C$
| Method 1:
Another approach using trigonometric substitutions
\begin{equation}
I =\int \frac{x^3}{\sqrt {x^2+1}}dx
\end{equation}
Here let $x = \tan(\theta)$ we arrive at:
\begin{align}
I &=\int \frac{\tan^3(\theta)}{\sqrt {\tan^2(\theta)+1}}\sec^2(\theta)\:d\theta = \int \frac{\tan^3(\theta)}{\sec(\theta)}\sec^2(\theta)\:d\theta\\
&= \int \tan^3(\theta)\sec(\theta)\:d\theta = \int \sec(\theta)\tan(\theta) tan^2(\theta)\:d\theta \\
&= \int \sec(\theta)\tan(\theta) \left(\sec^2(\theta) - 1\right)\:d\theta
\end{align}
Now let $u = \sec(\theta)$ to yield:
\begin{equation}
\int \sec(\theta)\tan(\theta)\left(u^2 - 1\right) \frac{\:d\theta}{\sec(\theta)\tan(\theta)} = \int u^2 - 1 \:du = \frac{u^3}{3} - u + C
\end{equation}
Where $C$ is the constant of integration.
Now $u = \sec(\theta)$ and $x = \tan(\theta)$ and so $u = \sec(\arctan(x)) = \sqrt{x^2 + 1}$
Thus,
\begin{equation}
I =\int \frac{x^3}{\sqrt {x^2+1}}dx = \frac{1}{3}\left( \sqrt {x^2+1}\right)^3 - \sqrt {x^2+1} + C = \frac{1}{3}\left(x^2+1\right)^{\frac{3}{2}} - \sqrt {x^2+1} + C
\end{equation}
Method 2:
\begin{align}
I &=\int \frac{x^3}{\sqrt {x^2+1}}dx = \int x \cdot \frac{x^2}{\sqrt {x^2+1}}dx \\
&= \int x \cdot \frac{x^2 + 1 - 1}{\sqrt {x^2+1}}dx = \int x \left( \sqrt {x^2+1} - \frac{1}{\sqrt{x^2+1}}\right)dx
\end{align}
Here let $u = x^2 + 1$:
\begin{align}
I &= \int x \left( \sqrt {u} - \frac{1}{\sqrt{u}}\right)\frac{du}{2x} = \frac{1}{2}\left[\int \left( u^{\frac{1}{2}} - u^{-\frac{1}{2}}\right)du\right] = \frac{1}{2}\left[\frac{ u^{\frac{3}{2}}}{\frac{3}{2}} + \frac{ u^{\frac{1}{2}}}{\frac{1}{2}} \right] + C \\
&= \frac{1}{3}u^{\frac{3}{2}} + u^{\frac{1}{2}} + C = \frac{1}{3}\left(x^2 + 1\right)^{\frac{3}{2}} + \left(x^2 + 1\right)^{\frac{1}{2}} + C
\end{align}
Where $C$ is the constant of integration.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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exponential equation has non positive roots
Find real values of $a$ for which the equation
$4^x-(a-3)\cdot 2^x+(a+4)=0$
has non positive roots
Try: Let $2^x=y\in (0,1].$ Then equation convert into
$y^2-(a-3)y+(a+4)=0$
For real roots its discriminant $\geq 0$
So $$(a-3)^2-4(a+4)\geq 0$$
$$a^2-10a-7\geq 0$$
$$a\in \bigg(-\infty,5-4\sqrt{2}\bigg]\cup \bigg[5+4\sqrt{2},\infty\bigg).$$
but answer is different from that , i did not know where i
am missing. could some help me to solve it. thanks
| Not only the equation in $y$ must have real roots, but these roots have to be $\le 1$, since it's asked that the equation in $x$ has non-positive roots, so that $y=2^x\le 2^0=1$.
To test whether the roots are less than $1$, we can place $1$ w.r.t. the roots $y_1,y_2$ of $\;p(y)=y^2-(a-3)y+a+4$. The standard method is to determine the sign of $p(1)=8>0$. Therefore $1$ does not separate the roots of $p(y)$, and $1$ is either less than or greater than both of them. Furthermore, the half-sum of the roots is $\;\frac{a-3}2$ by Vieta's relations, so that
$$y_1, y_2 \le 1\iff \frac{y_1+y_2}2=\frac{a-3}2 \le 1\iff a\le 5. $$
So the solution is
$$a\in \Bigl\{\Bigl(-\infty,5-4\sqrt{2}\Bigr]\cup \Bigl[5+4\sqrt{2},\infty\Bigr)\Bigr\}\cap (-\infty,5]=\Bigl(-\infty,5-4\sqrt{2}\Bigr].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3048594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An exercise on the calculation of a function of operator The operator is given by
$$A=\begin{pmatrix}
1 & 0 & 0\\
1 & 1 & 0\\
0 & 0 & 4
\end{pmatrix}$$
I have to write down the operator $$B=\tan(\frac{\pi} {4}A)$$
I calculate $$\mathcal{R} (z) =\frac{1}{z\mathbb{1}-A}=\begin{pmatrix}
\frac{1}{z-1} & 0 & 0\\
\frac{1}{(z-1)^2} & \frac{1}{z-1} & 0\\
0 & 0 & \frac{1}{z-4}\end{pmatrix} $$
Now the B operator is given by:
$$B=\begin{pmatrix}
Res_{z=1}\frac{\tan(\frac{\pi}{4}z)}{z-1} & 0 & 0\\
Res_{z=1}\frac{\tan(\frac{\pi}{4}z)}{(z-1)^2} & Res_{z=1}\frac{\tan(\frac{\pi}{4}z)}{z-1} & 0\\
0 & 0 & Res_{z=4}\frac{\tan(\frac{\pi}{4}z)}{z-4}
\end{pmatrix} $$
For me the result should be
$$ B=\begin{pmatrix}
1 & 0 & 0\\
\frac{\pi}{2} & 1 & 0\\
0 & 0 & 0\end{pmatrix}$$
But the exercise gives as solution:
$$ B=\begin{pmatrix}
1 & 0 & 0\\
\frac{\pi}{4} & 1 & 0\\
0 & 0 & 1\end{pmatrix}$$
Where is the error?
Thank you and sorry for bad English
| Your calculation seems plausible. Let's try it the other way: the Taylor expansion of $\tan(\pi x/4)$ at $x=1$ is
$$
\tan(\pi x/4)=1+\frac{\pi}{2}(x-1)+\ldots
$$
Then for the Jordan block
$$
J=\begin{bmatrix}1 & 0\\1 & 1\end{bmatrix}
$$
we have
$$
\tan(\pi J/4)=I+\frac{\pi}{2}\begin{bmatrix}0 & 0\\1 & 0\end{bmatrix}=\begin{bmatrix}1 & 0\\\color{red}{\frac{\pi}{2}} & 1\end{bmatrix}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I solve $\lim \limits_{x \to \frac{Ο}{3}} \frac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$ Alright, I know, there are easier ways to solve this, like L'hopitals Rule etc.
But I'm not solving it for the answer, just doing it for the fun so I tried using substitution method.
Put $t= x- \dfrac{Ο}{3}$
$\lim \limits_{x \to \frac{Ο}{3}} \dfrac{2 \sin x - \sqrt{3}}{\cos \frac{3x}{2}}$
$= \lim\limits_{t \to 0} \dfrac{2 \sin \left(t+\frac{Ο}{3} \right) - \sqrt{3}}{\cos \left( \frac{3t}{2} + \frac{Ο}{2}\right)}$
$= \lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}}$
Where do I go from here, I'm not able to eliminate the $t$ fully from the Nr and Dr, any help?
Or any other alternative way that uses only the fact that $\lim\limits_{x \to 0} \dfrac{ \sin x}{x} = 1$?
Thanks :)
| $$\lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}}=\lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ \sin \frac{3t}{2}}\cdot\frac{3t/2}{3t/2}$$
$$=\frac{2}{3} \lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{ t}\ \ \ \ \ \ \ (1)$$
Now, $1-\cos t=2\sin^2(t/2)$
$$\lim\limits_{t \to 0} \dfrac{\sqrt{3}- \sin t - \sqrt{3} \cos t}{t}= \lim\limits_{t \to 0}( \frac{\sqrt{3}(1-\cos t)}{t}-\frac{\sin t}{t})$$
$$=\lim\limits_{t \to 0}( \frac{2\sin^2(t/2)}{t}-\frac{\sin t }{t})$$
$$=\lim\limits_{t \to 0}( \frac{2\sin^2(t/2)}{t^2/4}\frac{t^2/4}{t}-\frac{\sin t}{t})$$
$$=0-1=-1$$
Put $-1$ in (1) to get the final limit, $-2/3$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof that $x^{(n+4)} \bmod 10 = x^n \bmod 10\,$ for $\,n\ge 1$ While solving a programming challenge in which one should efficiently compute the last digit of $a^b$, I noticed that apparently the following holds (for $n > 0$)
$x^{(n+4)} \mod 10 = x^n \mod 10$
How can this be proven?
| Hint
$$ x^{n} \left( x^4 - 1 \right) \equiv x^{n} \left( x^2 - 1 \right) \left( x^2 + 1 \right)\equiv \left( x - 1 \right) x \left(x+1\right) x^{n-1} \left( x^2 + 1 \right) \pmod{2}\\
x^{n} \left( x^4 - 1 \right) \equiv x^{n} \left( x^2 - 1 \right) \left( x^2 + 1 \right)\equiv x^{n} \left( x^2 - 1 \right) \left( x^2 -4 \right)\pmod{5}\\
\equiv x^{n-1} \left( x- 2 \right) \left( x- 1 \right) x\left( x+2 \right) \left( x+ 2 \right) \pmod 5
$$
Now use the fact that one of $x-1$ or $x$ is even, and that among 5 consecutive integers one is a multiple of 5.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3050972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 7
} |
Trigonometric parametric system I have a very specific system of two trigonometric equations
$$\left( 3A^2\sin x \cos x - A \sin x \right) + \left( 3B^2\sin y \cos y - B \sin y \right) = AB \sin (x+y)$$
$$\left( 3A^2\cos^2 x - A \cos x \right) + \left( 3B^2\cos^2 y - B \cos y \right) = AB \cos (x+y) - \frac{3}{2}\left(1-A^2-B^2\right)$$
I need to solve it with respect to angles $x$, $y$ with $A$ and $B$ being parameters. If one of the parameters is equal to zero or if $A=B$ it is actually possible to solve it, but I lost all hopes to get an adequate closed-form solution in a general case.
Are there any ways to solve it? Maybe for some special values of $A$, $B$, or maybe it is provably unsolvable for specific parameters?
| For the first equation,
$$\left( 3A^2\sin x \cos x - A \sin x \right) + \left( 3B^2\sin y \cos y - B \sin y \right) = AB \sin (x+y)$$
set $A=0$,
you get
$$ 3B^2\sin y \cos y - B \sin y = 0$$
Let $\sin y=\alpha$
$$3B^2\alpha\sqrt{1-\alpha^2}-B\alpha=0$$
$$9B^4\alpha^2(1-\alpha^2)=B^2\alpha^2$$
Now we can see one of the solutions is $\alpha=0,~\color{red}{y=n\pi}$,where $n$ is an integer.
Reduce both sides, the equation becomes
$$9B^2(1-\alpha^2)-1=0$$
Using the quadratic formula
$$\alpha=\frac{\pm\sqrt{4(9B^2)(9B^2-1)}}{18B^2}$$
$$\alpha=\frac{\pm\sqrt{9B^2-1}}{3B}$$
$$\color{red}{y=\sin^{-1}\left(\frac{\pm\sqrt{9B^2-1}}{3B}\right)}$$
For the second equation,
$$\left( 3A^2\cos^2 x - A \cos x \right) + \left( 3B^2\cos^2 y - B \cos y \right) = AB \cos (x+y) + \frac{3}{2}\left(1-A^2-B^2\right)$$
also set $A=0$, and you get
$$\left( 3B^2\cos^2 y - B \cos y \right) = \frac{3}{2}\left(1-B^2\right)$$
Set $\cos y= \beta$
$$\left( 3B^2\beta^2 - B\beta \right) = \frac{3}{2}\left(1-B^2\right)$$
Again, use the quadratic formula
$$\beta=\frac{B\pm\sqrt{B^2+18B^2(1-B^2)}}{6B^2}$$
$$\beta=\frac{1\pm\sqrt{1+18(1-B^2)}}{6B}$$
$$\color{red}{y=\cos^{-1}\left(\frac{1\pm\sqrt{19-18B^2}}{6B}\right)}$$
Since the equations you give is symmetric, I would conclude that
$$\small\color{green}{x=\cos^{-1}\left(\frac{1\pm\sqrt{19-18A^2}}{6A}\right)}\large{,~}\small\color{green}{\sin^{-1}\left(\frac{\pm\sqrt{9A^2-1}}{3A}\right)}\large{,~}\small \color{green}{n\pi}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Integral between 2 spheres
Let D be the set of all points $(x, y, z)$ satisfying
$ 1 \leq x^2 + y^2 + z^2 \leq 2 $ and $ z \geq 0 $
, find $\int_{D}x^2 $
how do i solve this question through triple integral and spherical co-ordinates ?
should i calculate $ \int_{r=1}^{r=\sqrt{2}}\int_{\theta = 0}^{\theta = 2\pi}\int_{\phi = 0}^{\phi = \frac{\pi}{2}} r^4cos(\theta)^2sin(\phi)^3{dr}\ d{\theta}\ d{\phi} $
imaginging the volume it is between a ball with radius $1$ and a ball with radius $\sqrt{2}$
| First, notice that, due to symmetry
$$\iiint\limits_{1<r<\sqrt 2,z>0}x^2 dxdydz=\frac{1}{2}\iiint\limits_{1<r<\sqrt 2}x^2 dxdydz$$
Now
$$\iiint\limits_{1<r<\sqrt 2}x^2 dxdydz = \iiint\limits_{1<r<\sqrt 2}y^2 dxdydz = \iiint\limits_{1<r<\sqrt 2}z^2 dxdydz$$
Thus,
$$\iiint\limits_{1<r<\sqrt 2,z>0}x^2 dxdydz = \frac{1}{6}\iiint\limits_{1<r<\sqrt 2}(x^2+y^2+z^2) dxdydz = \frac{1}{6}\iiint\limits_{1<r<\sqrt 2}r^2 dxdydz $$
Now, since the integrand only depends on $r$, we don't need full spherical coordinates, but only the surface area of a sphere of radius $r$, which is $4 \pi r^2$. So,
$$\frac{1}{6}\iiint\limits_{1<r<\sqrt 2}r^2 dxdydz =\frac{1}{6} \int\limits_{1<r<\sqrt 2}4 \pi r^4 dr = \frac{2 \pi}{15}r^5 \biggr\rvert_1^\sqrt 2=\frac{2 \pi}{15}(4\sqrt 2-1)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3051974",
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Find $x$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$ Find all $x \in (-1, +\infty)$ such that $(x^2 + 4x + 3)^x + (2x + 4)^x = (x^2 + 4x + 5)^x$.
What I have done so far was a substitution $y = x + 2$ which results in a nicer form of the equation:
$(y - 1)^{y - 2}(y + 1)^{y - 2} + 2^{y - 2}y^{y - 2} = (y^2 + 1)^{y - 2}$ and $y > 1$, which can then be rewritten as:
$$ \left( \frac{y^2 + 1}{2y} \right)^{y - 2} - \left( \frac{y^2 - 1}{2y} \right)^{y - 2} = 1 $$
Also, my intuition is that $y = 4$ is the unique solution and initially I wanted to prove that this function of $y$ is injective, but it didn't work either.
Do you have any suggestion on how I can continue?
| Hint:
$$(x^2+4x+5)^2-(x^2+4x+3)^2=2(2x^2+8x+8)=(2x+4)^2$$
$$\iff1=\left(\dfrac{2x+4}{x^2+4x+5}\right)^2+\left(\dfrac{x^2+4x+3}{x^2+4x+5}\right)^2$$
We have
$$\left(\dfrac{2x+4}{x^2+4x+5}\right)^x+\left(\dfrac{x^2+4x+3}{x^2+4x+5}\right)^x=1$$
WLOG $\dfrac{2x+4}{x^2+4x+5}=\cos t,\dfrac{x^2+4x+3}{x^2+4x+5}=\sin t$ with $0<t<\dfrac\pi2$ for $-1<x<\infty$
Clearly, $$(\cos t)^x+(\sin t)^x$$ is a decreasing function to forbid multiple solutions
Generalization : $$\left(\dfrac{2x+4}{x^2+4x+5}\right)^y+\left(\dfrac{x^2+4x+3}{x^2+4x+5}\right)^y=1$$
$\implies y=2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3052915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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If $\lim\limits_{x\to\alpha}\frac{x-2}{x^3-2x+m}=-\infty$, then what are the possible values for $\alpha$ and $m$?
If $\lim\limits_{x\to\alpha}\dfrac{x-2}{x^3-2x+m}=-\infty$, then what are the possible values for $\alpha$ and $m$?
A student I'm tutoring came to me with this problem. I believe the limit is not one-sided, so that the expression approaches $-\infty$ as $x\to\alpha$ from either direction.
I believe this would require $x=\alpha$ to be a zero of the denominator, so that
$$x^3-2x+m=(x-\alpha)(x^2+\alpha x+\alpha^2-2)+\alpha^3-2\alpha+m$$
The remainder term should vanish, so
$$\alpha^3-2\alpha+m=0$$
But in order for the limit to diverge to the "same" $-\infty$ from either side of $x=\alpha$, I'm under the impression that $x=\alpha$ should actually be a zero of multiplicity $2$. (I'm picturing the behavior of $-\dfrac1{x^2}$ around $x=0$.) Then we'd have
$$x^2+\alpha x+\alpha^2-2=(x-\alpha)(x+2\alpha)+3\alpha^2-2$$
Again the remainder should be $0$, so that
$$3\alpha^2-2=0\implies\alpha=\pm\sqrt{\frac23}\implies m=\pm\frac43\sqrt{\frac23}$$
When I check the limits for either pair of $(\alpha,m)$, I find
$$\lim_{x\to-\sqrt{\frac23}}\frac{x-2}{x^3-2x-\frac43\sqrt{\frac23}}=\color{red}+\infty$$
$$\lim_{x\to\sqrt{\frac23}}\frac{x-2}{x^3-2x+\frac43\sqrt{\frac23}}=-\infty$$
It seems that the sign of $\dfrac{x-2}{x+2\alpha}$ dictates whether the limit diverges to positive or negative infinity.
This explanation seems a bit too hand-wavy and perhaps too verbose for a high-school-level calculus student. Is there a more straightforward or concise argument that can be made to show that $\alpha=\sqrt{\dfrac23}$ and $m=\dfrac43\sqrt{\dfrac23}$ is the answer?
| Using the given condition we can see that $(x^3-2x+m)/(x-2)$ is negative and tends to $0$ as $x\to a$ (replaced $\alpha$ with $a$ to simplify typing). Note that the given condition also excludes $a=\pm\infty$ and hence we assume $a\in\mathbb {R} $. Then by multiplication with $(x-2)$ we get $$\lim_{x\to a} (x^3-2x+m)=0$$ so that $$a^3-2a+m=0\tag{1}$$ If $a=2$ then $m=-4$ and then $$x^3-2x-4=(x-2)(x^2+2x+2)$$ and then $(x^3-2x+m)/(x-2)$ does not tend to $0$. Hence $a\neq 2$.
From $(1)$ we can see that $$x^3-2x+m=x^3-2x-a^3+2a=(x-a)(x^2+ax+a^2-2)$$ and we need to ensure that the expression above maintains a constant sign opposite to that of $(a-2)$ as $x\to a$. This is possible only when the factor $x^2+ax+a^2-2$ has a single root $a$ ie $3a^2-2=0$ or $a=\pm\sqrt{2/3}$ and $$x^2+ax+a^2-2=(x-a)(x+2a)$$ so that $x=a$ is indeed a single root.
Next note that we have $$x^3-2x+m=(x-a)^2(x+2a)$$ and we want its sign to be opposite to that of $a-2$ as $x\to a$ so that $a=\sqrt{2/3}$ is the only option.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3055415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Factors in a different base $\ 2b^2\!+\!9b\!+\!7\,\mid\, 7b^2\!+\!9b\!+\!2$
Two numbers $297_B$ and $792_B$, belong to base $B$ number system. If the first number is a factor of the second number, then what is the value of $B$?
Solution:
But base cannot be negative. Could someone please explain where I am going wrong?
| $$2B^2+9B+7\mid 7B^2+9B+2$$
Let's write $aB^2+bB + c$ as $[a,b,c]_B$ to emphasis that $a,b,c$ are digits base $B$.
Then $[2,9,7]_B \mid [7,9,2]_B-[2,9,7]_B$ and we are assuming that $2,9,7 < B$
Writing this out "subtraction-style", we get
$\left.\begin{array}{c}
& 7 & 9 & 2 \\
-& 2 & 9 & 7 \\
\hline
\phantom{4}
\end{array}
\right.
\implies
\left.\begin{array}{c}
& 6 & (B+8) & (B+2) \\
-& 2 & 9 & 7 \\
\hline
& 4 & (B-1) & (B-5)
\end{array}
\right.
$
So $[4,B-1,B-5]_B$ is a multiple of $[2,9,7]_B$.
We must therefore have $[4,B-1,B-5]_B = 2[2,9,7]_B = [4,18,14]_B$ which implies $B-1=18$ and $B-5=14$. Hence $B=19$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})\cdots(1+\frac{1}{a_n})$, where $a_1=1$, $a_n=n(1+a_{n-1})$
Evaluate $\lim_{n\to\infty}(1+\frac{1}{a_1})(1+\frac{1}{a_2})\cdots(1+\frac{1}{a_n})$,
where $a_1 = 1$, $a_n = n(1+a_{n-1})$
\begin{align*}
&\lim_{n\to\infty} \left(1+\frac{1}{a_1}\right)\left(1+\frac{1}{a_2}\right)\cdots\left(1+\frac{1}{a_n}\right) \\
&= \lim_{n\to\infty} \frac{1}{a_1} \cdot \frac{1}{2} \cdot \frac{1}{3} \cdots \frac{1}{n} \cdot (a_n + 1) \\
&= \lim_{n\to\infty} \frac{a_n + 1}{n!}.
\end{align*}
Then I'm stuck. How to proceed?
| From where you stuck you can proceed as
\begin{align}
\frac{a_n+1}{n!}&=\frac{a_n}{n!}+\frac{1}{n!}=\frac{1}{(n-1)!}\frac{a_n}{n}+\frac{1}{n!}=\frac{1}{(n-1)!}(a_{n-1}+1)+\frac{1}{n!}=\\
&=\frac{a_{n-1}}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!}=\ldots=
1+\frac{1}{1!}+\ldots+\frac{1}{n!}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3058562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
If $A=A^2$ is then $A^T A = A$? I know that for a matrix $A$:
If $A^TA = A$ then $A=A^2$
but is it if and only if? I mean:
is this true that "If $A=A^2$ then $A^TA = A$"?
| The answer is no.
Consider $A = \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}$. We have
$$A^2 = \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix} = A$$
but $$A^TA = \begin{bmatrix} 1 & 0 \\ -1 & 0 \end{bmatrix}\begin{bmatrix} 1 & -1 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} 1 & -1 \\ -1 & 1 \end{bmatrix} \ne A$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3059796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating an improper integral $\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$ I tried to solve the integral: $$\int_{0}^{\infty}\frac{x^2}{(x^4+1)^2}dx$$
using $ x = \sqrt{\tan(u)}$ and $dx = \frac{ \sec^2(u)}{2\sqrt{\tan(u)}} du,$
but I ended up with an even worse looking integral $$ \int_{0}^{\frac{\pi}{2}}\frac{\sqrt{\tan(u)}}{\sec^2(u)}du.$$
Wolfram gave an answer of $ \dfrac{\pi}{8\sqrt{2}},$ but how would one get to that answer?
| We first reduce the power 2 in the denominator by integration by parts.
$$
\displaystyle \begin{aligned}
\int \frac{x^{2}}{\left(x^{4}+1\right)^{2}} d x &=-\frac{1}{4} \int \frac{1}{x} d\left(\frac{1}{x^{4}+1}\right) \\
&\stackrel{ IBP }=-\frac{1}{4 x\left(x^{4}+1\right)}-\frac{1}{4} \int \frac{1}{x^{2}\left(x^{4}+1\right)} d x \\
&=-\frac{1}{4 x\left(x^{4}+1\right)}-\frac{1}{4} \int\left(\frac{1}{x^{2}}-\frac{x^{2}}{x^{4}+1}\right) d x \\
&=-\frac{1}{4 x\left(x^{4}+1\right)}+\frac{1}{4 x}+\frac{1}{4} \int \frac{x^{2}}{x^{4}+1} d x \\
&=\frac{x^{3}}{4\left(x^{4}+1\right)}+\frac{1}{4} \int \frac{x^{2}}{x^{4}+1} d x
\end{aligned}
$$
$$
\therefore \int_{0}^{\infty} \frac{x^{2}}{\left(x^{4}+1\right)^{2}} d x =\frac{1}{4}\left[\frac{x^{3}}{x^{4}+1}\right]_{0}^{\infty}+\frac{1}{4} \int_{0}^{\infty} \frac{x^{2}}{x^{4}+1} d x
= \frac{1}{4} \int_{0}^{\infty} \frac{x^{2}}{x^{4}+1} d x
$$
Now we are going to play a little trick on the last integral. $$
\begin{aligned}
\int_{0}^{\infty} \frac{x^{2}}{x^{4}+1} d x&=\int_{0}^{\infty} \frac{1}{x^{2}+\frac{1}{x^{2}}} d x \\
&=\frac{1}{2} \int_{0}^{\infty} \frac{\left(1+\frac{1}{x^{2}}\right)+\left(1-\frac{1}{x^{2}}\right)}{x^{2}+\frac{1}{x^{2}}} d x \\
&=\frac{1}{2}\left[\int_{0}^{\infty} \frac{d\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^{2}+2}+\int_{0}^{\infty} \frac{d\left(x+\frac{1}{x}\right)}{\left(x+\frac{1}{x}\right)^{2}-2}\right] \\
&=\frac{1}{2 \sqrt{2}}\left[\tan ^{-1}\left(\frac{x-\frac{1}{x}}{\sqrt{2}}\right)+\frac{1}{4 \sqrt{2}} \ln \left|\frac{x+\frac{1}{x}-\sqrt{2}}{x+\frac{1}{x}+\sqrt{2}}\right|\right]_{0}^{\infty} \\
&=\frac{\pi}{2 \sqrt{2}}
\end{aligned}
$$
We can now conclude that
$$ \boxed{\int_0^{\infty} \frac{x^{2}}{\left(x^{4}+1\right)^{2}} d x =\frac{\pi}{8 \sqrt{2}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3061224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
How to show that $\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\frac{\pi^2}{6}$ Wolfram Alpha shows that
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\zeta(2)=\frac{\pi^2}{6}$$
I want to prove this.
Attempt:
I tried to treat this as a telescoping series:
$$\begin{align}
\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}&=\sum_{n=1}^{\infty}H_n\left(\frac{1}{n}-\frac{1}{n+1}\right)\\
&=H_{1}\left(1-\frac{1}{2}\right)+H_{2}\left(\frac{1}{2}-\frac{1}{3}\right)+H_{3}\left(\frac{1}{3}-\frac{1}{4}\right)\\
&=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{6}+\frac{1}{3}-\frac{1}{4}+\frac{1}{6}-\frac{1}{8}+\frac{1}{9}-\frac{1}{12}
\end{align}$$
I think this method is not quite useful, so I tried another one:
$$H_n=\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
Then,
$$\sum_{n=1}^{\infty}\frac{H_n}{n^2+n}=\sum_{n=1}^{\infty}\frac{1}{n^2+n}\int_{0}^{1}\frac{1-t^n}{1-t}dt$$
At this point, I do not know how to proceed.
| applying summation by parts
\begin{align}
\sum_{n=1}^N\frac{H_n}{n(n+1)}&=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^N\frac{H_n}{n+1}\\
&=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^{N+1}\frac{H_{n-1}}{n}\\
&=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^{N+1}\frac{H_{n}}{n}+\sum_{n=1}^{N+1}\frac{1}{n^2}\\
&=\sum_{n=1}^N\frac{H_n}{n}-\sum_{n=1}^{N}\frac{H_{n}}{n}-\frac{H_{N+1}}{N+1}+\sum_{n=1}^{N+1}\frac{1}{n^2}\\
&=-\frac{H_{N+1}}{N+1}+\sum_{n=1}^{N+1}\frac{1}{n^2}
\end{align}
letting $N$ approach $\infty$, we get,
$$\sum_{n=1}^\infty\frac{H_n}{n(n+1)}=0+\sum_{n=1}^{\infty}\frac{1}{n^2}=\zeta(2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3063965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 2
} |
Prove the identity for $\tan3\theta$
Prove the identity for $$\tan3\theta= \frac{3\tan\theta - \tan^3 \theta}{1-3\tan^2 \theta}$$
Using de Moivre's theorem I have found that:
$$\cos3\theta = 4\cos^3\theta - 3\cos \theta$$
$$\sin 3\theta = 3\sin \theta-4\sin^3\theta$$
therefore:
$$\tan 3\theta = \frac{\sin 3\theta}{\cos 3 \theta}=\frac{3\sin \theta-4\sin^3\theta}{4\cos^3\theta - 3\cos \theta}$$
To then try and get the whole expression in terms of $\tan\theta$ I multiplied top and bottom of the fraction by $(4\cos^3\theta)^{-1}$. This gave me the following but I'm not now sure how to finish it off
$$\tan3\theta =\frac{\frac{3\sin \theta}{4\cos^3\theta} - \tan^3\theta}{1-\frac{3\cos \theta}{4\cos^3 \theta}}$$
| Alternatively, to do it using what you've done so far, rather than applying the double-angle formula twice:
Note that $\frac{\sin\theta}{\cos^3\theta} = \tan\theta\sec^2\theta = \tan\theta(1 + \tan^2\theta) = \tan\theta + \tan^3\theta$. Alternatively, to do it using what you've done so far, rather than applying the double-angle formula twice:
note that $\frac{\sin\theta}{\cos^3\theta} = \tan\theta\sec^2\theta = \tan\theta(1 + \tan^2\theta) = \tan\theta + \tan^3\theta$. That will fix up your numerator nicely. Using $\sec^2 \theta = 1 + \tan^2\theta$ directly (since $\frac{\cos\theta}{\cos^3\theta} = \sec^2\theta$) will similarly fix up the numerator, so what you have so far is equal to
$$\frac{\frac{3}{4}\tan\theta-\frac{1}{4}\tan^3\theta}{\frac{1}{4} - \frac{3}{4}\tan^2\theta} = \frac{3\tan\theta-\tan^3\theta}{1-3\tan^2\theta},$$
as required
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3067903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Choosing an integer that is most likely to be the sum of die rolls Choose any positive integer $n$. Then roll an unbiased six-sided die as many times as needed until the sum of the results is either exactly equal to $n$ (you win) or greater than $n$ (you lose).
If you were to play this game; what number $n$ would you select in order to maximise the chance of winning and what is the probability of it?
(For example, choose $n=7$, then you would roll a die and if you get $6$ on the 1st throw and $1$ on the 2nd throw you would stop since you reached $n$).
| Denote by $p_{k}$ the probability that a sum of $k$ is observed. Clearly
$$p_{1}=\frac{1}{6}$$
For $p_{2}$, there is a $\frac{1}{6}$ chance of rolling a $2$ and a $\frac{1}{6}$ chance of rolling a $1$, followed by another one. This can be written as
$$p_{2}=\frac{1}{6}+\frac{p_{1}}{6}=\frac{1}{6}(1+p_{1})$$
For $p_{3}$, there is a $\frac{1}{6}$ chance of rolling a $3$ and winning. The only other ways to win are to roll a $1$ or $2$, each with probability $\frac{1}{6}$. If we roll a $1$, then our odds of winning are $p_{2}$ and if we roll a $2$, our odds of winning are $p_{1}$. Therefore
$$p_{3}=\frac{1}{6}+\frac{p_{1}}{6}+\frac{p_{2}}{6}=\frac{1}{6}(1+p_{1}+p_{2})$$
Similarly we have
$$p_{4}=\frac{1}{6}(1+p_{1}+p_{2}+p_{3})$$
$$p_{5}=\frac{1}{6}(1+p_{1}+p_{2}+p_{3}+p_{4})$$
$$p_{6}=\frac{1}{6}(1+p_{1}+p_{2}+p_{3}+p_{4}+p_{5})$$
Now, for $k>6$ we cannot win in one roll so we have
$$p_{k}=\frac{1}{6}(p_{k-1}+p_{k-2}+p_{k-3}+p_{k-4}+p_{k-5}+p_{k-6})$$
Observe that for any $k>6$,
$$p_{k}=\frac{1}{6}\sum_{m=1}^{6}p_{k-m}<\frac{1}{6}\cdot 6\max_{m=1,2,3,4,5,6}p_{k-m}=\max_{m=1,2,3,4,5,6}p_{k-m}$$
This proves that the winning choice is not bigger than $6$, and in fact must be the max of $p_{1},p_{2},...p_{6}$. Clearly this is $p_{6}$, so we should choose $n=6$ to maximize the odds of winning.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3070185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I create a function which describes the length between a parabola and a point on the y axis?
This is what I'm thinking:
The distance of the line segment is between points (0,y) and$ (x,y)$ and the y coordinate of the parabola is $4-x^2
so using he distance formula it is:
$\sqrt{(x-0)^{2} +((4-x^2)-y)^{2}} = \sqrt{x^{2}+((4-x^2)-y)^2}$
Now what do I do? Do I take the derivative? Please help.
| Distance form point $(x,y)$ on the graph and $(0,2)$ can be parametrized by;
$L(x) = \sqrt{x^2 + (|y-2|)^2} = \sqrt{x^2 + (2-x^2)^2} = \sqrt{4 - 3 x^2 + x^4}$ (assuming $y>2$)
To find the minimum use the given point $\sqrt{3/2}$ and put it into $L(X)$
$L(\sqrt{3/2}) = \sqrt{4 - 3 (\sqrt{3/2})^2 + (\sqrt{3/2})^4} = \sqrt{4 - 3 \cdot 3/2 + 9/4} = \sqrt{ 16 - 18 + 9}/2 = \sqrt{7}/2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3073719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a, b \in \mathbb{R^+}$ such that $a-b=10$, find smallest value of constant $K$ for which $\sqrt{(x^2 + ax)}- \sqrt {(x^2 +bx)}0$. Let $a, b$ are positive real numbers such that $a - b = 10$ , then the smallest value of the constant $K$ for which $\sqrt{(x^2 + ax)} - \sqrt {(x^2 +bx)} < K$ for all $x>0$.
My try:
Unable to solve further
| Hint.
$$
\sqrt{(x^2 + ax)} - \sqrt {(x^2 +bx)} = x\left(\sqrt{\frac{a}{x}+1}-\sqrt{\frac{b}{x}+1}\right)
$$
and for large $x$
$$
\sqrt{\frac{a}{x}+1}-\sqrt{\frac{b}{x}+1} = \frac{a-b}{2 x}+\frac{1}{8} \left(\frac{1}{x}\right)^2
\left(b^2-a^2\right)+O\left(\left(\frac{1}{x}\right)^3\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3075896",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integrate $\int\frac{\cos^2(x)-x^2\sin(x)}{(x+\cos(x))^2}dx$ I had to integrate the following integral:
\begin{equation}
\int\frac{\cos^2(x)-x^2\sin(x)}{(x+\cos(x))^2}dx
\end{equation}
but I can't find a suitable substitution to find a solution. Nothing I try works out and only seems to make it more complicated. Does anyone have an idea as to how to solve this?
I also tried to get help from WolframAlpha but it just says that there is no step-by-step solution available.
The sollution by wolfram alpha is:
\begin{equation}
\int\frac{\cos^2(x)-x^2\sin(x)}{(x+\cos(x))^2}dx = \frac{x\cos(x)}{x+\cos(x)} + c
\end{equation}
| \begin{equation}
\int\frac{\cos^2 x-x^2\sin x }{(x+\cos x)^2}dx
\end{equation}
Divide both numerator and denominator by $x^2\cos^2 x$
\begin{equation}
\int\frac{\cos^2 x-x^2\sin x }{(x+\cos x)^2}dx
=\int\frac{\dfrac{1}{x^2}-\dfrac{\sin x}{\cos^2 x}}{(\dfrac{1}{\cos x}+\dfrac{1}{x})^2}dx
=\frac{1}{\dfrac{1}{\cos x}+\dfrac{1}{x}} + C
=\frac{x\cos x}{x + \cos x} + C
\end{equation}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3078486",
"timestamp": "2023-03-29T00:00:00",
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How do I solve for $a$ and $b$ in $\lim\limits_{x \to β} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$? Given $\lim\limits_{x \to β} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2$
I need to solve for $a$ and $b$, so here we go,
$\lim\limits_{x \to β} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right)$
$= \lim\limits_{x \to β} x \left(2 +(3+x) \left(\ln (1+\frac{a}{x}) - \ln(1+\frac{b}{x}) \right) \right)$
$=\lim\limits_{x \to β} x \left(2 +(3+x)\left( \dfrac{a-b}{x} \right) \right)$
$=\lim\limits_{x \to β} \left(2x +(3+x)\left( a-b \right) \right)$
$=\lim\limits_{x \to β} \left(2x + 3(a-b) + x(a-b) \right) $
Since the limit exists, $2x$ and $x(a-b)$ must cancel out so $a-b = 2$
But if I do this, I'm left with $3(a-b) = 2$ which gives me a contradiction since $a-b$ is supposed to be $2$.
What do I do now? And where exactly have I gone wrong?
Thank you!
|
Since the limit exists, $2x$ and $x(aβb)$ must cancel out so $aβb=2$.
You wanted to say $2+a-b=0 \Rightarrow \color{red}{a-b=-2}$.
And you cannot go from the difference of logarithms (line $2$) to the difference of arguments (line $3$), because it is not a factor, but a sum $(2+\cdots)$, which is multiplied by $x$. In other words, you can not apply the asymptotics to an addend, but must apply to a factor.
Alternatively:
$$\lim\limits_{x \to β} x \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 2 \Rightarrow \\
\lim\limits_{x \to β} \frac{2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right)}{\frac 1x} = 2 \ \ \ (*)$$
For the limit to exist:
$$\lim\limits_{x \to β} \left(2 +(3+x) \ln \left( \dfrac{x+a}{x+b} \right) \right) = 0 \Rightarrow \cdots \Rightarrow a-b=-2.$$
Now use L'Hospital's rule for $(*)$ twice:
$$\lim\limits_{x \to β} \frac{\ln (x+a) +\frac{3+x}{x+a}- \ln (x+b) -\frac{3+x}{x+b}}{-\frac 1{x^2}} = \\
=\lim\limits_{x \to β} \frac{\frac1{x+a}+\frac{a-3}{(x+a)^2}-\frac1{x+b}-\frac{b-3}{(x+b)^2}}{\frac 2{x^3}} = \\
=\lim_{x\to\infty} -\frac{x^3 (a - b) (2 a b + a x - 3 a + b x - 3 b - 6 x)}{2 (x+a)^2 (x+b)^2}=2 \Rightarrow \\
\color{red}{a+b-6=2}.$$
Hence:
$$\begin{cases}a-b=-2 \\ a+b-6=2 \end{cases} \Rightarrow a=3, b=5.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3079016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
} |
Finding all roots of $x^6 + 2 x^5 + 2 x^4 + 3 x^2 + 6 x + 6$, knowing that one root is $x=-1+i$.
Find all roots of
$$V(x) = x^6 + 2 x^5 + 2 x^4 + 3 x^2 + 6 x + 6$$
knowing that one root is $x=-1+i$.
Sorry for the picture. I found two roots of the polynomial and also found an equation which can help me to find all of the rest. Can someone help me what do I need to do next?
| One factor is $$x^2+2x+2=0$$ which can be solved by the quadratic formula. Solving $$x^4+3=0$$ we get
$$\frac{1}{2}\sqrt{2}\cdot3^{1/4}(1+i),\frac{1}{2}\sqrt{2}\cdot3^{1/4}(1-i),\frac{1}{2}\sqrt{2}\cdot3^{1/4}(-1+i),\frac{1}{2}\sqrt{2}\cdot3^{1/4}(-1-i)$$
Substituting $$x^2=t$$ then we get $$t^2+(\sqrt{3})^2=0$$ By the formula $$a^2+b^2=(a+bi)(a-bi)$$ you will get the solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find $a, b\in\mathbb{R}$ so that $\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2}$ is finite.
Find $a, b\in\mathbb{R}$ so that
$$\lim_{x\to 1} \frac{ax^6+bx^5+1}{(x-1)^2}$$
is finite.
I tried to use polynomial division, but the computations get tedious really fast. Any suggestions?
| First of all, you need the numerator to be zero at $x=1$, which implies $a+b+1=0$. When you substitute, say, $a=-b-1$, the numerator becomes $-bx^5(x-1)-(x^6-1)$ which, when factoring out $(x-1)$ gives $(x-1)(-bx^5-x^5-x^4-x^3-x^2-x-1)$. After canceling $x-1$ we still have $x-1$ in the denominator, and for the limit to be finite we require $-bx^5-x^5-x^4-x^3-x^2-x-1$ to be zero at $x=1$. This gives $-b-6=0$ and then $b=-6$ and $a=5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Proving a strict inequality in the limit I want to prove that
$$ \lim_{k \to \infty} \left( 1 + \frac{1}{2} \right) \left(1 + \frac{1}{4} \right)...\left( 1 + \frac{1}{2^k} \right) < e .$$
Using the $AM-GM$ inequality we arrive at
$$\left( 1 + \frac{1}{2} \right) \left(1 + \frac{1}{4} \right)...\left( 1 + \frac{1}{2^k} \right) < \left(\frac{k + 1 - \frac{1}{2^k} }{k} \right)^k = \left( 1 + \frac{1}{k} - \frac{1}{k2^{k}}\right)^k < \left(1 + \frac{1}{k} \right)^k < e.$$
The first inequality is strict because the terms are different.However, I know that in the limit, strict inequalities can transform into equalities. Since the limit of $\left( 1 + \frac{1}{k} - \frac{1}{k2^{k}}\right)^k$ when $k$ goes to infinity is also $e$, how could I prove a strict inequality?
| You can just ignore the first term and carry on with your method from $i=2$.
Then you get $\prod\limits_{i=2}^ka_i\le\cdots\le\left(1+\dfrac 1{2(k-1)}\right)^{k-1}\le e^{0.5}$
And then $\prod\limits_{i=1}^\infty a_i=\left(1+\dfrac 12\right)\prod\limits_{i=2}^\infty a_i\le 1.5\,e^{0.5}<e$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3080771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Determine Fourier series expansion for $f(\theta)=\cos^4\!\theta$ Q: The function $f(\theta) = \cos^4\! \theta$ is a nice smooth function, so will have a Fourier series expansion. That is, it will have an expansion as a sum of functions $\cos j \theta$ and $\sin j \theta$ with real coefficients. Determine what the expansion is.
For reference, the Fourier coefficients are:
\begin{align*}
c_n &= \frac{1}{2\pi} \int_0^{2 \pi} f(t) e^{-int} \, dt = \frac{1}{2\pi} \int_0^{2 \pi} f(t) \left( \cos (nt) - i \sin (nt) \right) \, dt \\
\end{align*}
The coefficients give the Fourier Series expansion:
\begin{align*}
f(t) &= \sum\limits_{n=0}^\infty c_n e^{int} = \sum\limits_{n=0}^\infty c_n \left( \cos (nt) + i \sin (nt) \right) \\
\end{align*}
One route is to plug $f(\theta)$ into the equation for the Fourier series coefficients:
\begin{align*}
c_n &= \frac{1}{2\pi} \int_0^{2 \pi} \cos^4 t \cdot \left[ \cos (nt) - i \sin (nt) \right] \, dt \\
\end{align*}
That integral is looking complex. I suspect there is an easier solution to this problem?
| Use this ^.^
$\displaystyle \begin{array}{{>{\displaystyle}l}}
For\ odd\ n:\ cos^{n}( x) =\left(\frac{e^{ix} +e^{-ix}}{2^{n}}\right)^{n} =\sum _{k=0}^{\left\lfloor \frac{n}{2}\right\rfloor }\binom{n}{k} \cdotp \frac{cos( x( n-2k))}{2^{n-1}}\\
\\
For\ even\ n:\ cos^{n}( x) =\left(\frac{e^{ix} +e^{-ix}}{2^{n}}\right)^{n} =\sum _{k=0}^{\left\lfloor \frac{n-1}{2}\right\rfloor }\binom{n}{k} \cdotp \frac{cos( x( n-2k))}{2^{n-1}} +\frac{\left( 2\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right)\right) !}{\left(\left( n-\left\lfloor \frac{n}{2}\right\rfloor \right) !\right)^{2}} \cdot \frac{1}{2^{n}}
\end{array}$
Hopefully you're familiar with binomial coefficient function and the floor funtion
| {
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"url": "https://math.stackexchange.com/questions/3082089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the numerical value of this expression If $x$ is a complex number such that $x^2+x+1=0$, then the numerical value of $(x+\frac{1}{x})^2+(x^2+\frac{1}{x^2})^2+(x^3+\frac{1}{x^3})^2+\ldots+(x^{27}+\frac{1}{x^{27}})^{2}$ is equal to?
A) 52 . B) 56 . C) 54. D)58 . E)None of these
Where is this question from? I'm pretty sure it comes from one high school math contest, does anyone one know which math contest and of course i can't solve, I have an answer key but I don't know the solution.
| Thank you for all your answers, I just find the other solution to solve this, so I want to post it here.
Since $x^2+x+1=0$, so we can find
$x+\frac{1}{x}=-1$
and $x^2+\frac{1}{x^2}=(x+\frac{1}{x})^2-2=(-1)^2-2=-1$. Let's keep going
$x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2+\frac{1}{x^2})-(x+\frac{1}{x})=(-1)(-1)-(-1)=2$
$x^4+\frac{1}{x^4}=(x^2+\frac{1}{x^2})^2-2=(-1)^2-2=-1 $
$x^5+\frac{1}{x^5}=(x^2+\frac{1}{x^2})(x^3+\frac{1}{x^3})-(x^2+\frac{1}{x^2})=(-1)(-1)-2=-1 $
$x^6+\frac{1}{x^6}=(x^5+\frac{1}{x^5})(x+\frac{1}{x})-(x^4+\frac{1}{x^4})=(-1)(-1)-(-1)=2$
......
So it has a period,-1,-1,2,-1,-1,2,-1,-1,-2.......until the last term.
The sum of the one period is $(-1)^2+(-1)^2+(2)^2=6$, and we have 27$\div$ 3=9, so the final value is equal to 6$\times$9=54.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3082719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
When does the first repetition in $\;\lfloor x\rfloor, \lfloor x/2 \rfloor, \lfloor x/3\rfloor, \lfloor x/4\rfloor, \dots\;$ appear? Let $\lfloor x\rfloor$ denote the floor of $x$.
When does the first repetition in $\lfloor x\rfloor$, $\lfloor x/2\rfloor$, $\lfloor x/3\rfloor$, $\lfloor x/4\rfloor$, ... approximately appear, as a function of $x$?
It seems to be around ~ $c \sqrt x$.
Example: $x = 2500$:
2500, 1250, 833, 625, 500, 416, 357, 312, 277, 250, 227, 208, 192, 178, 166, 156, 147, 138, 131, 125, 119, 113, 108, 104, 100, 96, 92, 89, 86, 83, 80, 78, 75, 73, 71, 69, 67, 65, 64, 62, 60, 59, 58, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 45, 44, 43, 43, 42, 41, 40, 40, 39, 39, 38, 37, 37, 36, 36, 35, 35, ...
| $\color{brown}{\textbf{Preliminary Notes.}}$
$\dfrac xN > \dfrac x{N+1},$ so the required equality
$$\left\lfloor\dfrac xN\right\rfloor = \left\lfloor\dfrac x{N+1}\right\rfloor,\tag1$$
where $\lfloor a\rfloor = \mathrm{floor}\,(a),$
has solutions iff
$$\left\{\dfrac xN\right\} > \left\{\dfrac x{N+1}\right\}.\tag2$$
Taking in account that
$$k\left\{\dfrac xk\right\} = x\hspace{-12mu}\mod k,$$
inequality $(2)$ can be presented in the form of
$$\dfrac{x\hspace{-12mu}\mod N}N > \dfrac{x\hspace{-12mu}\mod (N+1)}{N+1},\tag3$$
and from $(3)$ should
$$x\hspace{-12mu}\mod N \ge x\hspace{-12mu}\mod (N+1).\tag4$$
Formula $(4)$ can be used for the testing, instead the issue one.
$\color{brown}{\textbf{Decision.}}$
The least solution $N$ of equality $(1)$ belongs to the interval $x\in(n(n-1),n(n+1)],$ so
$$x=n(n-1)+m,\quad m=x-(n^2-n),\quad m\in[1,2n],\tag5$$
wherein
$$n = \left\lceil\dfrac{\sqrt{4x-3}+1}2\right\rceil\tag6,$$
$\lceil a\rceil = \mathrm{ceil}\,(a).$
Let
$$N=n+k,\quad k\in\mathbb N\tag7,$$
$$\dfrac xN = \dfrac{n^2-n+m}{n+k}=n-k-1+\dfrac{k(k+1)+m}{n+k},$$
$$\dfrac x{N+1} = \dfrac{n^2-n+m}{n+k+1}=n-k-2+\dfrac{(k+1)(k+2)+m}{n+k+1},$$
If $\underline{m\in[1,n-1]}$ then the least solution of $(3)$ can be achieved iff
$$(k+1)(k+2)+m\ge n+k+1,\quad k =\left\lceil\sqrt{n-m}-1\right\rceil,$$
$$N = n - 1 + \left\lceil\sqrt{n^2-x}\right\rceil.$$
If $\underline{m\in[n,2n-1]}$ then the least solution of $(3)$ can be achieved iff
$$(k+1)(k+2)+m\ge 2(n+k+1),\quad k^2+k-(2n-m)\geq 0,\quad k = \left\lceil\dfrac{\sqrt{8n-4m+1}-1}2\right\rceil,$$
$$N = n + \left\lceil\dfrac{\sqrt{4(n^2+n-x^2)+1}-1}2\right\rceil.$$
If $\underline{m=2n}$ then
$$\dfrac xN = \dfrac{n^2+n}{n+k}=n-k+1+\dfrac{k(k-1)}{n+k},$$
$$\dfrac x{N+1} = \dfrac{n^2+n}{n+k+1}=n-k+\dfrac{k(k+1)}{n+k+1},$$
and the least solution of $(3)$ can be achieved iff
$$k(k-1)<n+k,\quad (k-1)^2<n+1,\quad k = \left\lceil\sqrt{n+1}\right\rceil,$$
$$N = n + \left\lceil\sqrt{n+1}\right\rceil.$$
$\color{brown}{\textbf{Result.}}$
Therefore, the least solution of $(1)$ is
$$\boxed{\begin{align}
&N=\begin{cases}
n - 1 +\left\lceil\sqrt{n^2-x}\right\rceil,\quad\text{if}\quad x-n(n-1)\in[1,n-1]\\[4pt]
n + \left\lceil\dfrac{\sqrt{4(n^2+n-x)+1}-1}2\right\rceil,\quad\text{if}\quad x-n(n-1)\in[n,2n-1]\\[4pt]
n + \left\lceil\sqrt{n+1}\right\rceil,\quad\text{if}\quad x = n(n+1),
\end{cases}\\
&\text{where}\\
&n = \left\lceil\dfrac{\sqrt{4x-3}+1}2\right\rceil.
\end{align}}\tag8$$
$\color{brown}{\textbf{Examples.}}$
$\underline{x=2475,\quad n=50,\quad x-n(n-1)=25}.$
There is the first case of $(8).$
Result is $N=54,$ with $\left\lfloor\dfrac{2475}{54}\right\rfloor = \left\lfloor\dfrac{2475}{55}\right\rfloor=45.$
$\underline{x=2500,\quad n=50,\quad x-n(n-1)=50}.$
There is the second case of $(8).$
Result is $N=57,$ with $\left\lfloor\dfrac{2500}{57}\right\rfloor = \left\lfloor\dfrac{2500}{58}\right\rfloor=43.$
$\underline{x=2450,\quad n=49,\quad x=n(n+1)}.$
There is the third case of $(8).$
Result is $N=57,$ with $\left\lfloor\dfrac{2450}{57}\right\rfloor = \left\lfloor\dfrac{2450}{58}\right\rfloor=42.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3083192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 4,
"answer_id": 0
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Cannot find angle for trigonometry problem
A right angle triangle is entrapped within a circle. The triangle entraps within it a circle of its own. The ratio between the big radius and the little radius is $\frac{13}{4}$. What are the angles of the triangle?
Here is the problem as I understand it, given that:
1. $\measuredangle ABC = 90^\circ \rightarrow AC = 2R$
2. The smaller radii are perpendicular to the triangle.
3. The origin of the smaller circle is the intersection between the angle bisectors of triangle ABC.
Let $\measuredangle ACB = 2\alpha \rightarrow \measuredangle ACG = \alpha$
$\measuredangle BAC = 90 - 2\alpha \rightarrow \measuredangle GAC = 45 - \alpha$
From here we can construct the following system:
$$\begin{cases} 2R = \frac{r}{\tan \alpha} + \frac{r}{\tan(45 - \alpha)} \\ \frac{R}{r} = \frac{13}{4} \rightarrow R = \frac{13r}{4} \end{cases}$$
$$\downarrow \\ \frac{26r}{4} = r\left(\frac{1}{\tan \alpha} + \frac{1}{\tan(45 - \alpha)}\right) \quad / \div r \\ 6.5 = \frac{\cos \alpha}{\sin \alpha} + \frac{\cos(45 - \alpha)}{\sin(45 - \alpha)} \quad / {\sin(\alpha \pm \beta) = \sin\alpha\cos\beta \pm \sin\beta\cos\alpha\\ \cos(\alpha \pm \beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta} \\ 6.5 = \frac{\cos \alpha}{\sin \alpha} + \frac{\cos \alpha \cos 45 + \sin \alpha \sin 45}{\sin 45 \ cos \alpha - \sin \alpha \cos 45} \\ 6.5 = \frac{\cos \alpha}{\sin \alpha } + \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} \cdot \frac{\cos \alpha + \sin \alpha}{\cos \alpha - \sin \alpha} \\ 6.5 = \frac{\cos^2 \alpha - \sin\alpha \cos \alpha + \cos \alpha \sin \alpha + \sin^2 \alpha}{\sin(\cos\alpha - \sin \alpha)} \quad / \sin^2\alpha + \cos ^2 \alpha = 1\\ 6.5 = \frac{1}{\sin\alpha (\cos\alpha - \sin \alpha)}$$
From this point on, I have no identity that I can think of which can be applied here to help me solve for $\alpha$. Amusingly enough, I got an answer which was very close to the correct one, but that was only because I made a mistake in the signs of the identity, and the second fracture turned into 1.
I am aware that they may be an identity which can solve this, but I am restricted to only use the identities available here. I have either made a mistake somewhere, or just don't know how to use the available identities to solve the problem.
| $$\dfrac4{13}\sin(45^\circ -\alpha+\alpha)=2\sin\alpha\sin(45^\circ-\alpha)$$
$$=\cos(45^\circ-2\alpha)-\cos45^\circ$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3084842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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How to find the value this sum converges to?$\sum_{n=2}^{\infty}\frac{n+1}{n(2n-1)(2n+1)} $ How to find the value this sum converges to?$$\sum_{n=2}^{\infty}\frac{n+1}{n(2n-1)(2n+1)} $$
I've tried separating it like
$$\sum_{n=2}^{\infty}\frac{1/2}{(2n+1)}+\frac{3/2}{(2n-1)}-\frac{1}{n}$$
and writing some terms and I get
$$1/6Β·(1+1/2+...+1/6n+3)+1/2Β·(1/3+1+1/2+...+1/6n-3)-(1+1/2+...1/n)$$
but I don't know how to end summing it all! Any hint!
FYI I haven't learnt integration and differentation.
| Here is a way to find those two sums @Kavi Rama Murthy refers to in his answer. They are found by first converting them to double integrals.
Let
$$S = \sum_{n = 2}^\infty \frac{n + 1}{n(2n - 1)(2n + 1)} = \frac{3}{4} \sum_{n = 2}^\infty \frac{1}{n (2n - 1)} - \frac{1}{4} \sum_{n = 1}^\infty \frac{1}{n(2n + 1)} + \frac{1}{12}. \qquad (*)$$
For the first of the sums appearing to the right in ($*$), observe that
$$\frac{1}{n} = \int_0^1 x^{n - 1} \, dx \quad \text{and} \quad \frac{1}{2n - 1} = \int_0^1 y^{2n - 2} \, dy.$$
The sum may now be rewritten as
\begin{align}
\sum_{n = 2}^\infty \frac{1}{n(2n - 1)} &= \sum_{n = 2}^\infty \int_0^1 \int_0^1 x^{n - 1} y^{2n - 2} \, dx \, dy\\
&= \int_0^1 \int_0^1 \frac{1}{xy^2} \sum_{n = 2}^\infty (xy^2)^n \, dx \,dy \tag1\\
&= \int_0^1 \int_0^1 \frac{1}{xy^2} \cdot \frac{(xy^2)^2}{1 - xy^2} \, dx \, dy \tag2\\
&= \int_0^1 \frac{1}{y^2} \int_0^{y^2} \frac{u}{1 - u} \, du \, dy \tag3\\
&= -\int_0^1 \frac{1}{y^2} \int_0^{y^2} \left (1 - \frac{1}{1 - u} \right ) \, du \, dy\\
&= -\int_0^1 \left (1 + \frac{\ln (1 - y^2)}{y^2} \right ) \, dy\\
&= -1 + 2 \int_0^1 \frac{1 - y}{1 - y^2} \, dy \tag4\\
&= -1 + 2 \int_0^1 \frac{dy}{1 + y}\\
&= -1 + 2 \ln 2.
\end{align}
Explanation
(1) Interchange of the sum and integral signs is justified by the dominated convergence theorem.
(2) Summing of a geometric series.
(3) Change of variable $u = xy^2$ is made.
(4) Integration by parts has been used.
The second of the sums appearing in ($*$) can be found in a similar manner (see here). The result is:
$$\sum_{n = 1}^\infty \frac{1}{n(2n + 1)} = 2 - 2\ln 2.$$
Thus
$$S = \frac{3}{4} (2 \ln 2 - 1) - \frac{1}{4} (2 - 2 \ln 2) + \frac{1}{12},$$
or
$$\sum_{n = 2}^\infty \frac{n + 1}{n(2n - 1)(2n + 1)} = 2 \ln 2 - \frac{7}{6}.$$
| {
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"url": "https://math.stackexchange.com/questions/3086481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the set of the real numbers $x$ satisfying the inequalities $|x+4|<|2x-1|$ and $|x|+|x+1|<3$
Find the set of the real numbers $x$ satisfying the given inequality and sketch them on the real line:
1) $|x+4|<|2x-1|$
*
*If $x<-4$, we have $x+4<0$ and $2x-1<0$, Then $|x+4|=-x-4$ and $|2x-1|=1-2x$
hence, in this case:
$|x+4|<|2x-1| \iff -x-4<1-2x \iff x<5$.
and we have that, $(-\infty,4)\cap (-\infty,5)=(-\infty,4)$.
So, for any $x \in (-\infty,-4)$, $|x+4|<|2x-1|$.
*
*If $-4\leq x <\frac{1}{2}$, we have that $x+4\geq 0$, and $2x-1<0$, so $|x+4+x+4$ and $|2x-1|=1-2x$
hence, in this case:
$|x+4|<|2x-1| \iff x+4<1-2x \iff 3x<-3 \iff x<-1$.
Thus, for any $x \in [-4,\frac{1}{2}) \cap (-\infty,-1)=[-4,-1)$, $|x+4|<|2x-1|$.
*
*If $x\geq \frac{1}{2}$, we have $x+4>0$, and $2x-1\geq 0$, then $|x+4|=x+4$ and $|2x-1|=2x-1$
Then, in this case
$|x+4|<|2x-1| \iff x+4<2x-1 \iff x>5$.
So, for any $x \in [\frac{1}{2},\infty) \cap (5,\infty)=(5,\infty)$, $|x+4|<|2x-1|$.
Therefore, The solution set of the inequality is $(-\infty,-4) \cup [-4,-1) \cup (5,\infty)=(-\infty,-1) \cup (5,\infty)$.
2) $|x|+|x+1|<3$
*
*If $x<-1$, we have that $x<0$ and $x+1<0$, Then from the Absolute value definition $|x|=-x$ and $|x+1|=-x-1$.
Thus, $|x|+|x+1|=-x-x-1=-2x-1$.
Hence in this case, $|x|+|x+1|<3 \iff -2x-1 < 3 \iff 2x>-4 \iff x>-2$
So, for any $x \in (-2,-1)$, $|x|+|x+1|<3$.
*
*If $-1 \leq x <0$, we have that $|x|=-x$ and $|x+1|=x+1$
we have, $|x|+|x+1|=-x+x+1=1<3$.
Hence, for any $x \in [-1,0)$, $|x|+|x+1|<3$.
*
*If $x \geq 0$, we have that $|x|=x$ and $|x+1|=x+1$
Then in this case, $|x|+|x+1|=x+x+1=2x+1<3 \iff 2x<2 \iff x<1$.
Thus, for any $x \in [0,1)$, $|x|+|x+1|<3$.
Therefore, The solution set of the inequality is $(-2,-1) \cup [-1,0) \cup
[0,1)=(-2,1)$
is it true, please?
| Hint: $$|x+4| < |2x-1| \iff (x+4)^2 < (2x-1)^2 \iff x^2 -4x-5 >0 \iff (x-2)^2 > 9.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3088763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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