Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the generating function and the number of integers solutions for $x_1 + x_2 + x_3 + x_4 = r$, where $-3 \leq x_i \leq 3$. Well, we have $x_1 + x_2 + x_3 + x_4 = r$ where $x_i\in\{-3,-2,-1,0,1,2,3\}$
Then the generating function is given by $f(x)=(x^{-3}+x^{-2}+x^{-1}+x^{0}+x^1+x^2+x^3)^4 = \frac{(x^{6}+x^{5}+x^{4}+x^{3}+x^2+x^1+x^0)^4}{(x^3)^4}=\frac{(x^{6}+x^{5}+x^{4}+x^{3}+x^2+x^1+x^0)^4}{x^{12}}= (x^{6}+x^{5}+x^{4}+x^{3}+x^2+x^1+x^0)^4x^{-12}= (\frac{(1-x^{7})}{(1-x)})^4x^{-12} $
Then
$(1-x^{7})^4=1-4x^7+6x^{14}-4x^{21}+x^{28}$
And
$(1-x)^{-4}=\sum_{r=0}^\infty \binom{-4}{r}x^r(-1)^{r}$
So
$\binom{-4}{r}-4 \binom{-4}{r-7}+6 \binom{-4}{r-14}-4 \binom{-4}{r-21}+ \binom{-4}{r-28}=\binom{r+4-1}{r}-4 \binom{r+4-1}{r-7}+6 \binom{r+4-1}{r-14}-4 \binom{r+4-1}{r-21}+ \binom{r+4-1}{r-28}=\binom{r+3}{r}-4 \binom{r+3}{r-7}+6 \binom{r+3}{r-14}-4 \binom{r+3}{r-21}+ \binom{r+3}{r-28}$
So this is what I did. Is correct? Shouldn't I be able to get to a number?
| First you need to know why generating function works, and why negative integers have a problem.
https://brilliant.org/wiki/generating-functions-solving-recurrence-relations/#:~:text=Generating%20Functions,in%20a%20sequence%20of%20numbers&text=an.,used%20for%20solving%20recurrence%20relations.
So your generating function is wrong. Each power of $x$ represents the number of times the object is being taken, hence $1$ or $x^0$ means that the the object is not taken.
since the values are negative, we have to take the powers negative too.
Let's take only two variables and $r=0$ Listing out all the possible cases, $$\{ (-3,3), (-2,2), (-1,1), (0,0), (1,-1), (2,-2), (3,-3)\}$$ i.e. $7$
Our function will be, $$\left[\dfrac{1-x^7}{x^3(1-x)}\right]^2$$
And the constant term here is 7
By,
https://www.wolframalpha.com/input/?i=Series+%28%281-x%5E7%29%2F%28x%5E%283%29%281-x%29%29%29%5E2
| {
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"url": "https://math.stackexchange.com/questions/3702491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Analytically solving $\frac{1}{\sin2x} + \frac{1}{\sin3x} = \frac{1}{\sin x}$ Given
$$ \frac{1}{\sin(2x)} + \frac{1}{\sin(3x)} = \frac{1}{\sin x}$$
I tried solving the equation above using the double and triple angle formulas and arrived at this cubic expression in $\cos x$
$$ 8\cos^3(x)-4 \cos^2(x)-4\cos(x) + 1$$
I ( and apparently wolfram alpha too) and unable to solve it analytically. But I when I take the inverse cosine of the “numerical” roots, i get exact answers, namely $\frac{\pi}{7}$, $\frac{5\pi}{7}$, and $\frac{3\pi}{7}$. How should I approach problems like these?
| Hint:
Remember $\sin x\ne0\implies\cos x\ne\pm1$
From
$$(2c)^3-(2c)^2-2(2c)+1=0$$
Replace $2c$ with $y+\dfrac1y$ where $y=e^{ix}$ to find
$$\left(y+\dfrac1y\right)^3-\left(y+\dfrac1y\right)^2-2\left(y+\dfrac1y\right)+1$$
$$=y^3+\dfrac1{y^3}+3\left(y+\dfrac1y\right)-y^2-\dfrac1{y^2}-2-2\left(y+\dfrac1y\right)+1$$
$$=\dfrac{y^7-1}{y^3(y-1)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Closed form of $\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx$
Can a closed form solution for the following integral be found:
$$\int_0^\infty \arctan^2 \left (\frac{2x}{1 + x^2} \right ) \, dx\,?$$
I have tried all the standard tricks such as integration by parts, various substitutions, and parametric differentiation (Feynman's trick), but all to no avail.
An attempt is letting
$$f(t):=\int_0^\infty\,\arctan^2\left(\frac{2tx}{1+x^2}\right)\,\text{d}x\,.$$
Therefore,
$$f'(t)=\int_0^\infty\,\frac{8x^2(x^2+1)}{\big(x^4+2(2t^2+1)x^2+1\big)^2}\,\left(1+x^2-4tx\arctan\left(\frac{2tx}{1+x^2}\right)^{\vphantom{a^2}}\right)\,\text{d}x\,.$$
This doesn't seem to go anywhere. Help!
| $$\begin{align*}
I &= \int_0^\infty \arctan^2\left(\frac{2x}{x^2+1}\right) \, dx \\[1ex]
&= \int_0^1 \left(1+\frac1{x^2}\right) \arctan^2\left(\frac{2x}{x^2+1}\right) \, dx \tag1 \\[1ex]
&= 2 \int_0^1 \frac{\arctan^2(x)}{x^2\sqrt{1-x^2}} \, dx \tag{2} \\[1ex]
&= 4 \int_0^1 \frac{\arctan(x) \sqrt{1-x^2}}{x(1+x^2)} \, dx \tag{3} \\[1ex]
&= 4 \left(\underbrace{\int_0^1 \frac{\sqrt{1-x^2}}x \arctan(x) \, dx}_{J_1} - \underbrace{\int_0^1 \frac{x\sqrt{1-x^2}}{x^2+1} \arctan(x) \, dx}_{J_2}\right) \tag{4}
\end{align*}$$
$$\begin{align*}
J_1 &= \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \int_0^1 x^{2n} \sqrt{1-x^2} \, dx & \tag{5} \\[1ex]
&= \frac12 \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \int_0^1 x^{n-\frac12} (1-x)^{\frac12} \, dx \tag{6} \\[1ex]
&= \frac12 \Gamma\left(\frac32\right) \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)} \frac{\Gamma\left(n+\frac12\right)}{\Gamma(n+2)} \tag{7} \\[1ex]
&= \frac{\sqrt\pi}4 \sum_{n=0}^\infty \frac{(-1)^n}{(n+1)(2n+1)} \frac{\Gamma\left(n+\frac12\right)}{\Gamma(n+1)} \tag{8} \\[1ex]
&= \frac{\sqrt\pi}2 \sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \frac{(2n)!}{2^{2n}(n!)^2} - \frac{\sqrt\pi}4 \sum_{n=0}^\infty \frac{(-1)^n}{n+1} \frac{(2n)!}{2^{2n}(n!)^2} \tag{4/9} \\[1ex]
&= \frac\pi2 (\operatorname{arsinh}(1) + 1 - \sqrt2) \tag{10}
\end{align*}$$
$$\begin{align*}
J_2 &= \int_0^1 \int_0^1 \frac{x^2 \sqrt{1-x^2}}{(x^2+1)(1+y^2x^2)} \, dy \, dx \tag{11} \\[1ex]
&= \frac\pi2 \int_0^1 \frac{(\sqrt2-1)y^2+1-\sqrt{y^2+1}}{y^2(y^2-1)} \, dy \tag{12} \\[1ex]
&= \frac\pi{2\sqrt2} \left(\sqrt2 - 2 + \log(2)\right)
\end{align*}$$
Putting these together, we arrive at
$$I = \boxed{\pi\bigg(2\operatorname{arsinh}(1)-\sqrt2 \log(2)\bigg)}$$
*
*$(1)$ : substitute $x\mapsto\dfrac1x$ in the integral over $[1,\infty)$
*$(2)$ : substitute $x\mapsto\dfrac{1+\sqrt{1-x^2}}x$
*$(3)$ : integrate by parts
*$(4)$ : partial fractions
*$(5)$ : series expansion of $\arctan(x)$
*$(6)$ : substitute $x\mapsto\sqrt x$
*$(7)$ : beta function
*$(8)$ : $\Gamma(n+1)=n\,\Gamma(n)$
*$(9)$ : convert gammas to factorials in preparation for ...
*$(10)$ : series expansion of $\operatorname{arsinh}(x)$
*$(11)$ : integral definition of $\arctan(x)$
*$(12)$ : the last integral is elementary
| {
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"url": "https://math.stackexchange.com/questions/3704882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\frac{1}{\sqrt[3]2}=\sqrt{\frac 5{\sqrt[3]4}-1}-\sqrt{(3-\sqrt[3]2)(\sqrt[3]2-1)}$ Playing around with denesting radicals, I arrived at the following formula which appears to be correct.
$$\frac 1{\sqrt[3]2}=\sqrt{\frac 5{\sqrt[3]4}-1}-\sqrt{(3-\sqrt[3]2)(\sqrt[3]2-1)}$$
If one were to prove this strictly from the given equation, say, as a contest math problem, how would one do it? I have literally no idea how to do this, and I only derive these nested radical equations backwards (e.g. substituting radical values for $a$, $b$ and $c$ in an expression like $(a+b-c)^2$ and hoping for an elegant result after some more or less tedious algebra).
Is there an official method by which to prove this, or is it a bit foggy? I have heard Galois theory is probably important here but that's all I know about it, pretty much, and the rest is vaguely known to me. I would love to see if there is some kind of process to solve/prove such problems, as it might shed light on how Ramanujan came across his several radical denestations and related general identities.
How it was discovered.
I noticed that $$1-\frac 1{\sqrt[3]2}+\frac 1{\sqrt[3]4}=\frac 12\Big\{1+\sqrt{(3-\sqrt[3]2)(\sqrt[3]2-1)}\Big\}$$ and $$1-\frac 1{2\sqrt[3]2}+\frac 1{\sqrt[3]4}=\frac 12\Bigg(1+\sqrt{\frac 5{\sqrt[3]4}-1}\Bigg)$$ and I put two and two together.
Of course, nobody just notices these things (except maybe Ramanujan). I was simply doing what I described earlier about deriving these backwards and merely experimenting and playing around with numbers for the fun of it. But I really want to know why these outputs do come out so nicely, and the essence of it all.
Any thoughts?
Thank you in advance.
| Well, let's do this step by step:
*
*Write:
$$\sqrt[3]{4}=\sqrt[3]{2^2}=2^\frac{2}{3}\tag1$$
*Write:
$$\frac{5}{2^\frac{2}{3}}=\frac{5}{2^\frac{2}{3}}\cdot\frac{\sqrt[3]{2}}{\sqrt[3]{2}}=\frac{5\sqrt[3]{2}}{2}\tag2$$
*Write:
$$\left(3-\sqrt[3]{2}\right)\left(\sqrt[3]{2}-1\right)=-3+3\sqrt[3]{2}-\left(-\sqrt[3]{2}\right)-\sqrt[3]{2}\sqrt[3]{2}=-3+3\sqrt[3]{2}+\sqrt[3]{2}-2^\frac{2}{3}=$$
$$-3+4\sqrt{3}{2}-2^\frac{2}{3}=1+2\sqrt[3]{2}-2^\frac{2}{3}-4+2\sqrt[3]{2}=$$
$$1+2\sqrt[3]{2}-\left(\sqrt[3]{2}\right)^2-2\left(\sqrt[3]{2}\right)^3+\left(\sqrt[3]{2}\right)^4=\left(1+\sqrt[3]{2}-2^\frac{2}{3}\right)^2\tag3$$
*Write:
$$\frac{5\sqrt[3]{2}}{2}-1=\frac{5\sqrt[3]{2}}{2}-\frac{2}{3}=\frac{5\sqrt[3]{2}-2}{2}\tag4$$
*Write:
$$5\sqrt[3]{2}-2=2+4\sqrt[3]{2}-4+\sqrt[3]{2}=\frac{4+8\sqrt[3]{2}-8+2\sqrt[3]{2}}{2}=$$
$$\frac{4+8\sqrt[3]{2}-4\left(\sqrt[3]{2}\right)^3+\left(\sqrt[3]{2}\right)^4}{2}=\frac{\left(2+2\sqrt[3]{2}-2^\frac{2}{3}\right)^2}{2}\tag5$$
I think you can finish.
| {
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"timestamp": "2023-03-29T00:00:00",
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find the value of $\sec^4(\pi/9) + \sec^4( 2\pi/9) + \sec^4 (4\pi/9)$ find the value of $\sec^4(\pi/9) + \sec^4( 2\pi/9) + \sec^4 (4\pi/9)$
I tried converting $\sec^2(\theta)$ to $\tan^2(\theta)$ but for no vain.
| Like Evaluate $\tan^{2}(20^{\circ}) + \tan^{2}(40^{\circ}) + \tan^{2}(80^{\circ})$
$$\cos3t=4\cos^3t-3\cos t$$
Observe that $\sec\left(3\cdot\dfrac{\pi}9\right)=2,\sec\left(3\cdot\dfrac{2\pi}9\right)=-2, \sec\left(3\cdot\dfrac{4\pi}9\right)=-2$
Put $3t=\dfrac\pi3$
So, the roots of $4c^3-3c-\dfrac12=0\iff8c^3-6c-1=0$ are $\cos\dfrac{(6n+1)\pi}9$ where $n=0,1,2$
As $\cos(\pi\pm y)=-\cos y$
$\cos\dfrac{(6\cdot1+1)\pi}9=\cdots=-\cos\dfrac{2\pi}9$ and $\cos\dfrac{(6\cdot2+1)\pi}9=\cdots=-\cos\dfrac{4\pi}9$
Set $\dfrac1c=s$ to find the roots of $s^3+6s^2-8=0$ are $$ \sec\dfrac\pi9, -\sec\dfrac{2\pi}9, -\sec\dfrac{4\pi}9$$
Square both sides of $s^3=8-6s^2$ and replace $s^2=t$ to find the roots of $$t^3-36t^2+96t-64=0$$ are $$a=\sec^2\dfrac\pi9,b=\sec^2\dfrac{2\pi}9,c=\sec^2\dfrac{4\pi}9$$
We need $$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ca)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How can I solve $f(x) = \int \frac{\cos{x}(1+4\cos{2x})}{\sin{x}(1+4\cos^2{x})}dx$? $$f(x) = \int \frac{\cos{x}(1+4\cos{2x})}{\sin{x}(1+4\cos^2{x})}dx$$
I have been up on this problem for an hour, but without any clues.
Can someone please help me solving this?
| Let $x=\tan^{-1}(y)$
$$I= \int \frac{\cos{(x)}(1+4\cos{(2x)})}{\sin{(x)}(1+4\cos^2{(x)})}\,dx=\int \frac{5-3 y^2}{y^5+6 y^3+5 y}\,dy$$ Now, partial fraction decomposition
$$\frac{5-3 y^2}{y^5+6 y^3+5 y}=\frac{1}{y}-\frac{2 y}{y^2+1}+\frac{y}{y^2+5}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3710804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Minimize area of ellipse that passes through given point The ellipse formula is:
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
$a$ and $b$ are assumed to be positive.
The point $(6, -7)$ is given as a point on the ellipse, with the area of the ellipse equal to $\pi ab$.
How would one find the values of $a$ and $b$ that minimize the area of the ellipse?
| Note that $$\frac{6^2}{a^2}+\frac{7^2}{b^2}=1$$
Using $AM\ge GM$,
\begin{align}
\frac{1}{2}
\left(
\frac{6^2}{a^2}+\frac{7^2}{b^2}
\right) & \ge \sqrt{\frac{6^2}{a^2} \times \frac{7^2}{b^2}} \\
\frac{1}{2} & \ge \frac{42}{ab} \\
\pi ab & \ge 84\pi
\end{align}
Equality holds if and only if $\dfrac{6^2}{a^2}=\dfrac{7^2}{b^2}$, therefore $\dfrac{a}{6}=\dfrac{b}{7}=\sqrt{2}$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$\frac{1}{4} (a^2+ 3 b^2)$ is of the form $(c^2+ 3 d^2)$ If $ 2 \mid (a^2+ 3 b^2)$ and $(a,b)=1$ then $4\mid (a^2+ 3 b^2)$. How can I show $\frac{1}{4} (a^2+ 3 b^2)$ is also of the form $(c^2+ 3 d^2)$?
Here, clearly $ a$ and $b$ are both odd.
Let $a=2m+1$ and $ b=2n+1$
$\implies\frac{1}{4} (a^2+ 3 b^2)= m^2 + m+1 +3n^2 +3n$.
I am stuck here. Can anyone please help how to approach from here. Any help would be appreciated. Thanks in advance.
| \begin{eqnarray*}
\frac{(c+3d)^2+3(c-d)^2}{4} =c^2+3d^2.
\end{eqnarray*}
Edit:
If $ a \equiv b \pmod{4}$ then
\begin{eqnarray*}
\frac{a^2+3b^2}{4} = \left( \frac{a+3b}{4} \right)^2 +3 \left( \frac{a-b}{4} \right)^2.
\end{eqnarray*}
& If $ a \equiv -b \pmod{4}$ then
\begin{eqnarray*}
\frac{a^2+3b^2}{4} = \left( \frac{a-3b}{4} \right)^2 +3 \left( \frac{a+b}{4} \right)^2.
\end{eqnarray*}
and the values in the brackets on the RHS will be whole numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
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Can any cyclic polynomial in $a, b, c$ be expressed in terms of $a^2b+b^2c+c^2a$, $a+b+c$, $ab+bc+ca$ and $abc$? Problem. Let $f(a,b,c)$ be a cyclic polynomial in $a, b, c$.
Can $f$ always be expressed as $g(a+b+c, ab+bc+ca, abc, a^2b+b^2c + c^2a)$ for some polynomial $g(p, q, r, Q)$?
Motivation: If we want to prove $g(a+b+c, ab+bc+ca, abc, a^2b+b^2c + c^2a)\ge 0$, and $g(p, q, r, Q)$ is non-increasing with $Q$,
by using the known inequality $a^2b + b^2c + c^2a \le \frac{4}{27}(a+b+c)^3 - abc$(see How to prove this inequality? $a^{2}+b^{2}+c^{2}\leq 3$),
sometimes, we may prove $g(a+b+c, ab+bc+ca, abc, \frac{4}{27}(a+b+c)^3 - abc)\ge 0$.
As a result, we deal with a symmetric inequality rather than the original cyclic inequality.
Is it known? Is it easily to prove?
Let us see some examples. Denote $p = a+b+c, q = ab+bc+ca, r = abc$ and $Q = a^2b+b^2c+c^2a$.
1) $ab^2 + bc^2 + ca^2 = (a+b+c)(a^2+b^2+c^2) - a^3-b^3-c^3 - (a^2b+b^2c+c^2a)$
2) From the known identity $a^3b+b^3c+c^3a +(ab+bc+ca)^2 = (a+b+c)(a^2b+b^2c+c^2a+abc)$, we have
$a^3b+b^3c+c^3a = p(Q + r) - q^2$.
3) $a^3b^2+b^3c^2+c^3a^2 = Qq - r(p^2-2q) - rq$.
4) $(a^2+b)(b^2+c)(c^2+a) = \cdots$
Any comments and solutions are welcome.
| Four years ago, I also encountered this problem. I can't prove but I can write $$f(a+b+c,\,ab+bc+ca,abc,a^2b+b^2c+c^2a) \to F(p,q,r,Q)$$
by my pqr tool run on Maple.
Similar to $a^3b+b^3c+c^3a,\,(a-b)(b-c)(c-a), \ldots$ The appropriate time (after my book is published) I will publish all the source code. Please forgive me.
| {
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If $a, b, c, d\in\mathbb R^+, $ then prove that $\displaystyle\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}\ge 0.$ My approach: We have:
$\displaystyle\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{d+a}+\frac{d-a}{a+b}$
$\displaystyle=\frac{a+c}{b+c}-1+\frac{b+d}{c+d}-1+\frac{c+a}{d+a}-1+\frac{d+b}{a+b}-1$
$\displaystyle=(c+a)\left(\frac1{b+c}+\frac1{d+a}\right)+(b+d)\left(\frac1{c+d}+\frac1{a+b}\right)-4\ge (c+a) \left(\frac{2\cdot 2}{b+c+d+a}\right)+(b+d)\left(\frac{2\cdot 2}{c+d+a+b}\right)-4$
(on applying AM $\ge$ HM.)
$\displaystyle=\left(\frac4{a+b+c+d}\right) (c+a+b+d)-4=0.$
Hence the inequality.
But I would love to know is there any other elegant method to prove the same? Please mention. Thanks in advance.
| We can use C-S one time only:
$$\sum_{cyc}\frac{a-b}{b+c}=-4+\sum_{cyc}\left(\frac{a-b}{b+c}+1\right)=-4+\sum_{cyc}\frac{a+c}{b+c}=$$
$$-4+\sum_{cyc}\frac{(a+c)^2}{(b+c)(a+c)}\geq-4+\frac{\left(\sum\limits_{cyc}(a+c)\right)^2}{\sum\limits_{cyc}(b+c)(a+c)}=-4+\frac{4(a+b+c+d)^2}{\sum\limits_{cyc}(ab+ab+ac+a^2)}=0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $f(x)=2|2x-5|+3$, with $x\geq0$, find all $k$ such that $f(x) = kx + 2$ has exactly two roots
If $f(x) = 2|2x-5| + 3 $, $x\geq0$, find the values of $k$ such that the equation $f(x) = kx + 2$ has exactly two roots.
My attempt :-
$2(2x -5)+3 = kx + 2$ ,
$x \geq 2.5$
$x =\frac{9}{4-k}$
Then
$\frac{9}{4-k} \geq \frac{2}{5}$ , $k<4$
Then
$\frac{2}{5} < k < 4$
$2(-2x +5) + 3 = kx +2$ , $ x < \frac{2}{5}$ , $x\geq 0$
$x = \frac{11}{k+4}$
Then
$ \frac{11}{k+4} < \frac{5}{2}$
Then
$k > \frac{2}{5}$
Then
$\frac{2}{5} < k < 4$
Is there any easier procedure to find the values of $k$ ?
| The equation is equivalent to:
$$
2\vert 2x−5\vert +3=kx+2 \land x\ge0\\
2\vert 2x−5\vert=kx-1 \land x\ge0\\
(2\vert 2x−5\vert)^{2}=(kx-1)^{2} \land kx-1\ge0 \land x\ge0\\
(2 (2x−5))^{2}-(kx-1)^{2}=0 \land kx-1\ge0 \land x\ge0\\
(2(2x−5)+kx-1)(2(2x−5)-(kx-1))=0 \land kx-1\ge0 \land x\ge0\\
((4+k)x−11)((4-k)x−9)=0 \land kx-1\ge0 \land x\ge0\\
(x=\frac{11}{4+k} \lor x=\frac{9}{4-k})\land kx-1\ge0 \land x\ge0\\
$$
So the equation have two solutions iff
$$
\frac{11}{4+k}\neq\frac{9}{4-k}\land kx-1\ge0 \land x\ge0\\
9(4+k)\neq\ 11(4-k)\land kx-1\ge0 \land x\ge0\\
k\neq\ \frac{2}{5}\land kx-1\ge0 \land x\ge0$$
now $kx-1\ge0$, but $(x=\frac{9}{4-k} \lor x=\frac{11}{4+k})$ so
$$
\frac{9}{4-k}k-1\ge0 \land \frac{11}{4+k}k-1\ge0 \\
\frac{2}{5}\le \:k<4 \land (\:k<-4\lor k\ge \frac{2}{5})
$$
we also have that $x\ge0$, but $(x=\frac{9}{4-k} \lor x=\frac{11}{4+k})$ so
$$
\frac{9}{4+k}\ge0 \land \frac{11}{4-k}\ge0\\
k\gt-4 \land 4\gt k
$$
then
$$
\frac{2}{5}\le \:k<4 \land (\:k<-4\lor k\ge \frac{2}{5})\land 4\gt k\gt-4 \land k\neq\ \frac{2}{5}\\
\frac{2}{5}\lt k<4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Deriving Law of Cosines from Law of Sines
How to eliminate $\alpha$ from the Law of Sines of plane trigonometry
$$ \dfrac{a}{\sin \alpha}= \dfrac{c}{\sin \gamma} =\dfrac{b}{\sin (\gamma+\alpha)} =2R $$
in order to arrive at the Law of Cosines
$$ c^2= a^2+b^2-2 a b \cos \gamma \;?$$
Starting to isolate $\alpha$
$$ \alpha =\sin^{-1}\big(\dfrac{b-c}{2R} +\sin \gamma \big)-\gamma$$
$$ =\sin^{-1}( b \sin \gamma/c) -\gamma$$
involve $c$ that we are finding and so on, how best to simplify?
| $$\frac {b}{\sin(\gamma + \alpha)}= 2R.$$
$$\frac {b}{\sin(\gamma) \;\cos(\alpha)+ \sin\alpha \; \cos\gamma} = 2R.$$
From Law of Sines: $\sin\alpha = \frac {a}{2R} \;\; and\;\; \sin(\gamma)= \frac {c}{2R}.$
Then
$$\frac {b}{\frac{c}{2R} \;\cos\alpha+ \frac{a}{2R} \; \cos\gamma} = 2R,$$
$$b= c\;\cos\alpha + a\;\cos\gamma,$$
$$b= c\;\sqrt{1-\sin^2\alpha} + a\;\cos\gamma.$$
From Law of Sines: $\sin\alpha = \frac{a\;\sin\gamma}{c}.$
Then
$$b= c\;\sqrt{1-\frac{a^2\;\sin^2\gamma}{c^2}} + a\;\cos\gamma \;=\; \sqrt{c^2-a^2\;\sin^2\gamma} + a\;\cos\gamma$$
$$(b-a\;\cos\gamma)^2 = c^2-a^2\;\sin^2\gamma,$$
$$b^2+a^2\cos^2\gamma - 2ab\;\cos\gamma = c^2-a^2\;\sin^2\gamma,$$
$$b^2+a^2\cos^2\gamma - 2ab\;\cos\gamma + a^2\;\sin^2\gamma = c^2,$$
$$b^2+a^2(\cos^2\gamma + \sin^2\gamma) - 2ab\;\cos\gamma = c^2.$$
Since $\cos^2\gamma + \sin^2\gamma =1,$
$$b^2+a^2 - 2ab\;\cos\gamma = c^2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3718421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Simplification of $\sqrt{2\zeta^2-1+2\zeta\sqrt{\zeta^2-1}}+\sqrt{2\zeta^2-1-2\zeta\sqrt{\zeta^2-1}}$ If I try to evaluate $\sqrt{2\zeta^2-1+2\zeta\sqrt{\zeta^2-1}}+\sqrt{2\zeta^2-1-2\zeta\sqrt{\zeta^2-1}}$ numerically for real $\zeta$, it looks like it is just equal to $2|\zeta|$ for $\zeta \ne 0$ and $2j$ for $\zeta=0$, but I can't figure out how to simplify to get there...
It's of the form $\sqrt{b+c} + \sqrt{b-c}$ with $b=2\zeta^2-1$ and $c=2\zeta\sqrt{\zeta^2-1}$. I can write:
$$\sqrt{b+c} + \sqrt{b-c} = \frac{(b+c) - (b-c)}{\sqrt{b+c} - \sqrt{b-c}}$$
but that doesn't seem to help either....
| Oh, I figured it out:
$$\begin{align}
(\sqrt{b+c}+\sqrt{b-c})^2 &= (b+c)+2\sqrt{b^2-c^2}+(b-c) \\
&= 2b+2\sqrt{b^2-c^2}
\end{align}$$
and in this case $b^2 - c^2 = 4\zeta^2-4\zeta+1 - 4\zeta^4 +4\zeta^2 = 1$
so
$$\begin{align}
(\sqrt{b+c}+\sqrt{b-c})^2 &= (b+c)+2\sqrt{b^2-c^2}+(b-c) \\
&= 2b+2 \\
&= 4\zeta^2
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3724205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the value of $a_{0}^{2}-a_{1}^{2}+a_{2}^{2}- \dots+a_{2 n}^{2}$ Let n be a positive integer and $$\left(1+x+x^{2}\right)^{n}=a_{0}+a_{1} x+\cdots+a_{2 n} x^{2 n}$$
then the value of $a_{0}^{2}-a_{1}^{2}+a_{2}^{2} - \dots+a_{2 n}^{2}$ is
My approach:-
Replacing $x$ by $(-1 / x),$ we get
$$
\begin{array}{r}
\left(1-\frac{1}{x}+\frac{1}{x^{2}}\right)^{n}=a_{0}-\frac{a_{1}}{x}+\frac{a_{2}}{x^{2}}+\cdots-a_{2 n-1} \cdot \frac{1}{x^{2 n-1}}+\frac{a_{2 n}}{x^{2 n}} \\
\text { or, }\left(1-x+x^{2}\right)^{n}=a_{0} x^{2 n}-a_{1} x^{2 n-1}+a_{2} x^{2 n-2}+\cdots+a_{2 n}..... \tag{1}
\end{array}
$$
And given $\left(1+x+x^{2}\right)^{n}=a_{0}+a_{1} x+\cdots+a_{2 n} x^{2 n} \ldots \ldots \ldots \ldots \ldots . \tag{2}.$
Multiplying corresponding sides of (1) and $(2),$ we have
$$
\left(1+x^{2}+x^{4}\right)^{n}=\left(a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{2 n} x^{2 n}\right) \times\left(a_{0} x^{2 n}-a_{1} x^{2 n-1}+a_{2} x^{2 n-2}+\cdots+\right.
$$
$\left.a_{2 n}\right) \ldots \ldots...\tag{3}$
$$
\left(1+x^{2}+x^{4}\right)^{n}=\left(a_{0}+a_{1} x^{2}+a_{2} x^{4}+\cdots+a_{n} x^{2n}+\cdots+a_{2 n} x^{4 n}\right) \ldots \ldots\tag{4}
$$
Equating coefficient of $x^{2 n}$ on both sides of (3) and (4)
$$
a_{0}^{2}-a_{1}^{2}+a_{2}^{2} -\cdots +a_{2 n}^{2}=a_{n}
$$
But this method seems very tedious to me.
Any other approach would be greatly appreciated
| The following is essentially the idea in your proof which is conceptually simple. One has by the given, $$(1+x+x^2)^n=a_0+a_1x+\cdots+a_{2n}x^{2n}.\quad (1)$$
Replacing $x$ by $1/x$ and multiplying by $x^{2n}$ in (1), one sees that $$a_k=a_{2n-k}, 0\leq k\leq 2n.\quad (1)$$
Replacing $x$ by $-x$ in (1), one has $$(1-x+x^2)^n=a_0-a_1x+\cdots+a_{2n}x^{2n}.\quad (2)$$
Replacing $x$ by $x^2$ in (1), one has $$(1+x^2+x^4)^n=a_0+a_1x^2+\cdots a_n x^{2n}+\cdots+a_{2n}x^{4n}.\quad (3)$$
Since $1+x^2+x^4=(1+x+x^2)(1-x+x^2)$, multiplying (2) and (1) and comparing coefficients of $x^{2n}$ with (3), one has $$a_0a_{2n}-a_1a_{2n-1}+a_2a_{2n-2}+\cdots+a_{2n}a_0=a_n,$$ which after applying (1) yields $$a_0^2-a_1^2+a_2^2-\cdots+a_{2n}^2=a_n,$$ as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3726741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
The set of all real number x such that $||3-x|-| x+2||=5$ The set of all real number x such that
$$||3-x|-| x+2||=5$$
is??
My approach:-
$(|| 3-x|-| x+||2)^{2}=25$
$\Leftrightarrow(3-x)^{2}+(x+2)^{2}-2|3-x||x+2|=25$
$\Rightarrow x^{2}-x-\left|-x^{2}+x+6\right|=6$
What to do next?.....
Edit
after seeing the comments I got the idea,
So, it is clear that $-x^{2}+x+6<0,$ i.e. $-x^{2}+x+6 \geq 0$
$(x-3)(x+2) \geq 0$
So, $x \leq-2 \& x \geq 3$
$\therefore \mathrm{x} \in(-\infty,-2] \cup[3, \infty)$
Correct me ,If I am wrong
| One can also approach the absolute-value expression "from the inside out". The definition of an absolute-value function leads us to write
$$ |3 \ - \ x| \ \ = \ \ \left\{ \begin{array}{cc} 3 \ - \ x \ , & \text{for} \ \ x \ < \ 3 \\ x \ - \ 3 \ , & \text{for} \ \ x \ \ge \ 3 \end{array}\right. $$
and
$$ |x \ + \ 2| \ \ = \ \ \left\{ \begin{array}{cc} -(x \ + \ 2) \ , & \text{for} \ \ x \ < \ -2 \\ x \ + \ 2 \ , & \text{for} \ \ x \ \ge \ -2 \end{array}\right. \ \ . $$
The "real-number line" is thus divided into three intervals (the "cases" for the inequality). The difference of terms can then be expressed as
$$ |3 \ - \ x| \ - \ |x \ + \ 2| \ \ = \ \ \left\{ \begin{array}{cc} \ (3 \ - \ x) \ - \ [ \ -(x \ + \ 2) \ ] \ = \ 5 \ \ , & \text{for} \ \ x \ < \ -2 \\ \ (3 \ - \ x) \ - \ (x \ + \ 2) \ = \ 1 \ - \ 2x \ \ , & \text{for} \ \ -2 \ \le \ x \ < 3 \\ \ (x \ - \ 3) \ - \ (x \ + \ 2) \ = \ -5 \ \ , & \text{for} \ \ x \ \ge \ 3 \end{array}\right. \ \ \ . $$
We see immediately from this that $ \ x < -2 \ $ and $ \ x \ge 3 \ $ are solution intervals for $ \ | \ |3 \ - \ x| \ - \ |x \ + \ 2| \ | \ = \ 5 \ \ . $ For the second interval, we can check that $ \ 1 - 2·[-2] \ = \ 5 \ \ , $ so $ \ x \ = \ -2 \ $ is a solution as well, and also that
$$ -2 \ < \ x \ < \ 3 \ \ \Rightarrow \ \ 4 \ > \ -2x \ > \ -6 \ \ \Rightarrow \ \ 5 \ > \ 1 \ - \ 2x \ > \ -5 \ \ , $$
so there are no other solutions. Thus, the complete solution is $ \ x \le -2 \ $ and $ \ x \ge 3 \ \ . $
We could also consider this graphically. The function $ \ |3 \ - \ x| \ = \ | \ -(x \ - \ 3) \ | \ $ is equivalent to $ \ |x \ - \ 3| \ \ , $ so we can plot $ \ y \ = \ |3 \ - \ x| \ $ and $ \ y \ = \ |x \ + \ 2 | \ $ as versions of the absolute-value function $ \ y \ = \ |x| \ $ "shifted" horizontal "to the right" by 3 units [the blue curve in the graph below] and "to the left" by 2 units [the green curve], respectively. We see that for $ \ x \ \le \ -2 \ \ , $ the blue curve is "above" the green one by 5 units [ $ \ |3 \ - \ x| \ - \ |x \ + \ 2| \ \ = \ +5 \ $ ] and that the blue curve is "below" the green one by the same amount $ [ \ |3 \ - \ x| \ - \ |x \ + \ 2| \ \ = \ -5 \ $ ] for $ \ x \ \ge \ 3 \ \ . $ (The red curve is the graph of the difference of absolute-value terms.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all positive integers $x$, that satisfy $29x^{33} \equiv 27\pmod {11} $
Find all positive integers $x$, that satisfy $29x^{33} \equiv 27 \pmod {11}$.
I approached this the following way:
Since from $29x^{33} \equiv 27 \pmod {11}$ we get that $7x^{33} \equiv 5 \pmod {11}$ and since $\gcd(7,5)=1$ we would get that $\phi(11)=10$ which would imply that $7x^{10} \equiv 5 \pmod {11}$.
How should I continue from here, it doesn't seem to be quite clear.
| $$29\equiv7,27\equiv5\pmod{11}$$
and $$\phi(11)=10$$
$$\implies7x^3\equiv5\pmod{11}$$
Now as $7\cdot8\equiv1\pmod{11},$
$$ x^3\equiv5\cdot8\equiv(-4)\pmod{11}$$
Finally as $1=10-3\cdot3,$
$$x=x^{10}(x^3)^{-3}\equiv1\cdot(-4)^{-3}\equiv-2^{-6}\equiv-2^4\equiv6$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3727589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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How do I show that $\lfloor{(\sqrt 2+1)^{2020}}\rfloor\equiv 1\left(\text{mod}~ 4\right)$? I think I can use that $$(a\sqrt b+c)^{2n}+(a\sqrt b-c)^{2n}\in\mathbb{Z},$$ but I have no idea about the next step.
| Expanding
Robert Israel's suggestion
(always a good thing to do):
If
$a_n
=u^n+v^n
$
then
$\begin{array}\\
a_n(u+v)
&=u^{n+1}+v^{n+1}+uv^n+vu^n\\
&=u^{n+1}+v^{n+1}+uv(v^{n-1}+u^{n-1})\\
&=a_{n+1}+uva_{n-1}\\
\end{array}
$
so
$a_{n+1}
=a_1a_n-uva_{n-1}
$.
If
$u = 1+\sqrt{2}$
and
$v = 1-\sqrt{2}
$
(trying
$u = \sqrt{2}+1,
v = \sqrt{2}-1$
doesn't work)
then
$uv = -1$
and
$u+v = 2$
so
$\begin{array}\\
a_{n+1}
&=2a_{n}+a_{n-1}\\
&=2(2a_{n-1}+a_{n-2})+a_{n-1}\\
&=5a_{n-1}+2a_{n-2}\\
&=5(2a_{n-2}+a_{n-3})+2a_{n-2}\\
&=12a_{n-2}+5a_{n-3}\\
\end{array}
$
so,
mod 4,
$a_{n+1}
=a_{n-3}
=a_{n+1-4k}
$
for $k \ge 0$.
Since
$a_0 = 2,
a_1 = 2,
a_2 = 6,
a_3 = 14
$,
we have
$\begin{array}\\
a_{2020}
&=a_{2020-4\cdot 505} \bmod 4\\
&=a_0 \bmod 4\\
&=2\\
\text{and}\\
u^{2020}
&=a_{2020}-v^{2020}\\
\text{so}\\
\lfloor u^{2020}\rfloor
&=\lfloor a_{2020}-v^{2020}\rfloor\\
&=a_{2020}-1
\qquad\text{since } 0 < v^{2020} < 1\\
&= 2-1 \bmod 4\\
&= 1 \bmod 4\\
\end{array}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3728455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
For what value of $m$ the is sum $\sum_{i = 0}^{m} {10 \choose i}{20 \choose m - i}$ where ${p\choose q} = 0$, if $pFor what value of $m$ the is sum
$$\sum_{i = 0}^{m} {10 \choose i}{20 \choose m - i}\text{where ${p\choose q}$} = 0\text{, if $p<q$, a maximum}$$
My approach
$$\sum_i^{m} {10 \choose i}{20 \choose m - i} = {10 \choose 0}{20 \choose m} + {10 \choose 1}{20 \choose m - 1} + \dots + {10 \choose m}{20 \choose 0}$$
$$(1 +x)^{20} = {20 \choose 0} + {20 \choose 1}x + \dots + {20 \choose m-1}x^{m-1} + {20 \choose m}x^{m} + \dots + {20 \choose 20}x^{20}$$
$$(1 +x)^{10} = {10 \choose 0} + {10 \choose 1}x + \dots + {10 \choose 10}x^{10}$$
Later what to do??
Any other method or hint will be greatly welcomed.
| $$(1+x)^{10}= 10_{C_{0}}+10_{C_{1}} x+\cdots+10_{C_{10}} x^{10} \ldots \ldots \ldots \ldots \ldots .(1)$$
$$(1+x)^{20}= 20_{C_{20}}x^{20}+20_{C_{19}} x^{19}+\cdots+20_{C_{0}} \ldots \ldots \ldots \ldots \ldots .(2)$$
On multiplying these two...
$$(1+x)^ {30}= (\cdots)x^{30}+(\cdots)x^{29}+\cdots$$
The largest coefficient would be that of $x^{15}$ and that would be equal to ...
$$\sum_{i=0}^{m}\left(\begin{array}{l}10 \\ i\end{array}\right)\left(\begin{array}{l}20 \\ 15-i\end{array}\right),$$
Hope this helps!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3734305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Showing $x^4 + 2$ is irreducible in $F_5[x]$ There is a similar question like this here
but I don't understand the solution. Since this is a fourth degree and since it has no root in $F_5$, it can only have a quadratic reduction. But how do we rule out the possibility that $x^4 + 2 = (x^2 + ax + b)(x^2 + cx + d)?$ Where the quadratic is irreducible in $F_5$?
There are a lot of similar questions like this on my practice exams.
Thanks
| Assume $x^4+2$ factors in $\mathbb{Z}_5[x]$ as
$$x^4+2=(x^2+ax+b)(x^2+cx+d)$$
If the $\text{RHS}$ is expanded,
*
*The coefficient of the $x^3$ term is $a+c$, hence we must have $c=-a$.$\\[4pt]$
*The constant term is $bd$, hence we must have $d={\large{\frac{2}{b}}}$.
hence the factorization can be rewritten as
$$x^4+2=(x^2+ax+b)(x^2-ax+\frac{2}{b})$$
Expanding, we get
$$x^4+2=x^4+\left(b+\frac{2}{b}-a^2\right)x^2+\left(\frac{2a}{b}-ab\right)x+2$$
hence we must have
$$
\left\lbrace
\begin{align*}
&\frac{2a}{b}-ab=0&&(\text{eq}1)\\[4pt]
&b+\frac{2}{b}-a^2=0&&(\text{eq}2)\\[4pt]
\end{align*}
\right.
$$
If $a\ne 0$, $(\text{eq}1)$ yields $b^2=2$, and if $a=0$, $(\text{eq}2)$ yields $b^2=-2$.
Either way we have a contradiction since $2$ and $-2$ are not squares in $\mathbb{Z}_5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3738494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Three boxes with balls Box $U_1$ contains $1$ white ball and $2$ black balls. Box $U_2$ contains $2$ white balls and $2$ black balls. We extract without reinsertion two balls from every boxes. The four balls are put in a third box $U_3$ initially empty. We randomly extract a ball from $U_3$. Find the probability that the ball is white.
Well, I reasoned in this way. The possible combinations that ensure that $U_3$ contains at least one white ball are BNBB, NBBB, NNBB, BNBN, BNNB, BNNN, NBBN, NBNB, NBNN, NNBN, NNNB. Thus:
*
*$\mathbb{P}$($U_3$ contains $3$ white balls)$=\mathbb{P}($(BNBB)$\cap$(NBBB)$)=(\frac{1}{3}\cdot1 \cdot\frac{1}{2}\cdot\frac{1}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2}\cdot \frac{1}{3})=0,11$
*$\mathbb{P}(U_3$ contains $2$ white balls)$=\mathbb{P}($(NNBB)$\cap$(BNBN)$\cap$(BNNB)$\cap$(NBBN)$\cap$(NBNB)$)=(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})+(\frac{1}{3}\cdot 1\cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{1}{3}\cdot 1 \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{3}\cdot \frac{1}{2} \cdot \frac{2}{3})=0,46$
*$\mathbb{P}(U_3$ contains $1$ white ball)$=\mathbb{P}($(BNNN)$\cap$(NBNN)$\cap$(NNBN)$\cap$(NNNB)$)=(\frac{1}{3}\cdot 1 \cdot \frac{1}{2}\cdot \frac{1}{3})+(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})+(\frac{2}{3} \cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{2}{3})+(\frac{2}{3}\cdot \frac{1}{2}\cdot \frac{1}{2}\cdot \frac{2}{3})=0,33$
*$\mathbb{P}(U_3$ doesn't contain any white balls)$=2\mathbb{P}($(NNNN)$)=2(\frac{2}{3}\cdot \frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{3})=0,1$
Thus $\mathbb{P}($one white ball from $U_3)=\frac{3}{4}\cdot 0,11+\frac{2}{4}\cdot 0,46+\frac{1}{4}\cdot 0,33+\frac{0}{4}\cdot 0,11=0,395$
Is it correct? Particularly I'm interested in reasoning. Thanks in advance.
| It does not appear to be correct.
You're drawing two balls without replacement from $U_1$, leaving one behind. The probability that the leftover ball is white (meaning the two balls you drew are black) is $\frac 13$. Thus, the probability that you draw a black ball and a white ball from $U_1$ is $\frac 23$.
You're drawing two balls without replacement from $U_2$, leaving two behind. The probability that both leftovers are black is $\frac 16 = \frac {1}{\binom 42}$, and by symmetry the probability that both leftovers are white also is $\frac 16$. Therefore the probability that you draw a white ball and a black ball from $U_2$ is $\frac 23$.
The draws from $U_1$ and $U_2$ are independent, so we can multiply probabilities.
Probability of no white balls in $U_3$ is $\frac 13 \frac 16 = \frac{1}{18}$.
Probability of one white ball in $U_3$ is $\frac 23 \frac 16+ \frac 13 \frac 23 =\frac 19 + \frac 29 = \frac 13$.
Probability of two white balls in $U_3$ is $\frac 23 \frac 23 + \frac 13 \frac 16= \frac 49 + \frac {1}{18} = \frac 12$.
Probability of three white balls in $U_3$ is $\frac 23 \frac 16=\frac 19$.
Thus, the probability that you draw a white ball is $0 \frac {1}{18}+ \frac 13 \frac 14+ \frac 12 \frac 12 + \frac 19 \frac 34 = \frac {5}{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3739864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $P(x) \bmod (x^2-x-2)$
Let $P(x)$ be a polynomial.
If $P(x) \bmod (x+1)=0$ and $P(x) \bmod (x-2)=6$, then evaluate $P(x) \bmod (x^2-x-2)$.
My attempt:
$$P(x)=k(x)(x-2)+6=k(x)(x+1)-3(k(x)-2)$$
$$\Longrightarrow (k(x))-2) \bmod (x+1)=0$$
$$\Longrightarrow \left(\dfrac{P(x)-6}{x-2}-2\right) \bmod (x+1)=0$$
$$\Longrightarrow (P(x)-2x-2) \bmod ((x+1)(x-2)) =0$$
$$\Longrightarrow P(x) \bmod (x^2-x-2) \equiv (2x+2) \bmod ((x+1)(x-2)) = (2x+2) \bmod (x^2-x-2)=2x+2 $$
Is my solution correct?
| In as simple a case as this we can just proceed this way.
We have $P(x)= ax+b \mod (x^2-x-2)$ for some $a,b$.
That is $P(x)=ax+b+ q(x)(x+1)(x-2)$.
Putting $x=-1$ and $x=2$ gives us $-a+b=0$ and $2a+b=6$.
Hence $a=b=2$.
But in general I'd suggest that the Extended Chinese Remainder Algorithm is the way to go.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to prove $(n + \frac{1}{2})\log(1+\frac{1}{n})$ is increasing in $n$? Is $(n + \frac{1}{2})\log(1+\frac{1}{n})$ increasing in $n$?
I attempted to differentiate $p_n=(n + \frac{1}{2})\log(1+\frac{1}{n})$ with respect to $n$.
$p_n^{'} = \log(1+\frac{1}{n}) + \frac{(n+\frac{1}{2})}{(1+\frac{1}{n})}(-\frac{1}{n^2}) = \log(1+\frac{1}{n}) - \frac{1}{2}(\frac{1}{n}+\frac{1}{n+1})$.
I am stuck at this point and I cannot prove that $\log(1+\frac{1}{n}) - \frac{1}{2}(\frac{1}{n}+\frac{1}{n+1}) > 0$.
I know by the definition $\log (1+\frac{1}{n}) =\int_1^{1+\frac{1}{n}}\frac{1}{x}dx$.
But I can only reach to the point where $ \frac{1}{n+1} < \log (1+\frac{1}{n}) < \frac{1}{n}$.
Thanks for your help in advance.
| Follow your argument. If you can show that $f(x)=(x+\frac{1}{2}) ln (1+\frac{1}{x})$ is decreasing when $x\ge 1$, then your problem is solved. Define $g(x)=f^\prime (x)=ln (1+\frac{1}{x})-\frac{1}{2}(\frac{1}{x}+\frac{1}{x+1})=ln(x+1)-ln(x)-\frac{1}{2}(\frac{1}{x}+\frac{1}{x+1})$. Then $g^\prime (x)=\frac{1}{x+1}-\frac{1}{x}+\frac{1}{2x^2}+\frac{1}{2(x+1)^2}=\frac{1}{2x^2 (x+1)^2}>0$. Since $g(1)=ln(2)-0.75<0$ and $g(+\infty )=0$, $g(x)<0$ when $x\ge 1$, i.e., $f(x)$ is decreasing.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\int_{-1}^{1}8x^3-5x^2+4dx$ I've done the following so far:
$$\left.\int_{-1}^{1}\left(8x^3-5x^2+4\right)dx=\left(\frac{8}{4}x^4-\frac{5}{3}x^3+4x\vphantom{\int}\right)\right|_{-1}^{1}$$
$$=\left(\frac{8}{4}-\frac{5}{3}+4\right)=\frac{13}{3}$$
However, I double-checked on wolfram alpha and the solution is actually $-\dfrac{14}{3}$. Would you know where I went wrong? I have no idea where the negative came from, in the solution, or how it's one value above mine.
| As $x^3$ is odd, the integral of $\int_{-1}^{1}8x^3\,dx$ is zero. Therefore
\begin{align}\int_{-1}^{1}\left(8x^3-5x^2+4\right)dx&=\int_{-1}^{1}(-5x^2+4)\,dx\\&=\left(\frac{-5x^3}{3}+4x\right)\Bigg|_{-1}^{1}\\&=\left(\frac{-5}{3}+4\right)-\left(\frac{5}{3}-4\right)\\&=\frac{24-10}{3}\\&=\frac{14}{3}.\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to show that $J_{n+1} = \frac{3n-1}{3n} J_n$? Let $$J_n := \int_{0}^{\infty} \frac{1}{(x^3 + 1)^n} \, {\rm d} x$$
where $n > 2$ is integer. How to show that $J_{n+1} = \frac{3n-1}{3n} J_n$?
| Note
\begin{align}
J_{n+1}&= \int_{0}^{\infty} \frac{(x^3+1)-x^3}{(x^3 + 1)^{n+1}} dx
=J_n +\frac1{3n} \int_{0}^{\infty} x\>d\left(\frac{1}{(x^3 + 1)^n} \right)\\
&=J_n - \frac1{3n}J_n =\frac{3n-1}{3n}J_n
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{x\to+\infty} \frac{\sqrt{x}\cos{x}+2x^2\sin\left({\frac{1}{x}}\right)}{x-\sqrt{1+x^2}}$ Evaluate $$\lim_{x\to+\infty} \frac{\sqrt{x}\cos{x}+2x^2\sin({\frac{1}{x}})}{x-\sqrt{1+x^2}}$$
My attempt: $$\lim_{x\to+\infty} \frac{\sqrt{x}\cos{x}+2x^2\sin\left({\frac{1}{x}}\right)}{x-\sqrt{1+x^2}}=\lim_{x\to+\infty} \frac{x^2\sqrt{x}\left(\frac{\cos{x}}{x^2}+2\frac{\sin{\frac{1}{x}}}{\sqrt{x}}\right)}{x\left(1-\sqrt{1+\frac{1}{x^2}}\right)}$$$$=\lim_{x\to+\infty} x\sqrt{x}\cdot \frac{\left(\frac{\cos{x}}{x^2}+2\frac{\sin{\frac{1}{x}}}{\sqrt{x}}\right)}{1-\sqrt{1+\frac{1}{x^2}}}$$
Both numerator and denominator tend to zero, while $x\sqrt{x} \to +\infty$. Any help is appreciated.
| HINT:
Let $x=\frac1t$
$$\lim_{x\to+\infty} \frac{\sqrt{x}\cos{x}+2x^2\sin({\frac{1}{x}})}{x-\sqrt{1+x^2}}=\lim_{t\to 0}\frac{\sqrt{t}\cos\frac1t+\frac{2}{t}\sin t}{1-\sqrt{t^2+1}}$$
$$=\lim_{t\to 0}\frac{\left(\sqrt{t}\cos\frac1t+2\cdot \frac{\sin t}{t}\right)(1+\sqrt{t^2+1})}{(1-\sqrt{t^2+1})(1+\sqrt{t^2+1})}$$
$$=-\lim_{t\to 0}\frac{\left(\sqrt{t}\cos\frac1t+2\cdot \frac{\sin t}{t}\right)(1+\sqrt{t^2+1})}{t^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Verifying the period of $f(x)=\sin(x)+\cos(x/2)$ It seems clear from the graph of $f(x)=\sin(x)+\cos(x/2)$ that the period $p$ of the function is equal to $4\pi$.
To verify that $4\pi$ is a period of $f(x)$, note that
\begin{align}
\sin(x + 4\pi) + \cos\left(\frac{x + 4\pi}{2}\right) & =\sin(x)\cos(4\pi)+\cos(x)\sin(4\pi)+\cos(x/2)\cos(4\pi/2)-\sin(x/2)\sin(4\pi/2) \\
& =\sin(x)+\cos(x/2)
\end{align}
Thus $4\pi$ is indeed a period of $f$. My question is, how would one go about trying to prove that $4\pi$ is the smallest $p>0$ such that $f(x+p)=f(x)$?
| If $f$ is periodic with period $T$ then so is $f'$. This means that
$$ f'(x) = c \implies f'(x+T) = c.$$
When looking at the graph of $f$ it looks like the solutions of the equation
$$ f'(x) = f'(\pi)$$
are exactly the points $S = \{ \pi + 4k \pi : k \in \mathbb Z\}$ . If we manage to prove this then we are finished since for a smaller period $\tilde T,$ $\pi + \tilde T$ would not be a solution of the equation which contradicts the periodicity of $f'$.
We now solve $f'(x) = f'(\pi).$ By definition of $f$ we have $f'(x) = \cos(x) - \frac 12 \sin \frac x2$ and $f'(\pi) = -1.5$. Let $x = 2u$ and write $-1.5$ as $ -1.5 = -1 - 1/2$ and we have
$$ f'(x) = -1.5 \iff \cos(2u)- \frac 12\sin(u) = -1 - 1/2 \iff \cos(2u)+1 -\frac 1 2 \sin(u) + 1/2 = 0.$$
Using the identity
$$ \cos(2u) + 1 = 2 -2 \sin^2(u)$$
(which can be deduce from the double angle formula and the $\cos^2 u + \sin^2 u =1)$ we have
$$ -2 \sin^2 u - \frac 1 2 \sin u + 2.5 = 0 \iff -4 \sin^2(u)- \sin(u) + 5 = 0$$
This is a quadratic equation in $\sin(u)$ whose solutions are
$$ \sin(u) = 1, \sin(u) = - 5/4.$$
Since $-5/4 < -1$ we have
$$ f'(x) = -1.5 \iff \sin(x/2) = 1 \iff x = \pi + 4 k \pi, k \in \mathbb Z$$
Therefore $4\pi$ is the smallest possible period of $f$.
| {
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"timestamp": "2023-03-29T00:00:00",
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fibonacci recurrence relation proof I've been trying to prove the closed form solution of fibonacci recurrence sequence and achieve this
$a_n=\frac{1}{\sqrt{5}}[(\frac{1+\sqrt{5}}{2})^n−(\frac{1-\sqrt{5}}{2})^n]$
And so far I haven't achieve that, this is how I did it
$a_n=x(\frac{1+\sqrt{5}}{2})^n+y(\frac{1-\sqrt{5}}{2})^n$
$a_0=0=x+y$
$a_1=1=x(\frac{1+\sqrt{5}}{2})+y(\frac{1-√5}{2})$
thus, I was able to get $x=\frac{\sqrt{5}+5}{10}$ and $y=-\frac{\sqrt{5}+5}{10}$. Then plugging in $x$ and $y$ to the formula this is what I got
$a_n=\frac{\sqrt{5}+5}{10}(\frac{1+\sqrt{5}}{2})^n+(-\frac{\sqrt{5}+5}{10})(\frac{1-\sqrt{5}}{2})^n$
beyond that, I just can't prove the closed form from above, I'm stuck to this, since I can't or don't know how to further reduce $\frac{\sqrt{5}+5}{10}$.
Did I miss anything? or got something wrong?
| From $x+y=0$ we have $y=-x$ and substituting into the second we obtain $x(\frac{1+\sqrt{5}}{2})+(-x)(\frac{1-\sqrt{5}}{2})=x(\frac{1+\sqrt{5}-1+\sqrt{5}}{2})=x\sqrt{5}=1$ and thus $x=\frac{1}{\sqrt{5}}$ so $y=-\frac{1}{\sqrt{5}}$ as required.
Alternatively, you can use the generating function of the fibonacci sequence as @SeraPhim mentioned.
| {
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Proof that any number is equal to $1$ Before I embark on this bizzare proof, I will quickly evaluate the following infinite square root; this will aid us in future calculations and working:
Consider $$x=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}$$
$$x^2-2=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}=x \implies x^2-x-2=0\implies x=2$$
as $x>0$. Now for the proof:
I was attempting some different infinite expansions/square roots when trying to solve another question of mine (Evaluate $\sqrt{x+\sqrt{{x^2}+\sqrt{{x^3}+\sqrt{{x^4}...}}}}$ ) and I came across this:
$$x+\frac{1}{x}=\sqrt{(x+\frac{1}{x})^2}=\sqrt{2+x^2+\frac{1}{x^2}}=\sqrt{2+\sqrt{(x^2+\frac{1}{x^2}}})^2=\sqrt{2+\sqrt{2+x^4+\frac{1}{x^4}}}=\sqrt{2+\sqrt{2+\sqrt{(x^4+\frac{1}{x^4})^2}}}=\sqrt{2+\sqrt{2+\sqrt{2+x^8+\frac{1}{x^8}}}}=\sqrt{2+\sqrt{{2}+\sqrt{{2}+\sqrt{{2}...}}}}=2$$
if you keep on applying this and using the result found at the start of the question. So we have that for any real number $x$ that
$$x+\frac{1}{x}=2\implies x^2-2x+1=0\implies (x-1)^2=0$$
so we finally have:
$$x=1$$
Where have I gone wrong, for surely this cannot be correct?
| As other answers explained clearly, you cannot just replace repeated operation with ... and wave away the terms on the tail end without justification. If you still struggle to understand why this is wrong, what you did is roughly equivalent to:
$$
\begin{aligned}
x + \frac 1 x &= 2 + (x + \frac 1 x - 2) \\
&= 2 + 0 + (x + \frac 1 x - 2) \\
&= 2 + 0 + 0 + (x + \frac 1 x - 2) \\
&= 2 + 0 + 0 + 0 + (x + \frac 1 x - 2) \\
&= 2 + 0 + 0 + 0 + ... \\
&= 2
\end{aligned}
$$
Every partial sum equals to $x + \frac 1 x$ and we don't get to claim that the series converges to $2$ simply because we can insert an arbitrary number of repeated operation (here $+0$) in the middle.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the volume between $z=\sqrt{x^{2}+y^{2}}$ and $x^2+y^2+z^2=2$ in spherical cordinates I am asking to find the volume of the volume trap above the cone $z=\sqrt{x^{2}+y^{2}}$ and below the sphere $x^2+y^2+z^2=2$
When I checked the solution I noticed that it was writen as $$V=\int_{0}^{2 \pi} \int_{0}^{\frac{\pi}{4}} \int_{0}^{\sqrt{2}} r^{2} \sin \theta \,d r \,d \theta \,d \varphi$$ and my question is why the boundries of $\theta$ is between $0$ to $\frac{\pi}{4}$ and not $0$ to $\pi$.
why $0$ to $\pi$ is wrong? I just can't imagine the scenerio in my head
| Intersection of $x^2+y^2+z^2=2$ and $z=\sqrt{x^{2}+y^{2}}$ on $OXY$ plane gives $x^2+y^2=1$ so on Cartesian coordinates we have $$\int\limits_{-1}^{1}\int\limits_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}}\int\limits_{\sqrt{x^{2}+y^{2}}}^{\sqrt{2-x^2-y^2}}dxdydz = 4\int\limits_{0}^{1}\int\limits_{0}^{\sqrt{1-x^2}}\int\limits_{\sqrt{x^{2}+y^{2}}}^{\sqrt{2-x^2-y^2}}dxdydz$$
Now let's take spherical coordinates:
\begin{array}{}
x = r \sin \phi \cos \theta; \\
y = r \sin \phi \sin \theta; \\
z = r \cos \phi
\end{array}
From $z$ coordinate bounds
$$r \sin \phi \leqslant r \cos \phi \leqslant \sqrt{2-r^2 \sin^2 \phi}$$
Left inequality gives $\sin \phi \leqslant \cos \phi$, from which we can obtain $\phi \leqslant \frac{\pi}{4}$. Right inequality gives $r \leqslant \sqrt{2}$.
So for integral in spherical coordinates we have
$$4\int\limits_{0}^{\frac{\pi}{2}}\int\limits_{0}^{\frac{\pi}{4}}\int\limits_{0}^{\sqrt{2}} r^{2} \sin \theta d r d \phi d \theta$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $|z_1 + z_2 + z_3|$ on unit circle given $ \frac{ z_1^2}{z_2 z_3} + \frac{ z_2^2}{z_1 z_3} + \frac{ z_3^2}{z_1 z_2} +1 =0$
given three complex numbers $(z_1 , z_2 , z_3)$ lying on the unit circle and related by the equation $ \frac{ z_1^2}{z_2 z_3} + \frac{ z_2^2}{z_1 z_3} + \frac{ z_3^2}{z_1 z_2} +1 =0$, find the sum of all possible values of $|z_1 + z_2 + z_3|$
I have no idea how to solve it systematically but with some guessing I found that $(1,1,-1) $ is a solution triplet for this constraint. And, if $(1,1,-1)$ is a solution then by the symmetry of how the equation is, it means that $(-1,1,1)$ and $(1,-1,1)$ are also solutions.
But, I'm not sure how to rule out/ find more solutions. Ideally, the answer I"m looking for is a general method to approach these kind of questions.
| Let us define the angles $\alpha$ and $\beta$ as
\begin{eqnarray}
\alpha &=& \mathrm{Arg}\left[\frac{z_2}{z_1}\right] \, ,
\\
\beta &=& \mathrm{Arg}\left[\frac{z_3}{z_1}\right] \, .
\end{eqnarray}
Both the condition and the norm $|z_1 + z_2 + z_3|$ can be written in terms of only $\alpha$ and $\beta$ since
\begin{eqnarray}
0 &=& \frac{z^2_1}{z_2 z_3} + \frac{z^2_2}{z_3 z_1} + \frac{z^2_3}{z_1 z_2} + 1 \,
\\
&=& \frac{1}{\frac{z_2}{z_1} \frac{z_3}{z_1}} + \frac{\left(\frac{z_2}{z_1}\right)^2}{\frac{z_3}{z_1}} + \frac{\left(\frac{z_3}{z_1}\right)^2}{\frac{z_2}{z_1}} + 1 \, ,
\end{eqnarray}
and
\begin{equation}
|z_1 + z_2 + z_3| = |z_1|\left|1 + \frac{z_2}{z_1} + \frac{z_3}{z_1}\right| = \left|1 + \frac{z_2}{z_1} + \frac{z_3}{z_1}\right| \, .
\end{equation}
This means that for each pair of values of $\alpha$ and $\beta$ that solve the condition above, $z_1$ (or, equivalently, its $\mathrm{Arg}$) still remains an unconstrained degree of freedom, which is probably considered a redundancy when computing the sum that the exercise asks for.
An additional source or redundancy is the fact that $\alpha$ and $\beta$ that solve the condition above are interchangeable, given the symmetry of the expression.
If we focus on only the ranges $0 \leq \alpha < 2 \pi$ and $0 \leq \beta < 2 \pi$, the pairs of values that solve the condition are
\begin{equation}
\left( \pi, \pi \right), \left( 0, \pi \right), \left( \frac{\pi}{3} , \frac{2\pi}{3} \right) , \left( \frac{\pi}{3} , \frac{5\pi}{3} \right), \left( \frac{4\pi}{3} , \frac{5\pi}{3} \right) \, ,
\end{equation}
where, again, each can be either $\alpha$ or $\beta$. The corresponding values of $|z_1 + z_2 + z_3|$ are
\begin{equation}
1, 1, 2, 2, 2.
\end{equation}
From here it's not completely clear how much redundancy should we consider in order to compute the sum, but the fundamental components are there. If we are expected to sum each possible value of the norm once then we get $1+2 = 3$.
| {
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Distance between set and point, confused of partial derivatives. Let $H = \{(x,y,z)\ \in \mathbb{R}^{3}: x^2+y^2 - z^2 + 4 = 0$
Compute the shortest distance between H and point $p=(2,4,0)$.
I am a bit confused because I tried a direct approach.
$$ x^2+y^2 + 4 = z^2$$
Let $D(H,p) = \sqrt{(2-x)^{2}+(4-y)^{2} + x^{2} + y^{2}+4}$
So I tried compute $$\frac{\partial D}{\partial x} = (\sqrt{2} (-1 + x))/\sqrt{12 - 2 x + x^2 - 4 y + y^2}$$
$$\frac{\partial D}{\partial y} = (\sqrt{2} (-2 + y))/\sqrt{12 - 2 x + x^2 - 4 y + y^2}$$
It seems not nice to compare with zero.
Do you have another idea?
| You are after the minimum of $(x-2)^2+(y-4)^2+z^2$ (no, I did not use the square root) under the restriction that $x^2+y^2-z^2=-4$. It is natural to use here the method of Lagrange multipliers. So, one has to solve the system$$\left\{\begin{array}{l}2(x-2)=2\lambda x\\2(y-4)=2\lambda y\\2z=-2\lambda z\\x^2+y^2-z^2=-4.\end{array}\right.$$Its only solutions are $(x,y,z,\lambda)=(1,2,\pm3,-1)$. So, the distance from $H$ to $(2,4,0)$ is$$\sqrt{(1-2)^2+(2-4)^2+(\pm3)^2}=\sqrt{14}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim_{x\to 0} \frac{\sqrt{ax+b}-1}{x}=1$ My answer
Let $\sqrt{ax+b}=y$
Then
$$\lim_{y\to \sqrt b} \frac{(y-1)a}{y^2-b}$$
Let $b=1$
Then $$\lim _{y\to 1} \frac{a}{\frac{y^2-1}{y-1}}$$
$$=\frac a2 =1$$
$$a=2$$
The answer is correct, but this relies on assuming $b=1$, and that doesn’t seem appropriate. What is the correct answer for this?
| We can solve this by using series expansion for $\sqrt{ax+b}$. By expanding it, we have
$$
\sqrt{ax +b}= \sqrt{b} +\frac{ax}{2\sqrt{b}} - \frac{a^2x^2}{8b^{3/2}}+ {\large O} $$
(${\large O}$ means other higher powers of $x$ terms).
$$
\lim_{x\to 0} \frac{\sqrt{ax+b} -1}{x}= \frac{\sqrt{b}-1}{x} + \frac{a}{2\sqrt{b}}$$
Now, you can see that for limit to exist we have to have $\sqrt{b}=1 \implies b=1$. And by doing that we find
$$
1= a/2 \\
a=2$$
Hope it helps!
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}$ Can we find the solutions for this equation?
$$\sqrt{1+\sqrt{2x-x^2}} + \sqrt{1-\sqrt{2x-x^2}} = \sqrt{4-2x}, \quad x \in \mathbb{R}$$
I tried to amplify the second square root in the $LHS$ with the conjugate and then use AM-GM in order to find where $x$ can be.
Also, the existence conditions imply $x \leq 2$. I obtained $x \leq \frac{4}{3}$.
| You may first symmetrise $x(2-x)$ by setting $\boxed{x=y+1}$:
$$\sqrt{1+\sqrt{1-y^2}} + \sqrt{1-\sqrt{1-y^2}}= \sqrt{2(1-y)} \text{ with } -1\leq y\leq 1$$
Now, squaring gives
$$2+2|y|= 2(1-y) \Leftrightarrow |y| = -y$$
Hence, with $-1\leq y\leq 1 \stackrel{|y|=-y}{\Rightarrow} -1\leq y\leq 0 \stackrel{x=y+1}{\Rightarrow} \boxed{0 \leq x \leq 1}$ is the solution.
| {
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Partial Fraction Decomposition of $\frac{1}{x^2(x^2+25)}$ I have been reviewing some integration techniques and have been searching for tough integrals with solutions online. When I was going through the solution, however, I found a discrepancy between my solution and theirs and think what I did was correct instead.
I am trying to solve the indefinite integral: $\int\frac{dx}{x^2(x^2+25)}$. My first step was to break it into the fractions $$\frac{1}{x^2(x^2+25)}=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+25}$$ Then multiplying both sides by $x^2(x^2+25)$, we find our basic equation to be$$1=A*x(x^2+25)+B*(x^2+25)+(Cx+D)*x^2$$ Solving the system of linear equations, I found that $B=\frac{1}{25}$, $D=\frac{-1}{25}$, and $A=C=0$.
This is where I found the discepancy. The online solution has the basic equation as $$1=A*x(x^2+25)+B*(x^2+25)+(Cx+D)*x$$ so when they solve for coefficients they find that $B=\frac{1}{25}$, $C=\frac{-1}{25}$, and $A=D=0$.
Am I correct or are they? And if my answer is incorrect how does one of the $x$'s cancel out from the $(Cx+D)$ term? Thanks for any help!
| You are right. Here's an other proof:
If we put $X=x^2$ then
$$f(x)=\frac{1}{x^2(25+x^2)}$$
$$=\frac{1}{X(25+X)}$$
$$=\frac{B}{X}+\frac{D}{25+X}$$
$$Xf(x)=B+\frac{DX}{X+25}$$
with $ X=0$, it gives $ B=\frac{1}{25}$
$$(X+25)f(x)=\frac{B(X+25)}{X}+D$$
with $ X=-25$, we get $D=-\frac{1}{25}$
thus
$$f(x)=\frac{1}{25}\Bigl(\frac{1}{x^2}-\frac{1}{25+x^2}\Bigr)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3766754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to evaluate the following limit: $\lim_{x\to 0}\frac{12^x-4^x}{9^x-3^x}$? How can I compute this limit
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}\text{?}$$
My solution is here:
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\dfrac{1-1}{1-1} = \dfrac{0}{0}$$
I used L'H$\hat{\mathrm{o}}$pital's rule:
\begin{align*}
\lim_{x\to 0}\dfrac{12^x\ln12-4^x\ln4}{9^x\ln9-3^x\ln3}&=\dfrac{\ln12-\ln4}{\ln9-\ln3}
\\ &=\dfrac{\ln(12/4)}{\ln(9/3)}
\\ &=\dfrac{\ln(3)}{\ln(3)}
\\ &=1
\end{align*}
My answer comes out to be $1$. Can I evaluate this limit without L'H$\hat{\mathrm{o}}$pital's rule? Thanks.
| Yes, you can evaluate the limit without LHospital's rule as follows
$$\lim_{x\to 0}\dfrac{12^x-4^x}{9^x-3^x}=\lim_{x\to 0}\dfrac{4^x\left(\left(\frac{12}{4}\right)^x-1\right)}{3^x\left(\left(\frac93\right)^x-1\right)}$$
$$=\lim_{x\to 0}\dfrac{4^x\left(3^x-1\right)}{3^x(3^x-1)}$$
$$=\lim_{x\to 0}\left(\dfrac{4}{3}\right)^x$$
$$=\color{blue}{1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Solution of differential equation ${x^2}\frac{{dy}}{{dx}} + {y^2}{e^{\frac{{x\left( {y - x} \right)}}{y}}} = 2y\left( {x - y} \right)$
Solutions of differential equation
$$
{x^2}\frac{{dy}}{{dx}} + {y^2}{e^{\frac{{x\left( {y - x} \right)}}{y}}} = 2y\left( {x - y} \right)
$$
are given by
(A) $x(x + y) = y \ln (Ce^x– 1)$
(B) $x(x – y) = x \ln (Ce^x– 1)$
(C) $x(x + y) = x\ln (Ce^x + 1) $
(D) $x(x – y) = y \ln (Ce^x –1)$,
where $C$ is constant of integration.
My approach is as follow
\begin{align}
& \Rightarrow \frac{{{x^2}}}{{{y^2}}}\frac{{dy}}{{dx}} + {e^{\frac{{x\left( {y - x} \right)}}{y}}} = \frac{{2y\left( {x - y} \right)}}{{{y^2}}} \\
& t = \frac{1}{y} \\
& dt = - \frac{1}{{{y^2}}}dy \\
& \Rightarrow - \frac{{{x^2}dt}}{{dx}} + {e^{x\left( {1 - tx} \right)}} = 2\left( {tx - 1} \right) \\
& {e^{x\left( {1 - tx} \right)}} = g,\because \frac{{dg}}{{dx}} = g\left( {1 - 2tx - {x^2}\frac{{dt}}{{dx}}} \right) \Rightarrow \frac{{dg}}{{gdx}} = \left( {1 - 2tx - {x^2}\frac{{dt}}{{dx}}} \right) \Rightarrow \\
& \qquad- {x^2}\frac{{dt}}{{dx}} = 2tx - 2 + \frac{{dg}}{{gdx}} + 1 \\
& \Rightarrow 2tx - 2 + \frac{{dg}}{{gdx}} + 1 + g = 2\left( {tx - 1} \right) \Rightarrow \frac{{dg}}{{gdx}} + 1 + g = 0 \Rightarrow \frac{{dg}}{{g\left( {g + 1} \right)}} = - dx \\
& \Rightarrow \frac{{dg}}{g} - \frac{{dg}}{{\left( {g + 1} \right)}} = - x + C \\
& \Rightarrow \ln{e^{x\left( {1 - tx} \right)}} - \ln\left( {{e^{x\left( {1 - tx} \right)}} + 1} \right) = - x + C \Rightarrow \ln\frac{{{e^{x\left( {1 - tx} \right)}}}}{{{e^{x\left( {1 - tx} \right)}} + 1}} = - x + C
\end{align}
I am not able to proceed from here
| $${x^2}\frac{{dy}}{{dx}} + {y^2}{e^{\frac{{x\left( {y - x} \right)}}{y}}} = 2y\left( {x - y} \right)$$
Substitute $y=tx$
$$t'x+t+ t^2{e^{x(1-\frac 1t )}} = 2t\left( {1 - t} \right)$$
$$t'x+ t^2{e^{x(1-\frac 1t )}} = t -2 t^2$$
$$-\left(\dfrac 1t \right)'x+ {e^{x(1-\frac 1t )}} = \dfrac 1t -2 $$
$$-\left(\dfrac xt \right)'+ {e^{x(1-\frac 1t )}} = -2 $$
$$\left(\dfrac xt \right)'= {e^{x(1-\frac 1t )}} +2 $$
Substitute $u=\dfrac x t$ .
$$(e^u)'-2e^u=e^x $$
This is easy to solve.
$$(e^ue^{-2x})'=e^{-x}$$
Integrate
$$e^u=-e^{x}+Ce^{2x}$$
$$\dfrac xt=\ln |-e^{x}+Ce^{2x}|$$
$$\dfrac {x^2}y=x+\ln |-1+Ce^{x}|$$
$$x(x-y) =y\ln |-1+Ce^{x}|$$
Solution is option D
You can continue from the last line. Put the denominator on the right side and then factorize
$$z=Ke^{-x}(z+1)$$
$$ \implies z(1-ke^{-x})=ke^{-x}$$
$$e^{x(1-tx)}= \dfrac {ke^{-x}}{1-ke^{-x}}$$
$$e^{x(1-tx)}= \dfrac {ke^{-x}}{1-ke^{-x}}$$
Take log function on both sides and continue. Normally you should end with the right solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768701",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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} |
Linear law: how to determine three unknown constants from a non-linear equation?
Question:
The diagram shows part of a straight line graph drawn to represent the equation $y=\frac{ax^2+b}{cx}$, where $a$, $b$, and $c$ are integers. Given that the line passes through $(4,9)$ and has gradient $-\frac{1}{4}$, find
(i) the value of $\frac{y}{x}$ where the straight line cuts the horizontal axis, and
(ii) the value of $a$, of $b$ and of $c$.
Attempt:
(i)
$$Y=mX+c$$
Gradient, $m=-\frac{1}{4}$
At $(4,9)$,
$$9=-\frac{1}{4}(4)+c$$
$$c=10$$
Therefore, $Y=-\frac{1}{4}X+10$
When $Y=0$,
$$0=-\frac{1}{4}(X)+10$$
$$X=40$$
Since $X=\frac{x}{y}$, therefore $\frac{y}{x}=\frac{1}{40}$.
Is my solution correct?
(ii)
$$y=\frac{ax^2+b}{cx}$$
Divide the equation by $y$,
$$1=\frac{ax}{cy}+\frac{b}{cxy}$$
$$\frac{b}{cxy}=1-\frac{ax}{cy}$$
$$\frac{1}{xy}=-\frac{a}{b}\frac{x}{y}+\frac{c}{b}$$
Since $Y=-\frac{1}{4}X+10$,
$$-\frac{a}{b}=-\frac{1}{4}$$
$$\frac{c}{b}=10$$
After this step, I do not know how to continue. I have two equations, but three unknowns. How do I get a third equation so that I can solve for $a$, $b$, and $c$?
Any suggestions / help is greatly appreciated. Thank you!
| The equation $y=\frac{ax^2+b}{cx}$ can be written as
$ax^2-cxy+b=0$.
As you mentioned, $Y=-\frac{1}{4}X+10$
or $x^2-40xy+4 = 0$, by replacing $Y = \frac{1}{xy}, X = \frac{x}{y}$.
So, $a = 1, b = 4, c = 40$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $\lim_{x\to 0} \frac{1+a\cos 2x + b\cos 4x}{x^4}$ exists for all $x\in\mathbb R$ and is equal to $c$, find $\lfloor a^{-1} +b^{-1} + c^{-1}\rfloor$ $$\lim_{x\to 0} \frac{1+a(1-\frac{4x^2}{2!} + \frac{16x^4}{4!}-\cdots)+b(1-\frac{16x^2}{2!} + \frac{256x^4}{4!}-\cdots)}{x^4}$$
$$=\lim_{x\to 0} \frac{ (1+a+b) -\frac{x^2}{2!} (4a+16b) + \frac{x^4}{4!} (16a+256b)}{x^4}$$
For limit to exist, $1+a+b=0$ and $4a+16b=0$
So $a=-\frac 43$ and $b=\frac 13$
Now $$c=\lim_{x\to 0} \frac{\frac{x^4}{4!} (16a+256b)\cdots}{x^4}$$
$$c=\frac{16a+256b}{24}$$
$$c=\frac{2a}{3}+\frac{32}{3}b=\frac{24}{9}$$
Then $$\lfloor a^{-1} + b^{-1} + c^{-1} \rfloor = \left\lfloor 3-\frac 34 + \frac{9}{24}\right\rfloor = \lfloor 2.625\rfloor=2$$
The given answer is $8.$
What is wrong in my solution?
| Your result is correct
If we put $X=2x$ then it is equivalent to say that
$$\lim_0\frac{1+a\cos(X)+b\cos(2X)}{X^4}=\frac{c}{16}$$
$$=-\lim_0\frac{a\sin(X)+2b\sin(2X)}{4X^3}$$
$$=-\lim_0\frac{a\cos(X)+4b\cos(2X)}{12X^2}$$
$$=\lim_0\frac{a\cos(X)+16b\cos(2X)}{24}$$
So, we must have
$$a+b+1=0$$
$$a+4b=0$$
thus
$$b=\frac 13,\;\; a=-\frac 43$$
and
$$c=\frac 23(a+16b)=\frac 29(-4+16)=\frac{24}{9}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3769887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. So here is the Question :-
Given that $f(x)$ is a polynomial of degree $3$, and its remainders are $2x - 5$ and $-3x + 4$ when divided by $x^2 - 1$ and $x^2 - 4$ respectively. Find $f(-3)$ .
What I tried:- Since it's given that $f(x)$ is a polynomial of degree $3$ , I can assume $f(x) = ax^3 + bx^2 + cx + d$ for some integers $a,b,c,d$ and $a\neq 0$. Then we have :-
$$ax^3 + bx^2 + cx + d = (x^2 - 1)y + (2x - 5)$$
$$ax^3 + bx^2 + cx + d = (x^2 - 4)z + (-3x + 4)$$
This gives that $(x^2 - 1)y + (2x - 5) = (x^2 - 4)z + (-3x + 4)$ . But I am not sure how to proceed further since we have $3$ variables to deal with , and I am stuck here.
Any hints or explanations for this problem will be greatly appreciated !!
| You have (for instance) that $f(x) = Q(x)(x+1)(x-1) + R(x)$, where $R(x) = 2x-5$.
Putting $x=1$ then $x=-1$ gives you $f(1) = - 3$ and $f(-1)=-7$.
Doing the same with $x=2, x=- 2$ allows you to work out two more points on the cubic.
Four distinct points allow you to completely characterise a cubic, you have a system of four linear equations to solve for the coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3770874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
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} |
If $p$ and $q$ are primes such that $q \mid {\frac{x^p-1}{x-1}}$ then prove that $q\equiv 1 \pmod{p}$ or $q\equiv 0 \pmod{p}$.
QUESTION: If $p$ and $q$ are primes such that $$q \mid {\frac{x^p-1}{x-1}} , (x\in\Bbb{N}, x>1)$$ then prove that $q\equiv 1 \pmod{p}$ or $q\equiv 0 \pmod{p}$.
MY ANSWER: I came across this lemma, but couldn't prove the second part properly. Here's what I did -
By Fermat's Little Theorem we know that $x^p\equiv{x}\pmod{p}$. Therefore, $$\frac{x^7-1}{x-1}\equiv\frac{x-1}{x-1}=1\pmod{7}$$ Therefore, $q\equiv{1}\pmod{7}$.
Now, I cannot prove that $q\equiv{0}\pmod{7}$. Not simultaneously ofcourse, I know that ..
Here's my try -
We can write the above equation as $$x^6+x^5+x^4+x^3+x^2+x+1$$ But what after this? Even if I chose $q=7$, it does not divide the above equation for all values of $x$. Say $x=7$, then the equation can be rewritten as $$7^6+7^5+7^4+7^3+7^2+7+1$$ and $$7\nmid{7^6+7^5+7^4+7^3+7^2+7+1}$$ So, how do I rigorously prove that $q\equiv{0}\pmod{p}$ ? Or, for which cases is this true?
Any help will be much appreciated. Thank you :)
EDIT: Terrible Mistake :P The first proof of $q\equiv{1}\pmod{7}$ is wrong. So, now I am left with a full question to be proved °_°
| There are $2$ cases to consider.
Case $1$: $x \equiv 1 \pmod{q}$
Dividing $x - 1$ into $x^p - 1$, and using $x \equiv 1 \pmod{q}$, gives
$$0 \equiv \sum_{i=0}^{p-1}x^{i} \equiv \sum_{i=0}^{p-1}1^{i} = p \pmod{q} \tag{1}\label{eq1A}$$
Since $p$ and $q$ are primes, this means $p = q$, i.e., $q \equiv 0 \pmod{p}$.
Case $2$: $x \not\equiv 1 \pmod{q}$
In this case, you have
$$x^p - 1 \equiv 0 \pmod{q} \implies x^p \equiv 1 \pmod{q} \tag{2}\label{eq2A}$$
The multiplicative order of $x$ modulo $q$ divides any power of $x$ which is congruent to $1$. Since $x \not\equiv 1 \pmod{q}$, this means the multiplicative order must be $\gt 1$. As $p$ is prime, its only factors are $1$ and $p$, so this means the multiplicative order of $x$ modulo $q$ must be $p$.
However, Fermat's little theorem states
$$x^{q-1} \equiv 1 \pmod{q} \tag{3}\label{eq3A}$$
This means $p \mid q - 1$, i.e.,
$$q \equiv 1 \pmod{p} \tag{4}\label{eq4A}$$
In summary, this shows
$$q \equiv 1 \pmod{p} \; \; \text{ or } \; \; q \equiv 0 \pmod{p} \tag{5}\label{eq5A}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3771781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Write the polynomial of degree $4$ with $x$ intercepts of $(\frac{1}{2},0), (6,0)$ and $(-2,0)$ and $y$ intercept of $(0,18)$. Write the polynomial of degree $4$ with $x$ intercepts of $(\frac{1}{2},0), (6,0) $ and $ (-2,0)$ and $y$ intercept of $(0,18)$.
The root ($\frac{1}{2},0)$ has multiplicity $2$.
I am to write the factored form of the polynomial with the above information. I get:
$f(x)=-6\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$
Whereas the provided solution is:
$f(x)=-\frac{3}{2}(2x-1)^2(x+2)(x-6)$
Here's my working:
Write out in factored form:
$f(x) = a\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$
I know that $f(0)=18$ so:
$$18 = a\big(-\frac{1}{2}\big)^2(2)(-6)$$
$$18 = a\big(\frac{1}{4}\big)(2)(-6)$$
$$18 = -3a$$
$$a = -6$$
Thus my answer:
$f(x)=-6\big(x-\frac{1}{2}\big)^2(x+2)(x-6)$
Where did I go wrong and how can I arrive at:
$f(x)=-\frac{3}{2}(2x-1)^2(x+2)(x-6)$ ?
| You didn't go wrong anywhere.
$-6(x-\frac{1}{2})^2(x+2)(x-6) = -\frac{3}{2}(2x-1)^2(x+2)(x-6)$
That is because
$-6(x- \frac 1{2})^2 = -6(\frac 12[2x-1])^2= -6\cdot (\frac 12)^2(2x-1)^2 = -6\cdot \frac 14(2x-1) = -\frac 32(2x-1)$.
Is there some rule that that says fractions in the $(ax + b)$ parts are frowned upon?
If so, if you get $(x + \frac ab)^k$ and can replace it with $(\frac 1b)^k(bx + a)^k$ but I don't see why you should have to.
(In fact, I much prefer your notation as it directly indicates the roots and solutions... and indicates what the leading coeficient will actually be when expanded out. And what the heck kind of sense does removing a horrifying offensive fraction from one area make if you are just going to have to put an equally offensive fraction somewhere else?)
But.... weren't you supposed to expand this out? So far as I can tell neither answer has done that. If you expand it out, you will see both answers are exactly the same.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to obtain the sum of the series $\sum_{n=0}^{\infty}\frac{1}{2^{n}(3n+1)}$? How to prove what follows?
$$\sum_{n=0}^{\infty}\frac{1}{2^{n}(3n+1)}=\frac{2^{\frac{1}{3}}}{3}\ln\left(\frac{\sqrt{2^{\frac{2}{3}}+2^{\frac{1}{3}}+1}}{2^{\frac{1}{3}}-1}\right)+\frac{\sqrt[3]{2}}{3}\arctan\left(\frac{2^{\frac{2}{3}}+1}{\sqrt{3}}\right)-\frac{2^{\frac{1}{3}}\pi}{6\sqrt{3}}$$
My attempt:
$$\sum_{n=0}^{\infty}\frac{1}{2^n(3n+1)}=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{2^n(3n+1)}|_{x=1}$$
We put $$S(x)=\sum_{n=0}^{\infty}\frac{x^{3n+1}}{2^n(3n+1)}\implies S^{'}(x)=\sum_{n=0}^{\infty}\frac{x^{3n}}{2^n(3n+1)}$$
$$S^{'}(x)=\sum_{n=0}^{\infty}\frac{(\frac{x^3}{2})^n}{3n+1}=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(1-\frac{3n}{3n+1})=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n-\sum_{n=0}^{\infty}(\frac{x^3}{2})^n\frac{3n}{3n+1}=\frac{1}{2-\frac{x^3}{2}}-\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(\frac{3n}{3n+1})=\alpha-\beta$$
Where
$$\beta=\sum_{n=0}^{\infty}(\frac{x^3}{2})^n(\frac{3n}{3n+1})$$
So $$\beta=?$$
Waiting for your help to find a beta or prove equal above.
| Observe that
$$f(x):=\sum_{n=0}^\infty\frac{x^n}{3n+1}\implies f'(x)=\sum_{n=0}^\infty\frac{nx^{n-1}}{3n+1}$$ and
$$3xf'(x)+f(x)=\sum_{n=0}^\infty\frac{3n+1}{3n+1}x^n=\frac1{1-x}.$$
The solution of the homogeneous part of this linear ODE is
$$f_h(x)=\frac c{\sqrt[3]x},$$
and by variation of the constant
$$f(x)=\frac1{\sqrt[3]x}\left(\int\frac{\sqrt[3]x}{1-x}dx+c\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3774507",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $(1+a^2)(1+b^2)(1+c^2)\geq8 $ I tried this question in two ways-
Suppose a, b, c are three positive real numbers verifying $ab+bc+ca = 3$. Prove that $$ (1+a^2)(1+b^2)(1+c^2)\geq8 $$
Approach 1:
$$\prod_{cyc} {(1+a^2)}= \left({a^2\over2}+1+{a^2\over2}\right)\left({b^2\over2}+{b^2\over2}+1\right)\left(1+{c^2\over2}+{c^2\over2}\right)$$
$$ \geq\left(\sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4}\right)^3\geq8 $$
$$ \Rightarrow \sqrt[3]{(ab)^2\over4}+\sqrt[3]{(bc)^2\over4}+\sqrt[3]{(ca)^2\over4} \geq2 \Rightarrow \sqrt[3]{2(ab)^2}+\sqrt[3]{2(bc)^2}+\sqrt[3]{2(ca)^2}\geq4 $$
I reached till here but can't take it forward.
Approach 2:
$$ \prod_{cyc} {(1+a^2)}=\prod_{cyc} \sqrt{(1+a^2)(1+b^2)} \geq \prod_{cyc} {(1+ab)}\ge8 $$
but it failed as
$$ \sum_{cyc}{ab}=3 \Rightarrow \sum_{cyc}{(1+ab)}=6 \Rightarrow 8\ge \prod_{cyc} {(1+ab)} $$
This approach is surely weak, but I think that the first approach is unfinished.
Probably brute-force would help but other solutions are always welcome.
Thanks!
| Here is another solution using AM-GM and Cauchy-Schwarz. Our goal is to prove that:
$$(1+a^2)(1+b^2)(1+c^2)\geq8$$
Rewriting the L.H.S. expression a little, we have:
$$(a^2+1)(1+b^2)(1+c^2) \ge (a+b)^2(1+c^2)=S,$$ where we have applied C-S to the first two terms in the product.
Hence it suffices for us to show that:
$$S=a^2+a^2c^2+2ab+2abc^2+b^2+b^2c^2 \ge (ab+bc+ca)^2-1$$
$$\iff a^2+a^2c^2+2ab+2abc^2+b^2+b^2c^2 \ge a^2b^2+2a^2bc+a^2c^2+2ab^2c+2abc^2+b^2c^2-1$$
$$\iff a^2+2ab+b^2 \ge a^2b^2+2a^2bc+2ab^2c-1$$
$$\iff (a+b)^2+1 \ge ab(ab+2ac+2bc)$$
$$\iff (a+b)^2+1 \ge ab(3+ac+bc) $$
$$\iff a^2+b^2+1 \ge ab(1+ac+bc)=ab(4-ab)$$
$$\iff a^2+b^2+a^2b^2+1 \ge 4ab$$
But the last inequality is trivially true by AM-GM; since:
$$a^2+b^2+a^2b^2+1 \ge 4 \sqrt[^4]{a^4b^4}=4ab,$$
and we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 6
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How do I solve $(\cos2x+1)^2=1/2$? $x$ belongs to $$[0,\pi/2[$$
$$(\cos(2x+1))^2 = \frac{1}{2}$$
I tried to find $x$ using
$$2x+1=\frac{\pi}{4} +\frac{n\pi}{2}
\qquad\text{or}\qquad
2x+1=\frac{3\pi}{4} + n\pi$$
but I didn't find my answer in MCQ which is
$$\frac{3\pi-4}{8}
\qquad\text{and}\qquad
\frac{5\pi-4}{8}.$$
I just need a hint on how to solve such equation.
| $$
2x+1=\frac{3\pi}{4} + n\pi
$$
subtract $1$
$$
2x=\frac{3\pi}{4} - 1 + n\pi
$$
Multiply by $1/2$
$$
x=\frac{3\pi}{8} - \frac{1}{2} + n\frac{\pi}{2}
$$
The two answers shown correspond to $n=0$ and $n=1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find taylor series for $x/1+x$, at $x= -2$ using differentiation Attempt:
$\\f^ 1(x) = \frac{1}{(1+x)^2}$ ,
$\\f^ 2(x) = \frac{-2}{(1+x)^3}$ ,
$\\f^ 3 (x) = \frac{6}{(1+x)^4}$
$\\f^ n (x) = \frac{(-1)^{n+1}(n!)}{(1+x)^{n+1}}$,
$\\f^ n (-2) = \frac{(-1)^{n+1}(n!)}{(-1)^{n+1}}$
Using the taylor series formula,
$\frac {x}{1+x} = \sum_{n=0}^{\infty} \frac{f^n(-2)}{n!} (x+2)^n = \sum_{n=0}^{\infty} \frac{n!}{n!}(x+2)^n
= \sum_{n=0}^{\infty} (x+2)^n $
But the solution used geometric series and $\frac {x}{1+x} = 1 + \sum_{n=0}^{\infty} (x+2)^n$.
I am unsure as to what is wrong with my solution, is there a way to get this correct answer using taylor series formula directly instead of using geometric series?
| Hint
Make life simpler letting $x=y-2$. So
$$\frac x{1+x}=\frac{y-2}{y-1}=1+\frac{1}{1-y}$$
The derivatives wrt $y$ are very simple and, when finished, make $y= {x+2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
number of solutions of $f(f(f(f(x))))$ Let $f(x)={x^2+10x+20}$ then the number of real solutions of $f(f(f(f(x))))=0$
My try: I first tried finding the roots of $f(x)=0$ they came out to be
$-5+\sqrt{5}$ and $-5-\sqrt{5}$.
Let these roots be $\alpha$ and $\beta$
thus $f(\alpha)=0$ also if $f(f(f(f(x))))=0$ then $f(f(f(x)))=\alpha$
I was able to go only till this. I skipped the method of actually finding the composite function as it would be too cumbersome. Any ideas, helps will be appreciated.
| The multiple composition of the equation can also be viewed in this way (amounting to a different interpretation of what is described in many of the other posted answers). The function $ \ f(x) \ = \ x^2 + 10x + 20 \ = \ (x + 5)^2 - 5 \ \ $ has the two real zeroes $ \ r_{\pm} \ = \ -5 \ \pm \sqrt{5} \ \ , $ so the next level of composition, $ \ f( \ f(x) \ ) \ = \ 0 \ \ $ requires us to solve $ \ f(x) \ = \ -5 \ \pm \ \sqrt{5} \ \ ; $ further levels of composition seem to call for more daunting calculation.
We might instead look at how the function transforms numbers. The fixed points of the function are given by
$$ f(x) \ = \ x \ \ \Rightarrow \ \ x^2 \ + \ 10x \ + \ 20 \ \ = \ \ x \ \ \Rightarrow \ \ x^2 \ + \ 9x \ + \ 20 \ \ = \ \ (x + 5)·(x + 4) \ \ = \ \ 0 $$
$$ \Rightarrow \ \ x \ \ = \ \ -4 \ , \ -5 \ \ . $$
So $ \ f(-4) \ = \ -4 \ \ $ and $ \ f(-5) \ = \ -5 \ \ . $ We could then ask about how other numbers are "mapped" by the function relative to the fixed points. For $ \ x \ = \ -5 + \alpha \ \ , $ we obtain
$$ ( \ [-5 + \alpha] + 5)^2 - 5 \ \ = \ \ \alpha^2 \ - \ 5 \ \ , $$
We see from this that $ \ f(x) \ = \ 0 \ \ $ corresponds to the mapping of $ \ -5 + \alpha \ \rightarrow \ 0 \ \ $ such that $ \ \alpha^2 - 5 \ = \ 0 \ \ . $ Hence there are two such points, $ \ -5 \pm \sqrt5 \ \ , $ which are symmetrically located around the axis of symmetry of the parabola representing the quadratic polynomial (as we expect of the zeroes). What we now also see is that $ \ f( \ f(x) \ ) \ = \ 0 \ $ corresponds to $ \ -5 + \alpha' \ \rightarrow \ -5 + \sqrt5 \ \ $ such that $ \ (\alpha')^2 - 5 \ = \ ( \ \alpha^2 \ )^2 - 5 \ = \ 0 \ \ . $
Repeated applications of the function transformation maintain the symmetry of the two points about $ \ x = -5 \ \ $ while "compressing" them successively closer to this fixed point (which would be referred to as an "attractor"). As also described in comments and the other posted answers, we observe that $ \ f^3( x ) \ = \ 0 \ $ is solved by $ \ -5 + \alpha'' \ \rightarrow \ -5 + \alpha' \ \ $ with $ \ (\alpha'')^2 - 5 \ = \ ( \ ( \ \alpha^2 \ )^2 \ ) ^2 - 5 \ = \ 0 \ \ $ and finally $ \ f^4( x ) \ = \ 0 \ $ is solved by $ \ -5 + \alpha''' \ \ $ with $ \ (\alpha''')^2 - 5 \ = \ ( \ ( \ ( \ ( \ \alpha^2\ )^2 \ )^2 \ ) ^2 - 5 \ = \ 0 \ \Rightarrow \ \alpha''' \ = \ \pm 5^{1/16} \ \ . $ At any finite number of compositions then, we continue to obtain just two real zeroes symmetrically placed about $ \ -5 \ $ (and a whole slew of complex zeroes).
What about the other fixed point? If we use $ \ x \ = \ -4 + \beta \ \ , $ we obtain
$$ ( \ [-4 + \beta] + 5)^2 - 5 \ \ = \ \ ( \ 1 + \beta \ )^2 \ - \ 5 \ \ = \ \ 0 \ \ \Rightarrow \ \ \beta \ \ = \ \ -1 \pm \sqrt5 \ \ , $$
leading us again to $ \ x \ = \ -5 \pm \sqrt5 \ \ . $ (So the fixed point $ \ -4 \ $ acts as a "repellor".)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
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} |
Find the angles of a triangle given the area For a triangle $ABC$ if:
The area = $\frac{(a+b)^2}{8}$
Find its angles .
My attempt:
$\frac{1}{2} a b \sin{C} = \frac{(a+b)^2}{8}$
$2ab(2\sin{C} -1) = a^2 + b^2 $
$2\sin{C} -1 = \frac{a^2 +b^2}{2ab}$
$2\sin{C} -1$ =$ \frac{a^2 + b^2 -c^2}{2ab} + \frac{c^2}{2ab}$
= $\cos{C} + \frac{c^2}{2ab}$
But i could not go on any more i need a hint ?
| You are on the right track. I have edited out some calculation mistakes including the missing $2$ in front of the $\sin C$. I think it's this mistake that makes the progress hard.
Notice the following inequality:
$$2\sin{C} -1 = \frac{a^2 +b^2}{2ab} \geq1 \implies \sin C \geq 1$$
We can immediately infer that $C$ is $90$ degrees and $a^2+b^2=2ab$ which implies the other two angles are both $45$ degrees.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Diophantine Equation with Square Root I want to resolve the diophantine equation:
$\sqrt{x^2+5x+12} ≡ x-2\pmod 5$
I have thought 2 ways:
1.
$(\sqrt{x^2+5x+12})^2 ≡ (x-2)^2\pmod 5$
$ x^2+5x+12 ≡ x^2 -4x+4\pmod 5$
$ 9x+8 ≡ 0\pmod 5$
$ 4x+3 ≡ 0\pmod 5$
$ 4x+3 = 5y$
$ 4x-5y = -3$
I resolve:
$x=5n+3$,
n
∈
Z
$y=4n+3$,
n
∈
Z
2.
$\sqrt{x^2+5x+12} ≡ x-2\pmod 5$
$\sqrt{x^2+5x+12} = x-2 + 5y$
$(\sqrt{x^2+5x+12})^2 = (x-2 + 5y)^2$
$x^2+5x+12 = 10xy+x^2-4x-20y+25y^2+4$
$25y^2+10xy-9x-20y-8 = 0$ and resolve that diophantine equation.
My questions are: Are both procedures valid? If so, does the development of each procedure preserve the value of the initial x and y unknowns?
Thank you
| Then I have done:
$\sqrt{x^2+5x+12} ≡ x-2\pmod 5$
$\sqrt{x^2+2} ≡ x-2\pmod 5$
$\sqrt{x^2+2} = x-2+5y$
$x^2+2 = (x-2+5y)^2$ and that equation more recient dont have solutions and $\sqrt{x^2+5x+12} ≡ x-2\pmod 5$ has solutions, Why?
Thanks
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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how to solve $\int _0^1\frac{\ln \left(1+x\right)}{a^2+x^2}\:\mathrm{d}x$
how to solve $$\mathcal{J(a)}=\int _0^1\frac{\ln \left(1+x\right)}{a^2+x^2}\:\mathrm{d}x$$
i used the differentiation under the integral and got
\begin{align}
\mathcal{J(b)}&=\int _0^1\frac{\ln \left(1+bx\right)}{a^2+x^2}\:\mathrm{d}x
\\[3mm]
\mathcal{J'(b)}&=\int _0^1\frac{x}{\left(a^2+x^2\right)\left(1+bx\right)}\:\mathrm{d}x
\\[3mm]
&=\frac{a^2b}{1+a^2b^2}\int _0^1\frac{1}{a^2+x^2}\:\mathrm{d}x+\frac{1}{1+a^2b^2}\int _0^1\frac{x}{a^2+x^2}\:\mathrm{d}x-\frac{b}{1+a^2b^2}\int _0^1\frac{1}{1+bx}\:\mathrm{d}x
\\[3mm]
&=\frac{ab}{1+a^2b^2}\operatorname{atan} \left(\frac{1}{a}\right)+\frac{1}{2}\frac{\ln \left(1+a^2\right)}{1+a^2b^2}-\frac{\ln \left(a\right)}{1+a^2b^2}-\frac{\ln \left(1+b\right)}{1+a^2b^2}
\end{align}
But we know that $\mathcal{J}(1)=\mathcal{J(a)}$ and $\mathcal{J}(0)=0$
\begin{align}
\int_0^1\mathcal{J'(b)}\:\mathrm{d}b&=a\:\operatorname{atan} \left(\frac{1}{a}\right)\int _0^1\frac{b}{1+a^2b^2}\:\mathrm{d}b+\frac{\ln \left(1+a^2\right)}{2}\int _0^1\frac{1}{1+a^2b^2}\:\mathrm{d}b-\ln \left(a\right)\int _0^1\frac{1}{1+a^2b^2}\:\mathrm{d}b
\\
&-\int _0^1\frac{\ln \left(1+b\right)}{1+a^2b^2}\:\mathrm{d}b
\\[3mm]
\mathcal{J(a)}&=\frac{1}{2a}\operatorname{atan} \left(\frac{1}{a}\right)\ln \left(1+a^2\right)+\frac{1}{2a}\ln \left(1+a^2\right)\operatorname{atan} \left(a\right)-\frac{1}{a}\ln \left(a\right)\:\operatorname{atan} \left(a\right)-\underbrace{\int _0^1\frac{\ln \left(1+b\right)}{1+a^2b^2}\:\mathrm{d}b}_{\mathcal{I}}
\end{align}
but how to calculate ${\mathcal{I}}$, i tried using the same technique but it didnt work
| Repeating your calculations using Feynman's trick
$$J'(b)=\frac{-\log \left(a^2\right)+\log \left(a^2+1\right)+2 a b \cot ^{-1}(a)-2 \log
(b+1)}{2 (a^2 b^2+1)}$$
$$J(b)=\int\frac{-\log \left(a^2\right)+\log \left(a^2+1\right)+2 a b \cot ^{-1}(a)}{2 (a^2 b^2+1)}\,db-\int\frac{\log(b+1)}{ a^2 b^2+1}\,db$$ The first integral is simple
$$J_1(b)=\int\frac{-\log \left(a^2\right)+\log \left(a^2+1\right)+2 a b \cot ^{-1}(a)}{2 (a^2 b^2+1)}\,db$$
$$J_1(b)=\frac{\cot ^{-1}(a) \log \left(a^2 b^2+1\right)+\left(\log
\left(a^2+1\right)-\log \left(a^2\right)\right) \tan ^{-1}(a b)}{2 a}$$
$$K_1=\int_0 ^1\frac{-\log \left(a^2\right)+\log \left(a^2+1\right)+2 a b \cot ^{-1}(a)}{2 (a^2 b^2+1)}\,db$$
$$K_1=\frac{\left(\log \left(a^2+1\right)-\log \left(a^2\right)\right) \tan
^{-1}(a)+\log \left(a^2+1\right) \cot ^{-1}(a)}{2 a}$$
More tedious is the second integral
$$J_2(b)=\int\frac{\log(b+1)}{ a^2 b^2+1}\,db$$ $$J_2(b)=\frac{i \left(\text{Li}_2\left(\frac{a
(b+1)}{a-i}\right)-\text{Li}_2\left(\frac{a (b+1)}{a+i}\right)+\log (b+1)
\left(\log \left(1-\frac{a (b+1)}{a-i}\right)-\log \left(1-\frac{a
(b+1)}{a+i}\right)\right)\right)}{2 a}$$
$$K_2(b)=\int_0^1\frac{\log(b+1)}{ a^2 b^2+1}\,db=$$
$$\frac{i \left(-\text{Li}_2\left(\frac{a}{a-i}\right)+\text{Li}_2\left(\frac{2
a}{a-i}\right)+\text{Li}_2\left(\frac{a}{a+i}\right)-\text{Li}_2\left(\frac{2
a}{a+i}\right)+\log (2) \left(\log \left(-\frac{a+i}{a-i}\right)-\log
\left(-\frac{a-i}{a+i}\right)\right)\right)}{2 a}$$ which, for sure, is a real number.
Remark
I think that this could have benn done without Feynman's trick
$$\frac{\log \left(1+x\right)}{a^2+x^2}=\frac{\log \left(1+x\right)}{(x+i a)(x-i a)}$$
$$\int \frac{\log \left(1+x\right)}{a^2+x^2}\,dx=\frac i {2a}\left(\int \frac{\log \left(1+x\right)}{x+i a}\,dx-\int \frac{\log \left(1+x\right)}{x-i a}\,dx \right)$$
$$\int \frac{\log \left(1+x\right)}{x+i a}\,dx=\text{Li}_2\left(\frac{x+1}{1-i a}\right)+\log (x+1) \log \left(1-\frac{x+1}{1-ia}\right)$$
$$\int \frac{\log \left(1+x\right)}{x-i a}\,dx=\text{Li}_2\left(\frac{x+1}{1+i a}\right)+\log (x+1) \log \left(1-\frac{x+1}{1+ia}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3789171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
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Computing $\int_0^1 \frac{\arcsin \sqrt x}{x^2-x+1} dx$
Compute:
$$\int_0^1 \frac{\arcsin \sqrt x}{x^2-x+1} dx$$
Answer: $\frac{\pi^2}{6\sqrt 3}$
My Attempt:
The obvious substitution: $\arcsin \sqrt x=t$. This transforms my integral (say $I$) to:
$$I=\int_0^{\frac{\pi}{2}} \frac{t\cdot 2\sin t\cos t}{\sin^4-\sin^2+1}dt$$
Then the substitution $t\rightarrow \left(\frac{\pi}{2}+0\right)-t$. This leads to the denominator remaining the same.
Then, I added the "versions" of $I$ and on some simplifications, obtained:
$$I=\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\underbrace{\frac{\sin 2t}{1-\frac{\sin^2 2t}{4}}}_{f(t)}dt$$
Since $f(x)=f\left(\frac{\pi}{2}-x\right)$ we say:
$$I=\frac{\pi}{2}\int_0^{\frac{\pi}{4}}\frac{\sin 2t}{1-\frac{\sin^2 2t}{4}}dt$$
Now the substitution $2t=u$ transforms it to:
$$I=\frac{\pi}{4}\int_0^{\frac{\pi}{2}}\frac{\sin u}{1-\frac{\sin^2 u}{4}}du$$
Or,
$$I=\pi\int_0^{\frac{\pi}{2}}\frac{\sin u}{4-\sin^2 u}du$$
Then I tried to decompose it into partial fractions and integrate them individually, by converting them in terms of $\tan \frac u2$ but the answer I'm getting doesn't match the given one.
Thanks in advance!
| $$I=\pi\int_{0}^{\pi/2} \frac{\sin u}{4-\sin^2 u}= \pi \int_{0}^{\pi/2} \frac{\sin u}{3+\cos^2 u} du= \pi\int_{0}^{1} \frac{dv}{3+v^2}dv=\frac{\pi}{\sqrt{3}} \tan^{-1}(v/\sqrt{3})|_{0}^{1}= \frac{\pi^2}{6 \sqrt{3}}.$$
Lastly, we have used $v=\cos u$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$. Question: Find the minimum value of $x_1^2+x_2^2+x_3^2+x_4^2$ subject to $x_1+x_2+x_3+x_4=a$ and $x_1-x_2+x_3-x_4=b$.
My attempt: It can be easily seen that $x_1+x_3=\frac{a+b}{2}$ and $x_2+x_4=\frac{a-b}{2}$. Further, the expression $[x_1^2+x_2^2+x_3^2+x_4^2]$ can be written as $[(x_1+x_3)^2+(x_2+x_4)^2-2(x_1x_3+x_2x_4)].$ I'm having trouble eliminating $(x_1x_3+x_2x_4)$ from this expression. Failing to make any sense out of this, I manipulated the existing expressions to deduce $$x_1x_2+x_1x_4+x_2x_3+x_3x_4=\frac{a^2-b^2}{4}$$and $$(x_1^2+x_3^2)-(x_2^2+x_4^2)+2(x_1x_3-x_2x_4)=a\cdot b$$Beyond this, I cannot make sense of the expressions anymore. I have no idea how to proceed with simplifying the expressions further, and would appreciate hints in the same direction.
| Define $p=x_1+x_2$, $q=x_3+x_4$, $r=x_1-x_2$, $s=x_3-x_4$.
Restate the problem:
find the minimum of $\frac{p^{2}+q^{2}+r^{2}+s^{2}}{2}$ with constrain $p+q=a$ and $r+s=b$.
QM - AM inequality:
$\frac{p^{2}+q^{2}}{2}\geq\frac{(p+q)^{2}}{4}=\frac{a^{2}}{4}$
$\frac{r^{2}+s^{2}}{2}\geq\frac{(r^{2}+s^{2})^{2}}{4}=\frac{b^{2}}{4}$
$\frac{p^{2}+q^{2}+r^{2}+s^{2}}{2}\geq\frac{a^{2}+b^{2}}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3790537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Intersection of vectors to form a tetrahedron Suppose we have three unit vectors, namely $a,b,$ and $c,$ such that the angle between any of them is the acute angle $\theta.$ However, these vectors generate a tetrahedron with volume $\frac{1}{\sqrt{360}}.$ Find $$3\cos^2\theta - 2\cos^3\theta.$$
I've got little to no idea on how to start this problem due to the fact that it's a new concept to me. Can someone give me an overview of the basics and then a sketch of the problem for me to work on?
| Consider the following image
where the three angles in $V$ are $\theta$, develope the image in a plane, obtaining
Now we can observe that
\begin{align}
&\overline{AV}=\overline{BV}=\overline{CV}=1\\
&\overline{AB}=\overline{BC}=\overline{CA}=2\sin(\theta/2)\\
&\overline{HV}=\overline{KV}=\overline{LV}=\cos(\theta/2)\\
&\overline{AH}=\overline{BK}=\overline{CL}=\sqrt{3}\sin(\theta/2)\\
&\overline{H\Omega}=\overline{K\Omega}=\overline{L\Omega}=(\sqrt{3}/3)\sin(\theta/2)\\
\end{align}
then the height of the tetrahedron is
$$
h=\overline{V\Omega}=\sqrt{\overline{HV}^2-\overline{H\Omega}^2}=\sqrt{\cos^2(\theta/2)-\frac{1}{3}\sin^2(\theta/2)},
$$
while the base area is
$$
A_b=\frac{\sqrt{3}}{4}\overline{AB}^2=\sqrt{3}\sin^2(\theta/2).
$$
Finally, the volume is
$$
\mathscr{V}=\frac{1}{3}A_bh=\frac{\sqrt{3}}{3}\sin^2(\theta/2)\sqrt{\cos^2(\theta/2)-\frac{1}{3}\sin^2(\theta/2)}.
$$
Now observe that
\begin{align}
\mathscr{V}^2&=\frac{1}{3}\sin^4(\theta/2)\left(\cos^2(\theta/2)-\frac{1}{3}\sin^2(\theta/2)\right)=\\
&=\frac{1}{36}(1+2\cos\theta)(1-\cos\theta)^2=\\
&=\frac{1}{36}(1-3\cos^2\theta+2\cos^3\theta).
\end{align}
Now, from
$$
\mathscr{V}^2=\frac{1}{36}(1-3\cos^2\theta+2\cos^3\theta)=\frac{1}{360}
$$
we find
$$
3\cos^2\theta-2\cos^3\theta=\frac{9}{10}.
$$
| {
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"url": "https://math.stackexchange.com/questions/3791786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How does $x^2-8x+17=0$ have nonreal solutions?
The solutions of $x^2-8x+17=0$ are $4 + i$ and $4 - i$.
Well, I calculated and the results are different.
$$\begin{align}
x^2-8x+17 &= 0 \\
x^2-8x &=17 \\
x(x-8) &= 17
\end{align}$$
So the roots are $x=17$ or $x=17+8=25$.
Why $i$ comes from the problem? Could you please explain about it?
| Firstly, your rearrangement should have given $x(x-8)=-17$. Secondly, $x(x-8)=c\implies x=c\lor x-8=c$ only works for $c=0$, because non-zero $c$ have other factorizations. The correct observation is that $(x-4)^2=x^2-8x+16=-1=i^2$, so $x-4=\pm i$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
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Positive reals satisfy $ \sum_{i=1}^{24} x_i = 1 $, determine maximum of following quantity So, positive reals satisfy the following
$$ \sum_{i=1}^{24} x_i = 1 $$
And I need to find maximum of the following quantity.
$$ \left( \sum_{i=1}^{24} \sqrt{x_i}\right) \left(\sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i} } \right) $$
Now, using Cauchy Schwarz inequality, I got
$$ \left( \sum_{i=1}^{24} \sqrt{x_i}\right)^2 \leqslant \underbrace{(1+1+\cdots + 1)}_{\text{24 times}} \left( \sum_{i=1}^{24} x_i \right) $$
This leads to
$$ \left( \sum_{i=1}^{24} \sqrt{x_i}\right) \leqslant \sqrt{24} $$
I am stuck with other part. I can get the minimum of the following using similar technique.
$$ \left(\sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i} } \right) $$
But I need to have maximum of this quantity, so that I can combine the two. Any hints will help.
| We can bound the second sum as follows. Using Cauchy-Schwarz inequality, we have the following.
$$ \left( \sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i}} \right)^2 \leqslant \left( \sum_{i=1}^{24} \frac{1}{(1+x_i)} \right)\underbrace{(1+1+\cdots +1)}_{\text{24 times}} $$
$$ \left( \sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i}} \right)^2 \leqslant 24 \left( \sum_{i=1}^{24} \frac{1}{(1+x_i)} \right) \, \cdots \cdots \cdots(1)
$$
Now, I will use Hölder's inequality.
$$ \left( \sum_{i=1}^{24} \frac{1}{(1+x_i)} \right)^{1/2} \left( \sum_{i=1}^{24} (1+x_i) \right)^{1/2} \leqslant \left[ \sum_{i=1}^{24} \left(\frac{1}{\sqrt{1+x_i}}\right) \left(\sqrt{1+x_i}\right) \right] $$
$$ \left( \sum_{i=1}^{24} \frac{1}{(1+x_i)} \right)^{1/2} \sqrt{25} \leqslant 24 $$
$$ \left( \sum_{i=1}^{24} \frac{1}{(1+x_i)} \right) \leqslant \frac{24^2}{25} $$
$$ 24 \left( \sum_{i=1}^{24} \frac{1}{(1+x_i)} \right) \leqslant \frac{24^3}{25} $$
So, combining with equation $(1)$, I get,
$$ \left( \sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i}} \right)^2 \leqslant \frac{24^3}{25} $$
$$ \left( \sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i}} \right) \leqslant \frac{24^{3/2}}{5} $$
Finally, combining the two sums, I get
$$ \left( \sum_{i=1}^{24} \sqrt{x_i} \right) \left( \sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i}} \right) \leqslant \sqrt{24} \,\frac{24^{3/2}}{5} $$
$$ \left( \sum_{i=1}^{24} \sqrt{x_i} \right) \left( \sum_{i=1}^{24} \frac{1}{\sqrt{1+x_i}} \right) \leqslant \frac{24^{2}}{5} $$
Hope that helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to switch rows in matrix L when decomposing matrix A into PA = LU?
Find the permutation matrix $P$, the lower triangular matrix $L$ and the upper triangular matrix $U$ such that $$ PA=LU $$
Given $$ A= \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ -2 & -3 & -4 & -5 & -6 & -7 \\ 3 & 7 & 11 & 16 & 21 & 27 \\ -4 & -5 & -5 & -5 & -5 & -5 \end{pmatrix}$$
I came this far
$$ U = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 0 & 1 & 2 & 3 & 4 & 5 \\ 0 & 0 & 0 & 1 & 2 & 4 \\ 0 & 0 & 1 & 2 & 3 & 4 \end{pmatrix}$$
and
$$ L= \begin{pmatrix} 1 & 0 & 0 & 0 \\ -2 & 1 & 0 & 0 \\ 3 & 1 & 1 & 0 \\ -4 & 3 & 0 & 1 \end{pmatrix}$$
The last step I have to make is to change the fourth row with the third one but I do not know exactly how to change the entries in the lower triangular matrix L. Can anyone explain what exactly I have to switch in L?
| Using row reduction, we arrived at
$$U = \begin{pmatrix} 1 & 2 & 3 & 4 & 5 & 6 \\ 0 & 1 & 2 & 3 & 4 & 5 \\ 0 & 0 & 0 & 1 & 2 & 4 \\ 0 & 0 & 1 & 2 & 3 & 4 \end{pmatrix}$$
As an alternate approach to the excellent write up by @Ian (+1), you could have reversed the row reduction steps, including the swap, as $$A = E_1^{-1}E_2^{-1}E_3^{-1}E_4^{-1}U$$
This results in
$$L = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -2 & 1 & 0 & 0 \\ 3 & 1 &0 & 1 \\ -4 & 3 & 1 & 0 \end{pmatrix}$$
We see that $L$ is not lower triangular and we just need to swap row three and four, resulting in
$$L = \begin{pmatrix} 1 & 0 & 0 & 0 \\ -2 & 1 & 0 & 0 \\ -4 & 3 & 1 & 0 \\3 & 1 &0 & 1\end{pmatrix}$$
That swap requires a permutation matrix
$$P = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 &0 & 1 \\ 0 & 0 & 1 &0\end{pmatrix}$$
Now we can verify
$$PA = LU$$
You can also verify that $$A = PLU = P^T LU = P^{-1} LU$$
See, for example, How can LU factorization be used in non-square matrix?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3796441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit without de l'Hôpital: $\lim _{x\to 0\color{red}{\boldsymbol -}}\left(1+x^3\right)^{\frac{1}{\left(x^2+1\right)^4-1}}$ I have this limit of this form
$$f(x)^{g(x)}=e^{g(x)\ln(f(x))}$$
$$\lim _{x\to 0\color{red}{\boldsymbol -}}\left(1+x^3\right)^{1/\left((x^2+1)^4-1\right)}$$
In our case I can write in the exponent:
$${g(x)\ln(f(x))}=\frac{\ln(f(x))}{\frac1{g(x)}}$$
and I have an indeterminate form $(0/0)$ and I can apply de l'Hôpital rule. Right now I just thought to write
$$(1+x^3)=\left(1+\frac{1}{\frac1{x^3}}\right)$$ and I call $x^3=t$ but I think to obtain the exponent too long and it will be more complicated.
| We can use that
$$\large{\left(1+x^3\right)^{\frac{1}{\left(x^2+1\right)^4-1}}=\left[\left(1+x^3\right)^{\frac1{x^3}}\right]^{\frac{x^3}{\left(x^2+1\right)^4-1}}}\to e^0=1$$
indeed
$$\left(x^2+1\right)^4=1+4x^2+O(x^4) \implies \frac{x^3}{\left(x^2+1\right)^4-1}= \frac{x^3}{4x^2+O(x^4)}=\frac{x}{4+O(x^2)}\to 0$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Find all the $\bf{x}\in\mathbb{R}^3$, which has a property that $\displaystyle \lim_{ n \to \infty } A^n\bf{x} = \bf{x}$ $A$ is a matrix with elements
$$\begin{pmatrix}
\frac{2}{3} & -\frac{2}{3}& -\frac{1}{3} \\
-\frac{2}{3}& 1 & \frac{2}{3} \\
\frac{4}{3} & 0 & -\frac{1}{3}
\end{pmatrix}$$
Find all the $\bf{x}\in\mathbb{R}^3$, which has a property that $$\displaystyle \lim_{ n \to \infty } A^n\bf{x} = \bf{x}$$
A's eigenvalues are $-\frac{1}{3}$,1 and $\frac{2}{3}$ and eigenvectors are $\left(
\begin{array}{c}
0 \\
-\frac{1}{2} \\
1
\end{array}
\right),\left(
\begin{array}{c}
1 \\
-1 \\
1
\end{array}
\right),\left(
\begin{array}{c}
\frac{3}{4} \\
-\frac{1}{2} \\
1
\end{array}
\right)$ for each.
I tried to get $A^n$ at first then got $$\begin{pmatrix}
(\frac{2}{3})^n & -2+\frac{2^{n+1}}{3}& 0 \\
-\frac{2+2^{n+1}}{3^{n+1}}& \frac{1-2^{n+2}}{3^{n+1}}-2 & \frac{2}{3^n} \\
\frac{4+2^{n+2}}{3^{n+1}} & \frac{2-2^{n+3}}{3^{n+1}} & -\frac{1}{3^n}
\end{pmatrix}$$
Then I tried find $$\displaystyle \lim_{ n \to \infty }(\frac{2}{3})^nx_1+(-2+\frac{2^{n+1}}{3})x_2=x_1 \\\displaystyle \lim_{ n \to \infty }-\frac{2+2^{n+1}}{3^{n+1}}x_1+(\frac{1-2^{n+2}}{3^{n+1}}-2)x_2+\frac{2}{3^n}x_3=x_2\\\ \displaystyle \lim_{ n \to \infty }\frac{4+2^{n+2}}{3^{n+1}}x_1+\frac{2-2^{n+3}}{3^{n+1}}x_2-\frac{1}{3^n}x_3=x_3$$
But I could not lead meaningful conclusion from these equations but $(0,0,0)$.
Where did I get wrong?
| The answer of @AryamanMaithani explains very well how you can solve this question, but it uses the fact that the matrix is diagonalizable. I just would like to add a small precision, that you can prove directly and more easily that :
for every matrix $A \in \mathcal{M}_n(\mathbb{R})$ (diagonalizable or not),
$$\forall x \in \mathbb{R}^n, \text{ } \lim_{n\rightarrow +\infty} A^nx =x \text{ } \Longleftrightarrow \text{ }Ax=x$$
(so the set $\lbrace x \in \mathbb{R}^n \text{ }|\text{ } \lim_{n\rightarrow +\infty} A^nx =x \rbrace$ is always equal to the eigenspace associated to the eigenvalue $1$)
Indeed, if $A^nx \rightarrow x$, then you have $A^{n+1}x \rightarrow Ax$ and $A^{n+1}x \rightarrow x$, so $Ax=x$, so $x$ is an eigenvector associated to the eigenvalue $1$. Conversely, if $Ax=x$, it is immediate that $A^nx=x$ for all $n$, and then $A^nx \rightarrow x$ when $n$ tends to $+\infty$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solving the system $3-(y+1)^2 = \sqrt{x-y}$, $\;x+8y = \sqrt{x-y-9}$
How to solve these equations?
$$\begin{cases}
3-(y+1)^2 = \sqrt{x-y}\\
x+8y = \sqrt{x-y-9}
\end{cases}$$
I've tried solving this using the substitution and elimination methods without any success. I also tried plotting these equations and I got $x = 8$, $y = -1$.
Can someone show me the steps required to solve this?
Thanks in advance.
| Observing the right-hand sides, we get:
$$(x+8y)^2 + 9 = (3 - (y+1)^2 )^2$$
$$(x+8y)^2 -(3 - (y+1)^2 )^2 = -9$$
and now using the difference of two squares:
$$(x+8y+3-(y+1)^2)(x+8y-3 + (y+1)^2 )= -9$$
If there is a clean solution where $x, y$ are integers, then the two brackets must be integers themselves. There are only a few possibilities: which are: $$(\text{left}, \text{right}) = (-1, 9), (1, -9), (-3, 3), (3, -3), (-9, 1), (9, -1).$$
Some of these solutions are extraneous or have non-integer solutions (that can be expressed in radicals). With the pair $(3, -3)$, you get an integer solution $(x,y) = (1,8)$, and now substitute into the original equations to verify them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integrating to derive relation between binomial co-efficient
If $C_o,C_1...$ are the binomial coefficents in the expansion of $(1+x)^n$ and $$\sum_{r=0}^{n} (-1)^r \binom{n}{r} \frac{1}{(r+1)^2} = k \sum_{r=0}^{n} \frac{1}{r+1}$$
Find a k such that the above equation is satisfied
My attempt:
$$ (1+x)^n = \sum_{k=0}^{k=n} \binom{n}{k} x^k$$
integrate both sides
$$ \frac{ (1+x)^{n+1} }{n+1} =\sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{k+1} +C$$
$$ x= 0 $$
$$ \implies \frac{-n}{n+1} = C$$
Divide both sides by 'x' and integrate
$$ \int \frac{ (1+x)^{n+1} }{ (x) n+1} dx = \sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{(k+1)^2} - \frac{n}{n+1} \ln(x)+ C'$$
Or,
$$ \int \frac{ (1+x)^{n+1} }{ (x) n+1} dx - \frac{n}{n+1} \ln(x)+ C' = \sum_{k=0}^{n} \binom{n}{k} \frac{x^{k+1}}{(k+1)^2}$$
Not so sure what do here, like what to put as bounds. Slightly concerned I may have to evaluate a negative logarithm
| Here is my approach. Let denote that $H_n = \sum_{k=1}^{n}\frac{1}{k}$ is the n-th Harmonic series. Then it is easy to observe that:
$$H_n = \sum_{k=1}^{n}\frac{1}{k} = \sum_{k=1}^{n}\int_{0}^{1} x^{k-1}dx = \int_{0}^{1}\sum_{k=1}^{n} x^{k-1}dx = \int_{0}^{1} \frac{1-x^n}{1-x}~dx$$
For your approach above, instead of using $(1+x)^n$, why don't you use $(1-x)^n$?
$$(1-x)^n = \sum_{k=0}^{n} \binom{n}{k}(-1)^r x^r \Rightarrow \int_{0}^{t} (1-x)^n ~dx = \int_{0}^{t}\sum_{k=0}^{n} \binom{n}{k}(-1)^r x^r dx$$$$\Rightarrow \frac{1- (1-t)^{n+1}}{t(n+1)}= \sum_{k=0}^{n} \binom{n}{k}\frac{(-1)^r t^{r}}{r+1}$$
Now, notice that:
$$LHS=\sum_{r=0}^{n} \binom{n}{r}\frac{(-1)^r}{(r+1)^2}=\sum_{r=0}^{n} \binom{n}{r}\frac{(-1)^rt^{r+1}}{(r+1)^2}\Bigg|_0^1=\int_{0}^{1}\sum_{r=0}^{n} \binom{n}{r}\frac{(-1)^r t^{r}}{r+1}~dt$$$$=\int_{0}^{1} \frac{1- (1-t)^{n+1}}{t(n+1)} ~ dt$$
Here we will move to the RHS:
$$RHS=k \sum_{r=0}^{n} \frac{1}{r+1} = k\cdot H_{n+1} = k \cdot \int_{0}^{1} \frac{1-t^{n+1}}{1-t}~dt= k \int_{0}^{1} \frac{1-(1-t)^{n+1}}{t}~dt \cdot $$
By comparing $LHS$ with $RHS$, we obtain that $k=\frac{1}{n+1}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $r$ such that the equation $x^4+x^2(1-2r)-2x+1=0$ has only one real solution I'm looking for $r$ such that the equation $$x^4+x^2(1-2r)-2x+1=0$$
has only one real solution. I've made some attempts to this problem, but it seems that I even didn't get close to the solution. The approximation for $r$ is 0.3347498
Is it possible to find analytic solution for $r$ and if yes then how?
Thanks for all the help.
| Has noted, the polynomial should have a real double root and two conjugate complex roots, so its general form is
$$
(x-a)^2[(x+p)^2+q^2]
$$
Expanding this, and comparing with our initial polynomial we have the relations
\begin{alignat}{2}
&-&&a^2 p^2-a^2 q^2+1=0,\\
&-&&2 a^2 p+2 a p^2+2 a q^2-2=0,\\
&-&&a^2+4 a p-p^2-q^2-2 r+1=0,\\
&&&2 a-2 p=0
\end{alignat}
From forth equation we have $p=a$, so the system reduce to
\begin{alignat}{2}
&-&&a^4-a^2 q^2+1=0,\\
&&&2 a q^2-2=0,\\
&&&2 a^2-q^2-2 r+1=0
\end{alignat}
From the second we have $q^2=1/a>0$
so the system reduces to
\begin{alignat}{2}
&-&&a^4-a+1=0,\\
&&&2 a^2-\frac{1}{a}-2 r+1=0
\end{alignat}
So the problem is reduced to look for a positive solution to
$$
-a^4-a+1=0
$$
and once found, we have, from the second
$$
r=\frac{2 a^3+a-1}{2 a}
$$
The only positive solution for $a$ is
$$
a=0.7244919590005157
$$
so that
$$
r=0.3347498141075979.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3803011",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Inequality dealing with area and sides of a triangle I have to prove that:
$$\frac{a^6}{b^2+c^2}+\frac{b^6}{a^2+c^2}+\frac{c^6}{a^2+b^2}\geq 8S^2$$
$a$, $b$, $c$ are the sides of a triangle, and $S$ its area.
I tried using the Sine Theorem, different area formulas, and AM-GM but didn't get anywhere.
Thanks for your help!
| From How do I show that $\sum_{cyc} \frac {a^6}{b^2 + c^2} \ge \frac {abc(a + b + c)}2?$, we can show that
$$\frac {a^6}{b^2 + c^2} + \frac {b^6}{c^2 + a^2} + \frac {c^6}{a^2 + b^2} \ge \frac {abc(a + b + c)}2.$$
For your question, use Heron's formula:
$$S=\sqrt{s(s-a)(s-b)(s-c)}$$
$$s=\frac{a+b+c}{2}$$
So the right hand side of the inequality $8S^2$ can be turned into
$$8\left(\frac{a+b+c}{2}\right)\left(\frac{b+c-a}{2}\right)\left(\frac{a+c-b}{2}\right)\left(\frac{a+b-c}{2}\right)$$
which is equal to
$$\frac{1}{2}(a+b+c)(b+c-a)(a+c-b)(a+b-c)$$
Now, what's left is to prove
$$\frac {abc(a + b + c)}2\geq\frac{1}{2}(a+b+c)(b+c-a)(a+c-b)(a+b-c)$$
Or
$$abc\geq(b+c-a)(a+c-b)(a+b-c)$$,
which is true if we let $a\geq b\geq c$
So we are left with
$$\frac {a^6}{b^2 + c^2} + \frac {b^6}{c^2 + a^2} + \frac {c^6}{a^2 + b^2} \ge \frac {abc(a + b + c)}2\geq 8S^2$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 1,
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Inclusion-Exclusion Problem with equation Problem:
$x_1+x_2+x_3+x_4 = 22$. How many solutions are there if the $x_i$s are nonnegative integers and $1<x_1<7$, $3\leq x_2 \leq 5$, $x_3 \leq 7$, $1<x_4\leq 13$?
My work:
$y_1+y_2+y_3+y_4 = 15$
$y_1<5, y_2<3, y_3<8, y_4<12$
$x_1 = y_1+2, x_2=y_2+3, x_3=y_3,x_4=y_4+2$
Ignoring upper-bounds we have $C(15+4-1,15)=816$
Solutions when
$y_1 \geq 5, y_2\geq 0, y_3\geq 0, y_4 \geq 0$
$(y_1-5) + y_2 + y_3 + y_4 = 10$
$C(10+4-1,10) = 286$
Solutions when
$y_1 \geq 0, y_2 \geq 3, y_3 \geq 0, y_4 \geq 0$
$y_1 + y_2-3 + y_3 + y_4 = 12$
$C(12+4-1,12) = 455$
Solutions when $y_1 \geq 0, y_2 \geq 0, y_3 \geq 8,y_4 \geq 0$
$y_1 + y_2 + y_3-8 + y_4 = 7$
$C(7+4-1,7) = 120$
Solutions when $y_1 \geq 0, y_2 \geq 0, y_3 \geq 0, y_4 \geq 12$
$y_1 + y_2 + y_3 + y_4-12 = 3$
$C(3+4-1,3) = 20$
$816-286-455-120-20 = -65$
So I started some of the steps above. I am wondering if I am doing this correctly as well as what the next step is on how to calculate what I am under-counting because $-65$ is obviously not he answer.
| You have correctly reduced the problem to finding the number of solutions of the equation
$$y_1 + y_2 + y_3 + y_4 = 15 \tag{1}$$
subject to the restrictions $y_1 < 5, y_2 < 3, y_3 < 8, y_4 < 12$.
Let $A_1$ denote the set of outcomes in which $y_1 \geq 5$, $A_2$ denote the set of outcomes in which $y_2 \geq 3$, $A_3$ denote the set of outcomes in which $y_3 \geq 8$, and $A_4$ denote the set of outcomes in which $y_4 \geq 12$. By the Inclusion-Exclusion Principle, the number of outcomes in which none of the restrictions is violated is found by subtracting the number of solutions in which at least one of these restrictions is violated from the number of solutions of equation 1.
You correctly found that the number of solutions of equation 1 is
$$\binom{15 + 4 - 1}{4 - 1} = \binom{18}{3} = \binom{18}{15}$$
and that
\begin{align*}
|A_1| & = \binom{10 + 4 - 1}{4 - 1} = \binom{13}{3} = \binom{13}{10}\\
|A_2| & = \binom{12 + 4 - 1}{4 - 1} = \binom{15}{3} = \binom{15}{12}\\
|A_3| & = \binom{7 + 4 - 1}{4 - 1} = \binom{10}{3} = \binom{10}{7}\\
|A_4| & = \binom{3 + 4 - 1}{4 - 1} = \binom{6}{3}
\end{align*}
The reason you obtained a negative answer is that you have subtracted each case in which two restrictions are violated twice, once for each way you designated one of the restrictions as the restriction that is being violated. We only want to subtract such cases once, so we must add them to the total. In fact, by the Inclusion-Exclusion Principle, the number of solutions in which at least one condition is violated is
\begin{align*}
& |A_1 \cup A_2 \cup A_3 \cup A_4|\\
& \quad = |A_1| + |A_2| + |A_3| + |A_4|\\
& \qquad - |A_1 \cap A_2| - |A_1 \cap A_3| - |A_1 \cap A_4| - |A_2 \cap A_3| - |A_2 \cap A_4| - |A_3 \cap A_4|\\
& \quad \qquad + |A_1 \cap A_2 \cap A_3| + |A_1 \cap A_2 \cap A_4| + |A_1 \cap A_3 \cap A_4| + |A_2 \cap A_3 \cap A_4|\\
& \qquad \qquad - |A_1 \cap A_2 \cap A_3 \cap A_4|
\end{align*}
Notice that many of these terms are equal to zero. For instance, it is not possible for $y_1 \geq 5$ and $y_2 \geq 12$ since $5 + 12 > 15$.
Let's calculate $|A_1 \cap A_2|$. I will leave the calculations of the remaining terms to you.
$|A_1 \cap A_2|$: Then $y_1 \geq 5$ and $y_2 \geq 3$. Let $y_1' = y_1 - 5$ and $y_2' = y_2 - 3$. Then $y_1'$ and $y_2'$ are nonnegative integers. Substituting $y_1' + 4$ for $y_1$ and $y_2' + 3$ for $y_2$ in equation 1 yields
\begin{align*}
y_1' + 5 + y_2' + 3 + y_3 + y_4 & = 15\\
y_1' + y_2' + y_3 + y_4 & = 7 \tag{2}
\end{align*}
Equation 2 is an equation in the nonnegative integers with
$$\binom{7 + 4 - 1}{4 - 1} = \binom{10}{3} = \binom{10}{7}$$
solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Show that an inequality holds for all positive real numbers $a, b$ such that $ab \geq 1$ I found the following question on a past international competition:
Show that:
$\left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right) \geq 16$
for all positive real numbers $a, b$ such that $ab\geq 1$.
I solved it in the following way:
$\left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)$
$\displaystyle =ab+2a^2+\frac{2a}{b+1}+2b^2+4ab+\frac{4b}{b+1}+\frac{2b}{a+1}+\frac{4a}{a+1}+\frac{4}{(a+1)(b+1)}$
$\displaystyle \ge 5ab+2(a^2+b^2)+\frac{2(a+2b)}{b+1}+\frac{2(b+2a)}{a+1}+\frac{4}{(a+1)(b+1)}$
$\displaystyle \ge 5+2(a^2+b^2)+\frac{2(a+2b)}{b+1}+\frac{2(b+2a)}{a+1}+\frac{4}{(a+1)(b+1)}$
$\displaystyle \ge 9+\frac{2(a^2+2ab+a+2b)+2(b^2+b+2ab+2a)+4}{(a+1)(b+1)}$ (from AM-GM we have that $a^2+b^2\ge 2ab \ge 2$)
$\displaystyle \ge 9+4(a+1)(b+1)+\frac{2b^2+2a^2+4ab+2a+2b}{(a+1)(b+1)}$
$\displaystyle \ge 13+\frac{2b^2+2a^2+4ab+2a+2b}{(a+1)(b+1)}$
However we have that $a^2+b^2\ge (a+b)*\sqrt{a^2b^2} \ge a+b$ (this is true for the well known inequality that $x1^2+x2^2+...+xn^2\ge (x1+x2+...+xn)*\sqrt[n]{x1x2...xn}$), $a^2+b^2\ge 2ab\ge 2$. Hence:
$2b^2+2a^2+ab\ge a+b+3$ so $2b^2+2a^2+4ab+2a+2b\ge 3ab+3a+3b+3$
So, we have that $\displaystyle \left(a+2b+\frac{2}{a+1}\right)\left(b+2a+\frac{2}{b+1}\right)\ge 13+\frac{3ab+3a+3b+3}{(a+1)(b+1)}\ge 13+\frac{3(a+1)(b+1)}{(a+1)(b+1)} \ge 16$
I believe that my solution is correct, however I'm not completely certain, so could you please have a look at it and also share if there is an easier and simpler way of solving the problem?
| By AM-GM and C-S we obtain:$$\prod_{cyc}\left(a+2b+\frac{2}{a+1}\right)=\prod_{cyc}\left(\frac{a+1}{2}+\frac{2}{a+1}+2b+\frac{a}{2}-\frac{1}{2}\right)\geq$$
$$\geq\prod_{cyc}\left(2+2b+\frac{a}{2}-\frac{1}{2}\right)=\prod_{cyc}\left(2b+\frac{a}{2}+\frac{3}{2}\right)\geq$$
$$\geq\left(2\sqrt{ab}+\frac{1}{2}\sqrt{ab}+\frac{3}{2}\right)^2\geq\left(2+\frac{1}{2}+\frac{3}{2}\right)^2=16.$$
| {
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"url": "https://math.stackexchange.com/questions/3806429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the range of $y=\frac{x+5}{\sqrt{64-25x^2}}$? How to find the range of $y=\frac{x+5}{\sqrt{64-25x^2}}$?
I am sure that the range is $y \ge 0$ but I don't know the basic process of finding the range without using calculus.
Any comments or suggestion sill be much appreciated.
| Since for $64-25x^2>0$ we have $x+5>0$, by AM-GM we obtain:$$\frac{x+5}{\sqrt{64-25x^2}}=\frac{2\sqrt{17\cdot33}(x+5)}{2\sqrt{17(8-5x)\cdot33(8+5x)}}\geq$$
$$\geq\frac{2\sqrt{561}(x+5)}{17(8-5x)+33(8+5x)}=\frac{\sqrt{561}}{40}.$$
The equality occurs for $$17(8-5x)=33(8+5x)$$ or $$x=-\frac{64}{125},$$ which says that we got a minimal value.
Also, since $$\lim_{x\rightarrow\frac{8}{5}^-}\frac{x+5}{\sqrt{64-25x^2}}=+\infty$$ and we work with a continuous function on $\left(-\frac{8}{5},\frac{8}{5}\right),$ we got the range:
$$\left[\frac{\sqrt{561}}{40},+\infty\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Eigenvalues of $A^{2018}$ Find the eigenvalues and eigenvectors of $A^{2018}$.
$$
A=\begin{bmatrix} 1 & 3 & 4\\ 3 & 1 & 4\\ 0 & 0 & 4\end{bmatrix}
$$
My solution:
First, by substracting first row times three from second row we get:
$$
A\approx \begin{bmatrix} 1 & 3 & 4\\ 0 & -8 & -8\\ 0 & 0 & 4\end{bmatrix}
$$
We achieved the upper triangular matrix so the characteristic polynomial is:
$$
\chi_{A^{2018}}(\lambda)=det (\begin{bmatrix} 1 & 3 & 4\\ 0 & -8 & -8\\ 0 & 0 & 4\end{bmatrix}^{2018}-\lambda I)=(1^{2018}-\lambda)((-8)^{2018}-\lambda)(4^{2018}-\lambda)
$$
Therefore the set of eigevalues is $\{1,4^{2018},8^{2018},\}$.
Please verify if this the correct solution, and in case it isn't, help me find the correct one.
| Hint: The characteristic polynomial is $c_A(x)=(x-4)((x-1)^2-9)=(x-4)^2(x+2)$.
You can check that the Jordan normal form is $B=\begin{pmatrix} 4&1&0\\0&4&0\\0&0&-2\end{pmatrix}$.
Then $A^n$ is similar to $B^n$.
But $B^n=\begin{pmatrix}4^n&4+2\cdot4^n&0\\0&4^n&0\\0&0&(-2)^n\end{pmatrix}$.
Hence we get $\{4^n,(-2)^n\}$ as the eigenvalues.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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What is the probability that the total score after throwing darts is divisible by $3$.
To play a game of darts Michael throws three darts at the dart board shown. The number of points $(1,$ $5$ or $10)$ for each of the three regions is indicated. His score is the sum of the points for the three darts. If the radii of the three concentric circles are $1,$ $2$ and $3$ units, and each dart Michael throws hits this dart board at random, what is the probability that his score is evenly divisible by $3?$ Express your answer as a common fraction.
After taking the values modulo $3$, we have $1, 2, 1$. I am pretty sure that the only way we can get divisible by $3$ in this problem is if we have modulos $1, 1, 1$ or $2, 2, 2$ for the darts. This means that the probability is ${\left(\dfrac23\right)}^3+{\left(\dfrac13\right)}^3=\dfrac13$.
I feel as if I am missing something, or am I correct?
Thanks!
EDIT: "At random" means that the likelihood of a dart landing in a region is the total area of that region divided by the total area of the dart-board.
| The areas $A_1,A_5,A_{10}$ areas are
$$\begin{cases}
A_{10}: &\pi r^2 = pi\\
A_5: &4\pi-\pi=3\pi\\
A_1: &9\pi-4\pi=5\pi
\end{cases}$$
Out of $(1+3+5)\pi=9\pi$, the probabilities to get:
$$\begin{cases}
P(10) &= 1/9\\
P(5) &= 3/9\\
P(1) &= 5/9
\end{cases}$$
Any of the $27$ possible configurations $(a,b,c)$ has the probability
$$\dfrac{p}{27\cdot 3^3 = 3^6}$$
To reach a divisible by 3 score, there is
*
*only one way to make either $(10,10,10)$, $(5,5,5)$ or $(1,1,1)$ of respective probabilities $$\dfrac1{3^6}, \dfrac{27}{3^6}, \dfrac{125}{3^6}$$which sum is $\dfrac{153}{3^6}$
*three ways to make each of $(1,1,10)$ or $(1,10,10)$,
$$3\dfrac{25+5}{3^6}$$
Summing the whole gives
$$\bbox[5px,border:2px solid #ca9]{\dfrac{243}{3^6}=\dfrac{3^5}{3^6}=\dfrac13}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this alternative proof of the inequality $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\geq\frac{3}{2}$ correct? Prove that for all positive real numbers:
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}$$
This is same as this question but a different approach is used there whereas I want to verify my approach to this problem.
My Approach:
$$\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}=\Big(\dfrac{a}{b+c}+1\Big)+\Big(\dfrac{b}{c+a}+1\Big)+\Big(\dfrac{c}{a+b}+1\Big)-3$$
$$=(a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]-3$$
By AM-HM inequality:
$$\dfrac{3}{\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}}\leq\dfrac{2(a+b+c)}{3}\Rightarrow (a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]\geq \dfrac{9}{2}$$
$$(a+b+c)\Big[\dfrac{1}{b+c}+\dfrac{1}{c+a}+\dfrac{1}{a+b}\Big]-3\geq \dfrac{3}{2}$$
$\therefore \dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\geq\dfrac{3}{2}\space \forall\ a,b,c\in \mathbb R$ and $a,b,c>0$
Please check this approach and provide suggestions. Also please provide alternative solutions if available.
THANKS
| Solution by Buffalo Way method.
Let $a=\min\{a,b,c\},$ $b=a+u$ and $c=a+v$.
Thus, $$2\prod_{cyc}(a+b)\left(\sum_{cyc}\frac{a}{b+c}-\frac{3}{2}\right)=4(u^2-uv+v^2)a+(u+v)(2u^2-3uv+2v^2)\geq0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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SOS proof for $\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$ I need an SOS(sum of squares) proof for $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
if $a,b,c>0$
I already have a am-gm proof but is there a way to use SOS.
Am-gm proof :
$\frac{a^3}{bc}+b+c\ge 3a$ .....by(AM-GM ineq.)
thus $$\sum \frac{a^3}{bc}+2\sum a \ge 3\sum a$$
or $$\sum_{cyc}\frac{a^3}{bc}\ge a+b+c$$
| Proof 1. We have
$$\sum \left(\frac{a^3}{bc}-a\right) = \sum \left(\frac{a^2}{b} -2a + b \right) + \left (\frac{a^3}{4bc}+\frac{3b^3}{4ca}-\frac{b^2}{c} \right )$$
$$= \sum \frac{(a-b)^2}{b}+\frac{(a^2+2ab+3b^2)(a-b)^2}{4abc} \geqslant 0.$$
Proof 2. We write inequality as
$$a^4+b^4+c^4-abc(a+b+c) \geqslant 0.$$
We have
$$a^4+b^4+c^4 - a^2b^2-b^2b^2-b^2c^2 = \sum \frac{(a-b)^2(a+b)^2}{2},$$
and
$$a^2b^2+b^2c^2+c^2a^2-abc(a+b+c) = \sum \frac{c^2(a-b)^2}{2}.$$
Thefore inequality equivalent to
$$\sum \frac{(a-b)^2[(a+b)^2+c^2]}{2} \geqslant 0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Need explanation on a problem involving natural function Consider a positive integer $n$ and the function
$f:\mathbb{N}\to \mathbb{N}$ ($\mathbb N$ includes $0$) by
$$f(x) = \begin{cases} \frac{x}{2}
& \text{if } x \text{ is even} \\ \frac{x-1}{2} + 2^{n-1} & \text{if } x \text{ is odd} \end{cases} $$
Determine the set
$$ A = \{ x\in \mathbb{N} \mid \underbrace{\left( f\circ f\circ ....\circ f \right)}_{n\ f\text{'s}}\left( x \right)=x \}. $$
(Romania NMO 2013)
The solution starts by stating that $f(x)<x, \quad\forall x\ge 2^n-1$. This was easy enough to understand. However, they continue by saying this implies that $A\subset\{0,1,\dots,2^n-1\}$. Why is that?
Please help me understand! Thanks in advance!
| Hint:
Letting all divisions be integer and splitting the cases after the low order bits of $x$, we can expand the function iterates for, say, $n=4$, and the pattern becomes obvious.
$$f(x)=\begin{cases}\frac x2&\text{ if } x \text{ is even}\\\frac{x}2+8&\text{ if } x \text{ is odd}\end{cases}$$
$$f(f(x))=\begin{cases}\frac x4&\\\frac{x}4+4\\\frac{x}4+8\\\frac{x}4+12\end{cases}$$
$$f(f(f(x)))=\begin{cases}
\frac x8&\\\frac{x}8+2\\\frac{x}8+4\\\frac{x}8+6\\
\frac x8+8&\\\frac{x}8+10\\\frac{x}8+12\\\frac{x}8+14\end{cases}$$
$$f(f(f(f(x))))=\begin{cases}
\frac x{16}&\\\frac{x}{16}+1\\\frac{x}{16}+2\\\frac{x}{16}+3\\
\frac x{16}+4&\\\frac{x}{16}+5\\\frac{x}{16}+6\\\frac{x}{16}+7\\
\frac x{16}+8&\\\frac{x}{16}+9\\\frac{x}{16}+10\\\frac{x}{16}+11\\
\frac x{16}+12&\\\frac{x}{16}+13\\\frac{x}{16}+14\\\frac{x}{16}+15\end{cases}$$
| {
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"url": "https://math.stackexchange.com/questions/3810944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A hint on finding all solutions of $S = \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$? I've been working at this for awhile but I haven't been able to figure out the right approach. The question is to find all values of $S = \frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$ for $a,b,c,d > 0$. Anyone have a hint (no solutions, please) at a more principled way to approach the problem?
I've been just throwing stuff at the wall and trying to find something that sticks.
What I've tried so far:
Plugging in some quick values. Putting $a=b=c=d=1$ yields $S = 4/3$. Taking $a=c=1, b=d=2$ yields $7/5$ so $4/3$ is not the only answer. Note: if $(a,b,c,d)$ yields $k$ then $(sa,sb,sc,sd)$ yields $k$ for any $s > 0$.
If I fix $a=b=c=1$ then $S = 4/3$ regardless of the value of $d$.
If I fix $a=c, b=d$, I can write $b = ka$, which gives $\frac{2}{2k+1} + \frac{2k}{k+2} = S$. This is a quadratic in $k$, with positive discriminant for any value of $S$. But $k$ is not necessarily positive, so I can't claim that all values of $S$ are valid.
Getting some quick bounds: $S > \frac{a}{a+b+c+d} + \frac{b}{a+b+c+d} + \frac{c}{a+b+c+d} + \frac{d}{a+b+c+d} = 1$, and $S < a/a + b/b + c/c + d/d = 4$, so I know $1 < S < 4$. The thought is to either try and tighten these bounds somehow (not sure how to approach this) or figure out how to represent an arbitrary $x$ in this range with some choice of $a,b,c,d$.
Some substitutions: Imposing the constraint $a+b+c+d = 1$ or $u = a+b+d, v = a+b+c, w = b+c+d, x = a+c+d$ come to mind, both attempts at removing the sums from the denominator. Respectively those give $S = \frac{a}{1-c} + \frac{b}{1-d} + \frac{c}{1-a} + \frac{d}{1-b}$ and $S = \frac{u+v+w-2x}{3u} + \frac{u+v+w-2x}{3v} + \frac{v+x+w-2u}{3w} + \frac{u+w+x-2v}{3x}$.
Another way of looking at things: $\frac{a}{a+b+d} + \frac{b}{a+b+c} + \frac{c}{b+c+d} + \frac{d}{a+c+d}$ is continuous in $(a,b,c,d)$. And I think we can get arbitrarily close to 1 by setting $a$ very large, and then $c$ very small relative to $b + d$. So 1 is probably the $\inf$ of $S$. If we do the opposite and set $a$ large and $c$ large, we can get close to 2. My suspicion is that $2$ is the $\sup$ but I'm not sure how to prove it.
| Because
$$\frac{a}{a+b+d}>\frac{a}{a+b+c+d},$$
so
$$S>\frac{a+b+c+d}{a+b+c+d}=1.$$
And
$$\frac{c+a}{a+b+c+d}-\frac{a}{a+b+d} = \frac{c(b+d)}{(a+b+c+d)(a+b+d)}>0,$$
so
$$\frac{a}{a+b+d}<\frac{c+a}{a+b+c+d},$$
therefore
$$S<\frac{2(c+a)+2(b+d)}{a+b+c+d}=2.$$
We get the estimations $1<S<2.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the difference of square root of conjugate complex number Find the imaginary part of $\left( {{{\left( {3 + 2\sqrt { - 54} } \right)}^{\frac{1}{2}}} - {{\left( {3 - 2\sqrt { - 54} } \right)}^{\frac{1}{2}}}} \right)$
(1) $-\sqrt 6$
(2) $-2\sqrt 6$
(3) $\sqrt 6$
(4) $6$
My Approach is as follow and none of the answer is matching, I cross checked it
$T = 3 + 2\sqrt { - 54} = 3 + i6\sqrt { 6} \to {I^{st}} - Quadrant - Angle = \theta $
$U = 3 - 2\sqrt { - 54} = 3 - i6\sqrt { 6} \to I{V^{th}} - Quadrant - Angle = - \theta $
$\Rightarrow \left( {{{\left( {3 + i6\sqrt 6 } \right)}^{\frac{1}{2}}} - {{\left( {3 - i6\sqrt 6 } \right)}^{\frac{1}{2}}}} \right)$
$r\cos \theta = 3$ & $r\sin \theta = 6\sqrt 6 \Rightarrow {r^2} = 225 \Rightarrow r = 15 \Rightarrow \tan \theta = 2\sqrt 6 $
$ \Rightarrow \left( {\sqrt {15} {e^{\frac{{i\theta }}{2}}} - \sqrt {15} {e^{ - \frac{{i\theta }}{2}}}} \right) \Rightarrow \sqrt {15} \left( {{e^{\frac{{i\theta }}{2}}} - {e^{ - \frac{{i\theta }}{2}}}} \right) = i\sqrt {15} \left( {2\sin \frac{\theta }{2}} \right)$
$ \Rightarrow \frac{{2\tan \frac{\theta }{2}}}{{1 - {{\tan }^2}\frac{\theta }{2}}} = 2\sqrt 6 \Rightarrow {\tan ^2}\frac{\theta }{2} + \frac{2}{{\sqrt {24} }}\tan \frac{\theta }{2} + \frac{1}{{24}} = \frac{{25}}{{24}} \Rightarrow \left( {\tan \frac{\theta }{2} + \frac{1}{{\sqrt {24} }}} \right) = \frac{5}{{\sqrt {24} }} \Rightarrow \tan \frac{\theta }{2} = \frac{4}{{\sqrt {24} }} = \frac{{\sqrt 2 }}{{\sqrt 3 }}$
$\sin \frac{\theta }{2} = \frac{{\sqrt 3 }}{{\sqrt 5 }} \Rightarrow i\sqrt {15} \left( {2\sin \frac{\theta }{2}} \right) = i\sqrt {15} \left( {2 \times \frac{{\sqrt 2 }}{{\sqrt 5 }}} \right) = 2\sqrt 6 i$
| It is simpler to calculate the square roots of a complex number by hand: denote such a square root as $z=x+iy$. Then $z^2=x^2-y^2+2ixy$, so for the square roots of $3+2\sqrt{-54}=3+6i\sqrt 6$, you have to solve the system of equations
\begin{cases}
x^2-y^2=3, \\
xy=3\sqrt 6.
\end{cases}
Now there's a trick to make the computation faster:
$$|z|^2=x^2+y^2=\bigl|3+6i\sqrt 6\bigr|=\sqrt{225}=15,$$
so we have a linear system in $x^2$ and $y^2$: $\;\smash[b]{\begin{cases}
x^2-y^2=3, \\ x^2+y^2 =15,\end{cases}}$ and the 3rd equation tells us that $x$ and $y$ have the same sign.
Solving this linear system, we obtain
$$x^2=9,\enspace y^2=6,\quad\text{whence}\quad x+iy=\pm(3+i\sqrt 6).$$
For the square roots of $3-2\sqrt{-54}=3-6i\sqrt 6$, we obtain the same linear system, with the condition that $x$ and $y$ have opposite signs, so that the square roots are
$$x+iy=\pm(3-i\sqrt 6).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the basis for $W_1 \cap W_2$ $$W_1=\left \{\left[{a \atop a+c} \ {a+b\atop b+c}\right] \mid a,b,c \in F\right\}, W_2=\left \{\left[{x \atop -y} \ {x\atop y}\right] \mid x,y \in F\right\}$$
I know that
$$W_1 \cap W_2 = \left\{\left[{a \atop a+c} \ {a+b\atop b+c}\right] \mid \exists x,y \in F,\left[{a \atop a+c} \ {a+b\atop b+c}\right] = \left[{x \atop -y} \ {x\atop y}\right]\right\}$$
however I am having a hard time making this into something that I can find the basis of by writing it as a linear combination.
| Since $a=x$ and $a+b =x$, we have that $b=0$. Then $b+ c = c =y$. Then $a+c = a+y = -y$, hence $a=-2y$. Thus $x = 2y$. Thus we can see that any element of $W_1 \cap W_2$ must have that form:
$$\begin{pmatrix} -2y & -2y \\ -y & y \end{pmatrix}$$
Further by choosing $a = -2y$, $b =0$, $c=y$, we can see that:
$$\begin{pmatrix} a & a+b \\ a+c & b+c\end{pmatrix} = \begin{pmatrix} -2y & -2y \\ -y & y \end{pmatrix},$$
thus all of these matrices are in $W_1 \cap W_2$. Thus:
$$W_1 \cap W_2 = \left\{ \begin{pmatrix} -2y & -2y \\ -y & y \end{pmatrix} \mid y \in F\right\},$$
which has basis:
$$\left\{ \begin{pmatrix} -2 & -2 \\ -1 & 1 \end{pmatrix}\right\}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluation of $\int_{-2}^{-1} \tan^{-1}\sqrt{\frac{x+1}{x-1}} dx$ Evaluation of $$\int_{-2}^{-1} \tan^{-1}\sqrt{\frac{x+1}{x-1}} dx$$
requires a special care when done by hand. The question is: How to do it and what is the right answer.
| By letting $\;t=\sqrt{\cfrac{x+1}{x-1}}\;,\;$ we get $\;x=\cfrac{t^2+1}{t^2-1}\;$ and $$\int\limits_{-2}^{-1}x\cdot\frac{\left(\sqrt{\frac{x+1}{x-1}}\right)^\prime}{1+\left(\sqrt{\frac{x+1}{x-1}}\right)^2}dx=\int\limits_{\sqrt{\frac{1}{3}}}^0\frac{t^2+1}{t^2-1}\cdot\frac{1}{1+t^2}dt=\\= \int\limits_{\sqrt{\frac{1}{3}}}^0\frac{1}{t^2-1}dt=\int\limits_0^{\sqrt{\frac{1}{3}}}\frac{1}{1-t^2}dt=\frac{1}{2}\ln\left(\frac{1+t}{1-t}\right)\;\bigg|_0^{\sqrt{\frac{1}{3}}}=\\=\frac{1}{2}\ln\left(\frac{1+\frac{\sqrt{3}}{3}}{1-\frac{\sqrt{3}}{3}}\right)=\frac{1}{2}\ln\left(\frac{3+\sqrt{3}}{3-\sqrt{3}}\right)=\frac{1}{2}\ln\left(2+\sqrt{3}\right)\;.$$
Hence,
$$\int\limits_{-2}^{-1}\arctan\sqrt{\frac{x+1}{x-1}}dx=\\=x\arctan\sqrt{\frac{x+1}{x-1}}\;\bigg|_{-2}^{-1}-\int\limits_{-2}^{-1}x\cdot\frac{\left(\sqrt{\frac{x+1}{x-1}}\right)^\prime}{1+\left(\sqrt{\frac{x+1}{x-1}}\right)^2}dx=\\=2\arctan\sqrt{\frac{1}{3}}-\frac{1}{2}\ln\left(2+\sqrt{3}\right)=\\=2\arctan\frac{\sqrt{3}}{3}-\frac{1}{2}\ln\left(2+\sqrt{3}\right)=\frac{\pi}{3}-\frac{1}{2}\ln\left(2+\sqrt{3}\right)\;.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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prove $a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b)$
prove $$a^3+b^3+c^3+3abc\ge \sum_{cyc}ab(a+b),$$$a,b,c>0$
Obviously this is a direct consequence of the third degree schur's inequality.
I was wondering if this could be proved without this theorem ,or uvw but through basic methods like AM-GM,C-S etc.
| We write your inequality as$:$
$$\sum a(a-b)(a-c) \geqslant 0.$$
Due to symmetric$,$ by assuming $a :\neq \max\,\{a,b,c\},$ we have$:$
$$\sum a(a-b)(a-c) = a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b)$$
$$=a(a-b)(a-c) +[b(b-c)(b-a)+c(c-a)(c-b)]$$
$$=a(a-b)(a-c) +(b-c)^2(b+c-a) \geqslant 0,$$
which is SOS-Schur method.
Else method is$:$
$$(a+b+c) \sum a(a-b)(a-c) = \sum a^2(a-b)(a-c) +\sum (a^3b+b^3a-2a^2b^2)$$
which is just Schur degree $4$ and AM-GM.
Schur degree $4$ is easy prove.
| {
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Solving 2-degree equations in 3 variables. We are given 3 equations:
$x^2+\sqrt3 xy + y^2 = 25$
$y^2 + z^2 = 9$
$x^2 +xz+ z^2 = 16$.
$x,y,z$ are positive real numbers.
Then we have to find value of $xy + 2yz + \sqrt3 xz$.
| Note that you have $$\forall x,y,z \in \mathbb{R}^{+}: \left\{\begin{aligned} x^{2}+\sqrt{3} xy + y^2 &=& 25\\
y^{2} + z^{2} &=& 9\\
x^{2} +xz+ z^{2} &=& 16 \end{aligned} \right.$$ if, and only if, $$ \forall x,y,z \in \mathbb{R}^{+}: \left\{\begin{aligned} x^{2}+\sqrt{3} xy + y^2 &=& \color{blue}{5}^{2}\\
y^{2} + z^{2} &=& \color{blue}{3}^{2}\\
x^{2} +xz+ z^{2} &=& \color{blue}{4}^{2} \end{aligned} \right. $$
Now, we can approach this problem as an algebraic geometry problem. Indeed, consider a triangle $\bigtriangleup XYZ$ with side lengths $3,4,5$ and draw a point $P$ inside the triangle such that $XP=x$, $YP=y$, and $ZP=z$. Now, you can considerer he equations in the context of the law of cosines.
Can you continue from here?
| {
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Number of real roots $x^8-x^5+x^2-x+1=0$ Find the number of real roots of $x^8-x^5+x^2-x+1$.
My attempt: $f(x)$ has $4$ sign changes and $f(-x)$ has no sign changes, so the possibility of having real roots is $4+0=4$. Since this is a polynomial of degree $8$ it should have $8-4=4$ imaginary roots.
It is quite impossible to see that this equation does not have any real roots by observing the factor $x^8-x^5+x^2-x+1= (x-1)(x^7+x^6+x^5+x)+1>0$, hence it can't have any real roots. My question is how to prove using this equation doesn't have any real roots using Descartes' Rule? Please give some useful hints.
| $f(x)=x^8-x^5+x^2-x+1=0$ has no real root as $f(x)>0$ for all real values of $x$.
$$f(x)=x^3(x^3-1)+x(x-1)+1 >0, ~if~ x>1$$
$$f(x)=x^8+x^2(1-x^3)+(1-x) > ~if~x <1$$
And $f(1)=1>0$.
| {
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Are variance and unfairness sum perfectly related? This question is basically a part of an algorithmic problem about dynamic programming I was trying to solve.
You need to consider two things that I'm about to compare below:
*
*Variance
*Unfairness Sum - defined and explained below
Unfairness Sum-
Let's say we have a list of positive integers denoted as myList. The Unfairness Sum of myList is defined as the sum of the absolute differences of all the pairs (explained below) in myList.
To Explain
For example, if
myList = $\{1, 2, 5, 5, 6\}$
Then the Unfairness Sum will be (note that numbers are considered unique based on their index-or position- in list, not their values)
$$\text{Unfairness Sum}= |1-2| + |1-5| + |1-5| + |1-6| + |2-5| + |2-5| + |2-6| + |5-5| + |5-6| + |5-6|$$
What I want to know
Can I say that variance & Unfairness Sum are perfectly related (I know they are strongly related because this approach of variance has worked for half of my test cases - having up to a max of 9000 integers`)?
In other words,
Can I say that among a lots of lists of positive integers, a list with minimum variance will always be the list with Minimum unfairness sum?
| The list that minimizes the absolute sum does not always minimize the variance sum. Consider two lists:
$\{1, 1, x\}$ and $\{1, 2, 3\}$
where $x$ is a real number that satisfies $x \geq 1$.
We get
\begin{align}
a(1,2,3) &= |1-2| + |1-3| + |2-3| = 4\\
v(1,2,3) &= (1-2)^2 + (2-2)^2 + (3-2)^2 = 2
\end{align}
and
\begin{align}
a(1,1,x) &= |1-1| + |1-x| + |1-x| = 2(x-1)\\
v(1,1,x) &= (1 - \frac{(2+x)}{3})^2 + (1 - \frac{(2+x)}{3})^2 + (x - \frac{(2+x)}{3})^2
\end{align}
where we have used $x\geq 1$ to claim $|1-x|=x-1$.
We want to find $x\geq 1$ that satisfies:
\begin{align}
&a(1,2,3) > a(1,1,x)\\
&v(1,2,3)<v(1,1,x)
\end{align}
To find such a value $x\geq 1$, we have
$$a(1,2,3)>a(1,1,x) \iff 4 > 2(x-1) \iff x<3$$
However
\begin{align}
&v(1,2,3)<v(1,1,x) \\
&\iff 2<(1 - \frac{(2+x)}{3})^2 + (1 - \frac{(2+x)}{3})^2 + (x - \frac{(2+x)}{3})^2
\end{align}
There are values $x \geq 1$ that satisfy both inequalities, for example $x=2.8$.
| {
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Find all positive integers $N$ such that $N \cdot 2^{N+1}+1$ is a perfect square I just finished the following question:
Find all positive integers $N$ such that $N\times 2^{N+1}+1$ is a perfect square
I solved it in the following way:
$N\times2^{N+1}+1=k^2$ for $k\in Z$
$N\times2^{N+1}=(k-1)(k+1)$
So we have that $(k-1)(k+1)\equiv0 \pmod 2$, so $k+1 \equiv0\pmod2$ and $k-1\equiv0\pmod2$.
Since $k+1$ and $k-1$ differ by just 2, and they are both even, then one of the two is a multiple of $2^b$, where $b\in Z$ and $b>1$ and the other if divided by two, the result is odd. From this and the fact that $k+1>k-1$ we have that:
$k+1=2^N$ and $k-1=2N$ (since $2^N>2N$).
So $2N+2=2^N$ for which $N=3$ is the only solution. This can be proved as follows:
We have $f$ such that $f(x)=2x+2$, so $f'(x)=2$ and $g$ such that $g(x)=2^n$, so $g'(x)=\ln(2)\times2^x>2$
So the only solution is for $N=3$.
I find my logic overly complex and laborious, could you please suggest some alternative approaches?
| Other apporoach can be $N \times 2^{N+1} + 1 = k^2$. Hence,$k$ is in the form of $2m+1$ and we will have: $N \times 2^{N+1} + 1 = 4m^2 +4m + 1$. Then, $N \times 2^{N+1} = 4 \times m \times (m+1) \Rightarrow N \times 2^{N-1} = m \times (m+1)$. Now, we have a simple proof that $gcd(m, m+1) = 1$, and one of them is odd and te other is even. Now we have the following case:
$N$ is odd
*
*($m = N$) $N = 2^{N-1} - 1 \Rightarrow N+1 = 2^{N-1}$. As $\lim_{N\to\infty}\frac{N+1}{2^{N-1}} = 0 $, and for all $N > 3$, $N + 1 < 2^{N-1}$ (by induction), the only answer is $N = 3$.
*($m = 2^{N-1}$) $2^{N-1} = N-1$. The same as the foregoing case, for all $N > 3$, $N - 1 < 2^{N-1}$. Hence, we can cannot find any solution here.
$N$ is even
We can factorize $N = 2^{\alpha_0} q_1^{\alpha_1}q_2^{\alpha_2}\cdots q_i^{\alpha_i}$. However, like the previous cases, if we decompose the left side to two even and odd numbers, we will reach an inequality such that one side of it is greater than $2^{N-1}$, and the other side is less than $N-1$. Therefore, in that case, we will find $N \leq 3$ as well. The only even number in this range is 2, but it does not satisfy the condition. Hence, there is no answer when $N$ is even.
| {
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Inequality manipulation: $\frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} > \frac{1}{\sqrt{4n + 5}}$ I'm reading Analytic Inequalities by Nicholas D. Kazarinoff. On page 5, we are trying to use induction to prove the inequality
$$ \frac{1}{\sqrt{4n + 1}} < \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n} < \frac{1}{\sqrt{3n + 1}} . $$
For the inductive step, we want to show that it holds for $n + 1$, i.e.
$$ \frac{1}{\sqrt{4n + 5}} < \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 1}{2n} \cdot \frac{2n + 1}{2n + 2} < \frac{1}{\sqrt{3n + 4}} . $$
Kazarinoff says that this is true if
$$ \frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} > \frac{1}{\sqrt{4n + 5}} $$
is also true. I'm trying to figure out why this is the case, because it's not obvious to me.
Edit: this is completely nonsensical because it's circular.
What I've tried:
\begin{align}
\frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} &> \frac{1}{\sqrt{4n + 5}} \\
\implies \frac{1}{\sqrt{4n + 1}} &> \frac{1}{\sqrt{4n + 5}} \cdot \frac{2n + 2}{2n + 1} \\
&< \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n} \\
&> \frac{1}{\sqrt{4n + 1}}
\end{align}
That's a pretty useless result that hasn't gotten me anywhere. Any ideas?
| By induction assumption, you have
$$ \frac{1}{\sqrt{4n + 1}} < \frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n}.$$
Suppose you have
$$\frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2} > \frac{1}{\sqrt{4n + 5}}.$$
Then
$$\frac{1}{\sqrt{4n + 5}}<\frac{1}{\sqrt{4n + 1}} \cdot \frac{2n + 1}{2n + 2}<\frac{1}{2} \cdot \frac{3}{4} \cdots \frac{2n - 3}{2n - 2} \cdot \frac{2n - 1}{2n}\cdot \frac{2n + 1}{2n + 2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Direct common tangents to the given circles
Find the equation of common tangents to the circles $x^2+y^2-12x-8y+36=0$ and $x^2+y^2-4x-2y+4=0$ touching the circles in the distinct points.
The center of first circle $C_1$ is $(6,4)$. Its radius $r_1=4$.
The center of second circle $C_2$ is $(2,1)$. Its radius $r_2=1$.
Distance between $C_1$ and $C_2$ i.e $C_1C_2=5=r_1+r_2$. That means, circles touch each other externally.
Let the point of intersection of direct common tangents be $P$. And $P$ divides $C_1C_2$ externally in the ratio $4:1$. So $P$ is $(\frac23,0)$.
Now, equation of pair of tangents from an external point is $SS_1=T^2$. Using this on the first circle, I get equations as $y=0$ and $y=8$. Using the formula for second circle, I get equations as $y=0$ and $y=2$.
Now, I have two questions here.
a) Why are equations coming out to be different? Since it's the same pair of tangents, shouldn't it be the same?
b) The pair of tangents should pass through $P$. But $P$ has y-coordinate as zero. So, why am I getting $y=2$ and $y=8?$
Also, if we solve it differently i.e. by taking the slope of tangents as $m$, and then equating the perpendicular distance from center of second circle, onto the equation of line through P, with radius of that circle, we get $y=0$ and $24x-7y=16$. And if I do this on first circle, I get different equations.
What am I doing wrong here? Also, why am I getting so many equations here?
| You have already established there is a point of tangency between the two circles. Here is one way to locate the external tangents. Ask what kind of transformation would map the larger circle to the smaller. You know that the larger circle has radius $r_1 = 4$ and center $C_1 = (6,4)$, while the smaller has radius $r_2 = 1$ and center $C_2 = (2,1)$. So if we apply the transformation $(u,v) = (x/4, y/4)$, what happens to the first circle? The radius becomes $1$ but the center is now $(u,v) = (6/4, 4/4) = (3/2, 1)$. That's close, but not quite. We need to add $1/2$ to the horizontal coordinate. So a contraction of $1/4$ followed by a horizontal translation of $1/2$, or $$(u,v) = \left(\frac{x}{4} + \frac{1}{2}, \frac{y}{4} \right)$$ will map $C_1$ to $C_2$ and $r_1$ to $r_2$. Why does this help us? Because the intersection of the external tangent lines is the unique fixed point of this transformation. That is to say, where the tangent lines intersect, this point does not move when transformed in this way. So if we solve $$x = \frac{x}{4} + \frac{1}{2}, \\ y = \frac{y}{4},$$ we find this fixed point is $$(x,y) = \left(\frac{2}{3}, 0\right).$$ So both lines must pass through this point, hence have equation $$y - 0 = m(x - 2/3),$$ for some slope $m$. This allows us to substitute $x$ for $y$ in either equation of the circle; e.g., for $C_2$, we obtain
$$\begin{align}
0
&= x^2 + m^2(x-2/3)^2 - 4x - 2m(x-2/3) + 4 \\
&= (1+m^2)x^2 - \left(\frac{4}{3}m^2 + 2m + 4\right)x + \left(\frac{4}{9}m^2 + \frac{4}{3} m + 4\right).
\end{align}$$
We require the discriminant of this quadratic in $x$ to be zero, because otherwise there will not be a single unique intersection point for the tangent line and the circle. To avoid dealing with fractions we multiply the coefficients by $9$:
$$a = 9(1+m^2) \\ b = 12m^2 + 18m + 36 \\ c = 4m^2 + 12m + 36.$$ Then we must have $$0 = b^2 - 4ac = (12m^2 + 18m + 36)^2 - 36(m^2 + 1)(4m^2 + 12m + 36) = 36m(24 - 7m).$$ This gives us $m = 0$ and $m = \frac{24}{7}$ as solutions, and our desired tangent lines are $$y = 0, \quad y = \frac{27}{4}\left(x - \frac{3}{2}\right).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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solve the trigonometric equation: $2(1 - \cos(x))-x\sin(x)=0$ for find a global maximum I need to prove that the function $f: (0,\pi] \to \mathbb{R}$ given by $f(x):= \dfrac{x - \sin(x)}{1 - \cos(x)}$ has a global maximum in $x=\pi$.
Ok I did that: first I need find critical points but I have a problem with trigonometric equation $f^{'}(x)=\dfrac{(1-\cos(x))^2-(x - \sin(x))\sin(x)}{(1 - \cos(x))^2}$ doing $f^{'}(x)=0$ then $\dfrac{(1 - \cos(x)^{2}) - (x - \sin(x))\sin(x)}{(1 - \cos(x))^2}=0$.
In conclusion, I need to first solve the trigonometric equation: $2(1 - \cos(x))-x\sin(x)=0$ do you know other way to any viable solution for this problem
Here is my attempt:
$$\begin{align}
f'(x) &= (1-\cos(x))^2 + (-x\sin(x)+\sin^2(x))\\
&= 1 - 2\cos(x) + \cos^2(x) + \sin^2(x) - x\sin(x)\\
&= 1 - 2\cos(x) + 1 - x\sin(x)\\
&= 2 - 2\cos(x) - x\sin(x)\\
&= 2(1 - \cos(x)) - x\sin(x)\\
&= 0
\end{align}$$
| For $x\in[0,\pi)$ we have $$4\sin^2\frac{x}{2}-2x\sin\frac{x}{2}\cos\frac{x}{2}=0,$$ which gives
$$\sin\frac{x}{2}=0$$ or $x=0.$
Also, $$2\sin\frac{x}{2}=x\cos\frac{x}{2}$$ or $$\tan\frac{x}{2}=\frac{x}{2},$$ which gives $x=0$ again and we know that $$\tan\frac{x}{2}>\frac{x}{2}$$ for any $0<x<\pi.$
Id est, $f'$ has no zeros on $(0,\pi).$
But $f'$ is a continues on $(0,\pi]$ function and $f'\left(\frac{\pi}{2}\right)>0,$ which says that $f$ increases.
Id est, $$\max_{(0,\pi]}f=f(\pi).$$
| {
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Solve equation for $x$ given by a determinant I am looking for some interesting solution for this problem. Given $a\in \mathbb{C}$, solve for $x$ the following equation
$$
\begin{vmatrix}
1 & a & a+x & a+x^2\\
a & 1 & a+x^2 & a+x \\
a + x & a+ x^2 & 1 & a \\
a +x^2 & a+x & a & 1 \\
\end{vmatrix} =0\qquad (D_a)
$$
Set $Z_a = \lbrace x\in \mathbb{C} : x \text{ satisfies }(D_a)\rbrace$ to be the set of solutions of problem $D_a$.
Anyway, I am searching some trick or idea to solve the problem (i.e., find the set $Z_a$ for every value $a$). Obviously, I am not interested in computing explicitly the determinant, instead I wonder if there is a simple and elegant way to solve the problem (maybe using properties of determinant).
Note that the matrix is symmetric. This could be helpful (there are some beautiful results concerning symmetrics matrix).
Thanks in advance. Any idea is welcome!
| This is a block matrix of the form $\pmatrix{A&B\\B&A}$. Thus, we have
$$
\det \pmatrix{1 & a & a+x & a+x^2\\
a & 1 & a+x^2 & a+x \\
a + x & a+ x^2 & 1 & a \\
a +x^2 & a+x & a & 1} = \\
\det \left[
\pmatrix{1&a\\a&1}^2 - \pmatrix{a+x&a+x^2\\a+x^2 & a+x}^2
\right] = \\
\det\left[
\pmatrix{1&a\\a&1} - \pmatrix{a+x&a+x^2\\a+x^2 & a+x}
\right]
\det \left[
\pmatrix{1&a\\a&1} + \pmatrix{a+x&a+x^2\\a+x^2 & a+x}
\right] = \\
\det
\pmatrix{1-a-x&-x^2\\-x^2&1-a-x}\cdot
\det
\pmatrix{1+a+x&2a+x^2\\2a+x^2&1+a+x}.
$$
We see that a matrix of the form $\pmatrix{a&b\\b&a}$ is singular iff $a = b$ or $a = -b$. Thus, we obtain the desired values of $x$ by solving the following equations:
*
*$1 - a - x = -x^2$
*$1 - a - x = x^2$
*$1 + a + x = 2a + x^2$
*$1 + a + x = -2a - x$.
Another approach: in terms of Kronecker products, we could express your matrix as
$$
M = I + a I \otimes J + x J \otimes I + (a + x^2)J \otimes J,
$$
where
$$
I = \pmatrix{1&0\\0&1}, \quad J = \pmatrix{0&1\\1&0}.
$$
Thus, the eigenvalues of $M$ will be of the form $1 + a\lambda_1 + x\lambda_2 + (a + x^2) \lambda_1\lambda_2$, where each $\lambda_i$ is one of the eigenvalues of $J$ (which are $\pm 1$). $M$ has determinant $0$ iff one of these eigenvalues is $0$.
Another approach: we can diagoanlize the matrix by computing $U^TMU$, where
$$
U = \frac 12 \pmatrix{1&1\\1&-1} \otimes \pmatrix{1&1\\1&-1} =
\frac 12 \pmatrix{1&1&1&1\\1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1}.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate a definite integrals given in term of Bessel functions The problem is to solve the following integrals. The parameters are all positives.
\begin{align}
& \int \sqrt{ a^2 - x^2} \, J_1\left(r \sqrt{a^2-x^2}\right) \cos (x z) dx
\\
& \int x J_0\left(r \sqrt{a^2-x^2}\right) \sin (x z) dx
\end{align}
I try using trigonometric variable change and Euler formula. I think it makes them more complex to solve Here. I show how the integrals are transformed.
\begin{align}
& x=a sin\theta , \,\,\,\,\, dx=a\cos\theta d \theta\\ \\
& e^{ i a \sin \theta z}= \cos (a \sin \theta z) + i\sin (a \sin \theta z)
\\
\\
& a^2 \int \cos^2\theta J_1\left(r a \cos\theta \right) \cos (a \sin \theta z)
d\theta \\ & a \int \sin \theta J_0\left(r a \cos \theta\right) \sin (a \sin \theta z)
d\theta
\end{align}
The goal is to obtain close solutions for both integrals.
| You can evaluate them on $(0,a)$ in terms of spherical Bessel functions (which are expressible in terms of trigonometric functions) as follows:
$$
\int_0^a {\sqrt {a^2 - x^2 } J_1 \left( {r\sqrt {a^2 - x^2 } } \right)\cos (xz)dx} = a^2 r\frac{{\mathsf{j}_1 (a\sqrt {r^2 + z^2 } )}}{{\sqrt {r^2 + z^2 } }} \\ = r\frac{{\sin (a\sqrt {r^2 + z^2 } )}}{{r^2 + z^2}} - ar\frac{{\cos (a\sqrt {r^2 + z^2 } )}}{{\sqrt {r^2 + z^2 } }}
$$
and
$$
\int_0^a {xJ_0 \left( {r\sqrt {a^2 - x^2 } } \right)\sin (xz)dx} = a^2 z\frac{{\mathsf{j}_1 (a\sqrt {r^2 + z^2 } )}}{{\sqrt {r^2 + z^2 } }} \\ = z\frac{{\sin (a\sqrt {r^2 + z^2 } )}}{{r^2 + z^2 }} - az\frac{{\cos (a\sqrt {r^2 + z^2 } )}}{{\sqrt {r^2 + z^2 } }}.
$$
See (1.13.50) and (2.13.50) in A. Erdélyi, W. Magnus, F. Oberhettinger, F. G. Tricomi, Tables of Integral Transforms. Vol. I., McGraw-Hill Book Company, Inc., New York-Toronto-London, 1954.
| {
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length of side of a polygon having same area as a triangle Let P be a 12-sided regular polygon and T be an equilateral triangle with its incircle having radius $1$. I the area of P is the same as of area of T, then the length of side of P is
(A) $\sqrt{\sqrt{3} \cot15^\circ} \qquad$ (B) $ \sqrt{\sqrt{3} \tan15^\circ}\qquad$ (C) $ \sqrt{3\sqrt{2} \tan15^\circ}\qquad$ (D) $\sqrt{3\sqrt{2} \cot15^\circ} $
My try:
Since incenter of T is $1$, its height is $3$ and length of sides are $ 2\sqrt3 $. So area of T = $ 3\sqrt{3}$.
Area of $P=12*\delta.$ Where $\delta$ is the area of a triangle formed by any two adjacent vertices and the center of P. Let $x$ be length of sides of P, then $ \delta = \frac{x}{2}*h$. Here $\frac{\frac{x}{2}}{h} =\tan 15^ \circ \Rightarrow h = \frac{x\cot 15^\circ}{2} $. Therefore Area of P is $ 3x^2\cot15^\circ$
Equating areas of T and P, we have $$x^2\cot 15^\circ = \sqrt{3} \Rightarrow x^2 = \sqrt{3}\tan 15^\circ $$
So, my answer is $x = \sqrt{\sqrt{3} \tan15^\circ}$, option (B). But in my book, answer is given as (A). Can anyone suggest if my calculation of wrong somewhere?
| Equilateral triangle area is correct. Equating it to a 12 sided regular polygon area,
$$ 12\cdot \dfrac{x^2}{4}\cot 15^{\circ}= 3 \sqrt{3}\rightarrow x= \sqrt{\sqrt3 \tan 15^{\circ}}$$
Your answer is correct.
| {
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Three externally touching circles have their centers on the same line and have radii $a$,$b$ and $c$ (where $aThree circles have their centers on the same line and have radii $a$,$b$ and $c$ (where $a<b<c$).The circle with radius $b$ touches the other two circles but circles with radii $a$ and $c$ do not touch each other.The three circles also have a common tangent.Prove that $b^2=ac$.
I solved it with good deal of calculations. Is there an elementary way to do this problem
| Say common tangent length between circle A and B (IJ) is $x$, B and C (JK) is $y$ and so between A and C (IK) is $(x + y)$.
Applying Pythagoras, between circles of radius a and b,
$x^2 = (a+b)^2 - (b-a)^2 = 4ab$
Applying Pythagoras, between circles of radius b and c,
$y^2 = (b+c)^2 - (c-b)^2 = 4bc$
Applying Pythagoras, between circles of radius a and c,
$(x+y)^2 = (a+2b+c)^2 - (c-a)^2 = 4(b^2+ab+bc+ca)$
i.e $x^2 + y^2 + 2xy = 4b^2+4ca+x^2+y^2$
i.e $4b^2+4ca = 2xy = 8b\sqrt{ac}$
i.e $b^2 - 2b\sqrt{ac} + ca = 0$
i.e $b^2 = ca$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the cartesian equation of a plane given the vector equation? The vector equation is given by $r=\begin{pmatrix}-3\\ -5\\ -1\end{pmatrix}+k\begin{pmatrix}6\\ 6\\ -3\end{pmatrix}$
To find the cartesian equation, I have to consider a point on the plane:
$$\begin{pmatrix}x\\ y\\ z\end{pmatrix}=\begin{pmatrix}-3\\ -5\\ -1\end{pmatrix}+k\begin{pmatrix}6\\ 6\\ -3\end{pmatrix}$$
which gives:
$$\frac{x+3}{\:6}=\:\frac{y+5}{6}=\frac{z+1}{-3}$$
Now this is the part where I don't understand. Up till now my working follows with the answers provided by my book, but then the book did something that puzzle me:
$$\frac{x+3}{6}+\frac{y+5}{6}=2(\frac{z+1}{-3})$$
How did they reached here?
| You have $$\frac{x+3}{6}=\frac{z+1}{-3}$$ and $$\frac{y+5}{6}=\frac{z+1}{-3}$$
Now add these two equations.
| {
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Evaluating $\int_{0}^{\pi}\ln (1+b\cos x)\ \text{d}x$, $b$ is a parameter Evaluating $\int_{0}^{\pi}\ln (1+b\cos x)\ \text{d}x$ where $b$ is a parameter
I've tried Integration by parts which yield $$\int_{0}^{\pi}\ln (1+b\cos x)\ \text{d}x=\pi\ln(1-b)+b\int_{0}^{\pi}{x\sin x\over 1+b\cos x}\text{d}x$$ I cannot figure out what do next.
I also tried using Leibniz integral rule by putting $I(b)=\int_{0}^{\pi}\ln (1+b\cos x)\ \text{d}x$ to form a differential equation.
$${\text{d}I(b)\over \text{d}b}=\int_{0}^{\pi}{\cos x\over 1+b\cos x}\text{d}x$$ but I'm not able to solve the integral on right.
I've looked similar questions like this one Evaluating $\int_{0}^{\pi}\ln (1+\cos x)\, dx$ to no avail. Also I'm high school student so I don't understand advanced calculus stuff yet.
| \begin{align*}
\frac{\mathrm{d} I(b)}{\mathrm{d} b}=\int_{0}^{\pi}{\cos x\over 1+b\cos x}\; \mathrm{d}x &= \frac{1}{b}\int_0^{\pi} \frac{ 1+b \cos{x}-1}{1+b\cos{x}} \; \mathrm{d}x \\
&= \frac{\pi}{b}-\frac{1}{b} \int_0^{\pi} \frac{1}{1+b \cos{x}} \; \mathrm{d}x\\
&= \frac{\pi}{b}-\frac{2}{b} \int_0^{\infty} \frac{1}{(t^2+1)+b(1-t^2)} \; \mathrm{d}t \tag{1}\\
&= \frac{\pi}{b}-\frac{2}{b} \int_0^{\infty} \frac{1}{(1-b)t^2+(1+b)} \; \mathrm{d}t\\
&= \frac{\pi}{b}-\frac{2}{b} \left(\frac{\pi}{2\sqrt{1-b^2}}\right) \\
&= \frac{\pi}{b}- \frac{\pi}{b\sqrt{1-b^2}} \\
I(b) &= \int \frac{\pi}{b}- \frac{\pi}{b\sqrt{1-b^2}} \; \mathrm{d}b \\
&= \pi \ln|b| + \pi \operatorname{artanh}{\left(\sqrt{1-b^2}\right)}+C \\ I(1)&=-\pi \ln{2} \implies C=-\pi \ln{2}\\
I(b) &= \pi \ln|b| + \pi \operatorname{artanh}{\left(\sqrt{1-b^2}\right)}-\pi \ln{2} \\
&= \pi \ln\bigg|\frac{b}{2}\bigg| -\frac{\pi}{2} \ln\left(1-\sqrt{1-b^2}\right)+\frac{\pi}{2}\ln \left(1+\sqrt{1-b^2}\right) \\
&= \boxed{\pi \ln\left(\frac{1+\sqrt{1-b^2}}{2}\right)}
\end{align*}
Additionally, note that $-1<b<1$.
$(1):$ Weierstrass substitution
| {
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Finding the curve on a surface with a specific curvature I wish to find a curve $\langle x(t), y(t), z(t) \rangle$ on the surface $x + y + z = (x-y)^2 + (y-z)^2$ whose curvature is 1/2 for all $t$.
I am struggling with how to proceed. I initially tried expanding the right side and attempted to parametrize but because of the $yz$ and $xy$ terms, I wasn't able to do so. I would greatly appreciate any approaches to this problem. Thank you!
| By inspection, if the given surface is cut by a plane where $y$ is constant, the equation of the resulting curve is a circle (possibly degenerate) parallel to the $xz$-plane.
Leaving $y$ as an unknown constant and completing the square in the usual way, the equation
$$
x+y+z = (x-y)^2+(y-z)^2
$$
can be rewritten as
$$
(x-h)^2+(z-k)^2=r^2
$$
where
$$
\left\lbrace
\begin{align*}
h&=y+\frac{1}{2}\\[4pt]
k&=y+\frac{1}{2}\\[4pt]
r^2&=3y+\frac{1}{2}
\end{align*}
\right.
$$
For the circle to have curvature ${\large{\frac{1}{2}}}$, we want $r=2$.
Solving the equation $r^2 = 4$ for $y$ yields $y={\large{\frac{7}{6}}}$, hence $h=k={\large{\frac{5}{3}}}$.
Thus the curve satisfying the system
$$
\left\lbrace
\begin{align*}
&y=\frac{7}{6}\\[4pt]
&\Bigl(x-\frac{5}{3}\Bigr)^2+\Bigl(z-\frac{5}{3}\Bigr)^2=4\\[4pt]
\end{align*}
\right.
$$
is a circle of radius $2$, hence curvature ${\large{\frac{1}{2}}}$, lying on the given surface.
The circle can be expressed parametrically as
$$
\left\lbrace
\begin{align*}
x&=\frac{5}{3}+2\cos(t)
\qquad\qquad\;\,
\\[4pt]
y&=\frac{7}{6}\\[4pt]
z&=\frac{5}{3}+2\sin(t)\\[4pt]
\end{align*}
\right.
$$
where $0\le t < 2\pi$.
| {
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question from South Korean selection exam 1998, about proving that an inequality holds true if $a+b+c=abc$ I was just doing the following question:
If $a,b,c>0$ such that $a+b+c=abc$, prove that:
$\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le \frac{3}{2}$
I think that this question can be solved through the use of homogenization, something which I attempted to do in the following way:
We have that $\frac{a+b+c}{abc}=1$ and hence also $\sqrt{\frac{a+b+c}{abc}}=1$. So $\frac{3}{2}*\sqrt{\frac{a+b+c}{abc}}=\frac{3}{2}$.
So all we have to prove now is $\frac{1}{\sqrt{1+a^2}}+\frac{1}{\sqrt{1+b^2}}+\frac{1}{\sqrt{1+c^2}}\le\frac{3}{2}*\sqrt{\frac{a+b+c}{abc}}$ which are homogenized and hence we do not need the original equality any more. This is where I couldn't continue from, and got stuck. Could you please explain to me how I could finish it off like this, or tell me why it can't and how it can be done using homogenization?
| The homogenization gives:
$$\sum_{cyc}\frac{1}{\sqrt{\frac{abc}{a+b+c}+a^2}}\leq\frac{3}{2}\sqrt{\frac{a+b+c}{abc}}$$ or
$$\sum_{cyc}\sqrt{\frac{bc}{(a+b)(a+c)}}\leq\frac{3}{2},$$ which is true by AM-GM:
$$\sum_{cyc}\sqrt{\frac{bc}{(a+b)(a+c)}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{c}{a+c}+\frac{b}{a+b}\right)=$$
$$=\frac{1}{2}\sum_{cyc}\left(\frac{c}{a+c}+\frac{a}{c+a}\right)=\frac{3}{2}.$$
| {
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How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for even $n$? How to prove $5^{2n+1}-3^{2n+1}-2$ is divisible by $30$ for all positive even integer $n$?
Here are my efforts:
Let $f(n)=5^{2n+1}-3^{2n+1}-2$.
For $n=2$,
$$f(1)=5^{2\cdot 2+1}-3^{2\cdot2+1}-2=2880=30\times 96$$
So, the statement is true for $n=2$.
Assume the statement is true for some integer $k\geq 2$.
$$f(k)=5^{2k+1}-3^{2k+1}-2=30t$$
For $n=k+2$,
\begin{align}
f(k+2)&=(5^{2(k+2)+1})-(3^{2(k+2)+1})-2\\
&=(5^4)\cdot (5^{(2k+1)})-(3^4)\cdot (3^{(2k+1)})-2
\\&=30(80t)+160+544\cdot (5^{2k+1})
\end{align}
I'm stuck here. Any help would be appreciated!
| Write $g(n)=f(2n)=5\cdot 625^n-3\cdot 81^n-2\cdot1^n$. Then
$$
g(n+3)= 707 g(n+2) - 51331 g(n+1) + 50625 g(n)
$$
because $(x-625)(x-81)(x-1)=0$ can be written $x^3=707 x^2 - 51331 x + 50625$. (The important point here is that the coefficients are integers, not their exact values.)
Therefore, the claim follows at once by induction if it holds for $n=0,1,2$:
$$
g(0)=0, \quad g(1)=2880 = 30 \cdot 96, \quad g(2)= 1933440 = 30 \cdot 64448
$$
(No need to compute the quotient: it's easy to see that they are all multiples of $3$ and $10$.)
Actually, $g(n)$ is divisible by $960=\gcd(g(0),g(1),g(2))$.
| {
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Double the value of a quadratic expression If I have this expression $f(x)=ax^2+bx+c$ and I want an expression for $2f(x)$ can I get that with a given change of $x$ ? For example, I have $f(x)=2x^2+3x+9$ and I want to find which factor for $x$ would make $f(x)$ twice the value at every point. Basically I'm looking for another function $g(x)$ so that I can use that and then $f(g(x))=2f(x)$.
| Let's try your example, $f(x)=2x^2+3x+9$.
You require that
$$ 2(g(x))^2+3g(x)+9 = f(g(x))= 2f(x)=4x^2+6x+18.$$
As a quadratic equation in $g(x),$ this is
$$ 2(g(x))^2 + 3g(x) - (4x^2+6x+9) = 0$$
and its solution by the quadratic formula is
\begin{align}
g(x) &= \frac{-3 \pm \sqrt{3^2 - 4(2)(- (4x^2+6x+9))}}{2(2)} \\
&= \frac{-3 \pm \sqrt{32x^2 + 48x + 81}}{4} \\
\end{align}
Pick your choice of either $+$ or $-$ for the $\pm$ sign and you have a function,
for example,
$$
g(x) = -\frac34 + \frac14 \sqrt{32x^2 + 48x + 81}.
$$
Note that $f(x)=2x^2+3x+9$ has no real roots.
For a quadratic polynomial with just one real root,
$g(x)$ can be simpler.
For example, if we let $f(x) = x^2 + 4x + 4,$ then
$$ (g(x))^2+4g(x)+4 = f(g(x)) = 2f(x)=2x^2+8x+8, $$
$$ (g(x))^2+4g(x)-(2x^2+8x+4) = 0, $$
\begin{align}
g(x) &= \frac{-4 \pm \sqrt{4^2 - 4(- (2x^2+8x+4))}}{2} \\
&= -2 \pm \frac14 \sqrt{8x^2 + 32x + 32} \\
&= -2 \pm \frac14 \sqrt{8(x^2 + 4x + 4)} \\
&= -2 \pm \frac{\sqrt2}{2} (x + 2) \\
\end{align}
and you can choose a sign from $\pm$ to get a single function such as
$$ g(x) = -2 + \frac{\sqrt2}{2} (x + 2)
= \frac{\sqrt2}{2} x + \left({\sqrt2} - 1\right). $$
If the quadratic polynomial has two roots, however, it will have either a minimum value that is negative or a maximum value that is positive.
If there is a minimum value, it is not possible to get twice the minimum value by substituting $g(x)$ for $x$ in the polynomial, because twice the minimum is less than the minimum (which is negative) and by definition there is nothing you can plug into the polynomial that will give you less than the minimum.
For similar reasons, if there is a maximum you cannot get twice the maximum.
The best you can do is to double the polynomial's value for values of $x$ outside some interval around the value where the minimum or maximum of the polynomial occurs.
| {
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$\lim_{n\to\infty}\left( \frac1{4\cdot 7}+\frac1{7\cdot 10}+\ldots+\frac1{(3n+1)(3n+4)} \right) $ I am having trouble finding the infinite sum
$$
\lim_{n\to\infty}\left( \frac1{4\cdot 7}+\frac1{7\cdot 10}+\ldots+\frac1{(3n+1)(3n+4)} \right).
$$
I know that
$$
\lim_{n\to\infty}\frac1{(3n+1)(3n+4)} =0,
$$
but I have no ideas for a further solution.
| $$S_n=\sum_{k=1}^{n} \frac{1}{(3k+1)(3k+4)}=\frac{1}{3}\sum_{k=1}^{n} \left(\frac{1}{3k+1}- \frac{1}{3k+3}\right)= \frac{1}{3} \sum_{k=0}^{n} \int_{1}^{1}[t^{3k}- t^{3k+3}] dt.$$
Use IGP: $$\sum_{k=1}^n r^k=\frac{r}{1-r}, |r|<1$$
$$S_{\infty}= \frac{1}{3} \int_{0}^{1} t^3\frac{1-t^3}{1-t^3} dt=\frac{1}{3}\int_{0}^{1} t^3 dt=\frac{1}{12}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the minimum of $P = (a - b)(b - c)(c - a)$
Given $a, b, c$ are real numbers such that $a^2 + b^2 + c^2 = ab + bc + ca + 6$. Find the minimum of
$$P = (a - b)(b - c)(c - a)$$
My solution:
*
*We have:
$$a^2 + b^2 + c^2 = ab + bc + ca + 6$$
$$\implies 2a^2 + 2b^2 + 2c^2 = 2ab + 2bc + 2ca + 12$$
$$\implies (a - b)^2 + (b - c)^2 + (c - a)^2 = 12$$
*
*Using AM-GM Inequality, we have:
$$(a - b)^2 + (b - c)^2 + (c - a)^2 \geq 3 \sqrt[3]{((a - b)(b - c)(c - a))^2}$$
$$\implies 3 \sqrt[3]{P^2} \leq 12$$
$$\implies -8 \leq P \leq 8$$
*
*Therefore, $\min P = -8$
Is this solution correct? If not, then why?
| We need to show that the value $-8$ occurs, otherwise, we can not say that it's a minimal value.
By the way, we can prove that the minimum is $-4\sqrt2.$
Indeed, we need to prove that
$$(a-b)(b-c)(c-a)\geq-4\sqrt2\left(\sqrt{\frac{\sum\limits_{cyc}(a^2-ab)}{6}}\right)^3,$$ which easy to get by AM-GM after substitution $a-b=x$, $b-c=y$.
| {
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Proving that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1\implies (a+1)(b+1)(c+1)\geq 64$ where $a,b,c>0$. I am trying to solve the following problem:
Let $a,b,c>0$ with $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} = 1$. Prove that:
$$(a+1)(b+1)(c+1)\geq 64$$
So far, I have gotten that by AM-GM, $(a+1)\geq 2\sqrt{a}$, $(b+1)\geq 2\sqrt{b}$ and $(c+1)\geq 2\sqrt{c}$ so:
$$(a+1)(b+1)(c+1)\geq 8\sqrt{abc} \tag{1}$$
Then using AM-GM on $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$, we get that:
\begin{align}
\frac{1}{a}+\frac{1}{b}+\frac{1}{c} & \geq 3\times \frac{1}{\sqrt[3]{abc}} \\
1 & \geq 3\times \frac{1}{\sqrt[3]{abc}} \\
\frac{1}{3} & \geq \frac{1}{\sqrt[3]{abc}} \\
\frac{1}{27} & \geq \frac{1}{abc} \\
abc & > 27
\end{align}
Substituting this into $(1)$, we get:
$$(a+1)(b+1)(c+1)\geq 8\sqrt{27}$$
So I clearly went wrong somewhere, though I'm not sure why. It would be best if you could provide a solution using AM-GM or Muirhead's Inequality. The question seems so simple, where equality happens at $a=b=c=3$, though I can't prove this. Thank you in advance for your solutions and hints!
| we use $a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}$,with $x+y+z=1$
we have to prove $(1+x)(1+y)(1+z)\ge 64xyz$
But $$1+x=x+y+z+x\ge 4{(x^2yz)}^{1/4}$$ similarly can you do it for $1+y$ and $1+z$ ?.
Multiplying the above results you get dezired inequality.
| {
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Bounds on $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}$ It's simple. What are the bounds on $\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}$ as $a,b,c>0$. Thanks in advance!
Edit: I need bound that can actually be touched i.e. $\alpha$ and $\beta$ such that $\alpha\leq\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}\leq\beta$
| If $a,\,b,\,c$ are positive real numbers then the answer is $2.$
Indeed, we have
$$\frac{c+a}{a+b+c}-\frac{a}{a+b}=\frac{bc}{(a+b)(a+b+c)} \geqslant 0,$$
so
$$\frac{a}{a+b} \leqslant \frac{c+a}{a+b+c},$$
therefore
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a} \leqslant \frac{(c+a)+(a+b)+(b+c)}{a+b+c}=2.$$
In addition, from
$$\frac{a}{a+b}>\frac{a}{a+b+c},$$
we get
$$\frac{a}{a+b}+\frac{b}{b+c}+\frac{c}{c+a}>1.$$
| {
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Please give an algebraic proof of the following inequality: $\frac{a}{a+b} \gt \frac{a-1}{a+b-1}$ for $a,b \gt 0$. For $ a,b > 0 $ and $\left(\frac{a}{a+b}\right)\cdot\left(\frac{a-1}{a+b-1}\right)=\frac{1}{3}$, show that $$\frac{a}{a+b} > \frac{a-1}{a+b-1}$$
I am not getting how to do this. Please enlighten me. What i need is an algebraic proof.
My thought, let $f(x) = \frac{x}{x+b}$, and $f'(x) = \frac{b}{(x+b)^2}$, this implies $f(x)$ is increasing, i.e, $f(x) > f(x-1)$. But can we solve it simply, I tried solving just using inequalities I can't find how.
| Multiplying both sides by $\frac{a}{a+b}$ you get because of your hypothesis :
$$\frac{a}{a+b} > \frac{a-1}{a+b-1} \implies (\frac{a}{a+b})^2 > \frac{1}{3} \implies a(\frac{\sqrt{3}-1}{\sqrt{3}}) > b\frac{1}{\sqrt{3}} \implies a>\frac{b}{\sqrt{3}-1}$$
But this is not always the case.
| {
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Show that $-1^2+3^2-5^2\mp ...+(2^n-1)^2=2^{2n-1}$ Playing with numbers, I construct following expression.
Can it be shown that
$$\sum_{i=1}^{2^{n-1}}(-1)^i(2i-1)^2=2^{2n-1}$$
attempt
We can construct following, using finite calculus as
$(-1)^2+3^2+7^2+...+(4k-5)^2 = \binom{k}1+8\binom{k}2+32\binom{k}3\quad\quad eq(1)$
$1^2+5^2+9^2+...+(4k-3)^2 = \binom{k}1+24\binom{k}2+32\binom{k}3\quad\quad eq(2)$
Let $4k-1=2^n-1$ so we can write above claim as
$eq(1)-eq(2)-1+(4k-1)^2=2^{2n-1}$
Which is equivalent to show
$(2^n-1)^2-16\binom{2^{n-2}}2-1= 2^{2n-1}$
Here I'm stuck. Thanks
| For $n\leftarrow 2$, the equality holds.
Let's assume it holds for $n\leftarrow k\in\Bbb{N}$, $k\ge 2$. Then
$$\color{black}{\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 = 2^{2k-1}}$$
We want to show that the equality holds for $k+1$, i.e., that
$$\color{red}{\sum_{i=1}^{2^{k}}(-1)^i(2i-1)^2 = 2^{2k+1}}$$
In fact
$\color{black}{\begin{aligned}
\sum_{i=1}^{2^{k}}(-1)^i(2i-1)^2 &&=\\
\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=2^{k-1}+1}^{2^k}(-1)^i(2i-1)^2 &=\\
\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=1}^{2^{k-1}}(-1)^{2^{k-1}+i}(2(2^{k-1}+i)-1)^2 &=\\
\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=1}^{2^{k-1}}(-1)^i(2^k+2i-1)^2 &=\\
\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=1}^{2^{k-1}}(-1)^i(2^{2k}+2^{k+1}(2i-1)+(2i-1)^2) &=\\
2\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1)^2 &+ \sum_{i=1}^{2^{k-1}}(-1)^i(2^{2k}+2^{k+1}(2i-1)) &=\\
2(2^{2k-1}) &+ \underbrace{\sum_{i=1}^{2^{k-1}}(-1)^i(2^{2k})}_0 + \sum_{i=1}^{2^{k-1}}(-1)^i(2^{k+1}(2i-1)) &=\\
2^{2k} &+ 2^{k+1}\sum_{i=1}^{2^{k-1}}(-1)^i(2i-1) &
\end{aligned}}$
It's well known (and easy to show) that $\begin{aligned}\sum_{i=1}^{m}(-1)^i(2i-1) = (-1)^m m\end{aligned}$, so it is proven that
$$\sum_{i=1}^{2^{k}}(-1)^i(2i-1)^2 = 2^{2k} + 2^{k+1}2^{k-1} = 2^{2k+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3855368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
If $a+b+c=3$, then show $\sqrt{a^2 + ab +b^2}+ \sqrt{b^2 + bc +c^2}+\sqrt{c^2 + ac +a^2} \geq \sqrt{3}$ If $a+b+c=3$, and $a,b,c$ are positive real numbers, then show $\sqrt{a^2 + ab +b^2} + \sqrt{b^2 + bc +c^2} +\sqrt{c^2 + ac +a^2}\geq \sqrt{3}$
Normally when I do inequalities I try to first find where equality would be achieved, but in this case I have no idea where to start to find it.
From $a^2 - 2ab +b^2 \geq 0$ we can obviously get $a^2 -ab +b^2\geq 3ab$, and after substituting that, we can change the inequality to showing that $\sqrt{ab} + \sqrt{bc} +\sqrt{ac} \geq 1$, but clearly if a=b=c, then equality is not achieved. I wanted to try to use Cauchy-Schwarz or something of the sort, but those inequalities generally are used to find the upper bound of sums of roots.
| By $a + b + c = 3$, the inequality is equivalent to
$$
\sum_{cyc}{\sqrt{(3-c)^2-ab}}\geqslant \sqrt{3}
$$
$$
\sum_{cyc}{9 - 6c + c^2 - ab}\geqslant 3
$$
Which is true because $\sum_{cyc}9-6c = 9$ and $a^2 + b^2 + c^2\geqslant ab + bc + ca$.
And $\sum_{cyc}{\sqrt{((3-c)^2-ab)((3-b)^2 - ca)}} \geqslant0$ is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3861614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
For $a,b,c>0$ proving $\frac{a^2}{b} + \frac{b^2}{c} + \frac{c^2}{a} \geqslant a + b + c + \frac{4(a - b)^2}{a + b + c}$ The problem with which I have a problem it's this:
For $a,b,c>0$ prove that
$$
\frac{a^2}{b} +
\frac{b^2}{c} +
\frac{c^2}{a} \geqslant
a + b + c +
\frac{4(a - b)^2}{a + b + c}
$$
Titu's Lemma and AM-GM work no good because this similar looking inequality is sharper.
After trying these, I decided to go for the following.
Here is my work:
Multiply $abc(a + b + c)$ to both the sides,
$$
(a + b + c)
(a^3c + ab^3 + bc^3) \geqslant abc(a^2 + b^2 + c^2 + 2ab + 2bc + 2ca) + 4abc (a - b) ^2
$$
After some work we are left to prove that:
$$
\sum_{cyc} {a^2b^3 + ab^4 - 2a^2 b^2 c} \geqslant abc(4 a^2 + 4 b^2 - 8ab)
$$
How to prove this or there is some better approach?
This inequality was the first in a list of many inequalities, but I don't think that means it is easy.
Thanks for help!
| By C-S
$$\sum_{cyc}\frac{a^2}{b}-\sum_{cyc }a=\sum_{cyc}\left(\frac{a^2}{b}-2a+b\right)=\sum_{cyc}\frac{(a-b)^2}{b}\geq$$
$$\geq\frac{(a-b+c-b+a-c)^2}{a+b+c}=\frac{4(a-b)^2}{a+b+c}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3870314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding specific values of $n$ such that $n^2+3n+3$ is factorisable with certain constraints on the factors I received the following question during a Maths Olympiad:
Let $n$ be a positive integer. When is it possible to express $n^2+3n+3$ into the form $ab$ with $a$ and $b$ being positive integers, and such that the difference between $a$ and $b$ is smaller than $2\sqrt{n+1}$?
My work so far:
WLOG $a<b$. We can factor this quadratic into $(n+1)(n+2)+1$. This means both $n+1$ and $n+2$ cannot divide $n^2+3n+3$, and $a$ can at most be $n$. If $a=n$, then $b=\frac{n^2+3n+3}n=n+3+\frac3n>n+3$ and $b$ must be equal to or larger than $n+4$. The difference between $a$ and $b$ is at least $4$. We need $4<2\sqrt{n+1}\Rightarrow2<\sqrt{n+1}\Rightarrow4<n+1\Rightarrow n>3$.
Apart from this, $n^2+3n+3\equiv1,3\pmod 6$. It also seems that apart from $3$, the only primes that divide $n^2+3n+3$ are of the form $6k+1$. However this is conjecture and based on the few test cases I checked.
Quadratic residues might come in handy in this problem but I'm not sure.
How could I do this problem?
| Let $c:=b-a$, so that $|c|<2\sqrt{n+1}$ and $a$ and $-b$ are roots of
$$X^2+(b-a)X-ab=X^2+cX-(n^2+n+3).$$
Because this quadratic has two integral roots its discriminant $\Delta$ is a perfect square, where
$$\Delta=c^2+4(n^2+3n+3).$$
Of course $c^2\geq0$ so
$$\Delta\geq4(n^2+3n+3)=(2n+3)^2+3,$$
which shows that $\sqrt{\Delta}>2n+3$. On the other hand, because $c<2\sqrt{n+1}$ it follows that
$$\Delta<4(n+1)+4(n^2+3n+3)=4(n^2+4n+4)=(2n+4)^2,$$
which shows that $\sqrt{\Delta}<2n+4$, a contradiction. Hence it is never possible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3872041",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
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