Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to approach this divisibility problem on polynomial Prove that the polynomial $x^{9999} + x^{8888} + x^{7777} + ... + x^{1111} + 1$ is divisible by $x^9 + x^8 + x^7 + ... + x + 1$.
I have no idea how to approach this problem, some help would be appreciated.
| $$ ( x^9 + x^8 + \ldots + x + 1) \left(x^{9990} - x^{9989} \right) = x^{9999} - x^{9989}$$
So,
$$ x^{9999} - x^{9979} = x^{9999} - x^{9989} + x^{9989} - x^{9979} = ( x^9 + x^8 + \ldots + x + 1) \left(x^{9990} - x^{9989} + x^{9980} - x^{9979} \right) $$
Continuing in this way, we see that:
$$ x^9 + x^8 + \ldots + x + 1\ \vert\ x^{9999} - x^9, $$
$$ x^9 + x^8 + \ldots + x + 1\ \vert\ x^{8888} - x^8, $$
$$ \ldots$$
$$ x^9 + x^8 + \ldots + x + 1\ \vert\ x^{1111} - x^1, $$
$$ x^9 + x^8 + \ldots + x + 1\ \vert\ 1 - 1, $$
Therefore,
$$ x^9 + x^8 + \ldots + x + 1\ \vert\ x^{9999} - x^9 + x^{8888} - x^8 + \ldots + x^{1111} - x^1 + 1 - 1,\quad \text{ i.e.,} $$
$$ x^9 + x^8 + \ldots + x + 1\ \vert\ x^{9999} + x^{8888} + \ldots + x^{1111} + 1 - \left( x^9 + x^8 + \ldots + x + 1 \right) $$
$$ \implies x^9 + x^8 + \ldots + x + 1\ \vert\ x^{9999} + x^{8888} + \ldots + x^{1111} + 1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3212481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Convert $\sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right)$ into an algebraic function
Convert trigonometric function into algebraic function.
$$\sin\left(2\cos^{-1}\left(\cot\left(2 \tan^{-1}x\right)\right)\right)$$
My approach is as follows:
$sin2\theta$ is to be calculated
$$\tan^{-1}x=\gamma \tag{1}$$
Hence
$$\begin{align}
\cos\theta&=\cot2\gamma \tag{2}\\[4pt]
\cos\theta&=\frac{\cot^2\gamma-1}{2\cot\gamma} \tag{3}\\[4pt]
\cos\theta&=\frac{1-x^2}{2x} \tag{4}
\end{align}$$
The value of $\sin\theta$ is coming in negative.
| If $\cos^{-1}y=u,\cos u=y,0\le y\le\pi$
If $\sin(2\cos^{-1}y)=\sin2u=2\sin u\cos u=\begin{cases}2u\sqrt{1-u^2} &\mbox{if }0\le y\le\dfrac\pi2 \\
-2u\sqrt{1-u^2} & \mbox{if } y>\dfrac\pi2 \end{cases} $
Here $u=\dfrac{1-x^2}{2x}$
For real $\cos^{-1}y,-1\le\cot(2\tan^{-1}x)\le1$
$\iff\dfrac\pi4\le2\tan^{-1}x\le\dfrac{3\pi}4$
$\iff\tan\dfrac\pi8\le x\le\tan\dfrac{3\pi}8$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve $f(-4)$ if $f(x)=\frac{x^3+64}{x+4}$? I need to solve $f(-4)$ if $f(x)=\frac{x^3+64}{x+4}$. I have changed the form of the function, but for some reason I'm still not getting the right answer.
My Steps:
$$\frac{x^3+64}{x+4}$$
apply $(a+b)^3=(a+b)(a^2+2ab+b^2)$ formula
$$=\frac{(x+4)(x^2+8x+16)}{x+4}$$
$$=x^2+8x+16$$
Insert $-4$ for $x$.
$$(-4)^2+8(-4)+16 = 16-32+16 = 0$$
| There is no value $f(-4)$ since $-4$ is not in the domain of the function by mere fact of the division by zero that results. Even if the expression simplifies, it doesn't change the removable discontinuity being present.
In any event, if you want to find an almost-equivalent expression for $f$ that gets rid of the discontinuity, note that you factor a sum of two cubes as so:
$$a^3 + b^3= (a+b)(a^2 - ab + b^2)$$
(You made a sign error, thus why I noted it. In general, you might see this written as $a^3 \pm b^3 = (a \pm b)(a^2 \mp ab + b^2)$ to account for the sign change.)
Thus, $x^3 + 64 = x^3 + 4^3$ factors as
$$x^3 + 64 = (x+4)(x^2 - 4x + 16)$$
Then dividing by $(x+4)$ gives you $g(x) = x^2 - 4x + 16$. In this sense, you have $g(-4)$ defined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Differentiating $\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$
Find the derivative with respect to $x$ of
$$\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)$$
The partial solution to this problem is given as follows:
$y=\left(\frac{x}{\sqrt{1-x^2}}\right)$
Then:
$\frac{dy}{dx} = (1-x^2)^{-3/2}$
Can you show the steps in arriving at this partial solution of $\frac{dy}{dx}$?
Product Rule with Chain Rule, Quotient Rule with Chain Rule, and I end up with additional terms. Notably, an $x^2$ multiplying the $(1-x^2)^{-3/2}$, plus an additional term of $(1-x^2)^{-1/2}$
| Continuing to work through the solution provided I have arrived at an answer. It is as follows:
$\frac{d}{dx} \frac{x}{\sqrt(1-x^2)}$
$\frac{d}{dx} x(1-x^2)^{-1/2}$
Using the Product Rule (u' * v) + (v' * u) where u = $x$ and v = $(1-x^2)^{-1/2}$ and combining with the Chain Rule for the derivation of v' yields the following:
$(1)*(1-x^2)^{-1/2} + (\frac{-1}{2}(1-x^2)^{-3/2}*(-2x)*(x)$
NOTE: The Chain Rule applied to v' gave us this: $\frac{-1}{2}(1-x^2)^{-3/2}*(-2x)$
Now, further simplifying ...
$(1-x^2)^{-1/2} + (\frac{2x^2}{2}(1-x^2)^{-3/2}$
$(1-x^2)^{-1/2} + (x^2(1-x^2)^{-3/2}$
The above was where I was stuck, but here is the breakthrough ...
$(1-x^2)^{-1/2} + \left(\frac{x^2}{(1-x^2)^{3/2}}\right)$
Get a common denominator ...
$\frac{(1-x^2)^{-1/2}(1-x^2)^{3/2} + x^2}{(1-x^2)^{3/2}}$
$\frac{(1-x^2)^{2/2} + x^2}{(1-x^2)^{3/2}}$
$\frac{1-x^2+x^2}{(1-x^2)^{3/2}}$
$\frac{1}{(1-x^2)^{3/2}}$
$=(1-x^2)^{-3/2}$
NOTE: This is only the partial solution to the problem. For those interested in the full solution read on ...
Without showing this derivation, I know
$\frac{d}{dx} tan^{-1}(x) = \frac{1}{1+x^2}$
We also know from the original problem that $x = \frac{x}{\sqrt(1-x^2)}$
So now substitute $x = \frac{x}{\sqrt(1-x^2)}$ into derivation of $\frac{d}{dx} tan^{-1}(x) = \frac{1}{1+x^2}$
This yields
$\frac{d}{dx} tan^{-1}(x) = \frac{1}{1+x^2} = \frac{1}{1+\left(\frac{x}{\sqrt(1-x^2)} \right)^2}$
Square the term in the denominator
$=\frac{1}{1+\frac{x^2}{1-x^2}}$
Get a common denominator
$=\frac{1}{\frac{1-x^2+x^2}{1-x^2}}$
Simplify
$=1-x^2$
Now, finishing up ...
The derivative with respect to $x$ of $tan^-1\left(\frac{x}{\sqrt(1-x^2)}\right)$ requires the Chain Rule. This is:
$\frac{d}{dx}tan^{-1}\left(\frac{x}{\sqrt(1-x^2)}\right) * \frac{d}{dx}\frac{x}{\sqrt(1-x^2)}$
So, the derivative of the outside function $\frac{d}{dx}tan^{-1}\left(\frac{x}{\sqrt(1-x^2)}\right)$, or think of it just like the derivative of the $tan^{-1}(x)$ where we later substitute back in $x = \frac{x}{\sqrt(1-x^2)}$, times the derivative of the inside function $\frac{d}{dx}\frac{x}{\sqrt(1-x^2)}$. We have already found both derivations above. Let's substitute the derivations in and finish solving this problem.
By Chain Rule:
$\frac{d}{dx}tan^{-1}\left(\frac{x}{\sqrt(1-x^2)}\right) = \frac{d}{dx}\left(tan^{-1}\left(\frac{x}{\sqrt(1-x^2)}\right)\right) * \frac{d}{dx}\left(\frac{x}{\sqrt(1-x^2)}\right)$
$=(1-x^2) * (1-x^2)^{-3/2}$
$=(1-x^2)^{2/2} * (1-x^2)^{-3/2}$
$=(1-x^2)^{-1/2}$
$=\frac{1}{\sqrt(1-x^2)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3214409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Integrate $\int \frac {\sin (2x)}{(\sin x+\cos x)^2}\,dx$ Integrate
$$\int \frac {\sin (2x)}{(\sin x+\cos x)^2} \,dx$$
My Attempt:
$$=\int \frac {\sin (2x)}{(\sin x + \cos x)^2} \,dx$$
$$=\int \frac {2\sin x \cos x}{(\sin x+ \cos x)^2} \,dx$$
Dividing the numerator and denominator by $\cos^2 x$
$$=\int \frac {2\tan x}{(1+\tan x)^2} \,dx$$
| Note that
$$\begin{align}
\int \frac {\sin (2x)}{(\sin(x)+\cos(x))^2} dx
&=\int \frac {2\sin(x)\cos(x)}{\cos^2(x)(1+\tan(x))^2} dx\\
&=\int\frac {2\tan(x)}{(1+\tan(x))^2} dx\\
&=\int\left(1-\frac {1+\tan^2(x)}{(1+\tan(x))^2}\right) dx\\
&=x+\frac{1}{1+\tan(x)}+c.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Solve $\frac {dy}{dx} + y\tan x=y^3 \sec x$ Solve
$$\dfrac {dy}{dx}+y\tan x= y^3 \sec x$$
My Attempt:
$$\dfrac {dy}{dx} + y\tan x= y^3 \sec x$$
Dividing both sides $y^3$,
$$y^{-3} \dfrac {dy}{dx} + y^{-2} \tan x= \sec x$$
Put $y^{-2} = z$
$$(-2)y^{-2} \dfrac {dy}{dx} = \dfrac {dz}{dx}$$
$$y^{-3} \dfrac {dy}{dx} = \dfrac {-1}{2} \dfrac {dz}{dx}$$
Then,
$$-\dfrac {1}{2} \dfrac {dz}{dx} + z.\tan x= \sec x$$
$$-\dfrac {dz}{dx} + (2\tan x) z =2\sec x$$
| $$\frac{y'}{y^3}+\frac{\tan x}{y^2}=\frac{1}{\cos x}$$
$$ z'-2z\tan x=-\frac{2}{\cos x}$$
The integrating factor would be now:
$$e^{-2\int \tan x}dx=e^{\ln(\cos^2 x)+C}$$
So multiply both sides by $\cos^2 x$
$$(z\cos^2 x)'=-2\cos x\Rightarrow z\cos^2 x=-2\sin x+C$$
$$y=\sqrt{\frac{\cos^2 x}{-2\sin x+C}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $ab+ac+bc=9$ and $a,b,c \geq 1,$ what is the maximum value of $a^2+b^2+c^2?$ If $ab+ac+bc=9$ and $a,b,c \geq 1,$ could anyone advise me how to find the maximum value of $a^2+b^2+c^2?$
I have shown that $(a+b+c)^2=a^2+b^2+c^2+18,$ so does it suffice to maximise $a+b+c ?$
Thank you.
| $$\sum_{cyc}(a-1)(b-1)\geq0$$ gives
$$a+b+c\leq6$$ and by your work we are done!
The equality occurs for $(a,b,c)=(1,1,4).$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find a limit for $\frac{({1+x})^\frac{1}{x} - e}{x}$ as x tends to 0 I have done it thus far:
$$\lim_{x \to 0}\frac{{(x+1)}^\frac{1}{x}-e}{x} = \bigg[\frac{0}{0}\bigg] = \frac{((x+1)^\frac{1}{x}-e)'}{x'}=({x+1})^\frac{1}{x} \cdot \left(\frac{\ln(x+1)}{x}\right)' = \\({x+1})^\frac{1}{x} \cdot \frac{\left(\frac{x}{x+1}-(\ln(x+1)\right)}{x^2} = ({x+1})^\frac{1}{x} \cdot \left(\frac{\frac{1}{x+1}}{x} - \frac{\ln(x+1)}{x^2}\right) $$
I don't know what to do next.
Also could someone elaborate as to why when I have to find a derivative for $\frac{\ln(x+1)}{x}$ I need to use the quotient rule, but when I first derived the fraction that is given I could derive numerator and denominator separately?
| $$\begin{align}
\lim_{x\to0}\left(\frac{\frac{x}{x+1}-\ln{(x+1)}}{x^2}\right)
&=\lim_{x\to0}\left(\frac{x-(x+1)(x-\frac{x^2}{2}+\frac{x^3}{3}+O(x^4))}{x^2(x+1)}\right)\\
&=\lim_{x\to0}\left(\frac{-x^2+\frac{x^2}{2}+\frac{x^3}{2}-\frac{x^3}{3}+O(x^4)}{x^3+x^2}\right)\\
&=\lim_{x\to0}\left(\frac{-\frac{x^2}{2}+\frac{x^3}{6}+O(x^4)}{x^3+x^2}\right)\\
&=\lim_{x\to0}\left(\frac{-x+\frac{x^2}{2}+O(x^3)}{3x^2+2x}\right)\\
&=\lim_{x\to0}\left(\frac{-1+x+O(x^2)}{6x+2}\right)\\
&=-\frac12\\
\end{align}$$
Hence the initial limit becomes
$$\overbrace{\lim_{x\to0}\left((x+1)^\frac1x\right)}^{\large{\to\, e}}\cdot\overbrace{\lim_{x\to0}\left(\frac{\frac{x}{x+1}-\ln{(x+1)}}{x^2}\right)}^{\large{\to\, -\frac12}}=-\frac{e}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Number of six digit numbers divisible by $3$ but none of the digits is $3$ Find number of six digit numbers divisible by $3$ but none of the digits is $3$
My try:
Let the six digits are $a,b,c,d,e,f$ such that
$$a+b+c+d+e+f=3p$$
where $1 \le p \le 18$
Now since $a \ge 1$ we have by Stars and Bars Technique number of solutions of the above equation as:
$$S=\sum_{p=1}^{18}\binom{3p+4}{5}$$
But if we use Exclusion method, its very lengthy.
Any hint?
| The digits we can use to form these numbers are $$\{0, 1,2,4,5,6,7,8,9\}\equiv \{0, 1, 2,1, 2,0,1,2,0\}\pmod3$$
Let $S$ be the sum of the last five digits,
$$
\begin{array}{c|lcr}
S\bmod3 & \color{blue}0 & \color{red}1 & \color{green}2 \\
\text{Probability} & 1/3 & 1/3 & 1/3 \\
\end{array}
$$
Observe that the first digit $\in\{1,2,4,5,6,7,8,9\}\equiv\{\color{green}1,\color{red}2,\color{green}1,\color{red}2,\color{blue}0,\color{green}1,\color{red}2,\color{blue}0\}\bmod3$.
Therefore, the probability, that a six digit number - from the ones we are considering - is divisible by $3$ is $$P=\color{blue}{\frac13\cdot\frac28}+\color{red}{\frac13\cdot\frac38}+\color{green}{\frac13\cdot\frac38}=\frac{8}{24}=\frac13$$ Therefore, exactly one third of all numbers - from the ones we are considering - is divisible by $3$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3217363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $CD\parallel AB$ with square $FEBD$ As the figure shows, $FEBD$ is a squre. $AE=GE$, $FA=FB$ and $CD=BA$. Show that $CD\parallel AB$.
I have found that $\triangle BGD$ must be an equilateral triangle, but I have no proof yet. Please help. It's better to offer a synthetical solution without much computation. Thanks in advance.
| Let $FEBD$ be the square with vertices $F=(-1,0)$, $E=(0,1)$, $B=(1,0)$, $D=(0,-1)$ and $A=(a,b)$ a point such that $AF = BF$.
$AF=BF$
By the formula of distance between two points
$\begin{array}{} (a+1)^{2}+b^2=4 & ⇒ & b^{2}=4-(a+1)^{2} & (I) \end{array}$
$\begin{array}{} CD∥AB & ⇒ & m_{CD}=m_{AB} \end{array}$
$\frac{y_{D}-y_{C}}{x_{D}-x_{C}}=\frac{y_{B}-y_{A}}{x_{B}-x_{A}}$
$\begin{array}{} x_{C}=\frac{a-1}{b} & (II) \end{array}$
$\begin{array}{} CD=BA & C∈BF &⇒ & y_{C}=0 \end{array}$
$\begin{array}{} x_{C}^{2}=3-4·a & (III) \end{array}$
$\begin{array}{} \text{From (I) (II) (III)} & \frac{(a-1)^{2}}{b^{2}}=3-4·a \end{array}$
$\frac{(a-1)^{2}}{4-(a+1)^{2}}=3-4·a$
$\begin{array}{} \text{solutions: } & a:\left( \begin{array}{} -1+\sqrt{3} \\ -1-\sqrt{3} \end{array} \right) \end{array}$
$AB=CD=\sqrt{6}-\sqrt{2}$
$\begin{array}{} x_{C}=-2+\sqrt{3} & y_{C}=0 \\ x_{G}=\frac{1-\sqrt{3}}{2} & y_{G}=\frac{\sqrt{3}-1}{2} \\ \end{array}$
$med(\angle{AFB})=\frac{ π }{6}$
$\begin{array}{} BG=DG=BD=\sqrt{2} & ⇒ & \triangle{BDG} & \text{ is an equilateral triangle} \end{array}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Evaluate $\lim_{x \to 0} \frac{\cos (\sin x) - \cos x}{x^4}$
Evaluate $\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$
The answer stated is $\displaystyle {1 \over 6}$.
What I've tried:
$$\displaystyle \lim_{x\to0} \frac{\cos (\sin x) - \cos x}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{\cos (\sin x) -1+1- \cos x}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{1- \cos x}{x^4} - \frac {1-\cos (\sin x)}{x^4}$$
$$=\displaystyle \lim_{x\to0} \frac{2 \sin^2(\frac {x}{2})}{x^4} - \frac {2 \sin^2(\frac {\sin x}{2})}{x^4}$$
$$=\displaystyle \lim_{x\to0} \left(\frac{\sin(\frac {x}{2})}{x} \right)^2. \left( \dfrac{1}{2x^2} \right) - \frac {2 \sin^2(\frac {\sin x}{2})}{x^4}$$
I'm not sure how I can evaluate the limit by proceeding this way. All help will be appreciated.
P.S. I'd prefer not using L'Hôpital's rule, it can get really messy.
EDIT: I should have mentioned that I would prefer if the solution does not use taylor series approximations (or any approximations) for that matter.
| Note that from the expansions $\sin(x)=x-\frac{x^3}{6}+O(x^5)$ and $\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O(x^6)$ we have
$$\begin{align}
\cos(\sin(x))&=\cos\left(x-\frac{x^3}{6}+O(x^5)\right)\\\\
&=1-\frac12\left(x-\frac{x^3}{6}+O(x^5)\right)^2+\frac{1}{24}\left(x-\frac{x^3}{6}+O(x^5)\right)^4+O(x^6)\\\\
&=1-\frac12x^2+\frac{5}{24} x^4+O(x^6)
\end{align}$$
Hence,
$$\frac{\cos(\sin(x))-\cos(x)}{x^4}=\frac16+O(x^2)$$
from which we find the coveted limit.
| {
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"url": "https://math.stackexchange.com/questions/3224598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find antiderivative $\int (2x^3+x)(\arctan x)^2dx $
Find antiderivative $$\int (2x^3+x)(\arctan x)^2dx $$
My try:
$$\int (2x^3+x)(\arctan x)^2dx =(\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\int \frac{2\arctan x}{1+x^2}(\frac{1}{2}x^4+\frac{1}{2}x^2)dx=(\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\int (\arctan x) (x^2)dx= (\arctan x)^2(\frac{1}{2}x^4+\frac{1}{2}x^2)-\arctan x\cdot \frac{1}{3}x^3-\int (\frac{1}{1+x^2}) (\frac{1}{3}x^3)dx$$And then I don't know how I can find $\int (\frac{1}{1+x^2}) (\frac{1}{3}x^3)dx$. Can you help me with it?
| $$\int\frac{x^3}{1+x^2}dx = \int\left(\frac{x^2}{2(1+x^2)}\right)2xdx$$
Using $u = x^2 \implies du = 2xdx$
$$ = \frac{1}{2}\int\left(1 - \frac{1}{1+u}\right)du = \frac{1}{2}(u - \ln(1+u)) = \boxed{\frac{1}{2}(x^2 - \ln(1+x^2))}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3224715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Continuity and differentiability of $f(x) = \frac{x}{1+x} + \frac{x}{(x+1)(2x+1)} + \frac{x}{(2x+1)(3x+1)}+...$ Here's the given function:
$$f(x) = \dfrac{x}{1+x} + \dfrac{x}{(x+1)(2x+1)} + \dfrac{x}{(2x+1)(3x+1)}+...$$
My question regarding this function: is this function considered a continuous and a differenciable function? Is it not continuous or differenciable at any specific points?
According to a question about this function, $f(x)$ is non-continuous. They haven't mentioned anything about it being differenciable or not.
My working:
Simplyfing the series on the right hand side of the equation is quite straightforward-
$$f(x) = \dfrac{x}{1+x} + \dfrac{x}{(x+1)(2x+1)} + \dfrac{x}{(2x+1)(3x+1)}+...$$
$$\implies f(x) = \dfrac{1+x-1}{1+x} + \dfrac{(2x+1)-(x+1)}{(x+1)(2x+1)} + \dfrac{(3x+1)-(2x+1)}{(2x+1)(3x+1)}+...$$
$$\implies f(x) = 1 -\dfrac{1}{1+x}+\dfrac{1}{1+x} -\dfrac{1}{2x+1} +\dfrac{1}{2x+1}-\dfrac{1}{3x+1}...$$
$$\implies f(x) =1 $$
Since $f(x)$ is a constant function, it appears to be both continuous and differenciable $\forall x \in \mathbb R$. This is clearly incorrect, what am I doing wrong?
EDIT:
Is there a way to graph a function like $f(x)$?
| Note that all the terms are $0$ when $x=0$!. Your calculation is correct if $x \neq 0$ and $x \neq -\frac 1 n$ for any $n$. Continuity and differentabilty are subject to intepretation. One interpretation is that since $f$ is not defined in any open interval containing $0$ it is not continuous, hence not differntiable, at $0$.
| {
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To find the sum of the given infinite series This is a question from the book 'Differential Calculus' by Joseph Edwards.
Prove that if $x$ be less than unity
$$\frac{1-2x}{1-x+x^2} + \frac{2x-4x^3}{1-x^2+x^4} + \frac{4x^3-8x^7}{1-x^4+x^8} \ldots = \frac{1+2x}{1+x+x^2}$$
As the numerators are negative derivative of the denominators, I tried differentiating the sum of the series
-log{1-x+x^2}-log{1-x^2+x^4}-log{1-x^4+x^8}....but was unable to arrive at the required answer.
| Start with the equality
$$
\frac{1+x}{1+x^3}\cdot \frac{1+x^2}{1+x^6}\cdot \frac{1+x^4}{1+x^{12} }\cdots=\frac{(1-x)^{-1}}{(1-x^3)^{-1}}\tag 1
$$
To understand the numerator, note that you can prove by induction that
$$
\prod_{i=0}^{n-1}(1+x^{2^i})=\sum_{i=0}^{2^n-1}x^i
$$
Letting $n\to\infty$, you get
$$
\prod_{i=0}^\infty (1+x^{2^i})=\sum_{i=0}^\infty x^i=\frac1{1-x}\tag2
$$ For the denominator, substitute $x^3$ into $(2)$.
Next, note that $(1+x)/(1+x^3)=1/(1-x+x^2)$, so $(1)$ can be rewritten as
$$
\frac1{1-x+x^2}\cdot \frac1{1-x^2+x^4}\cdot \frac1{1-x^4+x^8}\cdots =\frac{1-x^3}{1-x}=1+x+x^2\tag{3}
$$
Finally, take the logs of both sides in $(3)$, then differentiate.
| {
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Primes number $n,n+2,n+6,n+8,n+12,n+14$
Find all natural number $n$ such that all the following numbers are primes :
$$n,\;\; n+2,\;\;n+6,\;\;n+8,\;\;n+12,\;\;n+14$$
are all prime numbers
| As $n$ is a prime, $n\ge 2$.
When $n=2$, $n+2=4$ is not prime.
When $n=3$, $n+6=9$ is not prime.
When $n=5$, all of them are prime.
$(n+6)-n\equiv 1$ (mod $5$)
$(n+2)-n\equiv 2$ (mod $5$)
$(n+8)-n\equiv 3$ (mod $5$)
$(n+14)-n\equiv 4$ (mod $5$)
At least one of these integers is a multiple of $5$.
When $n>5$, at least one of the numbers is not prime.
The only possible answer is $5$.
| {
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Proof of $(px+1)^{p^{k}}=(p^{k+1}x+1)$ mod $p^{k+2}$ for any odd prime $p$ I need to prove this identity using only simple arithmetics and combinatorics (or to find a counter-example):
$(px+1)^{p^{k}}=(p^{k+1}x+1)$ mod $p^{k+2}$, which I verified using several values of $p$ and $k$ using GAP.
I tried the following (using Newton's binomial formula):
$$(px+1)^{p^k} = \sum_{i=0}^{p^k}{p^k\choose i}p^ix^i$$
where it seems sufficient to prove that only the terms with $i \in \{0,1\}$ survive. The term for $i=2$ becomes ${p^k \choose 2} p^2x^2 = \frac{p^k(p^k-1)}{2}p^2x^2$ which disappears mod $p^{k+2}$ since $p$ is odd. The term for $i=3$ already poses a problem: ${p^k \choose 3} p^3x^3 = \frac{p^k(p^k-1)(p^k-2)}{2.3}p^3x^3$. The case where $p=3$ needs special attention but the corresponding term can be proven to disappear also. The general term ${p^k\choose i}p^ix^i$ is more of a nightmare. To delve deeper into the problem I looked at the p-adic valuation of each of the factors. Using GAP I get the following result for $\operatorname{val}_p{p^k \choose i}$ where $p=5, k=3$:
gap> List([0..5^3], i->PValuation(Binomial(5^3,i),5));
[ 0, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 1, 3,
3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 1, 3, 3, 3,
3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 1, 3, 3, 3, 3, 2,
3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 1, 3, 3, 3, 3, 2, 3, 3,
3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 2, 3, 3, 3, 3, 0 ]
From which we could hypothesise that $\operatorname{val}_p{p^k \choose i} = k$ if $\gcd(i,k) = 1$, $\operatorname{val}_p{p^k \choose i} = k-1$ if $k$ is a multiple of $p$ but not of $p^2$, ..., $\operatorname{val}_p{p^k \choose i} = k-l$ if $k$ is a multiple of $p^l$ but not of $p^{l+1}$. So proving this would solve the problem since $\operatorname{val}_p{p^i} = i$. In order to prove this hypotesis I went for an expression for $\operatorname{val}_pn!$ which I found here, resulting in $\operatorname{val}_pn! = \left \lceil{n/p}\right \rceil + \left \lceil{n/p^2}\right \rceil + \left \lceil{n/p^3}\right \rceil + \ldots$.
And here I'm stuck in using this in finding the valuation of ${n \choose m} = \frac{n!}{m!(n-m)!}$.
| Lemma 1: For $k>0,$ if $x\equiv y\pmod{p^k}$ then $x^p\equiv y^p\pmod{p^{k+1}}.$
Proof: If $x\equiv y\pmod {p^k}$ then $x^{p-1}+x^{p-2}y+\cdots+xy^{p-2} + y^{p-1}\equiv px^{p-1}\pmod{p^k}$ so $$x^p-y^p=(x-y)\left(x^{p-1}+x^{p-2}y+\cdots+xy^{p-2} + y^{p-1}\right)$$ is divisible by $p^{k+1}$ and hence $x^p\equiv y^{p}\pmod{p^{k+1}}.$
Lemma 2: For $i=1,\dots,p-1$, $\binom{p}{i}$ is divisible by $p.$
Proof: Left to you.
Main proof: Let $a_k=(px+1)^{p^k}$ and $b_k=p^{k+1}x+1.$ We will prove by induction that for $k\geq 0,$ $a_k\equiv b_k\pmod{p^{k+2}}.$
When $k=0$ we have that $a_0=px+1\equiv px+1=b_0\pmod{p^{2}}$ because they are equal.
Now given $$a_k\equiv b_k\pmod{p^{k+2}}$$ we need to show that $$a_{k+1}\equiv b_{k+1}\pmod{p^{k+3}}.$$ But $a_{k+1}=a_k^p,$ and lemma (1) lets us deduce that:
$$a_{k+1}=a_k^p\equiv b_k^p\pmod{p^{k+3}}$$
If we prove that:
$$b_k^{p}\equiv b_{k+1}\pmod {p^{k+3}}$$
then we are done.
Now, $$b_k^p = (p^{k+1}x+1)^p = 1+\binom{p}{1} p^{k+1}x+\sum_{i=2}^{p}\binom{p}{i}p^{i(k+1)}x^i$$
If $i=2,\dots,p-1$, we hae that $p^{1+i(k+1)}$ divides the term, and $1+i(k+1)\geq 2k+3\geq k+3.$ And when $i=p,$ we get that $p^{p(k+1)}$ divides the term, and $pk+p\geq k+3$ since $p\geq 3.$
So this means that:
$$b_k^p\equiv p^{k+2}x + 1=b_{k+1}\pmod{p^{k+3}}$$
This finishes the proof.
Your approach, trying to prove bounds on $\operatorname{val}_p\left(\binom{p^k}m\right),$ can be done.
We will show the rule that for $1\leq m\leq p^k,$:
$$\operatorname{val}_p\left(\binom{p^k}m\right)=k-\operatorname{val}_p(m)\tag{1}$$
This is due to the recursion:
$$\binom{p^k}{i +1}=\frac{p^k-i}{i+1}\binom{p^k}{i}$$
Which lets us write, for $0<m<p^k,$ that:
$$\begin{align}\operatorname{val}_p\left(\binom{p^k}m\right) &= \sum_{i=0}^{m-1}\left(\operatorname{val}_p(p^k-i) - \operatorname{val}_p(i+1)\right)\\
&=\operatorname{val}_p(p^k)-\operatorname{val}_p(m)+\sum_{i=1}^{m-1}\left(\operatorname{val}_p(p^k-i)-\operatorname{val}_p(i)\right)
\end{align}$$
But, for $0<i<p^k$, $\operatorname{val}_p(p^k-i)=\operatorname{val}_p(i).$ So we have (1).
Now we have for $m>1$ that $$m-\operatorname{val}_p(m)\geq 2.\tag{2}$$
To prove this, you need to prove the inequality for $p\geq 3$ and $j\geq 1$ that: $$p^j\geq 2+j.$$
You can prove this by induction. $p^1\geq 2+1$ and $p^{j+1}\geq 2p+pj=2+2(p-1)+pj> 2+1+j.$
(1) and (2) together give you that, when $1<m\leq p^k$ then:
$$m+\operatorname{val}_p\left(\binom{p^k}m\right)=m+k-\operatorname{val}_p(m)\geq k+2$$
This shows that: $$(1+px)^{p^k}=1+\binom{p^k}{1}px+\sum_{m=2}^{p^k}\binom{p^k}{m}p^{m}x^m\equiv 1+p^{k+1}x\pmod{p^{k+2}}$$
since the terms in the sum are all divisible by $p^{k+2}.$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the coefficient of $x^5$ in $(1+x+x^2+x^3+x^4+x^5)^{17}$? I figured that $(1+x+x^2+x^3+x^4)^{17} = (1-x^6)^{17}*(1-x)^{-17}$ but don't know what else to do.
I would really appreciate any help
| Since
$$
\left( {1 + x + x^2 + x^3 + x^4 } \right)^{\,17} = \left( {{{1 - x^5 } \over {1 - x}}} \right)^{\,17}
$$
you might be interested to know that the general
case of what are somewhere called r-nomial coefficients corresponds to the O.G.F:
$$
F_b (x,r,m) = \sum\limits_{0\,\, \leqslant \,\,s\,\,\left( { \leqslant \,\,r\,m} \right)} {N_b (s,r,m)\;x^{\,s} }
= \left( {1 + x + \cdots + x^{\,r} } \right)^m = \left( {\frac{{1 - x^{\,r + 1} }}{{1 - x}}} \right)^m
$$
where
$$
N_b (s,r,m)\quad \left| {\;0 \leqslant \text{integers }s,m,r} \right.\quad =
\sum\limits_{\left( {0\, \leqslant } \right)\,\,k\,\,\left( { \leqslant \,\frac{s}{r+1}\, \leqslant \,m} \right)}
{\left( { - 1} \right)^k \binom{m}{k}
\binom
{ s + m - 1 - k\left( {r + 1} \right) }
{ s - k\left( {r + 1} \right)}\ }
$$
You can find a full explanation in this related post and in this paper
In your particular case, if it is $(1+ \cdots+x^4)^{17}$
$$
\eqalign{
& N_b (5,4,17)\quad = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{s \over {r + 1}}\, \le \,m} \right)} {\left( { - 1} \right)^k \left( \matrix{
17 \cr
k \cr} \right)\left( \matrix{
21 - 5k \cr
5 - 5k \cr} \right)} = \cr
& = \left( \matrix{
17 \cr
0 \cr} \right)\left( \matrix{
21 \cr
5 \cr} \right) - \left( \matrix{
17 \cr
1 \cr} \right)\left( \matrix{
16 \cr
0 \cr} \right) = \left( \matrix{
21 \cr
5 \cr} \right) - 17 = 20322 \cr}
$$
or for $(1+ \cdots+x^5)^{17}$
$$
\eqalign{
& N_b (5,5,17)\quad = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{s \over {r + 1}}\, \le \,m} \right)} {\left( { - 1} \right)^k \left( \matrix{
17 \cr
k \cr} \right)\left( \matrix{
5 + 17 - 1 - 6k \cr
5 - 6k \cr} \right)} = \cr
& = \sum\limits_{\left( {0\, \le } \right)\,\,k\,\,\left( { \le \,{s \over {r + 1}}\, \le \,m} \right)} {\left( { - 1} \right)^k \left( \matrix{
17 \cr
k \cr} \right)\left( \matrix{
21 - 6k \cr
5 - 6k \cr} \right)} = \left( \matrix{
21 \cr
5 \cr} \right) = 20349 \cr}
$$
| {
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Eliminate $\alpha,\beta,\gamma$ from the system of equations Eliminate $\alpha,\beta,\gamma$ from the following system of equations.
$$a\cos(\alpha)+b\cos(\beta)+c\cos(\gamma)=0$$
$$a\sin(\alpha)+b\sin(\beta)+c\sin(\gamma)=0$$
$$a\sec(\alpha)+b\sec(\beta)+c\sec(\gamma)=0$$
My try:
Squaring and adding first two equations, we get:
$$\cos(\alpha-\beta)=\frac{c^2-a^2-b^2}{2ab}$$
$$\cos(\gamma-\beta)=\frac{a^2-c^2-b^2}{2cb}$$
$$\cos(\alpha-\gamma)=\frac{b^2-a^2-c^2}{2ac}$$
Now the RHS of above all looks like negative cosines of triangle $\Delta ABC$.
But I am not sure whether it will help.
This question is taken from plane trigonometry part 1 by SL Loney book. Page number 264, question number 176.
| We have the following system of equations
\begin{align}
a\cos(\alpha)+b\cos(\beta)+c\cos(\gamma) &= 0, \tag{1}\\
a\sin(\alpha)+b\sin(\beta)+c\sin(\gamma) &= 0, \tag{2}\\
a\sec(\alpha)+b\sec(\beta)+c\sec(\gamma) &= 0. \tag{3}
\end{align}
From (1) and (2), we have
$$a^2 + b^2 + 2ab \cos \alpha\cos \beta + 2ab \sin \alpha \sin \beta = c^2$$
and thus
$$(2ab)^2\sin^2 \alpha \sin^2\beta = (c^2 - a^2 - b^2 - 2ab\cos \alpha\cos \beta)^2. \tag{4}$$
From (1) and (3), we have
$$a^2 + b^2 + ab\frac{\cos \alpha}{\cos \beta} + ab \frac{\cos \beta}{\cos \alpha} = c^2$$
and thus
$$(2ab)^2(\cos^2 \alpha + \cos^2\beta) = 2\cdot (c^2 - a^2 - b^2)\cdot 2ab \cos \alpha \cos \beta. \tag{5}$$
$(4) + (5)$ gives
$$(2ab)^2(1 + \cos^2\alpha \cos^2\beta ) = (c^2 - a^2 - b^2)^2 + (2ab)^2 \cos^2 \alpha \cos^2 \beta$$
and thus
$$(2ab)^2 = (c^2 - a^2 - b^2)^2$$
which can be written as
$$- a^4 - b^4 - c^4 + 2a^2b^2 + 2b^2c^2 + 2c^2a^2 = 0$$
or
$$(a + b + c)(a + b - c)(b + c - a)(c + a - b) = 0.$$
We are done.
| {
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Finding $\lim\limits _{x\to-\infty} \frac{\sqrt[3]{x^3+x}}{x}$
$$\lim\limits _{x\to-\infty} \frac{\sqrt[3]{x^3+x}}{x}$$
I have to resolve this limit, I tried factoring out x, I tried rewriting $x^3+x$ as
$x^3(1+ \frac{x}{x^3})$ and it doesn't seem to cancel. What should I do?
| $$\lim_{x\to-\infty} \frac{\sqrt[3]{x^3+x}}{x}=\lim_{x\to-\infty} \sqrt[3]{\frac{x^3+x}{x^3}} =\lim_{x\to-\infty} \sqrt[3]{1+\frac{1}{x^2}} = \sqrt[3]{1+\lim_{x\to-\infty}\frac{1}{x^2}}=1$$
| {
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Solve ODE $y'' = y^3 -y y'$ I want to solve $y'' = y^3 -y y'$ with boundary conditions $y(1) = 1/2$ and $y(2) = 1/3$ but not sure how to start. If someone could give me a complete step by step explanation, it would be greatly appreciated because I want to fully understand it.
Edit:
By the hint $(y'+\frac{y^2}{2})'=y^3$
$y' + \frac{y^2}{2} = \frac{y^4}{4} + c1$
$\frac{dy}{dx} = \frac{y^4}{4} - \frac{y^2}{2} + c1$
Integrate $\frac{dy}{\frac{y^4}{4} - \frac{y^2}{2} + c1} = dx$
Integration looks very messy.
I know that the answer is $y = \frac{1}{(x+1)}$
| Hint:
$$(y'+\frac{y^2}{2})'=y^3$$
| {
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If $\sin(18^\circ)=\frac{a + \sqrt{b}}{c}$, then what is $a+b+c$? If $\sin(18)=\frac{a + \sqrt{b}}{c}$ in the simplest form, then what is $a+b+c$?
$$ $$
Attempt:
$\sin(18)$ in a right triangle with sides $x$ (in front of corner with angle $18$ degrees), $y$, and hypotenuse $z$, is actually just $\frac{x}{z}$, then $x = a + \sqrt{b}, z = c$. We can find $y$ as
$$ y = \sqrt{c^{2}- (a + \sqrt{b})^{2}} $$
so we have
$$ \cos(18) = \frac{y}{z} = \frac{\sqrt{c^{2}- (a + \sqrt{b})^{2}}}{c}$$
I also found out that
$$b = (c \sin(18) - a)^{2} = c^{2} \sin^{2}(18) - 2ac \sin(18) + a^{2}$$
I got no clue after this.
The solution says that $$ \sin(18) = \frac{-1 + \sqrt{5}}{4} $$
I gotta intuition that we must find $A,B,C$ such that
$$ A \sin(18)^{2} + B \sin(18) + C = 0 $$
then $\sin(18)$ is a root iof $Ax^{2} + Bx + C$, and $a = -B, b = B^{2} - 4AC, c = 2A$.
Totally different. This question is not asking to prove that $sin(18)=(-1+\sqrt{5})/4$, that is just part of the solution.
| Hint This is a classic problem and probably a duplicate here, but I was unable to find another instance of this question as written. Edit Found one.
First, note that $\sin 18^\circ = \cos 72^\circ$; we find the latter.
Recall that (for $n \geq 2$) the sum of the $n$th roots of unity sum to zero, and taking real parts leaves $$\sum_{k = 0}^{n - 1} \cos \left(\frac{k}{n} \cdot 360^\circ\right) = 0 .$$
Setting $n = 5$ and using the symmetry of the cosine function gives
$$1 + 2 \cos 72^\circ + 2 \cos 144^\circ = 0 .$$
Using the double angle formula---in particular that $$\cos 144^\circ = 2 \cos^2 72^\circ - 1$$ ---and substituting in the previous display equation gives a quadratic expression in $\cos 72^\circ$:
$$4 \cos^2 72^\circ + 2 \cos 72^\circ - 1 = 0 .$$
By the quadratic equation,
$$\cos 72^\circ = \frac{-2 + \sqrt{(2)^2 - 4 (4) (-1)}}{2(4)} = \frac{-1 + \sqrt{5}}{4} .$$
(To resolve the $\pm$ ambiguity it's enough to know that $\cos 72^{\circ} > 0$.)
| {
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Is this valid when deriving quadratic equation? When deriving the quadratic formula, isn't the square root of $(x+\frac{b}{2a})^2$ the absolute value of $(x+\frac{b}{2a})$? It's usually just represented as $(x+\frac{b}{2a})$ without absolute value and then $\frac{b}{2a}$ is subtracted from left side and boom, theres the quadratic formula. I just don't understand why its not absolute value of $(x+\frac{b}{2a})$. For example the square root of $x^2$ is the absolute value of $x$, which is equal to $\pm{x}$. Sorry if this is confusing its more of a conceptual thing. Thank you in advance.
philalethesnew
| You can look at it differently:
$$ax^2+bx+c=0 \iff \\
x^2+\frac bax+\frac ca=0\iff \\
\left(x+\frac b{2a}\right)^2-\frac{b^2}{4a^2}+\frac{4ac}{4a^2}=0 \iff \\
\left(x+\frac b{2a}\right)^2-\left(\frac{\sqrt{b^2-4ac}}{2a}\right)^2=0 \iff \\
\left[\left(x+\frac b{2a}\right)-\frac{\sqrt{b^2-4ac}}{2a}\right]\cdot \left[\left(x+\frac b{2a}\right)+\frac{\sqrt{b^2-4ac}}{2a}\right]=0 \iff \\
\left(x+\frac b{2a}\right)-\frac{\sqrt{b^2-4ac}}{2a}=0 \ \ \text{or} \ \ \left(x+\frac b{2a}\right)+\frac{\sqrt{b^2-4ac}}{2a}=0 \iff \\
x=\frac{-b+\sqrt{b^2-4ac}}{2a} \ \ \text{or} \ \ x=\frac{-b-\sqrt{b^2-4ac}}{2a} \iff \\
x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
| {
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Does there exist a function $f_{\Box,\Box}(\Box)$ making the formula $a + (b \oplus c) = (f_{b,c}(a)+b) \oplus (f_{c,b}(a)+c)$ true? Let $a$ and $b$ denote the resistances of two resistors. If they're put in series, the total resistance is $a+b$. If they're put in parallel, the total resistance is $$a \oplus b := \frac{1}{\frac{1}{a}+\frac{1}{b}} = \frac{ab}{a+b}.$$
I suspect there's a formula describing how $+$ "distributes over" $\oplus$.
It should be of the form
$$a + (b \oplus c) = (f_{b,c}(a)+b) \oplus (f_{c,b}(a)+c)$$
for an appropriate choice of $f$. I haven't been able to find an $f$ that works, however.
Question. Does there exist an $f$ making the above formula true? If not, why not?
Remark. Unpacking the definitions, we're looking for a function $f$ such that
$$(ab+ac+bc)(f_{b,c}(a)+b + f_{c,b}(a)+c)$$
and
$$(b+c)(f_{b,c}(a)+b) (f_{c,b}(a)+c)$$
are equal.
| Let me assume that $b\ge0,c\ge0, b+c>0$.
I am looking for $f$ satisfying $f_{b,c}(a) = f_{c,b}(a)$.
The equation in the OP is then equivalent to
$$
(b+c) f_{bc}(a)^2 + ((b+c)^2-2ab-2bc-2ca)f_{bc}(a)- (b+c)(ab+bc+ca)=0.
$$
Setting $d:=ab+bc+ca$ and dividing by $b+c$ this is equivalent to
$$
f_{bc}(a)^2 + (b+c -\frac{2d}{b+c}) f_{bc}(a)- d=0.
$$
The roots of this equation are
$$\begin{split}
f_{bc}(a) &= -\frac{ (b+c)^2 - 2d}{2(b+c)} \pm \frac1{b+c} \sqrt{ \frac14((b+c)^2 - 2d)^2+d(b+c)^2}\\
&=-\frac{ (b+c)^2 - 2d \pm\sqrt{ (b+c)^4 +d^2}}{2(b+c)},
\end{split}
$$
one of them is positive, which gives you a formula for $f$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "9",
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Geometry: Find the radius of a circle given parallel secants Suppose a circle has two parallel chords of lengths $a$ and $b$, and the chords are separated by a distance of $c$. Using only the usual high school geometry theorems (i.e. no trig or calculus), can we derive a formula for the radius?
I've tried drawing radii in several places without making useful progress. I can't see how to find the radius intersecting the circle and a chord. I can draw the segment from one chord-circle intersection to another, but it need not pass through the center so I can't leverage this to get the radius.
| I just answered this question in chat, so I thought it would be worth posting the answer here.
The horizontal position of a vertical chord of length $c$ in a circle of radius $r$ would be $\pm\sqrt{r^2-\frac{c^2}4}$.
Thus, the distance, $d$, between chords of length $c_1$ and $c_2$ (with $c_1\ge c_2$) would be
$$
d=\sqrt{r^2-\tfrac{c_2^2}4}\pm\sqrt{r^2-\tfrac{c_1^2}4}\tag1
$$
The "$\pm$" depends on whether or not the chords appear on the same side of the center of the circle. In either case, we have
$$
\frac{c_1^2-c_2^2}{4d}=\sqrt{r^2-\tfrac{c_2^2}4}\mp\sqrt{r^2-\tfrac{c_1^2}4}\tag2
$$
To verify $(2)$, simply multiply $(1)$ and $(2)$.
Average $(1)$ and $(2)$ to get
$$
\frac{c_1^2-c_2^2+4d^2}{8d}=\sqrt{r^2-\tfrac{c_2^2}4}\tag3
$$
Thus,
$$
\begin{align}
r^2
&=\frac{\left(c_1^2-c_2^2+4d^2\right)^2+16c_2^2d^2}{64d^2}\tag{4a}\\
&=\frac{c_1^4+c_2^4+16d^4-2c_1^2c_2^2+8c_1^2d^2+8c_2^2d^2}{64d^2}\tag{4b}\\
&=\frac{\left(c_1^2+c_2^2\right)^2+8\left(c_2^2+c_1^2\right)d^2+16d^4-4c_1^2c_2^2}{64d^2}\tag{4c}\\
&=\frac{\left(c_1^2+c_2^2+4d^2\right)^2-4c_1^2c_2^2}{64d^2}\tag{4d}\\
&=\frac{\left((c_1+c_2)^2+4d^2\right)\left((c_1-c_2)^2+4d^2\right)}{64d^2}\tag{4e}
\end{align}
$$
Explanation:
$\text{(4a):}$ solve $(3)$ for $r^2$
$\text{(4b):}$ expand
$\text{(4c):}$ collect
$\text{(4d):}$ collect
$\text{(4e):}$ apply $a^2-b^2=(a+b)(a-b)$
Therefore,
$$
r=\frac{\sqrt{\left((c_1+c_2)^2+4d^2\right)\left((c_1-c_2)^2+4d^2\right)}}{8d}\tag5
$$
In $(5)$, it doesn't matter whether $c_1\ge c_2$ or not.
| {
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} |
integration of a gaussian with $x^2$ I need to integrate
$$\int_{-\infty}^{\infty} x^2 e^{-ax^2} \qquad \text{where } a\in R$$
The book does the following:
I don't understand what's happening. I tried solving the integral using integration by parts and this is what I got
$$
\begin{align}
\int_{-\infty}^{\infty} x^2 e^{-ax^2} &= x^2\sqrt{\frac{\pi}{a}} - \int_{-\infty}^{\infty} \sqrt{\frac{\pi}{a}} 2x dx && \text{as we are told } \int_{-\infty}^{\infty}dx e^{-ax^2}= \sqrt{\frac{\pi}{a}}\\
&=x^2\sqrt{\frac{\pi}{a}} - 2\sqrt{\frac{\pi}{a}}\frac{x^2}{2} && \text{as } \int xdx = \frac{x^2}{2} \\
&= 0
\end{align}
$$
What am I doing wrong?
Actually, I might have found a way of solving this
$$
\begin{align}
\int_{-\infty}^{\infty} x^2 e^{-ax^2} dx &= \int_{-\infty}^{\infty} x\cdot x e^{-ax^2} dx \\
&= - \frac{1}{2a} \int_{-\infty}^{\infty} x(-2ax)e^{-ax^2} dx \\
&= -\frac{1}{2a}\left(\left[e^{-ax^2}\right]_{-\infty}^{\infty} - \int_{-\infty}^{\infty} e^{-ax2} dx\right) \\
&= -\frac{1}{2a}\left(0 - \sqrt{\frac{\pi}{a}}\right) \\
&= \frac{1}{2a}\sqrt{\frac{\pi}{a}}
\end{align}
$$
| Good job for identifying the components used for integration by parts.
If you are familiar with normal distribution, here's an intepretation for $a>0$,
Let $\frac{1}{2\sigma^2}=a$
\begin{align}\int_{-\infty}^\infty x^2 \exp(-ax^2) \, dx &=\sqrt{2\pi \sigma^2}\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi\sigma^2}}x^2 \exp\left(-\frac{x^2}{2\sigma^2}\right) \, dx\\
&= \sqrt{2\pi \sigma^2}\left(\sigma^2+0^2 \right)\\
&=\frac1{2a}\sqrt{\frac{\pi}{a}}\end{align}
where I have used the property that $Var(X)=E(X^2)-E(X)^2$
| {
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} |
Range of rational expression having algebraic terms
Find the range of $\displaystyle f(x) = \frac{x^2+x-1}{x^2-x+2}$ subjected to $-1 \leq x\leq 1$.
Plan
\begin{align} y = \frac{x^2+x-1}{x^2-x+2}&\implies yx^2-yx+2y=x^2+x-1\\&\implies(y-1)x^2-(y+1)x+(2y+1)=0.\end{align}
For $y=1$, we have $x=3/2$.
For $y\neq 1$,
$$(y+1)^2-4(y-1)(2y+1)\geq 0$$
$$y^2+2y+1-4(2y^2-y-1)\geq 0$$
$$7y^2-6y-5\leq 0$$
$$y\in \bigg[\frac{3-2\sqrt{21}}{7},\frac{3+2\sqrt{21}}{7}\bigg]$$
But the solution is $\displaystyle \bigg[\frac{3-2\sqrt{21}}{7},\frac{1}{2}\bigg]\cup \{1\}$.
Help me please.
| I will propose an alternative solution using calculus (as this tag has been added by OP).
Rewrite $$f(x)=\frac{x^2+x-1}{x^2-x+2}=1+\frac{2x-3}{x^2-x+2}$$ and note that $f$ is defined in the interval $[-1,1]$.
Now we find the stationary points. Thus $$f'(x)=\frac{2(x^2-x+2)-(2x-3)(2x-1)}{(x^2-x+2)^2}=0\implies2x^2-6x-1=0$$ giving the stationary points $x_+,x_-=\frac12(3\pm\sqrt{11})$.
Checking using $f''$ or a nature table reveals that $f$ is at its maximum at $x=x_+$ and is at its minimum at $x=x_-$. However, only $x_-$ is contained within the interval $[-1,1]$, but we know that in the interval $(x_-,x_+)$, $f$ is strictly increasing.
Therefore, the range of $f$ in said interval is simply $$[f(x_-),f(1)]=\left[1+\frac{2\cdot\frac{3-\sqrt{11}}2-3}{\frac{(3-\sqrt{11})^2}4-\frac{3-\sqrt{11}}2+2},1+\frac{2(1)-3}{1^2-1+2}\right]=\left[\frac{3-2\sqrt{11}}7,\frac12\right].$$
Here is a plot which verifies this.
| {
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real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Number of real solution of equation $(x^2+6x+7)^2+6(x^2+6x+7)+7=x$ is
Plan
Put $x^2+6x+7=f(x)$. Then i have $f(f(x))=x$
For $f(x)=x$
$x^2+5x+7=0$ no real value of $x$
For $f(x)=-x$
$x^2+8x+7=0$
$x=-7,x=-1$
Solution given is all real solution
Help me please
| On completing the square, we have
\begin{align}
x= (x^2+6x+7)^2+6(x^2+6x+7)+7 &= ((x+3)^2-2)^2+6((x+3)^2-2)+7\\
&= (x+3)^4-4(x+3)^2+4+6(x+3)^2-5\\
&= (x+3)^4+2(x+3)^2-1= ((x+3)^2+1)^2-2 \geq 1-2=-1
\end{align}
Further, we alsoknow that $(x+3)^2$ is monotonically increasing for $x\geq -3$. Thus, we get that
\begin{align}
x= ((x+3)^2+1)^2-2\geq ((-1+3)^2+1)^2-2 = 23.
\end{align}
Further, $((x+3)^2+1)^2-2\geq 3^34x+3^4-2\geq x, \forall x\geq 23$.
| {
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} |
Evaluation of series $\sum_{n=0}^\infty\frac{5n+1}{(2n+1)!}$ How to evaluate series
$$\sum_{n=0}^\infty\frac{5n+1}{(2n+1)!}$$
I tried to split the summation...but I failed. Please help
| $$\sinh x=\sum_{n=0}^{\infty}\frac{x^{2n+1}}{(2n+1)!}$$
$$\frac{1}{x}\sinh x=\sum_{n=0}^{\infty}\frac{x^{2n}}{(2n+1)!}$$
let $x\rightarrow \sqrt{x}$
$$\frac{1}{\sqrt{x}}\sinh \sqrt{x}=\sum_{n=0}^{\infty}\frac{x^{n}}{(2n+1)!}$$
let $x\rightarrow x^5$
$$\frac{1}{\sqrt{x^5}}\sinh \sqrt{x^5}=\sum_{n=0}^{\infty}\frac{x^{5n}}{(2n+1)!}$$
multiply by $x$
$$\frac{x}{\sqrt{x^5}}\sinh \sqrt{x^5}=\sum_{n=0}^{\infty}\frac{x^{5n+1}}{(2n+1)!}$$
$$(\frac{x}{\sqrt{x^5}}\sinh \sqrt{x^5})'=\sum_{n=0}^{\infty}\frac{(5n+1)x^{5n}}{(2n+1)!}$$
then let $x=1$ to get the sum
| {
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} |
Rotation Matrix and Triple Angle Formulas? Define $R_{\theta}:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ as the rotation matrix by angle $\theta$, where
$$R_{\theta} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$
Observe that
$$
(R_\theta)^2 = \begin{pmatrix} \cos^2\theta-\sin^2\theta & -2\sin\theta\cos\theta \\ 2\sin\theta\cos\theta & \cos^2\theta-\sin^2\theta \end{pmatrix}
= \begin{pmatrix} \cos2\theta & -\sin2\theta \\ \sin2\theta & \cos2\theta \end{pmatrix}=R_{2\theta}
$$
This all makes sense of course since if you rotate a vector by $\theta$ twice, the net result should be a rotation by $2\theta$. The algebra of it all can be verified with the double angle formulas.
However, how do you prove that
$$
(R_\theta)^3 = R_{3\theta}
$$
or perhaps that
$$
(R_\theta)^n = R_{n\theta}
$$
Are there triple angle formulas that can be used to make the algebra work? n-tuple angle formulas?
| If $R_{\theta}:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is the rotation matrix by angle $\theta$, where
$$R_{\theta} = \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{pmatrix}$$
Then we can write
$$R_{\theta} = \begin{pmatrix} \cos\theta & 0 \\ 0 & \cos\theta \end{pmatrix} + \begin{pmatrix} 0 & -\sin\theta \\ \sin\theta & 0 \end{pmatrix}$$
$$ = \cos\theta\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} + \sin\theta\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$
Noting that $$I = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}$$ is the identity matrix and $$J=\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$ satisfies $$J^2=-\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix} = -I$$
Thus we have $$R_{\theta} =\cos{\theta}I+\sin{\theta}J$$ with $J^2=-I$
We can then make the analogy with De Moivre's equation $$e^{i \theta}=\cos{\theta}+i\sin{\theta}$$
This suggests that $$e^{I \theta}=R_{\theta}=\cos{\theta}I+\sin{\theta}J$$
This can be proved by considering power series as in the case of DeMoivre's theorem. Clearly then we have $$R_{n\theta}=e^{I n\theta}={(e^{I \theta})}^n=R_{\theta}^n$$
Alternatively you can avoid power series and use trig formulas or De Moivre's theorem directly by showing that $$R_{\theta+\alpha}=\cos{(\theta+\alpha)}I+\sin{(\theta+\alpha)}J = R_{\theta}R_{\alpha}$$ on expanding the sin and cosine terms. $R_{n\theta}=R_{\theta}^n$ then follows immediately.
| {
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"source": "stackexchange",
"question_score": "2",
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Evaluate $ \int \frac{x^4}{(2-x^2)^{3/2}}dx$
Evaluate the following integral:
$ \int \frac{x^4}{(2-x^2)^{3/2}}dx$
I've tried to apply Chebyshev theorem on the integration of binomial differentials.
We have $ m=4,a=2,b=-1,n=2,p=-3/2$.
$\frac{m+1}{n}+p$ is integer then we do the substitution
$t^2=2x^{-2}-1$, $x^2=\frac{2}{t^2+1}$
It go me there: $\int\frac{2}{t^2+1}^2(2-\frac{2}{t^2+1})^{-3/2}-\sqrt{2}t(\frac{1}{t^2+1})^{3/2}dt$ which is just more complicated expression. Where I went wrong?
| $\int \frac{x^4}{(2-x^2)^{3/2}}dx$
Comparing with $\int x^m (a+bx^n)^p dx$
$m=4,a=2,b=-1,n=2,p=-\frac{3}{2}$
Here $\frac{m+1}{n}+p=\frac{5}{2}-\frac{3}{2}=1$, integer.
So we first transform the integral $ x^m (a+bx^n)^p $ by factoring $x^n$ i.e., by $x^2$ in this way
$\int \frac{x^4}{(2-x^2)^{\frac{3}{2}}}dx=\int {x^4}{(2-x^2)^{-\frac{3}{2}}}dx=\int {x^4}{(x^2)^{-\frac{3}{2}}}{(2x^{-2}-1)^{-\frac{3}{2}}}dx=\int x{(2x^{-2}-1)^{-\frac{3}{2}}}dx$
Putting $2x^{-2}-1=z$ give $-4x dx =dz\implies x dx=-\frac{1}{4}dz$
Now $\int \frac{x^4}{(2-x^2)^{\frac{3}{2}}}dx=\int x{(2x^{-2}-1)^{-\frac{3}{2}}}dx=-\frac{1}{4}\int {z^{-\frac{3}{2}}}dz=-\frac{1}{4}\frac{z^{-\frac{1}{2}}}{-\frac{1}{2}}+c=\frac{1}{2} {(2x^{-2}-1)}^{-\frac{1}{2}}+c$
| {
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Question about the proof of simple algebra rule $\frac{1}{\frac{1}{a}} = a$ I have a question about a proof I saw in a book about basic algeba rules. The rule to prove is:
\begin{eqnarray*}
\frac{1}{\frac{1}{a}} = a, \quad a \in \mathbb{R}_{\ne 0}
\end{eqnarray*}
And the proof:
\begin{eqnarray*}
1 = a \frac{1}{a} \Longrightarrow 1 = \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow a = a \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow \frac{1}{\frac{1}{a}} = a
\end{eqnarray*}
Why is it allowed to just replace $a$ with $1/a$? What's the explanation behind it?
| Let $x=\frac{1}{a}$. Then:
\begin{eqnarray*}
1 = x \frac{1}{x} \Longrightarrow 1 = \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow a = a \frac{1}{a} \frac{1}{\frac{1}{a}} \Longrightarrow \frac{1}{\frac{1}{a}} = a
\end{eqnarray*}
Better? You're right, replacing $a\to\frac{1}{a}$ is a minor abuse of notation, because they're implicitly changing the variable without telling you. But you can fix that by just letting $x=\frac{1}{a}$ at the start.
| {
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"question_score": "4",
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Proving that, for an acute $\triangle ABC$, $\sin A + \sin B+\sin C\gt \cos A+\cos B+\cos C$
I need to prove or disprove that in any acute $\triangle ABC$, the following property holds:
$$\sin A + \sin B + \sin C \gt \cos A + \cos B + \cos C$$
To begin, I proved a lemma:
Lemma. An acute triangle has at most one angle which is less than or equal to $\dfrac{\pi}{4}$.
Proof:
Let there be an acute angled $\Delta ABC$ with the angles $A$ & $B \le \frac{\pi}{4}$. Then
$$ A + B \le \frac{\pi}{2}
\implies - (A + B) \ge -\frac{\pi}{2}
\implies C = \pi - (A+B) \ge \frac{\pi}{2}$$
thus contradicting that the triangle is obtuse. Hence, by contradiction, the lemma is proved. $\square$
Further, I used the identity that $\sin x - \cos x = \sqrt{2}\sin (x - \frac{\pi}{4})$ to rewrite the inequality as
$$\sin \biggr(A - \frac{\pi}{4}\biggr) + \sin \biggr(B - \frac{\pi}{4}\biggr) + \sin \biggr(C - \frac{\pi}{4}\biggr) \gt 0$$
Without loss of generality, I assumed that $A \le \frac{\pi}{4}$.
If $A = \dfrac{\pi}{4}$, then the inequality follows, since both $B$ and $C$ are strictly greater than $\dfrac{\pi}{4}$.
How do I prove the inequality if $A \lt \dfrac{\pi}{4}$?
Any help or hint will be appreciated.
| Your problem is equivalent to the following:
In any acute $ABC$ we have $AB+BC+CA>AH+BH+CH$ , where $H$ is the orthocenter of the triangle. But this inequality is true for all points from the tringle's interior.
| {
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Closed form for sum of powers I wonder if it is possible to evaluate explicitly the sum
$$S(N):=\sum_{j=1}^{\left\lfloor\frac{N-1}{2}\right\rfloor}\left(1-\frac{2j}{N}\right)^{N+1},\quad N\in\mathbb{N}.$$
In the large $N$ limit the result is straightforward if one approximates the argument of the sum with the exponential and then evaluates the sum as a geometric one
$$S(N)\approx\sum_{j=1}^{\left\lfloor\frac{N-1}{2}\right\rfloor}e^{-\frac{2(N+1)}{N}j}=\frac{1-e^{-N+\frac{1}{N}}}{e^{2+\frac{2}{N}}-1}\xrightarrow{N\to\infty}\frac{1}{e^2-1}.$$
In the general case I tried to manipulate the sum in some ways, considering separately odd or even $N$.
In the first case I wrote $S(N)$ in terms of the generalized Zeta function
$$\zeta(s,a)=\sum_{k=0}^{\infty}(k+a)^{-s},$$
starting as follows
$$\sum_{j=1}^{\frac{N-1}{2}}\left(1-\frac{2j}{N}\right)^{N+1}=\sum_{j=1}^{\frac{N-1}{2}}\left(\frac{2j-1}{N}\right)^{N+1}=\frac{1}{N^{N+1}}\left[\sum_{j=1}^{\infty}(2j-1)^{N+1}-\sum_{j=\frac{N+1}{2}}^{\infty}(2j-1)^{N+1}\right].$$
Note that the first passage holds because the summed terms are exactly the same, with only their order changed. The first term in the braket is $0$ for odd $N$ because it can be rewritten as
$$\sum_{j=1}^{\infty}j^{N+1}-\sum_{j=1}^{\infty}(2j)^{N+1}=\left(1-2^{N+1}\right)\zeta(-N-1)$$
but $\zeta(-2m)=0,\;m\in\mathbb{N}$. The change of variable $j\rightarrow j-(N+1)/2$ on the remaining term leads to the result
$$S(N)=-\left(\frac{2}{N}\right)^{N+1}\zeta\left(-N-1,\frac{N}{2}\right),\quad N\;\rm{odd}.$$
Another possibility is to use the generalized harmonic numbers
$$H_n^{(m)}=\sum_{k=1}^n\frac{1}{k^m}.$$
Doing so for odd $N$ it holds
$$S(N)=\frac{1}{N^{N+1}}\sum_{j=1}^{\frac{N-1}{2}}(2j-1)^{N+1}=\frac{1}{N^{N+1}}\left(\sum_{j=1}^{N-2}j^{N+1}-\sum_{j=1}^{\frac{N-3}{2}}(2j)^{N+1}\right)=\frac{1}{N^{N+1}}\left(H_{N-2}^{(-N-1)}-2^{N+1}H_{\frac{N-3}{2}}^{(-N-1)}\right).$$
Instead for even $N$
$$S(N)=\frac{1}{N^{N+1}}\sum_{j=1}^{\frac{N}{2}-1}(2j)^{N+1}=\left(\frac{2}{N}\right)^{N+1}H_{\frac{N}{2}-1}^{(-N-1)}.$$
I can't figure out if further manipulations are possible to simplify my results, or if a completely different approach may exist to treat such a sum.
| The finite sum can be expressed in terms of
Bernoulli polynomial $B_p(x)$. Bernoulli polynomials obey many interesting relations, the one we need
is
$$B_p(x+1) - B_p(x) = px^{n-1}$$
Let $J = \lfloor \frac{N-1}{2}\rfloor$ and using this relation, we find
$$\begin{align}S(N)
&=\sum_{j=1}^J\left(1-\frac{2j}{N}\right)^{N+1}\\
&= \left(\frac{2}{N}\right)^{N+1} \sum_{j=1}^J \left(\frac{N}{2}-j\right)^{N+1}\\
&= \frac{1}{N+2}\left(\frac{2}{N}\right)^{N+1} \sum_{j=1}^J \left[B_{N+2}\left(\frac{N}{2}-j+1\right) - B_{N+2}\left(\frac{N}{2} - j\right)\right]\\
&= \frac{1}{N+2}\left(\frac{2}{N}\right)^{N+1}\left[B_{N+2}\left(\frac{N}{2}\right) - B_{N+2}\left(\frac{N}{2}-J\right)\right]
\end{align}
$$
| {
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"source": "stackexchange",
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Integral test to find constant for which series is convergent Problem statement:
Use the integral test to find the value of $b$ for which the following series is
convergent
$$ \sum_{r=2}^{\infty}\frac{8}{3(br+1)}-\frac{3}{2(r+1)}+\frac{1}{6(r-1)}.$$
I evaluated the corresponding integral, which gave the following limit,
$$\lim_{m \to \infty}\left\{ \frac{8}{3b}\ln\left( \frac{mb+1}{2b+1} \right)+\frac{3}{2}\ln\left(\frac{3}{m+1}\right)+\frac{1}{6}\ln(m-1) \right\}.$$
I tried a few different things from there, but with no luck. How do I find the value of $b$ for which this limit exists?
| Hint $\sum_{r=2}^{\infty}\frac{8}{3(br+1)}-\frac{3}{2(r+1)}+\frac{1}{6(r-1)}$ is convergent implies
$$\lim_{r \to \infty} \frac{8}{3(br+1)}-\frac{3}{2(r+1)}+\frac{1}{6(r-1)} =0$$
This yields $b =\mbox{ something}$.
You can then use the integral test to decide if the series converges for this value of $b$.
Alternately
$$\frac{8}{3b}\ln\left( \frac{mb+1}{2b+1} \right)+\frac{3}{2}\ln\left(\frac{3}{m+1}\right)+\frac{1}{6}\ln(m-1) =\\
\ln \left(\frac{\sqrt{27}}{(2b+1)^\frac{8}{3b}} \frac{(mb+1)^\frac{8}{3b}(m-1)^\frac{1}{6}}{(m+1)^\frac{3}{2}}\right)$$
Now compare the powers:
*
*If $\frac{8}{3b}+\frac{1}{6}-\frac{3}{2}>0$
then
$$\ln \left(\frac{(mb+1)^\frac{8}{3b}(m-1)^\frac{1}{6}}{(m+1)^\frac{3}{2}}\right) \to \infty$$
*
*If $\frac{8}{3b}+\frac{1}{6}-\frac{3}{2}<0$
then
$$\ln \left(\frac{(mb+1)^\frac{8}{3b}(m-1)^\frac{1}{6}}{(m+1)^\frac{3}{2}}\right) \to -\infty$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3239535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find minimum value of $P=17x^2+17y^2+16xy$ Let $x,y$ be positive real numbers such that $4x^2+4y^2+17xy+5x+5y\ge 1$. Find minimum value of $$P=17x^2+17y^2+16xy$$
My idea:
I see that $x=y=\frac{\sqrt 2 -1}{5}\rightarrow P=2(3-2\sqrt 2)$
So I proved $P\ge 2(3-2 \sqrt 2)(4x^2+4y^2+17xy+5x+5y)\ge2(3-2\sqrt 2)$
Is my approachment is true?
| Yes, your answer is true.
The solution can be the following.
By AM-GM
$$1\leq4(x+y)^2+5(x+y)+9xy\leq4(x+y)^2+5(x+y)+\frac{9}{4}(x+y)^2,$$ which gives
$$x+y\geq\frac{2(\sqrt2-1)}{5}.$$
Thus, by C-S we obtain:
$$17(x^2+y^2)+16xy=8(x+y)^2+\frac{9}{2}(1^2+1^2)(x^2+y^2)\geq8(x+y)^2+\frac{9}{2}(x+y)^2=$$
$$=\frac{25}{2}(x+y)^2\geq\frac{25}{2}\cdot\frac{4(3-2\sqrt2)}{25}=2(3-2\sqrt2).$$
The equality occurs for $x=y=\frac{\sqrt2-1}{5},$ which says that we got a minimal value.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Gravity of a Circular Ring at a co-planar external point. I have a circular ring of unit mass and fixed radius R which lies in the XY plane at point $O$ with coordinates $O:(0,0)$.
I wish to find a formula for the gravitational force at a point $P: (D,0)$ which lies in the same plane as the ring and is at some variable distance D from the ring centre O.
(Note: There are many treatments of the case for a target lying on the axis of the ring. The reference so far found nearest to this co-planar case is Problems 5-12, 5-13, (no solutions given) p.127 in Classical Dynamics of Particles and Systems by Jerry B. Marion.
I expect that the formula should be of the form $F = GM*f(D)$ where $G$ is the gravitational constant, $M$ is the mass and $f$ is some function similar to the Newtonian spherical divergence function $f(D) = \frac{1}{ D^2}$ (where the factor $\frac{1}{4.\pi}$ is absorbed in the value of the constant $G$ ).
So far I have obtained an integral formula by initially modelling the ring as a series of $N$ small point masses of mass $\frac{1}{N}$ separated by angle $\delta\theta$, whose distance from target is $L$ where:
$$L^2 = (D-a)^2+b^2 = D^2-2aD+R^2 = D^2\left(1 -\frac{2a}{D} +\frac{R^2}{D^2}\right)$$
where $a (= R\cos\theta)$ and $b(=R\sin\theta)$ are the $x$ and $y$ coordinates of the point.
Due to symmetry and vector addition of forces there is no net force in the y-direction and so the effective force contribution (along $x$) for a point is given by multiplying by the cosine factor $(D-a)/L$ thus:-
$$ F = \frac{-GM}{N}\frac{1}{4\pi.L^2}\frac{D-a}{L}
= \frac{-GM}{ N} \frac{D-a}{L^3} $$
$$ F = \frac{-GM}{ N} \frac{D-R\cos\theta}{\left(D^2\left(1 -\frac{2a}{D} +\frac{R^2}{D^2}\right)\right)^{\frac{3}{2}}} $$
$$ F = \frac{-GM}{ N} \frac{D-R\cos\theta}{D^3 \left(1 -\frac{2a}{D} +\frac{R^2}{D^2} \right)^{\frac{3}{2}}} $$
$$ F = \frac{-GM}{ N} \frac{1-(R/D)\cos\theta}{D^2 \left(1 -\frac{2a}{D} +\frac{R^2}{D^2} \right)^{\frac{3}{2}}} $$
I then obtained the following integral formula for the force exerted on the target point by the ring:-
$$ F = \frac{-GM}{ D^2} \frac{1}{2\pi}\int_0^{2\pi}\frac{1-Q\cos\theta}{\left(1-2Q\cos\theta+Q^2\right)^{\frac{3}{2}}} \text{d}\theta$$
where $Q = R/D$.
$$ F = \frac{-GM}{ D^2} \frac{1}{2\pi} \frac{1}{(2Q)^{3/2}}\int_0^{2\pi}\frac{1-Q\cos\theta}
{\left(\frac{Q^2+ 1}{2Q} - \cos\theta\right)^{\frac{3}{2}}} \text{d}\theta$$
Defining $A = \frac{Q^2+ 1}{2Q}$, Wolfram Alpha gives...
$$ \int_0^{2\pi}\frac{ 1 - Q \cos x}{(A -\cos x)^{3/2}} dx $$
$$=\left[\frac{2}{(A^2-1)\sqrt{A - \cos x}}\left(A^2-1\right)Q\sqrt{\frac{A - \cos x}{A-1}} \operatorname{F}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)-AQ\sin x- (A-1)(AQ-1)\sqrt{\frac{A-\cos x}{A-1}}\operatorname{E}\left(\frac{x}{2}~\big|~\frac{2}{1-A}\right)
+\sin x\right]_0^{2\pi}$$
Where $E(x|m)$ is an elliptic integral of the 2nd kind with parameter $m=k^2$, and $F(x|m)$ is an elliptic integral of the 1st kind with parameter $m=k^2$.
Replacing $\cos x$ by $1$ and $\sin x$ by $0$...
$$=\frac{2}{(A^2-1)\sqrt{A -1}}*\left[(A^2-1)Q \operatorname{F}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)-(A-1)(AQ-1) \operatorname{E}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)\right]_0^{2\pi}$$
Cancelling $(A^2-1)$...
$$=\frac{2}{\sqrt{A -1}}\left[Q\operatorname{F}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right)
- \frac{(AQ-1)}{A+1} \operatorname{E}\left(\frac{x}{2}~\big|~\frac{-2}{A-1}\right) \right]_0^{2\pi}$$
Being unfamiliar with Elliptic Integrals, this is as far as I can comfortably go at present.
After reading the wikipedia article Elliptic Integral, proceeding tentatively, from the definitions of elliptic integrals I think that $E(x|k^2)$ and $F(x|k^2)$ both go to zero when $x$ is zero, thus...
$$=\frac{2Q}{\sqrt{A -1}}\left[\operatorname{F}\left(\pi~\big|~\frac{-2}{A-1}\right)-\frac{(AQ-1)}{AQ+Q} \operatorname{E}\left(\pi~\big|~\frac{-2}{A-1}\right)\right]$$
Next perhaps it would be helpful to reformulate the problem so that the amplitude(?) term in the elliptic integrals changes from $\pi$ to $\pi/2$, thereby making the elliptic integrals "complete" and permitting them to be expressed as power series. This reformulation could be done by modelling the gravitational effect ($Fx$ component only) of two half rings (positive $y$ and negative $y$), independently, and using respectively the angles $\theta_1$ and $\theta_2$ which both range from $0$ to $\pi/2$ but in different directions.
| Considering similar problems it is usually simpler to consider the potential rather than force. The latter can be found latter as the negative of the potential gradient. Assuming the masses of the test point-like body and the ring be $m$ and $M$, respectively, we have in spherical coordinates with the origin at the center of the ring and the polar axis directed perpendicular to the plane of the ring:
$$
U({\bf r})=-\frac{GmM}{2\pi}\int_0^{2\pi}\frac{d\theta}{\sqrt{r^2+R^2+2rR\sin\phi\cos\theta}},\tag1
$$
where (following the "maths" convention referred to in the Spherical Coordinates link and for consistency with the Question) $r,\phi,\theta $ are radial distance, polar angle, and azimuthal angle of the point ${\bf r}$, and $R$ is the radius of the circle.
The integral $(1)$ can be dealt in the following way:
$$\begin{align}
\int_0^{2\pi}\frac{d\theta}{\sqrt{r^2+R^2+2rR\sin\phi\cos\theta}}
&=2\int_0^{\pi}\frac{d\theta}{\sqrt{r^2+R^2+2rR\sin\phi\cos\theta}}\\
&=2\int_0^{\pi}\frac{d\theta}{\sqrt{(r^2+R^2+2rR\sin\phi)-4rR\sin\phi\sin^2\frac\theta2}}\\
&=\frac{4}{\sqrt{r^2+R^2+2rR\sin\phi}}
\operatorname{K}\left(\frac{4rR\sin\phi}{r^2+R^2+2rR\sin\phi}\right),
\end{align}
$$
where we used the convention
$$
\operatorname{K}(m)=\int_0^{\pi/2}\frac{d\theta}{\sqrt{1-m\sin^2\theta}}
$$
for the complete elliptic integral of the first kind.
Finally
$$
U({\bf r})=-\frac{2GmM}{\pi\sqrt{r^2+R^2+2Rr\sin\phi}}\operatorname{K}\left(\frac{4rR\sin\phi}{r^2+R^2+2rR\sin\phi}\right).\tag2
$$
In the plane of the circle $\phi=\frac\pi2$ and the above equation simplifies to:
$$
U({\bf r})=-\frac{2GmM}{\pi(R+r)}\operatorname{K}\left(\frac{4Rr}{(r+R)^2}\right).
$$
To obtain the expression for the acting force recall that:
$$
\nabla f={\partial f \over \partial r}\hat{\mathbf r}
+ {1 \over r}{\partial f \over \partial \phi}\hat{\boldsymbol \phi}
+ {1 \over r\sin\phi}{\partial f \over \partial \theta}\hat{\boldsymbol \theta}.
$$
As the potential $(2)$ does not depend on $\theta$ only two first terms remain.
Tedious but straightforward calculation reveals:
$$\begin{align}
{\bf F}_r&=\frac{GmM}{\pi}\frac{(R^2-r^2)\operatorname{E}\left(1-\frac {y^2}{x^2}\right)-y^2\operatorname{K}\left(1-\frac {y^2}{x^2}\right)}{rxy^2};\tag3\\
{\bf F}_\phi&=\frac{GmM}{\pi}\frac{(R^2+r^2)\operatorname{E}\left(1-\frac {y^2}{x^2}\right)-y^2\operatorname{K}\left(1-\frac {y^2}{x^2}\right)}{rxy^2}\cot\phi,\tag4\\
\end{align}
$$
where $x=\sqrt{R^2+r^2+2Rr\sin\phi},\ y=\sqrt{R^2+r^2-2Rr\sin\phi}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3243068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $ \int \frac{a^2\cos^2x+b^2\sin^2x}{a^4\cos^2x+b^4\sin^2x}\,dx$ Evaluate $$ \int \frac{a^2\cos^2x+b^2\sin^2x}{a^4\cos^2x+b^4\sin^2x}\,dx$$
I have tried Weierstrass substitution and tried to split into two integrations, but it gets really messy.
Is there a better way to approach this problem? I feel that complex numbers are the best way out, but I couldn't get anything using that as well.
| After substitution $t=\tan{x}$ use
$$\frac{a^2+b^2t^2}{(a^4+b^4t^2)(1+t^2)}=\frac{1}{a^2+b^2}\left(\frac{1}{1+t^2}+\frac{a^2b^2}{a^4+b^4t^2}\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find the minimum value of $\frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$ where $a, b > 0$ and $a + b \le 1$.
$a$ and $b$ are two positives such that $a + b \le 1$. Find the minimum value of $$\large \frac{a}{b + 1} + \frac{b}{a + 1} + \frac{1}{a + b}$$
We have that $\dfrac{a}{b + 1} + \dfrac{b}{a + 1} = \dfrac{a^2 + b^2 + a + b}{ab + a + b + 1} \ge \dfrac{\dfrac{(a + b)^2}{2} + (a + b)}{\dfrac{(a + b)^2}{4} + 2(a + b)} = \dfrac{2(a + b) + 4}{(a + b) + 8}$
$\implies \dfrac{a}{b + 1} + \dfrac{b}{a + 1} + \dfrac{1}{a + b} \ge \dfrac{2(a + b) + 4}{(a + b) + 8} + \dfrac{1}{a + b}$
Let $x = 1 - (a + b) \implies x \ge 0$.
Thus $\dfrac{a}{b + 1} + \dfrac{b}{a + 1} + \dfrac{1}{a + b} \ge \dfrac{2(1 - x) + 4}{(1 - x) + 8} + \dfrac{1}{1 - x} = \dfrac{2x^2 - 9x + 15}{x^2 - 10x + 9}$
$= \dfrac{x(x + 23)}{3(9 - x)(1 - x)} + \dfrac{5}{3} \ge \dfrac{5}{3}$
The equality sign happens when $a = b$ and $x = 0$ or $a + b = 1 \implies a = b = \dfrac{1}{2}$.
I want to ask if the above solution is correct and if there are any other solutions that are more rational and reasonable.
| Your answer is almost right - you just need to show there exists $a$ and $b$ where $\frac{5}{3}$ is attainable (this happens at $a = b = \frac{1}{2}).
Here is an alternative solution. Let $c = 1+a+b$. The expression becomes $\frac{a}{c-a} + \frac{b}{c-b} + \frac{1}{c-1}$. We claim that for fixed $c$, this is minimised when $a=b$. Let $f_c(x) = \frac{x}{c-x} = \frac{c}{c-x}-1$. This expression can be easily shown to be convex, and so by Jensen's inequality, if we let $d = \frac{a+b}{2}$, we have $\frac{2d}{c-d} \le \frac{a}{c-a} + \frac{b}{c-b}$, proving the claim.
We now need to minimise $\frac{2x}{1+x} + \frac{1}{2x} = 2-\frac{2}{1+x}+\frac{1}{x}$. It is simple to show that this is strictly decreasing on $(0,\frac{1}{2}]$, so the minimum is obtained at $x = \frac{1}{2}$ which gives $\frac{5}{3}$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3244950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Limit of matrix to a power and diagonalized matrix to a power give different results I have the following problem. I wish to find
$$
\lim_{n \to \infty} \psi M^n 1
$$
where
$$
\psi = \begin{bmatrix} 1/4 & 1/2 \end{bmatrix}, \quad
1=\begin{bmatrix} 1 & 1 \end{bmatrix}^T
$$
and
$$
M=
\begin{bmatrix}
\left( 1-\frac{L}{n} \right)^2
& 2 \left(\frac{L}{n} \right) \left(1-\frac{L}{n}\right) \\
\left(\frac{L}{n} \right) \left(1-\frac{L}{n}\right)
& \left( 1-\frac{L}{n} \right)^2 \\
\end{bmatrix}
$$
I diagonalized $M=SDS^{-1}$ with the eigensystem of $M$ and found
$$
\lim_{n \to \infty} \psi S D^n S^{-1}1
=\frac{1}{4}e^{-2L} \left(3 \cosh \left(\sqrt{2} L \right)
+ 2 \sqrt{2} \sinh \left( \sqrt{2} L \right) \right)
$$
I found this answer both by hand and by typing it into mathematica. However, when I type the original problem into mathematica, it gives
$$
\lim_{n \to \infty} \psi M^n 1=\frac{3}{4}e^{-2L}
$$
The latter answer is honestly more what I would have expected from the problem setup, but I don't know why the two answers differ nor how to perform the latter limit by hand. Being able to do it by hand is important, because this is actually a warm up to the bigger problem I'm working on. What am I missing? Is there good resource that could help me out, preferably in pdf form?
Edit: The eigensystem of $M$ also depends on $n$, as follows:
$$
\lambda_1 = \left(1-\frac{L}{n}\right) \left(1-\frac{L}{n} \left(1 - \sqrt{2} \right) \right),
\qquad
v_1 = \begin{bmatrix} \sqrt{2} & 1 \end{bmatrix}
$$
$$
\lambda_2 = \left(1-\frac{L}{n}\right) \left(1-\frac{L}{n} \left(1 + \sqrt{2} \right) \right),
\qquad
v_2 = \begin{bmatrix} -\sqrt{2} & 1 \end{bmatrix}
$$
giving $D=diag(\lambda_1, \lambda_2)$ and $S=[v1,v2]$.
Edit: Fixed $\sinh$ coefficient and $3/4$ coefficient.
| I got $$\frac{1}4e^{-2L}\left(3\cosh(L\sqrt{2})+2\sqrt{2}\sinh(L\sqrt{2})\right)$$ in two different ways.
This is close to your complicated example. I get a different coefficient for $\sinh(L\sqrt{2}).$ (Your coefficient was $\sqrt{2}.$)
In general, if $M=\begin{pmatrix}a&2b\\b&a\end{pmatrix},$ it has eigenvectors $\begin{pmatrix}\pm\sqrt{2}\\1\end{pmatrix}$ with eigenvalues $a\pm b\sqrt{2}.$ We can then write $$\begin{pmatrix}1\\1\end{pmatrix}=\left(\frac{1}{2}+\frac{\sqrt{2}}4\right)\begin{pmatrix}\sqrt{2}\\1\end{pmatrix}+\left(\frac{1}{2}-\frac{\sqrt{2}}4\right)\begin{pmatrix}-\sqrt{2}\\1\end{pmatrix}$$
So $$M^n 1 =\left(\frac{1}{2}+\frac{\sqrt{2}}4\right)\left(a+b\sqrt{2}\right)^n\begin{pmatrix}\sqrt{2}\\1\end{pmatrix}+\left(\frac{1}{2}-\frac{\sqrt{2}}4\right)\left(a-b\sqrt{2}\right)^n\begin{pmatrix}-\sqrt{2}\\1\end{pmatrix}$$
and hence:
$$\psi M^n 1=\left(\frac{3}{8}+\frac{\sqrt{2}}{4}\right)\left(a+b\sqrt{2}\right)^n+\left(\frac{3}{8}-\frac{\sqrt{2}}{4}\right)\left(a-b\sqrt{2}\right)^n$$
Here, we can write $$M_n = a_n\begin{pmatrix}a_n&2(1-a_n)\\1-a_n&a_n\end{pmatrix}$$ where $a_n=1-\frac{L}{n}.$ So $b_n=1-a_n=\frac{L}{n}.$
Then $$\psi M_n^n 1= a_n^n\left[\left(\frac{3}{8}+\frac{\sqrt{2}}{4}\right)\left(a_n+b_n\sqrt{2}\right)^n+\left(\frac{3}{8}-\frac{\sqrt{2}}{4}\right)\left(a_n-b_n\sqrt{2}\right)^n\right]$$
We know $a_n^n\to e^{-L}.$
And $$\left(a_n+b_n\sqrt{2}\right)^n = \left(1+\frac{L(\sqrt{2}-1)}{n}\right)^n\to e^{L(\sqrt 2-1)}$$ and $$\left(a_n-b_n\sqrt{2}\right)^n = \left(1+\frac{L(-\sqrt{2}-1)}{n}\right)^n\to e^{L(-\sqrt 2-1)}$$
So we get that:
$$\begin{align}\left(\frac{3}{8}+\frac{\sqrt{2}}{4}\right)\left(a_n+b_n\sqrt{2}\right)^n+\left(\frac{3}{8}-\frac{\sqrt{2}}{4}\right)\left(a_n-b_n\sqrt{2}\right)^n&\to \left(\frac{3}{8}+\frac{\sqrt{2}}{4}\right)e^{L(\sqrt{2}-1)}+\left(\frac{3}{8}-\frac{\sqrt{2}}{4}\right)e^{L(-\sqrt{2}-1)}\\
&=e^{-L}\left(\frac{3}{4}\cosh(L\sqrt{2})+\frac{\sqrt{2}}{2}\sinh(L\sqrt{2})\right)\end{align}$$
So I get the limit is:
$$e^{-2L}\left(\frac{3}{4}\cosh(L\sqrt{2})+\frac{\sqrt{2}}{2}\sinh(L\sqrt{2})\right)$$
Another approach.
Note that if $a_n=1-L/n$ and $\frac{1-a_n}{a_n}=\frac{L/n}{1-L/n}=\frac{1}{n/L-1}$. Then $$\begin{align}M_n&=a_n^2\left(I+\frac{1}{n/L-1}\begin{pmatrix}0&2\\1&0\end{pmatrix}\right)
\end{align}$$
Now, $a_n^{2n}\to e^{-2L}.$
But there is no reason to expect $$\left(I+\frac{1}{n/L-1}\begin{pmatrix}0&2\\1&0\end{pmatrix}\right)^n$$ to contribute nothing to this formula.
Indeed, I'd expect it to converge to $$\exp\left(L\begin{pmatrix}0&2\\1&0\end{pmatrix}\right)$$
3here $\exp(A)=\sum_{k=0}^{\infty}\frac{1}{k!}A^k$ is the matrix exponential.
This will be true at least when $L$ is an integer and and restricting to $n$ which are multiples of $L.$
Now, taking $A=\begin{pmatrix}0&2\\1&0\end{pmatrix},$ we have $A^2=2I$ and thus: $$\begin{align}\exp(LA)&=\left(\sum_{k=0}^{\infty} \frac{1}{(2k)!}2^kL^{2K}\right)I+\left(\sum_{k=0}^{\infty}\frac{1}{(2k+1)!}L^{2k+1}2^k\right)A\\
&=\cosh(\sqrt{2}L)I + \frac{1}{\sqrt{2}}\sinh(L\sqrt{2})A
\end{align}$$
Then $\psi I 1=\frac{3}{4}$ and $\psi A 1=1$ so you get the limit is:
$$e^{-2L}\left(\frac{3}{4}\cosh(L\sqrt{2})+\frac{1}{\sqrt{2}}\sinh(L\sqrt{2}\right)$$
which is $$\frac{1}{4}e^{-2L}\left(3\cosh(L\sqrt{2})+2\sqrt{2}\sinh(L\sqrt{2})\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3245540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integrating factor for $(x^2-y^2-y)dx-(x^2-y^2-x)dy=0$ I am having trouble finding the integrating factor for turning the below differential equation into an exact one (Tenenbaum and Pollard, exercise 10, problem 6). Any hints and suggestions would extremely helpful and lead me to the solution.
Solve the differential equation :
$$ (x^2-y^2-y)dx - (x^2-y^2-x)dy=0$$
My attempt:
The coefficients of $dx$ and $dy$ are not homogenous functions. Further,
$
\begin{align}
P(x,y) &= x^2 - y^2 - y \\
\frac{\partial P(x,y)}{\partial y}&=-2y-1 \\
Q(x,y) &= -(x^2-y^2-x)\\
\frac{\partial Q(x,y)}{\partial x}&=-(2x-1) \\
\therefore \frac{\partial P}{\partial y} \neq \frac{\partial Q}{\partial x}
\end{align}
$
The given differential equation is not exact. We have :
$\begin{align}
& \frac{\partial P}{\partial y} - \frac{\partial Q}{\partial x} \\
=& -2y-1+(2x-1)\\
=&(2x-2y-2)
\end{align}$.
Moreover,
$\begin{align}
& yQ-xP\\
= & -y(x^2-y^2-x)-x(x^2-y^2-y)\\
= & -x^2 y + y^3 + xy - x^3 + xy^2 + xy\\
= & y^3 - x^3 + xy - xy(x - y - 1)
\end{align}$
It doesn't look like $(\partial P / \partial y - \partial Q / \partial x)/(yQ-xP)$ will be a function of $u=xy$ alone.
Also,
$\begin{align}
& yQ+xP\\
= & -y(x^2-y^2-x)+x(x^2-y^2-y)\\
= & -x^2 y + y^3 + xy + x^3 - xy^2 - xy\\
= & y^3 + x^3 - x^2 y - x y^2 \\
= & (y + x)(y^2 + x^2 - xy) - xy(y + x) \\
= & (y + x)(y^2 + x^2 - 2xy)\\
= & (y + x)(y - x)^2
\end{align}$
It doesn't look like $y^2(\partial P / \partial y - \partial Q / \partial x)/(yQ+xP)$ will be a function of $u=x/y$ or $x^2(\partial P / \partial y - \partial Q / \partial x)/(yQ+xP)$ will be a function of $u=y/x$ alone.
| Start with
$$ (x^2-y^2-y)\,\mathrm{d}x - (x^2-y^2-x)\,\mathrm{d}y=0$$
We note that this is
$$
(x^2-y^2)\,\mathrm{d}(x-y)+(x\,\mathrm{d}y-y\,\mathrm{d}x)=0
$$
So $(u,v)=(x+y,x-y)$ is a reasonable choice to try. Simplifying the resulting equation gives
$$
(2u+1)v\,\mathrm{d}u=u\,\mathrm{d}v
$$
which is a separable equation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3247724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
If $\frac{2}{x}=2-x,$ Find $[x^9-(x^4+x^2+1)(x^6+x^3+1)]^3$ without entering $\Bbb C$ If $\frac{2}{x}=2-x,$ Find $[x^9-(x^4+x^2+1)(x^6+x^3+1)]^3$ without entering $\Bbb C$
After we solve for $x$ (its a quadratic), and find that $x=1\pm i$, it's trivial to see the powers of $x$ in the complex plane, but the problem must be solved without using the complex numbers. With complex numbers i found that the result is $1$.
Is there any way to solve this in $\Bbb R$ without entering $\Bbb C$?
| Hint
$$x^2=2x-2$$
$$x^3=2x^2-2x=2(2x-2)-2x=2x-4$$
$$x^4=(2x-2)^2=4-8x+4(2x-2)=-4$$
$$x^6=-4(2x-2)=?$$
$$x^9=(-4)^2x=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3250121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Solve $x^2+5x+6 \equiv 0 \pmod{\!11\cdot 17}$ Solve $x^2+5x+6 \equiv 187 \mod 187$
Solution
$$x^2+5x+6 \equiv 187 \mod 187$$
$$ (x+\frac{5}{2})^2 \equiv \frac{1}{4}$$
$$ 4(x+\frac{5}{2})^2 \equiv 1$$
$$ y:= x+\frac{5}{2} $$
$$ 4y^2 \equiv 1 \mod 11 \wedge 4y^2 \equiv 1 \mod 17 $$
$$ ( 2y \equiv 1 \mod 11 \vee 2y \equiv 10 \mod 11 ) \wedge ( 2y \equiv 1 \mod 17 \vee 2y \equiv 13 \mod 17) $$
$$ ( y \equiv 6 \mod 11 \vee y \equiv 5 \mod 11 ) \wedge ( y \equiv 9 \mod 17 \vee y \equiv 15 \mod 17) $$
Combining that from CRT I got:
$$ y \in \left\{49, 60,83,94 \right\} $$
and for example:
$$ x+\frac{5}{2} \equiv 94 \mod 187$$
$$ 2x \equiv 183 \mod 187$$
some calculus and get...
$$x \equiv 185 $$
And the same thing for each other case.
Question
Is there any faster (or smarter) way to solve equations like that?
| Solve $x^2+5x+6 \equiv (x+2)(x+3) \pmod {187}$.
Besides the two 'in your face' solutions, $x \equiv -2 \pmod{187}$ and $x \equiv -3 \pmod{187}$, we can buttress the argument given by José Carlos Santos to find all four solutions.
We want to find integers $x, k, j$ satifying
$\;\text{L1:}\quad x + 2 = 11k$
$\;\text{L2:}\quad x + 3 = 17j$
Subtracting $\text{L1}$ from $\text{L2}$ we write
$\tag 1 1 = 17j - 11k$
Bézout's identity gives us
$\quad 1 = 17(2) - 11(3)$
So $k = 3$ and, plugging into $\text{L1}$, $x \equiv 31 \pmod{187}$ is a solution.
For the last solution,
$\;\text{L1:}\quad x + 2 = 17j$
$\;\text{L2:}\quad x + 3 = 11k$
Subtracting $\text{L1}$ from $\text{L2}$ we write
$\tag 2 1 = -17j + 11k$
Bézout's identity gives us
$\quad 1 = 17(2) - 11(3)$
So $k = -3$ and, plugging into $\text{L2}$, $x = -36 \equiv 151 \pmod{187}$ is a solution.
It can also be argued that there are exactly four solutions.
| {
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"url": "https://math.stackexchange.com/questions/3250848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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For $a,b,c,d > 0$ and $abcd = 1$, show that $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{12}{a + b + c + d} \geq 7$
Question:
For $a,b,c,d > 0$ and $abcd = 1$, show that
$$\frac{1}{a} + \frac{1}{b} + \frac{1}{c} + \frac{1}{d} + \frac{12}{a + b + c + d} \geq 7$$
My Attempts:
Trivial to see that equality occurs for $a = b= c = d$ - so this indicates to use some inequality wherin the equality condition holds when all quantities are equal - I've tried AM-GM-HM, Power means, Rearrangement, Cauchy-Schwarz, Newton, Maclaurin, Weighted AM-GM - it doesn't help, and the reason is that these inequalities obtain a function of $a,b,c,d$ on the RHS whose minimum is $<7$ - and that minimum and the inequality can never hold simultaneously - giving a loose bound.
I've tried rearranging the inequality to the form $(\sum_{cyc} a )(\sum_{cyc} \frac{1}{a}) - 7 (\sum_{cyc} a) + 12 \geq 0 $ - This doesn't help. I've tried the substitution $(a,b,c,d) \rightarrow (\frac{1}{x}, \frac{1}{y}, \frac{1}{z}, \frac{1}{w} )$ - this is of no help either. Even tried the substitution $(a,b,c,d) \rightarrow (\frac{x}{y}, \frac{y}{z}, \frac{z}{w}, \frac{w}{x} )$ - of no help either. Writing it as $\frac{4}{M(-1)} + \frac{3}{M(1)} \geq 7$ where $M(x)$ is the power means function, and hoping to use some property of $M(x)$ doesn't work either.
My Analysis: Suppose $S_i$ denotes the sum taken $i$ at a time. It is trivial to see that $S_3 = \sum_{cyc} \frac{1}{a}$ is lower bounded by $S_4$ by Maclaurin or GM-HM. However $S_1$ is unbounded above even if $S_4 = 1$. However when $S_1$ reaches a very high value, so does $S_3$, so a minimum has to involve considering both terms simultaneously i.e. you can't minimize each separately and then hope to minimize the function.
Brute Force Solution: The brute force solution of considering it as a function of 4 variables and then optimizing obviously works, I'm hoping for something more smarter or elegant.
| Mixing Variables and $uvw$ help.
I'll post my solution, which I found eight years ago.
Let $f(a,b,c,d)=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}+\frac{12}{a+b+c+d}- 7$ and $a=\max\{a,b,c,d\}$.
Thus, $$f(a,b,c,d)-f\left(a,\sqrt[3]{bcd},\sqrt[3]{bcd},\sqrt[3]{bcd}\right)=$$
$$=\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{12\left(b+c+d-3\sqrt[3]{bcd}\right)}{(a+b+c+d)\left(a+3\sqrt[3]{bcd}\right)}\geq$$
$$\geq\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{12\left(b+c+d-3\sqrt[3]{bcd}\right)}{(\frac{b+c+d}{3}+b+c+d)\left(\frac{b+c+d}{3}+3\sqrt[3]{bcd}\right)}=$$
$$=\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}.$$
We'll prove that $$\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}\geq0.$$
Indeed, let $b+c+d=3u$, $bc+bd+cd=3v^2$ and $bcd=w^3$.
Thus, $$\frac{bc+bd+cd-3\sqrt[3]{b^2c^2d^2}}{bcd}-\frac{27\left(b+c+d-3\sqrt[3]{bcd}\right)}{(b+c+d)\left(b+c+d+9\sqrt[3]{bcd}\right)}\geq0$$ it's $ g(v^2)\geq0,$
where $g$ is a linear increasing function.
Id est, $g$ gets a minimal value, when $v^2$ gets a minimal value,
which happens for equality case of two variables.
Since $g(v^2)\geq0$ is homogeneous inequality, it's enough to check one case only:
$c=d=1$, which after substitution $b=x^3$ gives
$$(x-1)^2(2x^7+x^6+18x^5-10x^4-50x^3+36x^2+26x+4)\geq0,$$ which is true.
Id est, $$f(a,b,c,d)\geq f\left(a,\sqrt[3]{bcd},\sqrt[3]{bcd},\sqrt[3]{bcd}\right)$$ and it's enough to prove that $f(a,b,b,b)\geq0$, where $a=\frac{1}{b^3}$, which gives
$$(b-1)^2(3b^6+6b^5+9b^4-9b^3-5b^2-b+3)\geq0,$$ which is true.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Another way to calculate the integral $ \int_{0}^{1} (x-x^{2})^{3/2} dx $ $$\int_{0}^{1} (x-x^{2})^{3/2} dx=\dfrac{1}{8}\int_{0}^{1}[1-(2x-1)^{2}]^{3/2}dx=\dfrac{1}{8}\int_{-\pi/2}^{\pi/2}\dfrac{cos^{4}\theta}{2}d\theta=\dfrac{1}{8}\int_{0}^{\pi/2}cos^{4}d\theta=\dfrac{1}{8}(\dfrac{3\pi}{16})=\dfrac{3\pi}{128}$$
The change of variable $2x-1 = sin\theta$ was made.
How could it be integrated without the "trick" $(x-x^{2})^{3/2}=\dfrac{1}{8}[1-(2x-1)^{2}]^{3/2}$?
| There is also the following way.
Let $x=\cos^2t$, where $t\in\left[0,\frac{\pi}{2}\right]$.
Thus, $$\int\limits_0^1(x-x^2)^{\frac{3}{2}}dx=\int_{\frac{\pi}{2}}^0\sin^3t\cos^3t(-2\sin{t}\cos{t})dt=2\int\limits_0^{\frac{\pi}{2}}\sin^4t\cos^4tdt=$$
$$=\frac{1}{8}\int\limits_0^{\frac{\pi}{2}}\sin^42tdt=\frac{1}{32}\int\limits_0^{\frac{\pi}{2}}(1-\cos4t)^2dt=\frac{1}{32}\int\limits_0^{\frac{\pi}{2}}\left(1-2\cos4t+\frac{1+\cos8t}{2}\right)dt=$$
$$=\frac{1}{32}\left(1+\frac{1}{2}\right)\frac{\pi}{2}=\frac{3\pi}{128}.$$
| {
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"url": "https://math.stackexchange.com/questions/3252607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Interesting inequality $\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$
Prove that $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[k]{\frac{1+x^k}{2}}$$ for all real $x \ge 1$ and for all positive integers $m$ and $k \le 2m+1$.
My work. If $k \le 2m+1$ then $$\sqrt[k]{\frac{1+x^k}{2}}\le \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}.$$ I need prove that for all real $x \ge 1$ and for all positive integers $m$ the inequality $$\frac{x^{m+1}+1}{x^m+1} \ge \sqrt[2m+1]{\frac{1+x^{2m+1}}{2}}$$ holds.
| for all $m\in \mathbb{N}$ and $x\geq 1$ we have $x^{2m+1}\geq 1$ then $2x^{2m+1}\geq x^{2m+1}+1$
Notice that $\frac{x^{m+1}+1}{x^m+1} = x - \frac{x-1}{x^m+1} \geq x$ because $x\geq 1$
Then $$2\left(\frac{x^{m+1}+1}{x^m+1}\right)^{2m+1}\geq 2x^{2m+1}\geq x^{2m+1}+1$$
We obtain easily that $$\frac{x^{m+1}+1}{x^m+1} \geq \sqrt[2m+1]{\frac{x^{2m+1}+1}{2}}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $76^{2019} \equiv 45 \pmod{101}$ Using Euler's theorem I see that $76^{2019} = 76^{19} \pmod{101}$, how do I proceed?
| Here is an approach that requires minimum computation. We have:
$$76^{19} \equiv (-25)^{19} \equiv -5^{38} \pmod{101}$$
How we have that by quadratic reciprocity:
$$\left( \frac{5}{101} \right)=\left( \frac{5}{101} \right) \cdot \left( \frac{1}{5} \right) = \left( \frac{5}{101} \right) \cdot \left( \frac{101}{5} \right)=(-1)^{\frac{101-1}{2} \cdot \frac{5-1}{2}}=1$$
So $5$ is the quadratic residue modulo $101$ and therefore $5^{50} \equiv 1 \pmod{101}$. So we can write:
$$-5^{38} \equiv -5^{-12} \equiv -25^{-6}\pmod{101}$$
And since $25 \cdot (-4) \equiv -100 \equiv 1 \pmod{101}$ we have $25^{-1} \equiv -4 \pmod{101}$ and therefore:
$$-25^{-6} \equiv -(-4)^6 \equiv -4^6 \equiv -4096 \equiv 45\pmod{101}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.
Prove that $(x + y + z)^3 + 9xyz \ge 4(x + y + z)(xy + yz + zx)$ where $x, y, z \ge 0$.
This has become the norm now... This problem is adapted from a recent competition.
We have that $6(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) \ge 4(x + y + z)(xy + yz + zx)$
Furthermore, $$(x + y + z)^3 - 6(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) + 9xyz$$
$$ = x^3 + y^3 + z^3 - 3(x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2) + 15xyz$$
In addition, $x^3 + y^3 + z^3 + 3xyz \ge x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2$.
So now we need to prove that $6xyz \ge x^2y + xy^2 + y^2z + yz^2 + z^2x + zx^2$, which I don't even know if it is true or not.
| Use the fact that $ GM \leq AM $. For six numbers $ x_1, x_2, \dots, x_6 $, you have
$$
(x_1 x_2 \dots x_6)^{1/6} \leq \frac{x_1 + x_2 + \dots + x_6}{6}
$$
Then, you have
$$
x y z \leq \frac{x^2 y + x y^2 + y^2 z + y z^2 + z^2 x + z x^2}{6}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3254924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $\lim_{x,y \rightarrow 0,0} \frac{x \ln(1+y)}{2x^2+y^2}$
Find bounds for $\lim_{x,y \rightarrow 0,0} \frac{x \ln(1+y)}{2x^2+y^2}$
I am finding maximum and minimum for function and one of critical case is to find possible minimal and maximal value of given function in $0,0$. But how can I do this due to this limit doesn't exists (for example we can take $x,y = {1\over n},{1\over n}$ and $x,y = {2\over n},{1 \over n}$
| Substitute $x= r \cos \theta$ and $y= r \sin \theta$, then you have
$$\lim_{r \to 0}\frac{r \cos \theta \ln(1+r \sin \theta)}{2 r^2 \cos ^2 \theta + r^2 \sin^2 \theta}=\lim_{r \to 0}\frac{ \cos \theta \ln(1+r \sin \theta)}{2 r \cos ^2 \theta + r \sin^2 \theta}\stackrel{l'Hospital}{=}\lim_{r \to 0}\frac{\frac{\cos \theta \sin\theta}{1+r \sin \theta}}{2 \cos^2\theta + \sin^2 \theta}=\lim_{r \to 0}\frac{\cos \theta\sin \theta}{(2 \cos^2 \theta + \sin^2\theta)(1+r \sin \theta)}=\frac{\cos \theta\sin \theta}{2\cos^2 \theta+\sin^2 \theta}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3256578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\cos^4(\theta)-\sin^4(\theta)=\cos(2\theta)$ The problem is to prove that $\cos^4(\theta)-\sin^4(\theta)=cos(2\theta)$.
So, here are my steps so far:
$(\cos^2\theta)^2-(\sin^2\theta)^2=\cos(2\theta)$
$(\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta)=\cos(2\theta)$
$(\cos^2\theta+\sin^2\theta)(\cos\theta+\sin\theta)(\cos\theta-\sin\theta)=\cos(2\theta)$
I don't know where to go from here. Please help! I feel like I'm missing something super obvious but I don't know what...
| There are a couple of nice ways we can do it. Firstly, note that:
$$\cos^4(\theta)-\sin^4(\theta)=\bigl(\cos^2\theta-\sin^2\theta\bigr)\bigl(\cos^2\theta+\sin^2\theta\bigr)$$
and since:
$$\cos^2\theta+\sin^2\theta=1\tag{1}$$
we can say that:
$$\cos^4\theta-\sin^4\theta=\cos^2\theta-\sin^2\theta=\cos2\theta$$
which is the double angle formula for cosine. This can be proved geometrically a number of ways but also as shown below
We can also take advantage of De Moivre's theorem, which states that:
$$\left(\cos x+i\sin x\right)^n=\cos(nx)+i\sin(nx)\tag{2}$$
Using the fact that:
$$e^{ix}=\cos(x)+i\sin(x)\tag{3}$$
We can do this by saying:
$$\cos(2x)+i\sin(2x)=\left(\cos x+i\sin x\right)^2$$
$$\cos(2x)+i\sin(2x)=\cos^2x+2i\cos x\sin x+i^2\sin^2x$$
$$\cos(2x)+i\sin(2x)=\cos^2x-\sin^2x+2i\cos x\sin x$$
Now if we separate the real and imaginary parts we are left with:
$$\cos(2x)=\cos^2x-\sin^2x,\,\sin(2x)=2\sin(x)\cos(x)\tag{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding solutions to a system of equations using elementary symmetric polynomials Find the value of a, b, and c given:
$a^2 + b^2 + c^2 = 129$
$ab + ac + bc = -4$
$(a^2)(b^2) + (a^2)(c^2) + (b^2)(c^2)$ = 984
I attempted this problem using elementary symmetrical polynomials and found that the second equation can be written as $e_2(a, b, c)$. However, the squares in the other equations made me quite confused and I was unable to write them in the previous from. If I was able to write it in the form of elementary symmetrical polynomials, I should have been able to use simple substitution to find the value of $e_1, e_2$, and $e_3$. This then would have allowed me to form a cubic equation, where I could find the solutions by solving it.
However, as I mentioned before, I'm a little bit stuck on the part where I need to find the values of $e_1, e_2, e_3$. Any help would be extremely appreciated :) Also, is there any other way to solve for a, b, and c using elementary symmetric polynomials and substitution, but not cubic equations?
Update:
Thanks to @Gerry_Myerson, I've worked out that the first equation is:
$(e_1)^2 - 2e_2 = 129$
and the second equation is:
$e_2 = -4$
However, I still don't know how to work out the last one.
| \begin{align}
a^2 + b^2 + c^2 &= 129
\tag{1}\label{1}
,\\
ab + bc + ac &= -4
\tag{2}\label{2}
,\\
a^2b^2 + b^2c^2 + a^2c^2&=984
\tag{3}\label{3}
.
\end{align}
\begin{align}
\eqref{2}^2:\quad
a^2b^2+a^2c^2+b^2c^2+2abc(a+b+c)
&=16
,\\
abc(a+b+c)&=-484
,\\
\eqref{1}+2\cdot\eqref{2}:\quad
(a+b+c)^2&=121
.\\
a+b+c&=\pm 11
,\\
abc&=-\frac{484}{a+b+c}
=\mp 44
,
\end{align}
and we have two cubics
\begin{align}
x^3-11x^2-4x+44
,\\
y^3+11y^2-4x-44
\end{align}
with roots $(-2, 2, 11)$
and $(-2, 2, -11)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$. Please tell if the problem can be solved using telescoping technique or not.
If yes, how to prove $a^n − b^n = (a − b) \sum_{i=1}^{n}a^{n-i} b^{i-1}\le (a − b)na^{n−1}$ using that. It is given that $a,b \in \mathbb{R}{+},\, a\gt b,\, n \in \mathbb{N}.$
I tried as follows, but was unsuccessful to pursue:
$a^n − b^n = a^n+\sum_{i=1}^{n-1}(a^ib^{n-i}-a^ib^{n-i})-b^n=a^n+\sum_{i=1}^{n-1}a^ib^{n-i}-\sum_{i=1}^{n-1}a^ib^{n-i}-b^n$
Edit : based on the selected answer's comment.
Writing a few terms of the series, $\sum_{i=1}^n (a^{n+1-i}b^{i-1}-a^{n-i}b^i)$ get:
For $n =5$, get the terms as:
$i=1, \,\, a^{5+1-1}b^{1-1}-a^{5-1}b^1 = a^5-a^4b.$
$i=2, \,\, a^{5-1}b^{2-1}-a^{5-2}b^2 = a^4b-a^3b^2.$
$i=3, \,\, a^{5-2}b^{3-1}-a^{5-3}b^3 = a^3b^2-a^2b^3.$
$i=4, \,\, a^{5-3}b^{4-1}-a^{5-4}b^4 = a^2b^3-a^1b^4.$
$i=5, \,\, a^{5-4}b^{3-1}-a^{5-3}b^5 = a^1b^4-b^5.$
Adding all the terms, get:
$a^5-a^4b+ a^4b-a^3b^2+a^3b^2-a^2b^3+a^2b^3-a^1b^4+a^1b^4-b^5 = a^5 - b^5$
| Geometric progression
\begin{align}
(a-b)\sum_{i=1}^{n}a^{n-i}b^{i-1}&=(a-b)a^{n-1}\left(\frac{1-\left(\frac{b}{a}\right)^n}{1-\frac{b}{a}}\right)\\
&=a^n-b^n.
\end{align}
Can you complete the answer now?
| {
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$(x^3 + \frac{a}{x^2})^5 = -270$, find $a$ I am working on my scholarship exam practice and not sure how to begin. Please assume math knowledge at high school or pre-university level.
Let $a$ be a real constant. If the constant term of $(x^3 + \frac{a}{x^2})^5$ is equal to $-270$, then $a=$......
Could you please give a hint for this question? The answer provided is $-3$.
| From the binomial theorem, we know that the sum of the powers of $x^3$ and $a/x^2$ must add to $5$. So if our term in the expansion is $k (x^3)^p (\frac{a}{x^2})^q$, $p+q = 5$. Moreover, we want our term to be constant (no nonzero powers of $x$), so we have $3p - 2q = 0$. Solving these two expressions, we have $p = 2, q =3$. So we must find $a$ such that the coefficient of the $(x^3)^2 (\frac{a}{x^2})^3$ term is $270$.
The binomial theorem says that the coefficient of this power in the expansion is ${5 \choose 2} = 10$, so in particular, the coefficient of the constant term is $10a^3$, which we want to equal $270$. Therefore, $a = -3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that inequality $27+7\left(ab+bc+ca\right)\le 8\left(\sum \sqrt{a+3b}\right)$ For the positive reals $a,b$ and $c$ so that $a+b+c=3$. Show that $$27+7\left(ab+bc+ca\right)\le 8\left(\sqrt{a+3b}+\sqrt{b+3c}+\sqrt{c+3a}\right)$$
That inequality has been created by Imad Zak, i think it is interesting problem.
We can see that $$\text{L.H.S}=3(a+b+c)^2+7(ab+bc+ca)$$
$$=3(a^2+b^2+c^2)+13(ab+bc+ca)$$
$$=\sum_{cyc} (a+3b)(b+3c)$$
Let $\sqrt{a+3b}=x;\sqrt{b+3c}=y$ and $z=\sqrt{c+3a}$ where $x,y,z>0$ and $x^2+y^2+z^2=12$
We prove $$8(x+y+z)\ge x^2y^2+y^2z^2+z^2x^2$$
I am stuck here. I tried to homogenize the last inequality but i only get wrong inequalities. I also used C-S or Holder but failed.
| Let $a^2+b^2+c^2=k(ab+ac+bc).$
Thus, by Holder
$$\sum_{cyc}\sqrt{a+3b}=\sqrt{\frac{\left(\sum\limits_{cyc}\sqrt{a+3b)}\right)^2\sum\limits_{cyc}(a+3b)^2}{\sum\limits_{cyc}(a+3b)^2}}\geq$$
$$\geq\sqrt{\frac{\left(\sum\limits_{cyc}(a+3b)\right)^3}{\sum\limits_{cyc}(10a^2+6ab)}}=\sqrt{\frac{32(a+b+c)^6}{27\sum\limits_{cyc}(5a^2+3ab)}}.$$
Id est, it's enough to prove that
$$8\sqrt{\frac{32(a+b+c)^6}{27\sum\limits_{cyc}(5a^2+3ab)}}\geq3(a+b+c)^2+7(ab+ac+bc)$$ or
$$8\sqrt{\frac{32\left(\sum\limits_{cyc}(a^2+2ab)\right)^3}{27\sum\limits_{cyc}(5a^2+3ab)}}\geq\sum_{cyc}(3a^2+13ab)$$ or
$$8\sqrt{\frac{32(k+2)^3}{27(5k+3)}}\geq3k+13$$ or
$$2048(k+2)^3\geq27(5k+3)(3k+13)^2,$$ which is true by AM-GM:
$$27(5k+3)(3k+13)^2=\frac{27}{2}(10k+6)(3k+13)^2\leq$$
$$\leq\frac{27}{2}\left(\frac{10k+6+2(3k+13)}{3}\right)^3=2048(k+2)^3$$
and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Exercise 3 in Section 7.11 of Apostol's Calculus (Vol. 1) little o-notation The problem is stated as:
Find the polynomial $P(x)$ of minimal degree such that
$$
\sin(x-x^2)=P(x)+o(x^6)\quad \textrm{as }x\to 0.
$$
and the answer in the book is
$$
P(x)=x-x^2-\frac{x^3}{6}+\frac{x^4}{2}-\frac{59x^5}{120}+\frac{x^6}{8}
$$
I am wondering why the answer cannot be
$P(x) = x-x^2$
Since we can use the fact that $\sin(x-x^2) = x-x^2 + o((x-x^2)^2)$ as $x \to 0$
And we know that $o((x-x^2)^2) = o(x^5) = o(x^6)$ as $x$ goes to zero.
Then why is the answer from the back of the book not equal to $x-x^2$??
| A sketch of the correct method:
Since $x-x^2$ has order $1$, you have to expand $\sin u$ up to degree $6$, and substitute $x-x^2$ in $u, u^2,\dots u^6$, truncating the expansions of the different $(x-x^2)^k$ at degree $6$, which leads to a recursive computation:
\begin{align}
(x-x^2)^2&=x^2-2x^3+x^4& \quad(x-x^2)^3&=x^3-3x^4+3x^5-x^6 \\
(x-x^2)^4&=x^4-4x^5+6x^6-4x^7+x^8&(x-x^2)^5&=(x-x^2)^4(x-x^2)\\
&=x^4-4x^5+6x^6+o(x^6)&&=\bigl((x^4-4x^5+6x^6+o(x^6)\bigr)(x-x^2)\\
(x-x^2)^6&=(x-x^2)^5(x-x^2)&&=\begin{aligned}[t]&x^5-4x^6+6x^7+o(x^7)\\&\phantom{x5}-x^6+4x^7-6x^8+o(x^8)
&\end{aligned}\\&=\bigl(x^5-5x^6+o(x^6)\bigr)(x-x^2)&&=x^5-5x^6+o(x^6)\\
&=\dots=x^6+o(x^6)
\end{align}
Can you end the computations?
| {
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How to correctly solve $\sqrt{1+x}+\sqrt{1-x}>1$? $$\sqrt{1+x}+\sqrt{1-x}>1, x\geq-1 \wedge x\leq1$$
$$\sqrt{1+x}>1-\sqrt{1-x}$$
$$1+x>1-2\sqrt{1-x}+1-x$$
$$0>-2\sqrt{1-x}+1-2x$$
$$2\sqrt{1-x}>1-2x$$
*
*Lets say both sides of inequality are positive, then we can easily $2\sqrt{1-x}>1-2x$ square and arrange. What we get$\implies x^2<\frac{3}{4} \implies \frac{-\sqrt{3}}{2}<x<\frac{\sqrt{3}}{2}$Now, when we make set intersection with $x\geq-1 \wedge x\leq1$ we get the solution $\frac{-\sqrt{3}}{2}<x<\frac{\sqrt{3}}{2}$.
*We also have to look at the possibility that $1-2x$ can be negative: $0>1-2x\implies x>\frac{1}{2}$We determine the solution with $x\geq-1 \wedge x\leq1$ and we get the complete solution $\frac{1}{2}<x<1$.
Now I want to get the solution for the original inequality and therefore I unite the final solutions from 1. and 2.case. I get the interval $$(-\frac{\sqrt{3}}{2}, 1]$$
This solution is wrong. The correct one is $[-1,1]$. I would really appreciate if someone could check the complete process and point out where I made a mistake. Also, is this solving process good and effective or I complicate things too much?
| Note that the domain of $\sqrt {1+x}+\sqrt {1-x}$ is $[-1,1]$
On $(0,1]$ we have $\sqrt{1+x}>1$
On $[-1,0)$ we have $\sqrt {1-x}>1$
At $x=0$ we have $\sqrt {1+x}+\sqrt{1-x}=2>1$
Thus $\sqrt {1+x}+\sqrt {1-x}>1$ on its domain.
| {
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The sum of an infinite series containing a finite series in each denominator
Evaluate
$$\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{6^2}+\frac{1}{10^2}+\cdots+\frac{1}{\left[\frac{k(k+1)}{2}\right]^2}+\cdots$$
to $\infty$, where $k$ is the $k$th term of the series.
Using Microsoft Excel, I found that the sum of the first $100$ terms $=1.15947\dots$.
Does the convergence value have an exact form? If yes, what is it?
| $$\frac {1}{k(k+1)}= \frac {1}{k}-\frac {1}{k+1}$$
$$\frac {1}{k^2(k+1)^2}= \frac {1}{k^2}+\frac {1}{(k+1)^2} - \frac {2}{k(k+1)}$$
$$\sum_1^{\infty}\frac {1}{k^2(k+1)^2}= \sum_1^{\infty}\frac {1}{k^2}+\sum_1^{\infty}\frac {1}{(k+1)^2} -\sum_1^{\infty} \frac {2}{k(k+1)}=$$
$$\frac {\pi ^2}{6} +(\frac {\pi ^2}{6}-1)-2$$
$$\sum_1^{\infty} \frac {4}{k^2(k+1)^2} = 4(\frac {\pi^2 }{6} + \frac {\pi^2 }{6}-1 -2) = 4(\frac {\pi^2-9}{3})\approx 1.159472535...$$
| {
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Factoring out the expression I'm sorry, it must pretty basic, but...
How to factor out this expression $x^2 + 5x - 5 = 0$ ?
(The method I normally use to factor expressions doesn't work there because when we factor expression above so that format will be $(ax + b)(qx + c)$ $b$ and $c$ definitely won't be integers, and method that I use only works when $b$ and $c$ are integers)
| Since $x^2+5x-5$ is a monic polynomial whose roots are $\frac{1}{2} \left(-5+3\sqrt{5}\right)$ and $\frac{1}{2} \left(-5-3\sqrt{5}\right)$,$$x^2+5x-5=\left(x-\frac{1}{2} \left(-5+3\sqrt{5}\right)\right)\left(x-\frac{1}{2} \left(-5-3\sqrt{5}\right)\right).$$
| {
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Prove that the series is conditionally convergent $$ \sum_{n=1}^\infty (-1)^{n+1} \left( \frac{1.4.7\dots .(3n-2)}{2.3.8\dots .(3n-1)} \right)^2 $$
I have done the$ \sum_{n=1}^\infty \left( \frac{1.4.7\dots .(3n-2)}{2.3.8\dots .(3n-1)} \right)^2 $ part , and showed it divergent using Gauss test .
But i am not able to do this part $ \sum_{n=1}^\infty (-1)^{n+1} \left( \frac{1.4.7\dots .(3n-2)}{2.3.8\dots .(3n-1)} \right)^2 $ ,tried leibniz test to do , but could not do that.
I have no idea how to do this please help.
| $(\frac{1}{2})^2 < \frac{1}{2}$,
$(\frac{4}{5})^2 < \frac{3}{4}$,
$(\frac{7}{8})^2 < \frac{5}{6}$, $\dots \dots$
$(\frac{3n-2}{3n-1})^2 < \frac{2n-1}{2n}$
Multiplying all , we get
$\prod_{n=1}^n(\frac{3n-2}{3n-1})^2 <\prod_{n=1}^n \frac{2n-1}{2n}$$ ......(1)$
Now,
$(\frac{1}{2}) < \frac{2}{3}$,$(\frac{3}{4}) < \frac{4}{5}$,$\dots$ $(\frac{2n-1}{2n}) < \frac{2n}{2n+1}$
Multiplying all, we get
$(\frac{1.3.5. \dots 2n-1}{2.4.6. \dots 2n}) < \frac{2.4.6. \dots2n}{3.5.\dots.2n-1.2n+1}$
$(\frac{1.3.5. \dots 2n-1}{2.4.6. \dots 2n}) < \frac{2.4.6. \dots2n}{3.5.\dots.2n-1}.(\frac{1}{2n+1})$
$(\frac{1.3.5. \dots 2n-1}{2.4.6. \dots 2n})^2 < (\frac{1}{2n+1})$
$(\frac{1.3.5. \dots 2n-1}{2.4.6. \dots 2n})< (\frac{1}{\sqrt{(2n+1)}})$$......... (2)$
From $(1)$&$(2)$,we get
$\prod_{n=1}^n(\frac{3n-2}{3n-1})^2< (\frac{1}{\sqrt{(2n+1)}})$
Now,
$lim_{n\to \infty} (\frac{1}{\sqrt{(2n+1)}}) = 0$
Hence , $lim_{n\to \infty} \prod_{n=1}^n(\frac{3n-2}{3n-1})^2 = 0$
Hence by leibniz criterion, the given series $\sum_{n=1}^\infty (-1)^{n+1} \left( \frac{1.4.7\dots .(3n-2)}{2.3.8\dots .(3n-1)} \right)^2 $ is convergent.
Answer credit, phara narai
| {
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Dividing polynomial $f(x)$ by $(x-a)(x-b)(x-c)$
$f(x)$ is a polynomial with a degree greater than 3. When $f(x)$ is divided by $(x-a)(x-b)(x-c)$, prove remainder is
$$\frac{f(a)(x-b)(x-c)}{(a-b)(a-c)}+ \frac{f(b)(x-a)(x-c)}{(b-c)(b-a)} +\frac{f(c)(x-a)(x-b)}{(c-b)(c-a)}$$
My Try
I tried this using the conventional method,$$f(x)=Q(x)(x-a)(x-b)(x-c)+Ax^2+Bx+C$$
But then I got long answers for coefficients $A$, $B$ & $C$.
Is there a better way to solve this? Can anyone give me a hint to work this?
| We have:
$$f(x)=Q(x)(x-a)(x-b)(x-c)+Ax^2+Bx+C$$
as you mentioned. Now note that
$$f(a)=Aa^2+Ba+C\\
f(b)=Ab^2+Bb+C\\
f(c)=Ac^2+Bc+C$$
Now it is easy to solve this system of equations for $A$, $B$ and $C$.
| {
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If $f(x)=\frac{x}{\sqrt{x^2+1}}$ then find $ f^{-1}(\sin x)$ let $f(x)=\dfrac{x}{\sqrt{x^2+1}}$ then find $ f^{-1}(\sin x)$
my try : $x:=\tan t$ then $f(\tan t)=\dfrac{\tan t}{1+\tan^2 t}=\sin t \ \ \ \text{or} \ \ -\sin t$
Now what ?
| $f(x)=\dfrac{x}{\sqrt{x^2+1}}$
Let $y=f(x)$
Let's start by making x a function of y, which will give you inverse.
$y=\dfrac{x}{\sqrt{x^2+1}}$
This simplifies to,
$y^2(x^2+1)=x^2$
From here,
$x=\dfrac{y}{\sqrt{1-y^2}}$
Therefore,
$f^{-1}(x) = \dfrac{x}{\sqrt{1-x^2}}$
Hence,
$f^{-1}(\sin x) = \dfrac{\sin x}{\sqrt{1-\sin^2 x}}=\tan x$
| {
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Finding constants in partial fraction In an example for partial fractions we want to find $A$, $B$, $C$, $D$ and $E$ in the expression:
$$
\frac{x^4-x^3+2x^2-x+2}{(x-1)(x^2+2)^2} = \frac{A}{(x-1)} + \frac{Bx+C}{(x^2+2)} + \frac{Dx+E}{(x^2+2)^2}
$$
Multiplying through to clear the fractions I obtained:
$$x^4-x^3+2x^2-x+2 = A(x^2+2)^2 + (Bx+C)(x-1)(x^2+2) + (Dx+E)(x-1)$$
I found $A=\frac{1}{3}$ by letting $x=1$.
Now in the book they let me know that $B=\frac{2}{3}$, $C=-\frac{1}{3}$, $D=-1$ and $E=0$. But I would really like to figure out how I can find the values for $B, C, D, E$.
| Setting $x=\sqrt 2i$ (so $x^2=-2$) yields
$$-4D-E+\sqrt2(-2D+E)i=2+\sqrt 2 i,\quad\text{whence }\begin{cases}\;2D+E=-2,\\-D+E=1,
\end{cases}\iff D=-1,\;E=0.$$
Next, setting $x=0$, you get
$$2=4A-2C-E, \quad\text{ so }\; C=-\tfrac13.$$
Last, multiply both sides of the decomposition by $x$ and let $x\to+\infty$, obtaining:
$$1=A+B,\quad\text{ so }\;B=\tfrac23.$$
| {
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Prove that $x^2-5y=3$ has no integer solutions. Here is my attempt:
$x^2-5y=3$ is the same as $2x^2-10y=6$, which is equivalent to $2x^2=10y+6$.
This means that 2 divides $x^2$. Since 2 is prime, then 2 divides x, or $x=2k$ for some integer $k$.
Then we have that $(2k)^2-5y=3$, or $4k^2=5y+3$. This implies that 2 divides $5y+3$, but this is a contradiction.
This looks valid to me but any feedback is greatly appreciated.
| Hint: $x^2 - 5y = 3 \iff x^2 = 5y + 3$. Now observe that the units digit of a perfect square must be one of the following digits: $0, 1, 4, 5, 6, 9$.
| {
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Evaluating $\lim\limits_{x\to 2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}$ without l'Hospital's rule
How to evaluate the following limit?
$$
\lim\limits_{x\to 2} \dfrac{\sqrt[3]{x} - \sqrt[3]{2}}{x^3 - 8}
$$
I factored the denominator into $(x-2)(x^2+2x+4)$, but I couldn't go on from there.
| Hint: Consider $f(x)=\sqrt[3]{x}$. So $$f'(2)=\lim_{x \to 2}\frac{\sqrt[3]{x}-\sqrt[3]{2}}{x-2}=\frac{1}{3}x^{-\frac{2}{3}}\Bigg|_{x=2}=\frac{1}{3}\frac{1}{2^{\frac{2}{3}}}$$ So $$\frac{\sqrt[3]{x}-\sqrt[3]{2}}{x^3-8}=\frac{\sqrt[3]{x}-\sqrt[3]{2}}{x-2} \frac{1}{x^2+2x+4} \longrightarrow f'(2).\frac{1}{12}$$ as $ x \to 2$
| {
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What is the correct solution of $\sqrt[7]{(-\sqrt{3}-i)^5}$? $\sqrt[7]{(-\sqrt{3}-i)^5}=(-\sqrt{3}-i)^\frac{5}{7}=
2^\frac{5}{7}(\cos(\frac{5}{7}\alpha)+i\sin(\frac{5}{7}\alpha)=$
$\tan\alpha=\frac{-1}{-\sqrt{3}} \implies \alpha=\frac{\pi}{6}+2k\pi$
$=2^\frac{5}{7}(\cos(\frac{5\pi}{42}+\frac{2k\pi}{7})+i\sin(\frac{5\pi}{42}+\frac{2k\pi}{7})$
The problem I have currently is that I have no idea where I made an error. The correct solution is $2^\frac{5}{7}(\cos(\frac{5\pi}{6}+\frac{2k\pi}{7})+i\sin(\frac{5\pi}{6}+\frac{2k\pi}{7})$.
| You just divided twice by $7$.
$$\sqrt[7]{(-\sqrt3-i)^5}=\sqrt[7]{32\left(\cos(5\cdot210^{\circ})+i\sin(5\cdot210^{\circ})\right)}=$$
$$=2^{\frac{5}{7}}\left(\cos\left(150^{\circ}+\frac{360^{\circ}k}{7}\right)+i\sin\left(150^{\circ}+\frac{360^{\circ}k}{7}\right)\right),$$ where $k\in\{0,1,2,3,4,5,6\}$.
| {
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Find $a,b,c \in \mathbb{R^+}$ such that $a^2+b^2+c^2+abc=4$ and $a+b+c=3$
Find $a,b,c \in \mathbb{R^+}$ such that $$a^2+b^2+c^2+abc=4\;\;\;{\rm and}\;\;\;a+b+c=3$$
Seeing $a^2+b^2+c^2+abc=4$,I substituted $a=2\cos{A}$,$b=2\cos{B},c=\cos{C}$ with $A+B+C=180°$.
Therefore,$cosA+cosB+cosC=\frac 32$ have thousands of solution. What to do now? Please help me.
| If we put $c=3-a-b$ we get quadratic equation on $a$ with parameter $b$, if $b \ne 2$:
$$(2-b)a^2-a(b^2-5b+6)+(2b^2-6b+5)=0$$
which has a solution iff it discriminat is nonnegative: $$(b-2)(b+2)(b-1)^2\geq 0$$
so $b>2$ or $b=1$...
*
*Clearly if $b=1$ we get $a^2-2a+1=0$ so $a=1$ and then $c=1$.
*If $b>2$ then with similar reason say on $b$ we get $a>2$ or $c>2$ which is impossibile.
| {
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How to solve $2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0$?
How to solve $$2x^5+5\sqrt{2}x^4+20x^3+20\sqrt{2}x^2+20x+4\sqrt{2}=0?$$
I just have no idea and I'have some knowledge about the polynomial equations. Here, just nothing.
| Substitute $x = \sqrt{2}y$ and divide by $\sqrt{2}^5$. The resulting equation is
$$y^5 + (y+1)^5 = 0.$$
Now substitute $y = -1/(z+1)$ and multiply by $(z+1)^5$ and you get
$$z^5 - 1 = 0.$$
| {
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Solving $z^4*\overline{z}=-1+\sqrt{3}i$ need some help solving this problem:
$z^4*\overline{z}=-1+\sqrt{3}i$$\DeclareMathOperator{\cis}{cis}$
The answer is: $z_k=(2*{2^{1/5}})^{1/5}\cis(\frac{2\pi+6\pi*k}{15})$,
$k=0,1,2,3,4$.
I tried using De Moivre to solve it but I got the wrong answer :
I substituted $z=r*\cis(\alpha)$
$r^4*\cis(4\alpha)*r*\cis(-\alpha)=2\cis(\frac{2\pi}{3})$
$r^5*\cis(3\alpha)=2\cis(\frac{2\pi}{3})$
$r*\cis(\frac{3\alpha}{5})=2^{1/5}\cis(\frac{\frac{2\pi}{3}+2\pi*k}{5})=2^{1/5}\cis(\frac{2\pi+6\pi*k}{15})$
from here I just couldn't find a way to get to the right answer.
| Since $r^5\operatorname{cis}3\alpha=2\operatorname{cis}\frac{2\pi(1+3n)}{3}$ with $n\in\Bbb Z$, $r=2^{1/5}$ while $\alpha=\frac{2\pi(1+3n)}{9}$ with $0\le n\le4$.
| {
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Speed of convergence of a special sequence. Let $a\in(0,1)$.
Consider a following sequence:
$$b_{n}=(\sum_{k=1}^{n}\frac{(-1)^k\cos(\ln(k))}{k^{a}})^{2}+(\sum_{k=1}^{n}\frac{(-1)^k\sin(\ln(k))}{k^{a}})^{2}$$
How fast does this series converge to its limit $b$?How to estimate $|b_{n}-b|$?
I know that the limit exists and $b=(\sum_{k=1}^{\infty}\frac{(-1)^k\cos(\ln(k))}{k^{a}})^{2}+(\sum_{k=1}^{\infty}\frac{(-1)^k\sin(\ln(k))}{k^{a}})^{2}$
But apart of it i am not able to tell something else.
I hope for your help.
| Hint:
Consider the complement of $b_n$ to $b$, i.e.
$$
\eqalign{
& \bar b_{\,n} = \left( {\sum\limits_{k = n + 1}^\infty {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln k} \right)} \over {k^{\,a} }}} } \right)^{\,2}
+ \left( {\sum\limits_{k = n + 1}^\infty {{{\left( { - 1} \right)^{\,k} \sin \left( {\ln k} \right)} \over {k^{\,a} }}} } \right)^{\,2} = \cr
& = \left( {\sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)^{\,k + n} \cos \left( {\ln \left( {k + n} \right)} \right)} \over {\left( {k + n} \right)^{\,a} }}} } \right)^{\,2}
+ \left( {\sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)^{\,k + n} \sin \left( {\ln \left( {k + n} \right)} \right)} \over {\left( {k + n} \right)^{\,a} }}} } \right)^{\,2} = \cr
& = \left( {{{\left( { - 1} \right)^{\,n} } \over {n^{\,a} }}\sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln n + \ln \left( {1 + k/n} \right)} \right)}
\over {\left( {1 + k/n} \right)^{\,a} }}} } \right)^{\,2}
+ \left( {{{\left( { - 1} \right)^{\,n} } \over {n^{\,a} }}\sum\limits_{k = 1}^\infty {{{\left( { - 1} \right)^{\,k} \sin \left( {\ln n + \ln \left( {1 + k/n} \right)} \right)}
\over {\left( {1 + k/n} \right)^{\,a} }}} } \right)^{\,2} = \cr
& = {1 \over {n^{\,2a} }}\left( {\sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{\,k} {{\cos \left( {\ln n} \right)\cos \left( {\ln \left( {1 + k/n} \right)} \right)
- \sin \left( {\ln n} \right)\sin \left( {\ln \left( {1 + k/n} \right)} \right)} \over {\left( {1 + k/n} \right)^{\,a} }}} } \right)^{\,2} + \cr
& + {1 \over {n^{\,2a} }}\left( {\sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{\,k} {{\sin \left( {\ln n} \right)\cos \left( {\ln \left( {1 + k/n} \right)} \right)
+ \cos \left( {\ln n} \right)\sin \left( {\ln \left( {1 + k/n} \right)} \right)} \over {\left( {1 + k/n} \right)^{\,a} }}} } \right)^{\,2} = \cr
& = {1 \over {n^{\,2a} }}\left( {\sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{\,k} {{\cos \left( {\ln n} \right)\left( {1 - {{k^{\,2} } \over {2n^{\,2} }}
+ O\left( {{{k^{\,3} } \over {n^{\,3} }}} \right)} \right)
- \sin \left( {\ln n} \right)\left( {{k \over n} - {{k^{\,2} } \over {2n^{\,2} }} + O\left( {{{k^{\,3} } \over {n^{\,3} }}} \right)} \right)}
\over {\left( {1 + a{k \over n} + O\left( {{{k^{\,2} } \over {n^{\,2} }}} \right)} \right)}}} } \right)^{\,2} + \cr
& + {1 \over {n^{\,2a} }}\left( {\sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{\,k} {{\sin \left( {\ln n} \right)\left( {1 - {{k^{\,2} } \over {2n^{\,2} }}
+ O\left( {{{k^{\,3} } \over {n^{\,3} }}} \right)} \right) + \cos \left( {\ln n} \right)\left( {{k \over n}
- {{k^{\,2} } \over {2n^{\,2} }} + O\left( {{{k^{\,3} } \over {n^{\,3} }}} \right)} \right)} \over {\left( {1 + a{k \over n}
+ O\left( {{{k^{\,2} } \over {n^{\,2} }}} \right)} \right)}}} } \right)^{\,2} = \cr
& = \quad \cdots \cr}
$$
It looks that from here you can arrive to the answer to your question.
An alternative approach would be
(here, for simplicity, we consider the development of only the cosine component)
$$
\eqalign{
& c_{\,2n} = \left( {\sum\limits_{k = 1}^{2n} {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln k} \right)} \over {k^{\,a} }}} } \right)^{\,2} = \cr
& = \left( {\sum\limits_{k = 1}^n {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln k} \right)} \over {k^{\,a} }}}
+ \sum\limits_{k = n + 1}^{2n} {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln k} \right)} \over {k^{\,a} }}} } \right)^{\,2} = \cr
& = \left( {\sum\limits_{k = 1}^n {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln k} \right)} \over {k^{\,a} }}} + {{\left( { - 1} \right)^{\,n} } \over {n^{\,a} }}
\sum\limits_{k = 1}^n {{{\left( { - 1} \right)^{\,k} \cos \left( {\ln n + \ln \left( {1 + k/n} \right)} \right)} \over {\left( {1 + k/n} \right)^{\,a} }}} } \right)^{\,2} = \cr
& = \quad \cdots \cr}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3286768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
If $f(x) =5x^2 - 2kx + 1 < 0$ has exactly one integral solution, find the sum of all positive integral values of $k$. Question : If $f(x) =5x^2 - 2kx + 1 < 0$ has exactly one integral solution, find the sum of all positive integral values of $k$.
My Attempt
Corresponding Equation of Inequality, $f(x) =5x^2 - 2kx + 1 = 0$. Let $\alpha,\beta$ be the roots. $f(x)<0$ for exactly one integral value of $x$.
Possible Graphs
Conditions Required
*
*$D>0 \\ 4k^2-20>0 \\k^2-5>0\\k \in(-\infty,-\sqrt5) \;\cup \;(\sqrt5,\infty) \qquad\text{(1)}$
*$|\alpha-\beta|\leq2 \\ |\frac{\sqrt D}a|\leq 2 \\\\ 2\frac{\sqrt{k^2-5}}{5}\leq2 \;\;\&\;\; 2\frac{\sqrt{k^2-5}}{5}\geq-2 \\\sqrt{k^2-5}\leq{5}\;\;\&\;\;\sqrt{k^2-5}\geq{-5}\\$
Solving we get $k\in[-\sqrt{30},\sqrt{30}]\qquad{(2)}\\$
and $k\in(-\infty,-\sqrt5] \; \cup \;[\sqrt5,\infty) \qquad\text{(3)}$
From $(1),(2),(3)$, we get $\boxed{k\in[-\sqrt{30},-\sqrt5) \; \cup \;(\sqrt5,\sqrt{30}]}$
Hence, sum of all positive integral values of $x$ is $3+4+5=12$
Issue: I think I have solved this problem correctly. But the answer at the back of my book is: $4+5=9$.
I mean, $\{3\}$ does not belong to the solution set, which I find strange. I might have made a mistake somewhere while solving it, but couldn't find it. I am just asking why $\{3\}$ does not belong to the solution set. Because it should, from what I see. Please help me with it. Thanks.
| We have $f(x) =5x^2 - 2kx + 1=5(x^2-2\frac{k}{5}x +\frac{k^2}{25})-\frac{k^2}{5}+1=5(x-\frac{k}{5})^2 - \frac{k^2}{5}+1 \geq 1-\frac{k^2}{5}$
So it takes it's minimum at $\frac{k}{5}$ which is $1-\frac{k^2}{5}$
We want $1-\frac{k^2}{5}<0$ in order for the inequality to have solution, hence $k<-\sqrt{5}$ and $k>\sqrt{5}$ $(1)$
The solutions to $f(x)=0$ are
$x_1=\frac{k-\sqrt{k^2-5}}{5}$ and $x_2=\frac{k+\sqrt{k^2-5}}{5}$
As you mentioned there must be the following requirement:
$x_2-x_1<2$ which by solving gives $-\sqrt{30}< k < \sqrt{30}$ $(2)$
From $(1)$ and $(2)$ we get the required solutions which are $k=4$ and $k=5$
The problem with $k=3$ is that $f(x)<0$ then $x \in (0.2, 1)$ which has no integral solution
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3298508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\sum_{r=1}^n \frac{\tan(x/2^r)}{2^{r-1}\cos(x/2^{r-1})}$
I was asked to find the sum of
$$\sum_{r=1}^n \dfrac{\tan \dfrac{x}{2^r}}{2^{r-1}\cos\dfrac{x}{2^{r-1}}}$$
I proceeded by breaking $\tan x$ into $\sin x$ and $\cos x$ and writing numerator as follows
$$\sin\left(\frac{x}{2^{r-1}}-\frac{x}{2^r} \right)$$
then by opening brackets and simplifying I got
$$\frac{1}{2^{r-1}}\left(\tan\frac{x}{2^{r-1}}-\tan\frac{x}{2^r}\right)$$
But I couldn't proceed from here.
Any help will be appreciated
| Start with $$\cos(x/2) \cos(x/2^2) \cos(x/2^3)....\cos(x/2^n)= \frac{\sin x}{2^n\sin(x/2^{n})}$$
Take $\ln$ of both sides, then
$$\sum_{r=1}^{n} \ln \cos (x/2^{r}) = -\ln \sin (x/2^{n})+\ln \sin x-n\ln 2$$ D.w.r.t x and get
$$\sum_{r=1}^{n} \frac{1}{2^{r}} \tan (x/2^{r})=\frac{1}{2^{n}} \cot (x/ 2^{n})- \cot x=g(x) ~~~~(1)$$
Similarly $$\cos x \cos(x/2) \cos(x/2^2) \cos(x/2^3)....\cos(x/2^{n-1})= \frac{\sin 2x}{2^n\sin(x/2^{n-1})}$$
Take $\ln$ of both sides, then
$$\sum_{r=1}^{n} \ln \cos (x/2^{r}) = -\ln \sin (x/2^{n-1})+\ln \sin 2x-n\ln 2$$ D.w.r.t x and get
$$\sum_{r=1}^{n} \frac{1}{2^{r-1}} \tan (x/2^{r-1})=\frac{1}{2^{n-1}} \cot (x/ 2^{n-1})- 2\cot 2 x=f(x) ~~~~(2)$$
The required sum is
$$ S=\sum_{r=1}^{n} \frac{1}{2^{r-1}} [\tan (x/2^{r-1})- \tan (x/2^r)] = f(x)-2g(x)$$ $$\Rightarrow S =\frac{1}{2^{n-1}} \left( \cot (\frac{x}{2^{n-1}})- \cot(\frac{x}{2^n}) \right) -2 (\cot 2 x -\cot x) $$ $$ \Rightarrow S=-2^{1-n} \csc(x2^{1-n})+2\csc(2x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3300150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
ambiguity in solving algebraic equations. Given $x+\frac{1}{x}=2 \tag{1}$
$x^3+\frac{1}{x^4}=1 \tag{2}$
Find the value of $$x^4+\frac{1}{x^3}$$
If equation $\bf{2}$ is multiplied by $\bf x$ we get $x^4+\frac{1}{x^3}=x$ now, we just need to find the value of $x$ which from $\bf{1}$ is equal to $1$.
But the answer to this question is $\bf3$.
Which can be get by solving differently. Why my method is wrong is it because I am introducing new root into the equation.
Another approach
Given, $(x + 1/x) = 2$
Squaring both sides,
$$x^2 + 1/x^2 + 2 = 4$$
$$ x^2 + 1/x^2 = 2 $$
Let,
$A = x^3 + 1/x^4$
$B = x^4 + 1/x^3$
Now, add A and B,
$A + B = x^3 + 1/x^4 + x^4 + 1/x^3$
$A + B = x^3 + 1/x^3 + x^4 + 1/x^4$
$A + B = (x + 1/x)(x^2 + 1/x^2 – 1) + (x^2)2 + (1/x^2)2 + 2 – 2$
$A + B = 2(2 – 1) + (x^2 + 1/x^2)2 – 2 $
$A + B = 2 + (2)2 – 2$
$ A + B = 4$
Given that $A = 1$
Then, $B = 4 – 1 = 3$
$ x^4 + 1/x^3 = 3$
| The system is inconsistent because the first equation gives $x=1$ which does not satisfy the second equation.
Thus the system does not have a solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3300686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The sum: $\sum_{k=1}^{n}(-1)^{k-1}~ [(H_k)^2+ H_k^{(2)}]~ {n \choose k}=\frac{2}{n^2}$ This attractive identity that
$$\sum_{k=1}^{n}(-1)^{k-1}~ [(H_k)^2+ H_k^{(2)}]~ {n \choose k}=\frac{2}{n^2}~~~(*)$$
emerged while doing numerics at Mathematica with harmonic numbers, binomial coefficients
and sums involving them.
Here $$H_k=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{k},~~ H_k^{(2)}=1+\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{k^2}. $$
The question is: How to prove it $(*)$ by hand?
| Let $$S_n=\sum_{k=1}^{n}(-1)^{k-1}~ [(H_k)^2+ H_k^{(2)}] {n \choose k}.$$
Notice that $$H_k^2+H_k^{(2)} =\left(\sum_{j=1}^{k} \frac{1}{j}\right)^2+\sum_{j=1}^{k} \frac{1}{j^2}=2 \sum \sum_{1 \le i \le j \le k} \frac{1}{ij}$$
So $S_n$ can be re-written as
$$S_n=2\sum_{i=1}^{n} \frac{1}{i} \sum _{j=i}^{n} \frac{1}{j} \sum_{k=j}^{n} (-1)^{k-1} {n \choose k}=2 \sum_{i=1}^{n} \frac{1}{i} \sum _{j=i}^{n} \frac{1}{j} \sum_{k=0}^{j-1} (-1)^{k} {n \choose k}=
2 \sum_{i=1}^{n} \frac{1}{i} \sum _{j=i}^{n} \frac{1}{j} (-1)^j {n-1 \choose j-1}.$$
Here we have used $$\sum_{k=j}^{n} (-1)^{k-1} {n \choose k}= 0-\sum_{k=0}^{j-1} (-1)^{k-1} {n \choose k}, \sum_{k=0}^m (-1)^k {n \choose k} =(-1)^m {n-1 \choose m}~~~(1)$$
Next we use $${n-1 \choose m-1}=\frac{m}{n} {n \choose n}~~~~(2)$$ twice to get
$$S_n=2\sum_{i=1}^{n} \frac{1}{i} \sum_{j=i}^{n} \frac{1}{j} (-1)^{j}~ \frac{j} {n} {n \choose j}=\frac{2}{n}\sum_{i=1}^{n} \frac{1}{i} \sum_{j=i}^{n} (-1)^j {n \choose j}.$$
Again using (1) and (2), we get
$$ \Rightarrow S_n=-\frac{2}{n}\sum_{i=1}^{n} \frac{1}{i} \sum_{j=0}^{i-1} (-1)^j {n \choose j}=\frac{2}{n}\sum_{i=1}^{n} \frac{1}{i} (-1)^i {n-1 \choose i-1}=\frac{2}{n}\sum_{i=1}^{n} \frac{1}{i} (-1)^i \frac{i}{n}{n \choose i}$$ $$\Rightarrow S_n=\frac{2}{n^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3302482",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Prove that $11 | 10^{2n+1}+1$ for all $n\in \mathbb{N}\cup \{0\}$. $$11 | 10^{2n+1}+1\;\;\; \forall n\in \mathbb{N}\cup\{0\} \tag{$\star$}$$
My proof of $(\star)$ is as follows:
\begin{align}
10^{2n+1}+1
&= 10\cdot10^{2n}+1 \\
&= (11-1)\cdot10^{2n}+1 \\
&= 11\cdot10^{2n}-10^{2n}+1 \\
&= 11\cdot10^{2n}-\left(10^{2n}-1\right) \\
&= 11\cdot10^{2n}-\left(100^{n}-1\right) \\
&= 11\cdot10^{2n}-\left((99+1)^{n}-1\right) \\
&= 11\cdot10^{2n}-\left(1+\binom{n}{1}99+\binom{n}{2}99^2+\cdots+\binom{n}{n-1}99^{n-1}+99^n-1\right) \\
&= 11\cdot10^{2n}-\underbrace{99}_{11\cdot9} \left(\binom{n}{1}+\binom{n}{2}99+\cdots+\binom{n}{n-1}99^{n-2}+99^{n-1}\right) \\
&= 11\left(10^{2n}-9 \left(\binom{n}{1}+\binom{n}{2}99+\cdots+\binom{n}{n-1}99^{n-2}+99^{n-1}\right)\right)
\end{align}
Is there an easier way to prove $(\star)$? The expansion of $(99+1)^n$ seems unnecessarily complicated, but I wasn't sure how else to go from there. Easier proofs are welcome!
| $2n+1$ is odd and $10\cong -1\pmod {11}$. Thus $10^{2n+1}\cong (-1)^{2n+1}\cong -1\pmod{11}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
shortest distance of $x^2 + y^2 = 4$ and $3x + 4y = 12$ $x^2 + y^2 = 4$ and $3x + 4y = 12$
line from origin with gradient $\frac{4}{3}$ intersect circle at $(\frac{6}{5} , \frac{8}{5}) $.
intersect $3x + 4y = 12$ at $(\frac{36}{25}, \frac{36}{25})$.
so distance is $\sqrt{{(\frac{36}{25}- \frac{6}{5})}^2 + {(\frac{36}{25}- \frac{8}{5})}^2} = \frac{\sqrt{52}}{25}$
but my answer is wrong. Where am i wrong?
| Let $c$ be a real number such that $3x+4y=c$ has common points $(x,y)$ with the circle and parallel to $3x+4y=12$.
Thus, by C-S we obtain:
$$|c|=|3x+4y|\leq\sqrt{(3^2+4^2)(x^2+y^2)}=\sqrt{25\cdot4}=10,$$ which says that
$3x+4y=10$ is a tangent to the circle,
which is parallel to the line $3x+4y=12$ and nearest to this line.
Id est, the needed distance it's:
$$\frac{|12-10|}{\sqrt{3^2+4^2}}=\frac{2}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3304841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
Find $h\colon\Bbb R\setminus\{0\}\to \Bbb R$ with $h(x - \frac{1}{x})= x^2 - \frac{1}{x^2}$ for all $x\ne0$.
I saw instantly that $$h(x - \frac{1}{x})= x^2 - \frac{1}{x^2} = \left( x -\frac{1}{x} \right)\left( x + \frac{1}{x} \right),$$ but I don't know how to proceed. I tried something like $$\left(x -\frac{1}{x} \right) = a$$ and $$\left( x + \frac{1}{x} \right) = a + \frac{2}{x},$$ trying to apply $h(a)$ but it doesn't seem to work.
Any hints?
| Let
$$x=y-\frac1y$$
so that
$$y=\frac{x\pm\sqrt{x^2+4}}2.$$
Then
$$f(x)=f\left(y-\dfrac1y\right)=y^2-\frac1{y^2}=\left(\frac{x\pm\sqrt{x^2+4}}2\right)^2-\left(\frac2{x\pm\sqrt{x^2+4}}\right)^2\\
=\left(\frac{x\pm\sqrt{x^2+4}}2\right)^2-\left(\frac{x\mp\sqrt{x^2+4}}2\right)^2=\pm x\sqrt{x^2+4}
.$$
Note that, as raised by Martin, to define a function only one sign at a time is possible and we have to choose. Then the sign of $y$ is constant and the domain of $x$ is restricted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3306654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Proving a reversed AM-HM inequality
Problem:
Given $n \ge 3$, $a_i \ge 1$ for $i \in \{1, 2, \dots, n\}$. Prove the following inequality:
$$(a_1+a_2+\dots +a_n) (\frac{1}{a_1}+\frac{1}{a_2} + \dots +\frac{1}{a_n}) \le n^2+\sum_{1\le i<j\le n}|a_i-a_j|.$$
This inequality seems like a reversed form of AM-HM inequality (which states that $\operatorname{LHS}\ge n^2$). I know that Kantorovich inequality is of the same direction, but that is not helpful. I tried to apply Cauchy–Schwarz inequality, but failed.
Thanks for any help.
| I will prove this via induction. Without loss of generality, assume $a_1 \le \cdots \le a_n$. It turns out this inequality remains true for $n = 1, 2$ as well, reducing to $1 \le 1$ in the first case. We actually need to prove the $n = 2$ case in order to elicit intuition for the inductive step.
Base Case $n = 2$
If $a_1 = a_2$, the inequality is trivially true, so assume $a_1 < a_2$. Then we seek to prove
\begin{align*}
(a_1 + a_2)(a_1^{-1} + a_2^{-1}) \le 2^2 + |a_2 - a_1| &\iff \frac{a_1}{a_2} + \frac{a_2}{a_1} \le 2 + (a_2 - a_1) \tag{*}\\
&\iff 0 \le \frac{a_1^2}{a_2 - a_1} + a_1 - 1
\end{align*}
which is true from the conditions $a_i \ge 1$.
Inductive Step
Define
\begin{align*}
S^A_n = \sum_{i=1}^{n}a_i \qquad \text{and} \qquad S^H_n = \sum_{i=1}^{n}a_i^{-1}
\end{align*}
We will use the relation
\begin{align*}
\sum_{1 \le i < j \le n}|a_i - a_j| = 2\sum_{i=1}^{n}ia_i - (n+1)S^A_n
\end{align*}
So we have
\begin{align*}
S^A_{n+1}S^H_{n+1} &= S^A_n S^H_n + a_{n+1}S^H_n + a^{-1}_{n+1}S^A_n + 1 \\
&\le n^2 + \left[2\sum_{i=1}^{n}ia_i - (n+1)S^A_n\right] + a_{n+1}S^H_n + a^{-1}_{n+1}S^A_n + 1 & \text{(Inductive Hypothesis)} \\
&\overset{\text{def}}{=} M
\end{align*}
To conclude, we want to show that $M$ does not exceed
\begin{align*}
N &\overset{\text{def}}{=} (n+1)^2 + \left[2\sum_{i=1}^{n+1}ia_i - (n+2)S^A_{n+1}\right]
\end{align*}
Computing the difference,
\begin{align*}
N - M &= 2n + 1 + 2(n+1)a_{n+1} - S^A_n - (n+2)a_{n+1} - a_{n+1}S^H_n - a^{-1}_{n+1}S^A_n -1\\
&= 2n + na_{n+1} - S^A_n - a_{n+1}S^H_n - a^{-1}_{n+1}S^A_n
\end{align*}
Finally, we have
\begin{align*}
na_{n+1} - S^A_n &= \sum_{i=1}^{n}(a_{n+1} - a_i) \\
&\ge\sum_{i=1}^{n}\left(\frac{a_i}{a_{n+1}}+\frac{a_{n+1}}{a_{i}}-2\right) & \text{From (*) in base step}\\
&=a_{n+1}S^H_n + a^{-1}_{n+1}S^A_n - 2n
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3308492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Find all irreducible monic polynomials of degrees $2$ in $\Bbb{Z}_{4}[x]$. Premise: not a native speaker.
So: I really don't know how to find all the polynomials except to calculate them all:
\begin{equation}x^{2}+ax+b \end{equation} \begin{equation} a,b ∈ {(0, 1, 2, 3)} \end{equation}
\begin{equation}
\begin{split}
1)x^{2}\\
2)x^{2} + x\\
3)x^{2} + 1\\
4)x^{2} + x + 1\\
5)x^{2} + 2x\\
6)x^{2} + 2\\
7)x^{2} + 2x + 1\\
8)x^{2} + x + 2\\
9)x^{2} + 2x + 2\\
10)x^{2} + 3x\\
11)x^{2} + 3\\
12)x^{2} + 3x + 3\\
13)x^{2} + 3x + 1\\
14)x^{2} + 32 + 2\\
15)x^{2} + x + 3\\
16)x^{2} + 2x +3\\
\end{split}
\end{equation}
I know that only numbers 3, 4, 6, 9, 12, 13, 15, 16 are Irreducible, because they didn't have roots in $\Bbb{Z}_{4}[x]$.
My question is: is there a more orthodox method to find the irreducible polynomials? Because with larger numbers, this method is not that practical.
| It is easier to count all reducible monic polynomials of degree $2$ because they are just the product of two monic linear polynomials, which are $x,x-1,x-2,x-3$ in your cases. So there are ${4 \choose 2} = 6$ products of different linear polynomials. As Robert pointed out, there are also $2$ squares of them. There are $16$ possible monic polynomials in $\Bbb{Z}_{4}[x]$ so there are $8$ irreducible such polynomials. In general, I think this is the efficient way to find all reducible and thus, all irreducible monic polynomials. For polynomials with larger degree, we can start with those with smaller degree first. For example, to know all reducible monic polynomials of degree $3$, we need to know irreducible linear polynomials and for degree $4$, we need to know all irreducible polynomials of degree $1$ and $2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Area boundes by two circles
Find the area bounded by $r \le 3\sin(t)$ and $r \le -3 \sqrt3 \cos(t)$.
Well, I've found the intersection point, which is $t=\frac{2 \pi}{3}$ . Then, I set up two integrals
$$
A=\frac{1}{2} \int_{\pi/2}^{2\pi/3} (3\sin(t))^2 dt
$$
and
$$
A=\frac{1}{2} \int_{2\pi/3}^{\pi}(-3 \sqrt3 \cos(t))^2 dt
$$
Is that correct?
| The correct integral expressions are
$$
A_s=\frac{1}{2} \int_{2\pi/3}^{\pi} [3\sin(t)]^2 dt=\frac{3\pi}{4}-\frac{9\sqrt{3}}{16}
$$
$$
A_l=\frac{1}{2} \int_{\pi/2}^{2\pi/3}[-3\sqrt3 \cos(t)]^2 dt =\frac{9\pi}{8}-\frac{27\sqrt{3}}{16}
$$
$$Area=A_s+A_l =\frac{15}{8}\pi-\frac{9}{4}\sqrt{3}.$$
The integration ranges are specified incorrectly in the post.
———————
For the bounded area, it is often easier just to visualize and quantify how the two circles overlap, as opposed to performing integration.
The bounded area is enclosed by two arcs, one spanning $\pi/3$ on the larger circle and $2\pi/3$ on the smaller one. The area is made up of two parts, separated by the line joining the two intersecting points.
The contribution from the larger circle is the difference between 1/6 of the full circle area and an overlaid 60-60-60 equilateral triangle, i.e.
$$A_l=\frac{1}{6}\pi r_l^2 - \frac{\sqrt{3}}{4}r_l^2$$
Similar, the contribution from the smaller circle is 1/3 of the full circle area and an overlaid 30-30-120 isosceles triangle
$$A_s=\frac{1}{3}\pi r_s^2 - \frac{\sqrt{3}}{4}r_s^2$$
$r_l=3\sqrt{3}/2$ and $r_s=3/2$ are the radii of the two circles, respectively.
Adding the two contributions together,
$$Area=\frac{\pi}{6} (r_l^2+2r_s^2)-\frac{\sqrt{3}}{4} (r_l^2+r_s^2)=\frac{15}{8}\pi-\frac{9}{4}\sqrt{3}.$$
| {
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Given $x, y$ that $xy-\frac{x}{y^2}-\frac{y}{x^2}=3$, work out $xy-x-y$. Today I had a competition in Xiamen, China. I know how to do the questions except this strange equation.
Given $x, y$ that $xy-\dfrac{x}{y^2}-\dfrac{y}{x^2}=3$, work out $xy-x-y$.
Such a strange question, right? I have found the integral solution are $0$ when $x=y=2$ and $3$ when $x=y=-1$, but I think there are infinitely many solutions but I can’t really prove it at that time. Can you guys help me?
| It is, of course, possible to solve the cubic $$xy-\frac x{y^2}-\frac y{x^2}=3\implies y^3\color{red}{-\frac{3x^2}{x^3-1}}y^2\color{blue}{+0}y\color{green}{-\frac{x^3}{x^3-1}}=0$$ using Cardano's method, which however may not be suitable as a contest problem.
We have $$Q=\frac{3\cdot\color{blue}{0}-\left(\color{red}{-\dfrac{3x^2}{x^3-1}}\right)^2}9=-\frac{x^4}{(x^3-1)^2}\tag1$$ and $$\small R=\frac{9\cdot\left(\color{red}{-\dfrac{3x^2}{x^3-1}}\right)\cdot\color{blue}0-27\cdot\left(\color{green}{-\dfrac{x^3}{x^3-1}}\right)-2\left(\color{red}{-\dfrac{3x^2}{x^3-1}}\right)^3}{54}=\frac{x^3}{2(x^3-1)}\left(1+\frac{2x^3}{(x^3-1)^2}\right)\tag2.$$ Now $$\small\sqrt{Q^3+R^2}=\sqrt{-\frac{x^{12}}{(x^3-1)^6}+\frac{x^6}{4(x^3-1)^2}\left(1+\frac{4x^3}{(x^3-1)^2}+\frac{4x^6}{(x^3-1)^4}\right)}=\frac{x^3(x^3+1)}{2(x^3-1)^2}\tag3$$ and so for $x\ne-1$, \begin{cases}R-\sqrt{Q^3+R^2}=\frac{x^3}{2(x^3-1)^2}\left(x^3-1+\frac{2x^3}{x^3-1}-(x^3+1)\right)=\frac{x^3}{(x^3-1)^3}\\R+\sqrt{Q^3+R^2}=\frac{x^3}{2(x^3-1)^2}\left(x^3-1+\frac{2x^3}{x^3-1}+(x^3+1)\right)=\frac{x^9}{(x^3-1)^3}\tag4.\end{cases} Therefore, $$S=\sqrt[3]{R+\sqrt{Q^3+R^2}}=\frac{x^3}{x^3-1},\quad T=\sqrt[3]{R-\sqrt{Q^3+R^2}}=\frac{x}{x^3-1}\tag5$$ and since $S-T\ne0$, the only real solution is $$y=S+T-\frac13\cdot\left(\color{red}{-\dfrac{3x^2}{x^3-1}}\right)=\frac{x^3+x^2+x}{x^3-1}=\frac x{x-1}=1+\frac1{x-1}\tag6.$$ Finally, this yields $$xy-x-y=x+\frac x{x-1}-x-1-\frac1{x-1}=0.\tag7$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Limit of an oscillating sequence Let $a$ be any real number. Then, how do I compute the
$$
\limsup_n \left|\cos\big(a\sqrt{n^2+1}\big) \right|^{1/n}\:?
$$
Also, how do I compute the value when $a$ is any complex value? The sequence is harshly oscillating... so I cannot find a way to approach
| I'll find the limit in the most general case when $a$ is complex.
Let $a=x+iy$ and the $n^{th}$ term of the sequence be denoted by $s_n$
\begin{align*}
s_n &= \left| \cos \left( a \sqrt{n^2+1} \right) \right|^{1/n} \\
&= \left| \frac{ e^{ia\sqrt{n^2+1}} + e^{-ia\sqrt{n^2+1}} }{2} \right|^{1/n} \\
&=\left|e^{ia\sqrt{n^2+1}} \right|^{1/n} \, \left| \frac{1+e^{-2ia\sqrt{n^2+1}}}{2} \right|^{1/n} \\
&= e^{ \frac{-y \sqrt{n^2+1}}{n}} \, \left| \frac{1+e^{-2ia\sqrt{n^2+1}}}{2} \right|^{1/n} \\
\end{align*}
Let the second term (without the power) be denoted by $t_n$ .
\begin{align*}
t_n &= \left| \frac{1+e^{2y\sqrt{n^2+1}} e^{-2ix\sqrt{n^2+1}} }{2} \right|\\
&= \frac{1}{2} \left( 1+ e^{4y\sqrt{n^2+1}}+2e^{2y\sqrt{n^2+1}}\cos(2x\sqrt{n^2+1}) \right)^{1/2}
\end{align*}
Now observe that
\begin{equation*}
u_n\leq t_n \leq v_n
\end{equation*}
where
\begin{align*}
u_n &= \left| \frac{1-e^{2y\sqrt{n^2+1}}}{2} \right| \\
v_n &= \left| \frac{1+e^{2y\sqrt{n^2+1}}}{2} \right|
\end{align*}
so we get for our original sequence $s_n$
\begin{equation*}
\left( e^{-y \sqrt{n^2+1}}u_n \right) ^{1/n} \leq s_n \leq \left( e^{-y \sqrt{n^2+1}}v_n \right) ^{1/n}
\end{equation*}
which gives
\begin{equation*}
\sinh \left( y\sqrt{n^2+1} \right) ^{1/n} \leq s_n \leq \cosh \left( y\sqrt{n^2+1} \right) ^{1/n}
\end{equation*}
Now in the limit $n\to \infty$, both of the sequences that sandwich our original sequence tend to $exp(y)$. Hence we get our answer
\begin{equation*}
\lim_{n\to \infty} s_n = e^y
\end{equation*}
| {
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$n^4 + 4^n$ is a not a prime Prove that $n^4 + 4^n$ is not a prime for all $n > 1$ and $n \in \mathbb{N}$.
This question appeared in the undergrad entrance exam of the Indian Statistical institute.
When $n$ is even the proof is simple.
For $n = 2m+1$ I am utterly stuck.
| Firstly, I’ll show that we can factorise $x^4+4y^4$.
$$x^4+4y^4=x^4-4x^2y^2+4y^4-4x^2y^2= \left(x^2+2y^2\right)^2-\left(2xy\right)^2= \left(x^2+2xy+2y^2\right)\left(x^2-2xy+2y^2\right)$$
Back to the question, when $n=2m+1$ that $m$ is a positive integer, we can substitute $x=n$ and $y=2^m$ into the polynomial above, we’ll get
$$n^4+4\times\left(2^m\right)^4=n^4+4\times4^{2m}=n^4+4^{2m+1}=n^4+4^n=\left(n^2+2^{m+1}n+2^{2m+1}\right)\left(n^2-2^{m+1}n+2^{2m+1}\right)$$
Therefore, $n^4+4^n$ is not a prime for all odd number $n>1$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Taylor series of $(1+3x) \cdot \ln(1+x)$ I have to find Taylor series of $(1+3x) \cdot \ln(1+x)$. I know Taylor series of $(1+3x) \cdot \ln(1+x)$ but I do not know hot to simplify.
Any help?
| The Maclaurin series of the given function is the product of the Taylor series of $\ln(1+x)$ and $1+3x$, treating the former as if it were a regular polynomial:
$$\ln(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\dots$$
$$3x\ln(1+x)=3x^2-\frac{3x^3}2+\frac{3x^4}3-\frac{3x^5}4+\dots$$
$$(1+3x)\ln(1+x)=x+\left(3-\frac12\right)x^2-\left(\frac32-\frac13\right)x^3+\left(\frac33-\frac14\right)x^4-\dots$$
$$=x+\sum_{k=2}^\infty(-1)^k\left(\frac3{k-1}-\frac1k\right)x^k$$
$$=x+\sum_{k=2}^\infty(-1)^k\frac{2(k+1)}{k(k-1)}x^k$$
| {
"language": "en",
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Every sufficiently large positive integer is the average of $n$ distinct primes for certain $n \geq 2$? I want to generalize a stronger Goldbach's conjecture a little bit because that might help solve it.
I was thinking:
For all $n \geq 2$, every sufficiently large positive integer $x \geq b_n$ is the average of $n$ distinct primes?
Clearly this implies Goldbach's conjecture.
So, was wondering if it has already been proven for any $n\gt 2$?
I think Terry Tao proved something different to the above, since it explicitly says at most 5 primes, which is algebraically not the same thing.
Remark. In particular the bound $b_n \geq \dfrac{p_1 + \dots + p_n}{n}$, where $p_i$ is the $i$th prime number: $2, 3, 5, \dots $
Proof of remark. If $\dfrac{q_1 + \dots +q_n}{n} = x \lt \dfrac{\sum_{i=1}^n p_i}{n}$, then cancel out the $n$ and the result follows. The $q_i$ are some prime numbers averaging to $x$.
For example for $n = 2$, the lower bound of $x$, $b_n$ seems to be $b_n = 4$, where $\dfrac{3 + 5}{2} = 4$ shows a solution for $x = 4$.
Here are some small cases:
$$
n = 2: \\
\dfrac{5 + 3}{2} = 4 \\
\dfrac{7 + 3}{2} = 5 \\
\dfrac{5 + 7}{2} = 6 \\
\dfrac{11 + 3}{2} = 7 \\
\dfrac{11 + 5}{2} = 8 \\
\dfrac{11 + 7}{2} = 9 \\
\ \\
n = 3: \\
\dfrac{3 + 5 + 7}{3} = 5 \\
\dfrac{2 + 5 + 11}{3} = 6 \\
\ \\
n = 4: \\
\dfrac{3 + 5 + 7 + 13}{4} = 7\\
\vdots
$$
We will need some SymPy Python 3.x code. I might code it up (testing up to $x = N, n = M$), but not for a few days. So there is some opportunity to beat me to it.
| For $n$ large enough Vinogradov's theorem gives $\sim C \frac{n^2}{\ln^3 n}$ solutions to $2n+1 = p_1+p_2+p_3$ which gives $\sim B \frac{n^3}{\ln^4 n}$ solutions to $2n = p_1+p_2+p_3+p_4$.
And the same method as Vinogradov gives $\sim A \frac{n^2}{\ln^3 n}$ solutions to $2n = 2q_1+q_2+q_3$.
Thus we have $\sim B \frac{n^3}{\ln^4 n}$ solutions for $2n = p_1+p_2+p_3+p_4$ as the sum of distinct primes.
Vinogradov's theorem follows from a strong form of the PNT in arithmetic progressions. I am quite sure for large enough $k$ there is a three line proof to your $2n = \sum_{j=1}^k p_j$ distinct primes problem just from $\pi(x) \sim \frac{x}{\ln x}$.
| {
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Find the sum of the series $1^3 + 3\cdot 2^2 + 3^3 + 3\cdot 4^2 + 5^3 + 3\cdot 6^2...$ up to $n$ terms
Find the sum of first $n$ terms of the series $1^3 + 3\cdot 2^2 + 3^3 + 3\cdot 4^2 + 5^3 + 3\cdot 6^2...$
*
*When $n$ is even.
*When $n$ is odd.
This sum can be written as
$$\sum_{1}^n (2k-1)^{3} +3 \sum_{1}^n (2k)^{2} $$
I can handle the sum up to n terms when it is not specified that $n$ is even or odd.
In this problem I'm confused, what changes should be done to get sum for even or odd $n$.
In my textbook, $n$ is replaced by $2m$ and then they solved the problem for first $m$ terms and then substituted $m = n/2$ and same is done for odd case, by substituting $n=2m-1$.
I didn't get that solution.
Any suggestion would be helpful.
| Its a hint
Split the series to two. One is sum of cubic of odd. And other one is 3*(sum of square of even )
| {
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Show that the vectors $v_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix}$ and $v_2 = \begin{pmatrix} 2 \\ 3 \end{pmatrix}$ are linearly independent... Show that the vectors $v_1 = \begin{pmatrix}
1 \\
2
\end{pmatrix}$ and $v_2 = \begin{pmatrix}
2 \\
3
\end{pmatrix}$ are linearly independent, and find the unique coefficients $x$ with
\begin{equation*}
x_1v_1+x_2v_2 = [v_1, v_2]x = \begin{pmatrix}
a \\
b
\end{pmatrix}.
\end{equation*}
To show that the vectors are independent we just row reduce the matrix containing the two vectors to make sure we get the identity matrix right.
My question is how do we find the unique constants with that equation shown?
Thank you.
| Another way is just to note that two vectors are linearly independent as long as they're not multiples of each other.
The hint by @Integrate This leads to $\begin {pmatrix}x_1\\x_2\end{pmatrix}=\begin{pmatrix}-3&2\\2&-1\end{pmatrix}\begin {pmatrix}a\\b\end{pmatrix}$, after inverting the $2×2$ matrix.
| {
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"timestamp": "2023-03-29T00:00:00",
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Discrete Mathematics - Combinatorics proof $\\$ I need to prove $\displaystyle\sum\limits_{k=0}^n\left(\frac{ 1^k+(-1)^k}{2}\right)\cdot\left(\begin{array}{c}n\\ k\end{array}\right)4^k=\frac{5^n+(-3)^n}{2}$
my Attempt below
$\displaystyle\sum\limits_{k=0}^n\left(\frac{ 1^k+(-1)^k}{2}\right)\cdot \left(\begin{array}{c}n\\ k\end{array}\right)4^k=\frac{1^0+(-1)^0}{2}\cdot \frac{n!}{0!(n-0)!}\cdot4^0\\ +\frac{1^1+(-1)^1}{2}\cdot \frac{n!}{1!(n-1)!}\cdot4^1+\frac{1^2+(-1)^2}{2}\cdot \frac{n!}{2!(n-2)!}\cdot4^2+\frac{1^3+(-1)^3}{2}\cdot \frac{n!}{3!(n-3)!}\cdot4^3+...+\frac{1^n+(-1)^n}{2}\cdot \frac{n!}{n!(n-n)!}\cdot4^n=\\ =\frac{1+1}{2}\cdot \frac{n!}{1n!}\cdot1+\frac{1+1}{2}\cdot \frac{n!}{2(n-2)!}\cdot16+\dotso+\frac{1^n+(-1)^n}{2}\cdot \frac{n!}{n!(n-n)!}\cdot4^n=\\= \frac{n!}{n!}+ \frac{4n!}{(n-1)!}+1\cdot \frac{16n!}{2(n-2)!} +1\frac{64n!}{6(n-3)!}+1\dotsm\ \frac{4^nn!}{n!(n-n)!}+1=$
$\frac{1}{2}\left[\sum_{k = 0}^{n} \binom{n}{k}4^k + \sum_{k = 0}^{n} \binom{n}{k}(-4)^k\right]=\\$1/2$(1+x)^n=\frac{1}{2}((1+4)^n+(1-4)^n=\frac{5^n+(-3)^n}{2}$
| By the binomial of Newton it's just $$\frac{1}{2}((1+4)^n+(1-4)^n).$$
| {
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Is $\frac{\frac{1}{b}-\frac{1}{2}}{b-2}$ the same as $\frac{\frac{2-b}{2b}}{b-2}$? Have I subtracted 2 fractions correctly?
$\frac{\frac{1}{b}-\frac{1}{2}}{b-2}$
Starting with the numerator which is a difference of fractions, the least common denominator is 2b? So:
$\frac{2(1)}{2b}-\frac{b(1)}{2b}$ = $\frac{2}{2b}-\frac{b}{2b}$ = $\frac{2-b}{2b}$
So my newly simplified expression is:
$\frac{\frac{2-b}{2b}}{b-2}$
My question - is this correct? If yes, great - now how would I go about simplifying further by taking the denominator $b-2$ into account? If no, where did I go wrong?
| Yes, it's the same, in fact the first fraction can be rewritten as $\frac{-(b-2)}{2b}\cdot \frac{1}{b-2}=-\frac{1}{2b}$. The second, instead: $\frac{-(b-2)}{2b}\cdot \frac{1}{2b}=-\frac{1}{2b}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$ Find the remainder of $\frac{x^{2015}-x^{2014}}{(x-1)^3}$.
Let $P(x)=x^{2015}-x^{2014}=Q(x)(x-1)^3+ax^2+bx+c.$ If we put $x=1$ in $P(x)$ and $P'(x)$, we get $a+b+c=0$ and $2a+b=1$. Then: $c=a-1$. The second derivative won't help in finding $b$, so, what should I do? Thank you
| Another way:
Set $x-1=y$
$x^r=(1+y)^r\equiv1+\binom r1y+\binom r2y^2\pmod{y^3}$
$x^m-x^n\equiv y(m-n)+y^2\left(\binom m2-\binom n2\right)\pmod{y^3}$
Replace $y$ with $x-1$
| {
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In $\triangle ABC$, if angle bisectors $AE$ and $CD$ meet at incenter $F$, and $|FE|=|FD|$, then the triangle is isosceles or $\angle B=60^\circ$ I was screwing around lately in GeoGebra and I realized something.
Draw a $\triangle ABC$, and let the bisectors for $\angle A$ and $\angle C$ meet sides $BC$ and $AB$ at points $E$ and $D$, respectively. If the angle bisectors meet at the incenter, $F$, and if $FD \cong FE$, then either $\triangle ABC$ must be isosceles or $\angle B$ must be $60^\circ$.
However I was unable to prove why that is. Any help would be appreciated.
| In the standard notation by the angle bisector theorem we obtain:
$$\frac{FE}{AF}=\frac{CF}{AC}=\frac{\frac{ab}{b+c}}{b}=\frac{a}{b+c}.$$
Thus, $$\frac{FE}{AE-FE}=\frac{a}{b+c},$$ which gives
$$FE=\frac{a}{a+b+c}\cdot AE=\frac{a}{a+b+c}\cdot\frac{2bc\cos\frac{\alpha}{2}}{b+c}=\frac{2abc\cos\frac{\alpha}{2}}{(a+b+c)(b+c)}.$$
Similarly, $$FD=\frac{2abc\cos\frac{\gamma}{2}}{(a+b+c)(a+b)}.$$
Id est,
$$\frac{2abc\cos\frac{\alpha}{2}}{(a+b+c)(b+c)}=\frac{2abc\cos\frac{\gamma}{2}}{(a+b+c)(a+b)}$$ or
$$a(b+c-a)(a+b)^2=c(a+b-c)(b+c)^2$$ or
$$(a-c)(a+b+c)(a^2+c^2-ac-b^2)=0.$$
Can you end it now?
| {
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Find $a,b$ such that $a^2+b^2=85113$ and that $\text{lcm}\{a,b\}=1764$
Find two positive integers $a,b$ such that $a^2+b^2=85113$ and that $\text{lcm}\{a,b\}=1764$
I have completely no idea where to enter this question.
| Note that $\text{lcm}(a,b)=1764=(2\cdot 3\cdot 7)^2$ and $85113=3^2\cdot 7^2\cdot 193$. We may assume that $7^2$ divides $b$ then $a^2=85113-b^2$ is divisible by $7^2$ which means that $7$ divides $a$. Let $A:=a/7$, $B:=b/7$ and solve the reduced diophantine equation
$$A^2+B^2=85113/7^2=1737$$
with $\text{lcm}(A,B)=2^2\cdot 3^2\cdot 7$.
Similarly, we can operate with respect to the factor $3$ and solve the reduced diophantine equation
$$X^2+Y^2=1737/3^2=193$$
where $X:=a/21$ and $Y:=b/21$ with $\text{lcm}(X,Y)=2^2\cdot 3\cdot 7$.
Since $13^2<193<14^2$, now it is easy to find the solutions by hand. Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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how to prove this inequality $\sum\frac{x}{(n-1)y+1}+\frac{1}{\sum x}\ge\frac{n+1}{n}$ if $x_{i}>0$,show that
$$\sum_{i=1}^{n}\dfrac{x_{i}}{(n-1)x_{i+1}+1}+\dfrac{1}{\sum_{i=1}^{n}x_{i}}\ge\dfrac{n+1}{n}$$
I have proven $n=2$and $n=3$
$n=3$ case
let $x,y,z>0$.prove
$$\dfrac{x}{2y+1}+\dfrac{y}{2z+1}+\dfrac{z}{2x+1}+\dfrac{1}{x+y+z}\ge\dfrac{4}{3}$$
By Cauchy-Schwarz inequality we have
$$\sum_{cyc}\dfrac{x}{2y+1}\ge \dfrac{(x+y+z)^2}{2xy+2yz+2zx+x+y+z}\ge\dfrac{t^2}{\dfrac{2}{3}t^2+t}=\dfrac{3t}{2t+3}$$
where $t=x+y+z>0$
so we must prove
$$\dfrac{3t}{2t+3}+\dfrac{1}{t}\ge\dfrac{4}{3}$$
the last inequality it is clear true!
| A proof for $n=4$.
Let $x+y+z+t=4u$.
Thus, by C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{x}{3y+1}+\frac{1}{x+y+z+t}=\sum_{cyc}\frac{x^2}{3xy+x}+\frac{1}{x+y+z+t}\geq$$
$$\geq\frac{(x+y+z+t)^2}{3\sum\limits_{cyc}xy+x+y+z+t}+\frac{1}{x+y+z+t}=\frac{16u^2}{3(x+z)(y+t)+4u}+\frac{1}{4u}\geq$$
$$\geq \frac{16u^2}{3\left(\frac{x+z+y+t}{2}\right)^2+4u}+\frac{1}{4u}=\frac{4u}{3u+1}+\frac{1}{4u}.$$
Id est, it's enough to prove that:
$$\frac{4u}{3u+1}+\frac{1}{4u}\geq\frac{5}{4}$$ or
$$(u-1)^2\geq0$$ and we are done.
For $n\geq5$ this way does not help.
| {
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Trigonometric inequality $\tan\left(\frac{1}{1+x^2}\right)<\frac{1}{1-x+x^2}$ for all $x>1/2$ In Hobson's book "A treatise on plane trigonometry", the following exercise is to be found
If $x>1/2$ then
$$\frac{1}{1+x+x^2}<\tan\left(\frac{1}{1+x^2}\right)<\frac{1}{1-x+x^2}.$$
The first inequality is easy as it follows directly from $\alpha<\tan(\alpha)$ and the fact that $x$ is positive.
For the second one can produce the even tighter bound (on the given interval, checked only graphically)
$$\tan(\alpha)<\frac{\alpha}{1-\frac{\alpha^2}{2}}$$
but from here I cannot get to the exact given inequality.
I am aware of the fact that $\tan(\alpha)\sim \alpha$ for small $\alpha$, but I cannot manage to find the exact bound given in the exercise.
Any hints are welcome.
| The given inequality is equivalent to
$$ \frac{1}{1-\sqrt{z-z^2}}-1\geq\frac{\tan(z)}{z}-1\geq\frac{1}{1+\sqrt{z-z^2}}-1 $$
for $z\in\left(0,\frac{4}{5}\right]$. This is fairly simple to prove by concavity/convexity: the LHS lies above the line $y=\frac{z}{2}$, the central term lies below it. The central term lies above the line $y=0$, the RHS lies below it.
Here it is a simpler derivation: by Lagrange's theorem
$$ \arctan\left(\frac{1}{1+x+x^2}\right)=\arctan(x+1)-\arctan(x)=\frac{1}{\xi^2+1},\quad \xi\in(x,x+1) $$
hence
$$ \arctan\left(\frac{1}{1+x+x^2}\right) < \frac{1}{x^2+1}, \qquad \frac{1}{1+x+x^2}<\tan\left(\frac{1}{x^2+1}\right). $$
Similarly
$$ \arctan\left(\frac{1}{1-x+x^2}\right)=\arctan(x)-\arctan(x-1)=\frac{1}{\eta^2+1},\quad \eta\in(x-1,x) $$
hence, assuming $x\geq\frac{1}{2}$,
$$ \arctan\left(\frac{1}{1-x+x^2}\right) > \frac{1}{x^2+1}, \qquad \frac{1}{1-x+x^2}>\tan\left(\frac{1}{x^2+1}\right). $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Number of roots of a Cubic Polynomial Let $f(x) = x^3 + px +q$ prove that if $(\frac{p}{2})^2+(\frac{q}{3})^3 > 0$ and $p <0 $ Then the equation has 1 real solution
Getting $f'(x)$ and setting it to $0$
$$x = \pm\sqrt{\frac{-p}{3}}$$
using the positive value and substituting it to $f(x)$, we get that its $y$ value is
$$\frac{\sqrt{-27p^3}}{27} + p\frac{\sqrt{-3p}}{3} + q$$
The $y$ value of the inflection point $(q)$ must be larger than the difference between the $y$ value of the inflection point and the $y$ value of the critical point.
$$q > q-(\frac{\sqrt{-27p^3}}{27} + p\frac{\sqrt{-3p}}{3} + q)$$
and I get
$$(\frac{p}{4})^2+(\frac{q}{3})^3 > 0$$
Where did i go wrong?
| Let $q\geq0$.
Thus, $$f\left(\sqrt{\frac{-p}{3}}\right)=\sqrt{\frac{-p^3}{27}}+p\sqrt{\frac{-p}{3}}+q=$$
$$=\sqrt{\frac{-p^3}{27}}-3\sqrt{\frac{-p^3}{27}}+q=2\left(\frac{q}{2}-\sqrt{\frac{-p^3}{27}}\right)=$$
$$=\frac{2\left(\left(\frac{q}{2}\right)^2+\left(\frac{p}{3}\right)^3\right)}{\frac{q}{2}+\sqrt{\frac{-p^3}{27}}}>0,$$
which says that our equation has an unique real root.
The case $q<0$ is a similar, but we need to work with a minimum point.
| {
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Solving $ax^3+bx^2+cx+d=0$ using a substitution different from Vieta's? We all know, a general cubic equation is of the form
$$ax^3+bx^2+cx+d=0$$ where $$a\neq0.$$
It can be easily solved with the following simple substitutions:
$$x\longmapsto x-\frac{b}{3a}$$
We get,
$$x^3+px+q=0$$ where, $p=\frac{3ac-b^2}{3a^2}$ and $q=\frac{2b^3-9abc+27a^2d}{27a^3}$
Then, using the Vieta substitution,
$$x\longmapsto x-\frac{p}{3x}$$
We get,
$$(x^3)^2-q(x^3)-\frac1{27}p^3=0$$
which is can easily turn into a quadratic equation, using the substitution: $x^3 \longmapsto x.$
And here is my question:
In mathematics is there a substitution that is "different" from the substitution $x\longmapsto x-\frac{p}{3x}$ that can be used for the standard form cubic equation $x^3+px+q=0$ , which is can easily turn into a quadratic equation?
I'm curious, if there's a new substitute I don't know about.
Thank you!
| Possible substitution $x\to X+\frac{1}{z}$, where $X$ is some parameter and $z$ is new unknown, but additional need use some properties of cubic polynomial, which historically we received from Lagrange.
Thus we have solution of equation $f=ax+bx^2+cx^3$:
$\begin{cases}
X=\dfrac{-(a b + 9 c f) + \sqrt{(a b + 9 c f)^2-4 (b^2 - 3 a c) (a^2 + 3 b f)}}{2 (b^2 - 3 a c)}\\
C=a X + b X^2 + c X^3 - f\\
B_1=a + 2 b X + 3 c X^2\\
W=B_1^3-27cC^2\\
B_2=\left\{W^{1/3}\,,-(-1)^{1/3} W^{1/3}\,,(-1)^{2/3}W^{1/3}\right\}\\
x=X+\frac{3C}{B_2-B_1}
\end{cases}$
Deriving this solution see in Intro, and there is a code to verifing in Wolfram and Geogebra.
| {
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"timestamp": "2023-03-29T00:00:00",
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Combinatorics floral arrangement question A florist has to make a floral arrangement. She has 6 Banksias, 5 wattles and 4 Waratahs. All the flowers of each kind are different. In how many ways can the florist make a bunch of 10 flowers if she has to use at least 3 of each kind? Please show working.
I have tried to use the combinations formula but have gotten two different answers - 1200 and 4800 depending on whether I do the first 9 and then the last flower or all at once.
Just an interesting problem I know and have not been able to figure out.
| At least three of each kind means that $10=4+3+3=3+4+3=3+3+4$ and therefore the number of different bunches of $10$ flowers out of $6$ Banksias, $5$ Wattles and $4$ Waratahs should be
$$\binom{6}{4}\binom{5}{3}\binom{4}{3}+\binom{6}{3}\binom{5}{4}\binom{4}{3}+\binom{6}{3}\binom{5}{3}\binom{4}{4}=600+400+200=1200.$$
We get $4800$ if we choose the first $9$ and then we add the last one (the fourth of the same kind):
$$\binom{6}{3}\binom{5}{3}\binom{4}{3}\cdot (3+2+1)=4800$$
But here we overcount each bunch four times because each of the four flowers of the same kind can be the last one. So we have to divide $4800$ by $\mathbf{4}$ and we obtain again the correct number $1200$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Differentiating $\frac{dy}{dx}=3x+2y+xy$? What would $y''$ be of $\frac{dy}{dx}=3x+2y+xy$? In other words, what is the result ($y''$) if I differentiate $y'=3x+2y+xy$?
Can I do the following:
Set $$f(x,y) = 3x+2y+xy$$
then
$$f'(x,y) = \frac{df}{dx} + \frac{df}{dy}\frac{dy}{dx}$$
$$ f'(x,y) = (3 + y)+(2+x)(y')$$
$$ f'(x,y) = 3 + y + 2y'+ xy'$$
?
| $\frac {dy}{dx} = 3x + 2y + xy$
Differentiate both sides using the rules of implicit differentiation:
$\frac{d}{dx}\frac {dy}{dx} = \frac{d}{dx} (3x + 2y + xy)\\
\frac {d^2y}{dx^2} = \frac{d}{dx} 3x + \frac{d}{dx}2y + \frac{d}{dx}xy\\
\frac {d^2y}{dx^2} = 3 + 2\frac {dy}{dx} + y + x\frac{dy}{dx}\\
$
We can substitute $\frac {dy}{dx} = 3x + 2y + xy$ from the begining
$\frac {d^2y}{dx^2} = 3 + y + (2+x)(3x + 2y + xy)$
And simplify.
| {
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"timestamp": "2023-03-29T00:00:00",
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Angle between two vector
I tried to solve this by directly multiplies a-b and 6a+b equal zero, and substitute some of it with the |a| and |b| to get a.b. but somehow I get different answer from the answer sheet that is 60 degree
It is still not complete but after that I get some bizzare answer so I didn't write it
| Given $(\overrightarrow{a}-\overrightarrow{b})\perp(6\overrightarrow{a}+\overrightarrow{b})$, we know that
\begin{align}
(\overrightarrow{a}-\overrightarrow{b})\cdot(6\overrightarrow{a}+\overrightarrow{b})&=0\\
6\overrightarrow{a}\cdot\overrightarrow{a}-5\overrightarrow{a}\cdot\overrightarrow{b}-\overrightarrow{b}\cdot\overrightarrow{b}&=0\\
6|\overrightarrow{a}|^2-5\overrightarrow{a}\cdot\overrightarrow{b}-|\overrightarrow{b}|^2&=0\\
6(2)^2-5\overrightarrow{a}\cdot\overrightarrow{b}-(3)^2&=0\\
-5\overrightarrow{a}\cdot\overrightarrow{b}&=-15\\
\overrightarrow{a}\cdot\overrightarrow{b}&=3
\end{align}
From the definition of the dot product:
\begin{align}
\cos\theta&=\dfrac{\overrightarrow{a}\cdot\overrightarrow{b}}{|\overrightarrow{a}||\overrightarrow{b}|}\\
\cos\theta&=\dfrac{3}{(2)(3)}\\
\cos\theta&=\dfrac12\\
\theta&=60^{\circ}
\end{align}
Anything from here that needs further explanation?
| {
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Powers of two are not sums of squares of different integers greater than zero It seems that non-trivial sums of squares of different non-zero integers aren't powers of two. Tested for sums of $2,3,4$ terms.
Is there something known about this topic? Are there counter-examples?
| Couldn't find a counter-example for up to $4$ terms as well (maybe someone can show why), but for more terms there are for example
$$
1^2+5^2+7^2+9^2+10^2=2^8\\
2^2+3^2+5^2+7^2+13^2=2^8\\
3^2+5^2+6^2+9^2+19^2=2^9\\
\vdots \\
1^2+5^2+6^2+7^2+8^2+9^2=2^8\\
\vdots \\
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove $\int_0^1\frac{\ln x\ln(1+x)}{1-x}\ dx=\zeta(3)-\frac32\ln2\zeta(2)$ How to prove without using Euler sums that
$$I=\int_0^1\frac{\ln x\ln(1+x)}{1-x}\ dx=\zeta(3)-\frac32\ln2\zeta(2)$$
where $\zeta$ is the Riemann zeta function.
We can relate this integral to some Euler sum as follows:
\begin{align}
I&=-\sum_{n=1}^\infty\frac{(-1)^n}{n}\int_0^1\frac{x^n\ln x}{1-x}\ dx\\
&=-\sum_{n=1}^\infty\frac{(-1)^n}{n}(H_n^{(2)}-\zeta(2))\\
&=-\sum_{n=1}^\infty\frac{(-1)^nH_n^{(2)}}{n}-\ln2\zeta(2)
\end{align}
Also the integral $I$ can be related to $\sum_{n=1}^\infty\frac{(-1)^nH_n}{n^2}$. So I am looking for a different way to evaluate $I$ besides using these two sums.
| Start off with the substitution $x\to \frac{1-x}{1+x}$ to get:
$$\require{cancel} I=\int_0^1 \frac{\ln x\ln(1+x)}{1-x}dx=\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{2}{1+x}\right)}{x}dx-\int_0^1 \frac{\ln\left(\frac{1-x}{1+x}\right)\ln\left(\frac{2}{1+x}\right)}{1+x}dx$$
$$X=\int_0^1 \frac{\ln(1-x)\ln 2 -\ln(1-x)\ln(1+x)-\ln(1+x)\ln 2+\ln^2(1+x)}{x}dx$$
$$Y=\int_0^1 \frac{\ln(1-x)\ln 2 -\ln(1-x)\ln(1+x)-\ln(1+x)\ln 2+\ln^2(1+x)}{1+x}dx$$
$$I_1=\ln 2\int_0^1 \frac{\ln(1-x)}{x}dx=\color{blue}{-\ln 2 \zeta(2)}$$
$$I_2=-\int_0^1 \frac{\ln(1-x)\ln(1+x)}{x}dx=\color{red}{\frac{5}{8}\zeta(3)}$$
$$I_3=-\ln 2 \int_0^1 \frac{\ln(1+x)}{x}dx=\color{blue}{-\frac{\ln 2}{2}\zeta(2)}$$
$$I_4=\int_0^1 \frac{\ln^2(1+x)}{x}dx=\color{red}{\frac{\zeta(3)}{4}}$$
$$I_5=\ln 2\int_0^1 \frac{\ln(1-x)}{1+x}dx=\cancel{\frac{\ln^3 2}{2}}-\cancel{\ln 2\frac{\zeta(2)}{2}}$$
$$I_6=-\int_0^1 \frac{\ln(1-x)\ln(1+x)}{1+x}dx =\cancel{-\frac{\ln^3 2}{3}}+\cancel{\ln 2\frac{\zeta(2)}{2}}-\color{red}{\frac{\zeta(3)}{8}}$$
$$I_7=-\ln 2 \int_0^1 \frac{\ln(1+x)}{1+x}dx=\cancel{-\frac{\ln^3 2}{2}}$$
$$I_8=\int_0^1 \frac{\ln^2(1+x)}{1+x}dx=\cancel{\frac{\ln^3 2}{3}}$$
$$I=X-Y=(I_1+I_2+I_3+I_4)-(I_5+I_6+I_7+I_8)=\boxed{\zeta(3)-\frac32 \ln 2 \zeta(2)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$X^5 + 1$ into irreducible factors over $\mathbb{Q}$ and $\mathbb{Z}_5$ I need to split this polynomial into irreducible factors. I managed to do this for $\mathbb{C}$ and $\mathbb{R}$, but I can't figure out how to solve this problem for $\mathbb{Q}$ and $\mathbb{Z}_5$.
| This answer only addresses $\mathbb{Z}_5$.
For factoring over (small) finite fields, the easiest choice is to just try all the points in the field.
\begin{align*}
&x :& &x^5+1 \pmod{5} \\
&0 :& &1 \pmod{5} \\
&1 :& &2 \pmod{5} \\
&2 :& &3 \pmod{5} \\
&3 :& &4 \pmod{5} \\
&4 :& &0 \pmod{5}
\end{align*}
So, $x-4 \cong x+1 \pmod{5}$ is a factor and no other linear polynomial is a factor. In $\mathbb{Z}_5$,
$$ \frac{x^5+1}{x+1} \cong x^4 + 4x^3 + x^2+4x+1 \pmod{5} $$
and we should check to see if any of the linear factors found in the first table continue to divide this quotient. (This langauge seems a little odd because it also works when the table finds more than one linear factor -- we have to keep checking all the factors we found to see if they divide the subsequent quotients.) \begin{align*}
\frac{x^4 + 4x^3 + x^2+4x+1}{x+1} &\cong x^3 + 3x^2+3x+1 \pmod{5} \text{,} \\
\frac{x^3 + 3x^2+3x+1}{x+1} &\cong x^2 + 2x + 1 \pmod{5} \text{, and} \\
\frac{x^2 + 2x + 1}{x+1} &\cong x+1 \pmod{5} \text{.}
\end{align*}
So we have shown $x^5+1 \cong (x+1)^5 \pmod{5}$.
| {
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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.