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How continue with this proof? I'm trying to proof the following statement:
$7+6 \cdot (7+7^1+7^2+\dots+7^n) = 7^{n+1}$
I try the next:
$n=k$
$7+6 \cdot (7+7^1+7^2+\dots+7^k) = 7^{k+1}$
$7+6 \cdot (7+7^1+7^2+\dots+7^k)-7 = 7^{k+1}-7$
$\frac{6 \cdot (7+7^1+7^2+\dots+7^k)}{6} = \frac{7^{k+1}-7}{6}$
$7+7^1+7^2+\dots+7^k = \frac{7^{k+1}-7}{6}$
$n=k+1$
$7+7^1+7^2+\dots+7^k+7^{k+1} = \frac{7^{k+2}-7}{6}$
Proof:
$\frac{7^{k+1}-7}{6}+7^{k+1} = \frac{7^{k+2}-7}{6}$
$\frac{7^{k+1}-7+6\cdot7^{k+1}}{6}$
Any idea?
| Use the formula for the sum of a geometric sequence:
$$7+6\left(7+7^2+\ldots+7^n\right)=7+6\frac{7-7^{n+1}}{1-7}=7-7+7^{n+1}=7^{n+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1977790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Show that for all Harmonic numbers $H(k)$, the inequality $H(2^k) \leq 1 + k$ holds for all natural numbers. Hi so the question is prove that for all harmonic numbers $$H(k) = 1 + \frac{1}{2} + \frac13 + \dotsb + \frac{1}{k}$$ the inequality $H(2^k) \leq 1 + k$ holds for all natural numbers. I'm not going to put my entire answer here but was wondering if someone could tell me if I'm on the right track.
So I got this far:
Say $H(2^k) = k + 1$
and $$H(2^{k+1}) = k + 1 + 2^k \frac{1}{2^{k+1}} = k + 1 + \frac{1}{2} = k + \frac{3}{2}$$
and $k + \frac{3}{2} \leq 1 + (k + 1)$
Is this correct or at least am I on the right track??
| Let us prove something stronger through a powerful technique, creative telescoping.
We may recall that since in a neighbourhood of the origin we have
$$ \text{arctanh}(x) = \frac{1}{2}\log\left(\frac{1+x}{1-x}\right)=x+\frac{x^3}{3}+\frac{x^5}{5}+\frac{x^7}{7}+\ldots\tag{1} $$
the inequality
$$ \frac{1}{n}\leq 2\,\text{arctanh}\left(\frac{1}{2n}\right) = \log\left(\frac{2n+1}{2n-1}\right) = \frac{1}{n}+\frac{1}{12n^3}+\frac{1}{80n^5}+\ldots \tag{2}$$
surely holds for any $n\geq 1$. Additionally, $\log\left(\frac{2n+1}{2n-1}\right)$ is a telescopic term, wonderful:
$$ \color{red}{H_N} = \sum_{n=1}^{N}\frac{1}{n}\color{red}{\leq} \sum_{n=1}^{N}\log\left(\frac{2n+1}{2n-1}\right) = \color{red}{\log(2N+1)}\leq \log(N)+1\tag{3} $$
In particular, $H_{2^k}\leq 1+k\log 2$.
The super-tight inequality
$$ \log\left(N+\frac{1}{2}\right)+\gamma \leq H_N \leq \log\left(N+\frac{1}{2}\right)+\gamma+\frac{1}{24 N(N+1)} \tag{4}$$
where $\gamma$ is the Euler-Mascheroni constant can be proved through the same technique.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1978209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Assume that $(a_n)^{\infty}_{n=0}$ is a sequence given by $a_0=1$ and $a_{n+1}=2a_{n}+n$ for all $n \ge 0$, what is the formula for $a_n$? Assume that $(a_n)^{\infty}_{n=0}$ is a sequence given by $a_0=1$ and $a_{n+1}=2a_{n}+n$ for all $n \ge 0$, what is the formula for $a_n$?
we know that $F(x)=a_0+a_1x+a_2x^2+a_3x^3+\dots$
and our sequence is $1, 2, 5, 12, 27, 58, \dots$
to find the generating function for the sequence I started with
$F(x)=a_0+(\underline{2a_0}+\widetilde0)x+(\underline{2a_1}+\widetilde1)x^2+(\underline{2a_2}+\widetilde2)x^3+\dots$
after distributing the variables into the parenthesis, i separated the terms with $\underline{}$ and $\widetilde{}$
$\underline{} = 2a_0x+2a_1x^2+2a_2x^3+2a_3x^4+\dots$
$\underline{} = 2x(a_0+a_1x^1+a_2x^2+a_3x^3+\dots)$
$\underline{} = 2x \times F(x)$
$\widetilde{} = x^2 + 2x^3 + 3x^4 + 4x^5$
now I don't know what to do with this part, if I had used $n$ earlier instead of putting in the numbers I could just factor out $nx^2$ and get $\widetilde{} = nx^2(\frac{1}{1-x})$ but I don't think I can have $n$ when solving for the generating function of the sequence?
Any ideas of how to get past just this part?
| My preferred approach using generating functions starts by assuming that $a_n=0$ for $n<0$ and rewriting the recurrence so that it hold for all $n\ge 0$:
$$a_n=2a_{n-1}+n-1+2[n=0]\;.\tag{1}$$
Here $[n=0]$ is an Iverson bracket, and the last term of $(1)$ ensures that $a_0=1$. Now multiply $(1)$ by $x^n$ and sum over $n\ge 0$; if we let $g(x)=\sum_{n\ge 0}a_nx^n$ be the generating function, we have
$$\begin{align*}
g(x)&=\sum_{n\ge 0}a_nx^n\\
&=2\sum_{n\ge 0}a_{n-1}x^n+\sum_{n\ge 0}nx^n-\sum_{n\ge 0}x^n+2\\
&=2x\sum_{n\ge 0}a_nx^n+x\sum_{n\ge 0}nx^{n-1}-\frac1{1-x}+2\\
&=2xg(x)+\frac{x}{(1-x)^2}-\frac1{1-x}+2\;.
\end{align*}$$
Solving for $g(x)$, decomposing the resulting rational function into partial fractions, and rewriting the resulting terms as formal power series, we get
$$\begin{align*}
g(x)&=\frac{x}{(1-x)^2(1-2x)}-\frac1{(1-x)(1-2x)}+\frac2{1-2x}\\
&=\frac{x-(1-x)+2(1-x)^2}{(1-x)^2(1-2x)}\\
&=\frac{1-2x+2x^2}{(1-x)^2(1-2x)}\\
&=\frac2{1-2x}-\frac1{(1-x)^2}\\
&=2\sum_{n\ge 0}(2x)^n-\sum_{n\ge 0}(n+1)x^n\\
&=\sum_{n\ge 0}\left(2^{n+1}-n-1\right)x^n\;,
\end{align*}$$
which yields the closed form
$$a_n=2^{n+1}-n-1\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1979381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration of $\arcsin(f(x))$ $$\int\sin^{-1}\left(\dfrac{2x+2}{\sqrt{4x^2+8x+13}}\right)dx$$
$\text {attempt}$ Let $x+1=u$, then the internal function becomes $\frac {2u}{\sqrt{4u^2+9}}$, and $dx=du$. Then I attempted by parts but derivative of $sin^{-1}(f(x))$ is troubling me in second part of IBP. Any better approach, or am I missing on a trick? I only know that the answer contains $\ln$ and $\tan^{-1}$. Thanks!
| Let $$I = \int \sin^{-1}\left(\frac{2x+2}{\sqrt{(2x+2)^2+3^3}}\right)dx$$
Now Put $2x+2=3\tan \theta\;,$ Then $\displaystyle dx = \frac{3}{2}\sec^2 \theta d \theta$
So Using Integration by parts $$I = \frac{3}{2}\int \theta \sec^2 \theta d \theta = \frac{3}{2}\left[\theta \tan \theta+\ln |\sec \theta|\right]+\mathcal{C}$$
So $$I = \frac{3}{2}\cdot \frac{2x+2}{3}\tan^{-1}\left(\frac{2x+2}{3}\right)+\frac{3}{2}\ln \left(\frac{\sqrt{3^2+(2x+2)^2}}{3}\right)+\mathcal{C}$$
So $$I = (x+1)\cdot \tan^{-1}\left(\frac{2x+2}{3}\right)+\frac{3}{2}\ln \left(\frac{\sqrt{3^2+(2x+2)^2}}{3}\right)+\mathcal{C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1980943",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding area under a curve, limit approach, Riemann sums I need to find the area under: $f(x) = 64 - x^3$ in $[2,4]$.
I know the base is $\frac{2}{n}$ and the height is $64 - \left(2 + \frac{2i}{n}\right)^3$ when I distribute everything I get
$$\frac{2}{n}\cdot \left[ 64 -8 -\frac{16i}{n} -\frac{16i^2}{n^2}
-\frac{8i^3}{n^3}\right].$$
Am I having some algebra issue here?
| Hint. Let $x_k=2+\frac{2i}{n}$ for $i=0,\dots,n$, then
$$A:=\int_2^{4}(64-x^3)\,dx=\lim_{n\to\infty}\frac{2}{n}\sum_{i=1}^{n} (64-x_i^3)=\lim_{n\to\infty}\frac{2}{n}\sum_{i=1}^n \left(56-\frac{24i}{n}-\frac{24i^2}{n^2}-\frac{8i^3}{n^3}\right).$$
Now, by Asymptotic behaviour of sums of consecutive powers, it follows that $\sum_{i=1}^n i^k\sim \frac{n^{k+1}}{k+1}$, which implies
$$A=\lim_{n\to\infty}\frac{2}{n}\cdot\left(56n-\frac{24n^2}{2n}-\frac{24n^3}{3n^2}-\frac{8n^4}{4n^3}\right)=2(56-12-8-2)=68.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1982579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Express $\arg (e^{z - i / z})$ in terms of $x$ and $y$.
Express $\arg (e^{z - i / z})$ in terms of $x$ and $y$.
I start with $\arg (e^{z - i / z}) = \arg (\frac{ze^z-i}{z})$ and after some algebraic manipulations end up with $\arg (\space\space\space e^x(x\cos y - y \sin y) + i[e^x(x \sin y + y \cos y) - 1]\space\space\space) - \arg(z)$. At this point is taking inverse tangents appropriate or have I overlooked a simpler method?
| For $z=x+iy$ we have:
$$
z-\frac{i}{z}=x+iy-i\left(\frac{x-iy}{x^2+y^2} \right)= x-\frac{y}{x^2+y^2}+i\left( y-\frac{x}{x^2+y^2}\right)= a+ib
$$
with:
$$
a=x-\frac{y}{x^2+y^2} \qquad b= y-\frac{x}{x^2+y^2}
$$
We have $e^{a+ib}= e^a e^{ib}$ and $b$ is the argument ( without the $2 \pi$ periodicity) of $e^{a+ib}$, so the argument of $e^{z-\frac{i}{z}}$ is:
$$
b= y-\frac{x}{x^2+y^2}
$$
| {
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"url": "https://math.stackexchange.com/questions/1982949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Easier way to solve $1-\frac1{2^2}-\frac1{4^2}+\frac1{5^2}+\frac1{7^2}-\frac1{8^2}-\frac1{10^2}\ldots=\frac49s$?
Show that $1-\frac1{2^2}-\frac1{4^2}+\frac1{5^2}+\frac1{7^2}-\frac1{8^2}-\frac1{10^2}++--\ldots=\frac49s$, where $s=\sum k^{-2}$
I can divide the LHS in the sum of four series
$$1-\frac1{2^2}-\frac1{4^2}+\frac1{5^2}+\frac1{7^2}-\frac1{8^2}-\frac1{10^2}++--\ldots=\\=\sum_{k\ge 1}\frac1{(6k-1)^2}+\sum_{k\ge 1}\frac1{(6k-5)^2}-\sum_{k\ge 1}\frac1{(6k-2)^2}-\sum_{k\ge 1}\frac1{(6k-4)^2}=\frac49\sum_{k\ge 1}\frac1{k^2}$$
but manipulating the above expression is so tedious to show the equality. Do you know a different way to solve this problem? Thank you.
P.S.: I dont know exactly what kind of tags use for this question.
| Since $(1,-1,0,-1,1,0)=(1,-1,1,-1,1,-1)-(0,0,1,0,0,-1)$ we have:
$$\begin{eqnarray*} S&=&\sum_{k\geq 0}\left(\frac{1}{(6k+1)^2}-\frac{1}{(6k+2)^2}-\frac{1}{(6k+4)^2}+\frac{1}{(6k+5)^2}\right)\\&=&\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}-\sum_{n\geq 1}\frac{(-1)^{n+1}}{(3n)^2}\\&=&\frac{8}{9}\sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2}\end{eqnarray*}$$
and
$$ \sum_{n\geq 1}\frac{(-1)^{n+1}}{n^2} = \sum_{n\geq 1}\frac{1}{n^2}-2\sum_{n\geq 1}\frac{1}{(2n)^2} = \frac{1}{2}\sum_{n\geq 1}\frac{1}{n^2}$$
hence $S=\frac{4}{9}\zeta(2)=\color{red}{\large\frac{2\pi^2}{27}}$ as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1984966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the range of $\frac{1+\sin^4 x}{\sin^4 x}\cdot \frac{1+\cos^4 x}{\cos^4 x}$
Range of function $$f(x) = \frac{(1+\sin^4 x)}{\sin^4 x}\cdot \frac{(1+\cos^4 x)}{\cos^4 x}$$
Using $\bf{A.\geq G.M}$ Inequality,
$$1+\sin^4 x\geq 2\sin^2 x \qquad\text{and}\qquad 1+\cos^4 x\geq 2\cos^2 x$$
So $$\frac{(1+\sin^4 x)}{\sin^4 x}\cdot \frac{(1+\cos^4 x)}{\cos^4 x}\geq \frac{4}{\sin^2 x\cdot \cos^2 x} = \frac{16}{(\sin 2x)^2}\geq 16$$
But the answer given as $\left[25,\infty\right)$.
Please help me. How can I solve it using an inequality or any other way. Thanks.
| Let $a=\sin^2 x, b = \cos^2 x$. Then $a+b=1$ and hence $\sqrt{ab} \leq \frac{a+b}{2} = \frac{1}{2}$. Thus $\frac{1}{ab} \geq 4$. Now the given expression is
\begin{align*}
\left(\frac{1+a^2}{a^2}\right)\left(\frac{1+b^2}{b^2}\right) &= 1+\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{a^2b^2} \\
&\geq 1 + \frac{2}{ab} + \frac{1}{a^2b^2} \\
&\geq 1 + 8 + 16 = 25
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1985787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Surface area of a cone intersecting a horizontal cylinder I'm trying to find the surface area of the cone $x^2 + y^2 = z^2$ above the $xy$-plane and under the cylinder $y^2 + z^2 = 16$. I thought of using spherical coordinates $x = \rho \sin \phi \cos \theta,\quad y = \rho \sin \phi \sin \theta,\quad z= \rho \cos \theta$. In this case $\phi=\pi/4$. I know $0\leq \theta < 2\pi$ and $\rho$ ranges from 0 but the upper bound depends on $\theta$. To solve for this upper bound I calculate
$$x^2+y^2=z^2,\quad y^2+z^2 = 16 \Rightarrow x^2 + 2y^2 = 16$$
$$\frac{1}{2}\rho^2\cos^2\theta + \rho^2\sin^2\theta = 16$$
$$\rho =\frac{4}{\sqrt{\frac{1}{2}\cos^2\theta +\sin^2\theta}}$$
Now I think I integrate
$$\int_0^{2\pi}\int_0^{\frac{4}{\sqrt{\frac{1}{2}\cos^2\theta+\sin^2\theta}}}\rho^2\sin\theta\, d\rho d\theta$$
But I'm not sure this is the right integral or even the right approach since that integral looks like hell.
| An easier way (with the correct result!): by using cylindrical coordinates we have
$$\begin{align*}\text{Area}&=\int_SdS=\int_E\sqrt{1+f_x^2+f_y^2}\,dxdy\\
&=
\int_E\sqrt{2}\,dxdy=\sqrt{2}|E|=\sqrt{2}\left(\pi\cdot 4\cdot 2\sqrt{2}\right)=\boxed{16\pi}
\end{align*}$$
where $$S=\{(x,y,f(x,y)):(x,y)\in E\}$$ with $f(x,y)=\sqrt{x^2+y^2}$
and $E$ is the $xy$-domain bounded by the ellipse $x^2 + 2y^2 = 16$ which has semiaxes $4$ and $2\sqrt{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1986791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Matrix associated to a projection onto the plane $π$ with equation $x−y + 2z = 0$ Let $T \colon \mathbb{R}^3 \rightarrow \mathbb{R}^3$ be a linear transformation which projects vectors onto the plane $π$ with equation $x−y + 2z = 0$. Find a matrix $A$ such that $T = T_A$
My workings so far:
$$ A =
\left(\begin{bmatrix}
T(e_1)&T(e_2)&T(e_3)
\end{bmatrix}\right). $$
First I need to find $f_1,f_2$ that span the plane $\pi$ and they are orthogonal. So I chose
$$ f_1 =
\left(\begin{bmatrix}
1
\\1
\\0
\end{bmatrix}\right), f_2 =
\left(\begin{bmatrix}
-1
\\1
\\1
\end{bmatrix}\right). $$
Next I need to use this formula but I am not quite sure how to
$$T(u)= uf_1/[f_1f_1] f_1 + uf_2/[f_2f_2] f_2$$
Looking for some help with carrying out the rest of the problem. Sorry I did not know the proper formatting for the $T(u)$ formula
| You have
$$ T \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \frac{\left< \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \right>}{\left \| \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} \right \|^2 }\begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \frac{\left< \begin{pmatrix} x \\ y \\ z \end{pmatrix}, \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \right>}{\left \| \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \right \|^2 }\begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} = \frac{x+y}{2} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix} + \frac{-x + y + z}{3} \begin{pmatrix} -1 \\ 1 \\ 1 \end{pmatrix} \\
= x \begin{pmatrix} \frac{5}{6} \\ \frac{1}{6} \\ -\frac{1}{3} \end{pmatrix} + y \begin{pmatrix} \frac{1}{6} \\ \frac{5}{6} \\ \frac{1}{3} \end{pmatrix} + z \begin{pmatrix} -\frac{1}{3} \\ \frac{1}{3} \\ \frac{1}{3} \end{pmatrix} = \underbrace{\begin{pmatrix} \frac{5}{6} & \frac{1}{6} & -\frac{1}{3} \\ \frac{1}{6} & \frac{5}{6} & \frac{1}{3} \\ -\frac{1}{3} & \frac{1}{3} & \frac{1}{3} \end{pmatrix}}_{A} \begin{pmatrix} x \\ y \\ z \end{pmatrix}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1990895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Relation between $S_{n-1}$ , $S_n$ and $S_{n+1}$ We know $ \alpha $ and $ \beta $ are roots of $ax^2+bx+c = 0 $.
also $S_{n-1} = \alpha^{n-1} + \beta^{n-1}$ , $S_{n} = \alpha^{n} + \beta^{n}$ and $S_{n+1} = \alpha^{n+1} + \beta^{n+1}$.
How we can find relation between $S_{n-1}$ , $S_n$ and $S_{n+1}$ ?
Note : We know that relation has $a$ , $b$ and $c$.
| Note that $$\alpha^{n} + \beta^{n}=(\alpha + \beta)(\alpha^{n-1} + \beta^{n-1})-\alpha\beta\cdot(\alpha^{n-2} + \beta^{n-2})$$
Now using the relation between the coefficients of a quadratic equation and its roots, we get that
$$S_n=-\frac{b}{a}S_{n-1}-\frac{c}{a}S_{n-2}$$
Or, in other words, we can write that $$S_{n+1}=-\frac{b}{a}S_{n}-\frac{c}{a}S_{n-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1991498",
"timestamp": "2023-03-29T00:00:00",
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${\sqrt{2x+1}=1+\sqrt{x}}$ — I dont know if the solution is correct. Help?
${\sqrt{2x+1}=1+\sqrt{x}}$
${2x+1=1+2\sqrt{x}+x}$
${x=2\sqrt{x}}$
${x*\frac{1}{x^{1/2}}=2}$
${\sqrt{x}=2}$
${x=4}$
| For more information try WolframAlpha.
As for the solution:
$x=2\sqrt{x}$
$x-2\sqrt{x}=0$
$\sqrt{x}\left(\sqrt{x}-2\right)=0$
$\sqrt{x}=0 \quad \text{or} \quad
\sqrt{x}-2=0$
We get:
$x=0, \qquad x=4$ as solutions.
Understood: you made the mistake in the 4th line
| {
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"url": "https://math.stackexchange.com/questions/1991823",
"timestamp": "2023-03-29T00:00:00",
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An inequality that holds for every $\epsilon >0 $
For every $\epsilon > 0$ and $a,b \in \mathbb{R}$, we have
$$ |ab| \leq \epsilon a^2 + \frac{b^2}{4 \epsilon} $$
attempt:
I was thinking on using AM-GM inequality:
$$ \epsilon a^2 + \frac{b^2}{4 \epsilon} = \frac{ \frac{b^2}{2 \epsilon} + 2 \epsilon a^2 }{2} \geq \sqrt{ \frac{b^2}{2 \epsilon} (2 \epsilon a^2) } = \sqrt{a^2b^2} = |ab|$$
Is this corect? What other nice results can we derive from the inequality? thanks
| Yes, it is correct. An equivalent way would be to use $2|xy|\leq x^2+y^2$ for $x=a\sqrt{\epsilon}$ and $y=\frac{b}{2\sqrt{\epsilon}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1992358",
"timestamp": "2023-03-29T00:00:00",
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To find least possible value of a no. in A.P. Let $a, b, c, d, e$ be natural numbers in an arithmetic progression such that $a + b + c + d + e$ is the cube of
an integer and $b + c + d$ is the square of an integer. The least possible value of the number of digits of $c$ is
| Let the common difference be $\alpha$.
$$a+b+c+d+e = 5c$$
$$b+c+d=3c$$
Hence $5c$ is a cube and $3c$ is a square.
Hence $c=5^2k^3$ and $c=3l^2$.
$k$ must be divisible by $3$ and $l$ must be divisible by $5$.
$c=5^2.3^3.k_1^3$ and $c=5^2.3.l_1^2$
$$c=5^2.3.3^2k_1^3$$
Picking $k_1=1$ satisfies the condition.
| {
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} |
Complex number (cube roots of unity) If $w$ and $w^2$ are non real cube roots of unity, then what would be the value of
$\frac{2015+2016w+2017w^2}{2017+2015w+2016w^2}$+$\frac{2015+2016w+2017w^2}{2016+2017w+2015w^2}$
All I know is --> $1+w+w^2=0$ and $w^3=1$
Any tips and suggestions regarding how to solve these kinds of problems would be great.
| hint:$$\frac{2015+2016w+2017w^2}{2017+2015w+2016w^2}+\frac{2015+2016w+2017w^2}{2016+2017w+2015w^2}=\\
\frac{2015(1+w+w^2)+w+2w^2}{2015(w+1+w^2)+2+w^2}+\frac{2015(1+w+w^2)+w+2w^2}{2015(w^2+w+1)+1+2w}=\\
\frac{2015(0)+w+2w^2}{2015(0)+2+w^2}+\frac{2015(0)+w+2w^2}{2015(0)+1+2w}=\\
\frac{w+2w^2}{2+w^2}+\frac{w+2w^2}{1+2w}=\\
\frac{w(1+2w)}{2+w^2}+\frac{w(1+2w)}{1+2w}$$
so :now put$w^2=-w-1$
$$\frac{w(1+2w)}{2+w^2}+\frac{w(1+2w)}{1+2w}=\\
\frac{w+w^2+w^2}{2+w^2}+w=\\
\frac{w+w^2+(-w-1)}{2+(-w-1)}+w=\\
\frac{w^2-1}{1-w}+w=\\
\frac{(w-1)(w+1)}{-(w-1)}+w=\\
-(w+1)+w=\\-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1999329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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} |
Find the minimum of the value $\max\left(\frac{1}{ac}+b,\frac{1}{a}+bc,\frac{a}{b}+c\right)$ Let $a,b,c>0$ find the following minimum of the value
$$\max\left(\dfrac{1}{ac}+b,\dfrac{1}{a}+bc,\dfrac{a}{b}+c\right)$$
the book is hint:when $a=b=c=1$,then the minimum of the value is $2$
| Hint:Use AM-GM $$\max\left(\dfrac{1}{ac}+b,\dfrac{1}{a}+bc,\dfrac{a}{b}+c\right)\ge \dfrac{\frac{1}{ac}+b+\frac{a}{b}+c}{2}\ge \dfrac{4\sqrt[4]{\frac{1}{ac}\cdot b\cdot\dfrac{a}{b}\cdot c}}{2}=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2000158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Help evaluating $ \int_{0}^{π} \frac{1}{({3+cosθ})^2}dt $ $\int_{0}^{π} \frac{1}{({3+cosθ})^2}dt $
so I approached
$\frac{1}{2}\int_{0}^{2π} \frac{1}{({3+cosθ})^2}dt $
$z=e^{iθ}$,
$\\dθ=\frac{1}{iz}dz\\\cosθ = \frac{e^{iθ}+e^{-iθ}}{2} =\frac{z+z^{-1}}{2}$
$\frac{2}{i}\oint\frac{z}{(z^2+6z+1)^2}dz$
then I cannot solve the problem....
| A geometric approach: since $\rho(\theta)=\frac{1}{3+\cos\theta}$ is the polar equation of an ellipse with major axis $2a=\frac{1}{4}+\frac{1}{2}=\frac{3}{4}$ and semi-latus rectum $\frac{b^2}{a}=\frac{1}{3}$, we have:
$$ \int_{0}^{\pi}\rho(\theta)^2\,d\theta = \frac{1}{2}\int_{0}^{2\pi}\rho(\theta)^2\,d\theta = \pi a b = \pi\cdot\frac{3}{8}\cdot\frac{1}{2\sqrt{2}}=\color{red}{\frac{3\pi}{16\sqrt{2}}} $$
since $\int_{0}^{2\pi}\frac{\rho(\theta)^2}{2}\,d\theta$ is the area enclosed by such ellipse.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2003350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
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Integration of $I_n=\int_{0}^{1} \frac{x^n}{\sqrt{x^3+1}}dx$ Let $I_n=\int_{0}^{1} \frac{x^n}{\sqrt{x^3+1}}dx$. Show that $(2n−1)I_n+2(n−2)I_{n-3}= 2\sqrt{2}$ for all $n\geq3$. Then compute $I_8$.
How do I compute that integral? My idea was to reduce it to some known forms, as $\int\frac{1}{\sqrt{1-x^2}}dx$... Is there any way to solve the problem without computing the integral though?
| I doubt that it is a good idea to use the hypergeometric function. Also substitution is not working. Actually it is pretty easy to use integration by parts to get the answer. Define
$$ J_n=(2n−1)I_n+2(n−2)I_{n-3}, n\ge 3.$$
Note
\begin{eqnarray}
J_n&=&(2n−1)I_n+2(n−2)I_{n-3}\\
&=&\int_0^{1}\frac{(2n-1)x^n+(2n-4)x^{n-3}}{\sqrt{x^3+1}}dx\\
&=&\int_0^{1}\frac{x^{n-3}\bigg[(2n-1)x^3+(2n-4)\bigg]}{\sqrt{x^3+1}}dx\\
&=&\int_0^{1}\frac{x^{n-3}\bigg[(2n-1)(x^3+1)-3\bigg]}{\sqrt{x^3+1}}dx\\
&=&(2n-1)\int_0^{1}x^{n-3}\sqrt{x^3+1}dx-3\int_0^{1}\frac{x^{n-3}}{\sqrt{x^3+1}}dx\\
&=&\frac{2n-1}{n-2}\int_0^{1}\sqrt{x^3+1}d(x^{n-2})-3\int_0^{1}\frac{x^{n-3}}{\sqrt{x^3+1}}dx\\
&=&\frac{2n-1}{n-2}\left(\sqrt{x^3+1}x^{n-2}\bigg|_0^1-\frac{3}{2}\int_0^1\frac{x^{n}}{\sqrt{x^3+1}}dx\right)-3\int_0^{1}\frac{x^{n-3}}{\sqrt{x^3+1}}dx\\
&=&\frac{2n-1}{n-2}\sqrt2-\frac{3}{2n-4}J_n\\
\end{eqnarray}
and hence
$$ \left(1+\frac{3}{2n-4}\right)J_n=\frac{2n-1}{n-2}\sqrt2$$
which gives $J_n=2\sqrt2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2005257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Solve the irrational equation I can not solve the equation:$$\sqrt[4]{x^3-2x^2-5x+6}+\sqrt{x^2+5x+6}=0.$$ Can someone help me. Thanky veru much.
| First I solve the equation:$x^3-2x^2-5x+6=0.$ Becouse the sum of the coefficinet of the given equation is $0,$ $(1+(-2)+(-5)+6=0)$, one zero is 1. Now we have a divide $$(x^3-2x^2-5x+6):(x-1)=x^2-x-6$$
i.e.,
$$x^3-2x^2-5x+6=(x-1)(x^2-x-6)=(x-1)(x+2)(x-3),$$ becouse
$$x^2-x-6=0\Rightarrow x_{1,2}=\frac{-(-1)\pm\sqrt{(-1)^2-4\cdot1\cdot(-6)}}{2}=\frac{1\pm \sqrt{1+24}}{2}=\frac{1\pm5}{2}; x_1=-2;x_2=3.$$
Now we solve the equation: $x^2+5x+6=0\Rightarrow x_1=-3,x_2=-2\Rightarrow x^2+5x+6=(x+2)(x+3).$
Now we have:
$$\sqrt[4]{x^3-2x^2-5x+6}+\sqrt{x^2+5x+6}=0$$
$$\sqrt[4]{x^3-2x^2-5x+6}=-\sqrt{x^2+5x+6}/^4$$
$$x^3-2x^2-5x+6=(x^2+5x+6)^2$$
$$(x-1)(x+2)(x-3)=[(x+2)(x+3)]^2$$
$$(x-1)(x+2)(x-3)=(x+2)^2(x+3)^2$$
$$(x-1)(x+2)(x-3)-(x+2)^2(x+3)^2=0$$
$$(x+2)[(x-1)(x-3)-(x+2)(x+3)^2]=0$$
$$(x+2)(x^3+7x^2+25x+14)=0$$
$$x+2=0\Rightarrow x=-2$$
$$x^3+7x^2+25x+14=0$$
(you try to solve this equation, subs. $x=\frac{3y-7}{3},$ but will not earn real solution)
I hope you help
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why is this trigonometric identity true? Suppose $$\frac{{{{\sin }^4}(\alpha )}}{a} + \frac{{{{\cos }^4}(\alpha )}}{b} = \frac{1}{{a + b}}$$ for some $a,b\ne 0$.
Why does $$\frac{{{{\sin }^8}(\alpha )}}{{{a^3}}} + \frac{{{{\cos }^8}(\alpha )}}{{{b^3}}} = \frac{1}{{{{(a + b)}^3}}}$$
| The general expression is:$$\frac{\sin^{4n}\theta}{a^{2n-1}} + \frac{\cos^{4n}\theta}{b^{2n-1}} = \frac{1}{(a+b)^{2n-1}}~, ~~n\in \mathbb N$$
From the given relation, it can be written \begin{align}(a+b)\left(\frac{\sin^4\theta}{a}+ \frac{\cos^4\theta}{b}\right) &= \left(\sin^2\theta + \cos^2\theta\right)^2\\ \implies\sin^4\theta + \cos^4\theta + \frac ba\sin^4\theta + \frac ab\cos^4\theta &= \sin^4\theta + 2\sin^2\theta\cos^2\theta + \cos^4\theta\\ \implies~~~~~~~~~~~ \left(\sqrt{\frac ba}\sin^2\theta - \sqrt\frac ab\cos^2\theta \right)^2&= 0\\ \implies~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \frac{\sin^2\theta}{a}&= \frac{\cos^2\theta }{b}\\ \implies~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ \frac{\sin^2\theta}{a}= \frac{\cos^2\theta }{b} &= \frac{\sin^2\theta + \cos^2\theta}{a+b}\end{align}
From this, it can be concluded that $$\sin^2 \theta = \frac a{a+b};~~~\cos^2\theta = \frac{b}{a+b} \,.$$
It's a matter of substituting the values to prove the general and the desired expression.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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Evaluating integral $\int_a^\infty dx\ \frac{x}{\sqrt{(1+\frac{x^2}{b^2})^5(x^2-a^2)}}$ Where $a,b\in\mathbb{R}$ and we may consider $a\ge 0$,
\begin{align}
\mathrm{I}\left(a\right) & =
\int_{a}^{\infty}{x \over
\,\sqrt{\,\left(\, 1 + x^{2}/b^{2}\,\right)^{5}
\left(\, x^{2} - a^{2}\,\right)}\,}\,\,\mathrm{d}x =
{2b \over 3\left(\, 1 + {a^{2}/b^{2}}\,\right)^{2}}
\end{align}
Mathematica provided the solution. Any ideas on how to solve by hand ?.
I solved the case $\,\mathrm{I}\left(0\right) = 2b/3$ with some simple trig substitutions. I'm not sure how to approach the general solution. If the general solution is rather nasty, I'm open to cases where $a$ or $b$ or both are considered to be small parameters. My purpose is to expand my knowledge of integration techniques.
| Let $t=\sqrt{x^2-a^2}$, then
$$dt=\frac{x}{\sqrt{x^2-a^2}}dx,\quad x^2=t^2+a^2.$$
It follows that
$$\begin{aligned}
\int_a^\infty\frac{x}{\sqrt{(1+\frac{x^2}{b^2})^5(x^2-a^2)}}dx&=\int_0^\infty\frac{1}{\left(1+\frac{t^2+a^2}{b^2}\right)^{\frac{5}{2}}}dt.
\end{aligned}$$
Then use the substitution
$$t=\sqrt{a^2+b^2}\tan u,$$
then
$$dt=\sqrt{a^2+b^2}\sec^2udu,$$
so we have
$$\begin{aligned}\int_0^\infty\frac{1}{\left(1+\frac{t^2+a^2}
{b^2}\right)^{\frac{5}{2}}}dt&=\int_0^{\frac{\pi}{2}}\frac{\sqrt{a^2+b^2}\sec^2u}{\left(\frac{a^2+b^2}{b^2}\sec^2u\right)^{\frac{5}{2}}}du\\
&=\int_0^{\frac{\pi}{2}}\frac{b^5}{(a^2+b^2)^2}\cos^3udu\\
&=\frac{b^5}{(a^2+b^2)^2}\int_0^{\frac{\pi}{2}}\cos u\frac{1+\cos(2u)}{2}du\\
&=\frac{b^5}{(a^2+b^2)^2}\int_0^{\frac{\pi}{2}}\frac{\cos u}{2}+\frac{\cos(3u)+\cos(u)}{4}du\\
&=\frac{b^5}{(a^2+b^2)^2}\left.\left[\frac{\sin u}{2}+\frac{\sin(3u)}{12}+\frac{\sin u}{4}\right]\right|_0^{\frac{\pi}{2}}\\
&=\frac{b^5}{(a^2+b^2)^2}\cdot\frac{2}{3}\\
&=\frac{2b}{3\left(\frac{a^2}{b^2}+1\right)^2}
\end{aligned}$$
| {
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"url": "https://math.stackexchange.com/questions/2013001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit with sum of cube roots I am trying to evaluate
$$\lim \limits_{n \to \infty} \left(\sqrt[3]{{1 \over n^4}} + \sqrt[3]{{2 \over n^4}} + \sqrt[3]{{3 \over n^4}} + \dots + \sqrt[3]{{n \over n^4}} \right)$$
I simplify the expression to
$${\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{3} + \dots + \sqrt[3]{n} \over \sqrt[3]{n^4}}$$
$y_n=\sqrt[3]{n^4} \to \infty$, so we might try to use the Stolz theorem to get the limit.
$${x_{n+1}-x_n \over y_{n+1}-y_n} = {\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{3} + \sqrt[3]{4} + \dots + \sqrt[3]{n+1} - \left(\sqrt[3]{1} + \sqrt[3]{2} + \sqrt[3]{3} + \dots + \sqrt[3]{n}\right) \over \sqrt[3]{(n+1)^4}-\sqrt[3]{n^4}}={\sqrt[3]{n+1} \over \sqrt[3]{(n+1)^4} - \sqrt[3]{n^4}}$$
Whne I use the Stolz theorem, the limit seems to be $\infty$. Stolz works for $\infty$, so this would contradict the limit ${3 \over 4}$ obtained below by integration which I'm fairly confident is correct. Any suggestions on what I'm missing?
| Hint:
$$
\sqrt[3]{a}-\sqrt[3]{b}=\frac{a-b}{\sqrt[3]{a^2}+\sqrt[3]{a\,b}+\sqrt[3]{b^2}}.
$$
Use it with both numerator and denominator in the expression you get from Stolz's criterium.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that $\tan 3a = \tan 3b = \tan 3c$ if $\sin a + \sin b + \sin c = 0$ and $\cos a + \cos b + \cos c = 0$ Let $a, b$ and $c$ be three angles such that:
$$\sin a + \sin b + \sin c = 0$$
$$\cos a + \cos b + \cos c = 0$$
Prove that $\tan 3a = \tan 3b = \tan 3c$.
I haven't done anything meaningful yet on this problem because I have no idea how I should start.
Thank you in advance!
| By geometry:
$(\cos a,\sin a),(\cos b,\sin b),(\cos c,\sin c)$ are three unit vectors which sum to zero.
Then they must be the vertices of an equilateral triangle inscribed in the trigonometric circle, at angles $$k\frac{2\pi}3+\phi$$ for $k=0,1,2$.
Obviously,
$$\tan\left(3\left(k\frac{2\pi}3+\phi\right)\right)=\tan3\phi$$ for all $k$.
From this finding, we can derive a more analytical solution.
Combining the two initial equations, we must have
$$(\sin a+\sin b)^2+(\cos a+\cos b)^2=(-\sin c)^2+(-\cos c)^2=1$$ or after simplification $$2\sin a\sin b+2\cos a\cos b=2\cos(a-b)=-1$$
giving
$$a-b=\pm\frac{2\pi}3.$$
By symmetry, this holds for $b-c$ and $c-a$ and leads us to
$$b=a\pm\frac{2\pi}3,c=a\mp\frac{2\pi}3,$$
then
$$3b=3a\pm2\pi,3c=3a\mp2\pi.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2015438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Other ways to calculate this indefinite integral ($\int \frac{2\,dx}{(\cos(x) - \sin(x))^2}$)? I came across the following indefinite integral
$$ \int \frac{2\,dx}{(\cos(x) - \sin(x))^2} $$
and was able to solve it by doing the following:
First I wrote
$$\begin{align*}
\int \frac{2\,dx}{(\cos(x) - \sin(x))^2} &= \int \frac{2\,dx}{1 - 2\cos(x)\sin(x)} \\
&= \int \frac{2\,dx}{1 - \sin(2x)} \\
\end{align*}$$
Then setting $2x = z$,
$$\begin{align*}
&= \int \frac{dz}{1-\sin(z)}\\
&= \int \frac{1+\sin(z)}{\cos^2(z)}\,dz \\
&= \int \sec^2(z) + \tan(z)\sec(z) \,dz \\
&= \tan(z) + \sec(z) \\
&= \tan(2x) + \sec(2x).
\end{align*}$$
The solutions to the problem were given as $\tan(x + \pi/2)$ or $\frac{\cos(x) + \sin(x)}{\cos(x) - \sin(x)}$. I checked that these solutions are in face equivalent to my solution of $\tan(2x) + \sec(2x)$.
My question is, are there other ways to calculate this integral that more "directly" produce these solutions? Actually, any elegant calculation methods in general would be interesting.
| see my nice answer:
$$\int \frac{2\ dx}{(\cos x-\sin x)^2}=\int \frac{2\ dx}{\cos^2 x\left(1-\frac{\sin x}{\cos x}\right)^2}$$ $$=2\int \frac{\sec^2 x\ dx}{\left(1-\tan x\right)^2}$$
$$=-2\int \frac{d(1-\tan x)}{\left(1-\tan x\right)^2}$$
$$=-2 \frac{-1}{\left(1-\tan x\right)}+C$$$$=\frac{2\cos x}{\cos x-\sin x}+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2016580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Prove in as elementary a way as possible that $a^{1/(a-1)} < 2$ for integer $a \ge 3$ This inequality came up
as the final step
in an elementary proof
that the only
integer solution to
$x^y = y^x$
with $x > y > 1$
is
$4^2 = 2^4$.
Here are two proofs
I came up with that
if integer $a \ge 3$,
then $ a^{1/(a-1)} < 2$.
Both proofs do it
by noting that
this is true for
$a=3$,
and $a^{1/(a-1)}$
is decreasing for
$a \ge 3$.
One way to show this
is by looking at
the log of this,
which is
$\frac{\ln a}{a-1}$.
Its derivative is
$\frac{(a-1)/a - \ln a}{(a-1)^2}
=\frac{1-1/a - \ln a}{(a-1)^2}
< 0$
for $a \gt 1$.
If
$f(a) =1-1/a - \ln a
$
then
$f(1) = 0$
and
$f'(a)
=1/a^2-1/a < 0
$ for $a > 1$.
Another proof:
Suppose
$a^{1/(a-1)} \le (a+1)^{1/a}
$.
Then
$a^a \le (a+1)^{a-1}$
or
$(a+1)a^a \le (a+1)^a$.
Dividing by
$a^a$,
$a+1 \le (1+1/a)^a$.
But,
as has been proved here
by elementary means
many times,
$(1+1/a)^a < 3$,
so that
$a+1 < 3$
or
$a < 2$.
So,
is there a more elementary proof than these?
Thanks.
| The inequality $a^{1/(a-1)}<2$ for integers $a\ge 3$ is equivalent to $a<2^{a-1}$ for integers $a\ge 3$. This is easily proved by induction on $a$. It’s clearly true for $a=3$, and if $a<2^{a-1}$ and $a\ge 3$, then
$$a+1<2^{a-1}+1<2\cdot 2^{a-1}=2^a\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2017600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $27 + abc = (a +b +c)(ab + ac+ bc)$ so $\frac{a^2}{a+2b} + \frac{b^2}{b+2c} + \frac{c^2}{c+2a }\geq \frac{3}{2}$ Let $a$, $b$ and $c$ be positive numbers such that $27 + abc = (a +b +c)(ab + ac+ bc)$. Prove that:
$$\frac{a^2}{a+2b} + \frac{b^2}{b+2c} + \frac{c^2}{c+2a }\geq \frac{3}{2}$$
I have that $\frac{a^2}{a+2b} + \frac{b^2}{b+2c} + \frac{c^2}{c+2a} \geq \frac{a+b+c}{3}$ then if I proof that $a+b+c \geq \frac{9}{2}$ I finish. Also $27 + abc = (a +b +c)(ab + ac+ bc)$ can be factorized to $27=(a+b)(ab + ac + bc + c^2)$.
| By Holder $\sum\limits_{cyc}\frac{a^2}{a+2b}=\sum\limits_{cyc}\frac{a^3}{a^2+2ab}\geq\frac{(a+b+c)^3}{3\sum\limits_{cyc}(a^2+2ab)}=\frac{a+b+c}{3}$.
The condition gives $27=(a+b)(a+c)(b+c)$.
Thus, it remains to prove that $\frac{a+b+c}{3}\geq\frac{1}{2}\sqrt[3]{(a+b)(a+c)(b+c)}$ or
$\frac{a+b+a+c+b+c}{3}\geq\sqrt[3]{(a+b)(a+c)(b+c)}$, which is AM-GM.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Logarithmic formula By making as few assumptions as possible, prove that
$$\log \left({{1+x} \over {1-x}} \right)=2 \left(\frac{x}{1}+\frac{x^3}{3}+\frac{x^5}{5}+\cdots\right)$$
for $|x|<1$.
| Note that for $|x|<1$, we can write
$$\begin{align}
\log\left(\frac{1+x}{1-x}\right)&=\int_0^{x}\color{#034da3}{\left(\frac{1}{1+u}+\frac{1}{1-u}\right)}\,du\\\\
&=\int_0^x \color{#034da3}{\frac{2}{1-u^2}}\,du\\\\
&=\int_0^x \overbrace{\color{#034da3}{2\sum_{n=0}^\infty u^{2n}}}^{\color{#034da3}{\text{Geometric Series}}}\,du\\\\
&=2\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}\\\\
&=2\sum_{n=1}^\infty\frac{x^{2n-1}}{2n-1}
\end{align}$$
Alternatively, for $|x|<1$, we can write
$$\begin{align}
\log\left(\frac{1+x}{1-x}\right)&=\int_0^{x}\color{#034da3}{\left(\frac{1}{1+u}+\frac{1}{1-u}\right)}\,du\\\\
&=\int_0^x \overbrace{\color{#034da3}{\sum_{n=0}^\infty (1+(-1)^n)u^n}}^{\color{#034da3}{\text{Geometric Series of the integrand}}}\,dx\\\\
&=\underbrace{\sum_{n=0}^\infty\frac{(1+(-1)^n)x^{n+1}}{n+1}}_{\text{After integrating the power series term-by-term}}\\\\
&=2\sum_{n=0}^\infty \frac{x^{2n+1}}{2n+1}\\\\
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2019030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 1
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Prove that this series converges: ${(\frac{n}{n+1}})^{n^2}$ How to prove that this series converges: $\sum_{n = 1}^{\infty}{({\frac{n}{n+1}})^{n^2}}$ ?
I tried root test, ratio test, comparison test, Cauchy condensation test, but none works. Any hints?
|
I thought it might be instructive to present a way forward that uses the comparison test. To that end, we proceed.
It is straightforward to show that $a_n=\left(\frac{n}{n+1}\right)^n$ is monotonically decreasing (See the note at the end of this post) and bounded below trivially by $0$.
Thus, $a_n$ converges. Furthermore, for $n\ge 1$, $\left(\frac{n}{n+1}\right)^n\le \frac12$.
Obviously then, can write
$$\left(\frac{n}{n+1}\right)^{n^2}\le \left(\frac{1}{2}\right)^n$$
where $\sum_{n=1}^\infty \left(\frac{1}{2}\right)^n =1<\infty$.
Therefore, the series $\sum_{n=1}^\infty \left(\frac{n}{n+1}\right)^{n^2}$ converges by comparison to the geometric series $\sum_{n=1}^\infty \left(\frac{1}{2}\right)^n$.
NOTE:
To show that $a_n=\left(\frac{n}{n+1}\right)^n$ is monotone decreasing, we form the ratio $\frac{a_n}{a_{n+1}}$ and show that this ratio is greater than or equal to $1$. We write
$$\begin{align}
\frac{a_n}{a_{n+1}}&=\frac{\left(\frac{n}{n+1}\right)^n}{\left(\frac{n+1}{n+2}\right)^{n+1}}\\\\
&=\left(\frac{n(n+1)}{(n+1)^2}\right)^{n+1}\,\left(\frac{n+1}{n}\right) \\\\
&=\left(1-\frac{1}{(n+1)^2}\right)^{n+1}\,\left(\frac{n+1}{n}\right) \tag1\\\\
&\ge \left(1-\frac{1}{(n+1)}\right)\,\left(\frac{n+1}{n}\right) \tag 2\\\\
&=1
\end{align}$$
as was to be shown. In going from $(1)$ to $(2)$, we applied Bernoulli's Inequality.
| {
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"url": "https://math.stackexchange.com/questions/2020754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve for $n$ in $18^{n+1} = 2^{n+1} \cdot 27$
Solve for $n$: $$18^{n+1} = 2^{n+1} \cdot 27$$
I tried:
$$18^{n+1} = 2^{n+1} \cdot 27 \Leftrightarrow 18^n \cdot 18 = 2^n \cdot 54 \Leftrightarrow \frac{18^n}{54} = \frac{2^n}{18} \Leftrightarrow \frac{18 \cdot 18^n - 54 \cdot 2^n}{972} = 0 \Leftrightarrow \\ 18 \cdot 18^n - 54 \cdot 2^n = 0 \Leftrightarrow ???$$
What do I do next? Am I doing it right?
| $$18^{n+1} = 2^{n+1}\cdot27$$
$$\Leftrightarrow \frac{18^{n+1}}{2^{n+1}} = 27$$
$$\Leftrightarrow \left(\frac{18}{2}\right)^{n+1} = 27$$
$$\Leftrightarrow 9^{n+1} = 27$$
$$\Leftrightarrow 9^{n} = 3$$
$$\Leftrightarrow n =\log_9(3) =\frac {\log 3}{\log 9} =\frac {\log 3}{2\cdot \log 3} = \frac 1 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2022079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to show $\sum_{n=1}^{a+b} \frac{(-1)^n}{n}\binom{n}{n-a,n-b,a+b-n}=0$ I was trying to prove directly that, for $a,b$ positive integers:
$$\sum_{n=1}^{a+b} \frac{(-1)^n}{n}\binom{n}{n-a,n-b,a+b-n}=0$$
Alternatively:
$$\sum_{n=1}^{a+b}\frac{(-1)^n}{n}\binom{n}{a}\binom{a}{n-b}=0$$
This comes from trying to prove the formal power series identity:
If $f(x)=\sum_{n=1}^{\infty}\frac{x^n}{n}$ then $$f(x)+f(y)=f(x+y-xy)$$
which we know is true because $f(x)=-\log(1-x)$ in the complex numbers when $|x|<1$.
I would like to prove this directly, without referencing logarithm, and it reduces to the above identity.
(Note: $\binom{n}{i,j,k}=\binom{n}{i}\binom{n-i}{j}=\frac{n!}{i!j!k!}$ is the trinomial coefficient, defined when $i+j+k=n$)
| It is convenient to use the coefficient of operator $[z^k]$ to denote the coefficient of $z^k$ in a series. This way we can write e.g.
\begin{align*}
[z^k](1+z)^n=\binom{n}{k}
\end{align*}
We obtain
\begin{align*}
\sum_{n=1}^{a+b}&\frac{(-1)^n}{n}\binom{n}{a}\binom{a}{n-b}\\
&=\frac{1}{a}\sum_{n=1}^{a+b}(-1)^n\binom{n-1}{a-1}\binom{a}{n-b}\tag{1}\\
&=\frac{1}{a}\sum_{n=0}^{a+b-1}(-1)^{n+1}\binom{n}{a-1}\binom{a}{n+1-b}\tag{2}\\
&=\frac{1}{a}\sum_{n=0}^\infty(-1)^{n+1}[z^{a-1}](1+z)^n[u^{n+1-b}](1+u)^a\tag{3}\\
&=-\frac{1}{a}[z^{a-1}]\sum_{n=0}^\infty(-1)^n(1+z)^n[u^n]u^{b-1}(1+u)^a\tag{4}\\
&=-\frac{1}{a}[z^{a-1}](-(1+z))^{b-1}(1-(1+z))^a\tag{5}\\
&=(-1)^{a+b}\frac{1}{a}[z^{a-1}](1+z)^{b-1}z^a\tag{6}\\
&=0
\end{align*}
Comment:
*
*In (1) we use the binomial identity
\begin{align*}
\binom{n}{a}=\frac{n}{a}\binom{n-1}{a-1}
\end{align*}
*In (2) we shift the index $n$ by $1$.
*In (3) we apply the coefficient of operator twice. We also extend the limits of the series without changing anything since we are adding zeros only.
*In (4) we use the linearity of the coefficient of operator.
*In (5) we use the substitution rule of the coefficient of operator with $u:= -(1+z)$
\begin{align*}
A(z)=\sum_{n=0}^\infty a_n z^n=\sum_{n=0}^\infty z^n [u^n]A(u)
\end{align*}
*In (6) we observe the coefficient of $z^{a-1}$ is zero.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Number Theory equality
Suppose that the equation $x^4+y^4=z^2$ has solutions $(x,y,z)\in\mathbb{Z}^{3+}$, let $c$ be the smaller $z$ for which $(x,y,c)$ is a solution. (1) If $x$ is even, prove $x^2=m^2-n^2, y^2=2mn, c=m^2+n^2$, $m,n$ relatively prime. (2) Prove $x=r^2 - s^2, n=2rs, m=r^2+s^2$, $r,s$ relatively prime.
(1) I have the following theorem: if $x^2+y^2=z^2$ then $x^2=2ab, y^2=b^2-a^2, c=b^2+a^2$.
Taking the $x^2$ to be $y$ in the latter, $y^2$ to be $x$ and $c$ to be $z$, then there is $m,n$ such that $x^2=m^2-n^2, y^2=2mn, c=m^2+n^2$.
(2) $x^2+y^2=m^2-n^2+2mn=(m-n)^2$. By means of the same theorem, there is $r,s$ such that $x=r^2-s^2, y=2rs, m+n=r^2+s^2$. This gives the equation for $x$, but the other two don't seem two follow, because it needs too be $n=y$ and if $m+n=r^2+s^2$ then $m=(r-s)^2$ instead of $m=r^2+s^2$.
| you have $x^2+n^2=m^2$, so by the theorem, there are integers $r,s$ such that $x=r^2-s^2, n=2rs, m=r^2+s^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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A limit without invoking L'Hopital: $\lim_{x \to 0} \frac{x \cos x - \sin x}{x^2}$ The following limit
$$\ell=\lim_{x \rightarrow 0} \frac{x \cos x - \sin x}{x^2}$$
is a nice candidate for L'Hopital's Rule. This was given at a school before L'Hopital's Rule was covered. I wonder how we can skip the rule and use basic limits such as:
$$\lim_{x \rightarrow 0} \frac{\sin x}{x} \quad , \quad \lim_{x \rightarrow 0} \frac{\cos x -1}{x^2}$$
| Is it compulsory to exploit $\lim_\limits{x\to 0} \dfrac{\sin x}{x}$ and $\lim_\limits{x\to 0} \dfrac{\cos x -1}{x^2}$?
If not, why don't you just expand it, like: $$
\lim_\limits{x\to 0} \frac{ x \cos x -\sin x }{ x^2 }
=\lim_\limits{x\to 0} \frac{ x [1 -x^2/2 +\mathscr{O}(x^4)] -[x -x^3/6 +\mathscr{O}(x^5)] }{x^2}
=\lim_\limits{x\to 0} \frac{x -x^3/2 -x +x^3/6 +\mathscr{O}(x^5)}{x^2}
=\lim_\limits{x\to 0} \frac{-x^3/3}{x^2}
=\lim_\limits{x\to 0} -\frac{x}{3}
=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2027117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Evaluating $\lim_{x\rightarrow0}\frac{1-\frac{1}{2}x^2-\cos(\frac{x}{1-x^2})}{x^4}$ $$\lim_{x\rightarrow0}\frac{1-\frac{1}{2}x^2-\cos(\frac{x}{1-x^2})}{x^4}$$
I have no idea on how to face this limit. Its value at $0$ seems to be $0$, but its limit equals $\frac{23}{24}$. I tried l'Hôpital but the new expressions were as confusing. I think maybe using squeeze theorem would be useful, but I don't know how to apply it in this case.
Any help would be appreciated, thanks in advance.
| HINT:
As suggested using Series Expansion,
$$\cos\dfrac x{1-x^2}=1-\dfrac{\left(\dfrac x{1-x^2}\right)^2}{2!}+\dfrac{\left(\dfrac x{1-x^2}\right)^4}{4!}+\text{ the terms containing higher powers of } x$$
$$1-\dfrac{x^2}2-\cos\dfrac x{1-x^2}$$ $$=\dfrac{x^2}{2(1-x^2)^2}-\dfrac{x^2}2-\dfrac{x^4}{24(1-x^2)^4}+\text{ the terms containing higher powers of } x$$
Now $\dfrac{x^2}{2(1-x^2)^2}-\dfrac{x^2}2=\dfrac{x^2\{1-(1-x^2)^2\}}{2(1-x^2)^2}=\dfrac{2x^4-x^6}{2(1-x^2)^2}$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Peter planted a tree. Its growth is given by $h(t) = \frac{a}{1+be^{kt}}$ for $t$ years. Find $a$,$b$ and $k$
Peter planted a tree on his backyard. Over the course of some years he
measured the height of the tree. first directly, and later on using
his knowledge of trigonometry. When he thought he had enough data, he
defined as the model for the growth of the tree in meters, after $t$
years of being planted, the following function:
$$h(t) = \frac{a}{1+be^{kt}}$$
Determine $a$,$b$ and $k$, knowing that:
*
*the tree was 50 cm tall when it was planted
*the maximum height the tree can have is 5 meters (approximate value)
*after 2 years, the tree has twice the height it had when it was planted
(Note: " 5 meters (approximate value)" is an exact quote)
I did:
$50 cm = 0.5m$,
$$0.5 = \frac{a}{1+be^{-k\cdot0}} \Leftrightarrow 0.5 = \frac{a}{1+b}$$
so I conlcude that
$a = \frac{b+1}{2}$ and $b = 2a-1$
Then I used the information on the third item,
$h(2) = 1$
and I tried to isolate $k$
$$1 = \frac{\frac{b+1}{2}}{1+be^{-k\cdot2}} \Leftrightarrow 1 = \frac{b+1}{2(1+be^{-k2})} \Leftrightarrow 1 = \frac{b+1}{2+2be^{-k2}} \Leftrightarrow b+1 = 2+2be^{-2k} \Leftrightarrow \frac{b-1}{2} = be^{2k} \Leftrightarrow \frac{\frac{b-1}{2}}{b} = e^{-2k} \Leftrightarrow \frac{b-1}{2b} = e^{-2k} \Leftrightarrow \frac{b-1}{2b} = (\frac{1}{e^2})^k \Leftrightarrow k = \log_{\frac{1}{e^2}}{\frac{b-1}{2b}} \Leftrightarrow k = -\frac{1}{2} \ln(\frac{b-1}{2b}) \Leftrightarrow k = \ln(\sqrt{\frac{2b}{b-1}})$$
Then I tried the third sentence with $k = \ln(\sqrt{\frac{2b}{b-1}})$
$$1 = \frac{\frac{b+1}{2}}{1+be^{-2\ln(\sqrt{\frac{2b}{b-1}})}}$$
This took me an $\infty$ of time to solve only to find out it was wrong, so I will skip the steps and tell you that both sides are always equal to 1 except for $x \in [-1;1]$ which show "ERROR"
How do I solve this?
| Let us consider the function $$h(t) = \frac{a}{1+be^{kt}}$$ and let us use the different given informations
*
*the tree was 50 cm tall when it was planted
*the maximum height the tree can have is 5 meters (approximate value)
*after 2 years, the tree has twice the height it had when it was planted
As you wrote, the first one gives $$h(0)=\frac{a}{1+be^{k\times 0}}=\frac{a}{1+b}=\frac 12\tag 1$$ Assuming that $k$ is negative, the second one write $$h(\infty)=\frac{a}{1+be^{k\times \infty}}=\frac{a}{1}=5\tag 2$$ The third one gives $$h(2)=\frac{a}{1+be^{k\times 2}}=\frac{a}{1+b e^{2k}}=2h(0)=1.0\tag 3$$ So, from $(2)$ $a=5$; from $(1)$ $b=9$. Plugging these numbers in $(3)$ $$\frac 5 {1+9e^{2k}}=1\implies 4=9 e^{2k}\implies 2=3e^k\implies k=-\log \left(\frac{3}{2}\right)$$ All of that makes that we can rewrite the model as $$h(t)=\frac 5{1+9 e^{-\log \left(\frac{3}{2}\right)\,t}}$$
Using the formula, this would give the following table
$$\left(
\begin{array}{cc}
t & h(t) \\
0 & 0.50 \\
1 & 0.71 \\
2 & 1.00 \\
3 & 1.36 \\
4 & 1.80 \\
5 & 2.29 \\
6 & 2.79 \\
7 & 3.27 \\
8 & 3.70 \\
9 & 4.05 \\
10 & 4.32 \\
11 & 4.53 \\
12 & 4.68 \\
13 & 4.78 \\
14 & 4.85 \\
15 & 4.90 \\
16 & 4.93 \\
17 & 4.95 \\
18 & 4.97 \\
19 & 4.98 \\
20 & 4.99 \\
\infty & 5.00
\end{array}
\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$. Exercise
Prove that $a^n-b^n = (a-b)(a^{n-1} + a^{n-2}b + \cdots + b^{n-1})$.
I've posted my solution below. In case someone has a more clever solution, feel free to post it!
(TBH, I was surprised that there was no question on Math.SE regarding this equation!)
| $a^n - b^n =$
$= a^{n}b^{0} + (a^{n-1}b^{1} + a^{n-2}b^{2} + \cdots + a^{2}b^{n-2} + a^{1}b^{n-1}) \\ \space\space\space - a^{0}b^{n} - (a^{n-1}b^{1} + a^{n-2}b^{2} + \cdots + a^{2}b^{n-2} + a^{1}b^{n-1}) = $
$= (a^{n}b^{0} + a^{n-1}b^{1} + a^{n-2}b^{2} + \cdots + a^{2}b^{n-2} + a^{1}b^{n-1}) \\ \space\space\space - (a^{n-1}b^{1} + a^{n-2}b^{2} + \cdots + a^{2}b^{n-2} + a^{1}b^{n-1} + a^{0}b^{n}) = $
$= a(a^{n-1}b^{0} + a^{n-2}b^{1} + \cdots + a^{1}b^{n-2} + a^{0}b^{n-1}) \\ \space\space\space - b(a^{n-1}b^{0} + a^{n-2}b^{1} + \cdots + a^{1}b^{n-2} + a^{0}b^{n-1}) = $
$= (a - b)(a^{n-1}b^{0} + a^{n-2}b^{1} + \cdots + a^{1}b^{n-2} + a^{0}b^{n-1})= $
$= (a - b)(a^{n-1} + a^{n-2}b + \cdots + ab^{n-2} + b^{n-1})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2031059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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New Question: Find $x,y,z$ s.t. $x+y+z=1$ and $x^2+y^2+z^2=1$, and $x,y,z \in ( - 1,0) \cup (0,1)$ If in my earlier question [1], we relieve the constraints such that $x,y,z$ can also get values between $(−1,0)$, it seems that the system has a solution. Now, the question is how we can find this solution?
$$\begin{cases}&x + y + z = 1\\
&{x^2} + {y^2} + {z^2} = 1\\
&x \ne y \ne z, x \ne z\\
&x,y,z \in ( - 1,0) \cup (0,1)
\end{cases}$$
--
[1] Multivariate-Multi-objective Optimization Problem: $x+y+z = 1$ and $x^2+y^2+z^2 = 1$
| First observe that exactly one of the $x,y,z$ must be in $(-1,0)$. Without loss assume that $x\in(-1,0)$ and $y,z \in (0,1)$. Then, the first condition gives
\begin{align}y+z=1-x\implies& y^2+2yz+z^2=1-2x+x^2\\[0.2cm]\overset{(2)}\implies &1-x^2+2yz=1-2x+x^2\\[0.2cm]\implies& y=\frac{x(x-1)}z\end{align} Now, return and plug in the value of $y$ in $(1)$
\begin{align}x+\frac{x(x-1)}{z}+z=1\implies &z^2+z(x-1)+x(x-1)=0\\[0.2cm]\implies &\frac1{(x-1)}z^2+z+x=0\\[0.2cm]\implies& z_{1,2}=\left(-1\pm\sqrt{1-\frac{4x}{(x-1)}}\right)\cdot\frac{x-1}2\end{align} Now, the condition $z>0$ implies that we can drop the one solution (which is obviously negative) and hence obtain $$z=\left(-1+\sqrt{\frac{1+3x}{1-x}}\right)\cdot\frac{x-1}2$$ Since the term in the square root must me non-negative we get the restriction \begin{align}\frac{1+3x}{1-x}\ge0 \iff (1+3x)(1-x)\ge 0 \iff -\frac13\le x\le 1\end{align} which together with $x<0$ implies that $-\frac13\le x<0$. For these values of $x$ the given $z$ exists and is $>0$. With this value of $z$, $y$ becomes $$y=\frac{x(x-1)}{z}=\frac{2x}{-1+\sqrt{\frac{1+3x}{1-x}}}$$ Hence, it remains to find all $x$ such that $$x+\frac{2x}{-1+\sqrt{\frac{1+3x}{1-x}}}+\left(-1+\sqrt{\frac{1+3x}{1-x}}\right)\cdot\frac{x-1}{2}=1$$
| {
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"url": "https://math.stackexchange.com/questions/2035522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then $x^{2000}+\frac{1}{x^{2000}}= $? If $x+\frac{1}{x}=\frac{1+\sqrt{5}}{2}$ then
$$x^{2000}+\frac{1}{x^{2000}}=?$$
My try:
$$\left(x^{1000}\right)^2+\left(\frac{1}{x^{1000}}\right)^2=\left(x^{1000}+\frac{1}{x^{1000}}\right)^2-2$$
Continuation ?
| If $a_n=x^n+\frac{1}{x^n}$ then $a_n=a_1\cdot a_{n-1}+a_{n-2}=\phi\cdot a_{n-1}-a_{n-2}$ which is a second-order linear recurrence, where $\phi=\frac{1+\sqrt 5}{2}$. The initial conditions are $a_1=\phi$ and $a_2=a_1^2-2=\phi^2-2=\phi-1$, since $\phi^2=\phi+1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2039286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 7,
"answer_id": 3
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Need help on a probability calculation Following is a problem statement:
A furniture shop has six identical steel cabinets of brand A and four
identical steel cabinets of brand B. Three customers buy one cabinet
each. Then the probability that two or more cabinets of brand A have
been sold is
My answer was 1/2, but it's incorrect. My initial attempt was as follows 6/10*5/9 + 6/10*5/9*4/8 = 1/2.
| What is the probability that exactly two cabinets sold were of type A?
There are 3 possible situations:
*
*first customer buys A, second A, third buys B
*first customer buys B, second A, third buys A
*first customer buys A, second B, third buys A
The probability of 1. scenario is $\frac{6}{10}\cdot \frac{5}{9} \cdot \frac{4}{8}$
To determine the probability of the event that exactly two cabinets of type A have been bought, you have to add up the probability of all 3 scenarios, that is
$\frac{6}{10}\cdot \frac{5}{9} \cdot \frac{4}{8} +
\frac{4}{10}\cdot \frac{6}{9} \cdot \frac{5}{8}+
\frac{6}{10}\cdot \frac{4}{9} \cdot \frac{5}{8} = \frac{1}{2}$
This is why your calculation failed.
If you add the probability that all cabinets were of type A, you get the probability you want.
$\frac{1}{2} +\frac{6}{10}\cdot \frac{5}{9} \cdot \frac{4}{8} = \frac{2}{3}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Given that $^_ = \frac{3}{8} \times{}^_{+1}$ and $^_ ={}^_{+6}$, find and . Given that $^_ = \frac{3}{8} \times{}^_{+1}$ and $^_ ={}^_{+6}$, find $$ and $$.
Can anyone see a slick way to do this? The solution is $x=10$ and $y=2$, but I am struggling with the method.
| Hint: The first relation gives \begin{align}\dbinom{x}{y}=\frac38\dbinom{x}{y+1}\implies& \frac{x!}{y!(x-y)!}=\frac38\frac{x!}{(y+1)!(x-y-1)!}\\[0.3cm]\implies&\frac{x!}{y!(x-y)(x-y-1)!}=\frac38\frac{x!}{(y+1)y!(x-y-1)!}\\[0.3cm]\implies&\frac{1}{(x-y)}=\frac38\frac{1}{(y+1)}\implies -3x+11y=-8\end{align} For the second, recall that $$\dbinom{x}{y}=\dbinom{x}{x-y}$$ which gives that $x-y=y+6$.
| {
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"url": "https://math.stackexchange.com/questions/2040183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Which Matrix Equality is always true? Let $A_{r\times m}$and $B_{m\times r}$ be full matrices, $C_{r\times r}$is a diagonal matrix, and $D_{r\times r}$is an identity matrix. Which of the following is always true?
*
*$AB=(AB)^{T}$
*$ABD = DAB$
*$BC = CB$
*$A^{T}C=CA^{T}$
I don't think it would be $1$ because the $A^{T}$ and $B$ would not take into account the diagonal and identity matrix, and I do not think that it would be $3$ either. Any help with this one?
| (1) is false
$$\begin{pmatrix} 1 & 2\\ 3 & 4 \end{pmatrix}\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}\ne \left(\begin{pmatrix} 1 & 2\\ 3 & 4 \end{pmatrix}\begin{pmatrix} 1 & 0\\ 0 & 1 \end{pmatrix}\right)^T$$
(3) is false
$$\begin{pmatrix} 1 & 2\\ 3 & 4 \end{pmatrix}\begin{pmatrix} 2 & 0\\ 0 & 1 \end{pmatrix}\ne \begin{pmatrix} 2 & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 2\\ 3 & 4 \end{pmatrix}$$
Moreover, note that we need $m= r.$
(4) is false
$$\begin{pmatrix} 1 & 3\\ 3 & 4 \end{pmatrix}^T\begin{pmatrix} 2 & 0\\ 0 & 1 \end{pmatrix}\ne \begin{pmatrix} 2 & 0\\ 0 & 1 \end{pmatrix}\begin{pmatrix} 1 & 3\\ 2 & 4 \end{pmatrix}^T$$
Moreover, note that we need $m= r.$
(2) is true
Since $D$ is the identity matrix one has that $MD=DM=M.$ Thus
$$ABD=(AB)D=(AB)=D(AB)=DAB.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Applying Cauchy's integral formula I'm trying to compute $$\int_\gamma \frac{z^4+z^2+1}{z^3-1} \, dz$$ where $\gamma$ is the circle $|z-i|=1$, using Cauchy's integral formula
My solution is as follows:
The integral can be rewritten as $$\int_\gamma \frac{\frac{z^4+z^2+1}{(z-1)\left(z+\frac{1}{2}-\frac{\sqrt{3}}{2}i\right)}}{z+\frac{1}{2}-\frac{\sqrt{3}}{2}i} \, dz = I$$
Then Cauchy's integral formula can be applied, giving $$I=2\pi if\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}i\right)=0$$
Is this correct?
| It is almost correct. The only small problem is that, when you wrote$$\require{cancel}\frac{\ \frac{z^4+z^2+1}{(z-1)\left(z+\frac12\color{red}-\frac{\sqrt3}2i\right)}\ }{z+\frac12-\frac{\sqrt3}2i},$$you should have written$$\frac{\ \frac{z^4+z^2+1}{(z-1)\left(z+\frac12\color{red}+\frac{\sqrt{3}}{2}i\right)}\ }{z+\frac12-\frac{\sqrt3}2i}.\tag1$$But I suppose that that was a typo.
However there is an easier way of computing that integral, which avoids computing the value of the numerator of $(1)$ at $-\frac12+\frac{\sqrt2}2i$. Note that\begin{align*}\frac{z^4+z^2+1}{z^3-1}&=\frac{\frac{z^6-1}{z^2-1}}{z^3-1}\\&=\frac{\frac{z^6-1}{z^3-1}}{z^2-1}\\&=\frac{z^3+1}{z^2-1}.\end{align*}Since the distance from $\pm1$ to $i$ is greater than $1$, it follows from Cauchy's integral theorem that your integral is equal to $0$.
Note: That final rational function can be simplified further:\begin{align}\frac{z^3+1}{z^2-1}&=\frac{\cancel{(z+1)}(z^2-z+1)}{\cancel{(z+1)}(z-1)}\\&=\frac{z^2-z+1}{z-1}.\end{align}
| {
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Prove this Equation is Divisible By... Prove that among any three distinct integers we can find two, say $a$ and
$b$, such that the number $a^3b − ab^3$ is divisible by $10$.
Can anyone help me solve this?
| $a^3b-ab^3 = ab(a^2-b^2)$. If both $a$ and $b$ are odd, $(a^2-b^2)$ is even, so the expression is guaranteed to be even.
Now consider three arbitrary integers $\{c,d,e\}$. If any of these is divisible by $5$, we can chose this and either of the other two to make the target expression divisible by $5$. Otherwise we have $c^2 \equiv \{-1,1\} \bmod 5$, and similarly for $d$ and $e$. Since there are only two choices for the value $\bmod 5$, at least two of $\{c,d,e\}$ must have the same value $\bmod 5$ when squared. Then, choosing these two as $a$ and $b$, we see that $(a^2-b^2)\equiv 0 \bmod 5$, so $5$ divides $(a^2-b^2)$, and thus the target expression $a^3b-ab^3$ is divisible by $10$.
$ (5k+1)^2 = 25k^2+10k + 1 = 5(5k^2+2k) + 1 \\
(5k+2)^2 = 25k^2+20k + 4 = 5(5k^2+4k + 1) - 1 \\
(5k+3)^2 = 25k^2+30k + 9 = 5(5k^2+6k + 2) - 1 \\
(5k+4)^2 = 25k^2+40k + 16 = 5(5k^2+8k + 3) + 1
$
| {
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"timestamp": "2023-03-29T00:00:00",
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Implicit equation of a sheared ellipse I have the following implicit equation of an ellipse
$\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1.$
What will be the equation once it is sheared in the $x$ direction by an angle $\phi$ ?
In other words, suppose I wanted to apply the following shear transformation to the points of an ellipse
$\begin{bmatrix}
x'\\
y'
\end{bmatrix}
=
\begin{bmatrix}
1 & \tan(\phi)\\
0 & 1
\end{bmatrix}
\begin{bmatrix}
x\\
y
\end{bmatrix}$
How would this transformation be represented in an implicit ellipse equation?
| \begin{align*}
x' &= x+y\tan \phi \\
y' &= y \\
x &= x'-y'\tan \phi \\
\frac{x^2}{a^2}+\frac{y^2}{b^2} &=
\frac{(x'-y'\tan \phi)^2}{a^2}+\frac{y'^2}{b^2} \\
\frac{x'^2-2x'y'\tan \phi+y'^2\tan^2 \phi}{a^2}+\frac{y'^2}{b^2} &= 1 \\
\end{align*}
$$b^2x'^2\cos^2 \phi-b^2x'y'\sin 2\phi+(a^2\cos^2 \phi+b^2\sin^2 \phi)y'^2-a^2 b^2 \cos^2 \phi=0$$
| {
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Prove that $5^{2^{m-3}}\equiv 1+2^{m-1} \bmod 2^m$. Prove that $5^{2^{m-3}}\equiv 1+2^{m-1} \bmod 2^m$.
The hints given are induction and binomial theorem.
For $m=4$, $5^2=25\equiv 9 \bmod 2^4$ where $9=1+2^3$.
So the base case is done.
Suppose $5^{2^{m-3}}\equiv 1+2^{m-1} \bmod 2^m$.
Then $5^{2^{m-2}}=5\cdot 5^{2^{m-3}} \equiv 5(1+2^{m-1}) \bmod 2^m $.
Write $5=1+4$. So the term becomes $1+2^{m-1}+2^2+2^{m-2}$.
I am not sure how the binomial theorem can be applied here.
|
$5^{2^{m-2}}=5\cdot 5^{2^{m-3}}$
This is incorrect. We have $5^{2^{m-2}}=5^{2^{m-3}\times 2}=(5^{2^{m-3}})^2$.
Inductive step : Supposing that $5^{2^{m-3}}=1+2^{m-1}+k\cdot 2^m$ where $k\in\mathbb Z$ gives
$$\begin{align}5^{2^{m-2}}&=(5^{2^{m-3}})^2\\\\&=(1+2^{m-1}+k\cdot 2^m)^2\\\\&=1+2^{2m-2}+k^2\cdot 2^{2m}+2^m+k\cdot 2^{m+1}+k\cdot 2^{2m}\\\\&=1+2^m+(2^{m-3}+k^2\cdot 2^{m-1}+k+k\cdot 2^{m-1})\cdot 2^{m+1}\\\\&\equiv 1+2^m\pmod{2^{m+1}}\end{align}$$
| {
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A very different property of primitive Pythagorean triplets: Can number be in more than two of them? While playing with numbers, I thought about squares of numbers, and then the first thing that came to mind was Pythagorean triplets.
I observed a very interesting fact that any $x\in\mathbb N$ can never be a member of more than two Pythagorean triplets of pairwise coprime numbers, like $(3,4,5)$ and $(8,15,17)$.
For example $$16^2+63^2=65^2$$$$33^2+56^2=65^2$$
are the possible triplets for $x=65$, and $65$ cannot exist in any other triplet of co-primes.
Now I need to prove this conjecture.
So I thought that in a Pythagorean triplet, the three numbers are of the form $(2mn, m^2-n^2, m^2+n^2)$.
Let $x$ be a number . Then I have to show that $$x=2mn$$$$x=a^2+b^2$$$$x=y^2-z^2$$ are not simultaneously possible.
But I am stuck and don't know where to go from here.
Please help me prove this or help me disprove it by giving a counter example.
| Any prime which is congruent to $1$ modulo $4$ is the largest member of a Pythagoren triplet. (That is not obvious.)
If $P_1,..., P_n$ are $n$ distinct primes, each congruent to $1$ modulo 4, then $\prod_{j=1}^nP_j$ is the largest member of $2^{n-1}$ different primitive Pythagorean triplets. ( It is not obvious that the method, below, always yields that many.)
Use $(aa'+bb')^2+(ab'-ba')^2=(aa'-bb')^2+(ab'+ba')^2=(a^2+b^2)(a'^2+b'^2).$
Example : (For ease of typing let "$.$" denote "$\times$"). From $3^2+4^2=5^2$ and $5^2+12^2=13^2$ we have $$(3.5\pm 4.12)^2+(3.12\mp 4.5)^2=5^2.13^2.$$ That is, $63^2+16^2=33^2+56^2=65^2.$...... Since $17=1^2+4^2$ we have $$(63.1\pm 16.4)^2+(63.4\mp 16.1)^2=17^2.65^2=$$ $$= (33.1\pm 56.4)^2+(33.4\mp 56.1)^2=17^2.65^2.$$ That is, $$127^2+236^2=1^2+268^2=257^2+76^2=191^2+188^2=1105^2.$$
| {
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Show that $e^n>\frac{(n+1)^n}{n!}$ without using induction. I have got an inequality problem which is as follow:
Show that $e^n>\frac{(n+1)^n}{n!}$
I can do it by induction but I have been told to prove it without induction.
My Work:
$$e^n=1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........$$
$$e^n>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........+\frac{n^n}{n!}$$
$$e^n>\frac{n^n}{n!}+\frac{n^{n-1}}{(n-1)!}.......+\frac{n^2}{2!}+n+1$$
From here I can't go further.
I shall be thankful if you guys can provide me a complete solution/proof of this inequality. A hint will also work.
Thanks in advance.
| $$e^n=1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........$$
$$e^n>1+n+\frac{n^2}{2!}+\frac{n^3}{3!}+........+\frac{n^n}{n!}$$
$$e^n>\frac{n^n}{n!}+\frac{n^{n-1}}{(n-1)!}.......+\frac{n^2}{2!}+n+1$$ $$e^n>n^n\left[\frac{1}{n!}+\frac{1}{n(n-1)!}+\frac{1}{n^2(n-2)!}...+\frac{1}{n^{n-1}}+\frac{1}{n^n}\right] $$ $$e^n>\frac{n^n}{n!}\left[1+\frac{1}{n}n+\frac{1}{n^2}n(n-1)+\frac{1}{n^3}n(n-1)(n-2)...+\frac{n!}{n^n}\right] $$ $\because$ $$n(n-1)>\frac{n(n-1)}{2!}$$and $$n(n-1)(n-2)>\frac{n(n-1)(n-2)}{3!}$$and $$n!>1$$
$\therefore $ $$e^n>\frac{n^n}{n!}\left[1+n\frac{1}{n}+\frac{n(n-1)}{2!}\frac{1}{n^2}+...+\frac{1}{n^n}\right]$$ $$e^n>\frac{n^n}{n!}(1+\frac1n)^n$$ $$e^n>\frac{n^n}{n!}\frac{(n+1)^n}{n^n}$$ $$e^n>\frac{(n+1)^n}{n!}$$
| {
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How to find $\sum_{i=1}^n\left\lfloor i\sqrt{2}\right\rfloor$ A001951 A Beatty sequence: a(n) = floor(n*sqrt(2)). A001951 A Beatty sequence: a(n) = floor(n*sqrt(2)).
If $n = 5$ then
$$\left\lfloor1\sqrt{2}\right\rfloor+ \left\lfloor2\sqrt{2}\right\rfloor + \left\lfloor3\sqrt{2}\right\rfloor +\left\lfloor4 \sqrt{2}\right\rfloor+ \left\lfloor5\sqrt{2}\right\rfloor
= 1+2+4+5+7 = 19$$
Sequence from $1$ to $20$ is:
$S=\{1,2,4,5,7,8,9,11,12,14,15,16,18,19,21,22,24,25,26,28\}$
I want to find answer for $n = 10^{100}$.
| Let $S(\alpha,n) = \sum_{k=1}^n \lfloor \alpha k \rfloor$ for $\alpha$ some irrationnal positive number.
if $\alpha \ge 2$ we let $\beta = \alpha-1$ and you get
$S(\alpha,n) = S(\beta,n) + \sum_{k=1}^n k \\
= S(\beta,n) + n(n+1)/2$
if $1 < \alpha < 2$, there is a theorem that says if $\beta$ satisfies $\alpha^{-1} + \beta^{-1} = 1$, then the sequences $\lfloor \alpha n \rfloor$ and $\lfloor \beta n \rfloor$ for $n \ge 1$ partition $\Bbb N$ (not counting $0$)
Therefore, letting $m = \lfloor \alpha n \rfloor$, $S(\alpha,n) + S(\beta, \lfloor m/\beta \rfloor) = \sum_{k=1}^m k = m(m+1)/2$
Also, $\lfloor m/ \beta \rfloor = m - \lceil m/\alpha \rceil = m- n = \lfloor (\alpha-1)n \rfloor$.
Then, letting $n' = \lfloor (\alpha-1)n \rfloor $ you have
$S(\alpha,n) = (n+n')(n+n'+1)/2 - S(\beta,n')$
So those two formulas give you a very fast way to compute $S$ if you can compute $n' = \lfloor (\alpha-1) n \rfloor$
In your case, $\alpha = \sqrt 2$, so you begin in the second case where you get $\beta = 2+\sqrt 2$. Since the sequence of $\alpha$s you get is periodic, you can get a recurrence formula :
Let $n' = \lfloor (\sqrt 2 -1) n \rfloor$,
$S(\sqrt 2,n) = (n+n')(n+n'+1)/2 - S(2+\sqrt 2,n') \\
= (n+n')(n+n'+1)/2 - S(\sqrt 2,n') - n'(n'+1) \\
= nn'+n(n+1)/2-n'(n'+1)/2 - S(\sqrt 2,n')$
For example this tells you that $S(\sqrt 2,5) = 22 - S(\sqrt 2, 2) = 22 - 3 + S(\sqrt 2, 0) = 19.$
Since at each step $n$ is approximately multiplied by $\sqrt 2 - 1$, the arguments decrease exponentially. For $n = 10^{100}$ you need approximately $\lceil {100 \log {10}/\log ({1\over(\sqrt 2-1)})} \rceil = 262$ steps to complete the recursion. This is basically equivalent to computing the powers of $(\sqrt 2-1)$ with enough precision and should be doable quickly on any computer.
| {
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How can I solve: $\frac {3x-2}2 - \frac {4x-5}3=2$? I am a a student and I am having difficulty with answering this question. I keep getting $6$ as my value for x whereas it is wrong. Please may I have a step by step solution to this question so that I won't have difficulties with answering these type of questions in the future.
Question: Solve $$\frac {3x-2}2 - \frac {4x-5}3=2$$
This is what I did to get the answer:
$$2(3x-2) - 3(4x-5) = 12$$
$$6x-4 - 12x+15 = 12$$
$$6x - 12x = 12+4-15$$
$$-6x = 1$$
$$x = -6 $$
Kind Regards help would be appreciated
| Notice that $$\frac{ 3x-2}{2} - \frac{4x-5}{3} = 2$$ is the same as
$$\ \frac{3\times( 3x-2)}{3 \times 2} - \frac{2\times(4x-5)}{2\times3} = 2$$
or equivalently
$$\ \frac{ 9x-6 }{6} - \frac{ 8x-10 }{6} = 2$$
which is the same as
$$ \frac{ (9x-6)-(8x-10) }{6} = 2$$
so we get
$$(9x-6)-(8x-10) =12 $$
so $$x=8 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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The limit of a function with sum of two roots. I need to find the limit of the following function:
$$\lim_{x\to-\infty} \left(\sqrt{x^2+2x}+\sqrt[3]{x^3+x^2}\right).$$
I derived it to the form of:
$$\lim_{x\to-\infty}\frac{(x^2+2x)^3-(x^3+x^2)^2}{\left(\sqrt{x^2+2x}-\sqrt[3]{x^3+x^2}\right)\left((x^2+2x)^2+(x^2+2x)\sqrt[3]{(x^3+x^2)^2}+\sqrt[3]{(x^3+x^2)^4}\right)},$$
hoping that I'd be able to simplify something but still have $\frac{0}{0}$ and don't see how to do that.
| When $x$ is negative, then $\sqrt{x^2} = -x$, which is positive. So we have
\begin{align}
& \lim_{x\to-\infty} \left(\sqrt{x^2+2x}+\sqrt[3]{x^3+x^2}\right) \\[10pt]
= {} & \lim_{x\to-\infty} \left( -x\sqrt{1 + \frac 2 x} + x \sqrt[3]{1 + \frac 1 x} \, \right) \\[10pt]
= {} & \lim_{x\to-\infty} x\left( -\left( 1 + \frac 2 x \right)^{1/2} + \left( 1 + \frac 1 x \right)^{1/3} \right) \\[10pt]
= {} & \lim_{x\to-\infty} x\left( -\left( 1 + \frac 2 x \right)^{1/2} + \left( 1 + \frac 1 x \right)^{1/3} \right) \\[10pt]
= {} & \lim_{u\,\uparrow\, 0} \frac 1 u \left( - (1+2u)^{1/2} + (1+u)^{1/3} \right) \text{ where } u = \frac 1 x.
\end{align}
L'Hopital's rule can deal with that. (And remember the chain rule when finding $\dfrac d {du} (1+2u)^{1/2}.$)
| {
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Cartesian equation for $\pi \le t \le 2\pi$ The graph for: $x=\cos{t}, y=\sin{2t} $
Looks like a figure 8. I've found the Cartesian equation for the line by doing:
$$ t = \arccos{x} \rightarrow y = \sin{2\arccos{x}} = 2x \cdot \sin{\arccos{x}} = 2x\sqrt{1-x^2}$$
A later part of the question asks for the equation for the part of the line where $\pi \le t \le 2\pi$
Wouldn't this be the same equation?
The question I solved asked to prove the cartesian equation for the line where $0 \le t \le 2\pi$ was $ 2x\sqrt{1-x^2}$
So would I rework my solution to this replacing $t= \arccos{x}$ with $2\pi-t = \arccos{x}$
getting:
$$ t = 2\pi - \arccos{x} \rightarrow\\ y = \sin{(2\cdot(2\pi - \arccos{x}))} \\=\sin(4\pi-2\arccos{x})\\= \sin{4\pi}\cos{2\arccos{x}} + \cos{4\pi}\sin{2\arccos{x}}\\=\sin{2\arccos{x}}$$
... = same answer.
How would I solve for the new limit for t?
| You could solve the problem in this way:
\begin{equation}
\begin{cases}
x=\cos t\\y=\sin2t
\end{cases}
\begin{cases}
x^2=\cos^2t\\y^2=4\sin^2t\cos^2t=4(1-\cos^2t)\cos^2t=4x^2(1-x^2)
\end{cases}
\end{equation}
So we have $y^2=4x^2(1-x^2)\longrightarrow y=\pm2|x|\sqrt{1-x^2}$ for $-1\le x\le1$.
| {
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If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
If $\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}$ show that $a^a \cdot b^b\cdot c^c=1$ .
My Working:
$\frac{\log a}{b-c}= \frac{\log b}{c-a}$
$ (c-a)\log a=(b-c) \log b$
$ \log a^{c-a}=\log b^{b-c}$
$ \frac {a^c}{a^a}=\frac{b^b}{b^c}$
$ \frac {a^c \cdot{b^c}}{a^a} =b^b\qquad \text{(i)}$
Similarly, taking the next two terms we obtain,
$b^b=\frac{b^a \cdot c^a}{c^c}\qquad \text{(ii)}$
I tried to solve the two equations obtained to get to the desired statement but I couldn't. Is the way adopted correct or is there another way to reach the desired answer. Please help me proceed with this question
| We use following property
if $F=\frac{A_1}{B_1}=\frac{A_2}{B_2}$
then
$$F=\frac{A_1+A_2}{B_1+B_2}.$$
So, your equality implies
$$\frac{\ln(a^a)}{ab-ac}=\frac{\ln(b^b)}{bc-ba}=\frac{\ln(c^c)}{ca-cb}$$
$$=\frac{\ln(a^a)+\ln(b^b)}{cb-ca}$$
thus
$$a^a.b^b.c^c=1.$$
| {
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How can I solve this question on Quadratic Equations and Arithmetic Progression? Let $P (x) = ax^2 + bx + c$ where $a,b,c$ are in arithmetic progression and are positive constants. Let $P (x) = 0$ for $t$ and $r$, where $t$ and $r$ are integers. Find $t + r +tr$.
Answer is : $7$
I tried it as follows :
$t + r = -b/a$ ;
$tr = c/a$ ;
$t + r + tr = -b/a + c/a = (c-b)/a = (b-a)/a= b/a - 1$
Therefore,
$2t + 2r + rt = -1$
I could not figure out what to do next. Please help.
| $$p(x)=ax^2+bx+c=ax^2+(a+p)x+(a+2p)$$
$$t+r+r\cdot t=-\frac{a+p}{a}+\frac{a+2p}{a}=\frac{p}{a}$$
then $p=a\cdot k$, $k=(t+r+t\cdot r) \in \Bbb Z$.
$$p(x)=ax^2+a(1+k)x+a(1+2k)$$
$$t=\frac{-(1+k) \pm \sqrt{k^2-6k-3}}{2} \quad (1)$$
So
$$k^2-6k-3=q^2 \Rightarrow (k-3)^2-12=q^2 \Rightarrow (k+q-3)(k-q-3)=12$$
and split $12$ as a product of two integer and find all possible values for $k$.
Example:
\begin{cases}
k+q-3=6 \Rightarrow k+q=9 \\
k-q-3=2 \Rightarrow k-q=5
\end{cases}
Adding up both equations we get $2k=14 \Rightarrow k=7$
Backing to $(1)$ we get $(t,r)=(-3,-5)$ or $(t,r)=(-5,-3)$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/2059951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Solving $x - \frac{1}{x} < 0.$ I need to solve for x:
$$x - \frac{1}{x} < 0.$$
I realize that $x\neq 0$.
The following is always true:
$$x-\frac{1}{x} < 0 \iff x < \frac{1}{x}.$$
If $x > 0$:
$$x < \frac{1}{x} \iff x^2 < 1 \iff x < \pm 1$$
$$0<x<1$$
If $x<0$:
$$x < \frac{1}{x} \iff x^2 > 1 \iff x > \pm 1$$
$$-1<x<0.$$
So my answer is that $-1<x<1$ with $x \neq 0$. But this is not true. The real answer is $x<1$ and $0<x<1$. Why? Where is my logic wrong?
| We have:
$$x - \frac{1}{x} < 0$$
$$\frac{x^2 - 1}{x} < 0 \text{ (common denominator)}$$
$$\frac{(x-1)(x+1)}{x} < 0 \text{ (common denominator)}$$
Now you can make a chart:
And we get what we are looking for
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Use functions $f(x) =\frac{x}{8} + 1$ and $g(x) = x^3$ to find $(f o g)^-1(5)$ Right answer to the question is
$$
3\sqrt[3]{2}
$$
but i got
$$
2\sqrt[3]{4}
$$
got it by finding inverse of
$$
\frac{x^3}{8} + 1
$$
and then sustituting 5 into
$$
2\sqrt[3]{x-1}
$$
who is right?
| First of all recall that
$$\left( {{g}^{-1}}\circ {{f}^{-1}} \right)\left( x \right)={{\left( f\circ g \right)}^{-1}}\left( x \right)$$
it follows that you need to find the inverse functions for both
we have $f(x)=\frac{x}{8}+1$ then $8f(x)=x+8$ i.e $f^{-1}(x)=8(x-1)$
similarly for $g(x)=x^3$ then $g^{-1}(x)=\sqrt[3]{x}$
hence by above property we get
${{\left( f\circ g \right)}^{-1}}\left( x \right)=(g^{-1}(f^{-1}(x))=\sqrt[3]{8(x-1)}=2\sqrt[3]{x-1}$
or: you can find $$\left( {{f}}\circ {{g}} \right)\left( x \right)=f(g(x))=\frac{g(x)}{8}+1=\frac{x^3}{8}+1$$
then its easy to find its inverse function to get
$${{\left( f\circ g \right)}^{-1}}\left( x \right)=\sqrt[3]{8(x-1)}=2\sqrt[3]{x-1}$$
Seems the error is not $x=5$ to get that result
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2063034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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p-adic number for polynomial I have three polynomials:
$$f=x^5+x^4+4x^3+3x^2+3x$$
$$g=x^5+4x^3+2x^2+3x+6$$
$$p=x^2+3$$
Question: what is $v_p(fg)$?
First, I have multiplied f and g:
$h :=f \cdot g = x^{10}+x^9+8x^8+9x^7+24x^6+29x^5+36x^4+39x^3+27x^2+18x $
Then I have divided that $h$ with $x^2+3$ to get a factor to know how often $h$ contains $x^2+3 $:
$(x^2+3) \cdot (x^6+x^5+2x^4+3x^3+3x^2+2x)$
When I divide $x^6+x^5+2x^4+3x^3+3x^2+2x$ again by $x^2+3$ then I get a rest:
$(x^4+x^3-x^2+6) \cdot (x^2+3) + 2x-18$
Does that mean that $h$ contains $x^2+3$ only once and
$v_p(f \cdot g)=1$?
Is that the correct way to find the p-adic number? Or is there a better conventional one?
I appreciate every hint.
| Since $p$ is square-free, you can use $v_p(fg)=v_p(f)+v_p(g)$, which might be less work ...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Help to prove $F_n^5+F_{n+1}^5=F_{n+2}[(F_nF_{n+1}+F_{n-1}^2)^2+F_{n-1}^2F_nF_{n+1}]$ $0,1,1,2,3,5,8,...$ for $n=0,1,2,3,4...$ it is the n-th Fibonacci numbers.
$$F_n^5+F_{n+1}^5=F_{n+2}[(F_nF_{n+1}+F_{n-1}^2)^2+F_{n-1}^2F_nF_{n+1}]$$
How can we show that?
I try:
Expand and simplify to
$F_{n+2}[(F_nF_{n+1})^2+3F_{n-1}^2F_nF_{n+1}+F_{n-1}^4]=F_n^5+F_{n+1}^5$
Any hints would be helpful? Thank you!
I saw two to these identities from mathworld
$F_{n-1}F_{n+1}=F_n^2+(-1)^n$ and $F_n^4-F_{n-2}F_{n-1}F_{n+1}F_{n+2}=1$
May be these can help to simplify the above but I can't see it yet.
A hint from @Rohan
$(a+2b)[b^2(a+b)^2+3a^2b(a+b)+a^4]$
$=(a+2b)[4a^2b^2+2ab^3+3a^3b+a^4+b^4]$
$=(4a^3b^2+2a^2b^3+3a^4b+a^5+ab^4)+(8a^2b^3+4ab^4+6a^3b^2+2ba^4+2b^5)$
$=a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+2b^5$
| Hint: Consider $F_{n-1} = a, F_n =b$. Then, we get, $F_{n+1} = a+b$ and $F_{n+2} = a+2b$.
Now then the LHS, becomes $[b^5 + (a+b)^5] = a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4 + 2b^5$. Try factoring this and the result follows. Hope it helps.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is it possible to have three real numbers that have both their sum and product equal to $1$? I have to solve $ x+y+z=1$ and $xyz=1$ for a set of $(x, y, z)$. Are there any such real numbers?
Edit : What if $x+y+z=xyz=r$, $r$ being an arbitrary real number. Will it still be possible to find real $x$, $y$, $z$?
| There are infinite solutions. $z=1-x-y$ is a plane's equation and $z=\dfrac{1}{xy}$ is a complicated curve, but their intersection presents infinitely many real points.
At first I would try some trivial values like $x=0$ or $x=1$ or $x=-1$ and check which of them works and which doesn't.
For instance, if $x=0$, then $xyz=0$ doesn't work.
If $x=1$, $x+y+z=1\rightarrow y+z=0\rightarrow z=-y$ means that $xyz=1\rightarrow y^{2}=-1$ is a complex number also doesn't work.
Then $x=-1$ is a good attempt because $y+z=2$ and $yz=-1$ looks promising.
\begin{equation*}
yz=-1
\end{equation*}
\begin{equation*}
z=-\frac{1}{y}
\end{equation*}
then,
\begin{equation*}
y+z=2
\end{equation*}
\begin{equation*}
y-\frac{1}{y}=2
\end{equation*}
\begin{equation*}
y^{2}-1=2y
\end{equation*}
\begin{equation*}
y^{2}-2y-1=0
\end{equation*}
\begin{equation*}
\Delta=4+4=8
\end{equation*}
\begin{equation*}
y=\frac{2\pm2\sqrt{2}}{2}=1\pm\sqrt{2}
\end{equation*}
\begin{equation*}
\therefore z=1\mp\sqrt{2}
\end{equation*}
Thus, both $\left(-1,1-\sqrt{2},1+\sqrt{2}\right)$ and $\left(-1,1+\sqrt{2},1-\sqrt{2}\right)$ are valid answers.
I could not comment above for the lack of space. Here are some notes about their signals:
As noticed before none of the variables can be zero, because $xyz\neq0$.
Also the three variables cannot be all positive, because if $x>0$, $y>0$, $z>0$, then it would put them under the interval $0<x,y,z<1$, if one of them would be greater than $1$, for example, $x>1$ then the sum $x+y+z$ would be greater than $1$, since $x>1$, $y>0$, $ z>0$. The problem of that interval is $0<y<1$ would mean that $\dfrac{1}{y}>1$ and for the same reason $0<z<1$ would mean $\dfrac{1}{z}>1$. But $xyz=1$ would make $x=\dfrac{1}{y}\cdot\dfrac{1}{z}>1$ and that would be impossible, so the three variables cannot be all positive and they cannot be localized on that interval.
On the other hand, you cannot have just one negative variable or three negative variables because their product would be negative, so you must have exactly two negative and one positive variables.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "60",
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An Expansion Problem (Algebra) In the expansion of $(1+x)^n$, if the Coeff of $T_r$ : Coeff of $T_{r+1}$ : Coeff of $T_{r+2}$ = a:b:c, prove that $r = \frac{a(b+c)}{b^2-ac}$, where $b^2$ does not equal ac, and a, b, c $\in Z^+$
| We know that the coefficient of the $r$th term in the expansion of $(1+x)^n$ is : $\binom{n}{r-1}$. Similarly the coefficients of the $(r+1)$th and $(r+2)$th term are respectively: $\binom{n}{r} \text{and} \binom{n}{r+1}$.
Thus, we have, $$\frac{T_{r+1}}{T_r} =\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n-r+1}{r} =\frac{b}{a} ...(1)$$ $$\frac{T_{r+2}}{T_{r+1}} =\frac{\binom{n}{r+1}}{\binom{n}{r}} = \frac{n-r}{r+1} = \frac{c}{b} ...(2)$$ Then we have from $(1)$ and $(2)$, $$n = \frac{b}{a}r-(-r+1) = \frac{c}{b}(r+1) +r$$ Solving for $r$, we thus get,$$\boxed{r =\frac{a(b+c)}{b^2-ac}}$$ Hope it helps.
| {
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"url": "https://math.stackexchange.com/questions/2069301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the GCD of $10$ integers
Find the greatest common divisor of the following ten integers:
\begin{align}
& 2000^3+3\cdot2000^2+2\cdot2000,\ 2001^3+3\cdot2001^2+2\cdot2001,\ \dots, \\[5pt]
& \dots, \ 2008^3+3\cdot2008^2+2\cdot2008,\ 2009^3+3\cdot2009^2+2\cdot2009.
\end{align}
How are these numbers related to each other? Is there a formula for this?
| Let $n=2000$; the first number is $n^3+3n^2+2n$ and the second number is $(n+1)^3+3(n+1)^2+2(n+1)$, so their difference is
$$
(n+1)^3-n^3+3(n+1)^2-3n^2+2(n+1)-2n=3n^2+9n+6
$$
The third number is $(n+2)^3+3(n+2)^2+2(n+2)$ and the difference with the first number is
$$
(n+2)^3-n^3+3(n+2)^2-3n^2+2(n+2)-2n=6n^2+24n+24
$$
The gcd should be a divisor of
$$
(6n^2+24n+24)-2(3n^2+9n+6)=6n+12=6(n+2)
$$
The same computation from $n=2001$ shows the gcd must be a divisor of both $6\cdot2002$ and $6\cdot2003$, but $2002$ and $2003$ are coprime.
Hence the gcd is a divisor of $6$; now, $n^3+3n^2+2n=n(n+1)(n+2)$, so…
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2070076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Discrepancy in differentiating: $y=\tan^{-1}\left(\frac{\sqrt{1+\sin{x}}+\sqrt{1-\sin{x}}}{\sqrt{1+\sin{x}}-\sqrt{1-\sin{x}}}\right)$ $$y=\tan^{-1}\left(\frac{\sqrt{1+\sin{x}}+\sqrt{1-\sin{x}}}{\sqrt{1+\sin{x}}-\sqrt{1-\sin{x}}}\right)$$
$$\sqrt{1+\sin{x}}=\sin{\frac{x}{2}}+\cos{\frac{x}{2}}$$
$$\sqrt{1-\sin{x}}=\sin{\frac{x}{2}}-\cos{\frac{x}{2}}$$
$$y=\tan^{-1}\left(\tan{\frac{x}{2}}\right)$$Differentiating, we get: $$\frac{dy}{dx}=\frac{1}{2}$$
But taking: $$\sqrt{1-\sin{x}}=\cos{\frac{x}{2}}-\sin{\frac{x}{2}}$$
$$y=\tan^{-1}\left(\cot{\frac{x}{2}}\right)$$
Differentiating, we get: $$\frac{dy}{dx}=\frac{-1}{2}$$
I figured that I needed to use absolute value for the simplification of $\sqrt{1-\sin{x}}$, i.e. $\sqrt{1-\sin{x}}=\left|\sin{\frac{x}{2}}-\cos{\frac{x}{2}}\right|$.
Subsequently, I put the below functions into Wolfram Alpha's input box to differentiate: $$y_1=\tan^{-1}\left(\frac{\sqrt{1+\sin{x}}+\sqrt{1-\sin{x}}}{\sqrt{1+\sin{x}}-\sqrt{1-\sin{x}}}\right)$$
And
$$y_2=\tan^{-1}\left(\frac{\left|\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right|+\left|\sin{\frac{x}{2}}-\cos{\frac{x}{2}}\right|}{\left|\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right|-\left|\sin{\frac{x}{2}}-\cos{\frac{x}{2}}\right|}\right)$$
The answers should ideally match, but they dont. $\frac{dy_1}{dx}=\frac{-1}{2}\neq\frac{dy_2}{dx}$
Why don't they?
| At $x = 0$, $\sqrt{1-\sin x} = 1$, this means $\sqrt{1-\sin x} = \cos\frac{x}{2} - \sin\frac{x}{2}$ instead of $\sin\frac{x}{2} - \cos\frac{x}{2}$.
For $x \approx 0$, more precisely, $x \in (-\frac{\pi}{2},\frac{\pi}{2})$,
we have $\frac{dy}{dx} = -\frac12$.
Notice $\sqrt{1-\sin x}$ is not smooth at the boundary of $(-\frac{\pi}{2},\frac{\pi}{2})$ and it develop cusps there.
When one start increasing $x$ from $0$ to beyond $\pi/2$, above formula for $\sqrt{1-\sin x}$ start to fail at $x = \frac{\pi}{2}$. Instead, we have $\sqrt{1-\sin x} = \sin\frac{x}{2} - \cos\frac{x}{2}$ for $x \in ( \frac{\pi}{2},\pi )$. This means $\frac{dy}{dx} = \frac12$ over that interval.
In general, $y(x)$ is a piecewise linear function with jump discontinuities at
integer multiple of $\pi$. Its derivatives has jump discontinuities at
half-integer mulitples of $\pi$. To summarize,
$$\frac{dy(x)}{dx} =
\begin{cases}
\text{undefined}, & \frac{2x}{\pi} \text{ is a integer}\\
-\frac12, & {\rm round}(\frac{x}{\pi}) \text{ is even}\\
+\frac12, & {\rm round}(\frac{x}{\pi}) \text{ is odd}.
\end{cases}
$$
In certain sense, both answer $\pm \frac12$ are right and both are wrong. Both are right because each of them are valid over some subsets of $\mathbb{R}$. Both are wrong because $\frac{dy}{dx}$ is not a constant function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2071186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $ax^2+bx+c=0$ has rational roots then $(a+1)x^2+bx+c=0$ cannot have rational roots I was working on quadratic equations and found the following fact that:
"If $ax^2+bx+c=0$ has rational roots then $(a+1)x^2+bx+c=0$ cannot have rational roots where $a,b,c\in\mathbb N$"
And now I have to prove or disprove it
And I know the first comment will be "show your efforts"
So I have assumed if a quadratic equation have rational roots, then $b^2-4ac$ should be a perfect square.
Applying to both equation I got $k^2-4c=l^2$ for some integer $k$ & $l$.
How to proceed now??
Please help!!
| I believe I can show that there are infinite many values of $a$, $b$ and $c$ that satisfy the conditions of your question. We know that
$$x_{1;2}=\frac{-b\pm\sqrt{\Delta}}{2a}=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
is the formula for the two solutions $x_1$and $x_2$ of the second-degree equation $ax^2+bx+c=0$. The equation has rational roots if $\Delta_1 =b^2-4ac$ is a square number.
With $(a+1)x^2+bx+c=0$ we have that $\Delta_2=b^2-4ac-4c$ which also has to be a square number (otherwise the solutions wouldn't be rational). Basically we can say that $\Delta_1$ and $\Delta_2$ are two square numbers that differ by a multiple of 4.
Note that every two number that differ by a multiple of $2$, squared will differ by a factor of $4$; for example $3^2$ and $5^2$ differ by a multiple of $4$.
With these assumptions we can construct $\Delta_1$ and $\Delta_2$ such that their difference is a multiple of $4$ and are both square numbers. We can set, as an example, $\Delta_1=49$; it follows that $\Delta_2 = 25$, they are two square numbers that differ by a multiple of $4$. Since $\Delta_1 - \Delta_2=4c$
$$\Delta_1 - \Delta_2=4c;\ 4c=49-25 \rightarrow c=6$$
to obtain the other values we use the equation
$$\Delta_1= b^2-4ac=b^2-16c=49$$
Which can be satisfied by infinite values of $a$ and $b$; if we want $b$ to be a natural number we can see that if $b^2=64$ then
$$64-40a=49 \rightarrow a=\frac{64-49}{40}=\frac58$$
In the end we have $a=\frac58$, $b=8$ and $c=6$
$$\frac58 x^2+8x+6=0 \rightarrow x_1=-\frac{4}5, x_2=-12$$
$$\left(\frac58 +1\right)x^2+8x+6=0 \rightarrow x_1=-\frac{12}{13}, x_2=-4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2073721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How can I isolate $x$ in this equation? Simultaneous equations - $7^{x-1} = 49 \hspace{1 mm} (7^{y})$ --------- (i)
$\hspace{46 mm} 3^x+3^y = 84$ ---------- (ii)
The first I make x subject of equation in this way:
$\Longrightarrow 7^{x-1} = 49 \hspace{1 mm} (7^{y})$
$\Longrightarrow 7 \times 7^x = 7^2 \times 7^y$
$\Longrightarrow 7^x = 7^{y+1}$
$\Longrightarrow x = y + 1$
The second I tried with logarithms but could not separate x in simple form:
$\Longrightarrow 3^x + 3^y = 84$
$\Longrightarrow 3^x = 84 - 3^y$
$\Longrightarrow \lg3^x = \lg (84 - 3^y)$
$\Longrightarrow x \lg3 = \lg(84 - 3^y)$
$\Longrightarrow x = \dfrac{\lg(84 - 3^y)}{\lg3}$
How can I simplify this further?
| You have made an error in the first equation:
$\Rightarrow 7^{x-1}=49\hspace{1 mm}(7^{y})$
$\Rightarrow \dfrac{7^{x}}{7}=49\hspace{1 mm}(7^{y})$
$\Rightarrow 7^{x}=7^{3}\hspace{1 mm}(7^{y})$
$\Rightarrow 7^{x}=7^{y+3}$
$\Rightarrow x=y+3$
Then, we can substitute this expression for $x$ into the second equation:
$\Rightarrow 3^{x}+3^{y}=84$
$\Rightarrow 3^{y+3}+3^{y}=84$
$\Rightarrow 27\cdot{3^{y}}+3^{y}=84$
$\Rightarrow 28\cdot{3^{y}}=84$
$\Rightarrow {3^{y}}=3^{1}$
$\Rightarrow y=1$
Now, let's substitute this value of $y$ into the first equation:
$\Rightarrow x=(1)+3$
$\hspace{8.75 mm}=4$
Therefore, $x=4$ and $y=1$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $f(x) = ax^2$ For some function $f \in C(R)$ the following equality holds: $f(x) + f(y) = f(\sqrt{x^2 + y^2})$ for any $x, \ y \in R$.
Prove that $f(x) = ax^2 \ \forall x\in \mathbb{R}$, where $a = f(1)$.
| The functional equation
$$
f(x) + f(y) = f \left( \sqrt{x^2 + y^2} \right)
$$
implies
$$
f(0) + f(0) = f \left( \sqrt{0^2 + 0^2} \right) = f(0)
$$
so $f(0) = 0$. Further
$$
f(-\lvert x \vert) = f(-\lvert x \vert) + f(0)
= f \left( \sqrt{(-\lvert x \rvert)^2 + 0^2} \right) = f(\lvert x \rvert)
$$
which means $f$ is symmetric regarding the $y$-axis:
$$
f(x) = f(-x)
$$
So we can write
\begin{align}
f(x) + f(y) &= f(\lvert x \rvert) + f(\lvert y \rvert) \\
&= f(\sqrt{X}) + f(\sqrt{Y}) \\
&= f(\sqrt{X + Y}) \\
\end{align}
so $g = f \circ \sqrt{.}$ is subject to Cauchy's functional equation
$$
g(x+y) = g(x) + g(y) \quad (x,y \in \mathbb{R}, x,y \ge 0)
$$
which over the reals has many interesting solutions, among them the family
$$
g(x) = a x
$$
with $a \in \mathbb{R}$. If a continuous solution is requested, which is the case here, it is the solution. This means
$$
f(\sqrt{X}) = a X \iff \\
f(x) = a x^2
$$
for non-negative $x$. It is symmetric and of course $f(1) = a$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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If $a+b+c = 6$ and $a$, $b$, $c$ are nonnegative then $a^2+b^2+c^2 \geq 12$ Let $a,b,c$ be three positive real numbers such that $a+b+c = 6$. Prove that $a^2+b^2+c^2 \geq 12$.
I tried using the AM-GM inequality to solve the same, however I wasn't able to make any considerable progress.
| Proof:$$a^2+b^2\geq 2ab$$so:$$3(a^2+b^2+c^2)=a^2+b^2+c^2+2(a^2+b^2+c^2)\geq a^2+b^2+c^2+2(ab+bc+ac)=(a+b+c)^2=36$$Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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maximum value of $|a-b||b-c|+|b-c||c-a|+|c-a||a-b|$ is If $a,b,c$ are complex numbers and $|a|=|b| = |c| = 2$
finding maximum value of $|a-b||b-c|+|b-c||c-a|+|c-a||a-b|$ is
$|a-b|^2=|a|^2+|b|^2-2(a\cdot b) = 8-8\cos 2A = 16\sin^2 A$
$|b-c|^2=|b|^2+|c|^2-2(b\cdot c) = 8-8\cos 2B = 16\sin^2 B$
$|a-c|^2=|a|^2+|c|^2-2(a\cdot c) = 8-8\cos 2C = 16\sin^2 C$
maximum value of $16(\sin A\sin B+\sin B\sin C+\sin C\sin A)$
$2A,2B,2C$ are angle between vectors $a,b$ and $b,c$ and $c,a$
i want be able to proceed after that, could some help me
| $a,b,c$ are points on a circle of radius $2$, so $A=|b-c|,B=|c-a|,C=|a-b|$ are lengths of the sides of the triangle formed between them. Let the angles subtended at the origin by the chords $A,B,C$ be $\alpha,\beta,\gamma$ respectively.
By Cauchy–Schwarz,
$$AB+BC+CA\ \le\ \sqrt{A^2+B^2+C^2}\sqrt{B^2+C^2+A^2}\ =\ A^2+B^2+C^2$$
By the cosine rule,
$$A^2\ =\ 2^2+2^2-2\cdot2\cdot2\cos\alpha\ =\ 8-8\cos\alpha$$
Similarly $B^2=8-8\cos\beta$ and $C^2=8-8\cos\gamma$. Hence
$$AB+BC+CA\ \le\ A^2+B^2+C^2\ =\ 24-8(\cos\alpha+\cos\beta+\cos\gamma)$$
Now we have $\cos\alpha+\cos\beta+\cos\gamma\ge-\dfrac32$ (see below); hence
$$AB+BC+CA\ \le\ 24-8(\cos\alpha+\cos\beta+\cos\gamma)\ \le\ 24+8\cdot\frac32=36$$
So the maximum value is $36$, attained when, e.g.
$$a=2,\ b=-1+i\sqrt3,\ c=-1-i\sqrt3$$
To prove that $\cos\alpha+\cos\beta+\cos\gamma\ge-\dfrac32$, we have $\alpha+\beta+\gamma=2\pi$ $\implies$ $\cos\gamma=\cos(\alpha+\beta)=2\cos^2\dfrac{\alpha+\beta}2-1$. Also $\cos\alpha+\cos\beta=2\cos\dfrac{\alpha+\beta}2\cos\dfrac{\alpha-\beta}2$. So we want to prove
$$2\cos^2\frac{\alpha+\beta}2+2\cos\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2-1\ \ge\ -\frac32$$
$$\iff\quad4\cos^2\frac{\alpha+\beta}2+4\cos\frac{\alpha+\beta}2\cos\frac{\alpha-\beta}2+1\ \ge\ 0$$
$$\iff\quad\left(2\cos\frac{\alpha+\beta}2+\cos\frac{\alpha-\beta}2\right)^2+1\ \ge\ \cos^2\frac{\alpha-\beta}2$$
The last inequality must be true otherwise $\cos^2\dfrac{\alpha-\beta}2 > \left(2\cos\dfrac{\alpha+\beta}2+\cos\dfrac{\alpha-\beta}2\right)^2+1 \ge 1$.
| {
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Factorization of $(1+2i)z^2+6iz+2i-1$ to find the residues, but it doesn't work I'm doing complex analysis and I have the following $$\int_{C}\frac{2dz}{(2i+1)z^2+6iz+2i-1}$$ where $C$ is the unit circle. I tried factorizing this polynomial and find the residues but I really can't solve this.
This is my work:
$$z_{1,2} = \frac{-3i\pm \sqrt{-9-(-4-1)} }{2i+1}=\frac{-3i\pm 2i}{1+2i} \cdot \frac{1-2i}{1-2i}= \frac{-3i-6\pm 2i\pm 4}{5}$$ which gives $z_1 = -\frac{2}{5}-\frac{i}{5}$ and $z_2 = -2-i$. Now clearly $z_2$ is outside the unit circle so I just have to consider the residue of $z_1$.
However I think I'm either making a factorisation mistake or a mistake in the roots. Indeed if we check the roots, I don't get the same polynomial! $$(z+2+i)\left(z+\frac{2}{5}+\frac{i}{5}\right) = z^2 +z\left(\frac{12}{5}+\frac{6i}{5}\right)+\frac{4i}{5}+\frac{3}{5}$$ which is clearly not the same polynomial I started with. Also if I try to calculate the residue using the limit, I don't get the correct one. Where's my mistake?
EDIT
As requested, I'll add some more calculations. To calculate the residue I did the following: $$\lim_{z\to z_1}(z+\frac{2}{5}+\frac{i}{5})\frac{2}{(z+2+i)\left(z+\frac{2}{5}+\frac{i}{5}\right)} = \frac{2}{\frac{8}{5}+\frac{4i}{5}} = \frac{5}{4+2i} = \frac{20-10i}{20}=1-\frac{i}{2}$$ which is clearly wrong as this integral is the mapping of a real integral of the first type (rational function in sine and cosine). Indeed by Cauchy's Residue Theorem this should give that the value of the initial real integral is $2\pi i (1-\frac{i}{2}) = 2\pi i+\pi$ which is a complex value, so this is wrong
| Let's start with the factorization question. The given integrand is a fraction, and to use the residue theorem, we must find the roots of the denominator, i.e., the roots of
$$
(2i+1)z^2+6iz+(2i-1).
$$
To find the roots of this polynomial, it's easiest to use the quadratic formula to get that the roots are
\begin{align*}
\frac{-6i\pm\sqrt{(6i)^2-4(2i+1)(2i-1)}}{2(2i+1)}&=\frac{-6i\pm\sqrt{-36+20}}{4i+2}\\
&=\frac{-6i\pm\sqrt{(6i)^2-4(2i+1)(2i-1)}}{2(2i+1)}\\
&=\frac{-6i\pm\sqrt{-16}}{4i+2}\\
&=\frac{-6i\pm4i}{4i+2}
\end{align*}
Therefore, the roots are $\frac{-2i}{4i+2}=\frac{-i}{2i+1}$ and $\frac{-5i}{2i+1}$. Multiplying through by the conjugate of the denominator, $(1-2i)$ gives that the roots are $\left(-\frac{2}{5}-\frac{1}{5}i\right)$ and $(-2-i)$.
As the OP observes, the polynomial
$$
\left(z-\left(-\frac{2}{5}-\frac{1}{5}i\right)\right)(z-(-2-i))=z^2+\left(\frac{12}{5}+\frac{6}{5}i\right)z+\left(\frac{3}{5}+\frac{4}{5}i\right)
$$
is not the original denominator. This is another polynomial with the same roots, as the original polynomial. To make the polynomials the same, we should multiply by $(1+2i)$ to correct the coefficient of $z^2$. In fact
$$
(1+2i)\left(z^2+\left(\frac{12}{5}+\frac{6}{5}i\right)z+\left(\frac{3}{5}+\frac{4}{5}i\right)\right)=(2i+1)z^2+6iz+(2i-1),
$$
the original polynomial.
Therefore, the original polynomial factors as
$$
(2i+1)z^2+6iz+(2i-1)=(2i+1)\left(z-\left(-\frac{2}{5}-\frac{1}{5}i\right)\right)(z-(-2-i)).
$$
To deal with the residues, the OP correctly determines that $-2-i$ is outside the unit circle, so it doesn't matter for the integral (although it could be used to calculate the integral, but that is a story for another time).
Therefore, the value of the given integral is $2\pi i$ times the residue at $\left(-\frac{2}{5}-\frac{1}{5}i\right)$. Rewriting the integral as
$$
\int_\gamma \frac{2}{(1+2i)(z-(-2-i))}\cdot\frac{1}{z-\left(-\frac{2}{5}-\frac{1}{5}i\right)}dz,
$$
the value of the residue can be calculated by substituting $\left(-\frac{2}{5}-\frac{1}{5}i\right)$ for $z$ in $\frac{2}{(1+2i)(z-(-2-i))}$. This substitution results in $-\frac{1}{2}i$ and Cauchy's residue theorem gives that the integral is $\pi$.
The only mistake that the OP seems to make throughout is forgetting the factor of $(1+2i)$ when factoring the denominator. When that is included, all of the answers (and the limit approach in the question) agree.
| {
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Points of Intersection of Curves There are $4$ distinct parabolas, $5$ distinct circles and $3$ distinct lines in the $x-y$ plane. Find the maximum possible value of the number of their points of intersection.
My Attempt at the Question:
I first to try and figure out the maximum number of points of intersection. For that I figured that parabola and circle intersect at max. $4$ points. And straight line will intersect the circle and parabola at $4$ distinct points. But this lead me nowhere.
| First we find the kinds of intersection that take place b/w the lines, circles and parabolas, which are as follows:-
*
*Parabola with a parabola
*Parabola with a circle
*Parabola with a line
*Circle with a circle
*Circle with a line
*Line with a line
Now, lets count the maximum number of point of intersection in each of the above cases and consequently count the maximum number of intersections that cantake place given that there are $4$ distinct parabolas, $5$ distinct circles and $3$ distinct lines:-
CASE 1:- Intersection of two parabolas
As we can see in the above image there are maximum $4$ possible points of intersection for two distinct parabolas. Since there are $4$ distinct parabolas hence there are maximum $4\cdot\binom{4}{2}$ maximum distinct point of intersections.
$\therefore$ Maximum number of points of intersection of $4$ distinct parabola$=4\cdot\displaystyle\binom{4}{2}=24$
CASE 2:- Intersection of parabola with a circle
As can be seen in the above image there are at max. $4$ distinct points of intersection of a parabola and a circle.
Ways to select a parabola and a circle$= \displaystyle\binom{4}{1}\cdot\binom{5}{1}$
$\therefore$ Max. no. of distinct points of intersection$=4\cdot\displaystyle\binom{4}{1}\cdot\binom{5}{1}=80$
CASE 3:- Intersection of a parabola with a line
As can be seen in the above image there can be max. $2$ distinct points of intersection b/w a line and a parabola.
Ways to select a parabola and a line $=\displaystyle\binom{4}{1}\cdot\binom{3}{1}$
$\therefore$ Max. no. of distinct points of intersection$=2\cdot\displaystyle\binom{4}{1}\cdot\binom{3}{1}=24$
CASE 4:- Intersection of two circles
As can be seen in the above image there can be max. $2$ distinct points of intersection b/w two circles.
Ways to select two circles $=\displaystyle\binom{5}{2}$
$\therefore$ Max. no. of distinct points of intersection$=2\cdot\displaystyle\binom{5}{2}=20$
CASE 5:- Intersection of a circle with a line
As can be seen in the above image there can be max. $2$ distinct points of intersection b/w a line and a circle.
Ways to select a circle and a line $=\displaystyle\binom{5}{1}\cdot\binom{3}{1}$
$\therefore$ Max. no. of distinct points of intersection$=2\cdot\displaystyle\binom{5}{1}\binom{3}{1}=30$
CASE 6:- Intersection of two lines
As can be seen in the above image there can be max. $1$ distinct points of intersection b/w two lines.
Ways to select two lines $=\displaystyle\binom{3}{2}$
$\therefore$ Max. no. of distinct points of intersection$=1\cdot\displaystyle\binom{3}{2}=3$
Since all the above cases are disjoint, and when anyone of them happens an intersection takes place, hence the total no of intersections are the sum of all the disjoint events.
Hence, total number of points of intersection is given by
$$4\cdot\displaystyle\binom{4}{2}+ 4\cdot\displaystyle\binom{4}{1}\cdot\binom{5}{1} + 2\cdot\displaystyle\binom{4}{1}\cdot\binom{3}{1} + 2\cdot\displaystyle\binom{5}{2} + 2\cdot\displaystyle\binom{5}{1}\binom{3}{1} + 1\cdot\displaystyle\binom{3}{2}\\
=24+80+24+20+30+3=\boxed{181}$$
| {
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How do I express this expression using partial fractions? The expression I'm trying to express using partial fractions is: $\frac{64x}{(2x-1)^2(2x+1)^2}$
What I've tried so far:
$\frac{64x}{(2x-1)^2(2x+1)^2}=\frac{A}{(2x-1)^2}+\frac{B}{2x-1}+\frac{C}{(2x+1)^2}+\frac{D}{2x+1}$ where $A,B,C,D$ are constants
Multiplying each term by $(2x-1)^2(2x+1)^2$ gives
$64x=A(2x+1)^2+B(2x-1)(2x+1)^2+C(2x-1)^2+D(2x+1)(2x-1)^2$
Now:
$x=-\frac{1}{2} \implies -32=4C \implies C=-8$
$x=\frac{1}{2} \implies 32=4A \implies A=8$
$x=0 \implies 0=A-B+C+D \implies B=D$
$x=1 \implies 64=9A+9B+C+3D \implies 0=3B+D$
Solving the two simultaneous equations yields $B=D=0$
Hence:
$\frac{64x}{(2x-1)^2(2x+1)^2}=\frac{8}{(2x-1)^2}-\frac{8}{(2x+1)^2}$
But we know this is wrong since, checking with $x=1$ gives:
$\frac{64}{9}=\frac{1}{9}$ which is obviously wrong.
So my questions are:
(1) What method(s) can we use to express this in partial fractions?
(2) What went wrong above?
| I'm not sure how you get $\frac19$ when you checked for $x=1$. Notice that
$$\frac8{(2(1)-1)^2}=\frac8{(2-1)^2}=\frac81=8$$
$$\frac8{(2(1)+1)^2}=\frac8{(2+1)^2}=\frac89$$
Thus,
$$\frac8{(2(1)-1)^2}-\frac8{(2(1)+1)^2}=8-\frac89=\frac{64}9$$
Similarly,
$$\frac{64(1)}{(2(1)-1)^2(2(1)+1)^2}=\frac{64}{1^23^2}=\frac{64}9$$
Just as suspected.
As far as I can tell, your partial fractions are done correctly.
| {
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$a^4+b^4+c^4+d^4=4\implies\sum\limits_{cyc}\frac{a^3}{bc}\geq4$ Let $a$, $b$, $c$ and $d$ be positive numbers such that $a^4+b^4+c^4+d^4=4$. Prove that:
$$\frac{a^3}{bc}+\frac{b^3}{cd}+\frac{c^3}{da}+\frac{d^3}{ab}\geq4$$
I tried C-S, BW and more, but without success.
| We know that $$a^8 b^4 c^4 + b^8 c^4 d^4 + c^8 d^4 a^4+d^8 a^4 b^4 \leq 4.$$ (see this question for a proof)
By the generalized weighted mean inequality we have:
\begin{align}
\left(\frac{a^8 b^4 c^4 + b^8 c^4 d^4 + c^8 d^4 a^4+d^8 a^4 b^4}{a^4+b^4+c^4+d^4}\right)^{\frac{1}{4}} &\geq \frac{a^5bc+b^5cd+c^5da+d^5ab}{a^4+b^4+c^4+d^4}
\\\\ \Rightarrow 4 \left(\frac{a^8 b^4 c^4 + b^8 c^4 d^4 + c^8 d^4 a^4+d^8 a^4 b^4}{4}\right)^{\frac{1}{4}} &\geq a^5bc+b^5cd+c^5da+d^5ab
\\\\ \Rightarrow 4 &\geq a^5bc+b^5cd+c^5da+d^5ab
\\\\ \Rightarrow \frac{a^4+b^4+c^4+d^4}{a^5bc+b^5cd+c^5da+d^5ab} &\geq 1.
\end{align}
Let $f\colon \mathbb R_{> 0} \rightarrow \mathbb R$ be defined by $f(x):= \frac{1}{x}$. Since $f$ is convex, we have by Jensen's inequality:
\begin{align}
\frac{1}{4}\sum\limits_{cyc} \frac{a^3}{bc}
&=\sum\limits_{cyc} \frac{a^4}{a^4+b^4+c^4+d^4} f(a b c)
\\&\geq f\left(\sum\limits_{cyc}\frac{a^5 b c}{a^4+b^4+c^4+d^4}\right)
\\&= \frac{a^4+b^4+c^4+d^4}{a^5bc+b^5cd+c^5da+d^5ab}
\\&\geq 1.
\end{align}
| {
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Prove that $\frac{bc}{a(a-b)(a-c)}+\frac{ca}{b(b-c)(b-a)}+\frac{ab}{c(c-a)(c-b)}=\frac{ab+bc+ca}{abc}$
Prove that $$\frac{bc}{a(a-b)(a-c)}+\frac{ca}{b(b-c)(b-a)}+\frac{ab}{c(c-a)(c-b)}=\frac{ab+bc+ca}{abc}$$
To do this I unified three terms on the left side with a common denominator and then factored the numerator (with the aid of Wolfram Alpha... as the numerator is a 5th order expression).
$$\frac{b²c²(c-b)+c²a²(a-c)+a²b²(b-a)}{abc(a-b)(b-c)(c-a)}=\dots=\frac{(ab+bc+ca)(a-b)(b-c)(c-a)}{abc(a-b)(b-c)(c-a)}$$
My question is would there be a better or easier way to lead from the left side to the right side? Actually, the original question was to 'simplify' the left side without showing the right side. I am not sure if I could do so without Wolfram Alpha. So that's why I'm asking an easier way.
What I've thought of was, for example,
$$f(x):=(x-a)(x-b)(x-c)$$
$$\text{LHS}=abc\left(\frac1{a^2f'(a)}+\frac1{b^2f'(b)}+\frac1{c^2f'(c)}\right)$$
which didn't help.
| We have to prove that
$$ \sum_{cyc}\frac{bc}{a(a-b)(a-c)}=\sum_{cyc}\frac{1}{a} $$
or:
$$ \sum_{cyc}\frac{bc-(a-b)(a-c)}{a(a-b)(a-c)} = \sum_{cyc}\frac{b+c-a}{(a-b)(a-c)}=0.$$
Since
$$ 0 = \sum_{cyc}\left(\frac{1}{a-b}+\frac{1}{a-c}\right)=\sum_{cyc}\frac{2a-b-c}{(a-b)(a-c)} $$
it is enough to prove that
$$ \frac{a}{(a-b)(a-c)}+\frac{b}{(b-c)(b-a)}+\frac{c}{(c-a)(c-b)}=0$$
or, by multiplying both sides by $(a-b)(a-c)(b-c)$,
$$ a(b-c)-b(a-c)+c(a-b) = 0 $$
that is trivial.
| {
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Prove something similar to a variant of Cauchy-Shwarz inequality Cauchy-Shwarz Inequality is:
$$(a_1b_1 + a_2b_2 + \cdots + a_nb_n)^2 \leq (a_1^2 + a_2^2 + \cdots + a_n^2)(b_1^2 + \cdots + b_n^2)$$
However, it can be manipulated as:
$$\sqrt{(a_1-b_1)^2 + (a_2-b_2)^2 + \cdots + (a_n-b_n)^2} \leq \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2} + \sqrt{b_1^2 + b_2^2 + \cdots + b_n^2}$$
I'm tasked with proving the following inequality:
$$\sqrt{(a_1 + b_1 + \cdots + z_1)^2 + (a_2 + b_2 + \cdots + z_2)^2 + \cdots + (a_n + b_n \cdots + z_n)^2} \leq \sqrt{a_1^2 + a_2^2 + \cdots + a_n^2} + \sqrt{b_1^2 + b_2^2 + \cdots + b_n^2} + \cdots + \sqrt{z_1^2 + z_2^2 + \cdots + z_n^2}$$
I've proved both Cauchy-Shwarz and its manipulation, but am lost when it come to the inequality right above. Hints and/or solutions are welcome.
| A geometrical interpretation would be the following:
Consider an $m$-dimensional box and subdivide it in $n\times n\times\dots\times n$ grid. Then its diagonal length is at most the sum of the diagonals in a "diagonal chain" of boxes of the grid.
| {
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How to find the integral $\int{\frac{1}{\sin^2(z)-\frac{1}{4}}}dz$? I need to integrate the following function:$$\frac{\cos(x)}{1+\sin(x)}\sqrt{\frac{2-\sin(x)}{2+\sin(x)}}$$
For this I use the substitution $2+\sin(x)=y^2$
I get $$2\int{\frac{\sqrt{4-y^2}}{y^2-1}} dy$$.
Then I use $y=2\sin(z)$.
So I get:
$$I=8\int\frac{{\cos^2(z)}}{4\sin^2(z)-1}dz$$
$$\implies I=2\left[\frac{\cos^2(z)}{\sin^2(z)-\frac{1}{4}}\right]$$
$$-2z+\frac{3}{2}\int{\frac{1}{\sin^2(z)-\frac{1}{4}}}dz + C $$
After this, how to integrate $$\int{\frac{1}{\sin^2(z)-\frac{1}{4}}}dz$$ (without using any hypergeometric functions) ? You may use logarithmic functions.
| Well, multiply numerator and denominator by $4\csc^2(z)$:
$$\int\frac{1}{\sin^2(z)+\frac{1}{4}}\space\text{d}z=\int\frac{4\csc^2(z)}{4\csc^2(z)\left(\sin^2(z)+\frac{1}{4}\right)}\space\text{d}z=\int\frac{4\csc^2(z)}{\cot^2(z)+5}\space\text{d}z$$
Now, substitute $u=\cot(z)$:
$$\int\frac{4\csc^2(z)}{\cot^2(z)+5}\space\text{d}z=-4\int\frac{1}{u^2+5}\space\text{d}u=\text{C}-\frac{4\arctan\left(\frac{u}{\sqrt{5}}\right)}{\sqrt{5}}$$
| {
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Understanding the proof of $\displaystyle\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$ I need help understanding this:
Prove that $$\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$$
$$(\sqrt n)^\frac{1}{n}\geq
(\sqrt1)^\frac{1}{n}=1, \forall n\in\Bbb N$$
By binomial theorem we get for $n\geq 2$
$$n=((\sqrt n)^\frac{1}{n})^n=[1+((\sqrt n)^\frac{1}{n}-1)]^n=\sum_{k=0}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k$$
$$\geq 1+{{n}\choose {2}}(\sqrt n^\frac{1}{n}-1)^2=1+\frac{n(n-1)}{2}(\sqrt n^\frac{1}{n}-1)^2$$
$$\Rightarrow (\sqrt n^\frac{1}{n}-1)^2\leq \frac{2}{n}$$
$$\Rightarrow \sqrt n^\frac{1}{n}\leq 1+\frac{\sqrt 2}{\sqrt n}$$
$1+\frac{\sqrt 2}{\sqrt n}$ approaches $1$ for $n \rightarrow \infty$ so by the sandwich theorem we get $\lim_{n\to\infty}(\sqrt n)^\frac{1}{n}=1$
I don't understand this part:
$$
\sum_{k=0}^{n}{{n}\choose{k}}1^{n-k}(\sqrt n^\frac{1}{n}-1)^k\geq 1+{{n}\choose {2}}(\sqrt n^\frac{1}{n}-1)^2.$$ Could someone explain how they got that on the RHS?
| I'd like to offer another method, using $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x^{2}+x+1)$
$$n-1=(n^{\frac{1}{2n}})^{2n}-1=\\(n^{\frac{1}{2n}}-1)(n^{\frac{1}{2n}(2n-1)}+n^{\frac{1}{2n}(2n-2)}+...+n^{\frac{1}{2n}2}+n^{\frac{1}{2n}}+1)\geq$$
using AM-GM
$$\geq(n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt[2n-1]{n^{\frac{1}{2n}(2n-1)}\cdot n^{\frac{1}{2n}(2n-2)}\cdot ...\cdot n^{\frac{1}{2n}2}\cdot n^{\frac{1}{2n}}}+1\right)=\\
(n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt[2n-1]{n^{\frac{1}{2n}(2n-1+2n-2+...+1)}}+1\right)=
\\(n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt[2n-1]{n^{\frac{1}{2n}\frac{(2n-1)2n}{2}}}+1\right)=\\
(n^{\frac{1}{2n}}-1)\left((2n-1)\sqrt{n}+1\right)$$
Also $\ln{x}\leq x-1,x>0$ which means $$\ln{n^{\frac{1}{2n}}}=\frac{\ln{n}}{2n}\leq n^{\frac{1}{2n}}-1$$
Altogether
$$\frac{\ln{n}}{2n}\leq n^{\frac{1}{2n}}-1\leq \frac{n-1}{(2n-1)\sqrt{n}+1}<\frac{1}{2\sqrt{n}}$$
| {
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Integer solutions for $a+b+c+d=0$, $a^3+b^3+c^3+d^3=24$ How to get all integer solutions for $$a+b+c+d=0,$$ and $$a^3+b^3+c^3+d^3=24$$
So far I've put $a=-b-c-d$ into 2nd equation and try to factorise it, but didn't find anything useful.
| Since $d=-a-b-c$, you can check the following identity:
$$a^3+b^3+c^3+d^3=-3(a+b)(a+c)(b+c).$$
Thus $$(a+b)(a+c)(b+c)=-8,$$
and that easily gives you all integer solutions. For instance, the case
\begin{cases}
a+b=1 \\
a+c=1\\
b+c=-8
\end{cases}
gives you $(a,b,c,d)=(5,-4,-4,3),$ and the case
\begin{cases}
a+b=-2 \\
a+c=-2\\
b+c=-2
\end{cases}
gives you $(a,b,c,d)=(-1,-1,-1,3).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2087170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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For $abc=1$ prove that $\sum\limits_\text{cyc}\frac{1}{a+3}\geq\sum\limits_\text{cyc}\frac{a}{a^2+3}$ Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that:
$$\frac{1}{a+3}+\frac{1}{b+3}+\frac{1}{c+3}\geq\frac{a}{a^2+3}+\frac{b}{b^2+3}+\frac{c}{c^2+3}$$
I tried TL, BW, the Vasc's Theorems and more, but without success.
I proved this inequality!
I proved also the hardest version: $\sum\limits_{cyc}\frac{1}{a+4}\geq\sum\limits_{cyc}\frac{a}{a^2+4}$.
Thanks all!
| BW in the following version does not help.
Let $a=x^3$, $b=y^3$ and $c=z^3$.
Hence, we need to prove that
$$\sum_{cyc}\frac{1}{x^3+3xyz}\geq\sum_{cyc}\frac{x^3}{x^6+3x^2y^2z^2}$$ or
$$\sum_{cyc}\frac{1}{x^3+3xyz}\geq\sum_{cyc}\frac{x}{x^4+3y^2z^2}.$$
Now, we can assume that $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$
and these substitutions give inequality, which I don't know to prove.
But we can use another BW!
Let $a=\frac{y}{x}$, $b=\frac{z}{y}$ and $c=\frac{x}{z}$, where $x$, $y$ and $z$ are positives.
Hence, we need to prove that
$$\sum_{cyc}\frac{x}{3x+y}\geq\sum_{cyc}\frac{xy}{3x^2+y^2}$$ or
$$\sum_{cyc}\frac{x^3-x^2y}{(3x+y)(3x^2+y^2)}\geq0.$$
Now, let $x=\min\{x,y,z\}$, $y=x+u$ and $z=x+v$.
Hence, we need to prove that
$$128(u^2-uv+v^2)x^7+16(16u^3+23u^2v-15uv^2+16v^3)x^6+$$
$$+32(8u^4+27u^3v+12u^2v^2-11uv^3+8v^4)x^5+$$
$$+4(32u^5+193u^4v+266u^3v^2-42u^2v^3-33uv^4+32v^5)x^4+$$
$$+2(8u^6+178u^5v+435u^4v^2+152u^3v^3-99u^2v^4+30uv^5+8v^6)x^3+$$
$$+uv(45u^5+375u^4v+291u^3v^2-83u^2v^3+57uv^4+3v^5)x^2+$$
$$+2u^2v^2(24u^4+66u^3v-18u^2v^2+13uv^3+3v^4)x+$$
$$+u^3v^3(18u^3-6u^2v+3uv^2+v^3)\geq0,$$
which is obvious.
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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} |
Calculate $\int_0^{2\pi} \frac{1}{2+\sin(x)} \ dx$ $$\int_0^{2\pi} \frac{1}{2+\sin(x)} \ dx$$
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
$$z=\alpha(x)=e^{ix}$$
$$\sin(x)=\frac{z^2-1}{2zi}$$
$$2+\sin(x)=\frac{4zi+z^2-1}{2zi}$$
$$\frac{1}{2+\sin(x)}=\frac{2zi}{4zi+z^2-1}$$
$$\int_\alpha \frac{2zi}{4zi+z^2-1} \ \ \frac{1}{zi}$$
$$4zi+z^2-1=0$$
$$z_1=\frac{-4i+2\sqrt{3}i}{2}$$
$$z_2=\frac{-4i-2\sqrt{3}i}{2}$$
$$\rvert z_1 \rvert <1$$
$$\rvert z_2 \rvert >1$$
Using Residue theorem:
$$\int_\alpha \frac{2}{4zi+z^2-1}=2 \pi i \ \lim_{z\rightarrow-2i+\sqrt{3}i} \ \frac{2}{z+2i+\sqrt{3}i}=\frac{2 \pi}{\sqrt{3}}$$
Is it correct?
Thanks!
| Another way is
Set $u=tg\frac{x}{2}$ and $du=sec^2\frac{x}{2}*\frac{1}{2}dx \rightarrow dx=\frac{2}{1+u^2}dt$ and $sin(x)=\frac{2u}{u^2+1}$
$$\int_0^{2\pi} \frac{1}{2+\sin(x)} \ dx=\int_{-\pi}^{\pi} \frac{1}{2+\sin(x)} \ dx= \int_{-\infty}^{\infty} \frac{2}{(u^2+1)(\frac{2u}{u^2+1}+2)}=\int_{-\infty}^{\infty} \frac{1}{u^2+u+1}$$
And to solve this just complete the square and you will get something in function of $arctg(u)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Find indefinite integral $\int \frac{\arcsin (x)}{x^2}\, \mathrm{d}x$ $$\int \frac{\arcsin (x)}{x^2}\, \mathrm{d}x$$
$$\frac {1}{x^2\sqrt{1-x^2}}+2\int \frac{1}{x^3\sqrt{1-x^2}}\, \mathrm{d}x$$
I try to integrate by parts method, but its doesnt want to be solved. I try to substitue $x=\sin u \mathrm{d}x=\cos u$ but failed somewhere
| Consider
$$
\int\frac{\arcsin(x)}{x^2}dx.
$$
We may use integration by parts where
\begin{align*}
u&=\arcsin(x)&dv&=\frac{dx}{x^2}\\
du&=\frac{dx}{\sqrt{1-x^2}}&v&=-\frac{1}{x}.
\end{align*}
In this case,
$$
\int\frac{\arcsin(x)}{x^2}dx=-\frac{\arcsin(x)}{x}-\int-\frac{dx}{x\sqrt{1-x^2}}=-\frac{\arcsin(x)}{x}+\int\frac{dx}{x\sqrt{1-x^2}}.
$$
Focusing on the last term, we would like to use $u$-substitution, $u^2=1-x^2$ so that $2udu=-2xdx$. In other words, $dx=-\frac{udu}{x}$. Note that in this substitution, it only makes sense when $1-x^2\geq 0$, i.e., $|x|\leq 1$ and we may assume that $u$ is positive. Using this substitution, we get the integral
$$
\int\frac{1}{x\sqrt{u^2}}\cdot\frac{-udu}{x}=-\int\frac{u}{|u|}\frac{du}{x^2}=-\int\frac{du}{1-u^2}=-\int\frac{du}{1-u^2}.
$$
We can now use partial fractions decomposition noting that
$$
\frac{1}{1-u^2}=\frac{1}{2(1-u)}+\frac{1}{2(1+u)}
$$
This ends up being
$$
-\left(-\frac{1}{2}\ln|1-u|+\frac{1}{2}\ln|1+u|\right).
$$
Now, finally, combine everything and you have your integral (near $x=0$). After substitution, one gets
$$
-\frac{\arcsin(x)}{x}+\frac{1}{2}\ln\left|1-\sqrt{1-x^2}\right|-\frac{1}{2}\ln\left|1+\sqrt{1-x^2}\right|
$$
whose derivative is $\frac{\arcsin(x)}{x^2}$ (checked with Maple).
Comparing this with the accepted answer, we can use log rules to get
$$
-\frac{\arcsin(x)}{x}+\frac{1}{2}\ln\left|1-\sqrt{1-x^2}\right|-\frac{1}{2}\ln\left|1+\sqrt{1-x^2}\right|=-\frac{\arcsin(x)}{x}+\frac{1}{2}\ln\left|\frac{1-\sqrt{1-x^2}}{1+\sqrt{1-x^2}}\right|.
$$
Multiplying the numerator and denominator of the argument to the logarithm by the conjugate of the numerator, we get
$$
-\frac{\arcsin(x)}{x}+\frac{1}{2}\ln\left|\frac{1-(1-x^2)}{(1+\sqrt{1-x^2})^2}\right|=-\frac{\arcsin(x)}{x}+\frac{1}{2}\ln\left|\frac{x}{1+\sqrt{1-x^2}}\right|^2.
$$
Using log rules the $2$'s cancel and we can break up the logarithm, leaving
$$
-\frac{\arcsin(x)}{x}+\ln\left|\frac{x}{1+\sqrt{1-x^2}}\right|=-\frac{\arcsin(x)}{x}-\ln\left|1+\sqrt{1-x^2}\right|+\ln\left|x\right|
$$
Finally, remember $+C$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the minimum value of $\dfrac {9x^2\sin^2 x+4}{x\sin x}$
Question: What is the minimum value of $$\dfrac {9x^2\sin^2x+4}{x\sin x}\tag1$$For $0<x<\pi$.
I solved this, but somewhere, I did something wrong. My work is as follows:
First, set $y=x\sin x$. Therefore,
$$\begin{align*}\dfrac {9y^2+4}{y}=\dfrac 4y+9y\tag2\end{align*}$$
And by the AM-GM Inequality, we have$$\begin{align*}\dfrac {9y+\dfrac 4y}2 & \geq\sqrt{\dfrac 4y\cdot 9y}\tag3\\9y+\dfrac 4y & \geq12\tag4\end{align*}$$
$(4)$ is minimized only when the RHS is equal to zero. So$$9y+\dfrac 4y=0\implies 9y^2=-4\implies y^2=-\dfrac 49\tag5$$
However, if I square root both sides, I get an imaginary value. Where did I go wrong? The book says the solution is $12$ (received by plugging in $x=\dfrac 23$).
| Actually, $(4)$ is minimized with the LHS is equal to $12$, since it is $\geq 12$. We get
$$9y + \dfrac{4}{y} = 12$$
is true if and only if
$$9y^2 + 4 = 12y$$
or
$$9y^2 - 12y + 4 = 0\text{.}$$
Apply the quadratic formula to get
$$y = \dfrac{12 \pm \sqrt{12^2-4(9)(4)}}{2(9)} = \dfrac{12 \pm \sqrt{144-144}}{18} = \dfrac{12}{18} = \dfrac{2}{3}\text{.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
For $x+y=1$, show that $x^4+y^4\ge \frac{1}{8}$ As in the title. Let $x,y$ be two real numbers such that $x+y=1$. Prove that $x^4+y^4\ge \frac{1}{8}$.
Any hints? Basically, the only method I am aware of is plugging $y=1-x$ into the inequality and investigating the extrema of the function, but I don't think it's the best method. I'm looking for a cleverer way to prove that inequality.
| Apply Lagrangian multiplier method ! \begin{align}
\min \quad x^4 + y^4 \\ \text{s.t.} \quad x + y = 1 \end{align} gives $\mathcal{L}(x,y) = x^4 + y^4 - \lambda(x+y-1)$. FOC gives \begin{align} \mathcal{L}_x = 4x^3 - \lambda & = 0 \\ \mathcal{L}_y = 4y^3 - \lambda & = 0 \\ x+y & = 1. \end{align} Solving gives an optimum of $(x^*,y^*) = (\frac{1}{2},\frac{1}{2})$ with value $\frac{1}{8}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 9
} |
Finding number of non negative integral solutions of $a+2b+3c+4d=20$ How to find out the number of non negative integral solutions of an equation containing 4 variables, for eg, say, ${a+2b+3c+4d=20}$?
I mean, we can calculate it quite easily for equations containing 2 variables, but what about equations containing 4 variables.?
| See, that $d\leq 5$, $c\leq \left\lceil \frac{20-4d}{3}\right\rceil $, $a$ and $b$ can be then obtained in $n_{d,c}=\left\lceil\frac{20-4d-3c+1}{2}\right\rceil $ ways.
*
*$d=5$
*
*$c=0$ : $n_{5,0}=1$ ($[0,0,0,5]$)
*$d=4$
*
*$c=0$ : $n_{4,0} =3$ ($[0,2,0,4]$, $[2,1,0,4]$, $[4,0,0,4]$)
*$c=1$ : $n_{4,1} =1$ ($[1,0,1,4]$)
*$d=3$
*
*$c=0$ : $n_{3,0} =5$
*$c=1$ : $n_{3,1} =3$
*$c=2$ : $n_{3,2} =2$
*$d=2$
*
*$c=0$ : $n_{2,0} =7$
*$c=1$ : $n_{2,1} =5$
*$c=2$ : $n_{2,2} =4$
*$c=3$ : $n_{2,3} =2$
*$c=4$ : $n_{2,4} =1$
*$d=1$
*
*$c=0$ : $n_{1,0} =9$
*$c=1$ : $n_{1,1} =7$
*$c=2$ : $n_{1,2} =6$
*$c=3$ : $n_{1,3} =4$
*$c=4$ : $n_{1,4} =3$
*$c=5$ : $n_{1,5} =1$
*$d=0$
*
*$c=0$ : $n_{0,0} =11$
*$c=1$ : $n_{0,1} =9$
*$c=2$ : $n_{0,2} =8$
*$c=3$ : $n_{0,3} =6$
*$c=4$ : $n_{0,4} =5$
*$c=5$ : $n_{0,5} =3$
*$c=6$ : $n_{0,6} =2$
The number of ways is then equal to:
$$1+4+10+19+30+44=108$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097082",
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"source": "stackexchange",
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"answer_id": 0
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Given the quadratic equation $p(x^2 +9)= -5qx$ has two equal roots, find the ratio of $p:q$. Hence, solve the quadratic equation Given the quadratic equation $p(x^2 +9)= -5qx$ has two equal roots, find the ratio of p:q. Hence, solve the quadratic equation
so this is what i got so far :
$$px^2+9p+5qx=0$$
$$(5q)^2 - 4(p)(9p)=0$$
$$25q^2 - 36p^2 =0$$
$$(5^2 q^2) - (6^2 p^2)=0$$
$$5^2 q^2 = 6^2 p^2$$
$$5q =6p$$
I might be wrong
what to do next?
| $p(x^2 + 9) + 5qx = 0 \Rightarrow px^2 + 5qx + 9p = 0$
$x= \frac{-5q \pm \sqrt{5^2q^2 -4\cdot9p^2}}{2p}$
Then you want the discriminant equal to 0. This is:
$5^2q^2 -4\cdot9p^2 = 0 \Rightarrow 5^2q^2 = 4\cdot9p^2 \Rightarrow 5q = 6p \Rightarrow p=\frac{5q}{6} \Rightarrow \frac{p}{q} = \frac{5}{6}$
Now the roots of the equation are:
$x= \frac{-5q}{2p} = \frac{-5\cdot6}{2\cdot5} = -3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Integral of $\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$ I was asked to find the following integral:
$$\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$$
What I tried to replace $\sqrt{1-\frac{1}{x^2}}$ with $u$ so that: $$du=\frac{dx}{x^3\sqrt{1-\frac{1}{x^2}}} \Rightarrow du*x=\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$$
And:$$x=\sqrt{\frac{1}{1-u^2}}$$
And we can replace:
$$\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}=\int\frac{du}{\sqrt{1-u^2}}=\arctan(u)+C=\arctan(\sqrt{1-\frac{1}{x^2}})+C$$
The problem is, when that result is derived we don't get the original expression. I just can't find my mistake, so some help would be appreciated.
| The substitution $u=\sqrt{1-\frac{1}{x^2}}$ is equivalent to $|x|=\frac{1}{\sqrt{1-u^2}}$. For $x>1$, we have
$$\begin{align}
\int \frac{1}{x^2\sqrt{1-\frac1{x^2}}}\,dx&=\int \frac{1}{\sqrt{1-u^2}}\,du\\\\
&=\arcsin(u)+C\\\\
&=\arcsin\left(\frac{\sqrt{x^2-1}}{x}\right)+C\\\\
&=-\arcsin(1/x)+C'
\end{align}$$
where we used the identity $\arcsin(1/x)+\arcsin\left(\frac{\sqrt{x^2-1}}{x}\right)=(\pi/2)\text{sgn}(x)=\pi/2$ when $x>1$.
One can proceed similarly for the case in which $x<-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
proving an inequality involving three terms The inequality is as follows:
$a^2+b^2+c^2+3\ge2(a+b+c)$
I was thinking about using the factorization $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$
But I can't get anywhere. I don't have any idea at the moment.
| From given equation we have,
$a^2 + b^2 + c^2 − 2a − 2b − 2c + 3 \ge 0$
$(a^2 - 2a +1) + (b^2 - 2b + 1) + (c^2 − 2c + 1) \ge 0$
$(a−1)^2 + (b−1)^2 + (c−1)^2 \ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2098274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving that $19\mid 5^{2n+1}+3^{n+2} \cdot 2^{n-1}$ How can I prove that $$5^{2n+1}+3^{n+2} \cdot 2^{n-1} $$ can be divided by 19 for any nonnegative n? What modulo should I choose?
| You can prove this by induction.
First, show that this is true for $n=1$:
$5^{2\cdot1+1}+3^{1+2}\cdot2^{1-1}=19\cdot8$
Second, assume that this is true for $n$:
$5^{2n+1}+3^{n+2}\cdot2^{n-1}=19k$
Third, prove that this is true for $n+1$:
$5^{2(n+1)+1}+3^{n+1+2}\cdot2^{n+1-1}=$
$5^{2+2n+1}+3^{1+n+2}\cdot2^{1+n-1}=$
$5^{2}\cdot5^{2n+1}+3^{1}\cdot3^{n+2}\cdot2^{1}\cdot2^{n-1}=$
$5^{2}\cdot5^{2n+1}+3^{1}\cdot2^{1}\cdot3^{n+2}\cdot2^{n-1}=$
$25\cdot5^{2n+1}+\color\green{6}\cdot3^{n+2}\cdot2^{n-1}=$
$25\cdot5^{2n+1}+(\color\green{25-19})\cdot3^{n+2}\cdot2^{n-1}=$
$25\cdot5^{2n+1}+25\cdot3^{n+2}\cdot2^{n-1}-19\cdot3^{n+2}\cdot2^{n-1}=$
$25\cdot(\color\red{5^{2n+1}+3^{n+2}\cdot2^{n-1}})-19\cdot3^{n+2}\cdot2^{n-1}=$
$25\cdot\color\red{19k}-19\cdot3^{n+2}\cdot2^{n-1}=$
$19\cdot25k-19\cdot3^{n+2}\cdot2^{n-1}=$
$19\cdot(25k-3^{n+2}\cdot2^{n-1})$
Please note that the assumption is used only in the part marked red.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2099097",
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"source": "stackexchange",
"question_score": "5",
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Solving two Cubic Equation on their Roots.
Let $x^{3}+ax+10=0$ and $x^{3}+bx^{2}+50=0$ have two roots in common. Let $P$ be the product of these common roots. Find the numerical value of $P^{3}$, not involving $a,b$.
My attempts:
Let roots of $x^{3}+ax+10=0$ be $\alpha,\beta,\gamma\implies$
\begin{align*}
&\alpha+\beta+\gamma =0 \\
&\alpha\beta+\alpha\gamma+\beta\gamma =a \\
&\alpha\beta\gamma = -10
\end{align*}
and of $x^{3}+bx^{2}+50=0$ be $\alpha,\beta,\gamma'\implies$
\begin{align*}
&\alpha+\beta+\gamma' =-b \\
&\alpha\beta+\alpha\gamma'+\beta\gamma' =0 \\
&\alpha\beta\gamma' = -50
\end{align*}
Few important equations:
*
*$\dfrac{\alpha\beta\gamma'}{\alpha\beta\gamma}=\dfrac{-50}{-10}=5$ $\implies \gamma'=5\gamma $
*$\gamma-\gamma'=b \because$ Substracting first eq
*$(\alpha+\beta)(\gamma-\gamma')=a$ Substracting second eq. from above
*$\alpha+\beta=\dfrac{a}{b} \ \ \because(2),(3)$ squaring gives: $(\alpha+\beta)^{2}=\dfrac{a^{2}}{b^{2}}$
*$\alpha\beta\gamma-\alpha\beta\gamma'=40\implies\alpha\beta(\gamma-\gamma')=40\implies\alpha\beta=\dfrac{40}{b}$
Also $(\alpha+\beta+\gamma')^{2} =b^{2}$
$\implies \alpha^2+\beta^2+\gamma'^2=b^2 \implies \alpha^2+\beta^2=b^2-\gamma'^2=b^2-\dfrac{25b^{2}}{16}=\dfrac{-9b^2}{16}\because (2),(1)$
Now, squaring first eq. of first set of eq.$\implies \alpha^2+\beta^2+\gamma^2+2a =0 \implies \alpha^2+\beta^2=\dfrac{-b^2}{16}-2a\because (2),(1)$
Equating this with previous equations, $\implies \alpha^2+\beta^2=\dfrac{-9b^2}{16}=\dfrac{-b^2}{16}-2a\implies a^2=\dfrac{b^4}{16}\rightarrow(7)$
Also, $\alpha^2 +\beta^2 +2\alpha\beta=\dfrac{a^2}{b^2} \because(4)$. And we calculated very thing in terms of $b$, putting all these,$\implies \dfrac{-9b^2}{16} +2\alpha\beta=\dfrac{b^4}{16b^2}\because(7)\implies \dfrac{80}{b}=\dfrac{10b^2}{16}\ \because (5)\implies b^3=128\rightarrow(6)$
Now, $(eq.5)^3\implies (\alpha\beta)^3=\dfrac{64000}{b^3}=\dfrac{64000}{128}=500=P^3\because(6)$
Is this correct, and if, then what are easiest/shortest method besides my GIANT method.
| With OP's notations, the common roots $\alpha,\beta$ must satisfy the quadratic that results by subtracting the two cubics: $\,b x^2 - ax + 40 = 0\,$. Then by Vieta's formulas $\,P = \alpha \beta = \cfrac{40}{b}\,$.
Again by Vieta's, the 3rd root of the 2nd cubic is $\,\gamma' = -\cfrac{50}{\alpha \beta} = -\cfrac{5}{4} \,b\,$. Substituting in the equation:
$$
-\frac{125}{64} \,b^3 + b \cdot \frac{25}{16} \,b^2 + 50 = 0 \quad\iff\quad \cfrac{25}{64} \,b^3 = 50 \quad\iff\quad b^3 = 128
$$
Therefore $P^3 = \cfrac{40^3}{b^3}=\cfrac{512\cdot125}{128}=500\,$.
[ EDIT ] The following answers @JeanMarie's interesting question of whether such polynomials do in fact exist which satisfy the given condition.
The answer above proved that $b^3 = 128 \iff b=4\sqrt[3]{2}$. Then turning back to the 1st equation, its 3rd root must be $\gamma = -\cfrac{10}{P}=-\cfrac{1}{4}b=-\sqrt[3]{2}$. Substituting $\gamma$ into the 1st equation gives $a = 4 \sqrt[3]{4}$. Then the two equations are determined to be:
$$
\begin{cases}
\begin{align}
x^3 + 4 \sqrt[3]{4} \,x + 10 & = 0 \\
x^3 + 4 \sqrt[3]{2} \,x^2 + 50 & = 0
\end{align}
\end{cases}
$$
It is straightforward to verify that they do in fact have the quadratic $
x^2 - \sqrt[3]{2}\,x + 5 \sqrt[3]{4}$ as a common factor, so they have the two common roots $ (1 \pm i \sqrt{19})/\sqrt[3]{4}\,$. (Of course, their product is $P = 10 / \sqrt[3]{2}$ so this could be construed as another, painstaking way to answer OP's question).
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the integral $\int \sqrt{1-\frac{3}{x}+\frac{1}{x^2}} \, dx$ How to find the integral $$\displaystyle\int \sqrt{1-\dfrac{3}{x}+\dfrac{1}{x^2}}\mathrm dx$$ ?
I cannot think of any suitable method. I tried trigonometric substitutions but they were of no use. Hints please!
| Use the Euler substitution of the first kind, let $\sqrt{x^2-3x+1}=x+t$,we have
\begin{align*}
\int{\frac{\sqrt{x^2-3x+1}}{x}}\text{d}x&=-2\int{\frac{\left( t^2+3t+1 \right) ^2}{\left( 1-t^2 \right) \left( 2t+3 \right) ^2}}\text{d}t \\
&=-2\int\left ( \frac{1}{2}\frac{1}{t+1}-\frac{3}{2}\frac{1}{2t+3}-\frac{5}{4}\frac{1}{\left ( 2t+3\right )^{2}}-\frac{1}{2}\frac{1}{t-1}-\frac{1}{4}\right ) \mathrm{d}t
\end{align*}
then you can take it from here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $x^2+xy+y^2+ \sqrt{3}y + 1=0$ for $xy$. I have a problem on my textbook:
$x,y \in\mathbb{R}$ and $x^2+xy+y^2+ \sqrt{3}y + 1=0$ is given. Find the value of $xy$.
I couldn't find.
| Multiply both sides on 4
$$x^2+xy+y^2+ \sqrt{3}y + 1=0$$
$$4x^2+4xy+y^2+3y^2+ 4\sqrt{3}y + 4=0$$
$$(2x+y)^2+(\sqrt3y+2)^2=0$$
Then $y=-\frac2{\sqrt3}, x=\frac1{\sqrt3}$
$$xy=-\frac2{3}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Known that $a + b + c = 0$ and $a^3 + b^3 + c^3 = 27$. Find ABC!
Known that
$$a + b + c = 0$$
$$a^3 + b^3 + c^3 = 27$$
What is the value of $abc?$
A.) 1
B.) 0
C.) 7
D.) 8
E.) 10
My Work:
$$a + b = -c$$
$$a + c = -b$$
$$b + c = -a$$
Then
$$(a + b + c)(a + b + c)(a + b + c) = 0$$
Expanded into:
$$a^3 + b^3 + c^3 + 3a^2b+3b^2c+3a^2c+3ab^2 + 3bc^2+ 3ac^2+6abc = 0$$
Putting the Values:
$$27+3a^2(-a)+3b^2(-b)+3c^2(-c)+6abc = 0$$
$$27 -3(a^3 + b^3 + c^3) = -6abc$$
$$27-3(27) = -6abc$$
$$-54=-6abc$$
$$abc = 9$$
$9$ wasn't an option in the question. Am I missing something?
| Your solution is correct.
$(a+b+c)^3=(a^3+b^3+c^3)+3a(b^2+c^2)+3b(a^2+c^2)+3c(a^2+b^2)+6abc$
$\implies 0=27+3ab^2+3ac^2+3ba^2+3bc^2+3ca^2+3cb^2+6abc$
$\implies 0 = 27+3b^2(a+c)+3a^2(b+c)+3c^2(a+b)+6abc$
$\implies 0 = 27+3b^2(-b)+3a^2(-a)+3c^2(-c)+6abc$
$\implies 0 = 27 - 3(a^3+b^3+c^3) +6abc$
$\implies 0 = 27 - 81 + 6abc$
$\implies 54 = 6abc$
$\implies abc = 9$
| {
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Find all $p, q$ (coprime) such that $\frac{(p-1)(q-1)}{2}$ is a prime number
Find all $p, q$ (coprime) such that $\frac{(p-1)(q-1)}{2}$ is a prime number.
I found that $(p,q) = (2,7),(7,2),(2,15),(3,8)$ gave prime numbers for $\frac{(p-1)(q-1)}{2}$, but how do we find all?
| If $p, q$ are coprime then at most one is even. Wolog $q$ is odd and $q-1$ is even.
So $(p-1)\frac{q-1}2$ is prime. So either $p-1 = \pm 1$ or $\frac {q-1}2 = \pm 1$.
Case 1: $p-1 = -1$
If $p-1 = -1$ then $p = 0$ and $0$ is not coprime to anything except $\pm 1$. (Everything divides 0) so $q = \pm 1$. $\frac {1-1}2 = 0$ is not prime so $q = -1$ and $\frac {q-1}2 = -1$ and ... $ (p-1)\frac{q-1}2=1$ is not prime.
Case 2: $p-1 = 1$
So $p = 2$ and $\frac {q-1}2 $ is prime. So $q = 2k + 1$ for any prime. i.e. Let $k$ be prime, the $\gcd(2, 2k+1) =1$ and $\frac{(2-1)(2k+1 -1)}2 = k$ is prime.
Case 3: $\frac {q-1}2 = -1$
$q = -1$ and $p= k + 1$ for any prime. i.e. $\gcd(k+1,-1)=1$ and $\frac{(k+1-1)(-1 -1)}2 = -k$ is prime.
Case 4: $\frac{q-1}2=1$
$q = 3$ and $p = k +1$ for any prime $k=3$ or $k \equiv 1 \mod 3$
So options are $(2,2k+1)$ for any prime $k$.
$(-1, k+1)$ for any prime.
$(3,4)$ and $(3,k+1)$ for any prime of the form $k \equiv 1 \mod 3$.
| {
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Write the quadratic function in the form $g(x)= a(x-h)^2 + k$ Write the quadratic function in the form $g(x)= a(x-h)^2 + k$
Then, give the vertex of its graph.
$g(x)= 2x^2-16x+35$
~
I tried, but still lost:
$(2x^2-16x)+35$
$2(x^2-4^2-16)+35$
$2(x^2-4^2-32+35$
| You have few mistakes.
It should be
$2(x^2 + 4^2 - 8x) + 35 - 2 \cdot 4^2$
= $2(x - 4)^2 + 35 - 32$
= $2(x - 4)^2 + 3$
| {
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"timestamp": "2023-03-29T00:00:00",
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Does $\int_{0}^{\pi/ n}{\sin^3(x)\over [1+\cos^2(x)]^2}\mathrm dx=F(n)$ have a closed form?
$$\int_{0}^{\pi/ n}{\sin^3(x)\over [1+\cos^2(x)]^2}\mathrm dx=F(n)\tag1$$
For $n\ge 1$
I have calculated out for:
$F(1)=1$
$F(2)={1\over 2}$
$F(3)={1\over 10}$
The general for $F(n)$ I wasn't able to find...
What is the closed form for $(1)?$
An attempt
Using $\cos^2(x)={1+\cos(2x)\over 2}$ then $1+\cos^2(x)={3+\cos(2x)\over 2}$
Calling upon a subsitition $u=3+\cos(2x)$ then $du=-2\sin(2x)dx$
Then $(1)$ becomes:
$$-\int_{1}^{\cos(2\pi/n)}{\sin^2(x)\over u^2\cos(x)}\mathrm du\tag2$$
$2+2\cos^2(x)=u$ then $\sin^2(x)={4-u\over 2}$
$\cos(x)=\sqrt{u-2\over 2}$
$${\sqrt{2}\over 2}\int_{1}^{\cos(\pi/n)}{u-4\over u^2\sqrt{u-2}}\mathrm du\tag3$$
$${\sqrt{2}\over 2}\int_{1}^{\cos{(\pi/n)}}{1\over u\sqrt{u-2}}\mathrm du+{2\sqrt{2}}\int_{1}^{\cos(\pi/n)}{1\over u^2\sqrt{u-2}}\mathrm du\tag4$$
From Standard integral table
$$\int{dx\over x\sqrt{ax+b}}={2\over \sqrt{-b}}\arctan\sqrt{ax+b\over -b}$$
$$\int{dx\over x^2\sqrt{ax+b}}=-{\sqrt{ax+b}\over bx}-{a\over 2b}\int{dx\over x\sqrt{ax+b}}$$
Substitute it in,
$${\sqrt{2}\over 2}\cdot{2\over\sqrt{2}}\arctan\sqrt{u-2\over 2}-2\sqrt{2}\left[-{\sqrt{u-2\over -2u}}+{1\over 4}\left({2\over\sqrt{2}}\cdot\arctan{\sqrt{u-2\over2}}\right)\right]\tag5$$
Break down to
$$\arctan\sqrt{u-2\over 2}-\sqrt{2}\cdot{\sqrt{u-2}\over u}-\arctan\sqrt{u-2\over 2}\tag6$$
$$=-\sqrt{2}\cdot{\sqrt{u-2}\over u}\tag7$$
I don't thing I am on the right track, any help?
| Hint:
\begin{align*}
F(n)&=-\int_{0}^{\pi/ n}{\sin^2(x)\over [1+\cos^2(x)]^2}\mathrm d\cos x\\
&=-\int_{0}^{\pi/ n}{1-\cos^2(x)\over [1+\cos^2(x)]^2}\mathrm d\cos x\\\tag1
&=\int_{\cos\left ( \frac{\pi }{n} \right )}^{1}\frac{1-x^{2}}{\left ( 1+x^{2} \right )^{2}}\, \mathrm{d}x\\
&=\frac{1}{2}-\frac{\cos\left ( \dfrac{\pi }{n}\right )}{1+\cos^2\left ( \dfrac{\pi }{n}\right )}
\end{align*}
$(1):$ $\displaystyle \int \frac{1-x^{2}}{\left ( 1+x^{2} \right )^{2}}\, \mathrm{d}x=\frac{x}{1+x^{2}}+C$
| {
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"timestamp": "2023-03-29T00:00:00",
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A cyclic, non-symmetric inequality If $x,y,z\geq 0$, prove
$$
(x^2+y^2+z^2)^2\geq 3(x^3y+y^3z+z^3x)
$$
With these nonsymmetric inequalities the usual tools like Muirhaed or Schur do not apply, and also AM-GM doesn't seem to be of any help. Also, it's non-factorizable.
| Because
$$(x^2+y^2+z^2)^2-3(x^3y+y^3z+z^3x)=\frac{1}{2}\sum\limits_{cyc}(x^2-z^2-2xy+xz+yz)^2\geq0$$
Also we have
$$(x^2+y^2+z^2)^2-3(x^3y+y^3z+z^3x)=\frac{1}{6}\sum\limits_{cyc}(x^2+y^2-2z^2-3xy+3xz)^2\geq0$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim\limits_{n\to\infty} ( \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+ \frac{1}{(2n)^2})$ Find $\lim\limits_{n\to\infty} ( \frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}+\dots+ \frac{1}{(2n)^2})$
Please help me find this limit .
I cannot use cauchys limit theorem on this problem . So I am stuck and clueless
| For $j\in \mathbb N$ and $x\in (j,j+1)$ we have $\frac {1}{(j+1)^2}<\frac {1}{x^2}$. Therefore $$0<\sum_{j=n}^{2n-1}\frac {1}{(j+1)^2}=\sum_{j=n}^{2n-1}\int_j^{j+1}\frac {1}{(j+1)^2}dx<\sum_{j=n}^{2n-1}\int_j^{j+1}
\frac {1}{x^2}dx=$$ $$=\int_n^{2n}\frac {1}{x^2}dx=\frac {1}{n}- \frac {1}{2n}=\frac {1}{2n}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{4})(1+\frac{1}{5})...$ Is there any well-known value (Or Approximation) for this?
$$(1-\frac{1}{2})(1+\frac{1}{3})(1-\frac{1}{4})(1+\frac{1}{5})...$$
we know that it converges as $$\sum_{i=2}^{\infty}\frac{(-1)^{i+1}}{i}=ln2-1$$
So there is a trivial upper bound $\frac{2}{e}$ for it. Is there any better result? In addition is there any similar result for
$$(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})...$$
or
$$(1+\frac{1}{2})(1+\frac{1}{4})(1+\frac{1}{8})(1+\frac{1}{16})...$$
| HINT:
For the first question,
$$\left(1+\dfrac1{2n+1}\right)\left(1-\dfrac1{2n+2}\right)=1$$
$$\prod_{r=1}^n\left(1-\dfrac1{2^r}\right)=\dfrac{\prod_{r=1}^n(2^r-1)}{2^{1+2+\cdots+n}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove by induction: $\sum_{k = 0}^n {{2n + 1}\choose{2k + 1}} = 4^n$ I'm having a hard time proving:
$$
\sum_{k = 0}^n {{2n + 1}\choose{2k + 1}} = 4^n
$$
The initial step is rather simple to prove, but I seem to miss something important in the induction step ($n \Rightarrow n + 1$).
This is what I got so far:
$$
\begin{alignat*}{3}
&\sum_{k = 0}^{n + 1} {{2(n + 1) + 1}\choose{2k + 1}} \\
= &\sum_{k = 0}^{n + 1} {{2n + 3}\choose{2k + 1}} \\
= &\sum_{k = 0}^{n} {{2n + 3}\choose{2k + 1}} + {{2n + 3}\choose{2n + 3}} \\
= &\sum_{k = 0}^{n} {{2n + 3}\choose{2k + 1}} + 1 \\
\end{alignat*}
$$
I cannot find a way to use the induction hypothesis, since the binomial coefficient inside the sum contains the term $2n + 3$ instead of $2n + 1$.
How can I proceed from here? Any tips?
| Here's an inductive proof.
Using $\binom{a+1}{b}=\binom{a}{b}+\binom{a}{b-1}$ twice, we get:
$$\begin{align}\binom{2n+3}{2k+1}&=\binom{2n+2}{2k+1} + \binom{2n+2}{2k}\\
&=\binom{2n+1}{2k+1}+2\binom{2n+1}{2k}+\binom{2n+1}{2k-1}\\
&=\binom{2n+1}{2k+1}+2\binom{2n+1}{2(n-k)+1}+\binom{2n+1}{2k-1}
\end{align}$$
The last equality because $\binom{a}{b}=\binom{a}{a-b}$.
Now, how many times does $\binom{2n+1}{2j+1}$ occur in:
$$\sum_{k=0}^{n+1}\binom{2n+3}{2k+1}=\sum_{k=0}^{n+1}\left(\binom{2n+1}{2k+1}+2\binom{2n+1}{2(n-k)+1}+\binom{2n+1}{2k-1}\right)?$$
It occurs once when $k=j$ once when $k=j+1$ and twice when $n-k=j$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem on Trigonometry If $ \sin x + \cos x = a$ , then find:
(i) $ \sin^6 x + \cos^6 x$
(ii) $ \lvert \sin x – \cos x \rvert $
I tried using the identity $a^3 + b^3$ for (i) but that didn't work and I ended up with an expression that was of no help. I have absolutely no idea on how to go about the problem. Any kind of help would be appreciated.
| We have $$\sin^6 x +\cos^6 x =(\sin^2 x +\cos^2 x)(\sin^4 x + \cos^4 x -\sin^2 x \cos^2 x) $$ $$=(1)((\sin^2 x + \cos^2 x)^2-3\sin^2 x \cos^2 x) $$ $$=(1-3\sin^2 x\cos^2 x) \tag {1}$$
Now we know that $$(\sin x + \cos x)^2 =1 +2\sin x \cos x$$ Use this result to find the value of $(1)$.
For $(2) $, use the fact that $$(\sin x -\cos x)^2 =1-2\sin x\cos x$$ Hope it helps.
| {
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Without using actual division method, find the quotient without using actual division method, find the quotient and remainder when $x^6-2x^4+x^2+5$ is divided by $x^2-2$.
Using Remainder Theorem I got the remainder as $7$ but how do I get the quotient..
| $x^6-2x^4+x^2+5=x^4(x^2-2)+x^2-2+7=(x^4+1)(x^2-2)+7$
so quotient is $x^4+1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\lim_{n\rightarrow \infty}\frac{n!\cdot e^n}{\sqrt{n}n^n}$ without stirling approximation
Evaluation of $\displaystyle \lim_{n\rightarrow \infty}\frac{n!\cdot e^n}{\sqrt{n}\cdot n^n}$
$\bf{My\; Try::}$we can write it as $$l=\lim_{n\rightarrow \infty}\frac{e^n}{\sqrt{n}}\cdot \left(\frac{1}{n}\cdot \frac{2}{n}\cdot \frac{3}{n}\cdots \cdots \frac{n}{n}\right)$$
$$\ln (l) = \lim_{n\rightarrow \infty}\bigg[n-\frac{1}{2}n+\sum^{n}_{r=1}\ln\left(\frac{r}{n}\right)\bigg]$$
Now how can i solve it, Help required, Thanks
| Let
$$
a_n : = \frac{{n!e^n }}{{n^n \sqrt n }}
$$
and $a:= \lim a_n$. I assume that this limit exists and is finite and positive. Then
\begin{align*}
a_{2n} & = \frac{{(2n)!e^{2n} }}{{(2n)^{2n} \sqrt {2n} }} = \frac{{(2n)!}}{{n!^2 2^{2n} }}\frac{{(n!)^2 e^{2n} }}{{n^{2n + 1} }}\sqrt {\frac{n}{2}} = \sqrt {\frac{{1 \cdot 1}}{{2 \cdot 2}}\frac{{3 \cdot 3}}{{4 \cdot 4}} \cdots \frac{{(2n - 1)(2n - 1)}}{{2n \cdot 2n}}} a_n^2 \sqrt {\frac{n}{2}}
\\ & = \sqrt {\frac{{1 \cdot 3}}{{2 \cdot 2}}\frac{{3 \cdot 5}}{{4 \cdot 4}} \cdots \frac{{(2n - 1)(2n + 1)}}{{2n \cdot 2n}}} a_n^2 \sqrt {\frac{n}{{4n + 2}}} .
\end{align*}
Taking the limit of both sides gives
$$
a = \sqrt {\mathop {\lim }\limits_{n \to + \infty } \frac{{1 \cdot 3}}{{2 \cdot 2}}\frac{{3 \cdot 5}}{{4 \cdot 4}} \cdots \frac{{(2n - 1)(2n + 1)}}{{2n \cdot 2n}}} a^2 \frac{1}{2}.
$$
The infinite product under the square root is the reciprocal of the famous Wallis product, whence
$$
a = \sqrt {\frac{2}{\pi }} a^2 \frac{1}{2},
$$
i.e., $a=\sqrt{2\pi}$.
| {
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Show that $45
If $x_0 = 5$ and $x_{n+1} = x_n + \frac {1}{x_n},$ show that
$45<x_{1000}<45.1$
This problem is taken from the list submitted for the $1975$ Canadian Mathematics Olympiad (but not used on the actual exam).
SOURCE : CRUX(Page Number 3 ; Question Number 162)
I tried writing out the first few terms :
$x_1 = 5+ \frac{1}{5} $
$x_2 = \big(5+\frac{1}{5}\big) + \big(5+\frac{1}{5}\big)^{-1} = \frac{x_0^2 + 1}{x_0} + \frac{x_0}{x_0^2 + 1} = \frac{(x_0^2 + 1)^2+x_0^2}{x_0(x_0^2+1)}$
$x_3 = \frac{(x_0^2 + 1)^2+x_0^2}{x_0(x_0^2+1)} + \frac{x_0(x_0^2+1)}{(x_0^2 + 1)^2+x_0^2} = Messy$
I tried a lot but could not find any general formula for the $n$th term. Does there even exist any?
Also it is clear that $\big(x_n + \frac{1}{x_n}\big)$ is an increasing function. So I think sequence diverges, but how can the $1000th$ term be calculated or aprroximated?
Any help would be gratefully acknowledged :).
| $x_n^2-x_{n-1}^2=2+\frac{1}{x_{n-1}^2}$ for $n\geq1$.
Thus, $$x_{1000}^2=2\cdot1000+25+\frac{1}{x_0^2}+\frac{1}{x_1^2}+...+\frac{1}{x_{999}^2}>2025$$
and $$x_{1000}^2=2\cdot1000+25+\frac{1}{x_0^2}+\frac{1}{x_1^2}+...+\frac{1}{x_{999}^2}<2025+\frac{100}{x_0^2}+\frac{900}{x_{100}^2}<$$
$$<2025+4+\frac{900}{225}=2033<45.1^2$$
Because $x_{100}^2>225$.
| {
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Any hints on how to show that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $ I am trying to prove that $$\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $$
My attempt was to suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10}$, and show that it was absurd.
Here's what I did:
Suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10} $, then $2 - \sqrt{3} \geq \frac{2\sqrt{3}}{10}$ thus:
$$\frac{20 -10 \sqrt{3}}{10} \geq \frac{2 \sqrt{3}}{10} \implies 20 -\sqrt{300} \geq 2 \sqrt{3} \implies 10 - \sqrt{\frac{300}{4}}\geq \sqrt{3} \implies 10 - \sqrt{75} \geq \sqrt{3} \implies \frac{10}{\sqrt{3}}- \sqrt{25} \geq 1 \implies \frac{10}{\sqrt{3}}- 5 \geq 1$$
Now I elevate everything to the square:
$$\frac{100}{3} - \frac{100}{\sqrt{3}} + 25 \geq 1 \implies \frac{100}{3} - \frac{100 \sqrt{3}}{3} + \frac{75}{3} \geq 1 \implies \frac{100(1-\sqrt{3}) +75}{3} \geq \frac{3}{3}$$
All is left to show is that $$100(1-\sqrt{3}) +75 \geq 3$$
And here, I don't know where to go, and it doesn't seem as I simplified the problem.
Any hints of help will be appreciated
| $\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10}
\iff 2-\sqrt{3} < \sqrt{3}/5
\iff 2 < 6\sqrt{3}/5
\iff 4 < 108/25
$
which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.