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Coefficients of a polynomial function from $\Bbb{N}$ to $\Bbb{N}$ Let $f$ be a polynomial from $\Bbb{N}$ to $\Bbb{N}$.
So let $f(x) = a_0+a_1x+a_2x^2+..a_rx^r$
Then for every natural number $k$, $f(k) \in \Bbb{N}$.
But how does it imply that all the coefficients of $f(x)$ i.e. $a_0, a_1,..a_r$ will belong to $\Bbb{Q}$?
|
It results that
$\begin{cases}
a_0+a_1+\ldots+a_i+\ldots+a_r=f(1)\\
a_0+2a_1+\ldots+2^ia_i+\ldots+2^ra_r=f(2)\\
a_0+3a_1+\ldots+3^ia_i+\ldots+3^ra_r=f(3)\\
......................................\\
a_0+ra_1+\ldots+r^ia_i+\ldots+r^ra_r=f(r)\\
a_0+(r+1)a_1+\ldots+(r+1)^ia_i+\ldots+(r+1)^ra_r=f(r+1)
\end{cases}$
Since all the coefficients of the unknowns $a_i$, $0\le i\le r$, and all the constant terms $f(j)$, $1\le\ j\le r+1$, are natural numbers, then
$\Delta=
\begin{vmatrix}
1 & 1 & \cdots & 1 & \cdots & 1\\
1 & 2 & \cdots & 2^i & \cdots & 2^r\\
1 & 3 & \cdots & 3^i & \cdots & 3^r\\
\cdots & \cdots & \cdots & \cdots & \cdots &\cdots\\
1 & r & \cdots & r^i & \cdots & r^r\\
1 & (r+1) & \cdots & (r+1)^i & \cdots & (r+1)^r
\end{vmatrix}$
and
$\Delta a_i=
\begin{vmatrix}
1 & 1 & \cdots & f(1) & \cdots & 1\\
1 & 2 & \cdots & f(2) & \cdots & 2^r\\
1 & 3 & \cdots & f(3) & \cdots & 3^r\\
\cdots & \cdots & \cdots & \cdots & \cdots &\cdots\\
1 & r & \cdots & f(r) & \cdots & r^r\\
1 & (r+1) & \cdots & f(r+1) & \cdots & (r+1)^r
\end{vmatrix}$
are integer numbers and $\Delta\ne0$.
Moreover $$a_i=\frac{\Delta a_i}{\Delta}$$ are all rational numbers.
|
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"url": "https://math.stackexchange.com/questions/3780124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Evaluate $\lim_{h\to 0}\frac{1}{h^2}\begin{vmatrix}\tan x&\tan(x+h)&\tan(x+2h)\\\tan(x+2h)&\tan x&\tan(x+h)\\\tan(x+h)&\tan(x+2h)&\tan x\end{vmatrix}$ Evaluate
$$
\lim_{h\to 0}\frac{\Delta}{h^2}=\lim_{h\to 0}\frac{1}{h^2}\begin{vmatrix}
\tan x&\tan(x+h)&\tan(x+2h)\\
\tan(x+2h)&\tan x&\tan(x+h)\\
\tan(x+h)&\tan(x+2h)&\tan x
\end{vmatrix}
$$
Attempt
$$
\lim_{h\to 0}\frac{\Delta}{h^2}=\begin{vmatrix}
\lim_{h\to 0}\tan x&\lim_{h\to 0}\dfrac{\tan(x+h)-\tan x}{h}&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}\\
\lim_{h\to 0}\tan(x+2h)&\lim_{h\to 0}\dfrac{\tan x-\tan(x+2h)}{h}&\lim_{h\to 0}\dfrac{\tan(x+h)-\tan(x+2h)}{h}\\
\lim_{h\to 0}\tan(x+h)&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}&\lim_{h\to 0}\dfrac{\tan x-\tan(x+2h)}{h}
\end{vmatrix}\\
=\begin{vmatrix}
\lim_{h\to 0}\tan x&\lim_{h\to 0}\dfrac{\tan(x+h)-\tan x}{h}&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}\\
\lim_{h\to 0}\tan(x+2h)&-2.\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan x}{h}&-1.\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}\\
\lim_{h\to 0}\tan(x+h)&\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}&-2.\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan x}{2h}
\end{vmatrix}\\
$$
$$
\lim_{h\to 0}\dfrac{\tan(x+h)-\tan x}{h}=\frac{d}{dx}\tan x=\sec^2x\\
\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan x}{2h}=\frac{d}{dx}\tan x=\sec^2x\\
\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}=\frac{d}{dx}\tan(x+h)=\sec^2(x+h)
$$
But my reference gives the solution $9\tan x.\sec^4x$, I think by taking $\lim_{h\to 0}\dfrac{\tan(x+2h)-\tan(x+h)}{h}=\sec^2x$. Will that make a difference ?
It might be silly but could anyone clarify this confusion in my attempt ?
|
For
$$M=\left(
\begin{array}{ccc}
a & b & c \\
c & a & b \\
b & c & a
\end{array}
\right)\implies |M|=a^3+b^3+c^3-3 a b c$$ So, replacing
$$\Delta=\tan ^3(x)+\tan ^3(x+h)+\tan ^3(x+2 h)-3 \tan (x) \tan (x+h) \tan (x+2 h)$$ Now, composing Taylor series around $h=0$ gives
$$\Delta=9 \left(\tan ^5(x)+2 \tan ^3(x)+\tan (x)\right)h^2+O\left(h^3\right)=9\tan (x) \sec ^4(x)\,h^2+O\left(h^3\right)$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
toom-cook algorithm matrix G For this toom-cook algorithm at https://arxiv.org/pdf/1803.10986v1.pdf#page=6 , how do I get the value 4/2 in the matrix G ?
|
We have
$$G = \begin{bmatrix} x_0^0 N^0 & x_0^1 N^0 & x_0^2 N^0 \\ x_1^0 N^1 & x_1^1 N^1 & x_1^2 N^1 \\ x_2^0 N^2 & x_2^1 N^2 & x_2^2 N^2 \\ x_3^0 N^3 & x_3^1 N^3 & x_3^2 N^3 \end{bmatrix} = \begin{bmatrix} (1)( -\dfrac{1}{6}) & (1)(-\dfrac{1}{6}) & (1)(-\dfrac{1}{6})\\ (1)(\dfrac{1}{2}) & (2)(\dfrac{1}{2}))& (4)(\dfrac{1}{2})) \\ (1)(-\dfrac{1}{2}) & (3)(-\dfrac{1}{2}) & (9)(-\dfrac{1}{2}) \\ (1)(\dfrac{1}{6}) & (4)(\dfrac{1}{6}) & (16)(\dfrac{1}{6}) \end{bmatrix}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/3780963",
"timestamp": "2023-03-29T00:00:00",
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|
Calculate the minimum value of $\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right|$.
Given positives $a, b, c$ such that $abc = 1$, if possible, calculate the minimum value of $$\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right|$$
Without loss of generalisation, assume that $a \le b \le c$.
We have that $$\left|\frac{a^2 - bc}{b - c}\right| + \left|\frac{b^2 - ca}{c - a}\right| + \left|\frac{c^2 - ab}{a - b}\right| \ge \frac{c^2 - ba}{b - a} + \frac{a^2 - bc}{b - c}$$
$$ = \frac{(c + a)(a^2 + b^2 + c^2 - bc - ca - ab)}{(c - b)(b - a)} \ (1)$$
Let $c' = b - a, a' = c - b \iff c = a' + b, a = b - c'$, $(1)$ becomes $$\frac{(2b - c' + a')(c'^2 + c'a' + a'^2)}{c'a'}$$
and $(b - c')b(b + a') = b^3 - (c' - a')b^2 - c'a'b = 1$
$$\iff (2b - c' + a')b^2 = b^3 + c'a'b + 1 \iff 2b - c' + a' = \frac{b^3 + c'a'b + 1}{b^2}$$
Another idea I had was that $\left|\dfrac{a^2 - bc}{b - c}\right| + \left|\dfrac{b^2 - ca}{c - a}\right| + \left|\dfrac{c^2 - ab}{a - b}\right|$
$$ = \frac{1}{2}\sum_{\text{cyc}}\left(|c - a|\left|\frac{2(b^2 - ca)}{(c - a)^2}\right|\right) = \frac{1}{2}\sum_{\text{cyc}}\left(|c - a|\left|\frac{2b^2 - c^2 - a^2}{(c - a)^2} + 1\right|\right)$$
$$ = \frac{1}{2}\left[(c - b)\left(\left|\frac{2(b^2 - a^2)}{(b - c)^2} + \frac{c^2 - b^2}{(b - c)^2} - 1\right| + \left|\frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2} + 1\right|\right)\right.$$
$$\left. + (b - a)\left(\left|\frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2} + 1\right| + \left|\frac{b^2 - a^2}{(a - b)^2} + \frac{2(c^2 - b^2)}{(a - b)^2} + 1\right|\right)\right]$$
$$ \ge \frac{1}{2}\left[(c - b)\left(\frac{2(b^2 - a^2)}{(b - c)^2} + \frac{c^2 - b^2}{(b - c)^2} + \frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2}\right)\right.$$
$$\left. + (b - a)\left(\frac{b^2 - a^2}{(c - a)^2} - \frac{c^2 - b^2}{(c - a)^2} + \frac{b^2 - a^2}{(a - b)^2} + \frac{2(c^2 - b^2)}{(a - b)^2} + 2\right)\right]$$
$$ = \frac{1}{2}\left[(c^2 - b^2)\left(\frac{1}{c - b} - \frac{1}{c - a} + \frac{2}{b - a}\right) + (b^2 - a^2)\left(\frac{2}{c - b} + \frac{1}{c - a} + \frac{1}{b - a}\right)\right] + (b - a)$$
$$ = \frac{1}{2}\left(\frac{b^2 + c^2 - 2a^2}{c - b} + \frac{2b^2 - c^2 - a^2}{c - a} + \frac{2c^2 - a^2 - b^2}{b - a}\right) + (b - a)$$
There must have been something wrong, but that's all I have for now.
|
We have
$$
\frac{bc-a^2}{c-b}+\frac{|b^2-ca|}{c-a}+\frac{c^2-ab}{b-a}\geq \frac{bc-a^2}{c-b}-\frac{|b^2-ca|}{c-a}+\frac{c^2-ab}{b-a}
$$
with equality only if $b=1$ and $c=\frac{1}{a}$. Now insert $b$ and $c$ to obtain
$$
\frac{\frac{1}{a}-a^2}{\frac{1}{a}-1}+\frac{\frac{1}{a^2}-a}{1-a}=\frac{1-a^3}{1-a}+\frac{1-a^3}{a^2-a^3}
$$
Differentiation suggest a minimum at $a=1$ but we are not allowed to set $a=1$ however we can take the limit $a\to 1^-$. Do it and use L' Hosptial.
$$
\lim_{a\to 1^-}\frac{-3a^2}{-1}+\frac{-3a^2}{2a-3a^2}=6.
$$
So, minimum value 6 is obtained when we set $b=1$ and take limits $a\to 1^-$ and $c\to 1^+$ with $c=\frac{1}{a}$.
EDIT: Adding more stuff for clarity
There can only be equality in the inequality if $|b^2-ca|=-|b^2-ca|$ which means equality is when $b^2=ca$, i.e. both sides are zero. Multiply by $b$ and we have $b^3=cab=1$ which gives $b=1$.
Since $b=1$ we get $ac=1$. Choose $a=0.5$ and $c=2$ then we have $\frac{2-0.25}{2-0.5}+\frac{4-0.5}{1-0.5}=\frac{49}{6}>\frac{36}{6}=6$.
Clearly $a<1<c$ otherwise we get division by zero.
Next define $P(a)=\frac{1-a^3}{1-a}+\frac{1-a^3}{a^2-a^3}=a^2+a+2+\frac{1}{a}+\frac{1}{a^2}$ which has extremum at $a=1$. Since we cannot choose $a=1$ we instead take the limit $\lim_{a\to 1^-}P(a)=6$ which is OK since the function $P(a)$ is continuous for $0<a<1$. And we have shown that $P(0.5)>6$ so therefore, because of continuity, we do not need to take second derivative of $P(a)$.
What if $|b^2-ca|<-|b^2-ca|$? Well, it cannot be true since a positive number is larger than a negative number. We have now proven global infimum is 6.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Substitution goes wrong while evaluating $\int _0^1\frac{\ln \left(1+x^2\right)}{1+x^2}\:dx$ Im aware this integral has been evaluated here before, i started with
$$\int _0^{\infty}\frac{\ln \left(1+x^2\right)}{1+x^2}\:dx=2\int _0^1\frac{\ln \left(1+x^2\right)}{1+x^2}\:dx+2G$$
Solving the integral on the left is very easy, then solving for the other one gets its value but the main point of my question comes from the substitution $ x=\frac{1-t}{1+t}$
$$I=\int _0^1\frac{\ln \left(1+x^2\right)}{1+x^2}\:dx=\int _0^1\frac{\ln \left(1+t^2\right)}{1+t^2}\:dt+\ln \left(2\right)\int _0^1\frac{1}{1+t^2}\:dt-2\int _0^1\frac{\ln \left(1+t\right)}{1+t^2}\:dt$$
And $I$ cancels out, can anyone tell me whats happening here?
|
Letting $x=\tan \theta$ yields
$$
\begin{aligned}
\int_0^1 \frac{\ln \left(1+x^2\right)}{1+x^2} d x &=\int_0^{\frac{\pi}{4}} \ln \left(\sec ^2 \theta\right) d \theta \\
&=-2 \int_0^{\frac{\pi}{4}} \ln (\cos \theta) d \theta \\
&=-2\left(-\frac{\pi}{4} \ln 2+\frac{1}{2} G\right) \\
&=\frac{\pi}{2} \ln 2-G
\end{aligned}
$$
By my post, $$\int_0^1 \frac{\ln \left(1+x^2\right)}{1+x^2} d x =\frac{\pi}{2}\ln2-G$$
where $G$ is the Catalan’s constant.
|
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|
Evaluate :- $\frac{(2020^2 - 20100)(20100^2 - 100^2)(2000^2 + 20100)}{10(2010^6 - 10^6)}$
Evaluate :- $\frac{(2020^2 - 20100)(20100^2 - 100^2)(2000^2 + 20100)}{10(2010^6 - 10^6)}$
What I Tried :- I couldn't think of any ways to factorise this expression . The denominator can be written as $10(2016^3 - 10^3)(2016^3 + 10^3)$ , but I can't understand how it will help here . I have absolutely no idea how to factorise the numerator except that it can be $(2020^2 - 20100)(20100 - 100)(20100 + 100)(2000^2 + 20100)$, other than that I got no idea, and it seems to me the only way to get it is to open the brackets , which will contain a lot of calculations .
Wolfram Alpha gives the answer to be $10$ . But I am looking for some clever way so that this expression gets factorised and I can get my answer in less calculations .
Can anyone help?
|
Let $2010=x$ and $10=y$.
Thus, for our expression we obtain: $$\frac{((x+y)^2-xy)(x^2y^2-y^4)((x-y)^2+xy)}{y(x^6-y^6)}=$$
$$=\frac{(x^2+xy+y^2)y^2(x^2-y^2)(x^2-xy+y^2)}{y(x^6-y^6)}=$$
$$=\frac{y(x^2-y^2)(x^4+x^2y^2+y^4)}{x^6-y^6}=y=10.$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Minimizing $x^2+y^2+z^2$ subject to $xy -z + 1 = 0$ via Lagrange multipliers $$\begin{array}{ll} \text{minimize} & f(x,y,z) := x^2 + y^2 + z^2\\ \text{subject to} & g(x,y,z) := xy - z + 1 = 0\end{array}$$
I tried the Lagrange multipliers method and the system resulted from has no solution. So I posted it to see if the question is wrong by itself or I'm missing something.
So I made the Lagrangian equation $L(x,y,z,λ)=x^2 + y^2 + z^2 + λ(xy -z+1)$
and then
$θL/θx = 2x + λy =0$
$θL/θy = 2y + λx =0$
$θL/θz = 2z - λ =0$
$θL/θλ = xy -z +1 =0 $
The obvious solution for that system is x=0 , y=0 , z=1 and λ=2
But solving it in an online solver for nonlinear systems of equation the answer I get is that it's unsolvable.
So my question is: What I'm doing wrong
|
This can be solved in at least two methods.
First, let's solve without Lagrange, using convenient changes of variables. Let $u=x+y, v=xy$. This results in $u^2=x^2+y^2+2xy=x^2+y^2+2v$.
We now need to minimize $u^2-2v+z^2$ under the constraint $v-z+1=0$. We can rearrange this constraint to be of the form $z=1+v$ and therefore $z^2=1+2v+v^2$. Substituting this, we need to minimize $u^2+v^2$. The minimum of this is for $u=0, v=0$, which returns $x=0, y=0, z=1$.
Solving this using Lagrange:
$$L=x^2+y^2+z^2-\lambda(xy-z+1)=x^2+y^2+z^2-\lambda xy-\lambda z-\lambda$$
$$\frac{\partial L}{\partial z}=2z-\lambda\rightarrow\lambda=2z$$
$$\frac{\partial L}{\partial x}=2x-\lambda y=0\rightarrow x=\frac{\lambda y}{2}=yz$$
$$\frac{\partial L}{\partial y}=2y-\lambda x=0\rightarrow2y-2yz^2=0$$$$\rightarrow y=0, x=0\cup z=1,\lambda=2,x=y\cup z=-1,\lambda=-2,x=-y$$
We have three possible solutions to this. We will plug each into the equation for the constraint $xy-z+1=0$
If $x=0, y=0$, our constraint becomes $-z+1=0$, which has the solution $x=0, y=0, z=1$, with the value of $x^2+y^2+z^2=1$
If $z=1, x=y$, our constraint becomes $x^2-1+1=0$, which has the exact same solution
If $z=-1, x=-y$, cour constraint becomes $-y^2+1+1=0$, which has the solutions $x=\pm\sqrt{2}, y=\mp\sqrt{2}, z=-1$. The value here is $x^2+y^2+z^2=5$, which is not the minimum
|
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|
Find all values of $a$ for which the maximum value of $f(x)=\frac{ax-1}{x^4-x^2+1}$equals $1$. Find all values of $a$ for which the maximum value of $$f(x)=\frac{ax-1}{x^4-x^2+1}$$equals $1$.
I equated $f(x)$ equal to $1$ to obtain a polynomial $x^4-x^2-ax+2=0$. Now this must have at least one repeated root.But I could not get any further.
Is there a calculus way to do it
|
Assume $a>0$. Since the denominator is always positive, maximum value of $f$ must occur when $x>0$.
Hence for all positive $x$'s,
$$\frac{ax-1}{x^4-x^2+1} \le 1 \implies x^3-x+\frac{2}{x} \ge a$$
Rewrite it as
$$2+\left(2+x+\frac{2}{x}\right)(1-x)^2 \ge a$$
The equality holds for $x=1$ and $a=2$.
Similar argumentation works for the case $a<0$ that yields $a=-2$.
|
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|
What is the minimum value of $x+y$?
Suppose $x,y$ are positive real numbers that satisfy $$xy(x+2y)=2$$ What is the minimum value of $x+y$?
My Thoughts
I’ve attempted using arithmetic-geometric mean inequality and got:
$\frac{x+y+x+2y}{3} \geq \sqrt[3]{2}$
Therefore $2(x+y)+y \geq 3\sqrt[3]{2}$, then I got trapped.
Feels like I’m in the wrong way, I need a hint.
|
Let $k$ be a minimal value of $x+y$.
Thus, $$x+y\geq k$$ or
$$\frac{2(x+y)^3}{xy(x+2y)}\geq k^3$$ or for $x=ty$
$$\frac{2(t+1)^3}{t^2+2t}\geq k^3$$ and since $$\min_{t>0}\frac{2(t+1)^3}{t^2+2t}=3\sqrt3,$$ which occurs for $t=\sqrt3-1,$ we obtain that $k=\sqrt3$.
Can you get now a full solution?
By the way, we can get the last result without derivatives because
$$2(t+1)^3-3\sqrt3(t^2+2t)=(t-\sqrt3+1)^2(2t+2+\sqrt3)\geq0.$$
|
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|
Roots of the Cubic equation The Cubic formula:
$ax^3+bx^2+cx+d=0$
With under the following conditions:
*
*$a \neq 0 $
*$a,b,c,d \in \Bbb{R} $
We can derive the following formula as a the root of $x$:
$u= \sqrt[3] {{bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3}+\sqrt{({bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3})^2+({c\over 3a}-{b^2\over 9a^2})^3}} $
$v= \sqrt[3] {{bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3}-\sqrt{({bc\over 6a^2}-{d\over 2a}-{b^3\over 27a^3})^2+({c\over 3a}-{b^2\over 9a^2})^3}} $
$x_1= u + v-{b\over 3a}$
But the proof of derivation of the formula mentioned above was only limited to 1 root. In addition, that formula had no complex conjugates. However as we know, there must be 2 other roots which includes complex conjugates in their formulas.
In reference, Wikipedia: Cubic equation also says that there should be 2 other roots at maximum.
So, altogether the 3 roots are:
$i= \sqrt {-1} $
$ \omega = -{1\over 2} + {{\sqrt 3}i\over 2} $
*
*$x_1= u + v-{b\over 3a}$
*$x_2= {\omega }u + {\omega}^2v -{b\over 3a} $
*$x_3= {\omega }^2 u + {\omega}v -{b\over 3a} $
And unfortunately, I didn't find or know know the proof of any other 2 roots, i.e $x_2$ & $x_3$.
So would you please show me the proof of the other 2 roots of Cubic formula?
Note: Please, no synthetic division. I need proof by formula
|
Starting with the "depressed" cubic:
$y^3 + py + q = 0$
If you don't have a diminished cubic you can substitute $x = y - \frac {b}{3a}$ which will eliminate the $bx^2$ term from the original cubic.
Next we do a similar substitution:
$y = z-\frac {p}{3z}$
$(z - \frac {p}{3z})^3 + p(z-\frac {p}{3z}) + q = 0\\
z^3 - pz + \frac {p^2}{3z} - (\frac {p}{3z})^3 + pz-\frac {p^2}{3z} + q = 0\\
z^3 - (\frac {p}{3z})^3 + q = 0\\
z^6 + qz^3 - (\frac {p}{3})^3 = 0\\
z^3 = -\frac {q}{2} \pm \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}$
We can choose $z^3 = -\frac {q}{2} + \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}$
with $\frac {1}{z^3} = \frac {-\frac {q}{2} - \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}}{-(\frac {p}{3})^3}$
Or, $-\frac {q}{2} - \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3} = (-\frac {p}{3z})^3$
Note that $\omega^3 = 1$ has $3$ roots
$\omega_1 = -\frac 12 + \frac {\sqrt 3}{2} i\\
\omega_2 = -\frac 12 - \frac {\sqrt 3}{2} i = \omega_1^2 = \frac {1}{\omega_1}\\
\omega_3 = 1$
$z = \omega {\sqrt[3]{-\frac {q}{2} + \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}}}$
with $z$ equal to any of the values of $\omega$ above, and
$y = z - \frac {p}{3z} = \omega {\sqrt[3]{-\frac {q}{2} + \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}}} + \frac {1}{\omega}{\sqrt[3]{-\frac {q}{2} - \sqrt {(\frac {q}{2})^2 + (\frac {p}{3})^3}}}$
If $(\frac {q}{2})^2 + (\frac {p}{3})^3 >0 $ there is only one real root.
If $(\frac {q}{2})^2 + (\frac {p}{3})^3 <0 $ the complex terms ultimately cancel out and there are 3 real roots.
|
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|
Evaluate $\frac{1+3}{3}+\frac{1+3+5}{3^2}+\frac{1+3+5+7}{3^3}+\cdots$ It can be rewritten as
$$S = \frac{2^2}{3}+\frac{3^2}{3^2}+\frac{4^2}{3^3}+\cdots$$
When $k$ approaches infinity, the term $\frac{(k+1)^2}{3^k}$ approaches zero. But, i wonder if it can be used to determine the value of $S$. Any idea?
Note: By using a programming language, i found that the value of $S$ is $3.5$
|
$$\left(\frac13+\frac1{3^2}+\frac1{3^3}+\cdots \right)\left(1+3+\frac53+\frac7{3^2}+\cdots\right)$$
\begin{aligned}=\left(\frac12\right)\left(1+3+ \frac33+\frac3{3^2}+...\\
+\frac23+\frac2{3^2}+\cdots\\+\frac2{3^2}+\cdots\\+\cdots\right)\\
=\frac12\left(4+\frac32\\+1\\+\frac13\\+\cdots\right)\\
=\frac12\left(4+\frac32+\frac32\right)\end{aligned}
|
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|
Check the convergence of the series : Which test do we apply? Check the convergence of the following series:
*
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^2+1}{n^4+n}}$
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{n^{n^2}}{(n+1)^{n^2}}}$
For the first series do we use the comparison test? But with which sequence do we compare $\frac{n^2+1}{n^4+n}$ ? Do we maybe do the following? $$\frac{n^2+1}{n^4+n}<\frac{n^2+n^2}{n^4+n}<\frac{n^2+n^2}{n^4}=\frac{2n^2}{n^4}=\frac{2}{n^2}$$
As for the second series I thought that we could use the ratio test but I am not sure if that works because we get $$\frac{a_{n+1}}{a_n}=\frac{\frac{(n+1)^{(n+1)^2}}{(n+2)^{(n+1)^2}}}{\frac{n^{n^2}}{(n+1)^{n^2}}}=\frac{(n+1)^{(n+1)^2}(n+1)^{n^2}}{(n+2)^{(n+1)^2}n^{n^2}}=\frac{(n+1)^{(n+1)^2+n^2}}{(n+2)^{(n+1)^2}n^{n^2}}$$ So is it better to apply an other test here?
|
Also, note that the terms of the second sum decay exponentially: Using $\ln(1-x)=-x+O(x^2)$ for $x\to0$, $$\frac{n^{n^2}}{(n+1)^{n^2}}=\left(1-\frac1{n+1}\right)^{n^2}=\exp\left(n^2\ln\left(1-\frac1{n+1}\right)\right)=O(1)\exp\left(-n\right).$$
|
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|
What is the meaning of "due to the symmetry of the coefficients, if $x=r$ is a zero of $x^4+x^3+x^2+x+1$ then $x=\frac1r$ is also a zero" I was studying this answer about factoring $x^4+x^3+x^2+x+1$:
https://socratic.org/questions/how-do-you-factor-x-4-x-3-x-2-x-1
The author says: "A cleaner algebraic approach is to notice that due to the symmetry of the coefficients, if $x=r$ is a zero of $x^4+x^3+x^2+x+1$ then $x= {1\over r}$ is also a zero"
And eventually he writes $x^4+x^3+x^2+x+1=(x^2+ax+1)(x^2+bx+1)$
Question $1$: What is the meaning of symmetry of the coefficients?
Question $2$: Can we do the same approach for $x^4-x^3+x^2-x+1$? ( I ask because it is relevant to my other question: Problem with factoring $x^4-x^3+x^2-x+1$)
|
The list of coefficients of$$x^4+x^3+x^2+x+1$$is $(1,1,1,1,1)$, which is symmetric (if you reverse it, you will get the same list). In other words, it is a list of the type $(a,b,c,b,a)$. And if $r(\ne0)$ is a root of$$ax^4+bx^3+cx^2+bx+a,\tag1$$then$$ar^4+br^3+cr^2+br+a=0,$$and therefore$$a+\frac br+\frac c{r^2}+\frac b{r^3}+\frac a{r^4}=0$$too; in other words, $\frac1r$ is also a root of $(1)$. So, unless one of the roots is $\pm1$ (which are the only numbers equal to their own inverses), $(1)$ can be written as\begin{multline}a(x-r)\left(x-\frac1r\right)(x-r')\left(x-\frac1{r'}\right)=\\=a\left(x^2-\left(r+\frac1r\right)x+1\right)\left(x^2-\left(r'+\frac1{r'}\right)x+1\right).\end{multline}
In particular, $x^4-x^3+x^2-x+1$ can be written as$$(x^2+ax+1)(x^2+bx+1)=x^4+(a+b)x^3+(ab+2)x^2+(a+b)x+1.$$In order to find $a$ and $b$, solve the system$$\left\{\begin{array}{l}a+b=-1\\ab+2=1.\end{array}\right.$$
|
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|
How to prove this equality of the determinant of matrix?
Prove that
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
=(a^2-bc)(b^2-ca)(c^2-ab)\end{equation*}
My attempt:
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
=\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
a(b-a) & b(c-b) & c(a-c) \\
(b+a)(b-a) & (c+b)(c-b) & (a+c)(a-c)
\end{pmatrix}
\end{equation*}
But I think my direction is incorrect. Can anyone give me some hints or the solution of this question?
|
Hint:
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}= (ab)(bc)(ca)\det\begin{pmatrix}
ab^{-1} & bc^{-1} & ca^{-1}\\
1 & 1 & 1\\
ba^{-1}& cb^{-1} & ac^{-1}
\end{pmatrix}
\end{equation*}
\begin{equation*}
= (ab)(bc)(ca)\det\begin{pmatrix}
ab^{-1} -ca^{-1} & bc^{-1} -ca^{-1} & ca^{-1}\\
0 & 0 & 1\\
ba^{-1} -ac^{-1} & cb^{-1} -ac^{-1} & ac^{-1}
\end{pmatrix}
\end{equation*}
$$ = -(ab)(bc)(ca)\left[ (ab^{-1} -ca^{-1} ) (cb^{-1} -ac^{-1}) -( ba^{-1} -ac^{-1})(bc^{-1} -ca^{-1})\right] $$
$$ = -(ab)(bc)(ca) \mathbf{(a^2-bc)}\left[ (ab)^{-1}(cb^{-1} -ac^{-1}) + (ac)^{-1}(bc^{-1} -ca^{-1})\right] $$
|
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|
Two inequalities with parameters $a,b,c>0$ such that $ca+ab+bc+abc\leq 4$
Let $a,b,c>0$ be such that $bc+ca+ab+abc\leq 4$. Prove the following inequalities:
(a) $8(a^2+b^2+c^2)\geq 3(b+c)(c+a)(a+b)$, and
(b) $\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{a^2b}+\dfrac{2}{b^2c}+\dfrac{2}{c^2a}\geq 9$.
Prove also that the unique equality case for both inequalities is given by $a=b=c=1$.
Below are some probably useful or relevant results.
*
*https://artofproblemsolving.com/community/c6h1241430p6342224
*https://artofproblemsolving.com/community/c6h284290p1535893
*https://artofproblemsolving.com/community/c6h608971p3619202
*https://artofproblemsolving.com/community/c6h1804479p11995588
*If $ab+bc+ca+abc=4$, then $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c$
Techniques used in solving the inequalities in these links may prove useful in proving our inequalities.
Attempt. In the simplest case, $a=b=c=:t$, we have $t^3+3t^2-4\leq 0$, whence $0<t\leq 1$. Therefore, the inequalities (a) and (b) become
$$24t^2\geq 24t^3$$
and
$$\frac{3}{t^2}+\frac{6}{t^3}\geq 9\,,$$
which are obviously true. How to prove these inequalities in general?
|
The source of the problem is: https://math.stackexchange.com/questions/2825783/problems-regarding-inequality. Below are solutions for both parts, some were written by other users in the old thread above.
Part (a): Solution by Truth.
Let $a = \dfrac{2kx}{y+z}$, $b = \dfrac{2ky}{z+x}$, and $c = \dfrac{2kz}{x+y}$, where $x,y,z,k>0$. From the given condition, we get $k \leqslant 1$ and the required inequality is equivalent to $$96k^2\,\sum_\text{cyc}\, \frac{x^2}{(y+z)^2} \geqslant 72k^3 \prod_\text{cyc}\, \left(\frac{x}{y+z}+\frac{y}{z+x}\right)\,.$$
This is equivalent to $$4\,\sum_\text{cyc}\, \frac{x^2}{(y+z)^2} \geqslant 3k\,\prod_\text{cyc}\, \left(\frac{x}{y+z}+\frac{y}{z+x}\right)\,.$$
Because $ k \leqslant 1$, it suffices to show that $$4\,\sum_\text{cyc}\, \frac{x^2}{(y+z)^2} \geqslant 3\, \prod_\text{cyc}\, \left(\frac{x}{y+z}+\frac{y}{z+x}\right)\,,$$ However, the last inequality is equivalent to
$$\sum_\text{cyc}\, \frac{\big(2x^4+16xyz^2+6y^2z^2+(2x^2+9xz+9yz+6z^2)(x+y-z)^2\big)(x-y)^2}{(x+y)^2(y+z)^2(z+x)^2} \geqslant 0\,.$$
Part (b): Solution by Michael Rozenberg. (@Michael, if you want to use this proof in your own separate answer, then you can remove this solution from my answer here and add it into your own answer.)
By the AM-GM Inequality,
$$4\geq3\sqrt[3]{a^2b^2c^2}+abc\,,$$ which gives $$abc\leq1\,.$$
Thus, by the AM-GM Inequality again, we obtain
$$\sum_\text{cyc}\,\frac{1}{a^2}+2\,\sum_\text{cyc}\,\frac{1}{a^2b}\geq\frac{3}{\sqrt[3]{a^2b^2c^2}}+\frac{6}{abc}\geq9\,.$$
Part (b): Alternative Proof by Me.
Use the AM-GM Inequality with the constraint inequality, we have
$$4\geq bc+ca+ab+abc\geq 4\,\sqrt[4]{(bc)(ca)(ab)(abc)}=4\,(abc)^{\frac{3}{4}}\,,$$
whence
$$abc\leq 1\,.$$
Now, we have by the AM-GM Inequality that
$$\sum_\text{cyc}\,\frac{1}{a^2}+2\,\sum_\text{cyc}\,\frac{1}{a^2b}\geq 9\, \sqrt[9]{\left(\prod_\text{cyc}\,\frac{1}{a^2}\right)\,\left(\prod_\text{cyc}\,\frac{1}{a^2b}\right)^2}=\frac{9}{(abc)^{\frac{8}{9}}}\,.$$
As $abc\leq 1$, the required inequality follows immediately.
|
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|
Proving $\int_{0}^{1} \frac{\tanh^{-1}\sqrt{x(1-x)}}{\sqrt{x(1-x)}}dx=\frac{1}{3}(8C-\pi\ln(2+\sqrt{3}))$ for an identity of Srinivasa Ramanujan Ramanujan is supposed to have given more than five thousand elegant results, a good number of them are yet to be proved or disproved.
Yesterday in the comment section of
Proving that $ \sum_{k=0}^\infty\frac1{2k+1}{2k \choose k}^{-1}=\frac {2\pi}{3\sqrt{3}} $
A wonderful Ramanujan identity
$$S=\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}{2k \choose k}^{-1}=\frac{1}{3}(8C-\pi\ln(2+\sqrt{3}))~~~~(1)$$ was showcased, Mathematica also gives this out.
My Attempt to prove (1) by hand:
Note the integral representation of the reciprocal of the binomial co-efficient:
$${n \choose j}^{-1}=(n+1)\int_{0}^{1} x^j (1-x)^{n-j}~ dx~~~~(2)$$
$$S=\sum_{k=0}^{\infty} \frac{1}{(2k+1)^2}{2k \choose k}^{-1}= \int_{0}^{1} \sum_{k=0}^{\infty} \frac{[x(1-x)]^{k}}{(2k+1)} dx= \int_{0}^{1} \frac{\tanh^{-1}\sqrt{x(1-x)}}{\sqrt{x(1-x)}} dx~~~~(3)$$
The question is: How to get this integral (3) by hand ?
|
You can give Feynman's trick a shot.
\begin{align*}
I&=\int _0^1\frac{\operatorname{arctanh} \left(\sqrt{x\left(1-x\right)}\right)}{\sqrt{x\left(1-x\right)}}\:dx\\[3mm]
I\left(a\right)&=\int _0^1\frac{\operatorname{arctanh} \left(a\sqrt{x\left(1-x\right)}\right)}{\sqrt{x\left(1-x\right)}}\:dx\\[3mm]
I'\left(a\right)&=\int _0^1\frac{1}{1-a^2x\left(1-x\right)}\:dx=\frac{4}{a\sqrt{4-a^2}}\arctan \left(\frac{a}{\sqrt{4-a^2}}\right)\\[3mm]
\int _0^1I'\left(a\right)da&=4\underbrace{\int _0^1\frac{1}{a\sqrt{4-a^2}}\arctan \left(\frac{a}{\sqrt{4-a^2}}\right)\:da}_{t=\frac{a}{\sqrt{4-a^2}}}\\[3mm]
I&=8\underbrace{\int _0^{\frac{1}{\sqrt{3}}}\frac{\arctan \left(t\right)}{4t\sqrt{1+t^2}}\:dt}_{t=\tan\left(x\right)}=2\int _0^{\frac{\pi }{6}}\frac{x\sec \left(x\right)}{\tan \left(x\right)}\:dx\\[3mm]
&=2\int _0^{\frac{\pi }{6}}\frac{x}{\sin \left(x\right)}\:dx
\end{align*}
That integral has been evaluated here by Zacky, using its result yields
$$\boxed{I=\int _0^1\frac{\operatorname{arctanh} \left(\sqrt{x\left(1-x\right)}\right)}{\sqrt{x\left(1-x\right)}}\:dx=\frac{\pi}{3}\ln(2-\sqrt 3) +\frac{8}{3}G}$$
One can also find the last integral by using the Weierstrass substitution.
\begin{align*}
2\int _0^{\frac{\pi }{6}}\frac{x}{\sin \left(x\right)}\:dx&=4\underbrace{\int _0^{2-\sqrt{3}}\frac{\arctan \left(t\right)}{t}\:\:dt}_{\operatorname{IBP}}\\[3mm]
&=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)-4\underbrace{\int _0^{2-\sqrt{3}}\frac{\ln \left(t\right)}{1+t^2}\:dt}_{t=\tan\left(x\right)}\\[2mm]
&=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)-4\int _0^{\frac{\pi }{12}}\ln \left(\tan \left(x\right)\right)\:dx\\[3mm]
&=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)+8\sum _{k=1}^{\infty }\frac{1}{2k-1}\int _0^{\frac{\pi }{12}}\cos \left(2\left(2k-1\right)x\right)\:dx\\[3mm]
&=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)+4\sum _{k=1}^{\infty }\frac{\sin \left(\frac{\pi }{6}\left(2k-1\right)\right)}{\left(2k-1\right)^2}\\[3mm]
&=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)+\frac{8}{3}\sum _{k=1}^{\infty }\frac{\left(-1\right)^{k+1}}{\left(2k-1\right)^2}\\[3mm]
&=\frac{\pi }{3}\ln \left(2-\sqrt{3}\right)+\frac{8}{3}G
\end{align*}
|
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|
If $f(x)=\sin^{-1} (\frac{2x}{1+x^2})+\tan^{-1} (\frac{2x}{1-x^2})$, then find $f(-10)$ Let $x=\tan y$, then
$$
\begin{align*}\sin^{-1} (\sin 2y )+\tan^{-1} \tan 2y
&=4y\\
&=4\tan^{-1} (-10)\\\end{align*}$$
Given answer is $0$
What’s wrong here?
|
From showing $\arctan(\frac{2}{3}) = \frac{1}{2} \arctan(\frac{12}{5})$
$$2\arctan x=\begin{cases}\arctan\dfrac{2x}{1-x^2}\text{ if } x^2\le1\\\pi+\arctan\dfrac{2x}{1-x^2}\text{ if } x^2>1\end{cases}$$
Again, from Proof for the formula of sum of arcsine functions $ \arcsin x + \arcsin y $
$$\sin^{-1}(\sin2y)=\begin{cases}2y\text{ if } -\dfrac\pi2\le2y\le\dfrac\pi2\\\pi-2y\text{ if } 2y>\dfrac\pi2\iff \tan y>1\\ -\pi-2y\text{ if } 2y<-\dfrac\pi2\iff \tan y<-1\end{cases}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the smallest possible value of an equation, where $a+b+c=3$ We have the positive real numbers $a, b, c$ such that $a+b+c=3$. Find the minimum value of the equation:
$$A=\frac{2-a^3}{a}+\frac{2-b^3}{b}+\frac{2-c^3}{c}$$
I solved it in the following fashion:
$$
\begin{align}
A&=2\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-(a^2+b^2+c^2)\\
&\ge 2\cdot\frac{3^2}{a+b+c}-(a^2+b^2+c^2) \quad \textrm{(Andreescu inequality)}\\
&=2\cdot3-(a^2+b^2+c^2)\\
&=6-(a^2+b^2+c^2).
\end{align}$$
However, $9=(a+b+c)^2\ge 3(ab+bc+ac)$, so
$$3\ge ab+bc+ac.$$
From this we have that $A \ge 6-3=3$, where equality is true for $a=b=c=1$.
Due to the total simplicity of my solution, I have difficulties believing that it is correct, despite having checked it thoroughly many times. Could you please tell me if my solution is correct and suggest some alternative solutions?
|
Your solution is wrong because after your first step we need to prove that:
$$6-(a^2+b^2+c^2)\geq3,$$ which is wrong for $a=2$.
We'll prove that $3$ is a minimal value.
Indeed, we need to prove that$$\frac{2(ab+ac+bc)}{abc}-(a^2+b^2+c^2)\geq3$$ or
$$\frac{2(ab+ac+bc)(a+b+c)}{3abc}-\frac{9(a^2+b^2+c^2)}{(a+b+c)^2}\geq3$$ or
$$\sum_{cyc}(2a^4b+2a^4c+6a^3b^2+6a^3c^2-22a^3bc+6a^2b^2c)\geq0$$ or
$$\sum_{cyc}(2a^4b+2a^4c-2a^3b^2-2a^3c^2+8a^3b^2+8a^3c^2-16a^3bc-6a^3bc+6a^2b^2c)\geq0$$ or
$$2\sum_{cyc}(a^4b-a^3b^2-a^2b^3+ab^4)+8\sum_{cyc}c^3(a^2-2ab+b^2)-3abc\sum_{cyc}(a^2-2ab+b^2)\geq0$$ or
$$\sum_{cyc}(a-b)^2(2ab(a+b)+8c^3-3abc)\geq0,$$ which is true by AM-GM:
$$2ab(a+b)+8c^3\geq3\sqrt[3]{(ab(a+b))^2\cdot8c^3}\geq3\sqrt[3]{32}abc>3abc.$$
The equality occurs for $a=b=c=1,$ which says that we got a minimal value.
|
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|
Median of triangle and tangents Let $AD$ be an altitude of any triangle $ABC$. Consider a circle with $AD$ as its diameter, cutting $AB$ and $AC$ at $P$ and $Q$ respectively. Let the tangents at $P$ and $Q$ meet at $X$. Prove that $AX$ bisects $BC$.
I have attempted the proof, using angles in the alternate segment, and Apollonius’ Theorem, but to no avail. Are there some other theorems that I should be using?
|
Pure coordinate proof:
$D:=(0,0),\,A:=(0,a),\,B:=(b,0),\,C:=(c,0),$
$O:=\left(0,\frac{a}{2}\right),\,P:=pB+(1-p)A,\,
Q:=qC+(1-q)A,\,X:=(x,y)$
$$\begin{cases}
OP^2=\frac{a^2}{4}\\
OQ^2=\frac{a^2}{4}\\
\overrightarrow{OP}\cdot\overrightarrow{XP}=0\\
\overrightarrow{OQ}\cdot\overrightarrow{XQ}=0\\
\end{cases}$$
$$\begin{cases}
\left(pb,\left(\frac12-p\right)a\right)^2=\frac{a^2}{4}\\
\left(qc,\left(\frac12-q\right)a\right)^2=\frac{a^2}{4}\\
\left(pb,\left(\frac12-p\right)a\right)\cdot
\left(pb-x,\left(1-p\right)a-y\right)=0\\
\left(qc,\left(\frac12-q\right)a\right)\cdot
\left(qc-x,\left(1-q\right)a-y\right)=0\\
\end{cases}$$
$$\begin{cases}
p^2b^2+\frac{a^2}{4}-pa^2+p^2a^2=\frac{a^2}{4}\\
q^2c^2+\frac{a^2}{4}-qa^2+q^2a^2=\frac{a^2}{4}\\
p^2b^2-pbx+\frac{a^2}{2}-pa^2-\frac{pa^2}{2}+p^2a^2
-\frac{ay}{2}+pay=0\\
q^2c^2-qcx+\frac{a^2}{2}-qa^2-\frac{qa^2}{2}+q^2a^2
-\frac{ay}{2}+qay=0\\
\end{cases}$$
$$\begin{cases}
pb^2-a^2+pa^2=0\hbox{, as }p\ne 0\ (p=0\hbox{ gives }P=A),\\
qc^2-a^2+qa^2=0\hbox{, as }q\ne 0\ (q=0\hbox{ gives }Q=A),\\
-2pbx+(-a+2pa)y=-2p^2b^2-a^2+3pa^2-2p^2a^2\\
-2qcx+(-a+2qa)y=-2q^2c^2-a^2+3qa^2-2q^2a^2\\
\end{cases}$$
$$\begin{cases}
p=\frac{a^2}{a^2+b^2}\\
q=\frac{a^2}{a^2+c^2}\\
-2pbx+(-a+2pa)y=-2p^2b^2-a^2+3pa^2-2p^2a^2\\
-2qcx+(-a+2qa)y=-2q^2c^2-a^2+3qa^2-2q^2a^2\\
\end{cases}$$
From the 3ed and the 4th equations we find $x,y$ by elimination or by Cramer's rule, it appears rather cumbersome, so I'll jump to the result:
$$x = \frac{a^2(b + c)}{2(a^2 + bc)},\, y = \frac{abc}{a^2 + bc},$$
so it's only left to show that $\overrightarrow{AX}=t\left(\frac{B+C}{2}-A\right)$ for some $t$, and indeed after comparing expressions for $\overrightarrow{AX}=X-A$ and $\left(\frac{b+c}{2},-a\right)$ we find $t=\frac{a^2}{a^2 + bc}$, QED.
|
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"timestamp": "2023-03-29T00:00:00",
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|
How many even three-digit numbers have distinct digits and have no digit $5$? How many even three-digit numbers have distinct digits and have no digit $5$?
The answer my teacher gave was $252$, but I don't see how she got that. I thought it would be $6\times 8 \times 5=240$ because the $3\text{rd}$ digit must be even $(0,2,4,6,8)$, the $2\text{nd}$ digit can't be $5$ or the $3\text{rd}$ digit $(10-2=8$ options) and the first digit can't be $0$, $5$, the $2\text{nd}$ digit, or the $3\text{rd}$ digit ($10-4=6$ options). Any way I look at it, the last digit has to be even and there are $5$ options for even digits, so the final answer must end in a $5$ or a $0$, not a $2$. Please help!
|
So, one is asking for the number of injections $f:\{1,2,3\} \to \{0,1,2,3,4,6,7,8,9\}$ where $f(1) \neq 0$ and $f(3) \in \{0,2,4,6,8\}$.
Ignoring the last two conditions, one gets $_{9}P_{3}=504$ possible numbers (possibly odd and/or having a leading zero).
If one requires the first digit to be nonzero, then one would then need to subtract the number of injections $\{2,3\} \to \{1,2,3,4,6,7,8,9\}$ to get $504-{_{8}P_{2}}=504-56=448$.
Odd numbers still have not been eliminated yet. We have to eliminate the cases where the last digit is $1, 3, 7,$ or $9$.
Suppose that the last two digits ($f(2)$ and $f(3)$) are $0$ and $1$ respectively. Then, one must have $f(1) \in \{2,3,4,6,7,8,9\}$, eliminating $7$ odd numbers to reduce the count to $441$.
Similar considerations apply when the middle digit is still $0$, but the last digit is now $3, 7,$ or $9$. This eliminates $3 \cdot 7=21$ more odd numbers, reducing the count to $420$.
Now, let's move on to the case where the middle digit is nonzero. Fixing a last digit $d \in \{1,3,7,9\}$, one would then need to subtract the number of injections $\{1,2\} \to \{1,2,3,4,6,7,8,9\} \setminus \{d\}$.
Suppose that $d=1$. Then, subtracting the number of injections $\{1,2\} \to \{2,3,4,6,7,8,9\}$ gives $420-{_{7}P_{2}}=420-42=378$. Forty-two odd numbers with the last digit equal to $1$ have now been eliminated.
Similar considerations apply when $d$ is $3,7,$ or $9$. This eliminates $3 \cdot 42=126$ more odd numbers, giving a final count of $252$.
|
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|
Evaluating a limit without usage of Riemann sum
Evaluation of $$\lim_{n \rightarrow \infty}\bigg[\frac{1}{n}+\frac{1}{n+2}+\frac{1}{n+4}+\cdots \cdots +\frac{1}{3n}\bigg]$$
My work: Using Riemann sum
$$\lim_{n\rightarrow \infty}\sum^{n}_{r=0}\frac{1}{n+2r}=\lim_{n\rightarrow\infty}\sum^{n}_{r=0}\frac{1}{1+2\frac{r}{n}}\cdot \frac{1}{n}$$
Put $\displaystyle \frac{r}{n}=x$ and $\displaystyle \frac{1}{n}=dx$ and changing limits
$$\int^{1}_{0}\frac{1}{1+2x}dx=\frac{1}{2}\ln|1+2x|\bigg|^{1}_{0}=\frac{1}{2}\ln(3)$$
But would it be possible to do the problem without usage of the Rienmann sum? as in, could one solve such infinite sums in a method alternate to converting the sum into an integral.
|
For even $n=2m$ we have
\begin{align}
\sum_{r=0}^{2m}\frac{1}{2m+2r}
&=\frac{1}{2}\sum_{r=0}^{2m}\frac{1}{m+r}=\\
&=\frac{1}{2}\left(\sum_{r=1}^{3m}\frac{1}{r}-\sum_{r=1}^{m-1}\frac{1}{r}\right)=\frac{1}{2}(H_{3m}-H_{m-1}),
\end{align}
where
$$
H_n=\sum_{r=1}^n\frac{1}{r}
$$
are the Harmonic numbers.
Given the known relation
$$
\lim_{n\to\infty}(H_n-\log n)=\gamma
$$
we have
\begin{align}
&\lim_{m\to\infty}\sum_{r=0}^{2m}\frac{1}{2m+2r}
=\frac{1}{2}\lim_{m\to\infty}(H_{3m}-H_{m-1})=\\
&\qquad=\frac{1}{2}\lim_{m\to\infty}[(H_{3m}-\log(3m))+\log(3m)-(H_{m-1}-\log(m-1))-\log(m-1)]=\\
&\qquad=\frac{1}{2}\lim_{m\to\infty}[\gamma+\log(3m)-\gamma-\log(m-1)]=\\
&\qquad=\frac{1}{2}\lim_{m\to\infty}\log\left(\frac{3m}{m-1}\right)=\frac{1}{2}\log 3.
\end{align}
For odd $n=2m+1$, taking into account
\begin{align}
&\frac{1}{n}+\frac{1}{n+2}+\ldots+\frac{1}{3n-2}+\frac{1}{3n}=\\
&=\left(\frac{1}{n}+\frac{1}{n+1}+\ldots+\frac{1}{3n-1}+\frac{1}{3n}\right)-\left(\frac{1}{n+1}+\frac{1}{n+3}+\ldots+\frac{1}{3n-3}+\frac{1}{3n-1}\right)
\end{align}
we can write
\begin{align}
\sum_{r=0}^{2m+1}\frac{1}{2m+1+2r}
&= \sum_{s=0}^{4m+2}\frac{1}{2m+1+s}-\sum_{r=0}^{2m}\frac{1}{2m+2+2r}=\\
&= \sum_{s=0}^{4m+2}\frac{1}{2m+1+s}-\frac{1}{2}\sum_{r=0}^{2m}\frac{1}{m+1+r}=\\
&= H_{6m+3}-H_{2m}-\frac{1}{2}[H_{3m+1}-H_{m}]
\end{align}
and
\begin{align}
\lim_{m\to\infty}\sum_{r=0}^{2m+1}\frac{1}{2m+1+2r}
&= \lim_{m\to\infty}\left(H_{6m+3}-H_{2m}-\frac{1}{2}[H_{3m+1}-H_{m}]\right)=\\
&= \lim_{m\to\infty}\left(\log(6m+3)-\log(2m)-\frac{1}{2}[\log(3m+1)-\log(m)]\right)=\\
&= \lim_{m\to\infty}\left(\log\left(\frac{6m+3}{2m}\right)-\frac{1}{2}\log\left(\frac{3m+1}{m}\right)\right)=\frac{1}{2}\log 3
\end{align}
Alternative proof
Let's rewrite the sum as
$$
\frac{1}{2}\sum_{r=0}^n\frac{1}{\frac{n}{2}+r}=\frac{1}{2}\left[\psi\left(\frac{3n+2}{2}\right)-\psi\left(\frac{n}{2}\right)\right],
$$
where $\psi$ is the digamma function and where we used the difference equation
$$
\psi(x+N)-\psi(x)=\sum_{k=0}^{N-1}\frac{1}{x+k},
$$
see Digamma::Recurrence formula and characterization.
Now, taking into account the following inequality, valid for $x>0$
$$
\log x-\frac{1}{x}\leq\psi(x)\leq\log x-\frac{1}{2x},
$$
see Digamma::Inequalities, we have
$$
\log\left(\frac{3n+2}{n}\right)-\frac{2}{3n+2}+\frac{1}{n}\leq\psi\left(\frac{3n+2}{2}\right)-\psi\left(\frac{n}{2}\right)\leq \log\left(\frac{3n+2}{n}\right)-\frac{1}{3n+2}+\frac{2}{n}
$$
and by the squeeze theorem, we get the result.
|
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|
Evaluate $\int \frac{11xe^{2x}}{(1+2x)^2}dx$ Evaluate $\int \frac{11xe^{2x}}{(1+2x)^2}dx$
Can somebody check my solution? Thanks!!
Let $u=11xe^{2x}$
then $\frac{du}{dx} = \frac{d}{dx}(11xe^{2x})$
$=11(e^{2x}+2xe^{2x})$
and so $du=11e^{2x}(2x+1)dx$
Now, let $dv=\frac{1}{(2x+1)^2}dx$
Then $\int dv = \int \frac{1}{(2x+1)^2}dx$
To evaluate this, let $r=2x+1$ and so $dr = 2dx$
Then we have $\int dv = \int \frac{1}{r^2}\frac{dr}{2} = \frac{-1}{2r}=\frac{-1}{2(2x+1)}$
So $v=\frac{-1}{2(2x+1)}$
Wow... So that was already a lot of work to find $du$ and $v$ once we chose $u$ and $dv$... Integration by parts totally sucks!!
Anyway, we are now ready to use our VOODOO formula!!
$\int \frac{11xe^{2x}}{(1+2x)^2}dx = \int udv = uv - \int vdu$
$=(11xe^{2x})(\frac{-1}{2(2x+1)}) - \int \frac{-1}{2(2x+1)}(11e^{2x}(2x+1))dx$
Now, there is a $2x+1$ in the denominator and the numerator so the cancel. We also pull the coefficient $\frac{-11}{2}$ outside of the integral, which we can do.
$=\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{2}\int e^{2x}dx$
$=\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{2} \frac{e^{2x}}{2}+C$
$=\frac{-11xe^{2x}}{2(2x+1)} + \frac{11}{4} e^{2x}+C$
|
When you have a short time to compute antiderivatives looking like
$$I=\int \frac {e^{k x}\,P_n(x)} {[Q_m(x)]^r} \,dx$$ think that a possible solution could be of the form
$$\frac {e^{k x}\,R_p(x)} {[Q_m(x)]^{r-1}}$$ where $P,Q,R$ are polynomials of different degrees.
Differentiate both sides to get
$$\frac {e^{k x}} {[Q_m(x)]^r}\,P_n(x)=\frac {e^{k x}} {[Q_m(x)]^r}\left(Q_m(x) \left(k R_p(x)+R_p'(x)\right)+(1-r) R_p(x) Q_m'(x) \right)$$ Comparing the degrees, we then have
$$n=m+p+p(m-1)\implies p=\frac n m-1$$ So, in $n$ is a muliple of $m$, big hope !
Tring for your case $n=1$, $m-1$, then $p=0$ that is to say that $R_p(x)$ is just a constant $R_p(x)=a$. Ralacing all numbers, we then have
$$11x=(2x+1) (2 a+0)+(1-2)a 2=4a x\implies a=\frac {11}4$$
$$\int \frac{11xe^{2x}}{(1+2x)^2}dx=\frac {11 e^{2x}}{4(1+2x)}$$
As you can notice, I did not use any integration step to get the final result.
|
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|
$\cos\left(\frac{\pi}{5}\right)$ using De Moivre's Theorem This is an Exercise 3.2.5 from Beardon's Algebra and Geometry:
Show that $\cos\left(\frac{\pi}{5}\right)=\frac{\lambda}{2}$, where $\lambda$ = $\frac{1+\sqrt{5}}{2}$ (the Golden Ratio).
[Hint: As $\cos(5\theta) = 1$, where $\theta = \frac{2\pi}{5}$, we see from De Moivre's theorem that $P(\cos\theta) = 0$ for some polynomial $P$ of degree five. Now observe that $P(z)=(1-z)Q(z)^2$ for some quadratic polynomial $Q$.]
So, by De Moivre's theorem:
$$\left(\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)\right)^5=\cos(2\pi)+i\sin(2\pi)=\cos(2\pi)=1$$
And so:
$$\left(\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)\right)^5-1=0$$
Thus, $\left(\cos\left(\frac{2\pi}{5}\right)+i\sin\left(\frac{2\pi}{5}\right)\right)^5-1$ is our polynomial $P$ of degree five. But how can I get from here to $Q$ and from $Q$ to $\cos\left(\frac{\pi}{5}\right)$?
|
Start with something simpler, an expression for $e^{i\frac{\pi}5}$:
$$\left(\cos\frac{\pi}5+i\sin\frac{\pi}5\right)^5=-1$$
Write explicitly the terms of the expansion. We know that we can ignore therms in even powers of $i$. Using $z=\cos\frac{\pi}5$ and $\sin^2\frac{\pi}5=1-z^2$ one gets:
$$z^5-10z^3(1-z^2)+5z(1-z^2)^2+1=0$$
Hopefully you can continue from here. Just note that your final polynomial first term in the expansion might be $1+z$, not $1-z$.
|
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|
On the Convergence of Series Will Jagy has provided the answer for the following question, the series is convergent:
I come across with the following question about the convergence of series:
\begin{align*}
\sum\dfrac{1}{k^{\epsilon}}\left(\dfrac{1}{\log k}-\dfrac{1}{\log(k+1)}\right),
\end{align*}
where $\epsilon\in(0,1)$ is fixed.
The only thing I can do is no more than
\begin{align*}
\sum\dfrac{1}{k^{\epsilon}}\dfrac{1}{(\log k)(\log(k+1))}\log(1+1/k).
\end{align*}
I propose the series is actually divergent. The term $(\log k)(\log(k+1))$ is slower than any $k^{\eta}$ for $\eta\in(0,1)$, on one hand the series has the lower bound
\begin{align*}
\sum\dfrac{1}{k^{\epsilon}}\dfrac{1}{k^{\eta}}\log(1+1/k),
\end{align*}
which we can put $\epsilon+\eta\leq 1$. But I cannot put a lower bound of the rate of $\log(1+1/k)$, any idea?
EDIT:
Stefan Lafon has answered the following question, the answer is negative, the series will always be convergent:
So I am looking for a nonnegative increasing function $f$ such that
\begin{align*}
\sum\dfrac{1}{k^{\epsilon}}\left(\dfrac{1}{f(k)}-\dfrac{1}{f(k+1)}\right)
\end{align*}
is divergent, but I cannot find one. Initially I was thinking the $\log$ thing will do, but Will Jagy disproves that. So what is such an $f$ for instance?
|
For $0 < t < 1,$
$$ t - \frac{t^2}{2} < \log (1+t) < t $$
$$ \frac{1}{k} - \frac{1}{2k^2} < \log \left(1+ \frac{1}{k} \right) < \frac{1}{k} $$
$$ \frac{2k-1}{2k^2} < \log \left(1+ \frac{1}{k} \right) < \frac{1}{k} $$
$$ \frac{k}{2k^2} < \frac{2k-1}{2k^2} < \log \left(1+ \frac{1}{k} \right) < \frac{1}{k} $$
$$ \frac{1}{2k} < \log \left(1+ \frac{1}{k} \right) < \frac{1}{k} $$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find all $x\in \mathbb{R}$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$
Find all $x\in \mathbb{R}$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$
Letting $a=2^x$ and $b=3^x$ we get $$\frac{a^3+b^3}{a^2b+ab^2} = \frac{7}{6}$$
from the numerator we have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=7$$
since $7$ is a prime we can say that $$a+b=1, a^2-ab+b^2=7.$$
It follows that $$a=1-b$$
from where $$(1-b)^2-(1-b)b+b²=7$$
this quadratic has solutions $b=1, b=0.$
what I now did was consider cases. Firstly $a=b=0$ which has no solutions for $x$.
case $a=b=1$ has the solution $x=0$.
However the actual solutions for this were $x=1, x=-1$ which I don't see how they came up with. What is wrong with my approach?
|
We have that
$$\frac{8^x+27^x}{12^x+18^x}=\frac{(2^x+3^x)(2^{2x}-6^x+3^{2x})}{(6^x)(2^x+3^x)}=\frac{2^{2x}-6^x+3^{2x}}{6^x}=\frac76$$
$$\iff\left(\frac23\right)^x+\left(\frac32\right)^x=\frac{13}6$$
$$\iff y+\frac1y=\frac{13}6 \iff y^2-\frac{13}6y+1=0$$
which leads to $y=\frac23,\frac32$ and $x=\pm1$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Finding remainder of $123^{456}$ divided by 88 using Chinese Remainder Theorem I tried using Chinese remainder theorem but I kept getting 19 instead of 9.
Here are my steps
$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equiv (-2)^9 \equiv -512 \equiv -6 \equiv 5 \pmod{11}\\
x_2 &= 123^{456} \equiv 123^0 \equiv 1 \pmod{8}\\
y_2 &= 1^{-1} \equiv 1 \pmod{8} \\
123^{456}
&\equiv \sum_{i=1}^2 x_i\times\frac{M}{m_i} \times y_i
\equiv 9\times\frac{88}{11}\times5 + 1\times\frac{88}{8} \times1 \equiv 371
\equiv 19 \pmod{88}
\end{split}
$$
|
Modulo $88$ one has $$123^{456} = 35^{456} = (35^2)^{228} = (-7)^{228} = ((-7)^6)^{38} = (-7)^{38} = ((-7)^6)^6 \times 49 = (-7)^6 \times 49 = -7 \times 49 = -343 = 9 \quad [88]$$
|
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|
Evaluate integral $\int \frac{x^2}{(15+6x-9x^2)^{\frac{3}{2}}} \, dx$ Evaluate the integral
$$\int \frac{x^2}{(15+6x-9x^2)^{\frac{3}{2}}} \, dx$$
Edit:
Here is Quanto's brilliant method of doing this problem with the details filled out:
SOLUTION:
First, lets complete the square $-9x^2+6x+15$:
$-9x^2+6x+15$
$=-9(x^2-\frac{2}{3}x)+15$
$=-9(x^2-\frac{2}{3}x+(\frac{2}{6})^2)+15-(-9(\frac{2}{6})^2)$
$=-9(x-\frac{1}{3})^2+16$...
$=-(1-3x)^2+16$
Now we are going to do a few extra steps in completing the square then usual, so this will be easier to integrate.
$=(-(\frac{1-3x}{4})^2+1)16$
Thus our original integral becomes:
$\int \frac{x^2}{(-(\frac{1-3x}{4})^2+1)16)^{\frac{3}{2}}}dx$
Now set $\sin(t) = \frac{1-3x}{4}$. Thus $x=\frac{1-4\sin(t)}{3}$ and $dx = \frac{-4\cos(t)}{3} \, du$
Note the strategic choice of defining $\sin(t) = \frac{1-3x}{4}$, it has a perfect place to plug in in the denominator. Let's make these substitutions
$=\int \frac{(\frac{1-4\sin(t)}{3})^2}{((-\sin^2(t)+1)16)^{\frac{3}{2}}}(\frac{-4\cos(u)}{3} \, du)$
$=\frac{-1}{432}\int \frac{(1-4\sin(t))^2}{(-\sin^2(t)+1)^{\frac{3}{2}}}(\cos(u) \, du)$
$=\frac{-1}{432}\int \frac{(1-4\sin(t))^2}{\cos^2(t)} \, du)$
$=\frac{-1}{432}\int \sec^2(t)+16\tan^2(t)-\frac{8\sin(t)}{\cos^2(t)}$
Now, Applying the identity $\tan^2(x)=\sec^2(x)-1$ we arrive at:
$=\frac{-1}{432}\int 17\sec^2(t)-16-\frac{8\sin(t)}{\cos^2(t)} \, dt$
Let's evaluate each of these separately:
$\int 17\sec^2(t)\,dt = 17\tan(t)$
$\int 16 \, dt = 16t$
$\int \frac{8\sin(t)}{\cos^2(t)}$ can be solved with u-substitution after setting $u=\cos(t)$. to get:
$\int \frac{8\sin(t)}{\cos^2(t)}=\frac{8}{\cos(t)}$
Thus our answer is:
$=\frac{1}{432}(16t-17\tan(t)+\frac{8}{\cos(t)})+C$
|
Completing the square in the denominator actually yields
$$\int \frac{x^2}{\left(-\left(3x-1\right)^2+\color{red}{16}\right)^{\frac{3}{2}}}dx$$
Substituting $x = \frac{4\sin(u)+1}{3}$ yields $$\frac{4}{27}\int \frac{\left(4\sin(u)+1\right)^2}{\left( -16\sin^2(u) + 16 \right)^{3/2}} \cos(u) du$$
Or since $1-\sin^2(x) = \cos^2(x)$, this simplifies to $$\frac{1}{27 \cdot 16}\int \frac{\left(4\sin(u)+1\right)^2}{ \cos^2(u)}du$$
Then you have a couple ways to solve this: You can integrate by parts using $\left( 4\sin(u)+1 \right)^2$ and $\sec^2(u)$ or you can expand the $\left( 4\sin(u)+1 \right)^2$ and solve with that. I will go with the first approach.
$$\frac{1}{27 \cdot 16} \left( (4\sin(u)+1)^2 \tan(u) -\int 8 \cos(u)(1+4\sin(u)) \tan(u) du\right)$$
This last integral can be solved using $\int \sin(x) dx = -\cos(x)$ and $\int \sin^2(x) dx = \frac{x}{2} - \frac{\sin(x)\cos(x)}{2}$.
|
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|
Integrate $ \int \frac{1}{\sin^{4}x+\cos^{4}x}dx $ Show that$$ \int \frac{1}{\sin^{4}(x)+\cos^{4}(x)}dx \ = \frac{1}{\sqrt{2}}\arctan\left(\frac{\tan2x}{\sqrt{2}}\right)+C$$
I have tried using Weierstrass substitution but I can't seem to get to the answer... Should I be using the said method or is there another way I can approach the question? Since the integrand evaluates into an arctangent function I am assuming there is some trickery in the manipulation that can get me there. But I just can't seem to see it...
|
$$
\frac{1}{\sin^4x+\cos^4x}=
\frac{1}{(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x}=
\frac{1}{1-\frac{1}{2}(2\sin x\cos x)^2}=\\
\frac{2}{2-\sin^22x}=
\frac{2}{2-\frac{1}{2}(1-\cos4x)}=
\frac{4}{3+\cos4x}=
$$
If we now set
$$
t=\tan2x\qquad\implies\qquad dx=\frac{1}{2}\cdot\frac{1}{1+t^2}dt
$$
we have
$$
\int\frac{1}{\sin^4x+\cos^4x}dx=
\int\frac{4}{3+\frac{1-t^2}{1+t^2}}\frac{1}{2(1+t^2)}dt=
\int\frac{2}{3+3t^2+1-t^2}dt=\\
\int\frac{2}{2t^2+4}dt=
\int\frac{1}{t^2+2}dt=
\frac{1}{\sqrt{2}}\arctan\left(\frac{t}{\sqrt{2}}\right)+C=
\frac{1}{\sqrt{2}}\arctan\left(\frac{\tan(2x)}{\sqrt{2}}\right)+C=
$$
OLD answer
$$
\frac{1}{\sin^4x\cos^4x}=\frac{1}{(\sin x\cos x)^4}=16\frac{1}{\sin^42x}=16\frac{\cos^22x+\sin^22x}{\sin^42x}=\\=16\frac{\cot^22x+1}{\sin^22x}=-8\frac{d}{dx}\left(\frac{1}{3}\cot^32x+\cot2x\right)
$$
Note that this is different from your supposed answer.
|
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Maximum value of $\lfloor a \rfloor^2+\lfloor b \rfloor^2+\lfloor c \rfloor^2$ under given conditions Consider $a,b,c \in \mathbb{R}-\{0\}$. A real valued function is defined as $$f(x,y,z)=2^y3^za^xb^yc^z+2^z3^xa^yb^zc^x+2^x3^ya^zb^xc^y$$ where $x,y,z \in \mathbb{Z}$ and $f(1,0,0)=4$ and $f(2,0,0)=6$. Find the maximum value of $\lfloor a \rfloor^2+\lfloor b \rfloor^2+\lfloor c \rfloor^2$.
Using given equations I got
$a+2b+3c=4$ and $a^2+4b^2+9c^2=6$ and also $2ab+6bc+3ac=5$. I have also obtained $a^2+b^2+c^2 \geq \displaystyle\frac{8}{7}$ but how to proceed further? Could someone help me with this?
|
$a+2b+3c=4$ ...(i)
$a^2+4b^2+9c^2=6$ ...(ii)
Given we have to find the maximum value of $\lfloor a \rfloor^2+\lfloor b \rfloor^2+\lfloor c \rfloor^2$ which has equal of $a^2, b^2, c^2$, we should maximize $a^2$ in (ii).
The minimum value of $a^2, b^2, c^2$ can be $0$ each. So individually, max value of
$a^2 = 6, b^2 = \dfrac{3}{2}, c^2 = \dfrac{6}{9}$ ...(iii)
So $\lfloor c \rfloor^2$ cannot be any value other than $0$.
Now, let's maximize $\lfloor a \rfloor^2$. From (iii), max value of $a = \sqrt6$ but any value greater than $2$ (and less than $3$) does not help. So, let's go with $a^2 = 4$.
Now, $4b^2+9c^2 = 2$. Even with $c^2$ as $0$, $b \lt 1$.
So, maximum value of $\lfloor a \rfloor^2+\lfloor b \rfloor^2+\lfloor c \rfloor^2 = 4$
|
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|
Proving that inequality holds under condition.
Let $a$ and $b$ be positive numbers. Prove that inequality $$\frac{ax+by}{2} \leqslant \sqrt{\frac{ax^2+by^2}{2}}$$
holds for all real $x$ and $y$ only and only if $a+b \leqslant2$
Problem needs to be done using "basic" algebraic methods.
I tried expanding this into form $$2ax^2+2by^2-a^2x^2-b^2y^2-2abxy \geqslant 0$$
and then take oustide parenthesis $2-a-b$. Inequalities between means did not help either.
Can you give me some clues?
|
Let $a+b\leq2$.
Thus, by C-S $$\sqrt{\frac{ax^2+by^2}{2}}=\sqrt{\frac{(a+b)(ax^2+by^2)}{2(a+b)}}\geq\frac{|ax+by|}{\sqrt{2(a+b)}}\geq\frac{|ax+by|}{2}\geq\frac{ax+by}{2}.$$
Let $a$ and $b$ are positives and $$\sqrt{\frac{ax^2+by^2}{2}}\geq\frac{ax+by}{2}$$ is true for any reals $x$ and $y$.
Thus, for $x=y=1$ we obtain:
$$\sqrt{\frac{a+b}{2}}\geq\frac{a+b}{2},$$ which gives $$a+b\leq2.$$
|
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Solve $ay''+by+c=0$ using separation of variables How could be
$$ay''+by+c=0$$
(for constants $a,b,c$) solved using separation of variables?
I tried using the method on an easier equation, namely $ay''+by'+c=0$:
$$\begin{align}a\frac{d^2y}{dx^2}+b\frac{dy}{dx}+c&=0,\, u=\frac{dy}{dx}\\a\frac{du}{dx}+bu+c&=0\\ \frac{1}{a}\frac{dx}{du}&=-\frac{1}{bu+c}\\x&=-a\int \frac{du}{bu+c}=-a\left(\frac{\ln |bu+c|}{b}+C\right)\\-\frac{b}{a}(x+aC)&=\ln |bu+c|\\|bu+c|&=e^{-\frac{b}{a}(x+aC)}\\bu+c=b\frac{dy}{dx}+c&=C'e^{iC''}e^{-\frac{b}{a}(x+aC)},\, \color{red}{C'\ge 0,\, C''\in\mathbb{R}}\\y&=\frac{1}{b}\left(C_0\int e^{-\frac{b}{a}(x+aC)}\, dx-c\int dx\right)\\&=\frac{1}{b}\left(C_0\left(-\frac{ae^{-\frac{b}{a}(x+aC)}}{b}+C'''\right)-c(x+C'''')\right).\end{align}$$
When I tried to apply this method on $ay''+by+c=0$, I ran into problems, as the degrees of the derivatives differ by $2$, not $1$:
$$\begin{align}a\frac{d^2 y}{dx}+by+c&=0,\, \frac{dy}{dx}=u\\a\frac{du}{dx}+b\int u\, dx+c&=0\\ \frac{dx}{du}&=-\frac{a}{b\int u\, dx+c}\\x&=-a\int\frac{du}{b\int u\, dx+c}.\end{align}$$
I'm stuck here. The solution should be
$$y=Ce^{x\sqrt{\frac{b}{a}}}+C'e^{-x\sqrt{\frac{b}{a}}}-\frac{c}{b}.$$
|
$$ay''+by+c=0$$
Substitute $p=\dfrac {dy}{dx}=y'$ then you have:
$$y''=\dfrac {dy'}{dx}=\dfrac {dp}{dx}=\dfrac {dp}{dy}\dfrac {dy}{dx}=p'p$$
Where $p'=\dfrac {dp}{dy}$.
The differential equation becomes:
$$app'+by+c=0$$
This is now a first order DE that is separable:
$$a\int p\;dp =-\int (c+by) \; dy$$
$$ap^2=-2cy-by^2+C_1$$
$$\dfrac {dy}{dx}=\pm \sqrt {\dfrac { -2cy-by^2+C_1}a}$$
Note that:
$$-by^2-2cy+C_1=-b(y^2+2\dfrac cby +\dfrac {c^2}{b^2})+K_1$$
$$=-b(y+\dfrac cb)^2+K_1$$
This last differential equation is also separable:
$$\int \dfrac {dy}{\sqrt {C-(y+\dfrac cb)^2}}=\pm \int \sqrt {\dfrac {b}a}dx$$
And
$$(\arcsin x)'=\int \dfrac {dx}{\sqrt {1-x^2}}$$
|
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|
Can we find $ \sum_{n=1}^{\infty}\frac{1+2+\cdots +n}{n!} $? Consider the sequence $$ a_{n} = \sum_{r=1}^{n}\frac{1+2+\cdots +r}{r!} $$
Then we have, $$ a_{n} = \sum_{r=1}^{n}\frac{1}{r!} \ + 2\sum_{r=2}^{n}\frac{1}{r!} \ + 3\sum_{r=3}^{n}\frac{1}{r!} \ + \cdots + n\sum_{r=n}^{n}\frac{1}{r!} \ \geq \ 1 + \sum_{r=1}^{n}\frac{1}{r!}$$ for all $n \geq 2$.
$\implies \displaystyle \lim_{n \to \infty}a_{n} \ \geq \ e $
Now, in my book it says $\displaystyle \lim_{n \to \infty}a_{n} \ = \frac{3}{2}e $
How can I attack this problem? Anyone please?
|
$$\sum_{k=1}^n k = \frac{n(n+1)}{2}.$$ Thus the given sum is $$S = \frac{1}{2} \sum_{n=1}^\infty \frac{n(n+1)}{n!} = \frac{1}{2} \sum_{n=1}^\infty \frac{n+1}{(n-1)!} = \frac{1}{2} \left(\sum_{n=1}^\infty \frac{n-1}{(n-1)!} + 2\sum_{n=0}^\infty \frac{1}{n!}\right).$$ The second term in parentheses is simply $2e$; the first term simplifies further as $$\sum_{n=2}^\infty \frac{1}{(n-2)!} = e.$$ So the result is $$S = \frac{3e}{2}.$$
|
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|
How do i form the inequality $ -5\le 3\cos x - 4\sin x\le5$? We know that
$-1\le \cos x\le1$ so we if we multiply 3 so we get $-3\le 3\cos x\le 3$ and we also know that $-1\le \sin x\le 1$ and and again we multiply 4 we get $-4\le 4\sin x\le 4$ and we can add these two inequality we get $-7\le 3\cos x - 4\sin x\le 7$ but how in the above inequality is forming i m not able to understand at all.
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Note that\begin{align}3\cos x-4\sin x&=5\left(\frac35\cos x-\frac45\sin x\right)\\&=5\left(\cos\alpha\cos x-\sin\alpha\sin x\right),\end{align}where $\alpha\in\Bbb R$ is such that $\cos\alpha=\frac35$ and that $\sin\alpha=\frac45$. Therefore$$3\cos x-4\sin x=5\cos(\alpha+x)\in[-5,5].$$
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Proving $4\Big(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \Big)+\frac{81}{(a+b+c)^2}\geqslant{\frac {7(a+b+c)}{abc}}$ For $a,b,c>0.$ Prove$:$ $$4\Big(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \Big)+\dfrac{81}{(a+b+c)^2}\geqslant{\dfrac {7(a+b+c)}{abc}}$$
My proof is using SOS$:$
$${c}^{2}{a}^{2} {b}^{2}\Big( \sum a\Big)^2 \sum a^2 \Big\{ 4\Big(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2} \Big)+\dfrac{81}{(a+b+c)^2}-{\dfrac {7(a+b+c)}{abc}} \Big\}$$
$$=\dfrac{1}{2} \sum {a}^{2}{b}^{2} \left( {a}^{2}+{b}^{2}-2\,{c}^{2} +5bc-10ab+5\, ac \right) ^{2} +\dfrac{1}{2} \prod (a-b)^2 \left( 7\sum a^2 +50\sum bc \right) \geqslant 0.$$
From this we see that the inequality is true for all $a,b,c \in \mathbb{R};ab+bc+ca\geqslant 0.$
But we also have this inequality for $a,b,c \in \mathbb{R}.$ Which verify by Maple.
I try and I found a proof but I'm not sure$:$
If replace $(a,b,c)$ by $(-a,-b,-c)$ we get the same inequality.
So we may assume $a+b+c\geqslant 0$ (because if $a+b+c<0$ we can let $a=-x,b=-y,c=-z$ where $x+y+z \geqslant 0$ and the inequality is same!)
Let $a+b+c=1,ab+bc+ca=\dfrac{1-t^2}{3} \quad (t\geqslant 0), r=abc.$ Need to prove$:$
$$f(r) =81\,{r}^{2}-15\,r+\dfrac{4}{9} \left( t-1 \right) ^{2} \left( t+1 \right) ^{2
}\geqslant 0.$$
It's easy to see, when $r$ increase then $f(r)$ decrease. Since $r\leqslant \dfrac{1}{27} \left( 2\,t+1 \right) \left( t-1\right) ^{2} \quad$(see here). We get$:$
$$f(r)\geqslant f\Big(\dfrac{1}{27} \left( 2\,t+1 \right) \left( t-1\right) ^{2}\Big)=\dfrac{1}{9} {t}^{2} \left( 2\,t-1 \right) ^{2} \left( t-1 \right) ^{2} \geqslant 0.$$
Done.
Could you check it for me? Who have a proof for $a,b,c \in \mathbb{R}$?
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For $a,\,b,\,c$ are real numbers. We have
$$(a+b+c)^2 =(|a+b+c|)^2 \leqslant (|a|+|b|+|c|)^2,$$
$$\frac{a+b+c}{abc}=\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca} \leqslant \left |\frac{1}{ab} \right |+\left |\frac{1}{bc} \right |+\left |\frac{1}{ca} \right |=\frac{|a|+|b|+|c|}{|a||b||c|}.$$
So, we need to prove
$$4\left(\dfrac{1}{|a|^2}+\dfrac{1}{|b|^2}+\dfrac{1}{|c|^2} \right)+\dfrac{81}{(|a|+|b+|c|)^2} \geqslant \frac{7(|a|+|b|+|c|)}{|a||b||c|}.$$
Now, replace $(|a|,|b|,|c|) \to (a,b,c)$ then $a,b,c \geqslant 0.$ The inequality become
$$4\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\right)+\dfrac{81}{(a+b+c)^2}\geqslant{\dfrac {7(a+b+c)}{abc}}.$$
This is the original inequality.
|
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Let $a$ and $b$ be positive integers such that $(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20 \sqrt[3]{6}.$ Find $a + b.$ Let $a$ and $b$ be positive integers such that $$(\sqrt[3]{a} + \sqrt[3]{b} - 1)^2 = 49 + 20 \sqrt[3]{6}.$$ Find $a + b.$
We let $\sqrt[3]a, \sqrt[3]b$ be $x,y,$ respectively. We expand LHS to get $$x^2+y^2+2xy-2x-2y+1=49+20\sqrt[3]6.$$ Now, I'm stuck. Help? Please share a clear solution :).
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Let $$\alpha = \sqrt[3]{a} + \sqrt[3]{b} = 1 + \sqrt{49 + 20 \sqrt[3]{6}}$$
We can compute the minimal polynomial of the RHS:
(1 + sqrt(49 + 20 * (6)^(1/3))).minpoly()
x^3 - 72*x - 336
and then applying Cardano's formula we get $\alpha = 2\cdot (6^{1/3} + 6^{2/3})$.
We can transform this into $a,b$ form with $2^3 = 8$ and $6^2 = 36$:
$$\alpha = \sqrt[3]{48} + \sqrt[3]{288}$$
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proof that $ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} = \frac{3}{2} - \frac{1}{n} $
Proof that
$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} = \frac{3}{2} - \frac{1}{n} $$
by induction.
Proof
Base case: Statement clearly holds for $n = 1$. Now assume that statement holds for some $n = k$ and lets show that it implies $n = k + 1$ holds. The proof:
$$ \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \dots + \frac{1}{n(n-1)} + \frac{1}{n(n+1)} = \frac{3}{2} - \frac{1}{n} + \frac{1}{n(n+1)} = \frac{3}{2} - \frac{1}{n} + \frac{1}{n} - \frac{1}{n + 1} $$
$$ = \frac{3}{2} - \frac{1}{n+1} $$
Now the problem is I can't find the error. The statement doesn't clearly work for $ n = 2 $. However, the assumption seems to be correct since if I assume it's true for some $n = k$ and it is true for $ n = 1$? It shouldn't be possible to show that $p(n) \implies p(n+1)$ when $p(n)$ is true and $p(n+1)$ is false. This means that $p(n)$ has to be false in this case since when $p(n)$ is false then $p(n) \implies p(n+1)$ is tautology. The problem is I don't really see how? Isn't the whole point of induction to show that $p(n)$ is true for some specific $n = k$ (not all $n$) and then show $p(n+1)$ by assuming $p(n)$. Now when $p(n)$ is false you can show anything since it's tautology but how can you be sure $p(n)$ is true if you don't show it for all $n$? And wouldn't that defeat the purpose of induction (if you have already shown it's true for all $n$)?.
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For $n=1$ the last term on the lefthand side is $\frac1{1\cdot0}$, which is undefined. The induction has to start at $n=2$, and as you say, the statement is false for $n=2$. The fact that the induction step works (after you correct the sign error in your answer, which I suspect is a typo) means that the formula $\frac32-\frac1n$ is going to give the wrong answer for every $n\ge 2$.
In fact the lefthand side is a telescoping sum,
$$\left(\frac11-\frac12\right)+\left(\frac12-\frac13\right)+\ldots+\left(\frac1{n-1}-\frac1n\right)\,,$$
and the correct righthand side is $1-\frac1n$. The induction step works because the righthand side is offset from the correct value by a constant amount, $\frac12$, for every $n$.
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Collatz Conjecture: Is there a straightforward argument showing that there are no nontrivial 2 "step" repeats (where each "step" is an odd number) Let:
*
*$C(x) = \dfrac{3x+1}{2^w}$ where $w$ is the highest power of $2$ that divides $3x+1$
*$C_n(x) = C_1(C_2(\dots(C_n(x)\dots)) = \dfrac{3^n x + 3^{n-1} + \sum\limits_{i=1}^{n-1}3^{n-1-i}2^{\left(\sum\limits_{k=1}^i w_k\right)}}{2^{\left(\sum\limits_{i=1}^n w_i\right)}}$
It is well known that if $C_2(x)=x$, then $x=1$.
I am wondering if there is straight forward way to prove this. Based on the above equation, it seems like it should be very easy to find. I am unable to do it.
Here's what I have:
$$x = \frac{3 + 2^{w_1}}{2^{w_1 + w_2} - 9}$$
where:
*
*$w_1 + w_2 \ge 4$
*$w_1 \ge 1, w_2 \ge 1$
Would proof by infinite descent be the appropriate way to approach this?
|
You got
$$x = \frac{3 + 2^{w_1}}{2^{w_1 + w_2} - 9} \tag{1}\label{eq1A}$$
For $x$ to be a positive integer means the denominator must divide evenly into the numerator which means the denominator must also be less than or equal to the numerator.
For somewhat easier algebra, let
$$y = 2^{w_1} \tag{2}\label{eq2A}$$
so \eqref{eq1A} now becomes
$$x = \frac{y + 3}{2^{w_2}y - 9} \tag{3}\label{eq3A}$$
This gives
$$y + 3 \ge 2^{w_2}y - 9 \iff 12 \ge (2^{w_2} - 1)y \iff \frac{12}{2^{w_2} - 1} \ge y \tag{4}\label{eq4A}$$
The denominator must also be a positive integer, but since the smallest power of $2$ greater than $9$ is $16$, we get
$$2^{w_2}y \ge 16 \iff y \ge \frac{16}{2^{w_2}} \tag{5}\label{eq5A}$$
Combining \eqref{eq4A} and \eqref{eq5A} gives
$$\frac{16}{2^{w_2}} \le y \le \frac{12}{2^{w_2} - 1} \tag{6}\label{eq6A}$$
For $w_2 = 1$, \eqref{eq6A} gives $8 \le y \le 12$. Since \eqref{eq2A} states $y$ is a power of $2$, the only possible solution is $w_1 = 3$ giving $y = 8$. However, \eqref{eq3A} gives $x = \frac{11}{7}$, which is not an integer.
Next, if $w_2 = 2$, \eqref{eq6A} gives $4 \le y \le 4$, i.e., $w_1 = 2$. Substituting these into \eqref{eq3A} gives $x = 1$.
If $w_2 = 3$, then \eqref{eq6A} gives $2 \le y \le \frac{12}{7}$, i.e., there's no value of $y$. Likewise, any value of $w_2 \gt 3$ will not allow any value of $y$. Also, Steven Stadnicki's question comment gives another way to see $w_2 = 2$ is its maximum possible value.
This means the only valid positive integer solution of \eqref{eq1A} is $w_1 = w_2 = 2$ giving $x = 1$.
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How to deduce this factorization of $x^5+x+1$ by looking at $\int\frac{3x^4+2x^3-2x+1}{x^5+x+1}dx$? The question is:
$$\int\frac{3x^4+2x^3-2x+1}{x^5+x+1}dx$$
I tried a lot but couldn't solve it so I looked at the solution which is:
$$x^5+x+1=(x^2+x+1)(x^3-x^2+1)$$ and we can write $$3x^4+2x^3-2x+1=(x^3-x^2+1)+(3x^2-2x)(x^2+x+1)$$
which effectively reduces the integral to very simple ones.
My question is how they deduced the factorization of the denominator. After looking at the solution I think that if we put $x=1,x=\omega$ and $x=\omega^2$ we can deduce this but this was not immediately obvious to me. Is there some sort of hint you can get by looking at the integrand or is it simply a matter of less experience?
Any help would be appreciated.
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try :$$x^5+x+1=(ax^3+bx^2+cx+d)(Ax^2+Bx+C)$$and compare coefficients.!
Again intelligent guessing works the best
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|
How to find the $n$-th derivative of the function $\log(1-x(1-x))$ I know how to find the $n$ derivative of $\log (1-x)$ which is $-\frac{(n-1)!}{(1-x)^n}$, but how to find the $n$-th derivative of the function $\log(1-x(1-x))$?
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By purely symmetry conditions we get that $x(1-x)$ is a parabola with a vertex at $x=\frac{1}{2}$ which means we can rewrite it as
$$x(1-x) = \frac{1}{4}-\left(x-\frac{1}{2}\right)^2$$
Then the log becomes
$$\log\left(\frac{3}{4}+\left(x-\frac{1}{2}\right)^2\right)$$
Finding the $n$th derivative of this function at $\frac{1}{2}$ is equivalent to finding the $n$th derivative of
$$\log\left(\frac{3}{4}+x^2\right) = \log\frac{3}{4} + \log\left(1+\frac{4}{3}x^2\right)$$
at $0$. The function on the right has a known Taylor series
$$\log(1+x) = \sum_{k=1}^\infty \frac{(-1)^{k+1}x^k}{k}$$
Can you take it from here?
The answer when you want to check is
$$f^{(n)}\left(\frac{1}{2}\right) = \begin{cases} \log\frac{3}{4} & n=0 \\ 0 & n\text{ odd}\\ \frac{(-1)^{\frac{n}{2}+1}2^{n+1}(n-1)!}{\sqrt{3^n}} & n\text{ even}\\ \end{cases}$$
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Solving $\sin80^\circ\sin20^\circ\sin x = \sin 10^\circ\sin60^\circ\sin(30^\circ+x)$
Solve:
$$\sin80^\circ\sin20^\circ\sin x = \sin 10^\circ\sin60^\circ\sin(30^\circ+x)$$
I tried cancelling $\sin10^\circ$
$$\sin80^\circ\cdot 2\cos10^\circ\cdot\sin x=\sin60^\circ\sin(30+x)$$
$$\frac{2\sin^280^\circ}{\sin60^\circ} = \frac{\sin(30^\circ+x)}{\sin x}$$
Then I'm stuck.
|
We have $$\frac{\sin80^{\circ}\sin20^{\circ}}{\sin10^{\circ}\sin60^{\circ}}=\frac{1}{2}\cot{x}+\frac{\sqrt3}{2}$$ or
$$\frac{4\sin80^{\circ}\cos10^{\circ}}{\frac{\sqrt3}{2}}-\sqrt3=\cot{x}$$ or
$$\cot{x}=\frac{8\cos^210^{\circ}-3}{\sqrt3}$$ or
$$\cot{x}=\frac{1+4\cos20^{\circ}}{\sqrt3}.$$
Now, $$\frac{1+4\cos20^{\circ}}{\sqrt3}=\frac{\sin20^{\circ}+2\sin40^{\circ}}{\sqrt3\sin20^{\circ}}=\frac{\cos10^{\circ}+\sin40^{\circ}}{\sqrt3\sin20^{\circ}}=$$
$$==\frac{\sin80^{\circ}+\sin40^{\circ}}{\sqrt3\sin20^{\circ}}=\frac{\sqrt3\cos20^{\circ}}{\sqrt3\sin20^{\circ}}=\cot20^{\circ}.$$
Can you end it now?
|
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|
finding a relation in $p:p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots$
if $$p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots$$ and $$p^2+ap+c=0.$$ Find $a,c$ also $|c|=2$
My progress:The general term $$T_{m+1}=\frac{(1)(3)\cdots(2m+1)}{(3)(6)\cdots(3m+3)}$$ from here i tried to make it in terms of factorial but it does not help.
Is there any algorithm to solve these problems?
(This question came in a maths magazine
|
$p=\frac{1}{3}+\frac{1\cdot 3}{1 \cdot 2}\cdot (\frac{1}{3})^2+\frac{1\cdot 3\cdot 5}{1 \cdot 2\cdot 3}\cdot (\frac{1}{3})^3 + \cdots$
So $p+1=1+\frac{1}{3}+\frac{1\cdot 3}{1 \cdot 2}\cdot (\frac{1}{3})^2+\frac{1\cdot 3\cdot 5}{1 \cdot 2\cdot 3}\cdot (\frac{1}{3})^3 +\cdots$
Now notice that $$(1+x)^n=1+nx+\frac{n(n-1)}{2!}x^2+\cdots$$
We can assume that $$nx=\frac{1}{3}$$ and $$\frac{n(n-1)}{2!}x^2=\frac{1\cdot 3}{1 \cdot 2}\cdot (\frac{1}{3})^2$$
and we will get $$\frac{n-1}{n}=3\Longrightarrow n=-\frac{1}{2}\Longrightarrow x=-\frac{2}{3}$$
Therefore we have $$1+p=\frac{1}{\sqrt{1+x}}=\sqrt{3}$$ Can you end it now?
|
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|
Finding $\sum_{n=1}^{\infty} \frac{1}{f(n)}$ where $f$ is a real quadratic function? Let's consider the following function $f(x)=ax^2+bx+c$, where $a, b$ and $ c $ are all real constants. Is there any way to calculate the value of $$\sum_{n=1}^{\infty} \frac{1}{f(n)}\text{?}$$ I can find no sequences but that $\sum_{n=1}^{\infty}\frac{1}{a(x^2+(2b+1)x+b^2+b)} = \frac{1}{ab+a}$, but this only works for a few specific cases. Any help will be appreciated.
|
As @Mourad commented, rewrite
$$a n^2+b n +c=a (n-r)(n-s)$$ where $r$ and $s$ are the roots of the quadratic (hoping that they are not psitive integer numbers) and use partial fraction decomposition
$$\frac 1{a n^2+b n +c}=\frac 1{a(r-s)} \left(\frac 1{n-r}-\frac 1{n-s} \right)$$ So, for the partial sum
$$S_p=\sum_{n=1}^p \frac 1{a n^2+b n +c}=\frac 1{a(r-s)}\left(\psi (p-r+1)-\psi(1-r)-\psi (p-s+1)+\psi (1-s)\right)$$ Now, using the asymptotics
$$S_p=\frac 1{a(r-s)}\left((\psi (1-s)-\psi
(1-r))+\frac{s-r}{p}+O\left(\frac{1}{p^2}\right)\right)$$
$$S_\infty=\frac {\psi (1-s)-\psi
(1-r)}{a(r-s)}$$
Back to $a,b,c$
$$S_\infty=\frac{\psi \left(\frac{b+\sqrt{b^2-4 a c}}{2 a}+1\right)-\psi
\left(\frac{b-\sqrt{b^2-4 a c}}{2 a}+1\right)}{\sqrt{b^2-4 a
c}}$$ or
$$S_\infty=\frac{H_{\frac{b+\sqrt{b^2-4 a c}}{2 a}}-H_{\frac{b-\sqrt{b^2-4 a c}}{2
a}}}{\sqrt{b^2-4 a c}}$$
|
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|
Real roots of $x^7+5x^5+x^3−3x^2+3x−7=0$? The number of real solutions of the equation,
$$x^7+5x^5+x^3−3x^2+3x−7=0$$
is
$$(A) 5 \quad (B) 7 \quad (C) 3 \quad (D) 1.$$
Using Descartes rule we may have maximum no. of positive real roots is $3$ and negative real root is $0.$ So there can be either $3$ real roots or $1$ real root but how to conclude what will be the no. of real roots exactly. Can you please help me?
|
The first test of factors using the rational root theorem yields
$$(x - 1) (x^6 + x^5 + 6 x^4 + 6 x^3 + 7 x^2 + 4 x + 7) = 0$$
From this we can see that, for one root, $X=1$.
From the co-factor, we can see that the function never crosses the $x$-axis so there are no more real roots.
Answer: $D)1$
|
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If $\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$ ; $\frac{1}{a} + \frac{1}{c} + \frac{1}{a+y}$ ; $\frac{1}{a} + \frac{1}{x} + \frac{1}{y} = 0$.
If $a \neq 0$ , $b \neq 0$ , $c \neq 0$ and if :- $\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$ ; $\frac{1}{a} + \frac{1}{c} + \frac{1}{a+y}=0$ ; $\frac{1}{a} + \frac{1}{x} + \frac{1}{y} = 0$ , find $(a+b+c)$ .
What I Tried :- No information is given about $x$ and $y$ . So I thought of putting $x = y = 1$ , and this silly thing came out in the end .
Now, as $x = y = 1$ , I have $a = \frac{1}{-2}$ from the $3$rd equation .
So from the $1$st equation I get :- $$\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$$
$$ \rightarrow -2 + \frac{1}{b} + 2 = 0$$
$$ \rightarrow \frac{1}{b} = 0$$
This definitely looks absurd (also it's given that $b \neq 0$), so I guess putting $x = y = 1$ was a big mistake .
I don't have any other cool ideas for now as I see that doing it algebraically is going to include a lot of simplification and stuffs, and since there are $5$ variable there must be some shortcut of this .
Can anyone help?
|
Ok so by @player3236's hint , he said that taking $(x = y)$ will create a problem as then it would imply that $\frac{1}{a} + \frac{1}{a+x} = 0$ which is giving $\frac{1}{b} = 0$ and stopping me to move on.
For now, I am taking say $x = 1$ , $y = 2$ .
This gives :- $a = -\frac{2}{3}$ in the $3$rd equation .
Now in the $1$st equation, we have :-
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$$
$$\rightarrow \frac{-3}{2} + 3 + \frac{1}{b} = 0$$
$$\rightarrow \frac{1}{b} = \frac{-3}{2}$$
$$\rightarrow b = -\frac{2}{3}$$
Similarly solving for $c$ in the $2$nd equation we get :- $c = \frac{4}{3}$ .
This gives $(a + b + c) = 0$.
Fortunately I got my answer to this question, but can I get some generalization of why $(a + b + c) = 0$ every time, i.e. , instead of putting $x = 1$ , $y = 2$ can we find $(a + b + c)$ .
|
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What is the solution to $\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$ $$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$$
I've already tried this many times but still not get the answer, this was one of my solution which I can't go further anymore.
$\frac{{x}^{2}}{(x+1)({\sqrt{x+1}}-1)^{2}}< \frac{{x}^{2}+3x+18}{({x+1})^{2}}$
$\frac{{x}^{2}(x+1)}{({\sqrt{x+1}}-1)^{2}}< {x}^{2}+3x+18$
$\frac{{x}^{3}+{x}^{2}}{({\sqrt{x+1}}-1)^{2}}< \frac{({x}^{2}+3x+18)({\sqrt{x+1}-1})^{2}}{({\sqrt{x+1}-1})^{2}}$
$\frac{{x}^{3}+{x}^{2}}{({\sqrt{x+1}}-1)^{2}}< \frac{{x}^{3}+5{x}^{2}+24x+36-(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}-1})^{2}}$
$\frac{{x}^{3}+{x}^{2}-{x}^{3}-5{x}^{2}-24x-36+(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
$\frac{-4{x}^{2}-24x-36+(2{x}^2+6x+36)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
$\frac{-4(x+3)^{2}+2({x}^2+3x+18)\sqrt{x+1}}{({\sqrt{x+1}}-1)^{2}}<0$
Does anyone have a hint for the solution?
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We need
$$\left\{\begin{aligned} & x+1 > 0 \\& x+1 \ne \sqrt{x+1} \end{aligned}\right. \Rightarrow \left\{\begin{aligned} & x > -1 \\& x \ne 0\end{aligned}\right. \quad (1)$$
Setting $t = \sqrt{x+1} > 0,$ then $x=t^2-1,$ now
$$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} = \frac{(t+1)^2}{t^2},$$
and
$$\frac{{x}^{2}+3x+18}{({x+1})^{2}} = \frac{t^4+t^2+16}{t^4}.$$
Inequality become
$$\frac{(t+1)^2}{t^2}<\frac{t^4+t^2+16}{t^4},$$
or
$$\frac{2(2-t)[(t+1)^2+3]}{t^4}>0.$$
Therefore
$$0 < t < 2\Rightarrow 0 < \sqrt{x+1} < 2,$$
or
$$-1 < x < 3. \quad (2)$$
From $(1)$ and $(2)$ we get $x \in (-1,0)$ or $x \in (0,3).$
|
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In which base $b$ is $(374)_b$ a perfect square? If you convert this number to base 10, we can obtain the expression $$3b^2+7b+4 = (b+1)(3b+4).$$
Since $\gcd(b+1,3b+4) = 1$, we further conclude that both $b+1$ and $3b+4$ are perfect squares.
So the problem is equivalent to finding $b$ that satisfies the Diophantine equation $$3(b+1)^2+1 = L^2,$$ where $\gcd(b+1,L) = 1$.
I'm not sure how to find all solutions to the above equation.
|
You're off to a good start; if $b+1$ and $3b+4$ are both perfect squares, then
$$3b+4=x^2\qquad\text{ and }\qquad b+1=y^2,$$
for some integers $x$ and $y$, and hence
$$x^2-3y^2=1.$$
This is a Pell equation, and its solutions are well known. I suggest to start from the Wikipedia page to understand how to find all integral solutions. In particular there are infinitely many solutions.
One characterization of the integral solutions is that they are precisely the pairs of integers $(x,y)$ for which
$$x+y\sqrt{3}=\pm(2+\sqrt{3})^k,$$
for some integer $k$. Of course the choice of $\pm$ sign only changes the signs of $x$ and $y$, and the same is true if we replace $k$ by $-k$. So to find all solutions $b$ it suffices to consider $(2+\sqrt{3})^k$ with $k\geq0$. The first few solutions are:
$$\begin{array}{r|rr|rr}
k&x&y&b&(374)_b\\
\hline
0&1&0&\color{red}{-1}&\color{red}{0^2}\\
1&2&1&\color{red}{0}&\color{red}{2^2}\\
2&7&4&15&28^2\\
3&26&15&224&390^2\\
4&97&56&3135&5432^2\\
5&362&209&43680&75658^2
\end{array}$$
|
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|
Finding an epsilon-delta proof of a sequence I'd like to write an epsilon-delta proof that
$$
\lim_{x \to \infty} \frac{2x-3}{x^2-2x-5} = 0
$$
so what I've done is,
$\lvert \frac{2x-3}{x^2-2x-5} \rvert < \frac{2x}{x^2-2x-5}$ if $x>3/2$.
$$
\frac{2x}{x^2-2x-5} < \frac{2x}{x^2-7x} < \epsilon
$$
$$
x > \frac{2}{\epsilon} +7
$$
$
\therefore \lvert \frac{2x-3}{x^2-2x-5} \rvert < \epsilon $ if $ x>N(\epsilon) = Max [{\frac{3}{2}, \frac{2}{\epsilon}+7}]
$
Is this valid? Or is there a mistake that I made? If so, could I get an explanation? Thank you in advance!
|
You are essentially correct, though you are going to want $x > \sqrt{6}+1$ to avoid division by $0$ and to ensure $\frac{2x-3}{x^2-2x-5}>0$ Later you will want $x >7$ for the same reason. and since $\frac2\epsilon+7>7>\sqrt{6}+1> \frac32 >1$ you can slightly simplify your $N(\epsilon)$ expression. While your exploration is is sensible, a proof might better be in the opposite order.
So perhaps something like:
For any given $\epsilon>0$ and $x > N(\epsilon) = \frac2\epsilon+7$
you have $x>7$ and $x^2>7x$ and $5x > 5$ and $2x > 3$ and $x^2-2x-5>0$
in which case $0 < \frac{2x-3}{x^2-2x-5} < \frac{2x}{x^2-2x-5} < \frac{2x}{x^2-7x} = \frac{2}{x-7}\lt \epsilon$
so $\lim\limits_{x \to \infty} \frac{2x-3}{x^2-2x-5} = 0$
|
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|
Finding original function from composition of function If $f(f(x)) = x^2 + 2$, then find $f(11)$? Given that if $a>b$ then $f(a)>f(b)$
I got this question from a study group of which I am part of.
There the question was described as Let $x,f(x),a,b$ be positive integers and if $a>b$ then $f(a)>f(b)$ and $f(f(x)) = x^2 + 2$ then what is $f(11)$?
I tried by substituting $x= 1$ and $3$ and got $f(f(1)) = 3$ and $f(f(3))=11$ but don't know how to proceed further.
|
So the function has to be defined for positive integers. Then, we must have $f(x)>x$, because $f(x)<x$ would imply $f(f(x)<f(x)<x$, but $x^2+2>x$, and $f(x)=x$ is equally impossible. Replacing $x$ by $f(x)$ shows that $f(f(f(x))=f(x)^2+2=f(x^2+2)$. Since $x<f(x)<f(f(x))=x^2+2$, we must have $1<f(1)<1^2+2=3$, i.e. $f(1)=2$. Then, $f(2)=f(f(1))=1^2+2=3$, $f(3)=f(f(2))=2^2+2=6$ and $f(11)=f(3^2+2)=f(3)^2+2=38$.
|
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|
Dependent variable substitution of a differential equation. I am attempting to answer a question from a textbook. The question is as follows:
"Use the substitution $y = x^2$ to turn the differential equation $x\frac{d^2x}{(dt)^2} + (\frac{dx}{dt})^2 + x\frac{dx}{dt} = 0$ into a second order differential equation with constant coefficients involving y and t."
The answer according to the textbook is $\frac{d^2y}{(dt)^2} + \frac{dy}{dt} = 0$
I am unsure what to do with the $(\frac{dx}{dt})^2$ term in the original equation. Is it equivalent to $\frac{d}{dt}(x^2)$? If this is the case then my working gets as far as the following before I get stuck:
$x\frac{d^2x}{(dt)^2} + (\frac{dx}{dt})^2 + x\frac{dx}{dt} = 0$
$\Rightarrow x\frac{d^2x}{(dt)^2} + \frac{d}{dt}(x^2) + x\frac{dx}{dt} = 0$
$\frac{dy}{dt} =\frac{dy}{dx}\frac{dx}{dt}=2x\frac{dx}{dt} \Rightarrow \frac{d^2x}{(dt)^2}=\frac{d}{dx}(\frac{1}{2x}\frac{dy}{dt})$
$\Rightarrow x\frac{d}{dx}(\frac{1}{2x}\frac{dy}{dt}) + 2x + x\frac{1}{2x}\frac{dy}{dt} = 0$
$\Rightarrow \frac{1}{2x}\frac{dy}{dt} + \frac{1}{2}\frac{d^2y}{(dt)^2} + 2x + \frac{1}{2}\frac{dy}{dt} = 0$
Please can someone show me where I have gone wrong?
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In general, you are incorrect that $\left(\frac{dx}{dt}\right)^2 = \frac{d}{dt}\left(x^2\right)$ since $\frac{d}{dt}\left(x^2\right) = 2x\left(\frac{dx}{dt}\right)$, so they're equal only in the special case where $2x = \frac{dx}{dt}$. Instead, multiply the original equation on both sides by $2$ to get
$$2x\frac{d^2x}{(dt)^2} + 2\left(\frac{dx}{dt}\right)^2 + 2x\frac{dx}{dt} = 0 \tag{1}\label{eq1A}$$
Next, differentiating $y = x^2$ gives
$$\frac{dy}{dt} = 2x\left(\frac{dx}{dt}\right) \tag{2}\label{eq2A}$$
and then differentiating again, using the product rule this time, gives
$$\frac{d^2y}{(dt)^2} = 2\left(\frac{dx}{dt}\right)^2 + 2x\left(\frac{d^2x}{dt^2}\right) \tag{3}\label{eq3A}$$
As you can see, with the left side of \eqref{eq1A}, we have \eqref{eq3A} matching the first $2$ terms and \eqref{eq2A} matching the final third term. Thus, \eqref{eq1A} can be rewritten as
$$\frac{d^2y}{(dt)^2} + \frac{dy}{dt} = 0 \tag{4}\label{eq4A}$$
|
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|
Integral with the indicator function using spherical coordinates The integral of interest is:
$$I = \int_{\mathbb{R}^3}\int_{\mathbb{R}^3} \boldsymbol{1}\left(\frac{1}{2}\frac{(x_2^2 - x_1^2) + (y_2^2 - y_1^2) + (z_2^2 -z_1^2)}{x_2-x_1} \in [0,1]\right) \nonumber \\
\times \boldsymbol{1}\left(2\mathrm{arcsin}\left(\frac{\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2 + (z_2-z_1)^2}}{2\sqrt{z_1^2 + y_1^2 + \left(x_1 - \frac{1}{2}\frac{x_2^2 - x_1^2 + y_2^2 - y_1^2 + z_2^2 - z_1^2}{x_2-x_1}\right)^2}}\right) > \tau\right) \\ \times \exp\left(-C\left(z_1^2 + y_1^2 + \left(x_1 - \frac{1}{2}\frac{x_2^2 - x_1^2 + y_2^2 - y_1^2 +z_2^2 -z_1^2}{x_2-x_1}\right)^{2}\right)^{3/2}\right)\mathrm{d}Q_1\mathrm{d}Q_2,$$
where $C > 0$, $\tau \in (0, \pi]$, $\boldsymbol{1}(\cdot)$ is the indicator function, $Q_1 = (x_1, y_1, z_1)$ and $Q_2 = (x_2, y_2, z_2)$.
How does one solve this integral or at least simplify to have the expression with the least possible number of integrals?
My attempt: I tried the following transformations:
Substituting $t_1 = x_2 + x_1$, $t_2 = x_2 - x_1$, $t_3 = y_2 + y_1$, $t_4 = y_2 - y_1$, and $t_5 = z_2 + z_1$, $t_6 = z_2 - z_1$, and then eliminating the first indicator function, we get
$$I = \frac{1}{4}\int_{\mathbb{R}^5}\boldsymbol{1}\left(\mathrm{arcsin}\frac{\rho}{2 D} > \frac{\tau}{2}\right)\exp\left(-CD^3\right)\mathrm{d}t_2\mathrm{d}t_3\mathrm{d}t_4\mathrm{d}t_5\mathrm{d}t_6,$$
$\rho = \sqrt{t_2^2 + t_4^2 + t_6^2}$ and
$$D = \frac{1}{2}\sqrt{t_3^2 + t_5^2 + \frac{t_3^2t_4^2}{t_2^2}+\frac{t_5^2t_6^2}{t_2^2} + \frac{2t_3t_4t_5t_6}{t_2^2} + \rho^2}.$$
Now, we do the transformation to the spherical coordinates as $t_2 = \rho \sin \theta \cos \varphi$, $t_4 =\rho \sin \theta \sin \varphi$, and $t_6 = \rho\cos \theta$. Then, we have
$$D = \frac{1}{2}\sqrt{t_3^2\left(1 + \tan^2 \varphi\right) + t_5^2\left(1 + \frac{\cot^2 \theta}{\cos^2 \varphi}\right) + \frac{2t_3t_5 \cot \theta \tan \varphi}{\cos \varphi} + \rho^2}$$
and
$$\mathrm{d}t_2 \mathrm{d}t_4\mathrm{d}t_6 \mathrm{d}t_3 \mathrm{d}t_5 = \rho^2 \sin\theta \mathrm{d}\theta \mathrm{d}\varphi \mathrm{d}r\mathrm{d}t_3 \mathrm{d}t_5.$$
After doing the substitution, the resulting integration looks still difficult to further simplify.
|
You already made it to spherical coordinates. Using $r$ instead of $\rho$ and introducing polar coordinates for $t_3=\rho \cos \phi$ and $t_5=\rho \sin \phi$, the integral becomes $$I = \frac{1}{4} \int_0^{2\pi}\mathrm{d}\varphi\int_0^\pi\mathrm{d}\theta \int_0^\infty \mathrm{d}r\, r^2\sin \theta \int_0^{2\pi} \mathrm{d}\phi \int_0^\infty \rho \mathrm{d}\rho \boldsymbol{1}\left(\mathrm{arcsin}\frac{r}{2 D} > \frac{\tau}{2}\right)\exp\left(-CD^3\right)$$
where $$D=\frac{1}{2}\sqrt{\frac{1-\left(\sin\theta \sin\phi \sin\varphi - \cos\theta \cos\phi\right)^2}{\sin^2\theta \cos^2\varphi} \, \rho^2 + r^2} \equiv \frac{1}{2} \sqrt{a^2\rho^2+r^2} \, .$$
In order to carry out the $\rho$ integral, we replace the indicator function by bounds for $\rho$: $\rho=0$ leads to ${\rm arcsin}(1)=\frac{\pi}{2}>\frac{\tau}{2}$. The other bound follows from $$\frac{r^2}{a^2\rho^2+r^2}=\sin^2\frac{\tau}{2} \qquad \Rightarrow \qquad a\rho=r\cot\frac{\tau}{2} \, .$$
Hence the inner $\rho$ integral together with the $r$ integral becomes $$\int_0^\infty {\rm d}r \, r^2 \int_0^{\frac{r}{a}\cot \frac{\tau}{2}} {\rm d}\rho \, \rho \, e^{-\frac{C}{8}(a^2\rho^2+r^2)^{3/2}} \stackrel{u=\frac{C}{8}(a^2\rho^2+r^2)^{3/2}}{=} \frac{4}{3a^2C^{2/3}} \int_0^\infty {\rm d}r \, r^2 \int_{\frac{Cr^3}{8}}^{\frac{Cr^3}{8\sin^3\frac{\tau}{2}}} u^{-1/3} e^{-u} \, {\rm d}u \\ \stackrel{v=\frac{Cr^3}{8}}{=} \frac{32}{9a^2C^{5/3}} \int_0^\infty {\rm d}v \underbrace{\int_v^{\frac{v}{\sin^3\frac{\tau}{2}}} u^{-1/3} e^{-u} \, {\rm d}u}_{=f(v)} = \frac{32}{9a^2C^{5/3}} \left\{ v f(v) \Big|_0^\infty - \int_0^\infty v f'(v) \, {\rm d}v \right\} \\
=\frac{32}{9a^2C^{5/3}} \int_0^\infty {\rm d}v \left( v^{2/3} e^{-v} - \frac{v^{2/3}}{\sin^2 \frac{\tau}{2}} \, e^{-\frac{v}{\sin^3\frac{\tau}{2}}}\right) \\
\stackrel{\text{second term: }v\rightarrow v\sin^3\frac{\tau}{2}}{=} \frac{32}{9a^2C^{5/3}} \left( 1 - \sin^3 \frac{\tau}{2}\right) \int_0^\infty {\rm d}v \, v^{2/3} e^{-v} = \frac{32\,\Gamma(5/3)}{9a^2C^{5/3}} \left( 1 - \sin^3 \frac{\tau}{2}\right) \, .$$
It remains to calculate the angular integrals
$$I=\frac{8\,\Gamma(5/3)}{9\,C^{5/3}} \left(1-\sin^3\frac{\tau}{2}\right) \int_0^{2\pi} {\rm d}\varphi \int_0^\pi {\rm d}\theta \, \sin\theta \int_0^{2\pi} {\rm d}\phi \, \frac{1}{a^2} \\
=\frac{8\,\Gamma(5/3)}{9\,C^{5/3}} \left(1-\sin^3\frac{\tau}{2}\right) \int_0^{2\pi} {\rm d}\varphi \int_0^\pi {\rm d}\theta \, \sin\theta \int_0^{2\pi} {\rm d}\phi \, \frac{\sin^2\theta \cos^2\varphi}{1-\left(\sin\theta \sin\phi \sin\varphi - \cos\theta \cos\phi\right)^2} \\
=\frac{16\pi\,\Gamma(5/3)}{9\,C^{5/3}} \left(1-\sin^3\frac{\tau}{2}\right) \int_0^{2\pi} {\rm d}\varphi \int_0^\pi {\rm d}\theta \, \sin^2\theta \, |\cos \varphi| \\
=\frac{32\pi^2\,\Gamma(5/3)}{9\,C^{5/3}} \left(1-\sin^3\frac{\tau}{2}\right) \, .$$
|
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|
Show $\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx=\frac{5\pi^3}{64}+\frac\pi{16}\ln^22$ Tried to evaluate the integral
$$I=\int_0^\infty \frac{\ln^2x}{(x+1)^2+1} \, dx$$
and managed to show that
\begin{align}
I &= \int_0^1 \frac{\ln^2x}{(x+1)^2+1} \, dx + \int_0^1 \frac{\ln^2x}{(x+1)^2+x^2} \, dx\\
&= \int_0^1 \frac{\ln^2x}{(x+1+i)(x+1-i)} \, dx
+ \int_0^1 \frac{\ln^2x}{(x+1+ix )(x+1-ix )} \, dx\\
&= -2\operatorname{Im}\operatorname{Li}_3\left(-\frac{1+i}2\right)
-2\operatorname{Im} \operatorname{Li}_3(-1-i)
\end{align}
which is equal to $ \frac{5\pi^3}{64}+\frac\pi{16}\ln^22$.
It is perhaps unnecessary, though, to resort to evaluation in complex space. I would like to work out an elementary derivation of this integral result.
|
It is not bad if you write
$$(x+1)^2+1=(x-a)(x-b)$$
$$\frac 1{(x+1)^2+1}=\frac 1{a-b}\left(\frac 1{x-a}-\frac 1{x-b} \right)$$ making that you face two integrals
$$I_c=\int \frac {\log^2(x)}{x-c}\,dx=\log ^2(x) \log (x-c)-2\int \frac{ \log (x) \log (x-c)}{x}\,dx$$
$$I_c=-2 \text{Li}_3\left(\frac{x}{c}\right)+2 \log (x)
\text{Li}_2\left(\frac{x}{c}\right)+\log ^2(x) \log \left(1-\frac{x}{c}\right)$$
|
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|
Given $|b-c|\leq a\leq b+c$ show that $\frac{a}{1+a}\leq\frac{b}{1+b}+\frac{c}{1+c}$ I am trying to show the following:
Given that $|b-c| \leq a \leq b+c$ show that:
$$\frac{a}{1+a} \leq \frac{b}{1+b} + \frac{c}{1+c}$$
So far I have done the following:
$a/(1+a) \leq b/(1+b) + c/(1+c) \iff$
$1-1/(1+a) \leq 2 - 1/(1+b) -1/(1+c) \iff$
$1/(1+b) -1/(1+c) \leq 1 + 1/(1+a) \iff$
$ (2+b+c)/(1+b)(1+c) \leq (2+a)/(1+a) \iff$
$ (2+b+c)(1+a) \leq (2+a)(1+b)(1+c) \iff$
$ a \leq bc(2+a)$
I am not really sure how to proceed from here- any help or hints would be much appreciated.
|
I suppose $b,c\ge0$.
$$
\frac{b}{1 + b} + \frac{c}{1 + c} \ge \frac{a}{1 + a}
$$
$$
\Rightarrow \frac{b + c + 2bc}{1 + b + c +bc} \ge \frac{a}{1 + a}
$$
$$
\Rightarrow b + c + 2bc + ab +ca + 2abc\ge a + ab + ca + abc
$$
$$
\Rightarrow b + c + 2bc + abc \ge a
$$
Add
$$
b + c \ge a \quad \textrm{and} \quad bc(a +2)\ge0
$$
And we are done.
|
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|
Find the area between $r=1$ and $r=3\cos\theta$
Find the area between $r=1$ and $r=3\cos\theta$.
I squared both sides to get $r^2 = 1$, then did $r^2(\cos^2 \theta + \sin^2 \theta) = (r \cos \theta)^2 + (r \sin \theta)^2$$ = x^2+y^2 = 1$ to get $x^2+y^2=1$.
For $r = 3 \cos \theta$, I multiplied by $r$ on both sides to get $r^2 = 3r \cos \theta$, then substituted $x = r \cos \theta$ to get $x^2+y^2 =3x$. However, I don't know if it is easier to do it this way.
If not, how can I find this area?
|
Please always try and draw a sketch. It helps in ensuring you have the right bounds and also understand the easy way to find the area.
Now in the sketch, you can see that there are two circles -
i) Centered at $(0,0)$ with radius $1$
ii) Centered at $(3/2, 0)$ with radius $3/2$
As they are both symmetric to X-axis, they will intersect at the same angle in both first and fourth quadrant. Say that angle is $\alpha$. The Intersection points of both circles will be given by equating -
$3 \cos \alpha = 1$
$\alpha = cos^{-1} ({\frac{1}{3}}) \approx \frac{2\pi}{5}$ (I have taken as $2 \pi / 5$ but it is closer $1.231$. Use $1.231$ for more accurate area).
We have to find the area between two curves (thru points $O, A, B, C$).
a) Integrating the curve $r = 1$ over $\angle AOC$ will give us area bound by radii $OA, OC$ and arc $ABC$ (sector $OAC$ for circle $r = 1$).
b) Integrating the curve $r = 3 \cos \theta$ over angle between $Y$ axis and $OA$ and between $Y$ axis and $OC$ will give us area bound by chord $OA, OC$ and arc $AOC$.
If we add both, we get the area we desire.
a) $A_1 = \displaystyle \frac{1}{2} \int_{-\alpha}^{\alpha}d\theta = \frac{2\pi}{5}$
b) $A_2 = \displaystyle 2 \times \frac{1}{2} \int_{\alpha}^{\pi/2}(3 \cos \theta)^2 d\theta$
$ = \displaystyle \frac{9}{2} \int_{\alpha}^{\pi/2} 2 \cos^2 \theta \, d\theta$
$ = \displaystyle \frac{9}{2} \int_{\alpha}^{\pi/2} (1 + \cos2 \theta) \, d\theta$
$ = \displaystyle \frac{9}{2} [\theta + \frac{1}{2} \sin 2 \theta)]_{2\pi/5}^{\pi/2}$
$ = \displaystyle \frac{9}{2} [\frac{\pi}{2} - \frac{2\pi}{5} - \frac{1}{2}\sin \frac{4\pi}{5}]$
$A = A_1 + A_2 \approx 1.257 + 0.063 = 1.32$
|
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|
By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$
$\cos^6x + \sin^6x = -3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = (-3/2)(2\cos^2x\sin^2x)$
$\cos^6x + \sin^6x = (-3/2)(\sin^22x)$
How can I get it into $ 1 - (3/4)\sin^2(2x)$?
|
You start by writing
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$
but that's not a "sentence". It's just noun without a verb. What about $\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x\sin^4x) + \sin^6x$? what are you saying about it? Are you claiming it is equal to something? To what?
Then you go from this line
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$
which is also a noun without a verb to this line
$\cos^6x + \sin^6x = -3\cos^2x\sin^2x$
Which has a verb ("is equal to") and you moved the chunk "$(3\cos^2x\sin^2 x$" to the other side of the equation. But that assumes $\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$ had to have been equal to $0$ all along. Was $\cos^6x + (3\cos^2x\sin^2x) + \sin^6x= 0$?
I think you assumed that because you had the "noun" $\cos^6x + (3\cos^2x\sin^2x) + \sin^6x$ just sitting there by itself with nothing else when you moved "$(3\cos^2x\sin^2 x$" to "the other side" there was no other side (because you never made an equation sentence in the first place) you didn't think it matter.
In essence what you did was equivalent to the following:
$ 7$
$10- 3$
$10=3 $
Which .... would be true if we assume that $7=0$. But wouldn't be true otherwise.
So lets do this with full sentences.
We start with $(\cos^2 x + \sin^2 x)^3$. Will what about it. It's equal to something but to what? Well we know $(\cos^2 + \sin^2 x) = 1$ so:
$\cos^2 x + \sin^2 x = 1$
$(\cos^2 x + \sin^2 x)^3 = 1^3$
$\cos^6x + (3\cos^2x\sin^2x)(\cos^2x+\sin^2x) + \sin^6x = 1$
$\cos^6x + (3\cos^2x\sin^2x) + \sin^6x=1$
$\cos^6x + \sin^6x = 1-3\cos^2x\sin^2x$
Now we we can complete this but showing that $3\cos^2\sin^2 x = \frac 34 \sin 2x$.
And here made a second mistake. You figured that $sin^2 2x = 2\cos^2x\sin^2 x$
In actuality $\sin 2x = \sin(x+x) = \cos x\sin x + \sin x \cos x =$ \cos x \sin x$.
So $sin^2 2x = (2 \cos x \sin x) = 2^2 \cos^2 x \sin^2 x=4 \cos^2 sin^2 x$.
So....
$\cos^6x + \sin^6x = 1-3\cos^2x\sin^2x$
$\cos^6x + \sin^6x = 1-\frac 34(4\cos^2x\sin^2x)$
$\cos^6x + \sin^6x = 1 - \frac 34 \sin^2 x$.
|
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|
Collatz Conjecture: Does this argument show if a non-trivial cycle exists, the sum of powers of $2$ must be minimal power of $2 > 3^n$? What is wrong with this argument?
I am sure that I am misunderstanding something or there is a mistake in this argument. This argument is taken from the answer given to one of my questions about the Collatz Conjecture.
Let:
*
*$v_2(x)$ be the 2-adic valuation of $x$
*$C(x) = \dfrac{3x+1}{2^{v_2(3x+1)}}$
*$x_1>1, x_2>1, \dots, x_n>1$ be the sequence of $n$ distinct odd integers for each application of $C(x_i)$ so that for each $x_i$:
*
*for $i > 1$, $x_i = C(x_{i-1})$
*$x_i > 1$
*
*$x_{\text{min}}, x_{\text{max}}$ be the minimum and maximum value of $x_1, x_2, \dots, x_n$
*$C_1(x) = C(x)$
*$C_n(x) = C(C_{n-1}(x))$
Observations:
*
*$\left(3 + \dfrac{1}{x_{i-1}}\right) = \left(\dfrac{x_i}{x_{i-1}}\right)2^{v_2(3x_{i-1} + 1)}$
*
*$x_i = \dfrac{3x_{i-1}+1}{2^{v_2(3x_{i-1}+1)}}$
*$2^{v_2(3x_{i-1}+1)}x_i = 3x_{i-1} + 1$
*
*$\prod\limits_{k=1}^{n}\left(3 + \frac{1}{x_k}\right) = \left(\dfrac{x_{n+1}}{x_1}\right)\prod\limits_{k=1}^n2^{v_2(3x_k + 1)}$
This follows directly from the previous observation.
*
*$\left(3 + \dfrac{1}{x_{\text{max}}}\right)^{n} \le \left(\dfrac{x_{n+1}}{x_1}\right)\prod\limits_{k=1}^n2^{v_2(3x_k + 1)} \le \left(3 + \dfrac{1}{x_{\text{min}}}\right)^{n}$
This follows directly from the previous observation.
*
*if a non-trivial cycle exists, $n > 1$
$x = \dfrac{3x+1}{2^{v_2(3x+1)}}$ implies $x(2^{v_2(3x+1)} - 3) = 1$ which implies that $x=1$
Claim:
If there is a non-trivial cycle, the sum of the powers of $2$ in the cycle are the minimal integer power of $2$ greater than $3^n$
Argument:
(1) Assume that $x_1>1, x_2>1, \dots, x_n>1$ form an $n$-cycle such that:
*
*$x_i = C(x_{i-1})$
*$x_i = C_n(x_i)$ if $i \ge 1$
*Each $x_i$ is distinct. If $j < n$, $x_{i+j} \ne x_i$
(2) Let $m = \sum\limits_{k=1}^{n} v_2(3x_k + 1)$
(3) From the third observation and since each $x_i$ in the cycle is distinct and repeats:
$$2^m = \left(\dfrac{x_{\text{i+n}}}{x_{i}}\right)2^m < \left(3 + \dfrac{1}{x_{\text{min}}}\right)^{n}$$
(4) Assume that $2^{m-1} > 3^n$
(5) $2\times3^n < 2^m < \left(3 + \dfrac{1}{x_{\text{min}}}\right)^{n}$
(6) But then we have a contradiction because $x_{\text{min}} < 1$ which is impossible since all $x_i > 1$:
*
*$2^{\frac{1}{n}}3 < 3+ \dfrac{1}{x_{\text{min}}}$
*$x_{\text{min}}\left(3(2^{\frac{1}{n}} - 1)\right) < 1$
*$x_{\text{min}} < \dfrac{1}{3(2^{\frac{1}{n}} - 1)} < \dfrac{1}{3}$
|
You wrote in your last line
$x_{\text{min}} < \dfrac{1}{3(2^{\frac{1}{n}} - 1)} < \dfrac{1}{3}$
However, note for $n \gt 1$ that $2^{1/n} \lt 2 \implies 2^{1/n}
- 1 \lt 1$, so $3(2^{1/n} - 1) \lt 3$ and, thus, $\frac{1}{3\left(2^{1/n} - 1\right)} \gt \frac{1}{3}$. For example, $n = 10$ gives
$$\frac{1}{3\left(2^{0.1} - 1\right)} \approx 4.64 \tag{1}\label{eq1A}$$
Using
$$2^{1/n} = e^{\ln(2)(1/n)} \tag{2}\label{eq2A}$$
and the first few terms of the exponential Taylor series expansion, gives
$$\begin{equation}\begin{aligned}
\frac{1}{3\left(2^{1/n} - 1\right)} & = \frac{1}{3\left(\left(1 + \frac{\ln(2)}{n} + \frac{\ln(2)^2}{2n^2} + O\left(n^{-3}\right)\right) - 1\right)} \\
& = \frac{1}{3\left(\frac{\ln(2)}{n} + \frac{\ln(2)^2}{2n^2} + O\left(n^{-3}\right)\right)}\\
& = \frac{1}{3\left(\frac{\ln(2)}{n}\right)\left(1 + \frac{\ln(2)}{2n} + O\left(n^{-2}\right)\right)}\\
& = \frac{n}{3\ln(2)}\left(1 - \frac{\ln(2)}{2n} + O\left(n^{-2}\right)\right) \\
& = \frac{n}{3\ln(2)} - \frac{1}{6} + O\left(n^{-1}\right)
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
|
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|
If $n$ is odd, then $n/2 + 1/2$ is always even?
If $n$ is odd, prove that $n/2 + 1/2$ is even.
Context: I'm a Statistician and the term $n/2 + 1/2$ showed up in the index of a summation when deriving the pdf of some Order Statistic:
$$
\sum_{j = (n+1)/2}^{n}...
$$
I realized that $n/2 + 1/2$ is always even if $n$ is odd, but I couldn't prove the result to myself (well, I have no training in Number Theory).
What I've tried:
Suppose $n$ is odd. Then $n + 1$ is even (by the successor function?). Then $n + 1 = 2k$ for $k \in \mathbb{N} \Rightarrow (n+1)/2 = k $. But this doesn't show that $k$ is even.
|
If $n$ is odd then $n$ is of the form $2k - 1$.
The number $\frac n2 +\frac 12$ is simply $\frac {n+1}2 = \frac {2k-1 +1}2 = \frac {2k}2 = k$. The expression $\frac n2 + \frac 12$ is simply the the index of which odd number $n$ is.
If $n = 1,3,5,7,9,......, 2k -1 ,.....$ then $\frac n2 + \frac 12 = 1,2,3,4,5,.....,k,.....$ and $k$ is even only if $n$ is the in an even position on the list and odd if $n$ is in an odd position on the list.
....
There are $2$ types of odd numbers. Those where $n = 2k-1$ and $k$ is even, and those where $n=2k-1$ and $k$ is odd.
If $k = 2m$ is even then $n= 2(2m)-1 = 4m -1$ and $\frac n2 +\frac 1n = \frac {2k-1}2 + \frac 12 = k = 2m$ is even.
If $k = 2m -1$ then $n = 2(2m-1) -1 = 4m -3$ and $\frac n2 + \frac 1n = \frac {2k-1}2 +\frac 12= k = 2m-1$ is odd.
So $k$ will be even if $n$ is of the form $4m -1$ or $4m+3$ so for example if $n = 3,7,11,15,....$ then $\frac n2 + \frac 12$ will be even.
But $k$ will be odd if $n$ is of the form $4m -3$ or $4m + 1$ so for example if $n=1,5,9,13,17,..$ then $\frac n2 + \frac 12$ will be even.
|
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|
How to calculate combination with 2 variables? Combination formula with 1 variable is simple $nCr = n! / (r!(n-r)!)$
So to find a number of combinations for rolling a 6 ONCE out of 6 rolls is
$$\dfrac {6! }{ (1!(6-1)!)} = 6$$
But how do you find number of combinations for rolling a 6 ONCE AND rolling a 4 TWICE out of 6 rolls?
I can't just do 6 choose 3 because they're different events
Thank you
|
Assuming the die is fair, the probability of obtaining a 6, two fours, and three numbers other than 4 or 6 in that order is
$$\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)^2\left(\frac{4}{6}\right)^3$$
However, we have to multiply that number by the number of orders in which one 6, two 4s, and three numbers other than 4 or 6 could occur in six rolls, which is
$$\binom{6}{1, 2, 3} = \binom{6}{1}\binom{5}{2}\binom{3}{3} = \frac{6!}{1!5!} \cdot \frac{5!}{2!3!} \cdot \frac{3!}{3!0!} = \frac{6!}{1!2!3!}$$
since there are six possible positions for the six, two of the remaining five positions must be filled with 4s, and all three of the remaining three positions must be filled by a number other than 4 or 6. Hence, the desired probability is
$$\binom{6}{1, 2, 3}\left(\frac{1}{6}\right)\left(\frac{1}{6}\right)^2\left(\frac{4}{6}\right)^3$$
The number
$$\binom{6}{1, 2, 3}$$
is called a multinomial coefficient. Since the probabilities are the same for each roll of the die, this is a multinomial distribution problem.
|
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|
Find $\sqrt a + \sqrt b + \sqrt c$ only in terms of $p$ .
If :- $$a^2x^3 + b^2y^3 + c^2z^3 = p^5$$
$$ax^2 = by^2 = cz^2$$
$$\frac{1}{x} + \frac{1}{y} + \frac{1}{z} = \frac{1}{p}$$ Find $\sqrt a + \sqrt b + \sqrt c$ in terms of $p$ .
What I Tried :- How do you use the information here? I assumed that :-
$$ax^2 = by^2 = cz^2 = k$$
And then I can write the $1$st equation as :- $$k(ax + by + cz) = p^5$$
But I guess that just bring another extra variable into the scene so it does not help .
How can I use the $3$rd equation? I can write it as $\frac{xyz}{xy + yz + zx} = p$ . How do you use it?
|
proceeding from your method,$$a^2x^4=b^2y^4=c^2z^4=k^2$$
thus $a^2x^3+b^2y^3+c^2z^3=k^2(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=p^5$
$k=p^3$
can you end it now?
|
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"timestamp": "2023-03-29T00:00:00",
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|
find the largest integer $m$ such that $2^m$ divides $3^{2n+2}-8n-9$
find the largest integer $m$ such that $2^m$ divides $\space 3^{2n+2}-8n-9$ when $n$ is a natural number.
If the answer was known it will be easy induction.
I started out like this :
$\space 3^{2n+2}-8n-9=9(3^{2n}-1)-8n=9\underbrace{(3^n-1)(3^n+1)}-8n$
Now we have $\frac{3^n-1}{3-1}$ is some integer (sum of GP),or
$ 2|\space 3^n-1$
also we have $3^n+1$ is even ,or
$2|3^n+1....(3)$
From this we conclude $4|(3^n-1)(3^n+1) ...(1)$
Let n be even then $3^n-1=3^{2m}-1=(3^m-1)(3^m+1)$,
by $(1)$ :
$4|(3^m+1)(3^m-1)$ meaning $4|3^n-1...........(2)$
combining $(2),(3)$ we have $8|3^{2n+2}-8n-9$
Similarly i was able to work out the same when $n=2m+1$ by noting that $3^n+1=3^{2m+1}+1$ is divisible by $4$.
I got the largest integer as $3$.
But i am wrong as the MCQ did not have the option $m=3$
how do i proceed.
Note: I have not learnt about fermat's little theorem
Also i am looking for Hints rather than complete solutions .use of >! may help
|
Let $a_n=3^{2n +2}-8n -9$. Then the power series $f (z )=\sum_{k =0}^{\infty }a_k z^k$ can be written as
$$f(z)=\frac{b_0+b_1 z +b_2 z^2}{(1-c_0 z)(1-c_1 z)(1-c_2z)}$$
for some nonnegative integers $b_i$, $c_j$, and furthermore the common divisors of the $b_i$ are divisors of the $a_n$. Can you take it from there?
|
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|
$\frac{\tan8°}{1-3\tan^2 8°}+\frac{3\tan24°}{1-3\tan^2 24°} + \frac{9\tan72°}{1-3\tan^2 72°} + \frac{27\tan216°}{1-3\tan^2 216°}=x\tan108°+y\tan8°$ $$\frac {\tan 8°}{1-3\tan^2 8°}+\frac {3\tan 24°}{1-3\tan^2 24°} + \frac{9\tan 72°}{1-3\tan^2 72°} + \frac{27\tan 216°}{1-3\tan^2 216°} =x\tan108°+y\tan8° .$$
Find the value of $x$ and $y$.
I tried different approaches like using $$ \tan 3x=\frac{3\tan x-\tan^3 x}{1-3\tan^2 x}$$
and converting into $\sin$ and $\cos$, but I was unable to simplify.
Can anybody help?
|
$$\frac {\tan 8°}{1-3\tan^2 8°}+\frac {3\tan 24°}{1-3\tan^2 24°} + \frac{9\tan 72°}{1-3\tan^2 72°} + \frac{27\tan 216°}{1-3\tan^2 216°} =x\tan108°+y\tan8°$$
Now, $$\frac {\tan 8°}{1-3\tan^2 8°}=\frac{1}{8}\left(\frac {8 \tan 8°}{(1-3\tan^2 8°)}+\tan 8°-\tan8° \right) \\
=\frac{1}{8}\left(-\tan8° \right)+\frac{1}{8}\left(\frac {8 \tan 8°}{(1-3\tan^2 8°)}+\tan 8° \right) \\
=\frac{1}{8}\left(-\tan8° \right)+\frac{3}{8}(\tan 24°)$$
Solving in this way, we get the LHS as
$$\frac{81}{8}\tan(108^\circ)-\frac{1}{8}\tan(8^\circ)$$
From this, we get $$x+y=10$$
|
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|
Definite integrals of fractions of trig functions $$
\int _{\pi/4}^{\pi/3}\sin(x)\frac{\sin(x)+\cos^2(x)}{\sqrt{2}\sin(x)+\cos(x)}
\,\mathrm{d}x
$$
How do you solve for this integral?
I was considering using Weierstrass substitution or manipulating the integrand with respect to $\sec(x)$ or $\csc(x)$ but couldn't yield satisfactory results...
|
Using Weierstrass substitution, I suppose that you did arrive at
$$-4\int \frac{t \left(t^4+2 t^3-2 t^2+2 t+1\right) } {\left(t^2+1\right)^3 \left(t^2-2 \sqrt{2} t-1\right) }\,dt=-4\int \frac{t \left(t^4+2 t^3-2 t^2+2 t+1\right) } {\left(t^2+1\right)^3 \left(t+\sqrt 3-\sqrt 2\right)\left(t-\sqrt 3-\sqrt 2\right) }\,dt$$
Using partial fraction decomposition, the integrand is
$$\frac{-2 \sqrt{2} t-3 \sqrt{2}-4}{18 \sqrt{2} \left(t^2+1\right)}+\frac{\sqrt{2}
t-2 t+\sqrt{2}+4}{3 \sqrt{2} \left(t^2+1\right)^2}-\frac{\sqrt{2} \left(\sqrt{2}
t+2\right)}{3 \left(t^2+1\right)^3}+$$ $$\frac{3-2 \sqrt{3}}{36 \sqrt{3}
\left(-t-\sqrt{3}+\sqrt{2}\right)}+\frac{-3-2 \sqrt{3}}{36 \sqrt{3}
\left(-t+\sqrt{3}+\sqrt{2}\right)}$$ which seems to be more than practicable.
|
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|
Solve parameter function
Solve $4^{\sin{x}}+m×2^{\sin{x}}+m^2-1=0$ over parameter $m$
I dunno what to process but I want to show some of my work at least. I try to solve for $m$ firstly so I get
\begin{equation}
m_1=\frac{-2^{\sin{x}}+\sqrt{4-3×4^{\sin{x}}}}{2}\\m_2=\frac{-2^{\sin{x}}-\sqrt{4-3×4^{\sin{x}}}}{2}
\end{equation}
And then I try to find the range of $m_1$ and $m_2$ which give
\begin{equation}
\frac{-1+\sqrt{13}}{4}\geq m_1\geq -1\\ \frac{-1}{4}\geq m_2\geq \frac{-4-\sqrt{13}}{4}
\end{equation}
Because of $\Delta \geq 0$
Now I dunno what to do next. Please help
|
$4^{\sin{x}}+m×2^{\sin{x}}+m^2-1=0$
Say, $y = 2^{\sin x}$ and so we have $y^2 + my + (m^2-1) = 0$ ...(i)
For equation to have any real solution, its discrimant has to be $\ge 0$ i.e.
$m^2 - 4(m^2-1) \ge 0 \implies - \frac{2}{\sqrt 3} \le m \le \frac{2}{\sqrt 3}$ ...(ii)
Now solving (i), $ (y + \frac{m}{2})^2 = 1 - \frac{3m^2}{4} \implies y = -\frac{m}{2} \pm \sqrt {1 - \frac{3m^2}{4}}$ ...(iii)
$y = 2^{\sin x} \gt 0\,$ has a minimum value of $\frac{1}{2}$ (when $\sin x = -1$) and a max value of $2$ (when $\sin x = 1$).
That leads to further constraint on (ii) and $ - \frac{2}{\sqrt 3} \le m \le \frac{\sqrt 13 - 1}{4}$.
The upper bound comes from the fact that $y \ge \frac{1}{2}$. Further, any greater value of $m$ will mean modulus of $-\frac{m}{2}$ will increase (which is negative) whereas the value inside the square root will decrease. So $y$ will decrease which does not give a solution.
EDIT: Just some further remarks on the lower bound.
The lower bound $-\frac{2}{\sqrt 3}$ gives value of $y$ as $\frac{1}{\sqrt 3} (\gt \frac {1}{2})$ which works; $m = -\frac{\sqrt{13}+1}{4} (\gt - \frac{2}{\sqrt 3})$ gives $y = \frac {1}{2}$ but is not the lowest bound of $m$.
|
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|
Euclidean algorithm for greatest integers x and y for common divisor (GCD) I have a problem with finding the gcd of two numbers: gcd(4620, 8190) = 210.
I did the following:
8190 / 4620 = 1 with remainder: 3570
4620 / 3570 = 1 with remainder: 1050
3570 / 1050 = 3 with remainder: 420
1050 / 420 = 2 with remainder: 210
420 / 210 = 2 with remainder: 0
GDC = 210
So far so good, but I need to find x and y as integers that satisfy this condition:
4620x + 8190y
How can I achieve that? I did find that -9 and 16 satisfy this condition, but I don't know how to justify that.
Do I need to substitute the numbers in the steps from the algorithm?
|
$$ \frac{ 8190 }{ 4620 } = 1 + \frac{ 3570 }{ 4620 } $$
$$ \frac{ 4620 }{ 3570 } = 1 + \frac{ 1050 }{ 3570 } $$
$$ \frac{ 3570 }{ 1050 } = 3 + \frac{ 420 }{ 1050 } $$
$$ \frac{ 1050 }{ 420 } = 2 + \frac{ 210 }{ 420 } $$
$$ \frac{ 420 }{ 210 } = 2 + \frac{ 0 }{ 210 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccccc}
& & 1 & & 1 & & 3 & & 2 & & 2 & \\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 1 }{ 1 } & & \frac{ 2 }{ 1 } & & \frac{ 7 }{ 4 } & & \frac{ 16 }{ 9 } & & \frac{ 39 }{ 22 }
\end{array}
$$
$$ $$
$$ 39 \cdot 9 - 22 \cdot 16 = -1 $$
$$ \gcd( 8190, 4620 ) = 210 $$
$$ 8190 \cdot 9 - 4620 \cdot 16 = -210 $$
well, there you go
|
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|
Summing the first $n$-terms of the series whose general term is $nx^{n-1}$ I suppose several of you know some fancy ways to establish the formula for the sum of the first $n$ terms of the geometric series $$1+x+x^{2}+x^{3}+ \ldots $$
Can you share below some of your fave ways to sum the first $n$ terms of the series
$$ 1 + 2x+ 3x^{2} + \ldots $$
Naturally, for $x \neq 1$, one way to attain the goal is to take the derivative of both sides of $$\sum_{k=0}^{n-1} x^{k} = \frac{x^{n}-1}{x-1}.$$
Another possibility is to apply something that the authors of certain handbook of discrete and combinatorial math called "the perturbation method": if $S(n):= \sum_{k=1}^{n} kx^{k-1}$, then
$$ S(n) + (n+1)x^{n} = 1 + \sum_{k=2}^{n+1} kx^{k-1}$$
which can be rewritten as
$$ S(n) + (n+1)x^{n} = 1 + \sum_{k=1}^{n}(k+1)x^{k}.$$
Thus,
$$ (1-x)S(n) = \frac{x(x^{n}-1)}{x-1} + 1 - (n+1)x^{n}$$
or
$$S(n) = \frac{nx^{n+1}-(n+1)x^{n}+1}{(x-1)^{2}}.$$
What other cool ways to sum $1+2x+ \cdots + nx^{n-1}$ do you know or have heard of? Have you ever seen a "proof without words" of the result in question?
Thanks in advance for your attention.
|
$1 + 2x + 3x^2 + \cdots + nx^{n-1}$
= $1 + x + x^2 + \cdots + x^{n-1} $
$+ x + x^2 + \cdots + x^{n-1} $
$+ x^2 + \cdots + x^{n-1} $
$+ \cdots + x^{n-1}$
$= \dfrac{x^n-1}{x-1}$
$+\dfrac{x^n-x}{x-1}$
$+\dfrac{x^n-x^2}{x-1}$
$+\cdots + \dfrac{x^{n}-x^{n-1}}{x-1}$
$=\dfrac{nx^n-(1+x+x^2+\cdots+x^{n-1})}{x-1}$
$=\dfrac{nx^n-\dfrac{x^n-1}{x-1}}{x-1}$
|
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|
If $g(x)=\frac{P(x)}{x^2-4}$, $\lim_{x\rightarrow\infty}g(x)=5$, and $\lim_{x\rightarrow2}g(x)=\frac{9}{4}$, find P(x) (Sweden 1950) If $g(x)=\frac{P(x)}{x^2-4}$, $\lim_{x\rightarrow\infty}g(x)=5$, and $\lim_{x\rightarrow2}g(x)=\frac{9}{4}$, find P(x)
I have been trying to do this question but I have not succeeded. I have attempted to do it in the following way:
$\lim_{x\rightarrow\infty} \frac{P(x)}{x^2-4}=\lim_{x\rightarrow\infty}\frac{P'(x)}{2x}$ (L'Hospital rule)
$=\lim_{x\rightarrow\infty} \frac{P''(x)}{2}$
Hence we have that $P''(x)=10$.
I also attempted to do something with $\lim_{x\rightarrow2}g(x)=\frac{9}{4}$ however I didn't manage to find anything useful. After a lot of trial and error, I found that an equation which fits the given parameters is: $P(x)=5x^2-11x+2$. Could you please explain to me how to do this question and also explain the intuitive reason behind each step of your reasoning?
|
You have $g(x) = \frac{P(x)}{x^2-4}$ and you want to find $P$.
First condition about $\lim_{x \to \infty}$ gives us the rank of $P$ together with coefficient. Indeed, if $P$ is arbitrary polynomial $a_nx^n + ... + a_1x + a_0$ where $a_n \neq 0$ then
$$ \lim_{x \to \infty} \frac{P(x)}{x^2-4} = \begin{cases} 0 & n \in \{0,1\} \\ a_n & n = 2 \\ sign(a_n)\cdot \infty & n \ge 3 \end{cases} $$
Hence we need $n=2$ for limit to be finite, and moreover $a_2=5$.
Hence $P(x) = 5x^2 + bx + c$
The second condition tells us something about $\lim_{x \to 2}$, the point where denominator is $0$, so that (hence limit is finite), numerator needs to be $0$, too. It means that $2$ is a root of $P$, so that $P(x)=5(x-2)(x-d)$
And lastly, the value of that limit as $x\to 2$ gives us:
$$ \frac{9}{4} = \lim_{x \to 2} \frac{5(x-2)(x-d)}{(x-2)(x+2)} = \lim_{x \to 2}\frac{5(x-d)}{(x+2)} = \frac{5(2-d)}{4}$$
From there $2-d = \frac{9}{5}$ and $d = 2-\frac{9}{5} = \frac{1}{5}$
Hence $P(x) = 5(x-2)(x-\frac{1}{5}) = (x-2)(5x-1) = 5x^2 -11x + 2$
|
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Advice on integrating $\int \frac{x^2 +n(n-1)}{(x \sin x +n \cos x)^2} dx$ Our integral is:
$$\int \frac{x^2 +n(n-1)}{(x \sin x + n\cos x)^2} dx$$
I considered that this can be turned into some sort of quotient differential:
$$ d \frac{u}{v} = \frac{ v du - u dv}{v^2}$$
Now, comparing this with our integral:
$$ v = x \sin x + \cos x$$
And,
$$ dv = \sin x + x \cos x - n \sin x = (1-n) \sin x + x \cos x$$
Now the problem is I can't figure out $u$ / make the numerator of the form $ vdu - u dv$... what do I do next?
|
$$I=\int \frac{x^2+n(n-1)}{(x\sin x+n \cos x)^2} dx$$
Multiply up and down by $x^{2n-2} \cos x$, then
$$I=\int \frac{(x^2+n(n-1))x^{2n-2} \cos x dx}{(x^n \sin x+n x^{n-1} \cos x)^2 \cos x}$$
Let $$(x^n \sin x+n x^{n-1} \cos x)=t \implies x^{n-2} \cos x(x^2+n(n-1)) dx=dt$$
$$\implies I=\int \frac{(x^2+n(n-1))x^{n-2} \cos x }{(x^n \sin x+n x^{n-1} \cos x)^2} x^n \sec x dx$$
Do integration by part taking $x^n \sec x$ as first and remaining as second function, then
$$I=-\frac{x \sec x}{(x\sin x+n \cos x)}+\int \sec^2 x dx$$
$$I=-\frac{x \sec x}{x \sin x+ n \cos x)}+\tan x+C$$
|
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|
prove that $\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$
prove that $$\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$$ for positives $a,b,c$
Attempt: By C-S; $$\left(\sum_{cyc}\frac{a}{b^2+c^2} \right) \left(\sum_{cyc} a(b^2+c^2) \right)\ge {(a+b+c)}^2$$ .
or as inequality is homogenous we take $a+b+c=1$.
or we have to prove (i am skipping the steps as it is just algebra) :
$$ 5(ab+bc+ca-abc)\ge 4(1+ab+bc+ca)(ab+bc+ca-3abc)$$
But i am not able to prove this by expanding.
How do i proceed?
Other methods are welcome!
|
Also, $uvw$ helps here very well.
Indeed, let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Thus, after homogenization your last inequality it's $f(w^3)\geq0,$ where $f$ is a linear function,
which says that it's enough to prove it for the extreme value of $w^3$,
which by $uvw$ happens in the following cases.
*
*$w^3\rightarrow0^+$.
Let $c\rightarrow0^+$.
Thus, we need to prove that:$$5ab\geq4(1+ab)ab$$ or $$ab\leq\frac{1}{4},$$ which is true by AM-GM:
$$ab\leq\left(\frac{a+b}{2}\right)=\frac{1}{4}.$$
*Two variables are equal.
Let $b=a$ and $c=1-2a$, where $0<a<\frac{1}{2}.$
After this substitution we need to prove that: $$a(1-2a)(1-a)(18a^2-3a+1)\geq0,$$ which is obvious.
|
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|
Generating function of Trinomial Coefficients
Let $f(n)=\sum_{k=0}^{\lfloor\frac{n}{2}\rfloor}{n\choose k}{n-k\choose k}$. Show that $\sum_{n=0}^{\infty}f(n)x^n=\frac{1}{\sqrt{(1-3x)(1+x)}}$.
I managed to find the following recurrence relation for the trinomial coefficients in Wolfram: $(n+2)a_{n+2}=(2n+3)a_{n+1}+3(n+1)a_n$.
Now using $\sum_{n=0}^{\infty}(n+2)a_{n+2}x^n=\sum_{n=0}^{\infty}(2n+3)a_{n+1}x^n+\sum_{n=0}^{\infty}3(n+1)a_nx^n \implies \frac{dG}{G}=\frac{dx(x+3x^2)}{1-2x-3x^2}$ I can easily solve it except:
Question: How do I prove the relation $(n+2)a_{n+2}=(2n+3)a_{n+1}+3(n+1)a_n$?
|
Here is a direct computation of the generating function, using the binomial series:
\begin{align*}
\sum_{n=0}^\infty f(n)x^n\quad
&=\sum_{n=0}^\infty x^n\sum_{0\leqslant 2k\leqslant n}\frac{n! }{k!^2(n-2k)!}
=\sum_{k=0}^\infty\sum_{n=2k}^\infty\frac{n! \ x^n}{k!^2(n-2k)!}
\\\color{gray}{[\text{replace $n$ with $n+2k$}]}\quad
&=\sum_{k=0}^\infty\sum_{n=0}^\infty\frac{(n+2k)!}{k!^2 n!}x^{n+2k}
=\sum_{k=0}^\infty\binom{2k}{k}x^{2k}\sum_{n=0}^\infty\binom{n+2k}{2k}x^n
\\\color{gray}{[\text{use binomial series}]}\quad
&=\sum_{k=0}^\infty\binom{2k}{k}x^{2k}(1-x)^{-2k-1}
=\frac{1}{1-x}\sum_{k=0}^\infty\binom{2k}{k}\left(\frac{x}{1-x}\right)^{2k}
\\\color{gray}{[\text{... and one more time}]}\quad
&=\frac{1}{1-x}\left(1-\frac{4x^2}{(1-x)^2}\right)^{-1/2}
=\quad\bbox[2pt,border:2pt solid]{\begin{matrix}\text{expected}\\\text{result}\end{matrix}}
\end{align*}
(the last step uses $\sum_{k=0}^\infty\binom{2k}{k}z^k=(1-4z)^{-1/2}$, another instance of the binomial series).
Another way is to use $\delta_{mn}$${}=\frac{1}{2\pi}\int_{-\pi}^\pi e^{i(m-n)t}\,dt$ and the multinomial theorem: $$f(n)=\frac{1}{2\pi}\sum_{\substack{n_1,n_2,n_3\geqslant 0\\n_1+n_2+n_3=n}}\frac{n!}{n_1!n_2!n_3!}\int_{-\pi}^\pi e^{i(n_2-n_3)t}\,dt=\frac{1}{2\pi}\int_{-\pi}^\pi(1+e^{it}+e^{-it})^n\,dt,$$ giving $\displaystyle\sum_{n=0}^\infty f(n)x^n=\frac1\pi\int_0^\pi\frac{dt}{1-x(1+2\cos t)}$ which is evaluated easily.
|
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Determine whether a recurrence relation converges and, if yes, find its limit. I have the sequence defined as $$ a_{n+1} = \dfrac{5a_n -6}{a_n -2} $$ for $n \geq 2$, $a_n \neq 2$ for all $n \geq 1$ and $a_n$ is a real number.
I want to determine whether the sequence converges and if it does, find it's limit.
There are no initial conditions given and also I can't think of any way to solve the recurrence relation.
One thing I considered is the following:
If the sequence converges then
$$ \lim a_{n+1} =\lim a_n = L $$
Then substitute in the recurrence relation to get a polynomial whose solutions are $a=1$,or $a=6$ and take cases for $a_1$, but I couldn't end up with the solution.
Finally I couldn't find a pattern for the sequence either so I am stuck. How can I solve this?
|
Let's experiment with some values of $n$.
\begin{align}
a_2 &= \frac{5a_1 - 6}{a_1 - 2} \\
a_3 &= \frac{19a_1 - 18}{3a_1 - 2} \\
a_4 &= \frac{77a_1 - 78}{13a_1 - 14} \\
a_5 &= \frac{307a_1 - 306}{51a_1 - 50}.
\end{align}
The coefficient of $a_1$ in the numerator generates the sequence $5, 19, 77, 307, \dots$ and the coefficient of $a_1$ in the denominator generates the sequence $1,3,13,51,\dots$.
Notice that
\begin{align}
5\cdot 4 - 1 &= 19 \\
19 \cdot 4 + 1 &= 77 \\
77 \cdot 4 - 1 &= 307 \\
\vdots
\end{align} and
\begin{align}
1\cdot 4 - 1 &= 3 \\
3 \cdot 4 + 1 &= 13 \\
13 \cdot 4 - 1 &= 51 \\
\vdots
\end{align} Let's express one of the terms from one of our sequences, say $307$, using the above pattern $$307 = ((5 \cdot 4 - 1)(4) + 1)(4) - 1 = 5 \cdot 4^3 - 4^2 + 4^1 - 4^0.$$ Given that $307$ is the coefficient of $a_1$ in the numerator of the expression for $a_5$, it looks like the coefficient of $a_1$ in the numerator of the expression for $a_n$ is given by
\begin{align}
\alpha_n :&= 5\cdot 4^{n-2} + \sum_{k = 0}^{n - 3}(-1)^{n-k}4^k = \frac{6\cdot 4^{n-1} + (-1)^n}{5}.
\end{align} Similarly, it looks like the coefficient of $a_1$ in the denominator of the expression for $a_n$ is given by $$\beta_n := \frac{4^{n-1} + (-1)^n}{5}.$$ The constant in the numerator and the denominator appear to alternate between the negative of one greater than and one less than the coefficient of $a_1$ in the numerator and the denominator, respectively, and so we conjecture that
\begin{align}
a_n &= \frac{\alpha_n a_1 - (\alpha_n + (-1)^n)}{\beta_n a_1 - (\beta_n + (-1)^n)} = \frac{(6\cdot 4^{n-1} + (-1)^n)(a_1 - 1) - 5(-1)^n}{(4^{n-1} + (-1)^n)(a_1 - 1) - 5(-1)^n}
\end{align} and prove by induction.
Keep in mind that $\beta_n a_1 - (\beta_n + (-1)^n)$ cannot equal zero and, hence, $a_1$ cannot equal $\frac{\beta_n + (-1)^n}{\beta_n}$ (e.g., $n = 2$ $\implies$ $a_1$ cannot equal $2$ and $n = 3$ $\implies$ $a_1$ cannot equal $\frac{2}{3}$).
In summary, if $a_1 = 1$, then $a_n = 1$ for all $n \geq 1$ and if $a_1 \neq 1$ and $a_1 \neq \frac{\beta_n + (-1)^n}{\beta_n}$, then $a_n \rightarrow 6$ as $n \rightarrow \infty$.
|
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|
Computing $\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz$ using Cauchy integral formula Let $\alpha(t) = re^{it}$ where $|a|<r<|b|$ and $t \in [0,2\pi]$. I'd like to compute
$$\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz \ \ \ \ n, m \in \mathbb{N}.$$
It appears that the answer is
$$2\pi i (-1)^m{n + m -2 \choose n-1}\frac{1}{(b-a)^{n+m-1}}.$$
I try to compute it but I'm not sure how that's the final answer. Here is my attempt:
Let $f(z) = \frac{1}{(z-b)^m}$, then by Cauchy integral formula
$$\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz = \int_{\alpha}\frac{f(z)}{(z-a)^n}dz = \frac{2\pi i}{(n-1)!}f^{(n-1)}(a).$$
Trying to find an expression of $f^{(n-1)}(z)$
\begin{align*}
f^{(1)} &= -m\frac{1}{(z-b)^{m-1}}\\
f^{(2)} &= m(m-1)\frac{1}{(z-b)^{m-2}}\\
f^{(3)} &= -m(m-1)(m-2)\frac{1}{(z-b)^{m-3}}\\
&\vdots\\
f^{(n-1)} &= (-1)^{n-1}m(m-1)(m-2)\dotsc(m-n + 2)\frac{1}{(a-b)^{m-n+1}}
\end{align*}
I think this should be correct. So,
\begin{align*}
\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz &= \frac{2\pi i}{(n-1)!}(-1)^{n-1}m(m-1)(m-2)\dotsc(m-n + 2)\frac{1}{(a-b)^{m-n+1}}\\
&= \frac{2\pi i}{(n-1)!}(-1)^{2n-2-m}m(m-1)(m-2)\dotsc(m-n + 2)\frac{1}{(b-a)^{m-n+1}}\\
&=\frac{2\pi i}{(n-1)!}(-1)^{m}m(m-1)(m-2)\dotsc(m-n + 2)\frac{1}{(b-a)^{m-n+1}}.
\end{align*}
But the exponent on $(b-a)$ is not the same. Further, I have no clue how I can get the binomial coefficient. Writing out the binomial coefficient in the given answer doesn't seem to help me either.
$${n + m -2 \choose n-1} = \frac{(n+m-2)!}{(n-1)!(m-1)!} = \frac{(n+m-2)(n+m-3)\dotsc 2\cdot 1}{(n-1)!(m-1)!} $$
NOTE: I haven't learned residue theorem yet.
|
Answering my own question, thanks to the comments.
Let $f(z) = \frac{1}{(z-b)^m}$, then by Cauchy integral formula
$$\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz = \int_{\alpha}\frac{f(z)}{(z-a)^n}dz = \frac{2\pi i}{(n-1)!}f^{(n-1)}(a).$$
Trying to find an expression of $f^{(n-1)}(z)$
\begin{align*}
f^{(1)} &= -m\frac{1}{(z-b)^{m+1}}\\
f^{(2)} &= m(m+1)\frac{1}{(z-b)^{m+2}}\\
f^{(3)} &= -m(m+1)(m+2)\frac{1}{(z-b)^{m+3}}\\
&\vdots\\
f^{(n-1)} &= (-1)^{n-1}m(m+1)(m+2)\dotsc(m+n - 2)\frac{1}{(a-b)^{m+n-1}}
\end{align*}
So,
\begin{align*}
\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz &= \frac{2\pi i}{(n-1)!}(-1)^{n-1}m(m+1)(m+2)\dotsc(m+n - 2)\frac{1}{(a-b)^{m+n-1}}\\
&= \frac{2\pi i}{(n-1)!}(-1)^{n-1-m-n+1}m(m+1)(m+2)\dotsc(m+n - 2)\frac{1}{(b-a)^{m+n-1}}\\
&=\frac{2\pi i}{(n-1)!}(-1)^{-m}m(m+1)(m+2)\dotsc(m+n - 2)\frac{1}{(b-a)^{m+n-1}}\\
&=\frac{2\pi i}{(n-1)!}(-1)^{m}m(m+1)(m+2)\dotsc(m+n - 2)\frac{1}{(b-a)^{m+n-1}}.
\end{align*}
Note that
$${n + m -2 \choose n-1} = \frac{(n+m-2)(n+m-3)\dotsc m}{(n-1)!}.$$
Hence,
$$\int_{\alpha}\frac{1}{(z-a)^n(z-b)^m}dz = 2\pi i(-1)^m{n + m -2 \choose n-1}\frac{1}{(b-a)^{m+n-1}}.$$
|
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|
On odd perfect numbers $q^k n^2$ and the deficient-perfect divisor $q^{\frac{k-1}{2}} n^2$ - Part II (This question is an offshoot of this earlier post.)
Let $\sigma(x)$ be the sum of divisors of the positive integer $x$. Denote the deficiency of $x$ by $D(x)=2x-\sigma(x)$, and the sum of the aliquot divisors of $x$ by $s(x)=\sigma(x)-x$. Denote the abundancy index of $x$ by $I(x)=\sigma(x)/x$.
If $\sigma(m)=2m$ and $m$ is odd, then $m$ is called an odd perfect number. It is currently unknown whether there are any odd perfect numbers, although it is widely believed that there is none.
Euler proved that an odd perfect number $m$, if one exists, must have the so-called Eulerian form
$$m = q^k n^2$$
where $q$ is the special/Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Since the divisor sum function $\sigma$ is a multiplicative function and $m = q^k n^2$ is perfect, we obtain
$$2 q^k n^2 = 2m=\sigma(m)=\sigma(q^k n^2)=\sigma(q^k)\sigma(n^2)$$
which implies that
$$2 = \dfrac{\sigma(m)}{m} = \dfrac{\sigma\Bigg({q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}\Bigg)}{{q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}} \leq \dfrac{\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)}{q^{\dfrac{k+1}{2}}}\cdot\dfrac{\sigma\Bigg({q^{\dfrac{k-1}{2}}}{n^2}\Bigg)}{q^{\dfrac{k-1}{2}}{n^2}} = I\bigg(q^{\dfrac{k+1}{2}}\bigg)I(d),$$
where
$$d = q^{\dfrac{k-1}{2}}{n^2}.$$
We now compute
$$\dfrac{d}{D(d)} = \dfrac{q^{\dfrac{k-1}{2}}{n^2}}{2q^{\dfrac{k-1}{2}}{n^2} - \dfrac{\bigg(q^{\dfrac{k+1}{2}} - 1\bigg)\sigma(n^2)}{q - 1}} = \dfrac{q^{\dfrac{k-1}{2}}{n^2}}{2q^{\dfrac{k-1}{2}}{n^2} - \dfrac{\sigma(q^k)\sigma(n^2)}{q^{\dfrac{k+1}{2}} + 1}} = \dfrac{q^{\dfrac{k-1}{2}}{n^2}}{2q^{\dfrac{k-1}{2}}{n^2} - \dfrac{2q^k n^2}{q^{\dfrac{k+1}{2}} + 1}},$$
which simplifies to
$$\dfrac{d}{D(d)} = \dfrac{q^{\dfrac{k+1}{2}} + 1}{2\Bigg(q^{\dfrac{k+1}{2}} + 1\Bigg) - 2\Bigg(q^{\dfrac{k+1}{2}}\Bigg)} = \dfrac{q^{\dfrac{k+1}{2}} + 1}{2}.$$
This confirms a theorem of Holdener and Rachfal. (That is, the proper divisor
$$d = q^{\dfrac{k-1}{2}}{n^2}$$
of an odd perfect number
$$m = q^k n^2$$
must be deficient-perfect.)
So now we have the inequality
$$\dfrac{2}{I\bigg(q^{\dfrac{k+1}{2}}\bigg)} \leq I(d),$$
which implies that
$$\dfrac{D(d)}{d} = 2 - I(d) \leq 2 - \dfrac{2}{I\bigg(q^{\dfrac{k+1}{2}}\bigg)},$$
from which it follows that
$$\dfrac{q^{\dfrac{k+1}{2}} + 1}{2} = \dfrac{d}{D(d)} \geq \dfrac{I\bigg(q^{\dfrac{k+1}{2}}\bigg)}{2\Bigg(I\bigg(q^{\dfrac{k+1}{2}}\bigg) - 1\Bigg)}.$$
We therefore have to solve the resulting inequality
$$q^{\dfrac{k+1}{2}} + 1 \geq \dfrac{I\bigg(q^{\dfrac{k+1}{2}}\bigg)}{I\bigg(q^{\dfrac{k+1}{2}}\bigg) - 1} = \dfrac{\dfrac{q^{\dfrac{k+3}{2}} - 1}{q^{\dfrac{k+1}{2}} \bigg(q - 1\bigg)}}{\Bigg(\dfrac{q^{\dfrac{k+3}{2}} - 1}{q^{\dfrac{k+1}{2}} \bigg(q - 1\bigg)}\Bigg) - 1}$$
which implies that
$$\Bigg(q^{\dfrac{k+1}{2}} + 1\Bigg)\cdot\Bigg(q^{\dfrac{k+1}{2}} - 1\Bigg) \geq q^{\dfrac{k+3}{2}} - 1$$
from which it follows that
$$q^{k+1} - 1 \geq q^{\dfrac{k+3}{2}} - 1 \implies q^{k+1} \geq q^{\dfrac{k+3}{2}} \implies q^{2k+2} \geq q^{k+3} \implies 2k+2 \geq k+3 \implies k \geq 1.$$
INQUIRY
Is it possible to improve on the approach/attack as outlined in this post to hopefully produce a nontrivial lower bound for $k$?
|
Too long to comment :
You used an inequality
$$\sigma\Bigg({q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}\Bigg)\le \sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}{n^2}\Bigg)$$
This inequality is true, but we can have
$$\sigma\Bigg({q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}\Bigg)$$
$$=\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}{n^2}\Bigg)-\frac{q\Bigg(q^{\dfrac{k-1}{2}}-1\Bigg)\Bigg(q^{\dfrac{k+1}{2}}-1\Bigg)}{(q-1)^2}\sigma(n^2)$$
since we have
$$\begin{align}&\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}{n^2}\Bigg)-\sigma\Bigg({q^{\dfrac{k+1}{2}}\bigg(q^{\dfrac{k-1}{2}}{n^2}\bigg)}\Bigg)
\\\\&=\sigma\Bigg({q^{\dfrac{k+1}{2}}}\Bigg)\sigma\Bigg({q^{\dfrac{k-1}{2}}}\Bigg)\sigma({n^2})-\sigma(q^k)\sigma(n^2)
\\\\&=\Bigg(\frac{q^{\dfrac{k+3}{2}}-1}{q-1}\cdot\frac{q^{\dfrac{k+1}{2}}-1}{q-1}-\frac{q^{k+1}-1}{q-1}\Bigg)\sigma(n^2)
\\\\&=\frac{\Bigg(q^{\dfrac{k+3}{2}}-1\Bigg)\Bigg(q^{\dfrac{k+1}{2}}-1\Bigg)-(q^{k+1}-1)(q-1)}{(q-1)^2}\sigma(n^2)
\\\\&=\frac{q\Bigg(q^{\dfrac{k-1}{2}}-1\Bigg)\Bigg(q^{\dfrac{k+1}{2}}-1\Bigg)}{(q-1)^2}\sigma(n^2)\end{align}$$
|
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|
find function with condition using Leibniz okay, so we have $f:(0, +\infty) \to (0, +\infty)$, with $f(1)=2$, such as $f'(\frac{1}{x})=\frac{1}{f(x)}$.
Now, I know we're supposed to use Leibniz's binomial theorem. And the answer to the problem is $$f(x)=(1+x)e^\frac{1-x}{2+2x}$$
Further than that, I've tried working with the condition but i'm getting nowhere everytime and the answer in the textbook has no other suggestions. Any help is welcome at this point.
|
Hint:
If
$$
f'\left(\frac{1}{x}\right)=\frac{1}{f(x)}
$$
then
$$
f'(x)=\frac{1}{f\left(\frac{1}{x}\right)}
$$
and
\begin{align}
\frac{d}{dx} \left(
f(x)f\left(\frac{1}{x}\right)\right)&=f'(x) f\left(\frac{1}{x}\right)-\frac{f(x)f'\left(\frac{1}{x}\right)}{x^2}\\
&=1-\frac{1}{x^2}\ .
\end{align}
Therefore, by integrating this equation from $1$ to $\ x\ $, we get
\begin{align}
f(x) f\left(\frac{1}{x}\right)-f(1)^2&= f(x) f\left(\frac{1}{x}\right)-4\\
&=x+\frac{1}{x}-2\ ,\\
\end{align}
—that is,
\begin{align}x+\frac{1}{x}+2&=f(x) f\left(\frac{1}{x}\right)\\
&=\frac{f(x)}{f'(x)}\ ,\\
\end{align}
or
\begin{align}
\frac{x}{(1+x)^2}&= \frac{f'(x)}{f(x)}\\
&=\frac{d}{dx}\ln f(x)\ .
\end{align}
Can you finish it off from here?
|
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|
Find value of $\sec80^\circ-2\cos20^\circ-\dfrac{4}{\sec20^\circ}+\dfrac{1}{2\sin10^\circ}$ My attempt :
\begin{align*}
\dfrac{1}{\cos80^\circ}-2\cos20^\circ-4\cos20^\circ+\dfrac{1}{\sin10^\circ}\\
\left(\dfrac{1}{\sin10^\circ}+\dfrac{1}{2\sin10^\circ}\right)-6\cos20^\circ\\
\dfrac{3}{2\sin10^\circ}-6\cos20^\circ
\end{align*}
This seem like the value wouldn't be an integer, but after typing in WolframAlpha it returned 3
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Use Werner Formulas, $2\sin B\cos A=\sin(A+B)-\sin(A-B)$
$$\dfrac{3(1-4\sin10^\circ\cos20^\circ)}{2\sin10^\circ}$$
$$=\dfrac{3(1-2(\sin(20+10)^\circ-\sin(20-10)^\circ))}{2\sin10^\circ}$$
$$=?$$
Generalization:
$$\dfrac1{2\sin x}-2\cos y=\dfrac{1-2(\sin(x+y)+\sin(x-y))}{2\sin x}$$
We need $\sin(x+y)=\dfrac12$
and $\sin(x-y)=\pm\sin x\iff0=\sin^2x-\sin^2(x-y)=\sin(2x-y)\sin y$
Either $\sin y=0\implies \sin(x+180^\circ n)=\dfrac12, x=?$
Or $\sin(2x-y)=0\implies y=2x+m180^\circ\implies\dfrac12=\sin(3x+m180^\circ), x=?$
where $m,n$ are arbitrary integers
Here $x=10^\circ, m=0$
|
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|
How many real roots does $f(x) = 2x^5 - 3x^4 + x^3 - 4x^2 - 6x + 4$ have? A multiple choice questions asks me to find the number of real roots of $f(x) = 2x^5 - 3x^4 + x^3 - 4x^2 - 6x + 4$.
A brute force approach involving the factor theorem and algebraic long division gives $3$ real roots.
Is there a quicker way?
I was thinking Descartes rule of signs tells us that since there are $4$ sign changes, there are at most 4 positive roots. We can also use the rule on $f(-x)$ to say there is at most one negative root. However, I can't see how to continue along this path.
I tried the derivative and deduced that there at least two turning points on the curve $y=f'(x) = 10x^4 - 12x^3 + 3x^2 - 8x - 6$, since $10 > 0$ and the $y$ intercept is negative. But that doesn't rule out anything since the turning points could be above the axis.
Of course I could work this out in more detail, but I am looking for efficiency.
|
Using factorization it's very quickly here:
$$2x^5 - 3x^4 + x^3 - 4x^2 - 6x + 4=$$
$$=2x^5+2x^4-5x^4-5x^3+6x^3+6x^2-10x^2-10x+4x+4=$$
$$=(x+1)(2x^4-5x^3+6x^2-10x+4)=$$
$$=(x+1)(2x^4-4x^3-x^3+2x^2+4x^2-8x-2x+4)=$$
$$=(x+1)(x-2)(2x^3-x^2+4x-2)=(x+1)(x-2)(2x-1)(x^2+2),$$
which gives three real roots.
|
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|
Using differentiation to find the power series of a fairly tricky function!! (b) Use part (a) to find a power series for: $$f(x) = \frac{1}{(8+x)^3}$$
What is its radius of convergence?
From Using differentation to find a power series expression for a function We know that
$$\frac{1}{(8+x)^2}=\sum_{n=0}^\infty (-1)^{n} (n+1) x^{n} \frac{1}{8^{n+2}}$$
We do some reverse engineering to see that
$$\frac{1}{(8+x)^3}=\frac{-1}{2} \frac{d}{dx}\frac{1}{(8+x)^2}$$
Plugging in our answer from part (a):
$$= \frac{-1}{2} \frac{d}{dx}\sum_{n=0}^\infty (-1)^{n} (n+1) x^{n} \frac{1}{8^{n+2}}$$
Factoring out $\frac{1}{8^2}$ from our sum:
$$= \frac{-1}{2\cdot8^2} \frac{d}{dx}\sum_{n=0}^\infty (-1)^{n} (n+1) \left(\frac{x}{8}\right)^n$$
Taking the derivative yields:
$$= \frac{-1}{2\cdot8^2} \sum_{n=1}^\infty (-1)^{n} (n+1)n \left(\frac{x}{8}\right)^{n-1}\left(\frac{1}{8}\right)$$
The term when $n=0$ is $0$ when we take the derivative, so now our summations starts at $n=1$.
But we want it to start at $n=0$, so we reindex by adding $1$ to everywhere $n$ is:
$$= \frac{-1}{2\cdot8^2} \sum_{n=1}^\infty (-1)^{n+1} (n+2)(n+1) \left(\frac{x}{8}\right)^{n}\left(\frac{1}{8}\right)$$
Now let's pull that $\frac{-1}{8^2}$ back into the sum and combine all the $\frac{1}{8}$ terms:
$$= \frac{1}{2} \sum_{n=1}^\infty (-1)^{n+2} (n+2)(n+1) x^n \left(\frac{1}{8}\right)^{n+3}$$
Notice that the negative sign we brought in made the is what made the $(-1)^{n+1}$ become $(-1)^{n+2}$. Doing a ratio test will show that the radius of convergence is again $8$.
|
Looks good to me.
Maybe it's "nicer" to write it as $$f(x)=\sum_{n=0}^{\infty}(-1)^n(n+1)(n+2)\frac{x^n}{2^{3n+10}}$$
|
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|
Define $f(x) = x^6 + x^5 + 3x^4 +x^3 + 3x^2 + x + 1$. Find the largest prime factor of $f(19) + 1$ (Homework) Define $f(x) = x^6 + x^5 + 3x^4 +x^3 + 3x^2 + x + 1$. Find the largest prime factor of $f(19) + 1$ This problem is from a homework set of my class at source: Alphastar.academy. I believe there a number of ways to factor this and solve it, and I would appreciate it if I were able to see a couple methods on how to do this problem.
|
Conveniently, $f(x)+1$ factors as
$$(x^2-x+1)(x^2+x+1)(x^2+x+2)$$
With $x=19$ this produces the three factors $343×381×382$, from which we work out that the largest prime factor is $191$ (of $382$).
|
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|
Initial value problem with separation of variables
I have this initial value problem. Also I am a beginner on the topic.
As far as I know, I should separating the variables and integrate and with using initial condition I should have get rid off from the constant(C).
But the problem is, the equation kinda looks scary and messy.
Is there a way to verify that If the equation is separable or not ?
If not, how do I proceed ?
|
$$x^3y'-3x^2y=\dfrac {2x^7(4x^3-3y)}{3x^5+3x^3+2y}$$
Substitute $w=\dfrac y {x^3}$:
$$\left(\dfrac y {x^3}\right)'=\dfrac {2x(4-3y/x^3)}{3x^2+3+2y/x^3}$$
$$w'=\dfrac {2x(4-3w)}{3x^2+3+2w}$$
Then take $v=x^2+1$
$$\dfrac {dw}{dx}=\dfrac {2x(4-3w)}{3x^2+3+2w}$$
$$\dfrac {dw}{dx^2+1}\dfrac {dx^2+1}{dx}=\dfrac {2x(4-3w)}{3x^2+3+2w}$$
$$2x\dfrac {dw}{dx^2+1}=\dfrac {2x(4-3w)}{3x^2+3+2w}$$
$$\dfrac {dw}{dv}=\dfrac {4-3w}{3v+2w}$$
$$({3v+2w}) {dw}-(4-3w)dv=0$$
$$2w {dw}+3dwv-4dv=0$$
Integrate and unsubstitute $v,w$:
$$w^2+3wv-4v=C$$
$$\dfrac {y^2} {x^6}+\dfrac {3y(x^2+1)} {x^3}-4(x^2+1)=C$$
$$y(1)=1 \implies C=-1$$
$$y^2 +{3y(x^2+1)} {x^3}-4x^8-3x^6=0$$
|
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|
If $x$ and $y$ are two linearly independent column $n$-vectors how can I find all the eigenvalues of $xx^{T}-yy^{T}$ If $x$ and $y$ are two linearly independent column $n$-vectors where $n\geq2$ .find all the eigenvalues of $xx^{T}-yy^{T}$
I know that because the matrix $xx^T-yy^T$ has rank $2$. So $n-2$ of the eigenvalues are $0$, and the other two eigenvectors have to lie in the column space of $xx^T-yy^T$, which is $\text{span}\{x,y\}$.
I supposed $z = \alpha x + \beta y$ is an eigenvector of $xx^T-yy^T$ for some constants $\alpha$ and $\beta$ ,but I can’t find $\alpha$ and $\beta$ such that $(xx^T-yy^T)z = \lambda z$
|
Using the identity
\begin{align*}
\lambda^n\det(\lambda I_{(m)} - AB) = \lambda^m\det(\lambda I_{(n)} - BA)
\end{align*}
for $A \in F^{m \times n}$ and $B \in F^{n \times m}$,
we can calculate the characteristic polynomial of $xx^T - yy^T$ by setting $A = (x, y) \in F^{n \times 2}$ and $B = (x^T, -y^T)^T \in F^{2 \times n}$ directly as:
\begin{align*}
\varphi(\lambda) &= \det(\lambda I_{(n)} - (xx^T - yy^T)) =
\lambda^{n - 2}\det\left(\lambda I_{(2)} - \begin{pmatrix} x^T \\ -y^T \end{pmatrix}\begin{pmatrix} x & y \end{pmatrix}\right) \\
&=
\lambda^{n - 2}\begin{vmatrix}
\lambda - x^Tx & -x^Ty \\
y^Tx & \lambda + y^Ty
\end{vmatrix} \\
&= \lambda^{n - 2}[(\lambda - x^Tx)(\lambda + y^Ty) + (x^Ty)^2] \\
&=
\lambda^{n - 2}(\lambda^2 - (x^Tx - y^Ty)\lambda - (x^Txy^Ty - (x^Ty)^2))
\end{align*}
Since $x$ and $y$ are linearly independent, by Cauchy-Schwarz inequality $(x^Tx)(y^Ty) > (x^Ty)^2$ (that is, the equality of C-S inequality cannot hold), whence the determinant $\Delta$ of the quadratic equation $\lambda^2 - (x^Tx - y^Ty)\lambda - (x^Txy^Ty - (x^Ty)^2) = 0$ equals to
\begin{align*}
\Delta = (x^Tx - y^Ty)^2 + 4(\|x\|^2\|y\|^2 - (x^Ty)^2) > 0.
\end{align*}
Hence the two non-zero eigenvalues are two distinct real numbers
\begin{align*}
\lambda_1 = \frac{y^Ty - x^Tx + \sqrt{\Delta}}{2}, \quad
\lambda_2 = \frac{y^Ty - x^Tx - \sqrt{\Delta}}{2}.
\end{align*}
|
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|
The process of using taylor series to evaluate limits. For instance we want evaluate this simple limit using taylor series :
$$L=\lim_{x\to 0}\frac{\sin x}{x^5}=\lim_{x\to 0}\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+\cdots}{x^5}$$
In this case we only care about coefficient of $x^5$ to cancel out $x^5$s in the fraction and get $L=\frac{1}{5!}$. This is a simple example of what we usually do when evaluating limit using taylor series.
My question is why we ignore other terms of the numerator? I mean:
$$L=\lim_{x\to0}\color{red}{\frac{1}{x^4}-(\frac{1}{3!}\times\frac{1}{x^2})}+\color{green}{\frac{1}{5!}}\color{red}{-(\frac{1}{7!}\times x^2)+(\frac{1}{9!}\times x^4)+\cdots}$$
I don't understand why we can ignore other terms (showed in red color) and confirm $\frac1{5!}$ as the answer. It is obvious that terms appeared on the rightside of $\color{green}{\frac1{5!}}$ are equal to zero but what about other side?
|
We can't ignore the other terms, we obtain indeed
$$\frac{x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+\cdots}{x^5}=\frac1{x^4}-\frac1{3!x^2}+\frac1{5!}+O(x^2) \to \infty$$
the limit is finite for
$$\lim_{x\to 0}\frac{\sin x-\left(x-\frac{x^3}{3!}\right)}{x^5}=\lim_{x\to 0}\frac{\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^9}{9!}+O(x^{11})}{x^5}=\frac1{5!}$$
|
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|
Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
What I Tried: Here is a picture :-
I know the centroid divides each of the medians in the ratio $2:1$ . So $AD = 3\sqrt{3}$ , $BE = 3\sqrt{2}$ , $CF = 3$ .
From this site :- https://mathworld.wolfram.com/TriangleMedian.html, I find that the area of the triangle will be :- $$\frac{4}{3}\sqrt{s_m(s_m - m_1)(s_m - m_2)(s_m - m_3)}$$
Where $m_1,m_2,m_3$ are the medians of the triangle and $s_m = \frac{m_1 + m_2 + m_3}{2}$ .
After putting the respective values for the medians I get that $[\Delta ABC]$ is :-
$$\frac{4}{3}\sqrt{\Bigg(\frac{3(\sqrt{3} + \sqrt{2} + 1)}{2}\Bigg)\Bigg(\frac{3(\sqrt{2} + 1 - \sqrt{3})}{2}\Bigg)\Bigg(\frac{3(\sqrt{3} + 1 - \sqrt{2})}{2}\Bigg)\Bigg(\frac{3(\sqrt{3} + \sqrt{2} - 1)}{2}\Bigg)}$$
$$\rightarrow \frac{4}{3}\sqrt{\frac{81(\sqrt{3} + \sqrt{2} + 1)(\sqrt{2} + 1 - \sqrt{3})(\sqrt{3} + 1 - \sqrt{2})(\sqrt{3} + \sqrt{2} - 1)}{16}}$$
I am almost to the answer (assuming I made no mistake), but I think this simplification is getting complicated. How do I proceed next?
Can anyone help me?
|
I think, it's better to use $$S_{\Delta ABC}=\sqrt{p(p-a)(p-b)(p-c)}=$$
$$=\frac{1}{4}\sqrt{(a+b+c)(b+c-a)(a+c-b)(a+b-c)}=\frac{1}{4}\sqrt{\sum_{cyc}(2a^2b^2-a^4)}$$ because from the given easy to get $a^2$, $b^2$ and $c^2$.
Indeed, $$\frac{1}{3}\sqrt{2b^2+2c^2-a^2}=2\sqrt3,$$
$$\frac{1}{3}\sqrt{2a^2+2c^2-b^2}=2\sqrt2$$ and
$$\frac{1}{3}\sqrt{2a^2+2b^2-c^2}=2,$$
which gives $$\frac{1}{3}(a^2+b^2+c^2)=4(3+2+1)$$ or
$$a^2+b^2+c^2=72,$$ which gives
$$2(72-a^2)-a^2=108$$ or $$a^2=12.$$
By the similar way we obtain: $b^2=24$ and $c^2=36$, which gives $$S_{\Delta ABC}=6\sqrt2.$$
|
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|
How to solve the integral $\int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x$? I encountered a very difficult problem, to calculate the answer of this formula:
$$
\int \frac{1}{x^{8}\left(1+x^{2}\right)} \ \mathrm{d} x
$$
Can you help me to find out how it solved?
|
Using partial fraction decomposition (left as an exercise for the reader), notice that $$\frac{1}{x^8(x^2 + 1)} = \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1}.$$ Therefore, $$\int \frac{1}{x^8(x^2 + 1)} ~ dx = \int \left[ \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1} \right] ~ dx.$$ From here, one can notice that the first $4$ terms have simple anti-derivatives, and the anti-derivative of the last term is $\arctan x.$ One can do a sanity check by computing the derivative of $\arctan x$ to see that it is indeed the case. Therefore, we have that $$\int \left[ \frac{1}{x^8} - \frac{1}{x^6} + \frac{1}{x^4} - \frac{1}{x^2} + \frac{1}{x^2 + 1} \right] ~ dx = \frac{1}{x} - \frac{1}{3x^3} + \frac{1}{5x^5} - \frac{1}{7x^7} + \arctan x + C.$$
|
{
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|
Show that three numbers form an arithmetic progression
The numbers $a,b$ and $c$ form an arithmetic progression. Show that the numbers $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ also form an arithmetic progression.
We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2}\text{ } \forall \text{ }n\ge2$). I am stuck here and I would be very grateful if you could give me a hint.
|
Expressions like $ \ a^2+ab+b^2 \ , \ a^2+ac+c^2 \ , \ b^2+bc+c^2 \ \ $ make me think of "differences of two cubes", so let's see what happens. If we call the progression $ \ a \ = \ b - d \ , \ b \ , \ c \ = \ b + d \ \ , $ then we may write
$$ \ a^2 \ + \ ab \ + \ b^2 \ \ = \ \ \frac{b^3 \ - \ a^3}{b - a} \ \ = \ \ \frac{b^3 \ - \ (b - d)^3}{ d} \ \ = \ \ \frac{3·b^2d \ - \ 3·bd^2 \ + \ d^3}{d} \ \ , $$
$$ \ a^2 \ + \ ac \ + \ c^2 \ \ = \ \ \frac{c^3 \ - \ a^3}{c - a} \ \ = \ \ \frac{(b + d)^3 \ - \ (b - d)^3}{2d} \ \ = \ \ \frac{6·b^2d \ + \ 2·d^3}{2d} \ \ , $$
$$ \ b^2 \ + \ bc \ + \ c^2 \ \ = \ \ \frac{c^3 \ - \ b^3}{c - b} \ \ = \ \ \frac{(b + d)^3 \ - \ b^3}{d} \ \ = \ \ \frac{3·b^2d \ + \ 3·bd^2 \ + \ d^3}{d} \ \ . $$
The expressions in this sequence simplify to
$$ 3·b^2 \ - \ 3·bd \ + \ d^2 \ \ , \ \ 3·b^2 \ + \ d^2 \ \ , \ \ 3·b^2 \ + \ 3·bd \ + \ d^2 \ \ , $$
which differ sequentially by a constant amount $ \ 3·bd \ \ . $
|
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|
Prove by induction $2\left(n+1\right)\leq\left(n+2\right)^{2}$ Prove by induction $2\left(n+1\right)\leq\left(n+2\right)^{2}$
Case $S(1)$ is true:
$$2((1)+2)\leq((1)+2)^{2}$$
$$6\leq9$$
Case $S(n)$ is true for all $n=1,2,...$
$$2(n+2)\leq(n+2)^{2}(i)$$
Case $S\left(n+1\right)$
$$2(n+3)\leq(n+3)^{2}(ii)$$
From (i)
$$2(n+2+1-1)\leq(n+2+1-1)^{2}$$
$$2(n+3)-2(1)\leq(n+3)^{2}-2(n+3)+1$$
$$2(n+3)\leq(n+3)^{2}-2(n+3)+3$$
$$2(n+3)\leq(n+3)^{2}-(2n+3)$$
$$2(2n+3)+3\leq(n+3)^{2}$$
Thus $(i)$ is true for all $n=1,2,...$
My question: How can I get the same expression as (ii)?
I got to $2(2n+3)+3\leq(n+3)^{2}$ but it's clearly wrong
|
Note that$$2(n+2)-2(n+1)=2$$and that$$(n+3)^2-(n+2)^2=2n+5.$$So, if $2(n+1)\leqslant(n+2)^2$, you have\begin{align}2(n+2)&=2(n+1)+2\\&\leqslant(n+2)^2+2\\&\leqslant(n+2)^2+2n+5\\&=(n+3)^2.\end{align}
|
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|
Integrate $\int_{[0,1]}\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}d\lambda(x) d\lambda(y)$ This is question from Axler's Measure, Integration, & Real Analysis Problem 5.B.1, probability measures.
Let $\lambda$ denote Lebesgue measure on $[0,1]$. Show that
$$\int_{[0,1]}\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}d\lambda(x) d\lambda(y) = \frac{\pi}{4}$$ and
$$\int_{[0,1]}\int_{[0,1]}\frac{x^2-y^2}{(x^2+y^2)^2}d\lambda(y) d\lambda(x) = -\frac{\pi}{4}.$$
How does one calculate this integral? Since the answer is $\dfrac{\pi}{4}$ is there is a trig substitution involved?
I also graphed this function and don't understand what the meaning of the graph is from a geometric point of view.
|
Note that $$\frac {x^2-y^2}{(x^2+y^2)^2} = \frac{\partial}{\partial y} \left(\frac{y}{x^2 + y^2}\right) = -\frac{\partial}{\partial x} \left(\frac{x}{x^2 + y^2}\right), $$
Riemann and Lebesgue integrals coincide on finite intervals when the integrand is Riemann itegrable.
We can evaluate the inner integral as a Riemann integral using FTC to get for $y \in (0,1]$,
$$F(y) = \int_{[0,1]}\frac {x^2-y^2}{(x^2+y^2)^2}\, d\lambda(x)= \int_0^1 \frac {x^2-y^2}{(x^2+y^2)^2} \, dx = \left.\frac{-x}{x^2 + y^2}\right|_0^1= \frac{-1}{1+y^2}$$
Now $F$ is integrable over $(0,1]$ and easily evaluated as
$$\int_{[0,1]} F(y) \, d\lambda(y) = \int_0^1\frac{-1}{1+y^2} \, dy = -\arctan(1) = - \frac{\pi}{4}$$
|
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|
Solving $\lim_{x\to 0}\left(\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}\right)$ without series expansion or L'Hopital's rule I want to find the following limit without using series expansion or L'Hopital's rule. I tried replacing $x$ with $2x$.
$$\begin{aligned}L_{1}&=\lim_{x\to 0}\left(\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}\right)\end{aligned}$$
$$\begin{aligned}L_{2}&=\lim_{x\to 0}\left(\frac{4x+2x(\cos^2x-\sin^2x)-6\sin x\cos x}{32\sin x\cos x}\right)\end{aligned}$$
How to proceed. Any hints are appreciated. Thanks.
|
HINT
To simplify the derivation we can proceed starting to prove the followings
$$L_1: \frac{\cos x-1}{x^2}\to \frac12 \implies L_2: \frac{x\cos x-x+\frac12x^3}{x^5}\to \frac1{24}$$
$$L_3:\frac{x-\sin x}{x^3}\to \frac16 \implies L_4:\frac{3(x-\sin x)-\frac12 x^3}{x^5}\to -\frac1{40}$$
and then use that
$$\frac{2+\cos x}{x^3\sin x}-\frac{3}{x^4}=\frac{x}{\sin x}\left(\frac{x\cos x-x+\frac12x^3}{x^5}+\frac{3(x-\sin x)-\frac12 x^3}{x^5}\right)$$
Refer to the related
*
*Are all limits solvable without L'Hôpital Rule or Series Expansion
|
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|
Find $\int_0^bg(x,u)\,du$ given $\int_0^b (u^2+(b^2-1)u+(b-1)^2)g(x,u)du=\frac{x^2+b^2x+b^2-2b}{x-1}.$ I encounter the following equation, which holds for any $x\in[0,b]$:
$$\int_0^b (u^2+(b^2-1)u+(b-1)^2)g(x,u)du=\frac{x^2+b^2x+b^2-2b}{x-1}.$$
I would like to solve for $G(x,b)=\int_0^b g(x,u)du$, subject to the condition that $G(x,0)=0$. Intuitively, $G(x,b)$ is the column sum of an (infinite dimensional) symmetric matrix $g$ with entries given by the $g(x,y)$ function for $x,y\in[0,b]$, and the integral on the l.h.s. says that pre-multiplying the $x$-th column of $g$ by the vector $u^2+(b^2-1)u+(b-1)^2$ yields a known function of column sum in terms of $x$.
It seems that integration by parts won't really simplify the problem as the integral of $G$ then enters into the problem. I am not sure whether this question is solvable at all, and frankly, I do not even know where to start with. Maybe I have missed something very simple. Any hints or suggestions are highly appreciated.
|
We have \begin{align}\int_0^b (u^2+(b^2-1)u+(b-1)^2)g(x,u)\,du&=\frac{x^2+b^2x+b^2-2b}{x-1}\tag1\\\int_0^b (u^2+(b^2-1)u+(b-1)^2)g_x(x,u)\,du&=1-\frac{2b^2-2b+1}{(x-1)^2}\tag2\\\int_0^b (u^2+(b^2-1)u+(b-1)^2)g_{xx}(x,u)\,du&=\frac{2(2b^2-2b+1)}{(x-1)^3}\end{align} so assume that $(x-1)^3g_{xx}(x,u)$ is constant. Then $g_{xx}(x,u)=t(b)/(x-1)^3$ where \begin{align}t(b)=\frac{2(2b^2-2b+1)}{\int_0^bu^2+(b^2-1)u+(b-1)^2\,du}=\frac{12(2b^2-2b+1)}{b(3b^3+8b^2-15b+6)}\end{align} so $$g_x(x,u)=h_1(u)-\frac{t(b)}{2(x-1)^2}\implies g(x,u)=h_2(u)+xh_1(u)+\frac{t(b)}{x-1}$$ where $h_1,h_2$ can be determined (with considerable freedom) using $(1)$ and $(2)$.
|
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|
Elementary Ways to Solve System of Exponential Equation Is there any elementary way (or using Lambert-W maybe) to solve this system of the exponential equation:
$$ \begin{cases}
3^{x+y}+2^{y-1}=23, \\
3^{2x-1}+2^{y+1}=43.
\end{cases} $$
I have tried to eliminate the exponent of 2 but it gets me
$$ 12 \cdot 3^{x + y} + 3^{2x} = 405 $$
which is more complicated.
I have also tried to substitute $ 3^x = u $ and $ 2^y = v $ but there is still $ 3^y $.
Any advice is welcome (it's okay to use non-elementary method). Thanks :)
|
Hint:
$$(3^x)^2+(12\cdot3^y)3^x-405=0$$
The discriminant is $$(12\cdot3^y)^2+4\cdot405=16\cdot3^{2y+2}+3^4\cdot20=4\cdot3^2(4\cdot9^y+45)$$
For rational $3^x,$ we need $$(2\cdot3^y)^2+45$$ to be perfect square $=d^2, d\ge0$(say)
$$\implies45=d^2-(2\cdot3^y)^2=(d+2\cdot3^y)(d-2\cdot3^y)\le(d+2\cdot3^y)^2$$
$$\implies d+2\cdot3^y\ge\sqrt{45}>6$$
Again, $d+2\cdot3^y$ must divide $45,$ hence can be one of $$\{9,15,45\}$$
From here we can find $3^y$ and $d$ and hence $3^x$
|
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|
sum of binomial coefficients expansion to prove equation I want to prove that
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \binom{n+1}{3}
$$
I already expanded
$$
\binom{n+1}{3}
$$
to
$$
\binom{n+1}{3} = \frac{1}{6} * (n+1) *n*(n-1)
$$
and I know that the following equation must be right
$$
\sum_{i=1}^{n}{\binom{i}{2}} = \frac{1}{6} * (n+1) *n*(n-1)
$$
but I do not get the expansion right, I tried starting with writing the sum explicit
$$
\binom{1}{2}+\binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}
$$
since 1 over 2 is zero it can be shortened to
$$
\sum_{i=2}^{n}{\binom{i}{2}} = \binom{2}{2}+\ldots+\binom{n-1}{2}+\binom{n}{2}
$$
then I expanded the binomial coefficients to the corresponding factorial form
$$
\binom{n}{k} = \frac{n!}{k! * (n-k)!}
$$
but I do not get it right, could someone please help me?
|
$\begin{array}\\
\binom{n+1}{3}-\binom{n}{3}
&=\dfrac{(n+1)n(n-1)}{6}-\dfrac{n(n-1)(n-2)}{6}\\
&=\dfrac{n(n-1)((n+1)-(n-2))}{6}\\
&=\dfrac{3n(n-1)}{6}\\
&=\dfrac{n(n-1)}{2}\\
&=\binom{n}{2}\\
\end{array}
$
Therefore
$\sum_{i=1}^{n}\binom{i}{2}
=\sum_{i=1}^{n}(\binom{i+1}{3}-\binom{i}{3})
=\binom{n+1}{3}
$.
Note that
$\binom{n}{m} = 0$
for $m > n$.
In general,
since
$\binom{n}{m}
=\dfrac{\prod_{k=0}^{m-1}(n-k)}{m!}
$,
$\begin{array}\\
\binom{n+1}{m}-\binom{n}{m}
&=\dfrac{\prod_{k=0}^{m-1}(n+1-k)}{m!}-\dfrac{\prod_{k=0}^{m-1}(n-k)}{m!}\\
&=\dfrac{\prod_{k=-1}^{m-2}(n-k)}{m!}-\dfrac{\prod_{k=0}^{m-1}(n-k)}{m!}\\
&=\dfrac{(n+1)\prod_{k=0}^{m-2}(n-k)-(n-m+1)\prod_{k=0}^{m-2}(n-k)}{m!}\\
&=\dfrac{((n+1)-(n-m+1))\prod_{k=0}^{m-2}(n-k)}{m!}\\
&=\dfrac{m\prod_{k=0}^{m-2}(n-k)}{m!}\\
&=\dfrac{\prod_{k=0}^{m-2}(n-k)}{(m-1)!}\\
&=\binom{n}{m-1}\\
\end{array}
$
so that
$\sum_{i=1}^{n}\binom{i}{m-1}
=\sum_{i=1}^{n}(\binom{i+1}{m}-\binom{i}{m})
=\binom{n+1}{m}
$.
|
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|
Show $f(x,y)=\frac{x^3\sin(x-y)}{x^3-y^3}$ for $x\neq y$, $f(x,y)=0$ for $x=y$ is continuous at $(0,0)$. I know that once we bound $f$ for $x\neq y$ we are done because $f=0$ for $x=y$. I got that for $x\neq y$, $|f(x,y)|\le \left|\frac{x^3}{x^2+xy+y^2}\right|$ because $\left|\frac{\sin(a)}{a}\right|\leq 1$ for $a\neq 0$. But I don't know how to continue.
|
so following @Kelenner's hint, $x^2+xy+y^2 = \frac{x^2+y^2}{2} +\frac{(x+y)^2}{2} \geq \frac{x^2+y^2}{2}$, in particular it must be positive, and $|f(x,y)|\leq |\frac{x^3}{x^2+xy+y^2}| \leq |\frac{2x^3}{x^2+y^2}| \leq |\frac{2x^3}{x^2}| =2|x|$, which tends to $0$ as $(x,y)\rightarrow(0,0)$.
We can also say that $|f(x,y)|\leq |\frac{x^3}{x^2+xy+y^2}| \leq |\frac{2x^3}{x^2+y^2}| \leq \frac{2x^2(x^2+y^2)^{\frac{1}{2}}}{x^2+y^2} \leq 2x^2(x^2+y^2)^{\frac{1}{2}}$ which also tends to $0$.
|
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|
Evaluate $\int_{-\frac{1}{\sqrt 3}}^{\frac{1}{\sqrt 3}} \frac{x^4}{1-x^4} \cos^{-1} (\frac{2x}{1+x^2}) dx$ Note that the integral can(not) be simplified as
$$
2\int_0^{1/\sqrt 3} \frac{x^4}{1-x^4} \cos^{-1} \left(\frac{2x}{1+x^2}\right)dx
$$
Since $\cos^{-1}$ is not an even function. Let $x=\tan y$
$$
\implies
\int_0^{\pi/6}\frac{x^4}{1-x^4} \left(\frac{\pi}{2} -2y\right) \sec^2(y)\ dy
$$
So how do I solve the original problem?
|
Hint:
$$\int_{-a}^a\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}\ dx$$
$$=\int_{-a}^0\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}\ dx+\int_0^a\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}dx=I_1+I_2$$
For the first integral set $x=-y,dx=-dy$
$$I_1=\int_{-a}^0\dfrac{x^4}{1-x^4}\cos^{-1}\dfrac{2x}{1+x^2}\ dx=-\int_a^0\dfrac{y^4}{1-y^4}\cos^{-1}\dfrac{(-2y)}{1+y^2}dy$$
$$=\int_0^a\dfrac{y^4}{1-y^4}\cos^{-1}\dfrac{(-2y)}{1+y^2}dy$$
Using How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?,
$$I_1=\int_0^a\dfrac{y^4}{1-y^4}\left(\pi-\cos^{-1}\dfrac{2y}{1+y^2}\right)dy=\pi\int_0^a\dfrac{y^4}{1-y^4}dy-I_2$$
Hope you can take it from here!
|
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|
How to integrate $\frac{1}{3+\sin x+\cos x}$ I can easily do it with $\tan {\frac{x}{2}}$ substitution, but my problem requires $x\in(0,2\pi)$.
$\int \frac{1}{3+\sin x+\cos x}dx, x\in(0,2\pi)$
How do i solve it now ?
|
Take $\sin x=2t/(1+t^2), \cos x=(1-t^2)/(1+t^2), tan(x/2)=t \implies 2 \sec^2(x/2) dx=dt$. Then
$$I=\int \frac{dx}{3+\sin x+ \cos x}=\frac{1}{4}\int \frac{dt}{t^2+t+2}=\frac{1}{4}\int \frac{dt}{(t+1/2)^2+7/4}==\frac{1}{2\sqrt{7}} \tan^{-1} \frac{2t+1}{\sqrt{7}}+C.$$
where $t=\tan(x/2)$.
|
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|
Prove inverse trigonometric equation $2\tan^{-1}2=\pi-\cos^{-1}\frac{3}{5}$ The question is:
Prove that: $2tan^{-1}2=\pi-cos^{-1}\frac{3}{5}$ Hint: use the fact that $tan(\pi-x)=-tanx$
I did the question without using the Hint, but I don't know how to do it using the hint.
Quick working out of what I've done:
\begin{aligned}
\text { If } \theta &=\tan ^{-1} 2 \\
\tan \theta &=2 \\
0 & < \theta < \frac{\pi}{2}
\end{aligned}
\begin{aligned}
\cos 2 \theta &=2 \cos ^{2} \theta-1 \\
&=2 \times\left(\frac{2}{\sqrt{5}}\right)^{2}-1 \\
&=\frac{3}{5} \\
2 \theta =& \cos ^{-1} \frac{3}{5}, \quad \text { since } 0 < 2\theta < \pi
\end{aligned}
\begin{array}{l}
2 \tan ^{-1} 2=\cos ^{-1} \frac{3}{5} \text { . } \\
\text { Note: } \cos ^{-1} x \text { has point symmetry } \\
\text { in }\left(0, \frac{\pi}{2}\right) \text { . }
\end{array}
$$
\begin{array}{l}
\cos ^{-1} x+\cos ^{-1}(-x)=\pi \\
\cos ^{-1} \frac{3}{5}=\pi-\cos ^{-1}\left(-\frac{3}{5}\right) \\
\therefore \quad 2 \tan ^{-1} 2=\pi-\cos ^{-1}\left(-\frac{3}{5}\right)
\end{array}
$$
But I didn't use the Hint given in the question for this working out. How do I use the hint? Thank you !
|
$$\tan \alpha =2 \implies \tan 2\alpha = \frac{2\cdot 2}{1-2^2}=-\frac{3}{4} \implies tan(\pi-2\alpha)=\frac{3}{4} \\\implies \cos (\pi-2\alpha)=\frac 35. \blacksquare$$
|
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|
Minimum value of $\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$ If $a,b,c \in R$ find the minimum value of $$E=\left(\frac{a-b}{b-c}\right)^{4}+\left(\frac{b-c}{c-a}\right)^4+\left(\frac{c-a}{a-b}\right)^{4}$$
My attempt:
Let $x=a-b,y=b-c,z=c-a$
So $x+y+z=0$
We need to minimize $$E=\left(\frac{x}{y}\right)^{4}+\left(\frac{y}{z}\right)^{4}+\left(\frac{z}{x}\right)^{4}$$
$\implies$
$$E=\left(\frac{x}{y}\right)^{4}+\left(\frac{y}{x+y}\right)^{4}+\left(\frac{x+y}{x}\right)^4$$
Letting $\frac{x}{y}=t$ we get:
$$E=t^4+\frac{1}{(t+1)^4}+\frac{(t+1)^4}{t^4}$$
Now i tried using derivatives, but very tedious.
|
Hint
Define
$$
x={a-b\over b-c}\quad,\quad y={b-c\over c-a}\quad,\quad z={c-a\over a-b}
$$
and try to minimize $x^4+y^4+z^4$ constrained to $xyz=1$.
Alternatively, you can also define
$$
x=\ln\left|{a-b\over b-c}\right|\quad,\quad y=\ln\left|{b-c\over c-a}\right|\quad,\quad z=\ln\left|{c-a\over a-b}\right|
$$
and minimize $e^{4x}+e^{4y}+e^{4z}$ constrained to $x+y+z=0$.
Edit
Thanks to @RiverLi's comment below, also three more constraints should be included:
$$
{y(x+1)+1=0
\\
z(y+1)+1=0
\\
x(z+1)+1=0
}
$$
|
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|
Proof verification of $6^n \geq n3^n$ n is a natural number. Prove $6^n \geq n3^n$ holds for every natural number.
$n = 1:$
$$6 \geq 3 $$
$n \rightarrow n + 1:$
$$6 ^{n+1} = (3*2)^{n+1} = 3^{n+1}2^{n+1} \geq 3^{n+1}(n+1)2\geq 3^{n+1}(n+1) $$
Is my proof correct?
EDIT:
$$2^{n+1} \geq(n+1)2 \text{ because of Bernoulli inequality}$$
|
Suppose $6^n\ge n\cdot 3^n$ is true for $n$ and let's prove it for $(n+1)$.
$6^{n+1}=6\cdot 6^n\ge 6\left(n\cdot 3^n\right)=2\cdot 3\left(n\cdot 3^n\right)=2 n\cdot 3^{n+1}\ge (n+1)3^{n+1}$
proved.
|
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|
Find the average of the number $n \sin n^\circ$ for $n=2,4,6\cdots,180$ I have been asked in a exam to find the average of the number: $$n \sin n^\circ$$ for $n$=$2,4,6,\cdots,180$
I have tried a lot basically with sum product, or pairing the inputs...but at the end don't able to find any way to solve it, can someone help me with the approach?
|
Since $\sin(180^\circ - \theta) = \sin(\theta)$, $\sin{90^\circ} = 1$, and $\sin{180^\circ} = 0$, we can write the sum as
$$
(2 \sin{2^\circ} + 178 \sin{2^\circ}) + (4 \sin{4^\circ} + 176 \sin{4^\circ}) + \ldots + (88 \sin{88^\circ} + 92 \sin{88^\circ}) + 90\text.
$$
To get the average, divide by the number of terms, $90$, and get
$$
2 \sin{2^\circ} + 2 \sin{4^\circ} + \ldots + 2 \sin{88^\circ} + 1\text.\tag{*}
$$
Now, $\cos(\theta - 1^\circ) - \cos(\theta + 1^\circ) = 2 \sin\theta \sin 1^\circ$. Therefore,
$$
2\sin\theta = \frac{\cos(\theta - 1^\circ) - \cos(\theta + 1^\circ)}{\sin{1^\circ}}\text.\tag{**}
$$
When you plug $\text{(**)}$ into $\text{(*)}$, most of the $\cos$ terms cancel out and you are left with
$$
\frac{\cos{1^\circ} - \cos{89^\circ}}{\sin{1^\circ}} + 1 = \frac{\cos{1^\circ} - \sin{1^\circ}}{\sin{1^\circ}} + 1 = \color{red}{\cot{1^\circ}}\text.
$$
|
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|
What is the name of Fibonacci variation when $F(n) = a\cdot F(n-1) + b\cdot F(n-2) + c$, were $c$ is a constant, and $a >0, \ b >0, \ c>0$ I am trying to write $\log(n)$ algorithm for the above. I don't know if there is a specific name for the Fibonacci variation when:
$$F(n) = a\cdot F(n-1) + b\cdot F(n-2) + c$$
where: $a >0, \ b >0, \ c>0$
Could someone help me with the name of this variation?
|
Understanding that the function is null for negative $n$
$$
F(n) = aF(n - 1) + bF(n - 2) + c\quad \left| {\;F(n < 0) = 0} \right.
$$
so that its first values are
$$
\left\{ \matrix{
F(0) = c \hfill \cr
F(1) = \left( {a + 1} \right)c \hfill \cr
F(2) = \left( {a\left( {a + 1} \right) + b + 1} \right)c \hfill \cr
F(3) = \left( {a\left( {a\left( {a + 1} \right) + b + 1} \right)
+ b\left( {a + 1} \right) + 1} \right)c \hfill \cr
\quad \quad \vdots \hfill \cr} \right.
$$
then its o.g.f. will be derived as
$$
\eqalign{
& G(z) = \sum\limits_{0\, \le \,n} {F(n)z^{\,n} } = \cr
& = a\sum\limits_{0\, \le \,n} {F(n - 1)z^{\,n} } + b\sum\limits_{0\, \le \,n} {F(n - 2)z^{\,n} }
+ c\sum\limits_{0\, \le \,n} {z^{\,n} } = \cr
& = az\sum\limits_{1\, \le \,n} {F(n - 1)z^{\,n - 1} }
+ bz^{\,2} \sum\limits_{2\, \le \,n} {F(n - 2)z^{\,n - 2} }
+ c{1 \over {1 - z}} \cr}
$$
which gives
$$
\eqalign{
& G(z) = {c \over {\left( {1 - z} \right)\left( {1 - az - bz^{\,2} } \right)}}
= {{c/b} \over {\left( {z - 1} \right)\left( {z^{\,2} + {a \over b}z - {1 \over b}} \right)}} = \cr
& = {{c/b} \over {\left( {z - 1} \right)
\left( {z - \left( { - {a \over {2b}} + \sqrt {a^{\,2} + 4b} } \right)} \right)
\left( {z - \left( { - {a \over {2b}} - \sqrt {a^{\,2} + 4b} } \right)} \right)}} = \cr
& = {{c/b} \over {\left( {z - 1} \right)\left( {z - r} \right)\left( {z - s} \right)}} = \cr
& = {c \over b}\left( {{1 \over {\left( {s - 1} \right)\left( {s - r} \right)\left( {z - s} \right)}}
+ {1 \over {\left( {r - s} \right)\left( {r - 1} \right)\left( {z - r} \right)}}
+ {1 \over {\left( {r - 1} \right)\left( {s - 1} \right)\left( {z - 1} \right)}}} \right) = \cr
& = - {c \over b}
\left( {{1 \over {s\left( {s - 1} \right)\left( {s - r} \right)\left( {1 - {z \over s}} \right)}}
+ {1 \over {r\left( {r - s} \right)\left( {r - 1} \right)\left( {1 - {z \over r}} \right)}}
+ {1 \over {\left( {r - 1} \right)\left( {s - 1} \right)\left( {1 - z} \right)}}} \right) \cr}
$$
Therefore $F(n)$ will be
$$
F(n) = {A \over {s^{\,n} }} + {B \over {r^{\,n} }}
+C
$$
which is valid for $r$ and $s$ even complex, provided that they are not such as to
make null one of the denominators above.
|
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|
Factoring $4x^{4} - x^{2}$ into $4x^{2}(x-1/2)(x+1/2)$ So i was trying to make an integral but i'm stuck with the easiest part.
How do you get this?
$$4x^{4} - x^{2} = 4x^{2}(x-1/2)(x+1/2)$$
I know how to take the roots $x = 0$, $x = \pm 1/2$,
but when putting it all together I get
$$x(x−1/2)(x+1/2)$$ and I can't remember how to do it correctly.
|
I think you meant $4x^{4} - x^{2}$. In such case, you can factor $x^{2}$ out and apply the following identity:
\begin{align*}
a^{2} - b^{2} = (a+b)(a-b)
\end{align*}
At your case, you have $4x^{4} - x^{2} = x^{2}(4x^{2} - 1)$, where $a = 2x$ and $b = 1$. Consequently, we get that
\begin{align*}
4x^{4} - x^{2} = x^{2}(2x-1)(2x+1) = 4x^{2}(x-1/2)(x+1/2)
\end{align*}
|
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|
Find $k$ in $p(x) = 2x^3 - 6x^2 + kx -1$ such that its roots $x_1^2+x_2^2+x_3^2 = 6$ Let
$$p(x) = 2x^3 - 6x^2 + kx -1$$
and let $x_1, x_2$
and $x_3$ the $p(x)$ roots. What is the $k$ value such that
$$x_1^2+x_2^2+x_3^2 = 6$$
|
Rearrange $2x^3 - 6x^2 + kx -1=0$ as $x(2x^2+k)=6x^2+1$, and square both sides to get the equation in $x^2$
$$x^6 +(k-9)x^4+(\frac{k^2}4-3)x^2-\frac14=0$$
Then
$$x_1^2+x_2^2+x_3^2= 6= -(k-9)$$
which yields $k=3$.
|
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|
Is my method of solving equation correct? The problem in question is
$$\sqrt[5]{16+\sqrt{x}}+\sqrt[5]{16-\sqrt{x}}=2$$
using $$a+b=2$$ where $a=\sqrt[5]{16+\sqrt{x}}$ and $b=\sqrt[5]{16-\sqrt{x}}$
$$(a+b)^5=32$$ $$(a+b)^2(a+b)^3=32$$
$$a^5+5a^4b+10a^3b^2+10a^2b^3+5ab^4+b^5=32$$
$$a^5+b^5+5ab(a^3+b^3)+10a^2b^2(a+b)=32$$
$$a^5+b^5+5ab\biggl(\frac{32}{(a+b)^2}\biggr)+10a^2b^2(a+b)=32$$
Got $\frac{32}{(a+b)^2}$ from the fact that $(a+b)^2(a+b)^3=32$ and $a+b=2$
$$a^5+b^5+5ab\biggl(\frac{32}{(2)^2}\biggr)+10a^2b^2(2)=32$$
$$a^5+b^5+40ab+20a^2b^2=32$$
From when I defined a and b earlier, I substitute and get
$$\left(\sqrt[5]{16+\sqrt{x}}\right)^5+\left(\sqrt[5]{16-\sqrt{x}}\right)^5+40\sqrt[5]{\left(16-\sqrt{x}\right)\left(16+\sqrt{x}\right)}+20\sqrt[5]{\left(16-\sqrt{x}\right)^2\left(16+\sqrt{x}\right)^2}=32$$
$$\require{cancel}\cancel{16}\cancel{+\sqrt{x}}+\cancel{16}\cancel{-\sqrt{x}}+40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=\cancel{32} 0$$
$$40\sqrt[5]{256-x}+20\sqrt[5]{\left(256-x\right)\left(256-x\right)}=0$$
$$20\biggl(2\sqrt[5]{256-x}+\sqrt[5]{\left(256-x\right)\left(256-x\right)\biggr)}=0$$
Then let $u=\sqrt[5]{256+{x}}$,
$$20(2u+u^2)=0$$
$$u(u+2)=0$$
$$u=0,-2$$
Substituting u to get x from $u=\sqrt[5]{256+{x}}$, I get $$x=\cancel{-288},256$$
However, since the original equation has a $\sqrt{x}$, which can't be negative, I eliminate $x=-288$, leaving just $$x=256$$ as my answer.
So, this is how I arrived on my answer. Did I perform any mathematical errors or any illegal mathematical maneuvers? Please let me know.
Thank you!
|
There is another solution.
Let $\sqrt[5]{16+\sqrt{x}}=1+t$, where $t>0$; therefore, $\sqrt[5]{16-\sqrt{x}}=1-t$. Hence
$$(1+t)^5+(1-t)^5=32,$$
or
$$10\,t^4+20\,t^2+2=32.$$
Immediately we have
$$t^4+2\,t^2-3=0,$$
and $t^2=1$. Since $t>0$, $t=1$ is the single solution. So we have
$$\sqrt[5]{16-\sqrt{x}}=0\Rightarrow \sqrt{x}=16\Rightarrow x=256.$$
|
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|
Evaluate $\sum\limits_{n=1}^{+\infty} \frac{\left( \frac{3-\sqrt{5}}{2} \right)^{n}}{n^{3}}$ Evaluate
$$\sum\limits_{n=1}^{+ \infty} \frac{ \left( \frac{3-\sqrt{5}}{2} \right)^{n} }{n^{3}}$$
We can use the Fourier series to calculate this sum, because it converges.
Also, we know that $\frac{3-\sqrt{5}}{2} = \frac{1}{\varphi^{2}}$ where $\varphi = \frac{1+\sqrt{5}}{2}$ is the golden ratio. What is going on about this number $\frac{3-\sqrt{5}}{2}$ ? we know something else ?
Thank you for your answer but we know something else without about trilogarithm function?
|
$$\sum_{n=1}^\infty \frac{a^n}{n^3}=\text{Li}_3(a)$$ Making $a=\frac{3-\sqrt{5}}{2} = \frac{1}{\varphi^{2}}$, you just get a number
$$\text{Li}_3\left(\frac{1}{\varphi ^2}\right)=0.4026839629521090211599594481825111422197338\cdots$$
|
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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.