Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove that normals for surface of revolution $z=f(\sqrt{x^2+y^2})$ cross axis of revolution $z=f(\sqrt{x^2+y^2})$, $f'\not = 0$
My attempt: considering we have $F(x,y,z)=f(\sqrt{x^2+y^2}) - z$
And using normal equation: $$\frac{x-x_0}{\frac{\partial F}{\partial x}}=\frac{y-y_0}{\frac{\partial F}{\partial y}}=\frac{z-z_0}{\frac{\partial F}{\partial z}}$$
$\frac{\partial F}{\partial x} = \frac{x}{\sqrt{x^2+y^2}}\cdot F'$
$\frac{\partial F}{\partial x} = \frac{y}{\sqrt{x^2+y^2}}\cdot F'$
$\frac{\partial F}{\partial z} = -1$
I got $\overline{n} = \{\frac{x}{\sqrt{x^2+y^2}}, \frac{y}{\sqrt{x^2+y^2}},-1\}$ or
$\overline{n} = \{x, y,-\sqrt{x^2+y^2}\}$ and I got stuck 'cause I don't know how to connect that with axis of revolution
| The normal is not correctly stated. There are several errors. Correcting them,
$u=\sqrt{x^2+y^2}$
$\dfrac{\partial F}{\partial x}=\dfrac{\partial}{\partial x}(f(\sqrt{x^2+y^2})-z)=\dfrac{\partial}{\partial x}f(\sqrt{x^2+y^2})=\dfrac{df(u)}{du}\dfrac{\partial u}{\partial x}=\dfrac{x}{\sqrt{x^2+y^2}}\cdot f'(u)$
$\dfrac{\partial F}{\partial y}=\dfrac{\partial}{\partial y}(f(\sqrt{x^2+y^2})-z)=\dfrac{\partial}{\partial y}f(\sqrt{x^2+y^2})= \dfrac{df(u)}{du}\dfrac{\partial u}{\partial y}=\dfrac{y}{\sqrt{x^2+y^2}}\cdot f'(u)$
$\overline{n} = \{xf', yf',-\sqrt{x^2+y^2}\}$
Now, construct the line passing by the point in the surface $P(x,y,f(\sqrt{x^2+y^2}))$ and having as direction vector $\overline n$, the normal vector at this point $P$.
l:$\begin{cases}
\tilde x=x+txf'\\
\tilde y=y+tyf'\\
\tilde z=f(\sqrt{x^2+y^2})-t\sqrt{x^2+y^2}
\end{cases}$
The axis of revolution is the $z$ axis because $f$ is constant for points with constant distance to this axis ($d((x,y,z),(0,0,z))=\sqrt{x^2+y^2}$). Clearly, with the hypothesis of $f'\neq0$, $l$ passes thru the $z$ axis as $\tilde x=\tilde y=0$ for $t=-1/f'$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2247228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that the number $a_{n}=\sqrt{2+\sqrt{2+\sqrt{2+...}}}$, is irrational for every positive integer n. Define $a_{n}=\sqrt{2+\sqrt{2+\sqrt{2+...}}},\quad n=1,2,...$ where the radical contains $n$ twos.
Provide a recursive definition of $a_n$ and then show that the number $a_n$ is irrational for every positive integer n.
This is what I have so far-
$$a_0=\sqrt{2}, \quad a_{n+1}=\sqrt{2+a_n}$$
Assume $a_{n+1}$ is rational. Then
\begin{align*}
a_{n+1}&=\frac{b}{c} \quad grd(b,c)=1\\
(a_{n+1})^2&=\frac{b^2}{c^2}\\
2+a_n&=\frac{b^2}{c^2}\\
(2+a_n)c^2&=b^2 \quad =>(2+a_n)\big|b^2\\
\end{align*}
Is this correct so far? I am not sure what the next step is or if I am even on the right path?
| Using induction we suppose that $a_n$ is irrational. Suppose now that $a_{n+1}$ is rational. Then $a^2_{n+1}=2+a_n$, therefore $a_n$ is rational, contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2249529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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An object travelling on $x^2$ I have an object traveling along the curve $y=x^2$. $z$ is the distance from the origin and $dz/{dt}=1$ is the rate it's increasing per unit time. At what rate are my $x$ and $y$'s moving at the point (2,4)? In other words what is $dy/dt$ and $dx/dt$.
I've seen the geometry. Found $z$ as the hypotenuse of $x$ and $y$. Also, I infer that $dz=dt $ so they are interchangable. How do I solve this?
| Bu definition,
$ z = \sqrt{\left(x-0\right)^2 +\left(y-0\right)^2 } = \sqrt{x^2 + y^2}$.
First, let us compute $dx/dt:$
Since $y = x^2$ we can rewrite $z$ in the form $ z = z(x,y) = \sqrt{x^2 + x^4}$.
Then by chain rule we have
$$
1 = \frac{d z}{d t} = \frac{d z}{d x} \,\frac{d x}{d t} \implies
\frac{d x}{d t} = \frac{1}{dz/dx}
$$
Now,
$$
\frac{d z}{d x} = \left(\sqrt{x^2 + x^4}\right)'
=\frac12\, \frac{2x+4x^3}{\sqrt{x^2 + x^4}} = \frac{1+2x^2}{\sqrt{1+x^2}}
$$
Substituting $x=2$ into expression above will give you the speed of $x$ at the point $\left(2,4\right)$.
Second, $dy/dt$ can be computed in the same way. Hope you can pick it from here.
Alternatively, one can compute $dy/dt$ in the following manner:
$$
1 = \frac{d z}{d t} = \frac{d z}{d x} \,\frac{d x}{d t} + \frac{d z}{d y} \,\frac{d y}{d t}
= \frac{x}{\sqrt{x^2 + y^2}}\,\frac{d x}{d t} + \frac{y}{\sqrt{x^2 + y^2}}\,\frac{d y}{d t}
\\ \implies
\frac{d y}{d t} = \frac{\sqrt{x^2 + y^2}}{y} \left( 1 - \frac{x}{\sqrt{x^2 + y^2}}\,\frac{d x}{d t}\right)
= \frac{\sqrt{x^2 + y^2}}{y} - \frac{x}{y}\,\frac{d x}{d t},
$$
where $dx/dt$ is known from the first part.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2250641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Find a 3 x 3 orthogonal matrix Find a 3x3 orthogonal matrix whose first column is $[\frac1{\sqrt{3}},{- \frac 1{\sqrt{3}}}, \frac1{\sqrt{3}}]$.
I was thinking this has to do with: ${Q}^T•Q=I_{n}$, but I still don't understand how to go about this.
| You need to find an orthonormal basis of $\mathbb{R}^3$ whose first vector is the vector $v_1 = \left( \frac{1}{\sqrt{3}}, -\frac{1}{\sqrt{3}}, \frac{1}{\sqrt{3}} \right)^T$ given to you. This can be done in several ways:
*
*Complete $v_1$ arbitrary to a basis $v_1,v_2,v_3$ of $\mathbb{R}^3$ and perform Gram-Schmidt to get $v_1,v_2',v_3'$. Then you can take $Q = (v_1, v_2, v_3)$.
*Find two linearly independent vectors $v_2,v_3$ that are orthogonal to $v_1$ and perform Gram-Schmidt on them to get $v_2',v_3'$. Then you can take $Q = (v_1,v_2,v_3)$.
Let me demonstrate the second method. Since normalization is not important for orthogonality, we need to find two linearly independent vectors that are orthogonal to $(1,-1,1)^T$. Just by playing with vectors whose entries are $\pm 1$, we can get two such vectors
$$ v_2 = \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, v_3 = \begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix}. $$
If we perform Gram-Schmidt on $(v_2,v_3)$ we get
$$ v_2' = \frac{v_2}{\| v_2 \|} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}, \\
v_3' = \frac{v_3 - \left< v_3, v_2 \right> v_2}{\| v_3 - \left< v_3, v_2 \right> v_2 \|} = \frac{\begin{pmatrix} 0 \\ 1 \\ 1 \end{pmatrix} - \frac{1}{2} \begin{pmatrix} 1 \\ 1 \\ 0 \end{pmatrix}}{\dots} = \frac{\begin{pmatrix} -\frac{1}{2} \\ \frac{1}{2} \\ 1 \end{pmatrix}}{\| \begin{pmatrix} -\frac{1}{2} \\ \frac{1}{2} \\ 1 \end{pmatrix} \|} = \begin{pmatrix} -\frac{\sqrt{3}}{\sqrt{10}} \\ \frac{\sqrt{3}}{\sqrt{10}} \\ \frac{\sqrt{4}}{\sqrt{10}} \end{pmatrix} $$
and so we can take $Q$ to be
$$ Q = \begin{pmatrix} \frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & -\frac{\sqrt{3}}{\sqrt{10}} \\
-\frac{1}{\sqrt{3}} & \frac{1}{\sqrt{2}} & \frac{\sqrt{3}}{\sqrt{10}} \\
\frac{1}{\sqrt{3}} & 0 & \frac{\sqrt{4}}{\sqrt{10}} \end{pmatrix}. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2252292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $|f'(x)-\frac{\overline{f(x+h)}-\overline{f(x-h)}}{2h}|\le \frac{M_1h^2}{6}+\frac{\epsilon M_2}{h}$ Suppose $f$ is a $C^3$ function on $[x-2h,x+2h]$. Suppose that the computed values $\overline{f(x+h)}$ and $\overline{f(x-h)}$ satisfy $\overline{f(x+h)}=f(x+h)(1+e_1)$ and $\overline{f(x-h)}=f(x-h)(1+e_2)$, where $e_i$ is error coming
from floating-point arithmetic that satisfy $|e_i|\le \epsilon$.
Show that $|f'(x)-\frac{\overline{f(x+h)}-\overline{f(x-h)}}{2h}|\le \frac{M_1h^2}{6}+\frac{\epsilon M_2}{h}$, where $M_1 = \max_x |f'''(x)|, M_2 = \max_x |f(x)|$.
I tried taylor expansion for $f(x+h)$ and $f(x-h)$, but I did not get what we want.
| \begin{align*}
\left|f'(x) - \frac{\hat{f}(x+h) - \hat{f}(x-h)}{2h} \right|
&=
\left|f'(x) - \frac{f(x+h)(1+e_1) - f(x-h)(1+e_2)}{2h} \right| \\
&= \left|f'(x) - \frac{f(x+h) - f(x-h)}{2h} -\frac{f(x+h)e_1 - f(x-h)e_2}{2h} \right| \\
&\le
\left|f'(x) - \frac{f(x+h) - f(x-h)}{2h}\right| + \left|\frac{f(x+h)e_1 - f(x-h)e_2}{2h} \right| \\
&\le
\left|f'(x) - \frac{f(x+h) - f(x-h)}{2h}\right| + \frac{\epsilon M_2}{h} \\
\end{align*}
We also have
\begin{align*}
\left|f'(x) - \frac{f(x+h) - f(x-h)}{2h}\right|
&=
\left|f'(x) - \frac{f(x)+ hf'(x) + h^2f''(x)/2 + h^3f'''(\xi)/6 - (f(x) - hf'(x) +h^2f''(x)/2 - h^3f'''(\xi)/6}{2h}\right| \\
&=
\left|f'(x) - \frac{hf'(x) + h^3f'''(\xi)/6 + hf'(x) + h^3f'''(\xi)/6}{2h}\right| \\
&=
h^2|f'''(\xi)|/6 \\
&\le h^2 M_{1}/6 \qquad \blacksquare
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$n^{th}$ derivative of $y=(x \sqrt{1+x^2})^m$ I was trying to find the $n^{th}$ derivative and $y_n(0)$(nth derivative at zero) of the function
$$y=(x \sqrt{1+x^2})^m$$
Here is my approach:
$$y^2=(x^2 (1+x^2)^m$$
$$2yy_1=m(x^2 (1+x^2))^{m-1} \cdot(2x+4x^3)$$
$$2yy_1=m\frac{(x^2 (1+x^2))^{m}}{(x^2 (1+x^2)} \cdot(2x+4x^3)$$
$$2yy_1=m\frac{y^2}{(x^2 (1+x^2)} \cdot(2x+4x^3)$$
$$2yy_1((x^2 (1+x^2))=my^2(2x+4x^3)$$
$$y_1(x^2(1+x^2))=my(x+2x^3)$$
Derivating again I can write
$$y_2(x(1+x^2))+y_1(1+3x^2))=my_1(x+2x^3) + my(1+6x^2)$$
$$y_2(x(1+x^2))+y_1((1+3x^2)-m(x+2x^3))-my(1+6x^2)=0$$
Now my book says that the answer is
$$y_n(0)=m^2(m^2-2^2)(m^2-4^2).......[m^2-(2n-2)^2]; n=odd$$
$$y_n(0)=m^2(m^2-1^2)(m^2-3^2).......[m^2-(2n-1)^2]; n=even$$
But after applying Leibnitz theorem for the derivative, I don't end up with anything like this. Can anybody please tell me how to approach this question.
| You can use the series expansion:
$$
(1+x)^\alpha=\sum_{k=0}^\infty\pmatrix{\alpha \\ k}x^k
$$
where
$$
\pmatrix{\alpha \\ k}=\frac{\alpha\cdot(\alpha-1)\cdot\ldots\cdot(\alpha-k+2)\cdot(\alpha-k+1)}{k\cdot(k-1)\cdot\ldots\cdot2\cdot1}.
$$
Then, if $m$ is a positive integer,
$$
\left(x\sqrt{1+x^2}\right)^m=x^m\sum_{k=0}^\infty\pmatrix{m/2 \\ k}x^{2k}
=\sum_{k=0}^\infty\pmatrix{m/2 \\ k}x^{2k+m}
$$
Equating to the McLaurin series, we have, if $m$ is even
$$
y^{(n)}(0)=\begin{cases}
n!\pmatrix{m/2 \\ (n-m)/2} && n\ \text{even and}\ n\geq m\\
0 && n\ \text{odd}
\end{cases}
$$
and if $m$ is odd
$$
y^{(n)}(0)=\begin{cases}
0 && n\ \text{even} \\
n!\pmatrix{m/2 \\ (n-m)/2} && n\ \text{odd and}\ n\geq m
\end{cases}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2257800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{1}{5}$ and $abc = 5$, solve for $a^3 + b^3 + c^3$ I recently encountered this question and have been stuck for a while. Any help would be appreciated!
Q: Given that
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{5} \tag{1} \label{eq:1}$$
$$abc = 5 \tag{2} \label{eq:2}$$
Find $a^3 + b^3 + c^3$. It wasn't specified in the question but I think it can be assumed that $a, b, c$ are real numbers.
My approach:
$$ ab + ac + bc = \frac{1}{5} abc = 1 $$
$$ a^3 + b^3 + c^3 = (a+b+c)^3 - 3[(a + b + c)(ab + ac + bc) - abc] $$
$$ a^3 + b^3 + c^3 = (a+b+c)^3 - 3(a+b+c) + 15 $$
From there, I'm not sure how to go about solving for $a + b + c$.
Something else I tried was letting $x = \frac{1}{a}, y = \frac{1}{b}, z = \frac{1}{c}$, so we get $$ xyz = x + y + z = \frac{1}{5} $$Similarly, I'm not sure how to continue from there.
| As DanielV suspects, the question is indeed ill-posed: there are multiple possible outcomes for $a^3 + b^3 + c^3$. First of all notice that if $a$, $b$, and $c$ are positive, then in order for (1) to hold, each of them should be larger than 5. But this results in a contradiction for (2). It follows that at least one should be negative, but then (2) gives that at least two of $a,b,c$ should be negative. Without loss of generality, assume that $a > 0$ and $b, c < 0.$
Since $$\frac 1 b = \frac 1 5 - \frac 1 c - \frac 1 a,$$ we find
$$\frac{5}{ac} = b = \frac{1}{\frac 1 5 - \frac 1 c - \frac 1 a} = \frac{5ac}{ac-5a-5c},$$
so that $ (ac)^2 = ac - 5a - 5c.$ Solving this quadratic equality for $c$ gives
$$c = \frac {a-5}{2a^2} \pm\frac{\sqrt{(5-a)^2 - 20a^3}}{2a^2}.$$
By symmetry the same equality should hold for $b$,
$$b = \frac {a-5}{2a^2} \pm\frac{\sqrt{(5-a)^2 - 20a^3}}{2a^2}.$$
Supposing that $b$ and $c$ are not both equal, we can take $b$ with the + sign and $c$ with the minus sign. For all values of $a$ for which the argument of the square root is non-negative, the above choices for $b$ and $c$ solve the stated system. It can be checked that different choices of $a$, and corresponding choices for $b$ and $c$, yield different outcomes for $a^3 + b^3 +c^3$.
We assumed above that $b \neq c$. We can also assume that $b = c$, which gives the equations $a b^2 = 5$ and $ \frac 1 a + \frac 2 b = \frac 1 5$, resulting in the cubic equality $-b^3 - 10 + b = 0$. This may be solved to give (approximately) $b = c = -2.30891$ and $a = 0.937901$, yielding yet another value for $a^3 + b^3 + c^3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "5",
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Determine a constant that satisifies an inequality Determine the minimum real number $c$ such that $$\sum_ {k=1}^n \frac {a_k}{c^{a_1+a_2+ \dots +a_k}} \le 1,$$ for every positive integer $n$ and positive real numbers $a_1,a_2, \dots, a_n$.
| Let $x>1$ and $f(x)$ be the smallest real number $y$ satisfying that, for any integer $n$ and $n$ numbers $a_1, a_2, \cdots, a_n$, $\sum_ {k=1}^n \frac {a_k}{x^{a_1+a_2+ \dots +a_k}} \le y$. Let's find a closed form expression for $f(x)$.
First of all, $f(x) \ge \lim_{\epsilon \rightarrow 0}\frac{\epsilon}{x^\epsilon-1}=\frac{1}{\ln x}$ because the value of infinite sum $\sum_{k=1}^{\infty}\frac{\epsilon}{x^{k\epsilon}}$ is $\frac{\epsilon}{x^\epsilon-1}$ and this is strictly decreasing function of $\epsilon$ when $x>1$. We want to prove that $f(x)=\frac{1}{\ln x}$.
For any $x>1$, let's suppose that there are integer $M$ and $a_1, a_2, \cdots, a_M$ satisfying $\sum_ {k=1}^M \frac {a_k}{x^{a_1+a_2+ \dots +a_k}}>\frac{1}{\ln x}$. Let's also suppose that among these sequences, $a_1, a_2, \cdots, a_m$ is the shortest.
We know that $m>1$ because $\frac{a_1}{x^{a_1}}\le\frac{1}{e\ln x}$. By the assumption of shortest sequence, we also know that the sum generated by the sequence $a_2, a_3, \cdots, a_m$ does not exceed $\frac{1}{\ln x}$, in other words, $\sum_ {k=2}^m \frac {a_k}{x^{a_1+a_2+ \dots +a_k}}\le \frac{1}{\ln x}$. Let $A=\sum_ {k=2}^m \frac {a_k}{x^{a_1+a_2+ \dots +a_k}}$. By our assumption, $A<\frac{1}{\ln x}<\frac{A+a_1}{x^{a_1}}$ and $a_1$ maximizing $\frac{A+a_1}{x^{a_1}}$ is $a_1=\frac{1}{\ln x}-A$. Now we get$$\frac{1}{\ln x}<\frac{A+a_1}{x^{a_1}}\le\frac{x^A}{e\ln x}\le\frac{x^{1/\ln x}}{e\ln x}=\frac{1}{\ln x}$$which is contradiction. Therefore, our assumption of existence of $M$ and sequence $a_1, a_2, \cdots, a_M$ satisfying $\sum_ {k=1}^M \frac {a_k}{x^{a_1+a_2+ \dots +a_k}}>\frac{1}{\ln x}$ is wrong and it is true that $f(x)=\frac{1}{\ln x}$.
We are looking for smallest $x$ satisfying $f(x) \le 1$. The answer should be $e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Verifying the identity $\frac{1}{\cos x} - \cos x = \sin x \tan x$ $\frac{1}{\cos x} - \cos x = \sin x * \tan x$
I have tried a few things and nothing works
Left Side
$\frac{1}{\cos x} - \cos x$
$1 - (\cos x)^2$
$1 - 1 + \frac{\cos 2x}{2}$
$ \frac{\cos 2x}{2}$
$ \frac{\cos x}{2} \times \frac{\cos}{2}$
... And I am out in left field!
| $$\frac{1}{\cos x} - \cos x = \frac{1- \cos^2 x}{\cos x}$$
Now I think you can finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259676",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Does $\int _0^1 \cfrac{x}{x+\cfrac{x^{2}}{x^2+\cfrac{x^3}{x^3+\cdots}}} \, dx$ exist? $$\int_0^1 \cfrac x {x+\cfrac{x^2}{x^2+\cfrac{x^3}{x^3+\cdots}}} \, dx$$
Does this integral exist?
Any hint will be appreciated.
| $\int _{0}^{1} \frac{x}{x＋\frac{x^{2}}{x^2＋\frac{x^{3}}{x^{3}＋…}}}dx$
Let's call that monstrous integrand $u$. If you notice, $u=\frac{x}{x+xu}=\frac{1}{1+u}$.
$u+u^2=1$, so $u^2+u-1=0$. This means $u=\frac{\sqrt{5}-1}{2}$.
$\int_0^1 \frac{\sqrt{5}-1}{2} \,dx = \boxed{\frac{\sqrt{5}-1}{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2259965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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How can I solve $\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$ How can I solve the following integral?
$$\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx$$
$$\begin{align}
\int \frac{1}{\sqrt{x^2 + 7x} + 3}\, dx&=\int \frac{1}{\sqrt{\left(x+\frac72\right)^2-\frac{49}{4}} + 3}\, dx
\end{align}$$
$$u = x+\frac{7}{2}, \quad a = \frac{7}{2}$$
$$\int \frac{1}{\sqrt{u^2 - a^2} +3}\, \,du$$
Attempt I - By Trigonometric substitution
$$\sqrt{u^2 - a^2} = \sqrt{a\sec^2\varTheta - a^2} = \sqrt{a^2(\sec^2\varTheta - 1)} = \sqrt{a^2\tan^2\varTheta} = a\tan\varTheta $$
$$u = a\sec\varTheta \implies du = a\sec\varTheta \tan\varTheta d\varTheta$$
$$\int \frac{1}{a\tan\varTheta + 3}\, a\sec\varTheta \tan\varTheta d\varTheta$$
$$a\int \frac{\sec\varTheta \tan\varTheta}{a\tan\varTheta + 3}\, d\varTheta$$
How can I continue here ?
Attempt II - By First Euler Substitution
$$\sqrt{u^2 - a^2} = u - t$$
$\sqrt{u^2 - a^2} = u - t$
$ u^2 - a^2 = (u - t)^2$
$u^2 - a^2 = u^2 -2ut + t^2$
$- a^2 = -2ut + t^2$
$2ut = t^2 + a^2$
$$u = \frac{t^2 + a^2}{2t} \implies du = \frac{1}{2} - \frac{a^2}{2t^2}\, dt$$
Thus the integral takes the form
$$\int \frac{1}{u - t +3}\, \left( \frac{1}{2} - \frac{a^2}{2t^2}\right) \,du$$
Since $u = \frac{t^2 + a^2}{2t}$ and $\left( \frac{1}{2} - \frac{a^2}{2t^2}\right) = -\frac{a^2 - t^2}{2t^2}$ then
$$I = \int \frac{1}{ \frac{t^2 + a^2}{2t} - t +3} \left( -\frac{a^2 - t^2}{2t^2} \right) \, dt$$
$$-\int \frac{a^2 - t^2}{ ta^2 - t^3 +6t^2}\, dt$$
$$-\int \frac{a^2}{ ta^2 - t^3 +6t^2}\, dt + \int \frac{t^2}{ ta^2 - t^3 +6t^2}\, dt$$
How can I continue?
Attempt III -By First Euler Substitution
$$\sqrt{x^2 + 5x} = x + t$$
$\sqrt{x^2 + 5x} = x + t$
$ x^2 + 5x = (x + t)^2$
$x^2 + 5x = x^2 + 2ut + t^2$
$x^2 + 5x = x^2 + 2xt + t^2$
$5x = 2xt + t^2$
$5x -2xt = t^2$
$x(5 -2t) = t^2$
$$x = \frac{t^2}{(5 -2t)} \implies \frac{dt}{dx} = -\frac{2(t - 5)t}{(5 - 2t)^2 } \iff dx = -\frac{(5 - 2t)^2 }{2(t - 5)t}\,dt $$
$$\int \frac{1}{x + t + 3}\, \left( -\frac{(5 - 2t)^2 }{2(t - 5)t}\right)dt$$
$$-\int \frac{1}{\frac{t^2}{(5 -2t)} + t + 3}\, \left(\frac{(5 - 2t)^2 }{2(t - 5)t}\right)dt$$
$$-\int\frac{(5 - 2t)^2}{\frac{2(t - 5)t^3}{5 -2t} + 2(t - 5)t^2 + 6(t - 5)t}\,dt$$
How can I continue here?
| Start from your first result and multiply numerator and denominator with $\sqrt{u^2-a^2}-3$:
$$\int\frac{1\times \color{red}{\sqrt{u^2-a^2}-3}}{\sqrt{u^2-a^2}+3 \times \color{red}{\sqrt{u^2-a^2}-3}}du=\int\frac{\sqrt{u^2-a^2}-3}{u^2-a^2-9}du$$
Now we are going to split this integration into two parts. Each part will be calculated separately:
$$\int\color{blue}{\frac{\sqrt{u^2-a^2}}{u^2-a^2-9}du} -\int\color{purple}{\frac{3}{u^2-a^2-9}du}$$
First we calculate the blue part of the integration. We do this by multiplying numerator and denominator with $\sqrt{u^2-a^2}$. Then the integral can be broken as follows:
$$\int\frac{\sqrt{u^2-a^2}\times\color{red}{\sqrt{u^2-a^2}}}{u^2-a^2-9\times\color{red}{\sqrt{u^2-a^2}}}du=\int\frac{u^2-a^2}{(u^2-a^2-9)\sqrt{u^2-a^2}}du$$
$$=\int\frac{A}{\sqrt{u^2-a^2}}du+\int\frac{B}{u^2-a^2-9}du$$
Then the appropriate values of $A$ and $B$ are:
$$A\left(u^2-a^2-9\right)+B\left(\sqrt{u^2-a^2}\right)=u^2-a^2 \implies A=1, B=\frac{9}{\sqrt{u^2-a^2}}$$
The integral becomes:
$$\int\frac{1}{\sqrt{u^2-a^2}}du+\int\frac{\frac{9}{\sqrt{u^2-a^2}}}{u^2-a^2-9}du$$
This first part can be calculated using Euler substitution:
$$\int\frac{1}{\sqrt{u^2-a^2}}=\color{green}{\log\left(\sqrt{u^2-a^2}+u\right)} +C$$
We introduce $9u^2$ in the denominator and we multiply both numerator and denominator with $a^2$. Then we divide this with $u^2-a^2$.
$$\int\frac{\frac{9a^2}{\sqrt{u^2-a^2}}}{\color{red}{9u^2-9u^2}+u^2a^2-a^4-9a^2}du=\int\frac{\frac{9a^2}{(u^2-a^2)^\frac{3}{2}}}{\frac{9u^2}{u^2-a^2}-a^2-9}du$$
Then we introduce a variable $v$:
$$v=\frac{3u}{\sqrt{u^2-a^2}}\implies dv=\frac{3a^2}{(u^2-a^2)^\frac{3}{2}}\implies \int\frac{3}{v^2-a^2-9}dv=\int\frac{3}{v^2-\left(\sqrt{a^2+9}\right)^2}dv=\color{green}{\frac{3\times coth^{-1}\left(\frac{v}{\sqrt{a^2+9}}\right)}{\sqrt{a^2+9}}}+C$$
The purple part of the integration can be calculated identically:
$$\int\frac{3}{u^2-a^2-9}du=\int\frac{3}{u^2-\left(\sqrt{a^2+9}\right)^2}du=\color{green}{\frac{3\times coth^{-1}\left(\frac{u}{\sqrt{a^2+9}}\right)}{\sqrt{a^2+9}}}+C$$Finally:
$$\int\frac{1}{\sqrt{u^2-a^2}+3}=\color{green}{\log\left(\sqrt{u^2-a^2}+u\right)+}\color{green}{\frac{3\times coth^{-1}\left(\frac{v}{\sqrt{a^2+9}}\right)}{\sqrt{a^2+9}}}\color{green}{-\frac{3\times coth^{-1}\left(\frac{u}{\sqrt{a^2+9}}\right)}{\sqrt{a^2+9}}}+C$$
| {
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"url": "https://math.stackexchange.com/questions/2261818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Angles between lateral faces of any rectangular-based pyramid I was just wondering if anyone had any idea how to solve this problem:
What is the angle between lateral faces of a rectangular-based pyramid with length a, width b, and height h, in terms of a, b and h?
Any responses are much appreciated.
| @Aretino's coordinate/vector approach might be the most accessible, but it's worth mentioning a couple of others.
Given any three-faced "corner", one can determine the dihedral angles from the face angles (or vice-versa) using an appropriate Spherical Law of Cosines. For the problem at hand, with this configuration ...
... we can write ...
$$\cos\angle APB = \cos\angle APC \cos \angle BPC + \sin\angle APC \sin\angle BPC \cos\angle PC \tag{$\star$}$$
where "$\angle PC$" indicates the dihedral angle along $\overline{PC}$. In our rectangular pyramid, $\angle APB$ is a right angle, so that $(\star)$ reduces to
$$\cos\angle PC = -\cot\angle APC \cot\angle BPC \tag{1}$$
Labeling a few distances, including $m$ and $n$ as altitudes of faces $\triangle CPA$ and $\triangle CPB$, ...
... we have ...
$$\cos \angle PC = -\frac{a/2}{m} \cdot \frac{b/2}{n} = -\frac{ab}{4mn} \tag{2}$$
Invoking Pythagoras,
$$m^2 = \left(\frac{b}{2}\right)^2 + h^2 = \frac{1}{4}\left( b^2 + 4 h^2 \right) \qquad\qquad n^2 = \frac{1}{4}\left( a^2 + 4 h^2 \right)$$
we have
$$\cos\angle PC = -\frac{ab}{\sqrt{(a^2 + 4 h^2 )(b^2 + 4 h^2 )}} \tag{3}$$
I would be remiss not to also mention a solution using a hedronometric ("area-based") Law of Cosines for Tetrahedra. In particular:
$$\begin{align}
&|\triangle PAC|^2 + |\triangle PBC|^2 - 2 |\triangle PAC||\triangle PBC|\cos\angle PC \\
= &|\triangle ACB|^2 + |\triangle APB|^2 - 2|\triangle ACB||\triangle APB|\cos\angle AB
\end{align} \tag{$\star\star$}$$
As dihedral $\angle AB$ is a right angle, isolating $\cos \angle PC$ gives:
$$\cos\angle PC = \frac{|\triangle PAC|^2 + |\triangle PBC|^2 - |\triangle ACB|^2 - |\triangle APB|^2}{2|\triangle PAC||\triangle PBC|} \tag{4}$$
Then, since ...
$$
|\triangle PAC| = \frac12 a m = \frac14 a \sqrt{b^2 + 4 h^2} \qquad\qquad
|\triangle PBC| = \frac12 b n = \frac14 b \sqrt{a^2 + 4 h^2 }$$
$$
|\triangle ACB| = \frac12 h |\overline{AB}| = \frac12 h \sqrt{a^2+b^2} \qquad\qquad |\triangle APB| = \frac12 ab$$
... we find that $(4)$ ultimately reduces to $(3)$. $\square$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving simultaneous equations with complex numbers and moduli Would anyone be able to give me a starting point as to how to approach this? A friend suggested squaring both sides, but I read something that said you can't square both sides of an equation(?).
Complex numbers and simultaneous equations question
If $x$ and $y$ are both real numbers, find all the solutions $(x,y)$ of the simultaneous equations
$$
\begin{cases}
\lvert x + i y \rvert &= 1\\[3pt]
\big\lvert x + i y - \frac{3}{2} \big\rvert &= 2
\end{cases}
$$
Thank you in advance!
Edit: So, would you square the real and imaginary parts separately? Giving: $x^2+y^2=1$ for the first one? Then, $(x-3/2)^2+y^2=4$ for the second one?
| From the first equation,
$$
\color{blue}{\lvert x + i y \rvert = \sqrt{x^{2}+y^{2}} = 1} \qquad \Rightarrow \qquad \color{blue}{y_{1}=\pm \sqrt{1-x^2}}
$$
From the second equation,
$$
\color{red}{\lvert x -\frac{3}{2} + i y \rvert =\sqrt{\left(x-\frac{3}{2}\right)^2+y^2}}
\qquad \Rightarrow \qquad
\color{red}{y_{2} =\frac{1}{2} \sqrt{-4 x^2+12 x+7}}
$$
Solve
$$
\color{blue}{y_{1}(x)} = \color{red}{y_{2}(x)}
$$
to see that
$$
x= -\frac{1}{4}, \qquad y = \pm \frac{\sqrt{15}}{4}
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\sin(3x)$ is equivalent to $3\sin(x)\cos^2(x)-\sin^3(x)$
The expression $\sin(3x)$ is equivalent to:
A. ...
My book states the right answer is B which is $3\sin(x)\cos^2(x)-\sin^3(x)$.
I tried:
$$\sin(x)\cos(2x)+\cos(x)\sin(2x) = \\
\sin(x)(2\cos^2(x)-1)+\cos(x)\cdot2\sin(x)\cos(x) = \\
2\cdot\sin(x)\cos(x)\cos(x)-\sin(x)+\cos(x)2\sin(x)\cos(x) = \\
\sin(x)(2\cos^2(x)-1+2\cos^2(x)) = \\
\sin(x)(4\cos^2(x)-1) = \\
4\cos^2(x)\sin(x)-\sin(x) = ???$$
What do I do next?
| Notice that \begin{equation}\begin{split}\sin(3x) &= 4\cos^2 x \sin x - \sin x\\&= 3\sin x \cos^2 x +(\sin x \cos^2x -\sin x )\\&=3\sin x \cos^2 x + \sin x\cdot( \cos^2x - 1)\\&= 3\sin x \cos^2 x -\sin^3 x \end{split}\end{equation}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there a straight forward way to prove that $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{31} > 3$ Using brute force, it is straight forward to calculate that $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{31} > 3$
Is there a more elegant way to demonstrate this?
What if I want to find the smallest $n$ such that $\frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n} > 4$
Is there a standard way to solve for $n$ without using brute force?
| Claim: $$\sum_{i=1}^{3^n}\frac{1}{i} \geq \frac{7}{6}+\frac{2}{3}n$$
whenever $n\geq2$.
Proof: By Mathematical Induction.
Take $n=3$, then instantly $1+\frac{1}{2}+\frac{1}{3}+\cdots +\frac{1}{27}\geq \frac{7}{6}+2 > 3$.
Hope this is not a brutal way and this can be easily generalized.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find cubic function whose graph has horizontal tangents at two points Find the cubic function $y = ax^3 + bx^2 + cx + d$ whose graph has horizontal tangents at the points $(-2,7)$ and $(2,1)$.
I find the derivative and set equation to zero, but then that only gives me one solution. I can't find all solutions for some reason.
| First, find the derivative of $y$ with respect to $x$:
$$y'=3ax^2+2bx+c$$
And because of the tangent points given, we know that
$$3a(-2)^2+2b(-2)+c=0$$
$$3a(2)^2+2b(2)+c=0$$
And since the graph of the cubic must pass through those points,
$$a(-2)^3+b(-2)^2+c(-2)+d=7$$
$$a(2)^3+b(2)^2+c(2)+d=1$$
When we simplify these equations, we get as system of four equations with four variables to solve for:
$$12a-4b+c=0$$
$$12a+4b+c=0$$
$$-8a+4b-2c+d=7$$
$$8a+4b+2c+d=1$$
This system can be solved easily using matrices. After solving, we get
$$a=\frac{3}{16}$$
$$b=0$$
$$c=-\frac{9}{4}$$
$$d=4$$
So the cubic must be
$$y=\frac{3}{16}x^3-\frac{9}{4}x+4$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Composite Function value Given: $f(f(x)).[1+f(x)] = -f(x)$. Find $f(3)$
Approach to the question:-
$fof(x) + f(x).fof(x) = -f(x)$
$fof(x) + f(x) + f(x).fof(x)+1=1$
$(1+f(x))(1+f(f(x))=1$
This implies that $1+f(x)$ and $1+fof(x)$ are reciprocals.
This gives $f(x)=\frac{1}{x}-1$ or $ f(x)=\frac{-1}{x}-1$ as solutions. (By inspection)
Therefore, $f(3)=\frac{-2}{3}$ or $f(3)=\frac{-4}{3}$
But the answer given is $f(3)=\frac{-3}{4}$
What is my mistake?
| Well $1+f(x)$ and $1+f(f(x))$ are reciprocals of each other and not of $x$ which means $$1+f(x)=\frac1{1+f(f(x))}\\1+f(f(x))=\frac{1}{1+f(x)}$$
Consider this instead since $f(x)=-1$ doesn't satisfy the above condition we can divide by $1+f(x)$ to get
$$f(f(x))=\frac{-f(x)}{1+f(x)}$$
Now putting $f(x)=t$ (this step requires justification like that f is onto though I guess the problem assumes it) we get
$$f(t)=\frac{-t}{1+t}$$
So $f(3)=\frac{-3}{4}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$\cos\gamma+\sin\gamma$ in a 3-5-7 triangle ($\gamma$ is the greatest angle) Clearly $\gamma$ has to be the angle oposite to the side of lenght 7. The law of cosines gives me $$7^2 = 5^2+3^2 - 2\cdot 5\cdot3\cos{\gamma}\Longleftrightarrow\cos{\gamma}=-\frac{7^2-5^2-3^2}{2\cdot5\cdot 3}=-\frac{1}{2}.$$
This value of $\cos{\gamma}$ means there can be two values of $\gamma$ satisfying it in $0\leq\gamma \leq 2\pi$. So $\sin{\gamma}$ can be either negative or positive. Using the trigonometric identity $\sin^2{\theta}+\cos^2{\theta}=1$, I obtain $$\sin{\gamma}=\pm\sqrt{1-\cos{\gamma}}=\pm\sqrt{1-\left(-\frac{1}{2}\right)^2}=\pm\frac{\sqrt{3}}{2}.$$
Using the positive value I get the correct answer of $$\cos{\gamma} + \sin{\gamma}=-\frac{1}{2}+\frac{\sqrt{3}}{2}=\frac{\sqrt{3}-1}{2}.$$
Why is it incorrect to use the negative value of $\sin{\gamma}$ ?
| For $0^\circ<\gamma<180^\circ$ – the range of an angle in a non-degenerate triangle – $\sin\gamma$ must be positive.
| {
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System of two symmetric equations I'm trying to solve the following system of equations over positive reals, but I've got stuck.
$$
\left\{
\begin{array}{c}
ab + bc + ca = 12 \\
a + b + c + 2 = abc \\
\end{array}
\right.
$$
What I've tried: I've used Vieta's formulas to convert this system of equations to $x^3 + (a_0 + 2)x^2 + 12x + a_0$ by assuming that $a, b, c$ are roots of this polynomial ($a_0$ is just some parameter) and noticed that $abc \le 8$ due to $AM \ge GM$ for ${ab, bc, ca}$.
| Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the our conditions give $v^2=4$ and $w^3=3u+2$.
In another hand,
$$(a-b)^2(a-c)^2(b-c)^2=27(2u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6)\geq0,$$
which gives
$$(u-2)^2(2u+5)(6u+13)\leq0$$ or $u=2$, which gives $w^3=8$ and $a$, $b$ and $c$ are roots of the equation
$$x^3-6x^2+12x-8=0$$ or
$$(x-2)^3=0,$$
which gives the answer $a=b=c=2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Problem with square number I have a doubt with this problem :
"Show that there exist an infinite number of integers $n$ such that $$P(n)=\frac{\sum_{k=0}^{n}k^{2}}{n}=\frac{1^{2}+2^{2}+\cdots+n^{2}}{n}$$ is square number."
I find that $P(n)$ is integer if $n$ is prime number.
Sincerely,
| The formula for the sum of the first $n$ square numbers is given by
$$1^2+2^2+...+n^2=\frac{n(n+1)(2n+1)}{6}$$
Meaning that
$$P(n)=\frac{\frac{n(n+1)(2n+1)}{6}}{n}$$
$$P(n)=\frac{(n+1)(2n+1)}{6}$$
So now you just need to find out when
$$\frac{(n+1)(2n+1)}{6}$$
is a perfect square. Now let $a$ and $b$ be some two numbers (not necessarily integers) so that $ab=6$. Then $P(n)$ is a perfect square whenenver
$$\frac{n+1}{a}=\frac{2n+1}{b}$$
which can be solved for $n$ to get
$$n=\frac{b-a}{2a-b}$$
Now, by substitution, since $ab=6$, we have that
$$n=\frac{\frac{6}{a}-a}{2a-\frac{6}{a}}$$
$$n=\frac{6-a^2}{2a^2-6a}$$
Now all you need to do is show that there are an infinite number of numbers (not necessarily integers) $a$ so that
$$n=\frac{6-a^2}{2a^2-6}$$
is an integer.
| {
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Show monotonicity Originally, I want to show that
$$
\frac{\sqrt{a \cdot b + \frac{b}{a}x^2}\arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right)}{\sqrt{a \cdot b}\arctan \left(\frac{c}{\sqrt{a \cdot b}}\right)} \geq 1 \ \ \text{for} \ \ x, a,b,c > 0 \ .
$$
To do so, I figured it is sufficient to show that
$$
f(x) = \sqrt{a \cdot b + \frac{b}{a}x^2}\arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right)
$$
is monotonically increasing for $x > 0$.
Of course, I took the derivative $f'(x)$ and proceeded with the demand
$$
\frac{\frac{b}{a}x}{\sqrt{a \cdot b + \frac{b}{a}x^2}} \arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right) - \frac{c \frac{b}{a} x}{a \cdot b + \frac{b}{a} x^2 + c^2} > 0 \ .
$$
In the end, I got stuck with
$$
\arctan \left(\frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}\right) > \frac{c \sqrt{a \cdot b + \frac{b}{a}x^2}}{a \cdot b + \frac{b}{a} x^2 + c^2} \ .
$$
Inserting values for a,b, and c seems to work perfectly, but I can't manage to analytically solve the inequation.
Does anyone have an idea of how to approach this problem?
Edit:
I came across the Shafer-Fink inequality stating
$$
\frac{3y}{1+2\sqrt{1 + y^2}} < \arctan y < \frac{\pi y}{1 + 2\sqrt{1 + y^2}} \ .
$$
Can I substitute
$$
y = \frac{c}{\sqrt{a \cdot b + \frac{b}{a}x^2}}
$$
and therefore receive
$$
\arctan (y) > \frac{3y}{1+2\sqrt{1 + y^2}} > \frac{y}{1 + y^2} ?
$$
Is that a proper way?
| I'll first rearrange and reparametrise the original equality a little to make it easier to work with. Let $\alpha := c/\sqrt{ab}$, $z:= \sqrt{ 1 + (x/a)^2}$. Since $a,b,c >0$, it should be easy to see that the original inequality is equivalent to $$ z \arctan \frac{\alpha}{z} \overset{?}\ge \arctan \alpha$$ for every $\alpha >0, z> 1$ - much neater.
Note that the relation above is true for $z=1$. Now, going by your proof technique, we'll be done if we manage to show that $$ \arctan \frac{\alpha}{z} \ge \frac{\alpha/z}{1 + (\alpha/z)^2}$$ for every $\alpha >0, z>1$. But this is true since, for $\alpha/z>0$, $$\arctan \frac{\alpha}{z} = \int_0^{\alpha/z} \frac{1}{1+t^2} \mathrm{d}t \overset{(1)}\ge \int_0^{\alpha/z}\frac{1}{1+(\alpha/z)^2} \mathrm{d} t = \frac{\alpha/z}{1+(\alpha/z)^2},$$ where the inequality $(1)$ holds since $1+t^2$ is an increasing function for $t>0$.
| {
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About $142857$: proof that $\;3 \mid 1^n + 4^n + 2^n + 8^n + 5^n + 7^n$ Most of you already know the flabbergasting properties of cyclic numbers. Among these, $142857$ deserves to be the most famous and admired, being (in base $10$) the smallest and the only one without those spoilsport leading zeros. I personally love a couple of elegant features of its digits that aren't often underlined. Their sequence defines a graceful pattern on a calculator's keypad:
Another (partially related) feature is that they are exactly all the base-$10$ digits that are not divisible by $3$, each one taken once. Yet when you sum them up you get $1+4+2+8+5+7=27=3^3$, obviously divisible by $3$. I went on to discover that more generally
$$
\forall n,\qquad1^n + 4^n + 2^n + 8^n + 5^n + 7^n \equiv 0 \;\pmod 3 .
$$
Even if I'm not familiar with modular arithmetic (in the sense that I never really studied it — I've learned what I know here and there and I try to use it at my best following my instinct) I think I succeeded proving this equivalence. According to Stack Exchange's encouragement to answer own questions, I'll write my proof as an answer below, asking you to check it, improve it, or replace with a better one (in what case I won't hesitate accepting the new one as best answer). Thank you!
| I'll look at each addend one at a time.
$1^n$:
$$
1^n \equiv 1 \;\pmod 3
$$
$4^n$:
$$\begin{align}
4 &\equiv 1 &\pmod 3\\
4 \cdot 4 &\equiv 4 \equiv 1 &\pmod 3\\
&\;\;\vdots\\
4^n &\equiv 1 &\pmod3
\end{align}$$
$2^n$:
$$\begin{align}
2 &\equiv 2 &\pmod 3\\
2 \cdot 2&\equiv 4 \equiv 1 &\pmod 3\\
2 \cdot 2 \cdot 2 &\equiv 2 &\pmod 3\\
&\;\;\vdots\\
2^n &\equiv \begin{cases}
2 & \text{if $n$ is odd} \\
1 & \text{if $n$ is even}
\end{cases} &\pmod3
\end{align}$$
I'd rewrite this in a synthetic way so that in the following I won't need to examine odd and even cases separately:
$$
2^n \equiv 1+(n\bmod 2) \qquad\pmod 3
$$
$8^n$:
$$\begin{align}
8 &\equiv 2 &\pmod 3\\
8 \cdot 8&\equiv 16 \equiv 1 &\pmod 3\\
8 \cdot 8 \cdot 8 &\equiv 8 \equiv 2 &\pmod 3\\
&\;\;\vdots\\
8^n &\equiv \begin{cases}
2 & \text{if $n$ is odd} \\
1 & \text{if $n$ is even}
\end{cases} = 1+(n\bmod 2) &\pmod 3\\
\end{align}$$
$5^n$:
$$\begin{align}
5 &\equiv 2 &\pmod 3\\
5 \cdot 5&\equiv 10 \equiv 1 &\pmod 3\\
5 \cdot 5 \cdot 5 &\equiv 5 \equiv 2 &\pmod 3\\
&\;\;\vdots\\
5^n &\equiv \begin{cases}
2 & \text{if $n$ is odd} \\
1 & \text{if $n$ is even}
\end{cases} = 1+(n\bmod 2) &\pmod 3\\
\end{align}$$
$7^n$:
$$\begin{align}
7 &\equiv 1 &\pmod 3\\
7 \cdot 7 &\equiv 7 \equiv 1 &\pmod 3\\
&\;\;\vdots\\
7^n &\equiv 1 &\pmod3
\end{align}$$
Now I'm ready to sum them up:
$$\begin{align}
1^n + 4^n + 2^n + 8^n + 5^n + 7^n \equiv&\\
\equiv\ 1 + 1 + \left[1+(n\bmod 2)\right] + \left[1+(n\bmod 2)\right] + \left[1+(n\bmod 2)\right] + 1 =&\\
=3+3\cdot\left[1+(n\bmod 2)\right]=&\\
=6+3\cdot(n\bmod 2)\equiv& \ 0 \;\pmod 3\\
\end{align}$$
which is the wanted equivalence.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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} |
Multiple of $5$ If two natural nos. $x,y$ are selected at random then the probability that $x^2+y^2$ is a multiple of 5?
I think they should give a finite set for selecting natural nos. otherwise there are infinite ordered pairs of $x$ and $y$.
Is the question incomplete or am I missing some trick?
| Modulo $5$, the residues of $x^2$ take on the values $0, 1, 4, 4, 1$.
So $x^2 + y^2 \equiv 0 \bmod 5$ will be true if both residues are $0$ or if one is $1$ and the other is $4$.
This implies the probability is $\frac{1}{5} \cdot \frac{1}{5} + 2 \cdot \frac{2}{5} \cdot \frac{2}{5} = \frac{9}{25}$
Edit: Let's say we wanted to see the remainder after dividing integer $n$ by integer $k$. This means we could write $n = qk + r$ where $0 \leq r \lt k$ where $q$ is an integer. For example $20$ divided by $7$ can be modeled as $20 = 2 \cdot 7 + 6$.
Then:
$$\begin{align}
\frac{x^2+y^2}{5} &= \frac{(q_1 \cdot 5 + r_1) +(q_2 \cdot 5 + r_2 )}{5} \\
&= \frac{(q_1 + q_2) \cdot 5 + (r_1 + r_2) }{5} \\
&= (q_1 + q_2) + \dfrac{r_1 + r_2}{5}
\end{align}$$
In other words, after we divide $x^2 + y^2$ by $5$, we'll get some integer $q_1, q_2$ values (that we don't care about), but then we need the sum of the remainders to be divisible by $5$ if $\frac{x^2+y^2}{5}$ is going to be an integer.
| {
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"source": "stackexchange",
"question_score": "1",
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} |
Find tangent equation for a curve which is perpendicular to a line
How do I find the point on the curve $y=1/(2x−1)$ where the tangent line will be perpendicular to the line defined by $x−2y−1=0$?
What I have so far:
The equation of the given line is $y= 0.5x+1$
Since the slope is $-0.5$, the tangent slope must be $2$ .: $y=-2x+b$ (tangent line)
The differential of the curve is $y=-2/((2x-1)^2)$.
I can't figure out how to get the b value for the tangent, nor the point at which this happens.
| The line
$x - 2y - 1 = 0 \tag{1}$
may also be written
$y = \dfrac{1}{2}x - \dfrac{1}{2}; \tag{2}$
from this we see its slope is $\frac{1}{2}$. Thus the slope of a line normal or perpendicular to this line is $-2$.
For any value of $x$ other than $\frac{1}{2}$, the slope of the curve
$y(x)= \dfrac{1}{2x - 1} = (2x - 1)^{-1} \tag {3}$
is given by
$y'(x) = -2(2x - 1)^{-2}, \tag{4}$
We seek the points on the curve (3) where the slope is $-2$, yielding
$-2(2x - 1)^{-2} = -2, \tag{5}$
or
$(2x - 1)^{-2} = 1, \tag{6}$
or
$(2x - 1)^2 = 1; \tag{7}$
thus
$2x - 1 = \pm 1, \tag{8}$
whence
$x = 0, 1. \tag{9}$
That
$y'(0) = -2, y'(1) = -2 \tag{10}$
is easily checked using (4); thus the slope of the curve (3) is normal to the line (2) at the points $(0, -1)$ and $(1, 1)$.
Note added in edit: it will be observed that the value of $b$, that is, the $y$-intercept of the tangent lines to (3), does not enter in to the above calculations, by reason of the fact that the problem does not call for these lines, only for certain points on the given curve. However, having found these points where the slope is $-2$, the corresponding $b$-values are easily discovered. Each of these lines is of slope $-2$, and is represented by an equation of the form
$y = -2x + b; \tag{11}$
if $(0, -1)$ lies on such a line, we must have
$-1 = -2 \cdot 0 + b, \tag{12}$
or
$b = -1; \tag{13}$
the other $b$-value satisfies
$1 = -2 \cdot 1 + b, \tag{14}$
whence
$b = 3. \tag{15}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Riemann-Stieltjes sum Evaluate $$\lim_{n \to \infty} \sum_{k=1}^n \left[\left(\cos\left(\frac{2\pi(k+1)}n \right) +\cos\left(\frac{2\pi k}n\right) \right) \left(\sin\left(\frac{2\pi(k+1)}n \right) -\sin\left(\frac{2\pi k}n\right) \right) \right]$$
| First, by trigonometric identities involving sums and products, we have
$$\begin{align}
&\left(\cos\left(\frac{2\pi(k+1)}n \right) +\cos\left(\frac{2\pi k}n\right) \right) \left(\sin\left(\frac{2\pi(k+1)}n \right) -\sin\left(\frac{2\pi k}n\right) \right) \\ &=4\cos\left(\frac{2\pi(k+1/2)}n \right)\cos\left(\frac{2\pi(k+1/2)}n\right)\cos\frac{\pi}n\sin\frac{\pi}n \\
&=2\cos^2\left(\frac{2\pi(k+1/2)}n\right)\sin\frac{2\pi}n.\\
&=\left( 2\cos^2\left(\frac{2\pi(k+1/2)}n\right)-1+1\right)\sin\frac{2\pi}n \\
&=\cos\left(\frac{4\pi(k+1/2)}n\right)\sin\frac{2\pi}n+\sin\frac{2\pi}n.\end{align}
$$
By Euler's theorem, the sum over $k=1$ through $k=n$ becomes
$$
\Re \left( \sum_{k=1}^n e^{ 4\pi i\frac{k+1/2}n } \right) \sin\frac{2\pi}n + n\sin\frac{2\pi }n.
$$
By geometric sum formula, we have
$$
\Re \left( e^{2\pi i /n} e^{4\pi i/n} \frac{ e^{4\pi i }-1}{ e^{4\pi i/n} -1} \right)\sin\frac{2\pi}n + n\sin\frac{2\pi}n.
$$
We see that the quantity inside paranthesis is zero because $e^{4\pi i}=1$. Then as $n\rightarrow\infty$, we are left with
$$
n\sin \frac{2\pi}n = \frac{\sin\frac{2\pi}n}{\frac1n}\rightarrow 2\pi.
$$
Therefore, the limit is $2\pi$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculate the sum of $\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}$ I would like to calculate the sum for $x\in(-1,1)$ of this: $$\sum_{n=1}^{\infty} \frac{x^{n+1}}{n(n+1)}$$
So far I managed that
$$\int \frac{x^n}{n}dx = \frac{x^{n+1}}{n(n+1)}, $$
and
$$\sum_{n=1}^\infty \frac{x^n}{n}=-\log(1-x), $$
and
$$ \int -\log(1-x) dx \overset{c}{=} x+(1-x)\log(1-x). $$
But I really don't know if I can put it somehow together or not.
| When $|x| < 1$:
$$\begin{align*}\sum_{n=1}^\infty \dfrac{x^{n+1}}{n(n+1)} &= \sum_{n=1}^\infty \dfrac{x^{n+1}}{n} - \dfrac{x^{n+1}}{n+1} \\ &= x\sum_{n=1}^\infty \dfrac{x^n}{n} - \sum_{n=1}^\infty \dfrac{x^{n+1}}{n+1} \\ &= x\sum_{n=1}^\infty \dfrac{x^n}{n} - \sum_{n=1}^\infty \dfrac{x^n}{n} + x \\ & = -x\ln(x-1) + \ln(x-1) + x\\ &= x+(1-x)\ln (1-x)\end{align*}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
$(a^2+b^2)\cdot (c^2+d^2) = z^2+1$ then $z = ac+bd$ ? I have the following equation :
$$(a^2+b^2)\cdot (c^2+d^2) = z^2+1$$ with $a, b, c, d, z \in \mathbb{N}$
Then how can I prove that : $z = ac+bd$ ?
I tried Brahmagupta identity but it doesn't seem to work...
| The statement is false. Counter-example:
$$(9^2+7^2)(4^2+1^2) = 2210 = 47^2 + 1
\quad\text{ but }\quad
47 \ne
\begin{cases}
43 &= 9\cdot 4 + 7\cdot 1\\
37 &= 9\cdot 1 + 7\cdot 4
\end{cases}$$
Please note that $47$ is the smallest value of $z$ where $z^2+1$ contains three distinct prime factors of the form $4k+1$. This allow one to express one of
$a^2+b^2$ or $c^2+d^2$ as sum of two squares in more than one way.
As an example, the first factor above has following two representations:
$$9^2 + 7^2 = 11^2+3^2$$
and $11\cdot 4 + 3\cdot 1$ do equal to $47$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
} |
If $a,b,c \geq 1$ then prove $\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} < \sqrt{abc+c}$ If $a,b,c \geq 1$, then prove:
$$\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} < \sqrt{abc+c}$$
My try:
$$A=\sqrt{a-1} + \sqrt{b-1} + \sqrt{c-1} \\ A^2 =(a-1)+(b-1)+(c-1)+2\sqrt{(a-1)(b-1)}+2\sqrt{(a-1)(c-1)}+2\sqrt{(b-1)(c-1)} $$
$$B=\sqrt{abc+c} \\B^2=abc+c=c(ab+1) $$
Now what?
| Substitute $a=x^2+1$, $b=y^2+1$ and $c=z^2+1$ for $x,y,z\ge 0$ to get
$$x+y+z\le\sqrt{(z^2+1)\left(\left(x^2+1\right)\left(y^2+1\right)+1\right)}$$
Now after squaring and grouping we obtain
\begin{align*}x^2+y^2+z^2+2xy+2yz+2zx\le(z^2+1)\left((x^2+1)(y^2+1)+1\right) \\
0\le z^2(x^2+1)(y^2+1)-2z(x+y)+x^2y^2-2xy+2 \\
0\le z^2(x^2+1)(y^2+1)-2z(x+y)+(xy-1)^2+1
\end{align*}
We'll prove that $z^2(x^2+1)(y^2+1)-2z(x+y)+1\ge 0$. The determinant of this quadratic in $z$ is $\Delta =4(x+y)^2-4(x^2+1)(y^2+1)=-4(xy-1)^2\le0$, hence the inequality is true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that if $5^n$ begins with $1$, then $2^{n+1}$ also begins with $1$
Prove that if $5^n$ begins with $1$, then $2^{n+1}$ also begins with $1$ where $n$ is a positive integer.
In order for a positive integer $k$ to begin with a $1$, we need $10^m \leq k < 2 \cdot 10^m$ for some positive integer $m$. Thus since $5^n$ begins with a $1$, we have $10^m \leq 5^n < 2 \cdot 10^m$ for some positive integer $m$. Then multiplying by $\dfrac{2^{n+1}}{5^n}$ we get $10^m \cdot \dfrac{2^{n+1}}{5^n} \leq 2^{n+1} < 10^m \cdot \dfrac{2^{n+2}}{5^n}$, but I didn't see how this helped.
| Note that:
$$2\cdot10^n=5^n\cdot 2^{n+1}$$
If $5^n$ begins with $1$ then
$$10^m<5^n<2\cdot10^m\to10^m< \frac{2\cdot10^n}{2^{n+1}}<2\cdot10^m$$
what give us
$$2^{n+1}< 2\cdot 10^{n-m}\text{ and } 2^{n+1}>10^{n-m}\Leftrightarrow 10^{n-m}<2^{n+1}<2\cdot 10^{n-m}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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How to figure out either series are absolute or conditional convergence $\sum_1^\infty \cos(\frac{\pi n}{3})(n^{\frac{1}{\sqrt[6]{n+6}}}-1)$ $$\sum_1^\infty \cos\left( \frac{\pi n} 3 \right) \left(n^{\frac 1 {\sqrt[6]{n+6}}}-1\right)$$
I tried to use Dirichlet, but unsuccessfully. Please, give me hints. Thanks a lot.
| \begin{align}
& \Big( \cos \frac{1\pi} 3, \quad \cos\frac{2\pi} 3, \quad \cos\frac{3\pi} 3, \quad \cos\frac{4\pi} 3, \quad \cos \frac{5\pi} 3, \quad \cos\frac{6\pi} 3, \quad \ldots \Big) \\[10pt]
= \frac 1 2 & \Big( \underbrace{1,\ 1,\ 0,\ -1,\ -1,\ 0},\ \underbrace{1,\ 1,\ 0,\ -1,\ -1,\ 0},\ \ldots\ldots\ldots \Big)
\end{align}
The sequence $1,1,0,-1,-1,0$ repeats forever.
So you have a sum of two positive terms, then a sum of two negative terms, etc.
The alternating series test says that if the terms decrease and approach $0$ and have alternating signs, then the series converges.
Whether it converges absolutely takes more work to answer after that. Maybe I'll be back.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Computing $\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} $
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \space \text{ converges by the alternating series test, but what is its value?} $$
$1 - \sin(x) \space \text{ is an entire function} $, so by Weierstrass Factorization Theorem, it can be written as a product of its zeros, which occur at $x = \frac{(4n+1)\pi}{2} $ and $\frac{-(4n+3) \pi}{2}$
Equating this to its Taylor series, we get
$$1 - x + O(x^3) = (1 - \frac{2}{\pi}x)(1 + \frac{2}{3\pi}x)(1 - \frac{2}{5\pi}x)(1 + \frac{2}{7\pi}x)... $$
$$1 - x + O(x^3) = (1 - \frac{2}{\pi}x(1-\frac13)+O(x^2))(1 - \frac{2}{\pi}x(\frac15-\frac17)+O(x^2))... $$
$$1 - x + O(x^3) = 1 - \frac{2}{\pi}x(1-\frac13+\frac15-\frac17+...)+O(x^2)$$
$$x - O(x^3) = \frac{2}{\pi}x \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}-O(x^2)$$
Equating $x$ terms,
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = \frac{\pi}{2}$$
However, I seem to be off by a factor of $2$ (actual answer is $\frac{\pi}{4} $). Does anyone see where this went wrong?
| We can write $1-\sin{x}=2\cos^2{(x/2+\pi/4)}$, and I think we can now see the problem: all the roots have to be double roots. They'll be at $(4n + 1)\pi/2$ and $-(4n+3)\pi/2$, as you suggest, but all the factors have to be squared:
$$ 1-\sin{x} = A \prod_{n=0}^{\infty} \left(1-\frac{2x}{(4n+1)\pi}\right)^2 \left(1+\frac{2x}{(4n+3)\pi}\right)^2, $$
and we have to worry about the value of $A$ to make sure we have the right scaling, about which the zeros tell us nothing. Of course, we actually know that $A=1$ by putting $x=0$.
Okay, so
$$ 1-\sin{x} = \prod_{n=0}^{\infty} \left(1-\frac{2x}{(4n+1)\pi}\right)^2 \left(1+\frac{2x}{(4n+3)\pi}\right)^2 = \prod_{k=0}^{\infty} \left(1-(-1)^k\frac{2x}{(2k+1)\pi}\right)^2. $$
Now let's get to the series. The easiest way to differentiate a product is to differentiate the logarithm, since this turns it into a sum:
$$ \log{(1-\sin{x})} = 2\sum_{k=0}^{\infty} \log{\left(1-(-1)^k\frac{2x}{(2k+1)\pi}\right)}. $$
Now differentiate:
$$ -\frac{\cos{x}}{1-\sin{x}} = -\frac{4}{\pi}\sum_{k=0}^{\infty} \frac{(-1)^{k}}{4k+1} \frac{1}{1-(-1)^{k}2x/((2k+1)\pi)} $$
Putting $x=0$ and multiplying everything by $-1$ gives
$$ 1 = \frac{4}{\pi} \sum_{k=0} \frac{(-1)^k}{4k+1}, $$
as required.
| {
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"source": "stackexchange",
"question_score": "6",
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Calculate $\lim_{x\to a}(2-\frac{x}{a} )^{\tan( \frac{\pi x}{2a})}$
I would like to calculate
$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)},\quad a \in\mathbb{R}^* \,\,\text{fixed} $$
we've
$$\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}=e^{\tan\left( \dfrac{\pi x}{2a}\right)\ln\left(2-\dfrac{x}{a} \right)}.$$
Note that:
$$\ln\left(2-\dfrac{x}{a} \right)\sim_{a}1-\dfrac{x}{2}.$$
Now we have $\dfrac{\pi x}{2a}\underset{x\to a}{\longrightarrow} \dfrac{\pi}{2}$, i.e.$\dfrac{\pi x}{2a}-\dfrac{\pi}{2} \underset{x\to a}{\longrightarrow} 0$ and $\tan h \sim_{0}h.$
I'm stuck.
Update: here is another way :
\begin{aligned}
\left(2-\dfrac{x}{a} \right)^{\tan\left( \dfrac{\pi x}{2a}\right)}&=\exp\left[{\tan\left( \dfrac{\pi x}{2a}\right)\ln\left(2-\dfrac{x}{a} \right)}\right].\\
&=\exp\left[ \left(1-\dfrac{x}{a}\right)\tan\left(\dfrac{\pi x}{2a}-\dfrac{\pi}{2}+\dfrac{\pi}{2} \right).\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{1-\dfrac{x}{a}}\right]\\
&=\exp\left[ -\dfrac{\left(1-\dfrac{x}{a} \right)}{\tan\left(\dfrac{\pi x}{2a}-\dfrac{\pi}{2} \right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
&=\exp\left[ -\dfrac{\left(1-\dfrac{x}{a} \right)}{\tan\left(-\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)\right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
&=\exp\left[ \dfrac{2}{\pi} \dfrac{\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)}{\tan\left(\dfrac{\pi}{2}\left(1-\dfrac{x}{a} \right)\right)}.\dfrac{\ln\left(2-\dfrac{x}{a} \right)}{\left(1-\dfrac{x}{a}\right)}\right]\\
\end{aligned}
Thus
$$\lim\limits_{x\to a}\left(2-\dfrac{x}{a} \right)^{\tan\left(\dfrac{\pi x}{2a}\right)} =e^{\dfrac{2}{\pi}} $$
*
*Am i right beside i'm intersted in way which use equivalents
| Set $y=x-a$, then
$$
\lim_{x\to a}\left(2-\frac{x}{a}\right)^{\tan\left(\frac{\pi x}{2a}\right)}=\lim_{y\to0}\left(1-\frac{y}{a}\right)^{-\cot\left(\frac{\pi y}{2a}\right)}=\\=\lim_{y\to0}\exp
\left(-\frac{\log\left(1-\frac{y}{a}\right)}{\tan\left(\frac{\pi y}{2a}\right)}\right)=\lim_{y\to a}\exp{\left(\frac{y}{a}\cdot\frac{2a}{\pi y}\right)}=e^{2/\pi}
$$
| {
"language": "en",
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"question_score": "7",
"answer_count": 5,
"answer_id": 4
} |
Find $\tan x$ if $x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right)$ Find $\tan x$ if $$x=\arctan(2 \tan^2x)-\frac{1}{2}\arcsin\left(\frac{3\sin2x}{5+4\cos 2x}\right) \tag{1}$$
First i converted $$\frac{3 \sin 2x}{5+4 \cos 2x}=\frac{6 \tan x}{9+\tan^2 x}$$
So
$$\arcsin\left( \frac{6 \tan x}{9+\tan^2x}\right)=\arctan \left( \frac{6 \tan x}{9-\tan^2x}\right)$$ Now using above result $(1)$ can be written as
$$2x=2 \arctan(2 \tan^2 x)-\arctan \left( \frac{6 \tan x}{9-\tan^2x}\right) \tag{2}$$
But $$2\arctan (\theta)=\arctan \left(\frac{2 \theta}{1-\theta^2}\right)$$
So $(2)$ becomes
$$2x=\arctan \left(\frac{4 \tan^2x}{1-4 \tan^4x}\right)-\arctan \left( \frac{6 \tan x}{9-\tan^2x}\right)$$
Now using $$\arctan(a)-\arctan(b)=\arctan\left(\frac{a-b}{1+ab}\right)$$ i am getting a sixth degree polynomial in $\tan x$.
is there any better approach?
| Where you have left off,
$$\dfrac{6\tan x}{9-\tan^2x}=\dfrac{2\cdot\dfrac{\tan x}3}{1-\left(\dfrac{\tan x}3\right)^2}$$
Using my answer here, Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,
$$2\arctan\dfrac{\tan x}3=\begin{cases} \arctan\dfrac{6\tan x}{9-\tan^2x} &\mbox{if } \left(\dfrac{\tan x}3\right)^2\le1\iff-3\le\tan x\le3\\ \pi+\arctan\dfrac{6\tan x}{9-\tan^2x} & \mbox{if } \tan x>3\text{ or }\tan x<-3\end{cases}$$
Now the tougher case $\tan x>3$ or $\tan x<-3;$
$$x=\arctan(2\tan^2x)+\dfrac\pi2-\arctan\dfrac{\tan x}3$$
$$\tan x=\tan\left(\dfrac\pi2+\arctan(2\tan^2x)-\arctan\dfrac{\tan x}3\right)$$ $$=-\cot\left(\arctan(2\tan^2x)-\arctan\dfrac{\tan x}3\right)$$
$$\iff\tan x=-\dfrac{1+2\tan^2x\cdot\dfrac{\tan x}3}{2\tan^2x-\dfrac{\tan x}3}$$
On simplification, we should get a cubic equation in $\tan x$
Can you handle the simpler case $-3\le\tan x\le3$ to reach at $$\tan x(\tan^3x-3\tan x+2)=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295907",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof for sum of product of four consecutive integers I had to prove that
$(1)(2)(3)(4)+\cdots(n)(n+1)(n+2)(n+3)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}$
This is how I attempted to do the problem:
First I expanded the $n^{th}$ term:$n(n+1)(n+2)(n+3)=n^4+6n^3+11n^2+6n$.
So the series $(1)(2)(3)(4)+\cdots+(n)(n+1)(n+2)(n+3)$ will be equal to
$(1^4+2^4+\cdots+n^4)+6(1^3+2^3+\cdots+n^3)+11(1^2+2^2+\cdots+n^2)+6(1+\cdots+n)$
$=\frac{n(n+1)(2n+1)(3n^2-3n-1)}{30}+6*\frac{n^2(n+1)^2}{4}+11*\frac{(n+1)(2n+1)(n)}{6}+6*\frac{n(n+1)}{2}$
I tried to factor out $\frac{n(n+1)}{5}$ and tried to manipulate the expressions in ways which further complicated things. How should I proceed? Can anyone please give me some clues? Any help is appreciated.
| There is a mistake here:
$$1^4+2^4+\cdots+n^4=\frac{n(n+1)(2n+1)(3n^2-3n-1)}{30}$$
It should be
$$1^4+2^4+\cdots+n^4=\frac{n(n+1)(2n+1)(3n^2+3n-1)}{30}$$
You approach has no problem, except this minor mistake.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Help with simplification question: $\frac{x}{x-1} - \frac{x}{x+1}$ Write $\frac{x}{x-1} - \frac{x}{x+1}$ as a single fraction in its simplest form.
I got $\frac{x}{x-1}$ but according to the mark scheme this is incorrect. Could someone analyze my working out and see where I went wrong?
Working out:
Combined both fractions, then expanded brackets
$$\frac{x^2}{(x - 1)(x + 1)} = \frac{x^2}{x^ 2 - 1}$$
Further simplified
$$\frac{x(x)}{x(x) - 1}$$
Removed $x$ from top and bottom
$$\frac{x}{x-1}$$
Any help would be much appreciated.
| \begin{align}
\frac{x}{x-1}-\frac{x}{x+1} &= \frac{x(x+1)-x(x-1)}{(x-1)(x+1)} \\
&=\frac{x^2+x-(x^2-x)}{x^2-1} \\
&=\frac{x^2+x-x^2+x}{x^2-1} \\
&=\frac{2x}{x^2-1}
\end{align}
Remark: Without the answer scheme, we should know that soemthing went wrong if $$a-b=a$$ if $b \neq 0$ in general.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2301842",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve the $u'' + u + \sin (x) = 0$ Could someone help me to obtain the general solution to this differential equation?
I need to solve the DE: $$u'' + u + \sin (x) = 0$$
Note: "I got as far as the general solution for $\;u''+u=0,$ but it was the $\;\sin(x)$ term that confused me."
| You are looking at the 2nd order ODE with constant coefficients, which is linear. Hence, general solution comes from solving the homogeneous ODE $$u''+u=0$$ which you can do by substituting $u = e^{rx}$, which yields the equation $r^2+1=0$. You get $r = \pm i$ and hence
$$
u_h(x) = \alpha e^{ix} + \beta e^{-ix} = a \sin x + b \cos x
$$
is the general solution. To pick the particular solution for your free term, let
$$
u_p(x) = Ax \sin x + B x \cos x
$$
and note that
$$
\begin{split}
u_p'(x) &= A\sin x + Ax \cos x + B \cos x - Bx \sin x \\
&= (A-Bx) \sin x + (Ax+B) \cos x.
\end{split}
$$
Similarly,
$$
\begin{split}
u_p''(x) &= -B\sin x + (A-Bx) \cos x + A \cos x - (Ax+B) \sin x \\
&= (-Ax-2B) \sin x + (2A-Bx) \cos x.
\end{split}
$$
Hence
$$
\begin{split}
0 = u_p(x) + u_p''(x) + \sin x = (1-2B) \sin x + 2A \cos x.
\end{split}
$$
which implies $B = 1/2$ and $A = 0$, so $u_p(x) = x \cos (x) /2$ and hence the general solution would be
$$
u(x) = u_p(x) + u_h(x) = a \sin x + \left(b + \frac{x}{2} \right) \cos x
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2302900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Find $\lim_{x\to 0} \left\lfloor \frac{\tan 2x}{\sin x} \right\rfloor $
Find the limit
$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor $$
My try:
$$ \tan 2x =\dfrac{\sin 2x}{\cos 2x}$$
$$\sin 2x =2\cos x\sin x$$
So:
$$\dfrac{\tan 2x}{\sin x}=\dfrac{2\cos x}{\cos 2x}$$
So:
$$\lim_{x\to 0} \left\lfloor \dfrac{\tan 2x}{\sin x} \right\rfloor=
\lim_{x\to 0}\left\lfloor\dfrac{2\cos x}{\cos 2x}\right\rfloor $$
Now what?
| $x\neq0$ and around $0$ we have
$$\frac{2\cos{x}}{\cos2x}>2$$ because it's
$$\frac{2\cos{x}}{2\cos^2x-1}>2$$ or
$$(1-\cos{x})(1+2\cos{x})>0.$$
Thus, $$\lim_{x\rightarrow0}\left[\frac{2\cos{x}}{\cos2x}\right]=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303012",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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Having trouble integrating $ \ \ \int \frac{x^{3}+3x^{2}+2x+4}{x^{2}(x^{2}+2x+2)} dx $. I'm having trouble integrating $$\int \frac{x^{3}+3x^{2}+2x+4}{x^{2}(x^{2}+2x+2)} dx $$
My approach
*
*$ \frac{x^{3}+3x^{2}+2x+4}{x^{2}(x^{2}+2x+2)} =\frac{Ax+B}{x^{2}}+\frac{Cx+D}{x^{2}+2x+2} $ ,
*or, $ x^{3}+3x^{2}+2x+4=(Ax+B)(x^{2}+2x+2)+(Cx+D)x^{2} $ .
But is the process right ? If not, Any suggestions are welcome.
| $$\int \frac{x^3+3x^2+2x+4}{x^2(x^2+2x+2)}dx=\int \frac{x^3+3x^2+2(x+2)}{x^2(x^2+2x+2)}dx=\int (\frac{2x+3}{x^2+2x+2}-\frac{1}{x}+\frac{2}{x^2})dx=\int \frac{2x+3}{x^2+2x+2}dx-\int\frac{1}{x}dx+2\int\frac{1}{x^2}dx$$
Now solving: $\int \frac{2x+3}{x^2+2x+2}dx$ , write $2x+3$ as $2x+2+1$ and split it, we have:
$$\int \frac{2x+3}{x^2+2x+2}dx=\int (\frac{2x+2}{x^2+2x+2}+\frac{1}{x^2+2x+2})dx=2\int\frac{x+1}{x^2+2x+2}dx+\int\frac{1}{x^2+2x+2}dx$$
Now solving $\int\frac{x+1}{x^2+2x+2}dx$ ; substitute $u=x^2+2x+2\Rightarrow \frac{du}{dx}=2x+2$, so :
$$\int\frac{x+1}{x^2+2x+2}dx=\frac{1}{2}\int\frac{1}{u}du=\frac{ln(u)}{2}$$
Undo substitution $u=x^2+2x+2$ :
$$\int\frac{x+1}{x^2+2x+2}dx=\frac{ln(x^2+2x+2)}{2}$$
Now solving $\int\frac{1}{x^2+2x+2}dx=\int\frac{1}{(x+1)^2+1}dx$
Substitute $u=x+1\Rightarrow \frac{du}{dx}=1$
$$\int\frac{1}{(x+1)^2+1}dx=\int\frac{1}{u^2+1}du=arctan(u)=arctan(x+1)$$
Now we have:
$$\int\frac{1}{x}dx=ln(x)$$
$$\int\frac{1}{x^2}dx=-\frac{1}{x}$$
Plug in solved integrals we have:
$$\int \frac{x^3+3x^2+2x+4}{x^2(x^2+2x+2)}dx=ln(|x^2+2x+2|)-ln(|x|)+arctan(x+1)-\frac{2}{x}+C$$
Apply the absolute value function to arguments of logarithm functions in order to extend the antiderivatieve's domain!
Wish I haven't make a typing mistake ...
| {
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"url": "https://math.stackexchange.com/questions/2304492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
How prove this inequality with $\sqrt{\sum_{cyc}\frac{a^2}{b^2}}+\sqrt{3}\ge\sqrt{\sum\frac{a}{b}}+\sqrt{\sum\frac{b}{a}}$
Let $a,b,c>0$, show that
$$\sqrt{\dfrac{a^2}{b^2}+\dfrac{b^2}{c^2}+\dfrac{c^2}{a^2}}+\sqrt{3}\ge\sqrt{\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}}+\sqrt{\dfrac{b}{a}+\dfrac{c}{b}+\dfrac{a}{c}}$$
I tried C-S, AM-GM, Holder and more, but without success.
following maybe is idea:$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}=x\Longrightarrow x^2=\sum_{cyc}\dfrac{a^2}{b^2}+2\sum_{cyc}\dfrac{b}{a}$$
then How can prove this inequality?
| Let $\frac{a}{b}=x$, $\frac{b}{c}=y$ and $\frac{c}{a}=z$.
Hence, $xyz=1$ and we need to prove that
$$\sqrt{x^2+y^2+z^2}+\sqrt{3}\geq\sqrt{x+y+z}+\sqrt{xy+xz+yz}.$$
or
$$x^2+y^2+z^2+3+2\sqrt{3(x^2+y^2+z^2)}\geq x+y+z+xy+xz+yz+2\sqrt{(xy+xz+yz)}$$ or
$$x^2+y^2+z^2+3-2(xy+xz+yz)+2\sqrt{3(x^2+y^2+z^2)}-2(x+y+z)+$$
$$+\left(\sqrt{xy+xz+yz}-\sqrt{x+y+z)}\right)^2\geq0,$$
which is true because by C-S
$$\sqrt{3(x^2+y^2+z^2)}=\sqrt{(1^2+1^2+1^2)(x^2+y^2+z^2)}\geq x+y+z$$ and by Schur and by AM-GM we obtain:
$$x^2+y^2+z^2+3=\sum_{cyc}\left(\left(x^{\frac{2}{3}}\right)^3+x^{\frac{2}{3}}y^{\frac{2}{3}}z^{\frac{2}{3}}\right)\geq$$
$$\geq\sum_{cyc}\left(x^{\frac{4}{3}}y^{\frac{2}{3}}+x^{\frac{2}{3}}y^{\frac{4}{3}}\right)\geq\sum_{cyc}2xy=2(xy+xz+yz).$$
Done!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is there any way to simplify the product of cosines? I recently saw a problem: Estimate the following: $cos(\frac{\pi}{15})cos(\frac{2\pi}{15})\ldots cos(\frac{7\pi}{15})$, and the options were between different consecutive power of ten.
How would I do this, no calculator of course, and is there's any way to shorten any general products of sines or cosines or other trig functions?
| $$\prod_{k=1}^7\cos\frac{k\pi}{15}=
\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\left(-\cos\frac{8\pi}{15}\right)\cos\frac{\pi}{5}\cos\frac{\pi}{3}\cos\frac{2\pi}{5}=$$
$$=-\frac{16\sin\frac{\pi}{15}\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}}{16\sin\frac{\pi}{15}}\cdot\frac{4\sin\frac{\pi}{5}\cos\frac{\pi}{5}\cos\frac{2\pi}{5}}{8\sin\frac{\pi}{5}}=$$
$$=-\frac{\sin\frac{16\pi}{15}}{16\sin\frac{\pi}{15}}\cdot\frac{\sin\frac{4\pi}{5}}{8\sin\frac{\pi}{5}}=\frac{1}{128}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Find the inverse Laplace transform of $\frac{s^2}{s^3 -1}$
Find the inverse Laplace transform of
$$F(s)=\frac{s^2}{s^3 -1}$$
What I've done is to decompose into partial fractions, so I got:
$$\mathcal{L}^{-1}\{\frac{s^2}{s^3 -1}\}=\mathcal{L}^{-1}\{\frac{1}{3(s-1)}+\frac{2s+1}{3(s^2 +s+1)}\}$$
By the properties of the inverse Laplace transform:
$$=\mathcal{L}^{-1}\{ \frac{1}{3(s-1)}\}+\mathcal{L}^{-1}\{\frac{2s+1}{3(s^2 +s+1)}\}=\frac{1}{3}\mathcal{L}^{-1}\{ \frac{1}{(s-1)}\}+\frac{1}{3}\mathcal{L}^{-1}\{\frac{2s+1}{s^2 +s+1}\}$$
Where:
$$\mathcal{L^{-1}}\{ \frac{1}{(s-1)}\}=e^{t}$$
and
$$\mathcal{L}^{-1}\{\frac{2s+1}{s^2 +s+1}\}=\mathcal{L}^{-1}\{\frac{2s}{s^2 +s+1}\}+\mathcal{L}^{-1}\{\frac{1}{s^2 +s+1}\}$$
How can I finish finding the inverse transformation?
In Mathematica, the solution I've got is
$$f(t)=\frac{e^{t}}3{}+\frac{2}{3}e^{-t/2}\cos[\frac{\sqrt{3}t}{2}]$$
| $s^3-1$ vanishes at the third roots of unity, $1,\omega,\omega^2$. It follows that
$$ \frac{s^2}{s^3-1} = \frac{A}{s-1}+\frac{B}{s-\omega}+\frac{C}{s-\omega^2}\tag{1} $$
and by residues:
$$ A = \text{Res}\left(\frac{s^2}{s^3-1},s=1\right) = \lim_{s\to 1}\frac{s^2}{s^2+s+1}=\frac{1}{3}, $$
$$ B = \text{Res}\left(\frac{s^2}{s^3-1},s=\omega\right) = \lim_{s\to \omega}\frac{s^2}{(s-1)(s-\omega^2)}=\frac{1}{3}, $$
$$ C = \text{Res}\left(\frac{s^2}{s^3-1},s=\omega^2\right) = \lim_{s\to \omega^2}\frac{s^2}{(s-1)(s-\omega)}=\frac{1}{3}, \tag{2}$$
then since $\mathcal{L}^{-1}\left(\frac{1}{s-s_0}\right)(x)=e^{s_0 x}$ we have:
$$ \mathcal{L}^{-1}\left(\frac{s^2}{s^3-1}\right) = \frac{e^{x}+e^{\omega x}+e^{\omega^2 x}}{3} = \color{red}{\frac{1}{3}e^x+\frac{2}{3}e^{-x/2}\cos\left(\frac{x\sqrt{3}}{2}\right)}\tag{3}$$
as wanted.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7$, find $k$ I need help solving this question:
$$
\text{If }\frac{3-\tan^2\frac\pi7}{1-\tan^2\frac\pi7}=k\cos\frac\pi7\text{, find k.}
$$
I simplified this down to:
$$
\frac{4\cos^2\frac\pi7-1}{2\cos^2\frac\pi7-1}
$$
But am unable to proceed further. The value of k is given to be 4, but I am unable to derive that result.
Kindly provide me with some insight, or with a step-by-step solution.
Thanks in advance,
Abhigyan
| $$k\cos y=\dfrac{3-\tan^2y}{1-\tan^2y}=\dfrac{3\cos^2y-\sin^2y}{\cos^2y-\sin^2y}=\dfrac{4\cos^2y-1}{2\cos^2y-1}$$
$$\iff2k\cos^3y-4\cos^2y-k\cos y+1=0\ \ \ \ (1)$$
Now if $7y=(2n+1)\pi,$
$\cos4y=\cdots=-\cos3y$
Using $\cos2A=2\cos^2A-1,\cos3B=4\cos^3B-3\cos B$
The roots of $$8\cos^4y+4\cos^3y-8\cos^2y-3\cos y+1=0$$ are $$\cos\dfrac{(2n+1)\pi}7$$ where $n\equiv0,\pm1,\pm2,\pm3\pmod7$
As for $n=3,\cos\dfrac{(2n+1)\pi}7=-1,$ the roots of $$0=\dfrac{8\cos^4y+4\cos^3y-8\cos^2y-3\cos y+1}{\cos y+1}$$
$$\iff8\cos^3y-4\cos^2y-4\cos y+1=0\ \ \ \ (2)$$ will be $$\cos\dfrac{(2n+1)\pi}7$$ where $n\equiv0,\pm1,\pm2,3\pmod7$
Compare $(1),(2)$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Evaluating $\int_0^1 \sqrt{1 + x ^4 } \, d x $ $$
\int_{0}^{1}\sqrt{\,1 + x^{4}\,}\,\,\mathrm{d}x
$$
I used substitution of tanx=z but it was not fruitful. Then i used $ (x-1/x)= z$ and
$(x)^2-1/(x)^2=z $ but no helpful expression was derived.
I also used property $\int_0^a f(a-x)=\int_0^a f(x) $
Please help me out
| For an approximation, you could use a Padé approximant for the integrand. The simplest one would be
$$\frac{3 x^4+4}{x^4+4}=3-\frac{x+2}{x^2+2 x+2}+\frac{x-2}{x^2-2 x+2}$$
$$\int \frac{3 x^4+4}{x^4+4}\,dx=3x+\frac{1}{2} \log \left(\frac{x^2-2 x+2}{x^2+2 x+2}\right)+\tan ^{-1}(1-x)-\tan ^{-1}(1+x)$$ So, using the given bounds, an approximation is
$$\int_{0}^{1}\sqrt{\,1 + x^{4}\,}\,\,dx \approx 3-\frac{\log (5)}{2}-\tan ^{-1}(2)\approx 1.08813$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 4
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How to find expression for this tensor? First consider 1D case. Suppose I would like to compute
\begin{alignat}{2}
m_p = \sum_{x=1}^{N} x^p f(x), \qquad \forall\; p = 1, 2, \dots, P
\end{alignat}
I know it can be expressed as a matrix product as follows
\begin{alignat}{2}
\underbrace{ \begin{bmatrix}
m_1 \\
m_2 \\
\vdots \\
m_P
\end{bmatrix} }_{:=\, \boldsymbol{m}} = \underbrace{ \begin{bmatrix}
1^1 & 2^1 & \dots & N^1 \\
1^2 & 2^2 & \dots & N^2 \\
\vdots & \vdots & \ddots & \vdots \\
1^P & 2^P & \dots & N^P
\end{bmatrix} }_{ :=\, X} \underbrace{ \begin{bmatrix}
f(1) \\
f(2) \\
\vdots \\
f(N)
\end{bmatrix} }_{:=\, \boldsymbol{f}}
\end{alignat}
Now generalizing this to 2D case, I know that
\begin{alignat}{2}
m_{pq} = \sum_{x=1}^{N} \sum_{y=1}^{M} x^p y^q f(x, y), \qquad \forall\; p = 1, 2, \dots, P;\; q = 1, 2, \dots, Q
\end{alignat}
can be expressed as
\begin{alignat}{2}
&\underbrace{ \begin{bmatrix}
m_{11} & m_{12} & \dots & m_{1Q} \\
m_{21} & m_{22} & \dots & m_{2Q} \\
\vdots & \vdots & \ddots & \vdots \\
m_{P1} & m_{P2} & \dots & m_{PQ}
\end{bmatrix} }_{ :=\, M} =\\
&=\underbrace{ \begin{bmatrix}
1^1 & 2^1 & \dots & N^1 \\
1^2 & 2^2 & \dots & N^2 \\
\vdots & \vdots & \ddots & \vdots \\
1^P & 2^P & \dots & N^P
\end{bmatrix} }_{ :=\, X}
\underbrace{ \begin{bmatrix}
f(1,1) & f(1,2) & \dots & f(1,M) \\
f(2,1) & f(2,2) & \dots & f(2,M) \\
\vdots & \vdots & \ddots & \vdots \\
f(N,1) & f(N,2) & \dots & f(N,M)
\end{bmatrix} }_{ :=\, F}
\underbrace{ \begin{bmatrix}
1^1 & 2^1 & \dots & M^1 \\
1^2 & 2^2 & \dots & M^2 \\
\vdots & \vdots & \ddots & \vdots \\
1^Q & 2^Q & \dots & M^Q
\end{bmatrix}^{\top} }_{ :=\, Y^{\top}}
\end{alignat}
Now my question is how does this generalize to 3D case? That is how can I express in terms of vectors/matrices/tensors the tensor
\begin{alignat}{2}
\begin{pmatrix}
m_{pqr}
\end{pmatrix}_{p=1, q=1, r=1}^{p=P, q=Q, r=R}
\end{alignat}
whose elements are given as
\begin{alignat}{2}
m_{pqr} = \sum_{x=1}^{N} \sum_{y=1}^{M} \sum_{z=1}^{K} x^p y^q z^r f(x, y, z), \qquad \forall\; p = 1, 2, \dots, P;\; q = 1, 2, \dots, Q;\; r = 1, 2, \dots, R.
\end{alignat}
| People usually stop writing down matrices when dealing with rank-3 tensors and higher. In your case, using the Einstein summation convention,
1D: $m_p = X_{pi} f_i$ with $X_{pi} = i^p$, $f_i = f(i)$
2D: $m_{pq} = X_{pi} Y_{qj} f_{ij}$ with $X_{pi}$ as above, $Y_{qj} = j^q$, $f_{ij} = f(i,j)$.
3D: $m_{pqr} = X_{pi} Y_{qj} Z_{rk} f_{ijk}$ with $Z_{rk} = k^r$, $f_{ijk} = f(i,j,k)$.
For the 3D case, while the $X$, $Y$, and $Z$ objects could be written as matrices, the $m_{pqr}$ and $f_{ijk}$ are not easily expressed that way.
| {
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"timestamp": "2023-03-29T00:00:00",
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Sum $\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$ How can we find the sum
$$\sum_{n=1}^{\infty}\frac{3n+7}{n(n+1)(n+2)}.$$
WolframAlpha says it's $13/4$ but how to compute it? In general how to approach this kind of sums of rational functions where denominator has distinct roots?
| $$\frac{3n+7}{n(n+1)(n+2)}=\frac{3n+6+1}{n(n+1)(n+2)}=$$
$$=\frac{3}{n(n+1)}+\frac{1}{n(n+1)(n+2)}=\frac{3}{n}-\frac{3}{n+1}+\frac{1}{n}\left(\frac{1}{n+1}-\frac{1}{n+2}\right)=$$
$$=\frac{3}{n}-\frac{3}{n+1}+\frac{1}{n}-\frac{1}{n+1}-\frac{1}{2}\left(\frac{1}{n}-\frac{1}{n+2}\right)=$$
$$=\frac{7}{2n}-\frac{4}{n+1}+\frac{1}{2(n+2)}$$ and use the telescopic sum
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given $\triangle ABC$, find point $X$ that minimizes $|\overline{AX}||\overline{BX}|+|\overline{BX}||\overline{CX}|+|\overline{CX}||\overline{AX}|$ Given a $\triangle$$ABC$.Find with proof the point $X$ in the plane of the triangle,for which $$|\overline{AX}||\overline{BX}|+|\overline{BX}||\overline{CX}|+|\overline{CX}||\overline{AX}|$$ is as small as possible.
| Let $a = |BC|$, $b = |CA|$, $c = |AB|$, $R_1 = |XA|$, $R_2 = |XB|$, $R_3 = |XC|$.
WOLOG, we will assume $a \ge b \ge c$. We will show that
$$R_1R_2 + R_2R_3 + R_3R_1 \ge \min(ab,bc,ca) = bc \tag{*1}$$
This implies vertex $A$ is always a point that minimize $R_1R_2 + R_2R_3+R_3R_1$.
When $X \in \{ A, B, C \}$, $(*1)$ is trivially true.
Let us consider the case $X \notin \{ A, B, C \}$ which means $R_1, R_2, R_3 > 0$.
For any positive numbers $x, y, z$, it is well known the expression
$x R_1^2 + yR_2^2 + zR_3^2$ is minimized when and only when $X$ equals to the weighted centroid $\frac{xA + yB + zC}{x+y+z}$. By evaluating the value of this expression at that particular point, we obtain the so called "polar moment of inertia inequality" by M.S. Klamkin${}^{\color{blue}{[1]}}$,
$$(x+y+z)(xR_1^2 + yR_2^2 + zR_3^2) \ge a^2 yz + b^2 xz + c^2 xy
= (xyz)\left(\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z}\right)
$$
Using Engel's form of Cauchy Schwarz inequality, we obtain
$$
\frac{a^2}{x} + \frac{b^2}{y} + \frac{c^2}{z}
\ge \frac{(a+b+c)^2}{x+y+z}
\quad\implies\quad
\frac{R_1^2}{yz} + \frac{R_2^2}{xz} + \frac{R_3^2}{xy} \ge
\left(\frac{a + b + c}{x + y + z}\right)^2$$
The inequality on RHS is valid for any four distinct points $X, A, B, C$. If will continue to work if we replace $A, B, C$ by $A', B', C'$, the image of $A,B,C$ under inversion with respect to the unit circle centered at $X$.
For the new triangle $\triangle A'B'C'$, we have
$$a' = \frac{a}{R_2R_3},\; b' = \frac{b}{R_1R_3},\; c' = \frac{c}{R_1R_2},\; R_1' = \frac{1}{R_1},\; R_2' = \frac{1}{R_2},\; R_3' = \frac{1}{R_3}$$
This leads to
$$\begin{align} & \frac{(R'_1)^2}{yz} + \frac{(R'_2)^2}{xz} + \frac{(R'_3)^2}{xy} \ge
\left(\frac{a' + b' + c'}{x + y + z}\right)^2 \tag{*2}\\
\iff &
\frac{(R_2R_3)^2}{yz} + \frac{(R_1R_3)^2}{xz} + \frac{(R_1R_2)^2}{xy}
\ge \left(\frac{aR_1 + bR_2 + cR_3}{x + y + z}\right)^2
\end{align}
$$
Chose $x : y : z = aR_1 : bR_2 : cR_3$, we obtain an inequality first proved
by T. Hayashi${}^{\color{blue}{[2][3]}}$:
$$\frac{R_2R_3}{bc} + \frac{R_1R_3}{ac} + \frac{R_1R_2}{ab} \ge 1\tag{*3}$$
Since $a \ge b \ge c$, we have $bc \le ac \le ab$ and hence
$$R_2R_3 + R_1R_3 + R_1R_2 \stackrel{*}{\ge} bc\left(\frac{R_2R_3}{bc} + \frac{R_1R_3}{ac} + \frac{R_1R_2}{ab}\right) \ge bc$$
Please note that inequality marked by $(*)$ in last line will be strict unless $a = b = c$.
When $ABC$ is not equilateral, we find
*
*$\min_{X\in\mathbb{R}^2}(R_1R_2 + R_2R_3 + R_3R_1) = \min(ab,bc,ca) = bc$.
*If $a > b$, $X$ is unique and located at $A$.
*If $a = b > c$, $X$ can either be $A$ or $B$.
When $ABC$ is equilateral, it is clear $X$ can be $A,B,C$.
For $X \notin \{ A, B, C \}$, if the minimum is ever achieved, we need the inequality in $(*2)$ and the CS inequality become equality. This leads to two conditions
$$a' : b' : c' = x : y : z\quad\text{ and }\quad X = \frac{zA' + yB' + xC'}{x+y+z}$$
By plugging in the values, it is easy to see first condition is
trivially satisfied while the second condition reduces to
$$\frac{\overrightarrow{XA}}{|XA|} + \frac{\overrightarrow{XB}}{|XB|} + \frac{\overrightarrow{XC}}{|XC|} = \vec{0}$$
which is satisfied at and only at the centroid $G$.
As a summary, we have
$$\bbox[8pt,border:1px solid blue]{
\min_{X\in\mathbb{R}^2}\{ R_1R_2 + R_2R_3 + R_3R_1 \} = bc
\;\text{ achieved at }\;
\begin{cases}
A; & a > b\\
A, B; & a = b > c\\
A, B, C, G; & a = b = c
\end{cases}
}
$$
Notes
*
*$\color{blue}{[1]}$ M.S. Klamkin, Geometric inequalities via the polar moment of inertia, Mathematics Magazine,
48(1) (1975), 44–46.
*$\color{blue}{[2]}$ T. Hayashi, Two theorems on complex number, Tôhoku Math. J., 4 (1913-1914), 68–70.
*$\color{blue}{[3]}$ For the stronger inequality $(*3)$ by T. Hayashi, if $X$ is distinct from the vertices, equality holds when and only when $\triangle ABC$ is acute-angled and $X$ is the orthocenter.
| {
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"question_score": "2",
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} |
Find $\delta$ such that if $0 < \vert x - 2 \vert < \delta$, then $\vert f(x)-3 \vert < .2$ I'm looking for assistance with the following problem:
Find $\delta$ such that if $0 < \vert x - 2 \vert < \delta$, then $\vert f(x)-3 \vert < .2$ where $f(x)=x^2-1$.
My attempt:
We need $\delta$ such that $\vert f(x)-2 \vert < .2$. That is, we need $\vert x^2 - 1 - 3 \vert = \vert x^2 - 4 \vert = \vert x+2 \vert \cdot \vert x-2 \vert < .2$. We can control $\vert x - 2 \vert$ by $\delta$. Let us then restrict $x$ to the interval $(1,3)$. Thus $\vert x+2 \vert < 5$. Then we will let $\delta = \frac{.2}{5}$.
Now then, if $0 < \vert x - 2 \vert < \delta = \frac{.2}{5}$ then $\vert f(x)-3 \vert = \vert x^2 - 4 \vert = \vert x+2 \vert \cdot \vert x-2 \vert < 5 \cdot \frac{.2}{5}=.2.$ $\Box$
Does this work for a valid delta? I'm basically taking the more general case where $\delta$ would be $\min(\frac{\epsilon}{5},1)$ and trying to fit it to this specific case.
| The other solution is of course neat :-)
If $|x^2-4|<0.2$ then we have that $3.8<x^2<4.2$ that is either
*
*$\sqrt{3.8}<x<\sqrt{4.2}$ or
*$-\sqrt{4.2}<x<-\sqrt{3.8}$.
Consider case $1$. From here we get
\begin{align}
-0.05064\sim\sqrt{3.8}-2&<x-2<\sqrt{4.2}-2\sim0.04939
\end{align}
Now if we want $-\delta<x-2<\delta$, then we can choose $\delta<\sqrt{4.2}-2$.
For case $2$ we have
\begin{align}
-4.04939\sim-\sqrt{4.2}-2&<x-2<-\sqrt{3.8}-2\sim -3.94936
\end{align}
implying that $\delta$ shall be negative which is not possible.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve the equation $\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$ Find $x\in R$ satisfy $$\sqrt[3]{x+2}+\sqrt[3]{2x-1}+x=4$$
The root of equation very bad
My try 1:
Let $\sqrt[3]{x+2}=a;\sqrt[3]{2x-1}=b\Rightarrow a^3-b^3=3-x$
Have: $a+b=4-x$
=>Root of a system of equation bad, too
My try 2:
Use quality $\sqrt[3]{a}\pm \sqrt[3]{b}=\frac{a\pm b}{\sqrt[3]{a^2}\mp \sqrt[3]{ab}+\sqrt[3]{b^2}}$ :
$\sqrt[3]{x+2}-ax+\sqrt[3]{2x-1}-bx=4-x$
Need find $ax$, $bx$ but it's very bad, too
| (Too long for a comment.) Building upon this...
Let $\,\sqrt[3]{x+2}=a\,;\;\sqrt[3]{2x-1}=b$
The canonical way to get the polynomial equation in $x$ is to eliminate $a,b$ between:
$$
\begin{cases}
\begin{align}
a^3 &= x+2 \\
b^3 &= 2x-1 \\
a+b +x &= 4
\end{align}
\end{cases}
$$
This is a routine calculation using polynomial resultants, though generally not pretty to do by hand. In this case resultant[ resultant[ a+b+x-4, b^3-2x+1, b ], a^3-x-2, a ] gives $x$ as the root of:
$$
x^9 - 36 x^8 + 585 x^7 - 5589 x^6 + 34317 x^5 - 139563 x^4 + 374328 x^3 - 635553 x^2 + 615033 x - 253503 = 0
$$
Numerically, $x \;\;\simeq\;\; 1.3254785\dots$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
If $a^3+b^3+c^3=3$ so $2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$ I was looking at this question, and I derived this inequality from that
Let $a$, $b$ and $c$ be positive numbers such that $a^3+b^3+c^3=3.$ Prove that:
$$2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$$
I couldn't solve Rozenberg's inequality but I assumed it's true. So if $a^3+b^3+c^3=3$, then
$$\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}\tag {1}$$
and also by using this, if $b^3+a^3+c^3=3$, then
$$\frac{b^3}{b+a}+\frac{a^3}{a+c}+\frac{c^3}{c+b}\geq\frac{3}{2}\tag {2}$$
using $(1)$ and $(2)$, if $a^3+b^3+c^3=3$, then
$$\frac{a^3+b^3}{a+b}+\frac{b^3+c^3}{b+c}+\frac{c^3+a^3}{c+a}\geq3$$
$$\Leftrightarrow \sum\limits_{cyc}(a^2-ab+b^2)\geq 3$$
$$\Leftrightarrow 2(a^2+b^2+c^2)-(ab+bc+ca) \geq 3$$
If I'm not making any silly mistake, this inequality should be true. I couldn't solve yet, but I wanted to share with you.
$\mathbf {EDIT:}$ To be clear, I think that I derived an inequality weaker than Rozenberg's inequality by using his inequality, and I presented it for those of you who might interest trying to solve inequalities by ownselves
| Rozenberg's inequality is solved, find the solution at this link :
If $a^3+b^3+c^3=3$ so $\frac{a^3}{a+b}+\frac{b^3}{b+c}+\frac{c^3}{c+a}\geq\frac{3}{2}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove that $a \equiv b \pmod{1008}$
Suppose that the positive integers $a$ and $b$ satisfy the equation $$a^b-b^a = 1008.$$ Prove that $a \equiv b \pmod{1008}$.
Taking the equation modulo $1008$ we get $a^b \equiv b^a \pmod{1008}$, but I didn't see how to use this to get that $a \equiv b \pmod{1008}$.
| We have $1008 = 2^4 \cdot 3^2 \cdot 7$. First note that $a \equiv b \pmod{2}$ since $1008$ is even. Note that $2 \nmid a,b$ since if $2 \mid a,b$ then $a,b \leq 5$. Therefore, $a^b \equiv a \pmod{8}$ and $b^a \equiv b \pmod{8}$, so $a \equiv b \pmod{8}$. Then since $a^4 \equiv 1 \pmod{16}$, we have $a^b \equiv a^a \pmod{16}$ and so $a^a \equiv b^a \pmod{16}$. If $a \equiv 1 \pmod{4}$, then $a \equiv b \pmod{16}$. If $a \equiv 3 \pmod{4}$, then $a^3 \equiv b^3 \pmod{16}$ gives $a \equiv b \pmod{16}$.
We have $a^b \equiv a \pmod{3}$ and $b^a \equiv b \pmod{3}$ since $a$ and $b$ are both odd. Thus $a \equiv b \pmod{3}$. If $a \equiv b \equiv 1 \pmod{6}$, then $a^b \equiv a \pmod{9}$ and $b^a \equiv b \pmod{9}$ by Euler's Totient Theorem since $\varphi(9) = 6$. Thus $a \equiv b \pmod{9}$ in this case. If $a \equiv b \equiv 5 \pmod{6}$, then $a^b \equiv a^5 \pmod{9}$ and $a^5 \equiv b^5 \pmod{9}$. Thus $a \equiv b \pmod{9}$.
Similarly since $\varphi(7) = 6$ we get taking the cases $a \equiv b \equiv 1 \pmod{6}$ and $a \equiv b \equiv 5 \pmod{6}$ again that $a \equiv b \pmod{7}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f (x)$
If $f (x) +f'(x) = x^3+5x^2+x+2$ then find $f(x)$.
$f'(x)$ is the first derivative of $f (x)$.
I have no idea about this question, please help me.
| We can use an integrating factor, $u(x)$:
$$
\begin{align}
v(x)(f(x)+f'(x))
&=\frac{\mathrm{d}}{\mathrm{d}x}(u(x)\,f(x))\\
&=u'(x)\,f(x)+u(x)\,f'(x)\tag{1}
\end{align}
$$
If we set $\frac{u'(x)}{u(x)}=1$, we see that $(1)$ is satisfied. Thus, we get that we can use $u(x)=v(x)=e^x$. That is,
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}(e^x\,f(x))
&=e^x(f(x)+f'(x))\\
&=e^x\left(x^3+5x^2+x+2\right)\tag{2}
\end{align}
$$
Integrating $(2)$, we get
$$
e^x\,f(x)=e^x\left(x^3+2x^2-3x+5\right)+C\tag{3}
$$
and therefore,
$$
f(x)=x^3+2x^2-3x+5+Ce^{-x}\tag{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 5
} |
Prove that each divisor of $5^{(4n+1)}+5^{(3n+1)}+1 =$ $1 \pmod {10}$ Prove that if $d$ divides $5^{(4n+1)}+5^{(3n+1)}+1$ for any $n$, then $d =$ $1 \pmod {10}$.
A similar statement can be proven, where $d$ divides $(3^{(2n+1)}+1)/4$, for any $n$, then $d =$ $1 \pmod {6}$ by first showing that if $-3$ is a quadratic residue $\pmod p$, then $p = 1 \pmod 3$. For the base $5$ expression I would start by showing that if $-5$ is a quartic residue $\pmod p$, then $p = 1 \pmod {10}$. Any help on proving the original statement with base $5$ expression? Thanks!!!
| Each divisor of $5^{4n+1}+5^{3n+1}+1$ is odd, hence it is enough to show that any odd prime divisor of $5^{4n+1}+5^{3n+1}+1$ is $\equiv 1\pmod{5}$. Now we may consider the splitting field of $f(x)=5x^4+5x^3+1$ over $\mathbb{F}_p$ for any prime $p>5$. The complex roots of $f(x)$ are given by
$$ \frac{1}{4}\left(-1\color{red}{\pm}\sqrt{5}\color{blue}{\pm} i\sqrt{2\left(1\color{red}{\pm}\frac{1}{\sqrt{5}}\right)}\right) $$
hence if $p$ is a divisor of $5^{4n+1}+5^{3n+1}+1$ then $5$ is a quadratic residue $\!\!\pmod{p}$ and $p\mod{5}\in\{-1,1\}$. On the other hand, if $p\equiv -1\pmod{5}$ then $5x^4+5x^3+1$ factors in $\mathbb{F}_p$ as $5$ times the product of two quadratic irreducible polynomials. Long story short,
$$ p>5,\;\; p\mid 5^{4n+1}+5^{3n+1}+1\quad \Longrightarrow \quad p\equiv 1\pmod{5}$$
and the claim easily follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 1
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probability or rolling 1 1 2 with six dice What is the probability of rolling at least two ones and at least one two (order not important) with six 6-sided die?
Equivalently, to win a game, I need two 1s and a 2, (1,1,2). What is the probability of my rolling at least this with 6 dice.
There are $6^6=46656$ possible rolls. I believe that there are 6896 'winning' rolls among these as I wrote some code to check every roll. This makes the probability around 0.1478
I'm not sure how to get this without the help of a computer though.
| Approach via inclusion-exclusion and De Morgan's laws.
Let $A$ be the event that you roll at least two $1$'s. Let $B$ be the event that you roll at least one $2$.
We are attempting to calculate then $Pr(A\cap B)$
$Pr(A\cap B)=1-Pr((A\cap B)^c)=1-Pr(A^c\cup B^c) = 1-Pr(A^c)-Pr(B^c)+Pr(A^c\cap B^c)$
We calculate each term on the right now that they are in simpler form.
$Pr(A^c)$ refers to the probability of strictly fewer than two $1$'s occurring, i.e. at most one $1$ occurring. Either exactly one $1$ occurs or no $1$'s occur. In the case of exactly one $1$, pick its location and then pick each remaining digit for a total of $6\cdot 5^5$ possibilities. In the case of no $1$'s, pick each digit for a total of $5^6$ possibilities. Taking the ratio of this compared to the $6^6$ equally likely dice rolls, we calculate $Pr(A^c)=\frac{6\cdot 5^5+5^6}{6^6}$.
$Pr(B^c)$ refers to the probability of strictly fewer than one $2$ occurring, i.e. no $2$'s. This occurs with probability $Pr(B^c)=\frac{5^6}{6^6}$
$Pr(A^c\cap B^c)$ refers now to the probability of at most one $1$ and no $2$'s, which similarly to before we break into cases for either exactly one $1$ or no $1$'s. For exactly one $1$, first pick the location, then pick each remaining digit for $6\cdot 4^5$ possibilities and for no $1$'s pick each digit for $4^6$ possibilities for a probability of $Pr(A^c\cap B^c)=\frac{6\cdot 4^5+4^6}{6^6}$
This gives a final probability of:
$$Pr(A\cap B) = 1-\frac{6\cdot 5^5+5^6}{6^6}-\frac{5^6}{6^6}+\frac{6\cdot 4^5+4^6}{6^6}$$
$$=1-\frac{39760}{46656} = \frac{431}{2916}\approx 0.147805$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the Nth derivative using Maclaurin series The prompt is to find the 8th derivative of the function f(x) defined as,
$$f(x) = \frac{3}{1+x-2x^2}$$
$$f^{(8)}(0) = ?$$
To find the maclaurin series, I proceeded by finding the derivatives of the function at 0 as follows,
$$f^{(1)}(x) = \frac{ -3 + 12x }{ (1 + x -2x^2)^2 } $$
$$f^{(2)}(x) = \frac{18(4x^2-2x+1)}{(-2x^2+x+1)^3} $$
such that,
$$f(0) = 3$$
$$f^{(1)}(0) = -3$$
$$f^{(2)}(0) = 18$$
$$f^{(3)}(0) = -90$$
This makes the maclaurin series,
$$f(x) = 3 - 3x + \frac{18x^2}{3!} - \frac{90x^3}{4!} + ...$$
I understand from the series, we have to have $(-1)^n$ since the negative sign is alternating, also in the denominator we have n!
$$\sum_{n=0}^{\infty} \frac{(-1)^n(x)^n}{n!}$$ but it is incomplete, since I don't see the pattern, I would appreciate I someone can help with completing the series hence finding the 8th derivative.
| Suppose we want to do this without expanding in partial fractions. Then we can just perform a long division of the fraction to obtain the coefficient of $x^8$. But a faster division algorithm is Newton-Raphson division, $n$ steps will yield the coefficients up to $x^{2^n -1}$ This method can be generalized
such that it also works for arbitrary closed form functions, it is used in some computer algebra programs.
Newton-Raphson division for polynomials works in an analogous way as it does for real numbers. In the latter case this boils down to finding the reciprocal of some number $y$ by solving the equation $$\frac{1}{x} - y = 0$$ using the Newton-Raphson method, which yields a sequence of approximants $x_n$ satisfying the recurrence equation:
$$x_{n+1} = 2 x_{n} - y x_{n}^2\tag{1}$$
Then if $y$ is a polynomial and we take $x_0$ equal to the reciprocal of the constant term of that polynomial then iterating with Eq. (1) will yield successive approximations for the series expansion of the reciprocal of the polynomial, such that the number of correct terms doubles after each iteration.
If we put $p(x) = 1 + x - 2 x^2$ then the successive approximants $q_n(x)$ of the reciprocal of $p(x)$ satisfy the recurrence equation
$$q_{n+1}(x) = 2 q_n(x) - p(x) q_n(x)^2 \bmod x^{2^{n+1}}$$
where $q_0(x) = 1$. Here we compute modulo $x^{2^{n+1}}$ because only the coefficients of $x^r$ for $r<2^{n+1}$ will be correct. We then find:
$$
\begin{split}
q_1(x) &= &1-x\\
q_2(x) &= -&5 x^3+3 x^2-x+1\\
q_3(x) &= -&85 x^7+43 x^6-21 x^5+11 x^4-5 x^3+3 x^2-x+1 \\
q_4(x) &= -&21845 x^{15}+10923 x^{14}-5461 x^{13}+2731 x^{12}-1365 x^{11}+\\
& &683 x^{10}-341 x^9+171 x^8-85 x^7+43 x^6-21 x^5+11 x^4-5 x^3+3 x^2-x+1
\end{split}
$$
The 8th derivative of $\dfrac{3}{p(x)}$ is thus equal to $8!\times 3\times 171$
| {
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"url": "https://math.stackexchange.com/questions/2327947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
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Complex Integration - $\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx$
Exercise :
Show that :
$$\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx = \frac{4\pi}{5}\sin\bigg(\frac{2\pi}{5}\bigg)$$
Attempt :
$$\int_{-\infty}^{\infty} \frac{x-1}{x^5-1} dx = \int_{-\infty}^{\infty} \frac{1}{x^4 + x^3 + x^2 + x + 1}dx $$
$$x^4 + x^3 + x^2 + x + 1=0 \Leftrightarrow x = \{-(-1)^{1/5},(-1)^{2/5},-(-1)^{3/5},(-1)^{4/5} \}$$
So, the function $f(z) = \frac{1}{x^4 + x^3 + x^2 + x + 1}$ has poles at the points :
$$\{-(-1)^{1/5},(-1)^{2/5},-(-1)^{3/5},(-1)^{4/5} \}$$
Now, I know you have to integrate through a closed curve $C$ and on a line $γ_R$ and then continue on with residues for the poles that reside in this curve, but I am stuck on how to apply it here and I also miss it a bit on how to split the integral for the curve and the line. Most examples I've saw get simpler due to the even function trick, but that cannot be applied here.
I would really appreciate a thorough solution and explanation, since I've just started working on generalized integrals and I have to clear my mind on them.
Thanks for your time !
| Brute force also helps:
$$\int\limits_{-\infty}^{+\infty}\frac{x-1}{x^5-1}dx=\int\limits_{-\infty}^{+\infty}\frac{1}{x^4+x^3+x^2+x+1}dx=$$
$$=\int\limits_{-\infty}^{+\infty}\frac{1}{x^2\left(x^2+\frac{1}{x^2}+x+\frac{1}{x}+1\right)}dx=\int\limits_{-\infty}^{+\infty}\frac{1}{x^2\left(\left(x+\frac{1}{x}\right)^2+x+\frac{1}{x}-1\right)}dx=$$
$$=\int\limits_{-\infty}^{+\infty}\frac{1}{x^2\left(x+\frac{1}{x}-\frac{-1+\sqrt5}{2}\right)\left(x+\frac{1}{x}-\frac{-1-\sqrt5}{2}\right)}dx=$$
$$=\int\limits_{-\infty}^{+\infty}\frac{1}{\left(x^2-\frac{\sqrt5-1}{2}x+1\right)\left(x^2+\frac{\sqrt5+1}{2}x+1\right)}dx=$$
$$=\frac{1}{\sqrt5}\int\limits_{-\infty}^{+\infty}\left(\frac{x+\frac{\sqrt5+1}{2}}{x^2+\frac{\sqrt5+1}{2}x+1}-\frac{x-\frac{\sqrt5-1}{2}}{x^2-\frac{\sqrt5-1}{2}x+1}\right)dx=$$
$$=\frac{1}{\sqrt5}\int\limits_{-\infty}^{+\infty}\left(\frac{\frac{\sqrt5+1}{4}}{x^2+\frac{\sqrt5+1}{2}x+1}-\frac{\frac{\sqrt5-1}{4}}{x^2-\frac{\sqrt5-1}{2}x+1}\right)dx=$$
$$=\frac{\sqrt5+1}{4\sqrt5}\int\limits_{-\infty}^{+\infty}\frac{1}{\left(x+\frac{\sqrt5+1}{4}\right)^2+1-\frac{6+2\sqrt5}{16}}dx-\frac{\sqrt5-1}{4\sqrt5}\int\limits_{-\infty}^{+\infty}\frac{1}{\left(x-\frac{\sqrt5-1}{4}\right)^2+1-\frac{6-2\sqrt5}{16}}dx=$$
$$=\pi\left(\frac{\sqrt5+1}{4\sqrt5\sqrt{\frac{10-2\sqrt5}{16}}}-\frac{\sqrt5-1}{4\sqrt5\sqrt{\frac{10+2\sqrt5}{16}}}\right)=\frac{\pi}{\sqrt5}\left(\frac{\cos36^{\circ}}{\sin36^{\circ}}-\frac{\sin18^{\circ}}{\cos18^{\circ}}\right)=$$
$$=\frac{\pi\cos54^{\circ}}{\sqrt5\sin36^{\circ}\cos18^{\circ}}=\frac{\pi}{\sqrt5\cos18^{\circ}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2328207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that $2\cdot \cos \frac{72}{2}\cdot \cos \frac{24}{2}+2\cdot \sin \frac{96}{2}\cdot \sin \frac{72}{2}=0.5$. Additional data added To solve it I have tried some options, where in one of them I applied product to sum formulas, which seemed to be very helpful, but didn't get the answer.
Used these formulae:
$$\cos(a+b)+ \cos(a-b)=2\cdot \cos(a)\cdot \cos(b)$$
$$−\cos(a+b)+\cos(a−b)=2\sin(a)\sin(b)$$
And now, I am stuck at this:
$$\cos(48)+\cos(24)-\cos(84)+\cos(12)=0.5$$
Then tried to make the arguments similar using a double angle and a half angle formulae $$2\cos^2(24)-1+\cos(24)+\sqrt{\frac{1+\cos(24)}{2}}+\cos(84)=0.5$$ Anyway, I can't show that expression above really is 0.5.
| Using the formulae and supposing that you work on degrees :
$$ \cos(a+b)+\cos(a−b)=2\cos(a)\cos(b)$$
$$−\cos(a+b)+\cos(a−b)=2\sin(a)\sin(b) $$
we apply them and transform your given equation :
$$2\cos \frac{72^\circ}{2} \cos \frac{24^\circ}{2}+2\sin \frac{96^\circ}{2} \sin \frac{72^\circ}{2}= \cos\bigg(\frac{72^\circ}{2} + \frac{24^\circ}{2}\bigg) + \cos\bigg(\frac{72^\circ}{2} - \frac{24^\circ}{2}\bigg) + \cos\bigg(\frac{96^\circ}{2} - \frac{72^\circ}{2}\bigg) - \cos\bigg(\frac{96^\circ}{2} + \frac{72^\circ}{2}\bigg) $$
$$=$$
$$\cos(48^\circ) + \cos(24^\circ) +
\sin(12^\circ) - \sin(84^\circ) = \frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Prove that $\sum_{cyc}(1-x)^2\ge \sum_{cyc}\frac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}$
Let $x,y,z>0$. Show that
$$\sum_{cyc}(1-x)^2\ge \sum_{cyc}\dfrac{z^2(1-x^2)(1-y^2)}{(xy+z)^2}.$$
Source: In a blog entry posted in 2013, it was said that this problem was proposed by Dongyi Wei (full marks both at 50th IMO 2009 and 49th IMO 2008) and was solved by Zipei Nie (full marks at 51st IMO 2010).
Here's what I have done. The expression
$$\sum_{cyc}(1-x)\left((1-x)-\dfrac{z^2(1+x)(1-y^2)}{(xy+z)^2}\right)\ge 0$$
or
$$\sum_{cyc}(1-x)\cdot\dfrac{x^2y^2+y^2z^2+xy^2z^2+2xyz-2xz^2-2x^2yz-x^3y^2}{(xy+z)^2}\ge 0.$$
But this way does not help for the starting inequality.
| The idea to prove this inequality is to use the well-know Ky-Fan inequality :
Before that we make the following substitution:
$x=\frac{1}{a}$
$y=\frac{1}{b}$
$z=\frac{1}{c}$
We get the following inequality :
$$\sum_{cyc}\frac{(1-a)^2}{a^2}\geq \sum_{cyc} \frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
Now we have the Ky-fan inequality wich state that for $a,b,c$ belongs to $[\frac{1}{4};\frac{1}{2}]$:
$$\frac{(1-a)^2}{a^2}\geq\frac{b^2c^2}{(1-b)^2(1-c)^2}\Big(\frac{3-a-b-c}{a+b+c}\Big)^6
$$
So we get this refinement of the starting inequality if we sum each term of the previous inequality:
$$\sum_{cyc}\frac{(1-a)^2}{a^2}\geq \sum_{cyc}\frac{b^2c^2}{(1-b)^2(1-c)^2}\Big(\frac{3-a-b-c}{a+b+c}\Big)^6 \geq\sum_{cyc} \frac{(1-a^2)(1-b^2)}{(ab+c)^2}$$
So we prove the inequality for $a,b,c$ belongs to $[\frac{1}{4};\frac{1}{2}]$
To prove the remainder of the theorem you just have to play with a variable $u$ and remark that we always have for all positive $x,y,z$:
$$\sum_{cyc}\frac{(1-x)^2}{x^2}\geq u\sum_{cyc}\frac{(1-a)^2}{a^2}\geq u\sum_{cyc}\frac{b^2c^2}{(1-b)^2(1-c)^2}\Big(\frac{3-a-b-c}{a+b+c}\Big)^6\\ \geq u\sum_{cyc} \frac{(1-a^2)(1-b^2)}{(ab+c)^2}\geq \sum_{cyc}\frac{(1-y^2)(1-z^2)}{(yz+x)^2}$$
We make the following substitution on the first inequality :
$x=\cos^2 s$;
$y=\cos^2 v$;
$z=\cos^2 w$
And
$a=\cos^2 p$;
$b=\cos^2 r$;
$c=\cos^2 q$
With $p,q,r$ belonging to $[\arccos (\frac{1}{\sqrt{2}});\arccos(\frac{1}{2})]$ :
We get this :
$$\sum_{cyc}\tan^4 s\geq u\sum_{cyc}\tan^4 p$$
we put :
$$u=\frac{\sum_{cyc}\tan^2 s\tan^2 v}{\sum_{cyc}\tan^4 p}$$
Edit for Martin R :
We have build $u$ exactly to obtain the first inequality :
So we have :
$$u\sum_{cyc}\tan^4 p=\sum_{cyc}\tan^2 s\tan^2 v$$
And
$$\sum_{cyc}\tan^4 s\geq\sum_{cyc}\tan^2 s\tan^2 v$$
And to conclude it's just Am-Gm or $a^2+b^2\geq 2ab$
Now we need to prove that $u$ satisfy the last inequality :
With the previous substitution and combine with the well-know formula $(\cos a^2+\sin a^2)^2=1$ we find:
$$\frac{\sum_{cyc}\tan^2 s\tan v^2}{\sum_{cyc}\tan^4 p}\sum_{cyc}\frac{(\tan^4 p+2\tan^2 p)((\tan^4 r+2\tan^2 r)}{\Big(1+\frac{\cos^2 q}{\cos^2 r\cos^2 p}\Big)^2}\geq\sum_{cyc}\frac{(\tan^4 s+2\tan^2 s)((\tan^4 v+2\tan^2 v)}{\Big(1+\frac{\cos^2 w}{\cos^2 v\cos^2 s}\Big)^2}$$
| {
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"url": "https://math.stackexchange.com/questions/2330511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove: If n is a perfect square, then n+3 is not a perfect square Use either direct proof, proof by contrapositive, or proof by contradiction.
Using proof by contradiction method
Assume n is a perfect square and n+3 is a perfect square (proof by
contradiction)
There exists integers x and y such that $n = x^2$ and $n+3 = y^2$
Then $x^2 + 3 = y^2$
Then $3 = y^2 - x^2$
Then $3 = (y-x)(y+x)$
Then $y+x = 3$ and $y-x=1$
Then $x = 1, y = 2$
Since $x = 1$, that implies $n = 1$
this is how far I got
Anyone know what I should do now?
| First, as mentioned by others, $1$ is a perfect square and $1+3$ is a perfect square. So you need to prove "If $N > 1$ is a perfect square, then $n+3$ is not a perfect square.
Let $n = m^2$ where $m$ and $n$ are positive integers and $n > 1$. Then we must have $m > 1$. Since $m$ is an integer, we can say that $m \ge 2$. Hence $2m+1 \ge 5$. So
$$(m+1)^2 = m^2 + (2m+1) \ge n + 5 > n+3$$
Since $n < n+3 < n+5$, then $m^2 < n+3 < (m+1)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Log with $\sqrt x$ base I'd like to know how this simplification happened:
$$\frac{1}{2}\log _{\sqrt{2}}\left(x-2\right)=\log _2\left(x-2\right)$$
$$
\begin{array}{l}
\color{red}{2 \log _{2} x+\log _{\frac{1}{2}}(1-\sqrt{x})=\frac{1}{2} \log _{\sqrt{2}}(x-2 \sqrt{x}+2) \quad } \color{blue}{0<x<1} \\
\Leftrightarrow 2 \log _{2} x-\log _{2}(1-\sqrt{x})=\log _{2}(x-2 \sqrt{x}+2) \\
\Leftrightarrow \log _{2} x-\log _{2}(1-\sqrt{x})=\log _{2}(x-2 \sqrt{x}+2)-\log _{2} x
\end{array}
$$
| $$\log_a(x) = \frac {\log_b(x)} {\log_b(a)}$$ for any $b$.
therefore: $\frac 1 2 \log_{\sqrt 2}(x-2) = \frac {log_2(x - 2)} {2 \log_2(\sqrt 2)} = \frac {log_2(x - 2)} {2 \times \frac 1 2} = \log_2(x-2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Smith Normal Form of this polynomial non square matrix in $\mathbb Q[t]$
Given $$A=\begin{bmatrix}t^3-t^2&t^2-t\\2t^3&t^4-t\\t&2t^3-2t^2\end{bmatrix} \in M_{3\times 2}(\mathbb Q[t])$$
I'm looking for the matrices $S,P,Q$ such that $S=QAP$, where $Q$ are the row and $P$ are the column transformations and $S$ is its SNF.
I have tried using the euclidean algorithm, but I can't reach SNF, since there happen to be ridiculously high fractions. After that step I struggle to find any reasonable answer:
$$ \begin{bmatrix}t&0\\0&-4t^5+5t^4-t\\0&-2t^5+4t^4-2t^3+t^2-t\end{bmatrix}$$
Any help is appreciated.
| Since $-4t^5+5t^4-t=-t(t-1)(4t^3-t^2-t-1)$ and $-2t^5+4t^4-2t^3+t^2-t=-t(t-1)(2t^3-2t^2-1)$, you only need to find an integer multiple of the GCD of $4t^3-t^2-t-1$ and $2t^3-2t^2-1$ using Euclidean algorithm. This should be straightforward:
$$
\begin{align}
4t^3-t^2-t-1 &= 2(2t^3-2t^2-1) + (3t^2-t+1)\\
9(2t^3-2t^2-1) &= (6t-4)(3t^2-t+1) - 5(2t+1)\\
4(3t^2-t+1) &= (6t-5)(2t+1) + 9.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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show this inequality $f(\sqrt{x_{1}x_{2}})\ge \sqrt{f(x_{1})\cdot f(x_{2})}$ let $$f(x)=\dfrac{1}{3^x}-\dfrac{1}{4^x},x\ge 1$$
For any $x_{1},x_{2}\ge 1$,show that
$$f(\sqrt{x_{1}x_{2}})\ge \sqrt{f(x_{1})\cdot f(x_{2})}$$
or
$$\left(\dfrac{1}{3^{\sqrt{x_{1}x_{2}}}}-\dfrac{1}{4^{\sqrt{x_{1}x_{2}}}}\right)^2 \ge \left(\dfrac{1}{3^{x_{1}}}-\dfrac{1}{4^{x_{1}}}\right)\left(\dfrac{1}{3^{x_{2}}}-\dfrac{1}{4^{x_{2}}}\right)$$Any ideas?
Thanks.
| Let $$g(x)=\ln\left(\left(\frac{1}{3}\right)^x-\left(\frac{1}{4}\right)^x\right).$$
Thus, $$g'(x)=\frac{3^x\ln4-4^x\ln3}{4^x-3^x}<0$$
because $$3^x\ln4-4^x\ln3<0\Leftrightarrow\left(\frac{4}{3}\right)^x\geq\frac{\ln4}{\ln3},$$
for which it's enough to prove that
$$\frac{4}{3}\geq\frac{\ln4}{\ln3},$$
which is $3^4>4^3.$
Now, $$g''(x)=-\frac{12^x\ln^2\frac{4}{3}}{(4^x-3^x)^2}<0,$$
which says that $g$ is decreasing concave function.
Thus, by Jensen and AM-GM we obtain:
$$\frac{\ln\left(\left(\frac{1}{3}\right)^{x_1}-\left(\frac{1}{4}\right)^{x_1}\right)+\ln\left(\left(\frac{1}{3}\right)^{x_2}-\left(\frac{1}{4}\right)^{x_2}\right)}{2}\leq\ln\left(\left(\frac{1}{3}\right)^{\frac{x_1+x_2}{2}}-\left(\frac{1}{4}\right)^{\frac{x_1+x_2}{2}}\right)\leq$$
$$\leq\ln\left(\left(\frac{1}{3}\right)^{\sqrt{x_1x_2}}-\left(\frac{1}{4}\right)^{\sqrt{x_1x_2}}\right)$$
and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $z_0 = \left(a,b\right)$ find $z$ such that $z^2 = z_0$ where $z_0 , z$ are complex numbers. Let $z_0 = \left(a,b\right)$ find $z$ such that $z^2 = z_0$ where $z_0 , z$ are complex numbers.
Here is my attempt.
taking absolute values on both sides, we obtain
$\left|z\right|^2 = \left|z_0\right|^2$
we let $z = u +vi$ so that
$u^2 + v^2 = \sqrt{a^2 + b^2} $
Any help from here?
| It would be better to write the complex number $z_0$ in rectangular form:
$$z_0 = a + bi = \sqrt{a^2 + b^2}\cdot\bigg({\frac{a}{\sqrt{a^2 + b^2}} + \frac{b}{\sqrt{a^2 + b^2}}i}\bigg)$$
Then write the complex number $z_0$ in polar form:
$$z_0 = {r_0}\exp(i\arg{z_0})$$
where $r_0 = \sqrt{a^2 + b^2}$ and $\arg{z_0} = \arctan\frac{b}{a}$.
We now have to solve the equation
$$z^2 = z_0$$
for $z$.
But $z = r\exp(i\arg{z})$ means $z^2$ in polar form is
$$z^2 = {r^2}\exp\bigg(i(2\arg{z})\bigg).$$
Equating moduli and arguments for $z^2$ and $z_0$, we get
$$r^2 = r_0 = \sqrt{a^2 + b^2} \iff r = \pm \sqrt{r_0} = \pm \sqrt{\sqrt{a^2 + b^2}} = \pm (a^2 + b^2)^{1/4}$$
and
$$2\arg{z} = \arg{z_0} \iff \arg{z} = \frac{\arg{z_0}}{2}.$$
Thus, $z$ is given explicitly in polar form as:
$$z = {\pm (a^2 + b^2)^{1/4}}\cdot{\exp\bigg(i\frac{\arg{z_0}}{2}\bigg)} = {\pm (a^2 + b^2)^{1/4}}\cdot{\exp\bigg(i\frac{\arctan\frac{b}{a}}{2}\bigg)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Fourier sine series of $x^2$ When I am trying to compute $$\frac{2}{l}\int_{0}^{l}x^2\sin\left(\frac{n\pi x}{l}\right)dx$$
I let $u = x^2$, $du = 2x dx$, $dv = \sin\left(\frac{n\pi x}{l}\right)$, $v = -\frac{l}{n \pi}\cos\left(\frac{n\pi x}{l}\right)$. Thus applying integration by parts I get $$-\frac{2l^2}{n\pi}\cos(n\pi) + \frac{4l^2}{n^2 \pi^2}\sin(n\pi) + \frac{4l}{n^2 \pi^2}\cos(n\pi)$$
Just want to check if this is right.
| The integral in question is as follows.
\begin{align}
\int x^2 \, \sin\left(\frac{n \pi x}{l}\right) \, dx &= - \frac{l \, x^2}{n \pi} \cos\left(\frac{n \pi x}{l}\right) + \frac{2 l}{n \pi} \, \int x \, \cos\left(\frac{n\pi x}{l} \right) \, dx \\
&= \left( 2 \, \left(\frac{l}{n \pi}\right)^3 - \frac{l x^2}{n \pi} \right) \, \cos\left(\frac{n \pi x}{l}\right) + \frac{2 l^2 x}{(n \pi)^2} \, \sin\left(\frac{n \pi x}{l}\right).
\end{align}
Applying limits yields
\begin{align}
\int_{0}^{l} x^2 \, \sin\left(\frac{n \pi x}{l}\right) \, dx &=
\left( 2 \, \left(\frac{l}{n \pi}\right)^3 - \frac{l^3}{n \pi} \right) \, (-1)^n - 2 \, \left(\frac{l}{n \pi}\right)^3 \\
&= \frac{l^3 \, (-1)^{n-1}}{n \pi} - 2 \, \left(\frac{l}{n \pi}\right)^3 \, (1- (-1)^n).
\end{align}
This leads to
\begin{align}
\left(\frac{x}{l}\right)^2 &= \sum_{n=1}^{\infty} \left[\frac{2}{l^3} \, \int_{0}^{l} u^{2} \, \sin\left(\frac{n \pi u}{l}\right) \, du \right] \, \sin\left(\frac{n \pi x}{l}\right) \\
&= \frac{2}{\pi} \, \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{n} \, \sin\left(\frac{n \pi x}{l}\right) - \frac{4}{\pi^3} \, \sum_{n=0}^{\infty} \frac{1}{(2n+1)^3} \, \sin\left(\frac{(2n+1) \, \pi x}{l}\right).
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/2341377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do I solve $\int_0^\infty \frac{x^2}{x^4+1} \, dx$?
$$\int_0^\infty \frac{x^2}{x^4+1} \; dx $$
All I know this integral must be solved with beta function, but how do I come to the form $$\beta (x,y)=\int_0^1 t^{x-1}(1-t)^{y-1}\;dt \text{ ?}$$
| $$\begin{eqnarray*}I=\int_0^{+\infty}\frac{x^2\,dx}{x^4+1}\stackrel{\text{parity}}{=} \frac{1}{2} \int_{-\infty}^{+\infty}\frac{dx}{x^2+\frac{1}{x^2}}&=&\frac{1}{2} \int_{-\infty}^{+\infty}\frac{dx}{\left(x-\frac{1}{x}\right)^2+2} \\[8pt]
(\text{by Glasser's Master Theorem})&=&\int_0^{+\infty}\frac{dx}{x^2+2}=\color{red}{\frac{\pi}{2\sqrt{2}}}.\end{eqnarray*} $$
As a reference, http://mathworld.wolfram.com/GlassersMasterTheorem.html.
Through the Beta function, by setting $\frac{1}{x^4+1}=u$,
$$ I=\frac{1}{4}\int_{0}^{1}u^{-3/4}(1-u)^{-1/4}\,du=\frac{\Gamma\left(\frac{1}{4}\right)\,\Gamma\left(\frac{3}{4}\right)}{4}=\frac{\pi}{4\sin\frac{\pi}{4}} =\color{red}{\frac{\pi}{2\sqrt{2}}}$$
by the reflection formula for the $\Gamma$ function.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $ \left|x-1\right|<1$ implies $ \left|x^2-4x+3\right|<3$. Question:
Let $ x$ be a real number. Prove that if $ \left|x-1\right|<1$ then $ \left|x^2-4x+3\right|<3$.
We can write $ \left|x^2-4x+3\right|<3 $ as $ |x-3||x-1| < 3$. If I start with $ \left|x-1\right|<1$, how can I show that $ |x-3| < 3$ ?
| $|x-1|<1\implies |x^2-4x+3|=|(x-1)^2+2(1-x)|\leq $ $\leq |(x-1)^2|+|2(1-x)|<1^2+2\cdot 1=3.$
| {
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"url": "https://math.stackexchange.com/questions/2344577",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a,b$ are roots of $3x^2+2x+1$ then find the value of an expression It is given that $a,b$ are roots of $3x^2+2x+1$ then find the value of:
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3$$
I thought to proceed in this manner:
We know $a+b=\frac{-2}{3}$ and $ab=\frac{1}{3}$. Using this I tried to convert everything to sum and product of roots form, but this way is too complicated!
Please suggest a simpler process.
| Plug $x=\frac{1-y}{1+y}$ in the given equation. We get:
$$\frac{3 (1-y)^2}{(y+1)^2}+\frac{2 (1-y)}{y+1}+1=0$$
Expanding and collecting, we have:
$$y^2-2 y+3=0$$
whose solutions are $$y_1=\frac{1-a}{1+a};\;y_2=\frac{1-b}{1+b}$$
We also know that sum of roots is $s=y_1+y_2=2$ and product is $p=y_1y_2=3$.
The sum of cubes can be written as follows
$$y_1^3+y_2^3=\left(y_1+y_2\right)^3-3y_1y_2(y_1+y_2)=s^3-3ps=8-18=-10$$
so we have
$$\left(\dfrac{1-a}{1+a}\right)^3+\left(\dfrac{1-b}{1+b}\right)^3=-10$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Issue with Baby Rudin Example 7.3 Example 7.3 of Baby Rudin states that the sum
\begin{align}
\sum_{n=0}^{\infty}\frac{x^2}{(1 + x^2)^n}
\end{align}
is, for $x \neq 0$, a convergent geometric series with sum $1 + x^2$. This confuses me. As far as I know, a geometric series is a series of the form
\begin{align}
\sum_{n=0}^{\infty}x^n,
\end{align}
and such a series converges to $\frac{1}{1 - x}$ if $x \in [0,1)$. Now, it is certainly true that if we put $y = \frac{x^2}{1 + x^2}$, then $\frac{1}{1 - y} = 1 + x^2$. But in this case, shouldn't the series we're discussing be
\begin{align}
\sum_{n=0}^{\infty}\left[\frac{x^2}{1 + x^2}\right]^n?
\end{align}
What am I missing?
| You should write the first few terms to understand its behavior:
$$\sum_{n=0}^{\infty}\frac{x^2}{(1 + x^2)^n}=x^2+\frac{x^2}{1+x^2}+\frac{x^2}{(1+x^2)^2}+\frac{x^2}{(1+x^2)^3}+\cdots=$$
$$x^2\left(1+\frac{1}{1+x^2}+\frac{1}{(1+x^2)^2}+\frac{1}{(1+x^2)^3}+\cdots\right)\stackrel{x\ne 0}=$$
$$x^2\cdot \frac{1}{1-\frac{1}{1+x^2}}=1+x^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2348909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Are these two distinct series representations for $\frac{1}{1-x^2}$ correct? Let $f(x)=\frac{1}{1-x^2}$.
With a $u=x^2$ substitution, $f(x)$ becomes $g(u)=\frac{1}{1-u}$, which being the limit of a geometric series has the form $$g(u)=\sum_{k=0}^\infty u^k$$
Plugging the substitution back in yields a power series representation for $f(x)$
$$\frac{1}{1-x^2}=\sum_{k=0}^\infty (x^2)^k=\sum_{k=0}^\infty x^{2k}$$
A different approach would be to factor the denominator of $f(x)$
$$\frac{1}{1-x^2}=\frac{1}{(1-x)(1+x)}=\frac{1}{1+x}\frac{1}{1-x}=\frac{1}{1+x}\sum_{k=0}^\infty x^k=\sum_{k=0}^\infty \frac{x^k}{1+x}$$
Here we have two different series representations for the same function but the second one is not a power series representation, so I'm guessing there's no issues here (unless I've made some mistake in either of my derivations).
| We have
\begin{align}
\sum_{k \ge 0} \frac{x^k}{1 + x} &= \sum_{k \ge 0} \sum_{\ell \ge k} (-1)^{\ell-k}x^\ell \\
&= \sum_{\ell \ge 0} \sum_{k \le \ell} (-1)^{\ell-k}x^\ell \\
&= \sum_{\ell \ge 0} x^\ell \sum_{k \le \ell} (-1)^{\ell-k}
\end{align}
where
$$ \sum_{k \le \ell} (-1)^{\ell-k} = \begin{cases} 0 & \text{if $\ell$ is odd} \\ 1 & \text{if $\ell$ is even} \end{cases}. \tag{1} $$
In fact, the formal identity
$$ \sum_{\ell \ge 0} x^\ell \sum_{k \le \ell} (-1)^{\ell-k} = \frac{1}{1 - x^2} $$
is a great way to prove $(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2349503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Squares and Rationalization I was solving this question, and I'm hitting a wall.
If ${1\over{\sqrt{2011+\sqrt{2011^2-1}}}}=\sqrt{m}-\sqrt{n}$, where $m$ and $n$ are positive integers, what is the value of $m+n$?
I tried to solve this question with two approaches:
${\sqrt{m}-\sqrt{n}}={1\over{\sqrt{2011+\sqrt{2011^2-1}}}}$
$={\sqrt{2011+\sqrt{2011^2-1}}\over{2011+\sqrt{2011^2-1}}}$
$={({\sqrt{2011+\sqrt{2011^2-1}})(\sqrt{2011^2-1}-2011)}\over{2011^2-2012}}$
Squaring,
$m+n-2\sqrt{mn}$
$={{{{(2011+\sqrt{2011^2-1}})\{4022(2011-\sqrt{2011^2-1})-1\}}}\over{2011^4+2012^2-2\cdot{2011^2}\cdot{2012}}}$
This goes nowhere
And
${\sqrt{m}-\sqrt{n}}={1\over{\sqrt{2011+\sqrt{2011^2-1}}}}$
Squaring
${m+n-2\sqrt{mn}}$
$={{\sqrt{2011^2-1}-2011}\over{(2011)(2010)-1}}$
Which again, probably goes nowhere.
Can anyone help?
| HINT:
$$(a+\sqrt{a^2-1})(a-\sqrt{a^2-1})=a^2-(a^2-1)=?$$
$$\implies\sqrt{a+\sqrt{a^2-1}}\cdot\sqrt{a-\sqrt{a^2-1}}=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Confirming my solution to the expansion of $z/(z-1)(z+3)$ I would just like a confirmation of my work for the following problem.
Represent the function $$ \frac{z}{(z-1)(z+3)} $$ as a Laurent Series around $z=0$ when $1<|z|<3$.
Using partial fraction decomposition to simplify the problem a bit,
$$ \frac{1}{(z-1)(z+3)} = \frac{1}{4(z-1)} - \frac{1}{4(z+3)} \\ \Rightarrow \frac{z}{(z-1)(z+3)} = \frac{z}{4(z-1)} - \frac{z}{4(z+3)}$$
Tackling each fraction seperatly,
$$ \frac{1}{z-1} = \frac{1}{z}\cdot \frac{1}{1-z^{-1}} = \frac{1}{z} \sum_{n=0}^\infty z^{-n} = \sum_{n=0}^\infty z^{-n-1}= \sum_{n=1}^\infty z^{-n} $$
$$ \Rightarrow \frac{z}{4} \cdot \frac{1}{z-1} = \frac{z}{4} \sum_{n=1}^\infty z^{-n}
= \sum_{n=1}^\infty \frac{z^{-(n-1)}}{4} = \sum_{n=0}^\infty \frac{z^{-n}}{4} \\ $$
$$ \frac{1}{3+z} = \frac{1}{3z} \cdot \frac{1}{1-(-3z)^{-1}} = \frac{1}{3z} \sum_{n=0}^\infty \frac{(-1)^n z^{-n}}{3} = \sum_{n=0}^\infty \frac{(-1)^n z^{-n-1}}{9} = \sum_{n=1}^\infty \frac{(-1)^{n-1} z^{-n}}{9} \\ \Rightarrow \frac{z}{4}\cdot \frac{1}{3+z} = \frac{z}{4} \sum_{n=1}^\infty \frac{(-1)^{n-1} z^{-n}}{9} = \sum_{n=1}^\infty \frac{(-1)^{n-1} z^{-(n-1)}}{36} = \sum_{n=0}^\infty \frac{(-1)^{n} z^{-n}}{36} \\ $$
Now adding them back together,
$$ \frac{z}{(z-1)(z+3)} = \frac{z}{4(z-1)} - \frac{z}{4(z+3)} = \sum_{n=0}^\infty \frac{z^{-n}}{4} - \sum_{n=0}^\infty \frac{(-1)^{n} z^{-n}}{36} = \sum_{n=0}^\infty \frac{(9-(-1)^n)}{36} z^{-n} $$
Therefore,
$$ \frac{z}{(z-1)(z+3)} = \sum_{n=0}^\infty \frac{(9-(-1)^n)}{36} z^{-n} \ \
\ (1<|z|<3) $$
Does everything look correct?
Thank you!
| The function
\begin{align*}
f(z)&=\frac{z}{(z-1)(z+3)}\\
&=\frac{1}{4(z-1)}+\frac{3}{4(z+3)}\\
\end{align*}
has two simple poles at $z=1$ and $z=-3$.
Since we want to find a Laurent expansion with center $0$, we look at the poles $1$ and $-3$ and see they determine three regions.
\begin{align*}
|z|<1,\qquad\quad
1<|z|<3,\qquad\quad
3<|z|
\end{align*}
*
*The first region $ |z|<1$ is a disc with center $0$, radius $1$ and the pole $1$ at the boundary of the disc. In the interior of this disc all two fractions with poles $1$ and $-3$ admit a representation as power series at $z=0$.
*The second region $1<|z|<3$ is the annulus with center $0$, inner radius $1$ and outer radius $3$. Here we have a representation of the fraction with the pole at $1$ as principal part of a Laurent series at $z=0$, while the fraction with pole at $-3$ admits a representation as power series.
*The third region $|z|>3$ containing all points outside the disc with center $0$ and radius $3$ admits for all fractions a representation as principal part of a Laurent series at $z=0$.
A power series expansion of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{a}\cdot\frac{1}{1+\frac{z}{a}}\\
&=\sum_{n=0}^{\infty}\frac{1}{a^{n+1}}(-z)^n
\end{align*}
The principal part of $\frac{1}{z+a}$ at $z=0$ is
\begin{align*}
\frac{1}{z+a}&=\frac{1}{z}\cdot\frac{1}{1+\frac{a}{z}}=\frac{1}{z}\sum_{n=0}^{\infty}\frac{a^n}{(-z)^n}
=-\sum_{n=0}^{\infty}\frac{a^n}{(-z)^{n+1}}\\
&=-\sum_{n=1}^{\infty}\frac{a^{n-1}}{(-z)^n}
\end{align*}
We can now obtain the Laurent expansion of $f(z)$ at $z=0$ for all three regions. Here we have to consider
*
*Region 2: $1<|z|<3$
\begin{align*}
f(z)&=\frac{1}{4(z-1)}+\frac{3}{4(z+3)}\\
&=-\frac{1}{4}\sum_{n=1}^\infty\frac{(-1)^{n-1}}{(-z)^n}+\frac{3}{4}\sum_{n=0}^\infty \frac{1}{3^{n+1}}(-z)^n\\
&=\frac{1}{4}\sum_{n=1}^\infty\frac{1}{z^n}+\frac{1}{4}\sum_{n=0}^\infty\left(- \frac{1}{3}\right)^nz^n\\
\end{align*}
The Laurent expansion for the other regions can be calculated similarly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integration by trig substitution - why is my answer wrong? Attempting integral:
$$-\int \frac{dx}{\sqrt{x^2-9}}$$
Let $x = 3\ sec\ \theta$ so that under the square root we have:
$$\sqrt{9\ sec^2\ \theta - 9}$$
$$\sqrt{9(sec^2\ \theta - 1)}$$
$$\frac{dx}{d\theta} =3\ sec\ \theta\ tan\ \theta$$
The $1/3$ and the $3$ cancel eachother out outside the integral, so we have:
$$-\int \frac{sec\ \theta\ tan\ \theta\ d\theta}{tan\ \theta}$$
The $tan\ \theta$ terms cancel, so we're left to integrate $sec\ \theta$ which is equal to:
$$- \ln\ (tan\ \theta + \ sec\ \theta) + C$$
Since $tan\ \theta = \sqrt{x^2-9}\ $ since it replaced it in the integral, and $sec\ \theta = \frac{x}{3}$, the answer is:
$$-\ ln\ (\sqrt{x^2-9}\ + \frac{x}{3})+C$$
However, using an online calculator, the answer turned out to be:
$$-\ ln\ (\sqrt{x^2-9}\ + x)+C$$
It seemed to come about due to their substitution of $u = \frac{x}{3}$ we led them to get the standard integral of $sec^{-1}x$, which would imply that
$$\sqrt{\frac{x^2}{9} - 9}$$
becomes $$\sqrt{u^2 - 1}$$
And I just don't see how. Can someone explain why what they did is valid and/or where my mistake was?
| HINT:
If $x=3\sec\theta$
$$\tan^2\theta=\left(\dfrac x3\right)^2-1=\dfrac{x^2-9}9$$
$|\tan\theta|=\dfrac{\sqrt{x^2-9}}3$
$\sec\theta+|\tan\theta|=\dfrac{x+\sqrt{x^2-9}}3$
$\implies\ln(\sec\theta+|\tan\theta|)=\ln(x+\sqrt{x^2-9})-\ln3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2354524",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Multiplication of a continuous function with non continuous $$f(x) = \begin{cases} e^x e^{-1} & ,0 \leq x\leq 1
\\ \lfloor x \rfloor+ 1 & ,1< x < 3 \end{cases}$$
and $$g(x) = x^2 -ax + b $$ such that $f(x)g(x)$ is continuous in $[0,3)$ then find the ordered pair $(a,b)$
| You need to make it continuous at $x=1$. That is
\begin{align}
\lim_{x \to 1^-}f(x)g(x) &= \lim_{x \to 1^+}f(x)g(x) \\
e^{-2}(1-a+b) &= 2(1-a+b) \\
(e^{-2}-2)(1-a+b) &= 0 \\
(1-a+b) &= 0 \\
b &= a-1 \\
\hline
g(x) &= x^2 + ax +(a-1)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355066",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to find the value of this expression Suppose I have a general quadratic equation $ ax^2 + bx + c = 0 $
And I have $ \alpha , \beta $ as roots, then what is the simplest way to find $ (a\alpha + b)^{-2} + (a\beta + b)^{-2} $ ?
I know the formula for finding the sum and product of roots, but that doesn't helped me in finding the value.
Any good hint is welcomed.
| Alternative hint: derive the equation whose roots are $y_1=(a\alpha + b)^{-2}$ and $y_2 = (a\beta + b)^{-2}\,$.
Let $y=(ax+b)^{-2}\,$, then $\;\displaystyle ax+b=\pm \frac{1}{\sqrt{y}} \;\;\iff\;\; x = \frac{-b \pm \frac{1}{\sqrt{y}}}{a} = \frac{-b\sqrt{y}\pm1}{a\sqrt{y}}\,$.
Substituting back into the original equation:
$$\require{cancel}
\begin{align}
a\,\frac{b^2 y \mp 2b\sqrt{y}+1}{a^2 y} + b\,\frac{-b \sqrt{y} \pm 1}{a \sqrt{y}}+c = 0 \;\;&\iff\;\; \cancel{b^2 y} \mp 2b\sqrt{y}+1 -\cancel{b^2 y} \pm b\sqrt{y}+acy = 0 \\
&\iff\;\; acy \mp b\sqrt{y}+1=0 \\
&\iff\;\; (acy+1)^2 = b^2y \\
&\iff\;\; a^2c^2y^2 +(2ac-b^2)y+1=0
\end{align}
$$
Then $\displaystyle(a\alpha + b)^{-2} + (a\beta + b)^{-2} =y_1+y_2=\frac{b^2-2ac}{a^2c^2}$ by Vieta's relations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2355718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Small doubt on this trigonometry question If $\alpha, \beta$ are complementary angles such that $b \sin\alpha = a $, then find the value of $\sin \alpha \cos \beta - \cos \alpha \sin \beta$.
The question itself is straightforward, but while solving for $\cos \alpha $ and $ \sin \beta$ , which sign should I take? Or should I consider both signs?
The answer given is ${2a^2 - b^2 \over b^2}$
Am I missing something? Why is the solution only in terms of the positive values?
| Maybe the form of the answer : $X - 1$ should make you think about the following transformation
\begin{align*}\sin \alpha \cos \beta - \cos \alpha \sin \beta &= 2 \sin \alpha \cos \beta - (\sin \alpha \cos \beta + \cos \alpha \sin \beta) \\
& = 2 \sin \alpha \cos \big( \frac{\pi}{2} - \alpha \big) - \sin(\alpha+\beta) \\
& = 2 \sin^2(\alpha) - 1\quad \quad \quad \textrm{ as $\alpha+\beta=\frac{\pi}{2}$ } \\
& = 2 \frac{a^2}{b^2} - 1 = \frac{2a^2-b^2}{b^2}.
\end{align*}
$ $
(No need to use square roots, which bring sign problems : the result is true if $\alpha+\beta=\frac{\pi}{2}$ regardless of the signs of the $\cos$ and $\sin$)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2358461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Express the differential equation as power series
My attempt: $y=\sum^{\infty}_{n=0}a_n x^n \rightarrow y'=\sum^{\infty}_{n=0}na_n x^{n-1}\rightarrow y''=\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}\\
(2+x^2)y"+x^2y'+3y=(2+x^2)\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}+x^2\sum^{\infty}_{n=0}na_n x^{n-1}+3\sum^{\infty}_{n=0}a_n x^n\\
2\sum^{\infty}_{n=0}n(n-1)a_n x^{n-2}+\sum^{\infty}_{n=0}n(n-1)a_n x^{n}+\sum^{\infty}_{n=0}na_n x^{n+1}+3\sum^{\infty}_{n=0}a_n x^{n}$
from here can any help me
| Note:
$$(2+x^2)y''+x^2y'+3y=$$
$$(2+x^2)(2a_2+3\cdot 2a_3x+4\cdot 3a_4x^2+5\cdot 4a_5x^3+\cdots)+$$
$$x^2(a_1+2a_2x+3a_3x^2+\cdots)+$$
$$3(a_0+a_1x+a_2x^2+a_3x^3+a_4x^4+\cdots)=$$
$$4a_2+3a_0+(2(3\cdot 2)a_3+3a_1)x+$$
$$2(4\cdot 3)a_4+a_1+2a_2+3a_2)x^2+$$
$$(2(5\cdot 4)a_5+2a_2+3\cdot 2a_3+3a_3)x^3+$$
$$(2(6\cdot 5)a_6+3a_3+4\cdot 3a_4+3a_4)x^4+\cdots=$$
$$4a_2+3a_0+\sum_{n=1}^{\infty} [2(n+2)(n+1)a_{n+2}+(n-1)a_{n-1}+(3+(n-1)n)a_n]x^n.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2359382",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $y = \frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Prove, using an algebraic method,that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive. Hence, determine the minimum and maximum points $y=\frac{2x}{x^2 +1}$ .
What I tried:
Firstly, I thought of using partial fractions but since $x^2 +1 =(x-i)(x+i)$, I don't think it is possible to show using partial fractions.
Secondly, decided to use differentiation
$y=\frac{2x}{x^2 +1}$
$\frac {dy}{dx} = \frac {-2(x+1)(x-1)}{(x^2 +1)^2 }$
For stationary points:
$\frac {dy}{dx} = 0$
$\frac {-2(x+1)(x-1)}{(x^2 +1)^2 } = 0$
$x=-1$ or $x=1$
When $x=-1,y=-1$
When $x=1,y=1$
Therefore, this implies that $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
^I wonder if this is the correct method or did I leave out something?
The third way was using discriminant
Assume that $y=\frac{2x}{x^2 +1}$ intersects with $y=-1$ and $y=1$
For $\frac{2x}{x^2 +1} = 1$,
$x^2 -2x+1 = 0$
Discriminant = $ (-2)^2 -4(1)(1) = 0 $
For $\frac{2x}{x^2 +1} = -1$,
$x^2 +2x+1 = 0$
Discriminant = $ (2)^2 -4(1)(1) = 0 $
So, since $y=\frac{2x}{x^2 +1}$ touches $y=-1$ and $y=1$, $y=\frac{2x}{x^2 +1}$ lies between $-1$ and $1$ inclusive.
Is the methods listed correct?Is there any other ways to do it?
| We need to prove that$$\left|\frac{2x}{x^2+1}\right|\leq1$$ or
$$x^2-2|x|+1\geq0$$ or
$$\left(|x|-1\right)^2\geq0.$$
The equality occurs for $|x|=1$, which gives:
$(1,1)$ is a maximum point and $(-1-1)$ is a minimum point.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2361415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 9,
"answer_id": 0
} |
Math subject GRE Exam 9768 Q.16 The question is in the following picture:
And the answer is (D).
At first I agreed with this answer when I added the given three vectors component wise,but when I added the forth component in the three vectors I understood that it is impossible that we can get 5, so my opinion changed to the choice (A), so what is the misconception that I have or where is my wrong thinking, could anyone help me please?
| The question is the same as asking for what $m$ the following equation has a solution:
$$
\begin{pmatrix}
0&0&1\\
1&0&1\\
1&0&2\\
1&1&0
\end{pmatrix}
\begin{pmatrix}
x\\y\\z
\end{pmatrix}=
\begin{pmatrix}
1\\2\\m\\5
\end{pmatrix}\tag{1}
$$
Working on column vectors might make some observation easier. Note that (1) is equivalent to
$$
x\begin{pmatrix}
0\\1\\1\\1
\end{pmatrix}
+y\begin{pmatrix}
0\\0\\0\\1
\end{pmatrix}
+z\begin{pmatrix}
1\\1\\2\\0
\end{pmatrix}=\begin{pmatrix}
1\\2\\m\\5
\end{pmatrix}\tag{2}
$$
It is very easy to observe that when $m=3$, (2) has a solution: $x=z=1$, $y=4$. This rules out A,B,C. To see $m=3$ is necessarily true, note that (2) implies:
$$
0x+0y+1z=1\\
1x+0y+1z=2\\
1x+0y+2z=m
$$
To make the linear system consistent so that it has a solution, one must have $1+2=m$ by observing that the coefficients of the first two rows add up to those of the third row.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Integral $\int \frac{x^2}{x^4+x^2+1}\ dx$ I have taken $(x^4+x^2+1)=u$ to differentiate it w.r.t. $x$ to get
$$\int \frac{x}{2u(2x^2+1)}\ du$$
| $$((x^2+1)+x)((x^2+1) -x) = x^4 + 2x^2 + 1 - x^2 = x^4 + x^2 + 1$$
Then use partial fraction method.
Edit: More hints,
$$((x^2+1)+x) + ((x^2+1) -x) = 2(x^2+1)$$
$$\int\frac{dx}{a^2+x^2} = \frac{1}{a}\tan^{-1}\frac{x}{a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2363639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Explain why the graph of $y=\frac{4x}{x^2+1}$ and $y=2\sin(2\arctan x)$ are the same. Explain why the graph of $y=\frac{4x}{x^2+1}$ and $y=2\sin(2\arctan x)$ are the same.
The first equation is of the form of Newton's Serpentine. When you graph the second equation it appears to overlap the first equation.
I'm not sure whether these two equations are identities or just very close approximations.
I tried to manipulate both equations to get the other but failed.
How does one explain why these two equations are identical?
| We use the identity
$$\sin(2\theta) = \frac{2\tan\theta}{1 + \tan^2\theta}$$
Let $\arctan x = \theta$. Then $x = \tan\theta, -\frac{\pi}{2} < x < \frac{\pi}{2}$. Hence,
$$2\sin(2\arctan x) = 2\sin(2\theta) = 2 \cdot \frac{2\tan\theta}{1 + \tan^2\theta} = 2 \cdot \frac{2x}{1 + x^2} = \frac{4x}{1 + x^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2367735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove this trigonometric inequality about the angles of $\triangle ABC$
In $\Delta ABC$ show that
$$\cos{\frac{A}{2}}+\cos\frac{B}{2}+\cos\frac{C}{2}\ge \frac{\sqrt{3}}{2} \left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\frac{A-B}{2}\right)$$
since
$$\frac{\sqrt{3}}{2}\left(\cos\frac{B-C}{2}+\cos\frac{C-A}{2}+\cos\dfrac{A-B}{2}\right)=\frac{\sqrt{3}}{2}\sum\cos\frac{A}{2}\cos\frac{B}{2}+\sin\frac{B}{2} \sin\frac{C}{2}$$
| Denote the difference $\text{LHS}-\text{RHS}$ by $S(A,B,C)$. We need to show that $S(A,B,C)\ge0$. We begin with a simple
Proposition. For $\alpha\in[0,\pi/2]$ we have the following inequality
$$2\cos\frac{\pi-\alpha}4-\sqrt3\cos\frac{\pi-3\alpha}4<\sqrt3$$
Proof. $\frac{\pi-3\alpha}4\in[-\frac\pi8,\frac\pi4]$, thus
$$2\cos\frac{\pi-\alpha}4-\sqrt3\cos\frac{\pi-3\alpha}4\le2-\sqrt3\cos\frac\pi4<\sqrt3.\quad\square$$
Now suppose $A\le\pi/2$ (such an angle always exists in a triangle; e.g. take $A$ to be the smallest one among $A,B,C$). Reformulate the desired inequality as
$$\cos\frac A2+2\cos\frac{\pi-A}4\cos\frac{B-C}4\ge\sqrt3\cos^2\frac{B-C}4-\frac{\sqrt3}2+\sqrt3\cos\frac{\pi-3A}4\cos\frac{B-C}4$$
Let $t=\cos\frac{B-C}4\in(0,1]$. Consider the function
$$f(x)=-\sqrt3t^2+(2\cos\frac{\pi-A}4-\sqrt3\cos\frac{\pi-3A}4)x$$
By the proposition at the beginning of this proof, The axis of symmetry of $f$ is
$$x=\frac{2\cos\frac{\pi-A}4-\sqrt3\cos\frac{\pi-3A}4}{2\sqrt3}<\frac12$$
And since $-\sqrt3<0$, the parabola opens downwards. Consequently $f(t)\ge f(1)$. Furthermore, $f(t)>f(1)$ if $t\not=1$. In other words, $S(A,B,C)\ge S(A,\frac{B+C}2,\frac{B+C}2)$ with the equality if and only if $B=C$.
We're almost done. Observe that $S$ is continuous in $A, B, C$. We may extend the range of $A,B,C$ to $[0,\pi]^3\cap\{A+B+C=\pi\}$ without destroying the above argument. This is a compact set, on which $S$ attains its minimum $S_0$ at the point $(A_0, B_0, C_0)$. WLOG we may assume $A\le\pi/3$. If $B_0\not=C_0$, then $S_0>S(A_0,\frac{B_0+C_0}2,\frac{B_0+C_0}2)$, contradicting the minimality of $S_0$. Thus $B_0=C_0$. But $B_0+C_0=\pi-A_0\le\pi$, so $B_0\le\pi/2$. Applying the above argument once again, we see $A_0=C_0$. Now we conclude that $A_0=B_0=C_0=\pi/3$, hence $S_0=S(\pi/3,\pi/3,\pi/3)=0$. This shows that $S\ge S_0=0$.
| {
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"url": "https://math.stackexchange.com/questions/2368630",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Which area is larger, the blue area, or the white area? In the square below, two semicircles are overlapping in a symmetrical pattern. Which is greater: the area shaded blue or the area shaded white?
My Solution
Let the length of each side of the square be $2r$.
The area of the square is $4r^2$.
The two semi-circles have equal area.
Area of one semi-circle = $\frac{{\pi}r^2}{2}$.
${\times}2 = {\pi}r^2$
White area = ${\pi}r^2 - $ area of the intersection of the two circles.
Let the area of the intersection of the two circles be $t$.
White area = ${\pi}r^2 - t$.
The segments that make up $t$ are identical.
$t$ = area of segment ${\times}2$.
Area of segment = Area of sector - Area of triangle.
Angle of sector = $90^{\circ}$ (The circles both have radius $r$, and the outer shape is a square.
Angle of sector $ = \frac{1}{4} * {\pi}r^2$.
Area of triangle $ = \frac{1}{2} * r^2$.
Area of segment $ = \frac{{\pi}r^2 - 2r^2}{4}$.
$t = 2 {\times} \frac{{\pi}r^2 - 2r^2}{4}$.
$t = \frac{{\pi}r^2 - 2r^2}{2}$.
White area $ = {\pi}r^2 - \frac{{\pi}r^2 - 2r^2}{2}$.
White area $ = \frac{2{\pi}r^2 - {\pi}r^2 + 2r^2}{2}$.
White area $ = \frac{{\pi}r^2 + 2r^2}{2}$.
Blue area = $r^2\left(4 - \frac{{\pi} + 2}{2}\right)$.
Blue area = $r^2\left(\frac{8 - ({\pi} + 2)}{2}\right)$.
Blue area = $r^2\left(\frac{6 - {\pi}}{2}\right)$.
If White area $-$ Blue area $ \gt 0$, then the White area is larger.
$$r^2\left(\frac{{\pi}+2 - (6 - {\pi}}{2}\right)$$
$$r^2\left(\frac{2{\pi} - 4}{2}\right)$$
$$r^2(\pi - 2)$$
$\therefore$ the white area is larger.
My answer was wrong.
What is the error in my solution?
The provided solution:
| If you just divide the lower left part of the blue area and move each part $90$ degree up to join them to the main blue part, then the area of the new blue shape will be equal to the white part.
| {
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What is the largest 3-digit number which has all 3 digits different and is equal to 37 times the sum of it's digits?
What is the largest 3-digit number which has all 3 digits different and is equal to 37 times the sum of it's digits?
-Question 28, Junior Division, AMC 2016
I found the solution here, page 51, however I don't understand it from the very beginning. How does $100a+10b+c$ equal $37a+37b+37c$? Can somebody explain to me what's happened in this solution step by step please? Or can somebody give an alternate solution, suitable for Year 7 and 8? Thanks!
P.S. If anybody knows a more accurate tag for this question, please feel free to edit my question or comment so I can edit it if you can't.
| "What is the largest 3-digit number...."
Let $N = abc$ have three digits. $N = 100a + 10b + c$. That's what the writing numbers of digits mean. $593 = 500 + 90 +3$ and $abc = 100a + 10b + c$.
" which has all 3 digits different"
So $a \ne b; a\ne c; b \ne c$.
" and is equal to 37 times the sum of it's digits?"
So $N = 37*(a+b+c) = 37a + 37b + 37c$
And $N = 100a + 10b + c$.
So $100a + 10b + c = 37a + 37b + 37c$
So $(100a - 37a)= (37 - 10)b + (37-1)c$
So $63a = 27b + 36c$
So $\frac {63}9 a = \frac {27}9 b + \frac {36}9 c$
So $7a = 3b + 4c$.
Obviously $a = 1, b= 1, c=1$ is a solution but 1) It's probably not the largest and 2) The digits aren't all different.
$b \ne c$ so $c = b \pm k$ for some $k \ne 0$.
So $7a = 3b + 4c = 3b + 4(b \pm k) = 7b \pm 4k$
So $a = b \pm \frac 47k$ so $7$ must divide $k$. But $b$ and $c$ are single digits and $c = b \pm k$ so $k = 7$ and $a = b \pm 4$
The possible answers are $b =0; c = 0+7=7; a = 0+4=4$
$b=1; c = 1+7= 8; a = 1+4 =5$
$b=2; c = 2+7 =9; a= 2+4 =6$
$b = 7; c = 7-7=0 ; a=7-4 = 3$
$b = 8; c = 8-7; a = 8-4 = 4$
$b = 9; c = 9 -7 =2; a = 9-4=5$.
Of those $a=6;b=2; c= 9$ and $N = 629 = 37*17 = 37(6+2+9)$ is the largest such answer.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the probability that Cathy wins?
Alice, Bob and Cathy take turns (in that order) in rolling a six sided
die. If Alice ever rolls a 1, 2 or 3 she wins. If Bob rolls a 4 or a 5
he wins, and Cathy wins if she rolls a 6. They continue playing until
a player wins. What is the probability (as a fraction) that Cathy
wins?
This was I question I got when competing at a Mathematics competition earlier this week. I thought the answer was $\frac{1}{18}$ but it turns out the actual answer is $\frac{1}{13}$...
This is my working out of why I thought it was $\frac{1}{18}$:
For Cathy to win:
*
*Alice needs to get a 4, 5 or 6 and there is a $\frac{3}{6}$ chance of that happening when she rolls the die
*Bob needs to get a 1, 2, 3 or 6 and there is a $\frac{4}{6}$ probability of that happening when he rolls the die
*Cathy needs to get 6 and there is a $\frac{1}{6}$ probability of that happening when she rolls the die
Those three need to happen for Cathy to win so:
$$\frac{3}{6} \times \frac{4}{6} \times \frac{1}{6} = \frac{1}{18}$$
However, the answer was $\frac{1}{13}$ so something must be wrong! Could someone please demonstrate why it was $\frac{1}{13}$?
| What you are leaving out is the possibility that nobody wins the first round. There is a $\frac{1}{18}$ chance that Cathy wins AND does so in the first round. Her overall odds will be $\sum_{n=1}^{\infty}(\frac{1}{2})^n(\frac{2}{3})^n(\frac{5}{6})^{n-1}(\frac{1}{6})=\frac{6}{5}\frac{1}{6}\sum_{n=1}^{\infty}(\frac{5}{18})^n=\frac{1}{5}\frac{5}{13}=\frac{1}{13}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} =\sqrt \frac{x}{2}$ The value of $x$ (considering only the positive root) satisfying the equation $$\frac{\sqrt{5+2\sqrt{6}} + \sqrt{5-2\sqrt{6}}}{\sqrt{5+2\sqrt{6}} - \sqrt{5-2\sqrt{6}}} = \sqrt{\frac{x}{2}}$$ is?
Please, can you guys help me out because I can't understand which formula to use, Do we have to use the Discriminant formula ($b^2-4ac$)?
| Let
$ a = \sqrt{5+ 2\sqrt{6}}, \ \ b= \sqrt{5- 2\sqrt{6}}.$
$$ \frac{ a +b}{a-b} = \sqrt{\frac{x}{2}}$$
$$ \frac{(a+b)(a+b)}{(a-b)(a+b)} = \frac{(a+b)^2}{a^2 - b^2} = \sqrt{\frac{x}{2}}$$
$$ \frac{\left(\sqrt{5+ 2\sqrt{6}} + \sqrt{5-2\sqrt{6}}\right)^2}{\left(\sqrt{5+2\sqrt{6}}\right)^2-\left(\sqrt{5-2\sqrt{6}})\right)^2} = \sqrt{\frac{x}{2}}$$
$$ \frac{5+ 2\sqrt{6}+ 2\sqrt{(5+2\sqrt{6})(5 - 2\sqrt{6})}+5 -2\sqrt{6}}{5 +2\sqrt{6} -5 +2\sqrt{6}} = \sqrt{\frac{x}{2}}$$
$$ \frac{10 +2\sqrt{1}}{4\sqrt{6}} = \sqrt{\frac{x}{2}}$$
$$ \frac{12}{4\sqrt{6}}=\sqrt{\frac{x}{2}}$$
$$ \frac{3}{\sqrt{6}} = \sqrt{\frac{x}{2}}$$
$$ x = 3.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Difficult Homogeneous Differential Equation Solve the differential equation:
$$\frac{dy}{dx}=\frac{\sqrt{x^2+3xy+4y^2}}{x+2y}$$
I tried to solve it by putting $t=x+2y$ but that lead to a very complicated integral. The hint given is that equation is reducible to homogeneous form.
| The integral that Noé AC derived is expressible in terms of elementary functions although it is highly messy. Integration of any rational function $R(u,\sqrt{P_2(u)})$, where $P_2$ is a quadratic, can be reduced to the integration of a rational function.
The first step is the reduction to a rational function of $\sin\theta$ and $\cos\theta$. In order to do this we take the integral
$$
\int \frac{1+2u}{\sqrt{4 u^2 + 3 u + 1} - u - 2u^2} du
$$
and make the trig substitution $ 2 u + \frac34 = \frac{\sqrt{7}}{4}\tan\theta$. This leads to an integrand that is a rational function of $\sin\theta$ and $\cos\theta$. I got something like
$$
\int \frac{2\sqrt{7}\left(\cos\theta + \sqrt{7} \sin\theta\right)}{3 \cos^3\theta + 8 \sqrt{7} \cos^2\theta + 2 \sqrt{7} \sin\theta \cos^2\theta - 7 \sin^2\theta \cos\theta} d\theta
$$
The integral of any rational function of $\sin\theta$ and $\cos\theta$ can be converted to the integral of a rational function of $t$ through the change of variables $ t = \tan(\frac{\theta}{2});d\theta = \frac{2dt}{1+t^2}; \sin\theta=\frac{2t}{1+t^2}; \cos\theta=\frac{1-t^2}{1+t^2}$. This leads to the integration of a rational function of $t$. I get that the resulting integral is
$$
\int -\frac{16 \left(t^2+1\right)\left(t^2-2 \sqrt{7} t-1\right)}{\left(t^2-1\right)
\left(\left(8 \sqrt{7}-3\right) t^4+4 \sqrt{7} t^3+34 t^2-4 \sqrt{7} t-8
\sqrt{7}-3\right)}dt
$$
This integral is in principle doable with partial fractions, although the roots of the quartic don't seem to be anything nice. I stopped here because it did not seem pointful to me to do the partial fractions with numerical approximations for the roots.
Although this is ultimately expressible in terms of elementary functions I strongly suspect that there was a typo in this problem.
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
bisector theorem , cosine rule or other method? A fixed point $A(a,0)$, where $a>0$ and a straight line $l$ is given $x = -1$, $B$ is a moving point on the second quadrant and lies on the straight line $l$, the angle bisector of $\angle BOA$ intersects $AB$ at point $C$.
(a) Find the locus of point $C$ and state the range values of $x$ and $y$.
(b) Discuss the relationship between the values of $a$ and the type of the curve of the equation obtained in (a).
I have tried to use cosine rule to and bisector theorem to obtained the locus. but the equation seem too complicated. Any other good method please share with me !!
| Let $B=(-1,t)$, $C=(x,y)$.
Then
$$\frac{y}{x-a}=\frac{t}{-1-a}$$
Angle AOB can be obtained to be
$$\cos AOB=\frac{-a}{a\sqrt{1+t^2}}=\frac{-1}{\sqrt{1+t^2}}$$
Using
$$\cos^2 AOC = \frac{1+\cos AOB}{2}$$
we have
$$\frac{y^2}{x^2}=\tan^2 AOC = \sec^2 AOC - 1=\frac{2}{1-\frac{1}{\sqrt{1+t^2}}}-1=\frac{2\sqrt{1+t^2}}{\sqrt{1+t^2}-1}-1=\frac{\sqrt{1+t^2}+1}{\sqrt{1+t^2}-1}=\frac{\left(\sqrt{1+t^2}+1\right)^2}{t^2}$$
$$\frac{y}{x}=\frac{\sqrt{1+t^2}+1}{t}$$
$$\left(t\frac{y}{x}-1\right)^2=1+t^2$$
$$\left(ty-x\right)^2=\left(1+t^2\right)x^2$$
$$\left(\frac{(1+a)y^2}{x-a}+x\right)^2=\left(1+\frac{(1+a)^2y^2}{(x-a)^2}\right)x^2$$
$$((1+a)y^2+x(x-a))^2=((x-a)^2+(1+a)^2y^2)x^2$$
$$(1+a)^2y^4+2(1+a)x(x-a)y^2=(1+a)^2x^2y^2$$
$$y^2+\frac{2}{1+a}x(x-a)-x^2=0$$
$$\left(\frac{2}{1+a}-1\right)x^2-\frac{2a}{1+a}x+y^2=0$$
$$(1-a)x^2-2ax+(1+a)y^2=0$$
If $0<a<1$, then it is an ellipse.
If $a=1$, then it is a parabola.
If $a>1$, then it is a hyperbola.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "4",
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using Parseval's identity to estimate the value of $\Sigma_{n=1}^\infty \frac{1}{(2n-1)^2(2n+1)^2}$ There is a problem with two parts; The first part is asking to find Fourier series for $f(x)=|\sin(x)|$ on $[-\pi,\pi]$. And the second part wants to estimate the following using Parseval's identity:
$$
\sum_{n=1}^\infty \frac{1}{(2n-1)^2(2n+1)^2}
$$
I'm not sure it is necessary to use the answer of the first part or not. I don't know what to do!
The first part answer:
$$\forall n, b_n=0$$.
$$
a_0=\frac{1}{\pi}\int_{-\pi}^\pi |\sin(x)|dx=\frac{2}{\pi}\int_0^\pi \sin(x)dx=\frac{-2}{\pi}\cos(x)]_0^\pi=\frac{-2}{\pi}(-1-1)=\frac{4}{\pi}
$$
\begin{align*}
a_n=\frac{1}{\pi}\int_{-\pi}^\pi |\sin(x)|\cos(nx) dx
= \frac{2}{\pi}\int_0^\pi \sin(x)\cos(nx)dx = \frac{1}{\pi}\int_0^\pi [\sin(x+nx)+\sin(x-nx)]\\ =\frac{1}{\pi}[\frac{-\cos((n+1)x)}{2(n+1)}+\frac{-\cos((1-n)x)}{2(1-n)}]_0^\pi \\
a_n=\frac{-1}{\pi}(\frac{\cos(n\pi)+1}{n^2-1})
\end{align*}
| I guess I did it!
I want to know the value of:
$$
\sum_{n=1}^\infty \frac{1}{(4n^2-1)^2}=\sum_{n=1}^\infty \frac{1}{((2n)^2-1)^2}
$$
So let $m=2n$, then $\sum_{n=1}^\infty \frac{1}{(m^2-1)^2}$.
From the first part and Parseval's identity,
$$
\frac{2}{\pi}\int_0^\pi \sin^2(x)dx=\frac{16}{2\pi^2}\sum_{m=1}^\infty \frac{1}{\pi}\frac{((-1)^m+1)^2}{(m^2-1)^2}
$$
since $m=2n$,
$$
\sum_{m=1}^\infty \frac{1}{(m^2-1)^2}=\frac{\pi^3}{32}
$$
Is this correct?
| {
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Find the value of $\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}$
Let $a$, $b$ and $c$ be real numbers such that $a + b + c = 0$ and define:
$$P=\frac{a^2}{2a^2+bc}+\frac{b^2}{2b^2+ca}+\frac{c^2}{2c^2+ab}.$$
What is the value of $P$?
This question came in the regional maths olympiad. I tried AM-GM and CS inequality but failed to get a result. Please give me some hint in how to solve this question.
| plugging $c=-a-b$ in your term, we get $${\frac {{a}^{2}}{2\,{a}^{2}+b \left( -a-b \right) }}+{\frac {{b}^{2}}{
\left( -a-b \right) a+2\,{b}^{2}}}+{\frac { \left( -a-b \right) ^{2}
}{ab+2\, \left( -a-b \right) ^{2}}}
$$
simplifying this we get $1$
| {
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Floor functions of powers increase What is the smallest real number $x$ such that $\lfloor x^n\rfloor<\lfloor x^{n+1}\rfloor$ for all positive integers $n$?
In particular, we have $\lfloor x\rfloor<\lfloor x^2\rfloor<\lfloor x^3\rfloor<\dots$. For the first inequality to hold it must be that $x\geq\sqrt{2}$. But if $x=\sqrt{2}$ then $\lfloor x^2\rfloor=\lfloor x^3\rfloor=2$, so we need $x\geq \sqrt[3]{3}\approx 1.44$. From the first few powers it looks like this is the answer, but how can we prove it?
| Let $x=\sqrt[3]{3}$. First, you've already directly verified that $\lfloor x\rfloor<\lfloor x^2\rfloor<\lfloor x^3\rfloor$, as they are equal $1$, $2$, and $3$, respectively. Now, for all $n\ge3$:
$$x^{n+1}-x^n=x^n(x-1)=\left(\sqrt[3]{3}\right)^n\left(\sqrt[3]{3}-1\right)>3\cdot0.4>1,$$
so the numbers grow by more than $1$ each time, guaranteeing that their integer parts strictly increase.
| {
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show that $7\mid p^3-p$ if $p$ is a prime divisor of $n^3+n^2-2n-1$
Let $p$ be a prime number, and $n$ a positive integer such
$$p\mid n^3+n^2-2n-1, \quad n\ge 2.$$
Show that $$7\mid p^3-p.$$
It maybe can use Fermat's little theorem?
| An alternative way is to consider the prime divisors of
$a^3+a^2b-2ab^2-b^3$, and reduce the numbers by applying Thue's lemma.
Call a prime $p$ "bad" if $p\not\equiv0,\pm1\pmod7$, and there exist some co-prime integers $a,b$ such that $p$ divides $a^3+a^2b-2ab^2-b^3$. Suppose that there at least one bad prime, and take the least bad prime and the corresponding $a$ and $b$.
By Thue's lemma, there are some integers $0<k<p,c,d$ such that $ak\equiv c\pmod{p}$, $bk\equiv d\pmod{p}$, $|c|<\sqrt{p}$, and $|d|<\sqrt{p}$. Then $p$ divides $c^3+c^2d-2cd^2-d^3$. If $c$ and $d$ are not co-prime, we can divide them by their gcd. (Notice that at least one of $c$ and $d$ is not divisible by $p$, so this simplification does not destroy the divisibility by $p$.)
It can be verified that $c^3+c^2d-2cd^2-d^3\equiv\pm1\pmod7$ or $c^3+c^2d-2cd^2-d^3\equiv\pm7\pmod{49}$. Any case happens, there must be another "bad" prime divisor $q$ of $c^3+c^2d-2cd^2-d^3$. But $p$ is the least bad prime, so $q\ge p$; therefore $p^2\le pq\le|c^3+c^2d-2cd^2-d^3|<3\sqrt{p}^3$, so $p<9$.
The primes $2,3,5$ can be excluded manually.
| {
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Finding third row of orthogonal matrix?
Find a $3\times3$ orthogonal matrix whose first two rows are $\Big[\frac{1}{3},\frac{2}{3},\frac{2}{3}\Big]$ and $\left[0,\frac{1}{\sqrt{2}},-\frac{1}{\sqrt{2}}\right]$.
I tried two approaches.
One, finding vector cross product of given two rows.
Second, assuming third row as $[x,y,z]$ and applying the property of orthogonal matrix.
In each approach I got a different solution, which is not correct.
Even I suspect there is an error in given problem.
Please help me in finding the third row.
Adding cross product calculation below.
\begin{vmatrix}
\bar{\imath}&\bar{\jmath}&\bar{k}\\\frac{1}{3}&\frac{2}{3}&\frac{2}{3}\\0&\frac{1}{\sqrt{2}}&-\frac{1}{\sqrt{2}}
\end{vmatrix}
\begin{align}
&=\bar{\imath}(0)-\bar{\jmath}\left(-\frac{1}{3\sqrt{2}}\right)+\bar{k}\left(\frac{1}{3\sqrt{2}}\right) \\[1ex]
&=\frac{1}{\sqrt{\frac{1}{18}+\frac{1}{18}}}\left(0,\frac{1}{3\sqrt{2}},\frac{1}{3\sqrt{2}}\right) \\[1ex]
&=\left(0,\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)
\end{align}
| One approach that would be to find the cross product.
$(\frac 13, \frac 23, \frac 23)\times (0, \frac 1{\sqrt 2},-\frac 1 {\sqrt2}) = (-\frac 4{3\sqrt 2},\frac {1}{3\sqrt2},\frac {1}{3\sqrt2})$
Another would be to say by inspection any vector that is orthogonal to the second vector must have the from $(a,b,b)$
Choose an arbitrary value for $b$ (3 is convenient) and solve for $a.$
$(\frac 13,\frac 23,\frac23)\cdot(a,3,3) = 0\\
\frac {a}{3} + 4 = 0\\
(-12,3,3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2385989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Inequality $(xy+z)^2+(yz+x)^2+(zx+y)^2\geq\sqrt{2}(x+y)(y+z)(z+x)$ Let $x,y,z\geq 0$. Prove that
$$(xy+z)^2+(yz+x)^2+(zx+y)^2\geq\sqrt{2}(x+y)(y+z)(z+x).$$
Expanding gives
$$\sum x^2y^2+\sum x^2+6xyz\geq \sqrt{2}\sum x^2y+2\sqrt{2}xyz.$$
If we use AM-GM on the left-hand side, we get
$$\frac{1}{2}(x^2y^2+x^2)\geq x^2y$$
so the left-hand side is at least $\sum x^2y+6xyz$, but this is not enough.
| Let $x+y+z=3u$, $xy+xz+yz=3v^2$ and $xyz=w^3$.
Hence, our inequality it's
$$\sum_{cyc}(x^2y^2+2xyz+x^2)\geq\sqrt2\sum_{cyc}\left(x^2y+x^2z+\frac{2}{3}xyz\right)$$ or
$$9v^4-6uw^3+6w^3+9u^2-6v^2\geq\sqrt2(9uv^2-w^3),$$
which is a linear inequality of $w^3$,
which says that it's enough to prove our inequality for an extremal value of $w^3$.
Now,we see that $x$, $y$ and $z$ are non-negative roots of the following equation.
$$(X-x)(X-y)(X-z)=0$$ or
$$X^3-3uX^2+3v^2X-w^3=0$$ or
$$X^3-3uX^2+3v^2X=w^3,$$
which says that the graph of $f(X)=X^3-3uX^2+3v^2X$ and the line $Y=w^3$
have three common points: $(x,f(x))$, $(y,f(y))$ and $(z,f(z))$.
Now, let $u$ and $v^2$ be constants and $w^3$ changes.
We see that $w^3$ gets a maximal value, when the line $Y=w^3$ will touch to the graph of $f$,
which happens for equality case of two variables.
Also, we see that $w^3$ gets a minimal value, when the line $Y=w^3$ will touch to the graph of $f$,
which happens for equality case of two variables, or when $w^3=0$.
Thus, it's enough to prove our inequality in the following cases.
*
*$w^3=0$.
Let $z=0$.
Hence, we need to prove here that
$$x^2y^2+x^2+y^2\geq\sqrt2(x+y)xy,$$
which is C-S and AM-GM:
$$x^2y^2+x^2+y^2=x^2y^2+\frac{1}{2}(1^2+1^2)(x^2+y^2)\geq$$
$$\geq x^2y^2+\frac{1}{2}(x+y)^2\geq2\sqrt{x^2y^2\cdot\frac{1}{2}(x+y)^2}=\sqrt2(x+y)xy;$$
*$y=z$.
We need to prove
$$2(xy+y)^2+(x+y^2)^2\geq2\sqrt2y(x+y)^2,$$
which is AM-GM and C-S:
$$2(xy+y)^2+(x+y^2)^2=2y^2(x+1)^2+(x+y^2)^2\geq$$
$$\geq2\sqrt{2}y(x+1)(x+y^2)\geq2\sqrt2y(x+y)^2.$$
Done!
| {
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"url": "https://math.stackexchange.com/questions/2386218",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating limits Evaluate the limit without L’Hôpital rule:
$$
\lim_{x \to 0}\frac{\sin^2{x}+2\ln\left(\cos{x}\right)}{x^4}
$$
My work is:
\begin{align}
L&=\lim_{x \to 0}\frac{\sin^2{x}-x^2}{x^4}+\lim_{x \to 0} \frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\lim_{x \to 0}\frac{\sin{x}-x}{x^3}
\lim_{x \to 0}\frac{\sin{x}+x}{x}+
\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\frac{-1}{6}\left[\lim_{x \to 0}\frac{\sin x}{x}+1\right]
+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=\frac{-1}{6}\left(2\right)+\lim \limits_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}\\
&=
\frac{-1}{3}+\lim_{x \to 0}\frac{x^2+2\ln\left(\cos{x}\right)}{x^4}
\end{align}
I could not evaluate the second limit
| With Taylor expansion
$$\sin (x)=x-\frac {x^3}{6}+x^4\epsilon (x) $$
$$\sin^2 (x)=x^2-\frac {x^4}{3}+x^5\epsilon (x) $$
$$\cos (x)-1=-\frac {x^2}{2}+\frac {x^4}{24}+x^5\epsilon (x) $$
$$\ln (X+1)=X-\frac {X^2}{2}+X^2\epsilon (X) $$
$$\ln (\cos (x))=\ln \Bigl(\cos (x)-1+1\Bigr) $$
$$=-\frac {x^2}{2}+\frac {x^4}{24}-\frac {x^4}{8}+x^5\epsilon (x) $$
Replacing that, you should find
$$L=-\frac {1}{3}-\frac {1}{4}+\frac {1}{12}=-\frac {1}{2} $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that the roots of the equation $x^2-4abx+(a^2+2b^2)^2=0$ are imaginary. Show that the roots of the equation $x^2-4abx+(a^2+2b^2)^2=0$ are imaginary.
My Attempt:
$$x^2-4abx+(a^2+2b^2)^2=0$$
Comparing above equation with $Ax^2+Bx+C=0$, we get
\begin{align}
A&=1 \\
B&=-4ab \\
C&=(a^2+2b^2)^2
\end{align}
Now,
\begin{align}
B^2-4AC
&=(-4ab)^2 - 4\cdot 1\cdot (a^2+2b^2)^2 \\
&=16a^2b^2 - 4(a^4+4a^2b^2+4b^4) \\
&=-4a^4-16b^4. \end{align}
| If $a=b=0$ and the root is $0$.
However, if $(a,b) \neq 0$, then
the discriminant is equal to $-4(a^4+4b^4)<0$, hence the roots are imaginary.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2387278",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solution of first order nonlinear differential equation without inverse function The following differential equation is part of a path integral problem in quantum mechanics:
$$(\partial_{r} y)^2 = y^4 (-1+p y -q^2 y^2)^{3/2}$$
with $p>0$ and $q>0$ and $p,q \in \mathbb{R}$
Additionally it is required, that
$$\lim_{r\to \infty} y(r) = 0$$
To solve the equation we transform to
$$\int y^{-2} (-1+p y -q^2 y^2)^{-3/4} dy = \pm (r+C_1)$$
With mathematica we can solve the integral to
$$\frac{1}{8 x \left(p x-q^2 x^2-1\right)^{3/4}} \left(\left(-\sqrt{2}\right) p x \left(\frac{\sqrt{p^2-4 q^2}-p+2 q^2 x}{q^2 x}\right)^{3/4} \left(-\frac{\sqrt{p^2-4 q^2}+p-2 q^2 x}{q^2 x}\right)^{3/4} F_1\left(\frac{3}{2};\frac{3}{4},\frac{3}{4};\frac{5}{2};\frac{p+\sqrt{p^2-4 q^2}}{2 q^2 x},\frac{p-\sqrt{p^2-4 q^2}}{2 q^2 x}\right)-4 \left(\sqrt[4]{2} x \left(\sqrt{p^2-4 q^2}-p+2 q^2 x\right) \left(\frac{-2 q^2 \left(x \sqrt{p^2-4 q^2}+2\right)+p \sqrt{p^2-4 q^2}+p^2}{p^2-4 q^2}\right)^{3/4} \, _2F_1\left(\frac{1}{4},\frac{3}{4};\frac{5}{4};\frac{2 x q^2-p+\sqrt{p^2-4 q^2}}{2 \sqrt{p^2-4 q^2}}\right)-2 p x+2 q^2 x^2+2\right)\right)$$
We can see here Hypergeometric functions. The first is an Appell function. It is obviously not possible to find ad hoc the inverse function to determine $y(r)$.
I have the assumption, that it is possible to find a transformation of the integral, that the solution can be used to evaluate $y(r)$
A second option could be, that the limit of $y(r)$ can be used to find an equation for the values $p$ and $q$, which simplifies the Hypergeometric functions to arithmetic expressions which are good to handle to find $y(r)$
Does anyone has a tipp for me?
| I have an example for a simplification:
$$\int y^{-2} (-1+p y -q^2 y^2)^{-3/4} dy = \int y^{-2} \left(1-\frac{p^2}{4 q^2}+\left(q y-\frac{p}{2 q}\right)^2\right)^{-{3/4}} dy $$
now we substitute
$x = q y-\frac{p}{2 q}$ with $dy = \frac{1}{q} dx$ which leads to
$$\int y^{-2} (-1+p y -q^2 y^2)^{-3/4} dy = q \int \left( x + \frac{p}{2 q} \right)^{-2} \left(1-\frac{p^2}{4 q^2}+x^2\right)^{-{3/4}} dx $$
Now the integral has the solution
$$-\frac{8 \sqrt{2} q \left(\frac{2 x-\sqrt{\frac{p^2}{q^2}-4}}{p q+2 x}\right)^{3/4} \left(\frac{\sqrt{\frac{p^2}{q^2}-4}+2 x}{p q+2 x}\right)^{3/4} F_1\left(\frac{5}{2};\frac{3}{4},\frac{3}{4};\frac{7}{2};\frac{p q-\sqrt{\frac{p^2}{q^2}-4}}{p q+2 x},\frac{p q+\sqrt{\frac{p^2}{q^2}-4}}{p q+2 x}\right)}{5 (p q+2 x) \left(-\frac{p^2}{q^2}+4 x^2+4\right)^{3/4}}$$
The simplification is, that it depends now only on one Appell hypergeometric function
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2388084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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The Sum of 5 Consecutive Integers is 505, what is the Third number in this Sequence? So I've never covered this and I'm a bit confused.
$$x + (x+1) + (x+2) + (x+3) + (x+4)$$
So $5x + 10 = 505$ which equals $x = 99$. So isn't the third integer $2$? Which would mean that it equals to $1 + 2 + (99) = 102$? Why is the correct answer $101$, only taking into account the third integer ignoring its sequence @_@?
| In overly pedantic detail, you might think of the problem this way:
We have five consecutive integers that add to 505. Let the five numbers be $a$, $b$, $c$, $d$, and $e$. Since they are consecutive integers, we may assume that they are ordered $a < b < c < d < e$, from which it follows that
\begin{align}
b &= a+1 \\
c &= b+1 = (a+1) + 1 = a+2 \\
d &= c+1 = (a+2) + 1 = a+3 \\
e &= d+1 = (a+3) + 1 = a+4.
\end{align}
We want to find the third integer in this sequence, which means that we are trying to find the value of $c$. We already know that $c = a+2$, so we can, instead, try to find $a$, then add 2 to that in order to get $c$. But the numbers add to 505, so we have
\begin{align}
505 &= a + b + c + d + e \\
&= a + (a+1) + (a+2) + (a+3) + (a+4) \\
&= 5a + 10.
\end{align}
This implies that
$$ 5a = 495 \implies a = 99. $$
But remember that we were trying to find $c$, not $a$, and that $c=a+2$. Therefore the third integer in the sequence is
$$ c = a+2 = 99+2 = 101.$$
It might be simpler to work this way: $c$ is the third integer in the sequence. Since we want to find $c$, we should write the other four integers in terms of $c$. As above
\begin{align}
d &= c+1 \\
e &= d+1 = (c+1) + 1 = c+2,
\end{align}
since $d$ and $e$ are the next two consecutive integers. To get $a$ and $b$, we have to count down, instead of up, which gives us
\begin{align}
b &= c-1 \\
a &= b-1 = (c-1)-1 = c-2.
\end{align}
Then we have
\begin{align}
505 &= a + b + c + d + e \\
&= (c-2) + (c-1) + c + (c+1) + (c+2) \\
&= 5c,
\end{align}
which implies that $c = 101$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the number of divisors of $N=2^7\times3^5\times5^3$ Find the number of divisors of $N=2^7\times3^5\times5^3$ which are of the form $4t+1$, where $t$ is a natural number.
I understood the fact, that $5$ can be written in the form of $4t+1$. Also, every even power of $3$ can be written in the form of $4t+1.$
for example:-
$4(2)+1=9$ which is an even power of $3$
$4(0)+1=1$ which is an even power of $3$: $3^0$
$4(20)+1=81$ which is also an even power of $3$
Why are we multiplying power of $5$ that is $4$ powers of $5$ ($5^0$, $5^1$, $5^2$, $5^3$) and three powers of 3($3^0$, $3^2$, $3^4$)
That is $4\times3=12$.
And then why are we subtracting $-1$ from $12$. ( Final answer is $11$)
Why not adding $4+3=7$
| We multiply instead of adding because we're counting the number of ways to mix and match elements from a set of $4$ with elements from a set of $3$. Watch:
$5^03^0, 5^03^2,5^03^4$
That's $3$.
$5^13^0, 5^13^2,5^13^4$
That's another $3$.
$5^23^0, 5^23^2,5^23^4$
Another $3$.
$5^33^0, 5^33^2,5^33^4$
Another $3$.
That's $3$, a total of $4$ times. Thus, $4$ times $3$.
Finally, we subtract $1$ because one of these answers, $5^03^0$ equals $1$, which is $4t+1$ with $t=0$. We don't count $0$ as a natural number, so we have to exclude that one.
| {
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"timestamp": "2023-03-29T00:00:00",
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Isosceles triangle inscribed in an ellipse.
Find the maximum area of an isosceles triangle, which is inscribed inside an ellipse $\dfrac {x^2} {a^2} + \dfrac {y^2} {b^2} = 1$ with its (unique) vertex lying at one of the ends of the major axis of the ellipse.
The three vertices of the triangle would be $(a,0), (x,y), (x, -y)$.
The area of the triangle by Heron's formula is $$A^2 = (x-a)^2y^2 = (x-a)^2b^2\left( 1-
\dfrac{x^2}{a^2}\right) \tag{1}.$$
Hence $$\dfrac{dA}{dx} = 0 \implies (x-a)^2 \left( x + \dfrac{a}2\right) = 0.$$
We have minimum at $x = a$ and maximum at $x = -\frac{a}{2}$.
Substituting back in $(1)$ and taking square roots on both the sides gives $$A = \dfrac{\sqrt{3}ab}{4}$$.
The given answer is $3A$.
What mistake I made ? Is it possible to find the answer to this question if the triangle was scalene ?
| Everything is correct until you use the value of $x$ to calculate the area.
You should have $$A^2=(-\frac{3a}{2})^2b^2(1-\frac 14)$$ which will give you the correct answer $$A=\frac{3\sqrt{3}}{4}ab$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the equation $\sqrt{x^3+1}\left(4x-1\right)=2x^3+x^2+1$ Solve the equation $$\sqrt{x^3+1}\left(4x-1\right)=2x^3+x^2+1$$
$$\Leftrightarrow \sqrt{x^3+1}=\frac{2x^3+x^2+1}{4x-1}$$
$$\Leftrightarrow \sqrt{x^3+1}-\left(x+1\right)=\frac{2x^3+x^2+1}{4x-1}-\left(x+1\right)$$
$$\Leftrightarrow \frac{x^3+1-\left(x+1\right)^2}{\sqrt{x^3+1}+x+1}=\frac{2x^3-3x^2-3x+2}{4x-1}$$
$$\Leftrightarrow \frac{x^3-x^2-2x}{\sqrt{x^3+1}+x+1}-\frac{2x^3-3x^2-3x+2}{4x-1}=0$$
$$\Leftrightarrow \frac{x\left(x-2\right)\left(x+1\right)}{\sqrt{x^3+1}+x+1}-\frac{\left(x+1\right)\left(x-2\right)\left(2x-1\right)}{4x-1}=0$$
$$\Leftrightarrow \left(x-2\right)\left(x+1\right)\left(\frac{x}{\sqrt{x^3+1}+x+1}-\frac{2x-1}{4x-1}\right)=0$$
The equation $$\frac{x}{\sqrt{x^3+1}+x+1}-\frac{2x-1}{4x-1}=0$$ have a solution but it's very ugly and i can't solve this. Help me, thanks
| after squaring rearranging and factorizing we get
$$-(x-2) x (x+1) \left(4 x^3-8 x^2+9 x-4\right)=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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If $x+y+z+w=29$ where x, y and z are real numbers greater than 2, then find the maximum possible value of $(x-1)(y+3)(z-1)(w-2)$
If $x+y+z+w=29$ where x, y and z are real numbers greater than 2, then find the maximum possible value of $(x-1)(y+3)(z-1)(w-2)$.
$(x-1)+(y+3)+(z-1)+(w-2)=x+y+z+w-1=28$
Now $x-1=y+3=z-1=w-2=7$ since product is maximum when numbers are equal
My answer came out to be as $6*10*6*5=1800$ but the answer is $2401$. What am I doing wrong? And also, how we will get the answer $2401$
| $x=8, y=4, z=8, w=9$ is the solution.
edit:
your solution is also correct. I don't know why are you multiplying $6∗10∗6∗5$?
$x−1=y+3=z−1=w−2=7$ means 7*7*7*7=2401
| {
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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.