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The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$. Find the other two roots. The complex number $(1+i)$ is root of polynomial $x^3-x^2+2$.
Find the other two roots.
$(1+i)^3 -(1+i)^2+2= (1-i-3+3i)-(1-1+2i) +2= (-2+2i)-(2i) +2= 0$.
The other two roots are found by division.
$$
\require{enclose}
\begin{array}{rll}
x^2 && \hbox{} \\[-3pt]
x-1-i \enclose{longdiv}{x^3 -x^2 + 2}\kern-.2ex \\[-3pt]
\underline{x^3-x^2- i.x^2} && \hbox{} \\[-3pt]
2 +i.x^2
\end{array}
$$
$x^3-x^2+2= (x-1-i)(x^2) +2+i.x^2$
How to pursue by this or some other approach?
| One of the other roots is $1-i,$ since complex roots of real polynomials come in complex conjugate pairs.
The sum of the roots is $-1$ times the coefficient of $x^2.$ So, if $r$ is the third root, $(1+i)+(1-i)+r=1,$ or $r=-1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4464516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Proving contour integral equal to zero Let $G$ be the path traversed once as shown:
Show that $\displaystyle{\int_{G}{\dfrac{1}{v^4-1} \text{d}v} = 0}$.
By partial fraction decomposition,
$\dfrac{1}{v^4 -1} = \dfrac{1}{4} \left( \dfrac{1}{v-1} - \dfrac{1}{v+1} + \dfrac{i}{v-i} - \dfrac{i}{v+i} \right)$
The singular points $v = \pm 1, \pm i$ all lie inside the contour $G$. Thus, from this theorem (*), we have
\begin{align*}
\int_{G}{\dfrac{1}{v^4-1} \text{d}v} &= \dfrac{1}{4} \left( \int_{G}{\dfrac{1}{v-1}\text{d}v} - \int_{G}{\dfrac{1}{v+1}\text{d}v} + \int_{G}{\dfrac{i}{v-i}\text{d}v} - \int_{G}{\dfrac{i}{v+i}\text{d}v} \right) \\
&= \dfrac{1}{4}\left( 2\pi i - 2\pi i + i\left( 2\pi i \right) - i \left( 2\pi i \right) \right) \\
&= \dfrac{1}{4} \left( 0 \right) \\
&= 0
\end{align*}
(*) Theorem: Let $C$ be a simple closed contour with a positive orientation such that $v_0$ lies interior to $C$, then $\displaystyle{\int_{C} {\dfrac{dv}{(v-v_0)^n}} = 2\pi i}$ for $n =1$ and $0$ when $n \neq 1$ is an integer.
Is that proof correct? If so, could you also point out if there are still theorems I have to mention to make it more accurate?
I'm trying to solve (perhaps overthink) this with the other approach:
We see that it is analytic except at $\pm 1$ and $ \pm i$.
Also, we can apply deformation of the contour $G$ by forming a leaf-like contour and forming the respective circles $C_1, C_2, C_3,$ and $C_4$. As shown here:
The integration can then be evaluated as
$$ \int_{G}{\dfrac{1}{v^4-1} \text{d}v} = \int_{C_1}{\dfrac{1}{v^4-1} \text{d}v} + \int_{C_2}{\dfrac{1}{v^4-1} \text{d}v} + \int_{C_3}{\dfrac{1}{v^4-1} \text{d}v} + \int_{C_4}{\dfrac{1}{v^4-1} \text{d}v} $$
And,
$$\int_{C_n}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4} \left( \int_{C_n}{\dfrac{1}{v-1}\text{d}v} - \int_{C_n}{\dfrac{1}{v+1}\text{d}v} + \int_{C_n}{\dfrac{i}{v-i}\text{d}v} - \int_{C_n}{\dfrac{i}{v+i}\text{d}v} \right) $$
Note that when $v_n$ lies exterior to $C_n$, then by Cauchy-Goursat theorem, $\displaystyle{\int_{C_n}{\dfrac{dv}{v-v_n}} = 0}$.
Thus, for $n = 1,$,
$$\int_{C_1}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}(0-0 + i(2\pi i)- 0) = \dfrac{- \pi }{2} $$
for $ n = 2,$
$$\int_{C_2}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}( 2\pi i - 0 + 0-0) = \dfrac{ \pi i}{2}$$
for $ n = 3,$
$$\int_{C_3}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}(0 - 0 + 0 - i (2\pi i) ) = \dfrac{\pi }{2}$$
for $ n = 4,$
$$\int_{C_4}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{1}{4}(0 -(2\pi i) + 0-0 ) = \dfrac{ - \pi i }{2}$$
Therefore, $$ \int_{G}{\dfrac{1}{v^4-1} \text{d}v} = \dfrac{- \pi }{2} + \dfrac{ \pi i}{2} + \dfrac{\pi }{2} + \dfrac{ - \pi i }{2} = 0$$
Did I just overcomplicate it? Is my first proof already enough? If any of these proofs are correct, could you also point out if there are still theorems I have to mention for them to make it more accurate?
| The shortest proof is as follows:
Let $\displaystyle f(z)=\frac{1}{z^4-1}$. Let $\displaystyle I=\int \limits _Cf(z)\,dz$.
Note that both the contour $C$ and the function $f$ are invariant under the rotation operation $z\mapsto iz$. Therefore, $I=iI$. Then $I=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4465771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to solve $x\tan^2 x-\tan x+x-1=0$ One of my homework questions is to find where the gradient of $\frac{x}{1+\tan x}$ is zero. Using the quotient rule got me to $\frac{1+\tan x-x\sec^2 x}{(1+\tan x)^2}$. I know I can ignore the denominator when the equation is equal to $0$, thus resulting in $1+\tan x-x\sec^2 x$, which I rewrote as $x\tan^2 x-\tan x+x-1$ (using $\sec^2 A=\tan^2 A+1$, and multiplying through by $-1$)
So how do I solve this equation now?
| If you remember that, up to few weeks ago, $x=\cos(x)$ did not show explicit solution even with special functions, there is no hope to find the zero of function
$$f(x)=x\tan^2 (x)-\tan (x)+x-1$$and numerical methods would be required.
However, you can have quite good approximations.
First, inspection
$$f(0)=-1 \qquad f\left(\frac{\pi }{4}\right)=\frac{\pi }{2}-2 <0\qquad f\left(\frac{\pi }{3}\right)=\frac{4 \pi }{3}-1-\sqrt{3} >0$$ the solution is just above $\frac{\pi }{4}$.
Make a series expansion around this value
$$f(x)=\frac{1}{2} (\pi -4)+\pi \left(x-\frac{\pi }{4}\right)+2 (1+\pi )
\left(x-\frac{\pi }{4}\right)^2+O\left(\left(x-\frac{\pi }{3}\right)^4\right)$$
So, using more and more terms and solving the polynomial in $\left(x-\frac{\pi }{4}\right)$, you will have
$$x_{(1)}=\frac \pi 4+\left(\frac{2}{\pi }-\frac{1}{2}\right)=0.922018$$
$$x_{(2)}=\frac \pi 4+\frac{\sqrt{16+3 (4-\pi ) \pi }-\pi }{4 (1+\pi )}=0.892036$$
while the solution is $x=0.887829$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467163",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $(1+\frac{a^2+b^2+c^2}{ab+bc+ca})^{\frac{(a+b+c)^2}{a^2+b^2+c^2}} \leq (1+\frac{a}{b})(1+\frac{b}{c})(1+\frac{c}{a})$
Assuming $a,b,c>0$, show that
$$\Big(1+\frac{a^2+b^2+c^2}{ab+bc+ca}\Big)^{\frac{(a+b+c)^2}{a^2+b^2+c^2}} \leq \Big(1+\frac{a}{b}\Big)\Big(1+\frac{b}{c}\Big)\Big(1+\frac{c}{a}\Big).$$
I know from CS that $ab+bc+ca \leq a^2+b^2+c^2$ and $(a+b+c)^2 \leq 3(a^2+b^2+c^2)$ so the exponent is less than 3 and the second term in parenthesis greater than $1$, but I can't manage to convert this information, might work on the right hand side but seems like I'm missing a classical inequality since I'm a very beginner in this.
I noticed this inequality is symmetrical and homogeneous, maybe assuming $a+b+c=1$ could be useful...
| Let $x = \frac{a^2+b^2+c^2}{ab + bc + ca}$. Then $x \ge 1$.
Using Bernoulli inequality, we have
$$\mathrm{LHS} = (1 + x)^{1 + 2/x} = (1 + x)\Big((1 + x)^{1/x}\Big)^2
\le (1 + x)\left(1 + x\cdot \frac{1}{x}\right)^2 = 4 + 4x.$$
It suffices to prove that
$$4 + \frac{4(a^2+b^2+c^2)}{ab + bc + ca} \le \frac{(a + b)(b + c)(c + a)}{abc}$$
or (clearing the denominators)
$$(b-c)^2a^3 + (b^3 + c^3)a^2 - 2bc(b^2 + c^2)a + b^2c^2(b + c) \ge 0.$$
It suffices to prove that
$$ (b^3 + c^3)a^2 - 2bc(b^2 + c^2)a + b^2c^2(b + c) \ge 0$$
which is true since
$$(b^3 + c^3)a^2 + b^2c^2(b + c) \ge 2\sqrt{(b^3 + c^3)a^2 \cdot b^2c^2(b + c)}
\ge 2abc(b^2 + c^2)$$
where we have used AM-GM and Cauchy-Bunyakovsky-Schwarz.
We are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Proof $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le \lfloor{x + y}\rfloor $ I want to prove that $ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le \lfloor{x + y}\rfloor $, I started proving but I got stuck.
Let $ x, y \in \mathbb{R} $. Therefore:
$ \lfloor{x}\rfloor \le x $
$ \lfloor{y}\rfloor \le y $
$ \Rightarrow $
$ \lfloor{x}\rfloor + \lfloor{y}\rfloor \le x + y $
| Is this proof OK, using both (1) integers $n = \left\lfloor n \right\rfloor$, and (2) if $a\le b$ then $\left\lfloor a\right\rfloor \le \left\lfloor b \right\rfloor$?
$$\left\lfloor x\right\rfloor + \left\lfloor y \right\rfloor
= \big\lfloor \left\lfloor x\right\rfloor + \left\lfloor y \right\rfloor \big\rfloor
\le \left\lfloor x + y \right\rfloor$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Prove that $\lim_{n\to\infty} \sum_{k=1}^{n}$ $(\frac {k}{n^2})^{{k\over n^2}+1} = {1\over 2}$. To simplify the question I imagined what if $k\over n^2$= $x$ then it will look like $x^{x+1}$.Else I couldn't think on how to proceed ahead.
In one of my previously asked question viz. Prove that $\lim_{n\to\infty} n^2 \int _0^{1/n}{x^{x+1}}dx = {1\over2}$.. Can I relate both of these questions?
| We will show that
$$\lim_{n\to\infty}\frac{\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}}{1+\frac{1}{n}}=\frac{1}{2}$$
Your desired limit will then follow from $\lim_{n\to\infty}\left(1+1/n\right)=1$ and the product rule for limits.
$$\lim_{n\to\infty}\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}=\lim_{n\to\infty}\left[\left(1+\frac{1}{n}\right)\frac{\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}}{1+\frac{1}{n}}\right]=1\cdot\frac{1}{2}=\frac{1}{2}$$
Fix $\varepsilon>0$ arbitrarily. Since $\lim_{x\to 0^+}x^{x+1}/x=1$ (see my answer in your linked post), there is a $\delta>0$ satisfying
$$\left|\frac{x^{x+1}}{x}-1\right|<2\varepsilon\text{ whenever }0<x<\delta$$
which says the same thing as
$$\begin{equation}(1-2\varepsilon)x<x^{x+1}<(1+2\varepsilon)x\text{ whenever }0<x<\delta\end{equation}\tag{$\dagger$}$$
Pick $N\in\mathbb N$ sufficiently large for $1/N\leq \delta$. Then for any $n>N$ and any $k\in\mathbb N$ with $1\leq k\leq n$, we can see from division by $n^2$ that
$$\frac{1}{n^2}\leq\frac{k}{n^2}\leq\frac{n}{n^2}=\frac{1}{n}<\frac{1}{N}\leq\delta$$
so $0<k/n^2<\delta$ for every $n>N$ and $k\in\{1,\dots,n\}$. From $(\dagger)$, we get
$$(1-2\varepsilon)\frac{k}{n^2}<\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}<(1+2\varepsilon)\frac{k}{n^2}\text{ whenever }n>N\text{ and }1\leq k\leq n$$
Summing from $k=1$ to $n$ gives
$$\sum_{k=1}^n(1-2\varepsilon)\frac{k}{n^2}<\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}<\sum_{k=1}^n(1+2\varepsilon)\frac{k}{n^2}\text{ whenever }n>N$$
and, after evaluating the left and rightmost sums with $\sum_{k=1}^n k=n(n+1)/2$ and doing some algebra,
$$\left(\frac{1}{2}-\varepsilon\right)\left(1+\frac{1}{n}\right)<\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}<\left(\frac{1}{2}+\varepsilon\right)\left(1+\frac{1}{n}\right)\text{ whenever }n>N$$
Dividing through by $1+1/n$ and then subtracting by $1/2$ shows that
$$\left|\frac{\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}}{1+\frac{1}{n}}-\frac{1}{2}\right|<\varepsilon\text{ whenever }n>N$$
and since $\varepsilon>0$ was arbitrary, this implies
$$\lim_{n\to\infty}\frac{\sum_{k=1}^n\left(\frac{k}{n^2}\right)^{\frac{k}{n^2}+1}}{1+\frac{1}{n}}=\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $0
Show that $0<e-\sum\limits_{k=0}^n\frac{1}{k!}<\frac{1}{n!n}$, where $n>0$.
Hint:
Show that $y_m:=\sum\limits_{k=n+1}^{m+n}\frac{1}{k!}$ has the limit $\lim\limits_{m\to\infty}y_m=e-\sum\limits_{k=0}^n\frac{1}{k!}$ and use that $(m+n)!y_m<\sum\limits_{k=1}^m\frac{1}{(n+1)^{k-1}}$.
If I use the hint then the problem is pretty straightforward.
However, I don't understand why the inequality of the hint is true? A simple example shows that the inequality doesn't hold! There are no further information given.
What am I missing? Is $(m+n)!y_m<\sum\limits_{k=1}^m\frac{1}{(n+1)^{k-1}}$ simply nonsense? Or did the professor forgot to mention some more assumptions?
| We have
$$\begin{align}y_m &= \frac{1}{n!}\left[ \frac{1}{n+1} + \frac{1}{(n+1)(n+2)} + \ldots + \frac{1}{(n+1)(n+2)\cdots(n+m)}\right] \\&<\frac{1}{n!}\left[ \frac{1}{n+1} + \frac{1}{(n+1)^2} + \ldots + \frac{1}{(n+1)^m}\right]\end{align}$$
Evaluating the finite geometric sum on the RHS it follows that
$$y_m < \frac{1}{n!}\frac{\frac{1}{n+1}- \frac{1}{(n+1)^{m+1}}}{1 - \frac{1}{n+1}}= \frac{1}{nn!}\left[1 - \frac{1}{(n+1)^{m}} \right] \underset{ m \to \infty}\longrightarrow
\frac{1}{nn!} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find all positive integer solutions for $3^x-2^y=1$.
Find all positive integer solutions for $3^x-2^y=1$.
Quickly we can find the solutions $(1,1)$ and $(2,3)$. Now the claim is that for $x \ge 2$ and $y \ge 3$ there are no solutions.
The equation can be expressed as $3^x=2^y+1$ from where we can get to $3^x-3 = 2^y-2 \implies 3(3^{x-1}-1)=2(2^{y-1}-1)$ now from this clearly $2$ is not a multiple of $3$ so it must be that $$2^{y-1}-1 \equiv 0 \pmod 3 \implies 2^{y-1} \equiv 1 \pmod 3$$ and the order of $2$ modulo $3$ is $2$ so $y-1$ is a multiple of two say $y-2=2t$.
Plugging this back to the lhs of the equation we get $$3(3^{x-1}-1)=2(2^{2t}-1)=2(2^2-1)(2^{2(t-1)}+ \dots +1) = 6(2^{2(t-1)}+ \dots +1)$$ now $3$ is not a multiple of $6$ so we must have that $$3^{x-1}-1\equiv 0 \pmod 6 \implies 3^{x-1} \equiv1 \pmod 6$$ but noticing that the only way this can happen is if $x-1=0$ i.e. $x=1$, but this contradicts our assumption that $x \ge2$ thus there are no solutions for $x \ge 2$.
Is this a correct argument? I think there are multiple ways to do the problem, but this seemed most natural to me.
| We are going to assume there is a larger solution than $9-8=1$ and get a contradiction.
We have
$$ 3^u = 2^v + 1. $$ Subtract $9$ from both sides, we have
$$ 3^u - 9 = 2^v - 8. $$
Taking $u = x + 2$ and $v = 3 + y$ gives $9 \cdot 3^x - 9 = 8 \cdot 2^y - 8,$ or
$$ 9 (3^x - 1 ) = 8 (2^y - 1). $$ We think this is only possible with $x=y=0,$ so we assume there is a solution with $x,y > 0$ and get a contradiction.
First we have $9 |(2^y - 1),$ or $$ 2^y \equiv 1 \pmod 9.$$ This tells us that $ 6 | y. $ Meanwhile
$$ 2^6 - 1 = 63 = 3^2 \cdot 7. $$ Therefore $7 | (3^x - 1).$ In turn, we find $6 | x.$
Next
$$ 3^6 - 1 = 728 = 2^3 \cdot 7 \cdot 13. $$ Therefore $13 | (2^y - 1).$ In turn, we find $12 | y.$
Next
$$ 2^{12} - 1 = 4095 = 3^2 \cdot 5 \cdot 7 \cdot 13. $$ Therefore $5 | (3^x - 1).$ In turn, we find $4 | x.$
Finally, $3^x - 1$ is divisible by $3^4 - 1 = 80.$ In particular,
$3^x - 1$ is divisible by $2^4 = 16.$ However, this contradicts $2^y - 1 \neq 0$ and $y > 0.$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Minimum value of $ab+bc+ca$ Given that $a,b,c \in \mathbb{R^+}$ and $(a+b)(b+c)(c+a)=1$
Find the Minimum value of $ab+bc+ca$
My try: Letting $x=a+b, y=b+c, z=c+a$ we get
$$a=\frac{x+z-y}{2}, b=\frac{x+y-z}{2},c=\frac{y+z-x}{2}$$
$$xyz=1$$
Now the problem is to minimize $$ab+bc+ca=\frac{xy+yz+zx}{2}-\frac{x^2+y^2+z^2}{4}$$
Now by $A.M-G.M$ we have
$$\frac{xy+yz+zx}{2}\geq \frac{3}{2}\times \sqrt[3]{x^2y^2z^2}=\frac{3}{2}$$
But I am stuck for $\frac{-(x^2+y^2+z^2)}{4}$
| This doesn't have a well-defined minimum. Solve
$$(a+b)(b+c)(c+a)=1$$
for $c$ to get
$$
c=\frac12\left(-(a+b)\pm\sqrt{(a+b)^2-4\left(ab-{1\over a+b}\right)}\right)
$$
Let $a=b=1/N$ where $N$ is some
"big enough" number, then $a+b\approx0$ and $c\approx\sqrt N$ and $ab+bc+ca\approx {1/\sqrt N}\approx 0$, up to factors of two. Obviously, it never reaches the minimum though, since $a,b,c>0$, so we don't have a well-defined minimum value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Compute expected value from two dimensional normal distribution
Random vector (X, Y) has the two-dimensional normal distribution
with the density $ f(x,y)=\frac{1}{2\pi\sqrt{2}}\exp\left\{
-\frac{1}{8} \left[ 4x^2 +8x(y+3)+6(y+3)^2 \right] \right\} $
Compute $\mathbb{E} \left[ XY - 2X - 3Y^2 + 1 \right ]$
Find the covariance matrix of the random vector $(U, V) = (-2X+Y, X-Y)$
Solution:
$det \; C = 2$
$
f(x,y)=\frac{1}{2\pi\sqrt{2}}\exp\left\{ -\frac{1}{4} \left[ 2x^2 +4x(y+3)+3(y+3)^2 \right] \right\}
$
$
m = \begin{pmatrix}
0\\
-3
\end{pmatrix} = \begin{pmatrix}
\mathbb{E} X\\
\mathbb{E} Y
\end{pmatrix}
$
$
C = \begin{pmatrix}
3 & 2\\
2 & 2
\end{pmatrix} = \begin{pmatrix}
Var X & Cov(X, Y)\\
Cov(X, Y) & Var Y
\end{pmatrix}
$
$
\mathbb{E} \left[ XY - 2X - 3Y^2 + 1 \right ] = \mathbb{E}[XY] - 2\mathbb{E}X - 3\mathbb{E}[Y^2] + 1
$
$
Cov(X,Y)=\mathbb{E}[XY]-\mathbb{E}X\mathbb{E}Y \rightarrow \mathbb{E}[XY]=2
$
$
Var Y = \mathbb{E}(Y^2)-(\mathbb{E}Y)^2 \rightarrow (\mathbb{E}Y)^2 = 11
$
$
\mathbb{E} \left[ XY - 2X - 3Y^2 + 1 \right ] =2-2\cdot0-3\cdot11+1=-30
$
$
A \cdot \begin{pmatrix}
X\\
Y
\end{pmatrix} = \begin{pmatrix}
-2X & Y\\
X & -Y
\end{pmatrix} \rightarrow A = \begin{pmatrix}
-2 & 1\\
1 & -1
\end{pmatrix}
$
$
C_{UV} = A \cdot C \cdot A^T = \begin{pmatrix}
6 & -2\\
2 & 1
\end{pmatrix}
$
I am asking for verification of the solution and corrections if needed.
| The term in the exponent of $f(x,y)$ can be written as
$$\begin{split}-\frac 18\left[4x^2+8x(y+3)+6(y+3)^2\right] &= -\frac 12 \left[x^2+2x(y+3)+\frac 32(y+3)^2\right]\\
&=-\frac 12\begin{pmatrix}x\\
y+3\end{pmatrix}^T\begin{pmatrix}1&1\\1&3/2\end{pmatrix}\begin{pmatrix}x\\y+3\end{pmatrix}\end{split}$$
Thus $\Sigma^{-1}=\begin{pmatrix}1&1\\1&3/2\end{pmatrix}$ and the covariance matrix is $\Sigma=\frac 1{3/2-1}\begin{pmatrix}3/2&-1\\-1&1\end{pmatrix}=\begin{pmatrix}3&-2\\-2&2\end{pmatrix}$
Besides this, your calculations look correct, so just redo the parts that need changing.
| {
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"timestamp": "2023-03-29T00:00:00",
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Which one is the larger : $20!$ or $2^{60}$? Which one is the larger : $20!$ or $2^{60}$ ?
I am looking for an elegant way to solve this problem, other than my solution below. Also, solution other than using logarithm that uses the analogous inequalities below.
My solution:
Write $20!$ in prime factors and $2^{n}$:
$$ 20! = (2^{2} \cdot 5)(19)(2 \cdot 3^{2})(17)(2^{4})(3 \cdot 5)(2 \cdot 7)(13)(2^{2} \cdot 3)(11)(2 \cdot 5)(3^{2})(2^{3})(7)(2 \cdot 3) (5) (2^{2}) (3) (2) $$
$$ = 2^{18} (5)(19)(3^{2})(17)(3 \cdot 5)(7)(13)(3)(11)(5)(3^{2})(7)( 3) (5) (3) $$
so it is left to compare $2^{42}$ and $(5)(19)(3^{2})(17)(3 \cdot 5)(7)(13)(3)(11)(5)(3^{2})(7)( 3) (5) (3) $.
We write the prime factors nicely as:
$$ 3^{8}5^{4}7^{2}(11) (13)(17)(19) $$
Notice
$(3)(11) > 2^{5}$,
$(13)(5)>2^{6}$,
$(19)(7) >2^{7}$,
$17 > 2^{4}$, so we now focus on
$$3^{7}5^{3}7 = 2187(5^{3})7 > 2048(5^{3})7 = 2^{11}875 >2^{11}512 = 2^{20} $$
So we have that the prime factors is larger than $2^{42}$.
| This answer extends the comment of zwim.
First, see
Stirling approximation For factorials.
I know that as $n$ goes to $\infty$, that the geometric mean of
$\{1,2,\cdots,n\}$ approaches $~\displaystyle \frac{n}{e}~$ from above. Further, as $n$ increases, the ratio between $n$ and the geometric mean of $\{1,2,\cdots,n\}$ is strictly decreasing.
I also know, from involvement with a prior (similar) problem involving $(100)!$, that even with the number as large as $(100)$, the geometric mean of $\{1,2,\cdots,100\}$ is still greater than $(40)$.
This implies that since $(20) < (100)$, the geometric mean of $\{1,2,\cdots,20\}$ must be greater than $~(20 \times 0.4).~$
This surmise makes it game over, because $2^{60} = 8^{20}$. Therefore, since the geometric mean of $\{1,2,\cdots,20\}$ is greater than $(8)$, you must have that $(20)! > 8^{20}.$
| {
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"answer_count": 3,
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} |
What is wrong with my proof that $\int 2x dx= 2x^2$ by writing $2x=\underbrace{2+2+\cdots+2}_{x\;\text{times}}$? I know $\int 2x \,dx = x^2 + C$ (by the power rule) but why does the following proof not give the same answer?
\begin{align*}
\int 2x \,dx &= \int \underbrace{(2 + 2 + 2 + \dots + 2)}_{x \text{ times}} \, dx \\
&= \underbrace{\int{2} \, dx + \int{2} \, dx + \dots \ + \int{2}_ \, dx}_{x \text{ times}}\\
&= 2x + 2x + \dots + 2x + C \\
&= 2x \times x + C \\
&= 2x^2 + C
\end{align*}
(And I have the same question for this false proof that $\int{2^x} \, dx = 2^{x}x+ C$)
\begin{align*}
\int{2^x} \,dx &= \int \underbrace{(2 \cdot 2 \cdot 2 \cdot \dots \cdot 2)}_{x \text{ times}} \cdot 1 \, dx \\
&= 2 \cdot \int \underbrace{(2 \cdot 2 \cdot 2 \cdot \dots \cdot 2)}_{(x-1) \text{ times}} \cdot 1 \, dx && (\text{Constant Multipule Rule})\\
&= 2^2 \cdot \int \underbrace{(2 \cdot 2 \cdot 2 \cdot \dots \cdot 2)}_{(x-2) \text{ times}} \cdot 1 \, dx && (\text{Constant Multipule Rule})\\
&= 2^x \cdot \int{1} \, dx \\
&= 2^{x}x+ C \\
\end{align*}
I suspect that it has something to do with not being able to:
*
*Change integral of sums to sums of integrals for an arbitrary $x$, and
*Remove a constant out of an integral if there are variable numbers of those constants.
But I'm not sure why these do not hold. If this is the reason, is there a theorem stating it?
Thanks in advance!
| The first integral is wrong because $x\in\mathbb{R}$ need not be a natural number. So the multiplication cannot be split as you did. Similar reasoning applies to the second case, that is to say, since $x$ need not be a natural number, the interpretation for $2^{x}$ is wrong.
| {
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} |
Calculate the area of the surface $x^2 + y^2 = 1 + z^2$ as $z \in [- \sqrt 3, \sqrt 3]$
The question states the following: Calculate the area of the surface $x^2 + y^2 = 1 + z^2$ as $z \in [- \sqrt 3, \sqrt 3]$
My attempt
In order to solve this question, the first thing I think about is parametrize the surface so I can then just apply the definition of the area of a surface
$$A(S) = \iint_D || \Phi_x \times \Phi_y|| \ dx \ dy$$
I consider the parametrization $\Phi (x,y) = (x, y, \sqrt{x^2 + y^2 - 1}) \ $. Then $$\begin{cases} \Phi_x = (1,0,\displaystyle \frac{-x}{\sqrt{x^2 + y^2 - 1}}) \\ \Phi_y = (0,1,\displaystyle \frac{-y}{\sqrt{x^2 + y^2 - 1}})\end{cases} \Longrightarrow \Phi_x \times \Phi_y = (\frac{x}{\sqrt{x^2 + y^2 - 1}},\frac{y}{\sqrt{x^2 + y^2 - 1}},1)$$
Then
$$|| \Phi_x \times \Phi_y||= \displaystyle \sqrt{\frac{x^2}{x^2 + y^2 - 1} + \frac{y^2}{x^2 + y^2 - 1} + 1} = \sqrt{\frac{x^2 + y^2}{x^2 + y^2 - 1} + 1} $$
As we work in a symettric surface, we'll consider $z \in [0, \sqrt 3]$ and simply multiply the result by two. Then, the parametrization goes from $D$ to $\mathbb R^3$, $\Phi : D \subset \mathbb R^2 \rightarrow \mathbb R^3$, being $D$ the following domain
$$D = \lbrace (x,y) \in \mathbb R^2 : 1 \leq x^2 + y^2 \leq 4 \rbrace$$
Thus we get
$$A(S) = 2 \cdot \iint_D || \Phi_x \times \Phi_y|| \ dx \ dy = 2 \cdot \iint_D \sqrt{\frac{x^2 + y^2}{x^2 + y^2 - 1} + 1}\ dx \ dy $$
Using polar coordinates, $\begin{cases} x = r \cdot \cos \theta \\ y = r \cdot \sin \theta \end{cases} : r \in [1,2] \ \& \ \theta \in [0, 2\pi]$ we get the following integral
$$A(S) = 2 \cdot \int_0^{2\pi} \int_{1}^2 r \cdot \displaystyle \sqrt{\frac{r^2 \cos^2 \theta + r^2 \sin^2 \theta}{r^2 \cos^2 \theta + r^2 \sin^2 \theta - 1} + 1} \ dr \ d\theta = 2 \cdot \int_0^{2\pi} \int_{1}^2 r \cdot \sqrt{\frac{r^2}{r^2 - 1} + 1} \ dr \ d\theta$$
$$ = 4 \pi \cdot \int_{1}^2 r \cdot \sqrt{\frac{r^2}{r^2 - 1} + 1} \ dr $$
The problem is that I reach the integral above that I don´t know how to tackle. I think I may have done something wrong along the process since this is a question extracted from an university exam where no computers nor calculators were avilable. Any help?
| $x^2+y^2-z^2=1$, $-\sqrt{3}\le z\le\sqrt{3}$
Convert to cylindrical coordinates. Fitting given the symmetry, it's a hyperboloid of 1 sheet.
$z^2=r^2-1\implies r=\sqrt{z^2+1}\implies dr/dz=\frac{z}{\sqrt{z^2+1}}$
Lateral surface area of a cylinder is $2\pi rz$. To generalize, we keep the $2\pi r$ and the $z$ becomes the differential arc length along the curve being rotated about the z axis. So Area=$2\pi r \sqrt{1+(dr/dz)^2}dz$
$2\int_0^\sqrt{3}2\pi \sqrt{z^2+1} \sqrt{1+(dr/dz)^2}dz=4\pi\int_0^{\sqrt{3}}\sqrt{z^2+1}\sqrt{\frac{2z^2+1}{z^2+1}}dz=\int_0^\sqrt{3}4\pi\sqrt{2z^2+1}dz$
Let $\tan{\theta}=z\sqrt{2}$.$\sqrt{2z^2+1}=\sec{\theta}$. $dz=\sec^2{\theta}/\sqrt{2}$.
$\int 2\pi\sqrt{2}\sec^3{\theta}d\theta=2\pi\sqrt{2}(\frac{1}{2}\sec{\theta}\tan{\theta}+\frac{1}{2}\ln|\sec{\theta}+\tan{\theta}|)+C$
$\pi\sqrt{2}(z\sqrt{2}\sqrt{2z^2+1}+\ln|z\sqrt{2}+\sqrt{2z^2+1}|)|_0^\sqrt{3}$
$=\pi\sqrt{2}(\sqrt{42}+\ln|\sqrt{6}+\sqrt{7}|)$
| {
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} |
Prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{ab+bc+ca}{a^2+b^2+c^2}\ge4$
Let $a,b,c>0$. Prove that
$$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{ab+bc+ca}{a^2+b^2+c^2}\ge4$$
I know $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge 3$ but $\dfrac{ab+bc+ca}{a^2+b^2+c^2}\le1$. And then I try $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}\ge \dfrac{(a+b+c)^2}{ab+bc+ca}$ but still stuck. Any help please, thank you.
| Using Cauchy-Schwarz inequality,
\begin{aligned}
\frac ab+\frac b c+\frac ca+\frac{ab+bc+ca}{a^2+b^2+c^2}&\ge \frac{(a+b+c)^2}{ab+bc+ca}+\frac{ab+bc+ca}{a^2+b^2+c^2}\\
&=2+\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{ab+bc+ca}{a^2+b^2+c^2}\\
&\ge 4
\end{aligned}
| {
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} |
Given matrix $B$, find all possible matrices $A$ satisfying $ A(A-2B) = -(A-2B)A $
Let $A, B \in \Bbb R^{3 \times 3}$ such that $A(A-2B) = -(A-2B)A$. Given $$ B = \begin{pmatrix} 2 & -2 & 1 \\ -1 & 3 & -1 \\ 2 & -4 & 3 \end{pmatrix} $$ find all possible matrices $A$ satisfying the equation above.
I hope there is some way faster than compute directly...
| Following the hint of @P.Quinton, we can rewrite the problem as $(A-B)^2=B^2$, and that reduces the problem to finding all $C$ such that $C^2=B^2$. The difficulty here is that $B$ has a repeated eigenvalue, which means that finding all the square roots of $B^2$ is not quite straight forward. After all, any reflection will square to the identity, and so on that eigenspace we can take something that restricts to a reflection.
We first note that $B$ is diagonalizable, with $B=PDP^{-1}$, with $D=\operatorname{diag}(1,1,6)$.
Because of this, $B^2$ will satisfy the polynomial $f(x)=(x-1)(x-6^2)$, and hence so will $C$. But then $C$ will satisfy the polynomial $f(x^2)=(x^2-1)(x^2-6^2)=(x-1)(x+1)(x-6)(x+6)$ which has no repeated roots, so $C$ is also diagonalizable. If $v$ is an eigenvector of $C$, then it is also an eigenvector of $C^2=B^2$, The problem is that when we square $C$, things that weren't eigenvectors for $C$ can become eigenvectors for $C^2$, for example if $Cu=u$ and $Cv=-v$, then $C^2(u+v)=u+v$, but $C(u+v)=u-v\neq \lambda(u+v)$. But if we pick any two vectors that span $\ker(C^2-I)$, then we can pick them to be eigenvectors in $C$. If they both have eigenvalue $1$ or both have eigenvalue $-1$, then the matrix we get for $C$ will not depend on our choice because $C$ restricted to their span will be $\pm I$. If they have different signs, then the choices matter.
For the sake of computation, it is easier to find square roots of $D^2$ and then conjugate to turn those into square roots of $B^2$. If we follow the proceedure above, then we will get block diagonal matrices, $\operatorname{diag}(M,\pm 6)$ where the first block is $2\times 2$ and the last block is $1\times 1$. The first block, $M$ is a matrix satisfying $M^2=I_2$ (the $2\times 2$ identity matrix), which is either $\pm I$ or a matrix with eigenvalues $1$ and $-1$, which can be found easily by specifying a trace of $0$ and a determinant of $-1$. In particular, if you set the first column of $M$, those two conditions will uniquely determine the second column.
Putting everything together, $C$ is of the form $PNP^{-1}$ where $P$ was the matrix used in the diagonalization of $B$ and
$$N=\begin{pmatrix}
a & b & 0 \\
\frac{1-a^2}{b} & -a & 0 \\
0 & 0 & \pm 6
\end{pmatrix}$$
(except when $b=0$, in which case $a=\pm 1$ and the $(2,1)$ entry can be anything), or where $N$ is one of the other forms already discussed.
| {
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How can I show that $x$ and $y $ must have the same length where $ x+y $ and $x-y $ are non-zero vectors and perpendicular?
Problem: Let $x$ and $y$ be non-zero vectors in $\mathbb{R}^n$.
(a) Suppose that $\|x+y\|=\|x−y\|$. Show that $x$ and $y$ must be perpendicular.
(b) Suppose that $x+y$ and $x−y$ are non-zero and perpendicular. Show that $x$
and $y$ must have the same length.
Attempt:
(a)\begin{align*}\|x+y\|^2 & =\|x-y\|^2 \\
(x+y)\cdot (x+y) & =(x-y)\cdot (x-y) \\
\|x\|^2+\|y\|^2+2x\cdot y & =\|x\|^2+\|y\|^2-2x\cdot y \\
2x\cdot y & =-2x\cdot y \\
x\cdot y & =0.
\end{align*}(b)\begin{align*}(x+y)\cdot (x-y) & =0 \\
\|x\|^2-x\cdot y+x\cdot y-\|y\|^2 & =0 \\
\|x\|^2-\|y\|^2 & =0.
\end{align*}Are these attempts correct?
EDITED
| It's unclear from where you got the starting relations. That $x+y$ and $x-y$ are perpendicular means that
$$(x+y)\cdot(x-y) = 0 \tag 1$$
and you can start reasoning from there:
$$
0 \stackrel{(1)}= (x+y)\cdot(x-y) = x^2 +yx - xy -y^2 = \|x\|^2 - \|y\|^2
$$
Thus $\|x\|^2 = \|y\|^2$ and taking square root yields $\|x\| = \|y\|$ which means that $x$ and $y$ have same length.
Let $x$ and $y$ be non-zero vectors in $\Bbb R^n$. Suppose that $x + y = x − y$. Show that $x$ and $y$ must be perpendicular.
Subtracting $x$ yields $y = -y$ and hence $y=0$ which violates the precondition that $y\neq0$.
| {
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Show that $\lim_{n \to \infty} \sum_{k = 0}^\infty (-1)^k \frac{n}{n + k} = \frac{1}{2}$.
Show that
$$ \lim_{n \to \infty} \sum_{k = 0}^\infty (-1)^k \frac{n}{n + k} = \frac{1}{2}. $$
Progress:
I rewrote the inside as
$$ S_n = \sum_{k = 0}^\infty (-1)^k \frac{n}{n + k} = \sum_{k = 0}^\infty (-1)^k \left( 1 - \frac{k}{n + k} \right),$$
at which point I sought to find the (ordinary) generating function
$$A(x) = \sum_{k = 0}^\infty (-1)^k \left( 1 - \frac{k}{n + k} \right) x^k = \frac{1}{1 + x} - \sum_{k = 0}^\infty (-1)^k \left( \frac{k}{n + k} \right) x^k , $$
because then $S_n = A(1)$. Unfortunately, I have been trying to evaluate the last sum using the Taylor Series expansion of $\ln \left( 1 + x \right)$ at $x = 0$, but I am stuck.
While is true that
$$\ln \left( 1 + x \right) = \sum_{k = 1}^\infty (-1)^{k + 1} \left( \frac{1}{k} \right) x^k $$
(for $x \in (-1, 1)$ ; note this is irrelevant if we regard all g.f. as "formal"), I can't figure out how to transform this into $\sum_{k = 1}^\infty (-1)^{k + 1} \left( \frac{1}{n + k} \right) x^k$. The motivation is that once I can find that sum in explicit form, I can differentiate it w.r.t $x$ to find the desired sum. I guess my question is also: "How to shift backwards the coefficients of an ordinary generating function?", though I am eager to see other (possibly more elegant) solutions as well.
Note:
One attempt I tried at shifting the coefficients was dividing the coefficients by some power $x^k$, but this introduces negative coefficients, whereas I need those terms with negative coefficients to disappear.
| If you are willing to accept a solution that doesn't use generating functions, then here are such ones:
1st Solution. Fusing each consecutive even and odd terms into a single term,
\begin{align*}
S_n := \sum_{k=0}^{\infty} (-1)^k \frac{n}{n + k}
&= \sum_{k=0}^{\infty} \left[ \frac{n}{n + 2k} - \frac{n}{n + 2k+1} \right] \\
&= \sum_{k=0}^{\infty} \frac{n}{(n + 2k)(n + 2k+1)} \\
&= \sum_{k=0}^{\infty} \frac{1}{(1 + \frac{2k}{n})(1 + \frac{2k+1}{n})} \cdot \frac{1}{n}.
\end{align*}
The last sum looks like a Riemann sum of $\frac{1}{(1+2x)^2}$ over $[0, \infty)$, and so, it is not unreasonable to expect that $S_n$ will converge to $\int_{0}^{\infty} \frac{1}{(1+2x)^2} \, \mathrm{d}x = \frac{1}{2}$.
Indeed, noting that $x \mapsto \frac{1}{1+2x}$ is decreasing for $x > -\frac{1}{2}$, we find that
$$
\int_{\frac{k+1}{n}}^{\frac{k+2}{n}} \frac{1}{(1+2x)^2} \, \mathrm{d}x
\leq \frac{1}{(1 + \frac{2k}{n})(1 + \frac{2k+1}{n})} \cdot \frac{1}{n}
\leq \int_{\frac{k-1}{n}}^{\frac{k}{n}} \frac{1}{(1+2x)^2} \, \mathrm{d}x
$$
for $n \geq 3$ and $k \geq 0$. (This condition is just to ensure that $\frac{k-1}{n} > -\frac{1}{2}$.) Altogether,
$$
\int_{\frac{1}{n}}^{\infty} \frac{1}{(1+2x)^2} \, \mathrm{d}x
\leq S_n
\leq \int_{-\frac{1}{n}}^{\infty} \frac{1}{(1+2x)^2} \, \mathrm{d}x $$
Letting $n \to \infty$, we conclude:
$$ \lim_{n\to\infty} S_n = \int_{0}^{\infty} \frac{1}{(1+2x)^2} \, \mathrm{d}x = \frac{1}{2}. $$
2nd Solution. We have
\begin{align*}
S_n
&= \lim_{K \to \infty} \sum_{k=0}^{K-1} (-1)^k n \int_{0}^{1} x^{n+k-1} \, \mathrm{d}x \\
&= \lim_{K \to \infty} \int_{0}^{1} \sum_{k=0}^{K-1} n (-1)^k x^{n+k-1} \, \mathrm{d}x \\
&= \lim_{K \to \infty} \int_{0}^{1} n x^{n-1} \frac{1 - (-x)^K}{1 + x} \, \mathrm{d}x \\
&= \int_{0}^{1} \frac{n x^{n-1}}{1 + x} \, \mathrm{d}x \tag{DCT} \\
&= \int_{0}^{1} \frac{1}{1 + y^{1/n}} \, \mathrm{d}y \tag{$y = x^n$} \\
&\to \int_{0}^{1} \frac{1}{1 + y^{0}} \, \mathrm{d}y \quad \text{as } n \to \infty \tag{DCT} \\
&= \frac{1}{2}.
\end{align*}
3rd Solution. Let $f(x) = \frac{1}{1+x}$. Then
\begin{align*}
S_n
= \sum_{k=0}^{\infty} (-1)^k f\bigl(\tfrac{k}{n}\bigr)
&= \sum_{k=0}^{\infty} \left[ f\bigl(\tfrac{2k}{n}\bigr) - f\bigl(\tfrac{2k+1}{n}\bigr) \right] \\
&= - \sum_{k=0}^{\infty} \int_{0}^{\infty} f'(x) \mathbf{1}_{[\frac{2k}{n}, \frac{2k+1}{n}]}(x) \, \mathrm{d}x \\
&= -\int_{0}^{\infty} f'(x) k_n(x) \, \mathrm{d}x, \tag{Tonelli}
\end{align*}
where $k_n(x) = \sum_{k=0}^{\infty} \mathbf{1}_{[\frac{2k}{n}, \frac{2k+1}{n}]}(x)$. Now let $K_n(x) = \int_{0}^{x} k_n(t) \, \mathrm{d}t$. Clearly, $K_n(x) \leq x$ for all $x \geq 0$. Moreover, as $n \to \infty$ it is not hard to see that $K_n(x) \to \frac{1}{2}x$. Then by integration by parts,
\begin{align*}
S_n
&= \int_{0}^{\infty} f''(x) K_n(x) \, \mathrm{d}x \\
&\to \int_{0}^{\infty} f''(x) \cdot \frac{1}{2} x \, \mathrm{d}x \tag{DCT} \\
&= \left[ f'(x) \cdot \frac{1}{2} x \right]_{0}^{\infty} - \left[ \frac{1}{2}f(x) \right]_{0}^{\infty} \\
&= \frac{1}{2}f(0)
= \frac{1}{2}.
\end{align*}
This result generalizes fairly easily to a more general class of functions $f$.
| {
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Why is $(3, 1+\sqrt{-26})^3=( 1+\sqrt{-26})$ in $\mathbb Z[\sqrt{-26}]$? $a = 3, b = 1+\sqrt{-26}$ then $(a,b)^3=(a^3,b^3,a^2b,ab^2)$
each generator except $a^3$ has a $b$ factor and $\bar b b=27$, so $"\subseteq"$. Now the question is how to obtain $b$ using these generators. Is there a general formula, which states in which case this is possible or even is there an algorithm. I made it very complicated I think;
$a^3 = 27,\quad$
$b^3 = -77+23\sqrt{-26},\quad$
$a^2b = 9+9\sqrt{-26},\quad$
$ab^2 = -75+6\sqrt{-26}$
$3*a^3+ab^2=81+(-75+6\sqrt{-26})=6+6\sqrt{-26}$
$a^2b-ans = 3+3\sqrt{-26}$
are these correct so far ?
| To be explicit, you could use linear algebra:
$$\begin{align}
a^3&=27&&\mapsto\begin{bmatrix}27\\0\end{bmatrix}\\
b^3&=-77-23\sqrt{-26}&&\mapsto\begin{bmatrix}-77\\-23\end{bmatrix}\\
a^2b&=9+9\sqrt{-26}&&\mapsto\begin{bmatrix}9\\9\end{bmatrix}\\
ab^2&=-75+6\sqrt{-26}&&\mapsto\begin{bmatrix}-75\\6\end{bmatrix}\\
\end{align}$$
It would suffice to have an integer solution to the equation $$wa^3+xb^3+ya^2b+zab^2=b$$ The corresponding augmented matrix is
$$\begin{bmatrix}
27&-77&9&-75&1\\
0&-23&9&6&1
\end{bmatrix}$$
This reduces to
$$\begin{bmatrix}
23&0&-18&-81&-2\\
0&23&-9&-6&-1
\end{bmatrix}$$
You need choices for $y$ and $z$ such that $18y+81z-2$ and $9y+6z-1$ are each divisible by $23$. In other words, you want to solve
$$\begin{bmatrix}
18&81\\
9&6
\end{bmatrix}
\begin{bmatrix}
y\\
z
\end{bmatrix}
\equiv
\begin{bmatrix}
2\\
1
\end{bmatrix}$$
modulo $23$. So we can again use linear algebra (with some division mod $23$) to find
$$
\begin{bmatrix}
y\\
z
\end{bmatrix}
\equiv
\begin{bmatrix}
18&81\\
9&6
\end{bmatrix}^{-1}
\begin{bmatrix}
2\\
1
\end{bmatrix}
\equiv
\begin{bmatrix}
2\\
1
\end{bmatrix}$$
We can just go with $y=2$ and $z=1$. This gives $(w,x,y,z)=(5,1,2,1)$. So it seems that
$$5a^3+b^3+2a^2b+ab^2=b$$
| {
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Solving a $n\times n$ determinant. $$
\left|
\begin{array}{cccccc}
3&2&0&0&0&\ldots\\
1&3&1&0&0&\ldots\\
0&2&3&2&0&\ldots\\
0&0&1&3&1&\ldots\\
\vdots&\vdots&\vdots&\vdots&\vdots&\ddots
\end{array}
\right|
$$
I solved the determinant by realising it's matrix is tridiagonal and using the continuant, and have gotten a recursive formula $D_n = 2^{n+1} -1$, if $D_n$ is a determinant with respect to $n$. I'm having trouble solving the determinat using expansion on rows/columns, since this is how we've done similar ones in class. Appreciate any responses.
| If you develop along the first column, you get
$$\begin{align*} D_n &= 3\begin{vmatrix}3 & 1 & 0 & 0 & \cdots \\
2 & 3 & 2 & 0 & \cdots \\
0 & 1 & 3 & 1 & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots \end{vmatrix} -
\begin{vmatrix} 2 & 0 & 0 & 0 & \cdots \\
2 & 3 & 2 & 0 & \cdots \\
0 & 1 & 3 & 1 & \dots \\
\vdots & \vdots & \vdots & \vdots & \ddots
\end{vmatrix} \\
&= 3D_{n-1} - 2\begin{vmatrix}
3 & 2 & 0 & \cdots \\
1 & 3 & 1 & \dots \\
\vdots & \vdots & \vdots & \ddots
\end{vmatrix} \\
&= 3D_{n-1} - 2D_{n-2}
\end{align*}$$
where the $D_{n-1}$ term comes from the fact that the determinant of a matrix is the same as the determinant of its transpose (you really get the transposed determinant of $D_{n-1}$, but that is just $D_{n-1}$).
Now this is just a linear recursion formula, which you can solve using your favourite method, or you can just observe that $D_n = 2^{n+1}-1$ satisfies this formula.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Standard limit proof Everyone knows the standard result $$\boxed{\lim_{x\to\infty}\left(1+\dfrac{1}{x}\right)^x=e}$$ My friend gave a proof for this result using binomial expansion of $$\left(1+\dfrac{1}{x}\right)^n=1+nx+\dfrac{n(n-1)}{2!}x^2+\dfrac{n(n-1)(n-2)}{3!}x^3+\dots$$ which is valid only when $|1/x|<1$ and $n\in\mathbb{R}$. Here is how he proceeded-
$$\begin{align*}\lim_{x\to\infty}\left(1+\dfrac{1}{x}\right)^x=\lim_{x\to\infty}\left(1+x.\dfrac1x+\dfrac{x(x-1)}{2!}\dfrac1{x^2}+\dfrac{x(x-1)(x-2)}{3!}\dfrac1{x^3}+\dots\right)=\lim_{x\to\infty}\left(1+1+\dfrac{x(x-1)}{2!}\dfrac1{x^2}+\dfrac{x(x-1)(x-2)}{3!}\dfrac1{x^3}+\dots\right)=1+1+\lim_{x\to\infty}\dfrac{x(x-1)}{2!}\dfrac1{x^2}+\lim_{x\to\infty}\dfrac{x(x-1)(x-2)}{3!}\dfrac1{x^3}+\dots=1+1+\frac1{2!}+\frac1{3!}+\frac1{4!}\dots=\boxed{e}\end{align*}$$ The method of binomial expansion is clearly invalid as the exponent $x$ is variable not a constant. Yet the method ends up getting correct answer as $e$. Am I missing what is incorrect here? I am interested to know why this wrong approach leads to correct answer.
| This is correct for each $\vert x \vert > 1$:
$$ \left( 1+\frac{1}{x} \right) ^ x =
1+1+\dfrac{x(x-1)}{2!}\dfrac1{x^2}+\dfrac{x(x-1)(x-2)}{3!}\dfrac1{x^3}+\dots$$
It is simply Newton's Generalised Binomial Theorem with $x$ replaced by $1$, $y$ replaced by $\frac{1}{x}$ and $r$ replaced by $x.$
However, It is not obvious that the following limit even converges (although it probably does for some complicated reason):
$$\lim_{x\to\infty}\left(1+x\cdot\dfrac1x+\dfrac{x(x-1)}{2!}\dfrac1{x^2}+\dfrac{x(x-1)(x-2)}{3!}\dfrac1{x^3}+\dots\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Computing the integral of trig function under square root How can we solve this integral
$$\int \sqrt{\csc^2x -2}
\mathrm{d}x$$
My idea was substituting $\csc^2x=2\csc^2\theta$. Then the integral became $$\sqrt{2}\int \frac{\csc^2\theta-1}{\sqrt{2\csc^2\theta-1}} \mathrm{d}\theta$$ after a few simplifications. I don't know how to go about at this point. If i try breaking the numerator,still i would be left with an equivalent question of $$\int \sqrt{2\csc^2\theta-1}\mathrm{d}\theta$$ and $$\int \frac{\mathrm{d}\theta}{\sqrt{2\csc^2\theta-1}}$$
Could someone provide a cleaner approach to this problem or give new insights on how to further simply the above integrals?
| Letting $u=\sqrt{\csc ^{2} x-2}$ transform the integral into
$$\begin{aligned}
I&=-\frac{1}{2} \int \frac{u^{2} d u}{\left(u^{2}+2\right) \sqrt{u^{2}+1}}\\& = \frac{1}{2} \underbrace{\int \frac{d u}{\sqrt{u^{2}+1}}}_{\sinh ^{-1} u+C_1} + \underbrace{ \int \frac{d u}{\left(u^{2}+2\right) \sqrt{u^{2}+1}}}_{J}\end{aligned}
$$
For the integral $J$, letting $t=\frac{u}{\sqrt{u^{2}+1}} $ gives $u d u=\frac{t d t}{\left(1-t^{2}\right)^{2}}$ and
$$
\begin{aligned}
I &=\int \frac{d t}{\frac{2-t^2}{1-t^{2}}\left(1-t^{2}\right)} \\
&=\int \frac{d t}{2-t^{2}} \\
&=\frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{t}{\sqrt{2}}\right)+C_{2}\\&= \frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{u}{\sqrt{2} \sqrt{u^{2}+1}}\right)+C_2
\end{aligned}
$$
Now we can conclude that
$$
I=-\frac{1}{2} \sinh ^{-1} \left(\sqrt{\csc ^{2} x-2}\right) +\frac{1}{\sqrt{2}} \tanh ^{-1}\left(\frac{\sqrt{\csc ^{2} x-2}}{\sqrt{2} \cot x}\right)+C
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4498580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
How to prove that $\sqrt{\frac{x^2+1}{x+1}}+\frac{2}{\sqrt{x}+1}\ge2, \text{ }x\in \mathbb{R}_{>0}$? $$\sqrt{\frac{x^2+1}{x+1}}+\frac{2}{\sqrt{x}+1}\ge2, \text{ }x\in \mathbb{R}_{>0}$$
Equality seems to be when x = 1
I have managed to show that the derivative is 0 at x = 1, and that this is a minimum (by the second derivative test), but I am stuck on how to show that this is the only minimum.
If $f(x)$ is defined as the LHS, then we have
$$\begin{array}{l}f\left(x\right)=\sqrt{\frac{x^2-1+2}{x+1}}+\frac{2}{\sqrt{x}+1}=\sqrt{x-1+\frac{2}{x+1}}+\frac{2}{\sqrt{x}+1}\\
f^{\prime}\left(x\right)=\frac{1-\frac{2}{\left(x+1\right)^2}}{2\sqrt{x-1+\frac{2}{x+1}}}-\frac{2}{\left(\sqrt{x}+1\right)^2}\cdot\frac{1}{2\sqrt{x}}=0\\
\frac{1-\frac{2}{\left(x+1\right)^2}}{2\sqrt{x-1+\frac{2}{x+1}}}=\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)^2}\\
\sqrt{x}\left(\sqrt{x}+1\right)^2\left(\left(x+1\right)^2-2\right)=2\left(x+1\right)^2\sqrt{\frac{x^2+1}{x+1}}\\
x\left(x+1+2\sqrt{x}\right)^2\left(\left(x+1\right)^2-2\right)^2=2\left(x+1\right)^4\left(\frac{x^2+1}{x+1}\right)\end{array}$$
At this point the computations become unrealistic without enlisting the help of wolfram alpha to expand everything. I did this, and was left with a massive polynomial which had the root x = 1.
Here is a graph
| First, we can take the first derivative of this function
$$\frac{d}{dx}\sqrt{\frac{x^2+1}{x+1}}\;+\;\frac{2}{\sqrt{x}+1}$$
we get
$$\frac{x^2+2x-1}{2\sqrt{x^2+1}\left(x+1\right)^{\frac{3}{2}}}-\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)^2}$$
We then want to find the critical number by letting the first derivative be $0$, and we solve for x
$$\frac{x^2+2x-1}{2\sqrt{x^2+1}\left(x+1\right)^{\frac{3}{2}}}-\frac{1}{\sqrt{x}\left(\sqrt{x}+1\right)^2}=0$$
$$x=1$$
Next, we can use the first derivative again to see whether or not, at x=1, the function is a maximum or minimum. You can do this by using the first derivative test. After that, the only critical number is $1$, which tells us that when $x = 1$, it has to be a global minimum.
Sub $x=1$ into the original function, we obtain $2$, and since when $x=1$ is a global minimum,
$$\sqrt{\frac{x^2+1}{x+1}}+\frac{2}{\sqrt{x}+1}\ge 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve the inequality $3^{(x+3)^2}+\frac{1}{9}\leq 3^{x^2-2}+27^{2x+3}$ I tried to group the summands so that I could decompose them into multipliers, but nothing worked...
$$3^{(x+3)^2}+\frac{1}{9}\leq 3^{x^2-2}+27^{2x+3}\Leftrightarrow 3^{x^2+6x+9}+\frac{1}{9}\leq 3^{x^2-2}+3^{6x+9}$$
$$3^{x^2+6x+7}+1\leq 3^{x^2-4}+3^{6x+7}$$
How to solve it further?
| Multiplying both sides by $9=3^2$,
\begin{align*}
3^{(x+3)^2}+\frac{1}{9}\leq 3^{x^2-2}+27^{2x+3}&\Longleftrightarrow 3^{(x+3)^2\color{red}{+2}}+1\leq 3^{x^2-2\color{red}{+2}}+3^{3(2x+3)\color{red}{+2}}\\
&\Longleftrightarrow 3^{x^2+6x+11}+1\leq 3^{x^2}+3^{6x+11}\\
&\Longleftrightarrow \left(3^{x^2}-1\right)\left(3^{6x+11}-1\right)\leq 0.
\end{align*}
Since $x^2\geq 0$, we have $3^{x^2}-1\geq 0$, so the inequality is equivalent to
$$3^{x^2}-1=0\ \ \text{ or }\ \ 3^{6x+11}-1\leq 0\Longleftrightarrow x=0 \ \ \text{ or }\ \ 6x+11\leq 0\Longleftrightarrow x=0 \ \ \text{ or }\ \ x\leq-\frac{11}6.$$
So the solutions of the inequality are
$$x\in\{0\}\bigcup \left(-\infty,-\frac{11}6\right].$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Cholesky-like decomposition that works on singular matrices? Is there a variant of Cholesky-like decomposition that works for singular matrices?
My problem is that all implementations of Cholesky I found fail when the matrix is singular in machine precision.
One idea for such decomposition is to consider duality between Cholesky and Gram-Schmidt. Once you encounter vectors that are linear combinations of previous vectors, skip them. This for instance produces the following decomposition for rank-3 matrix below.
$$\left(
\begin{array}{ccccc}
9 & 1 & 0 & 0 & 0 \\
1 & 1 & 2 & 2 & 2 \\
0 & 2 & 5 & 5 & 5 \\
0 & 2 & 5 & 5 & 5 \\
0 & 2 & 5 & 5 & 5 \\
\end{array}
\right)=\left(
\begin{array}{ccc}
1 & 0 & 0 \\
\frac{1}{9} & 1 & 0 \\
0 & \frac{9}{4} & 1 \\
0 & \frac{9}{4} & 1 \\
0 & \frac{9}{4} & 1 \\
\end{array}
\right)\left(
\begin{array}{ccc}
9 & 0 & 0 \\
0 & \frac{8}{9} & 0 \\
0 & 0 & \frac{1}{2} \\
\end{array}
\right)\left(
\begin{array}{ccccc}
1 & \frac{1}{9} & 0 & 0 & 0 \\
0 & 1 & \frac{9}{4} & \frac{9}{4} & \frac{9}{4} \\
0 & 0 & 1 & 1 & 1 \\
\end{array}
\right)$$
notebook
| An approach using $QR$ decomposition: the following results in a decomposition of the form $A = MM^T$ with $M$ of full rank of size $n \times r$ (with $r$ equal to the rank of $A$), but $M$ is not necessarily in lower-triangular form. If it is desired, $M$ can be put into lower-triangular form with a further $QR$ decomposition $M^T = QR$.
Begin with a (symmetric) positive semidefinite $n \times n$ matrix $A$ of rank $r<n$. An exact $QR$ decomposition $A = QR$ must have the form
$$
Q = \pmatrix{Q_1 & Q_2}, \quad R = \pmatrix{R_1\\ 0}
$$
where $Q$ is square of size $n$, $Q_1$ has $r$ columns and $R_1$ has $R$ rows. Thus, the columns of $Q_1$ form an orthonormal basis for the column-space of $A$.
It follows that $Q_1^TAQ_1$ is strictly positive definite. Moreover, we have $Q_2^TA = (AQ_2)^T = 0$. Compute a Cholesky decomposition $LL^T = Q_1^TAQ_1$. From there, we may deduce that
$$
\pmatrix{L \\0} \pmatrix{L\\0}^T = \pmatrix{LL^T & 0\\0 & 0} = \pmatrix{Q_1^TAQ_1 & Q_1^TAQ_2\\ Q_2^TAQ_1 & Q_2^TAQ_2} = Q^TAQ.
$$
Conclude that
$$
A = Q\pmatrix{L\\0}\pmatrix{L \\ 0}^TQ^T = \left[Q\pmatrix{L \\0} \right]\left[ Q\pmatrix{L \\0}\right]^T
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:
If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$
In these types of questions generally I follow the following approach:
Since divisor is cubic so the remainder must be a constant/linear/quadratic expression.
$\Rightarrow F(x)=(x^3+x)Q(x)+ax^2+bx+c$
For $x=0$, we get $c=0$
But since $x^3+x$ has no other roots so I can't find $a$ and $b$.
Please help.
Answer:
Option (B)
| The polynomial $x^{k+2}+x^k$ is divisible by $x^3+x$ for $k\ge 1.$ Hence $[x^{13}+x^{14}+\ldots +x^{20}]$ and $x^{10}+x^{12}$ are divisible by $x^3+x.$ We are down to $x^{11}$ (thanks @Cathedral ) and $$\displaylines{x^{11}=x^{11}+x^9-(x^9+x^7)+(x^7+x^5)\\ -(x^5+x^3)+(x^3+x)-x}$$
The remainder is equal $-x.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 8,
"answer_id": 1
} |
Regular heptagon area formula Be a regular heptagon. Can anyone demonstrate this formula?
I know:
Let apothem = ap
side = l
$ S=\frac{p.ap}{2}\implies S = \frac{7l.ap}{2}\\
\triangle OBC: OC^2 = a^2+l^2\\
\triangle PAB: b^2+l^2 = PB^2\\
OC^2-a^2=PB^2-b^2$
but i can't go on
| Let $d$ be the diameter of the circumcircle of $ABCDEFG$. Note that $a = BC \sin \frac{2\pi}{7}=d\sin \frac{\pi}{7}\sin \frac{2\pi}{7}$ and $b=AD \sin \frac{2\pi}{7} = d \sin \frac{3\pi}{7} \sin \frac{2\pi}{7}$. On the other hand, the area of $ABCDEFG$ equals $7 \cdot \frac 12 \left(\frac{d}{2}\right) \sin \frac{2\pi}{7}$. The problem boils down to showing that
$$x=\sin \frac{\pi}{7}\sin \frac{2\pi}{7}\sin \frac{3\pi}{7} = \frac{\sqrt 7}{8}.$$
Using $2\sin^2 t = 1-\cos 2t$ we see that
\begin{align*}
8x^2 = 8\sin^2 \frac{\pi}{7}\sin^2\frac{2\pi}{7}\sin^2 \frac{3\pi}{7} &= \left(1-\cos \frac{2\pi}{7}\right)\left(1-\cos \frac{4\pi}{7}\right)\left(1-\cos \frac{6\pi}{7}\right) \newline
&= 1 - \left( \cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7}\right) + \newline
& \quad +\left( \cos \frac{2\pi}{7}\cos \frac{4\pi}{7}+\cos \frac{4\pi}{7}\cos \frac{6\pi}{7}+\cos \frac{6\pi}{7}\cos \frac{2\pi}{7}\right) - \newline
& \quad - \cos\frac{2\pi}{7}\cos \frac{4\pi}{7}\cos \frac{6\pi}{7}.
\end{align*}
Using $\cos t \cos u = \dfrac{\cos(t+u)+\cos(t-u)}{2}$ we get
\begin{align*}\cos \frac{2\pi}{7}\cos \frac{4\pi}{7}+&\cos \frac{4\pi}{7}\cos \frac{6\pi}{7}+\cos \frac{6\pi}{7}\cos \frac{2\pi}{7} \newline
&=\frac{\cos \frac{6\pi}{7}+\cos \frac{2\pi}{7}}{2}+\frac{\cos \frac{10\pi}{7}+\cos \frac{2\pi}{7}}{2}+\frac{\cos \frac{8\pi}{7}+\cos \frac{4\pi}{7}}{2}\newline
&=\cos \frac{2\pi}{7}+\cos \frac{4\pi}{7}+\cos \frac{6\pi}{7}
\end{align*}
since $\cos\frac{8\pi}{7}=\cos\frac{6\pi}{7}$ and $\cos\frac{10\pi}{7}=\cos\frac{4\pi}{7}$.
Moreover, using $\sin 2t = 2 \sin t \cos t$, we have
\begin{align*}
\cos\frac{2\pi}{7}\cos \frac{4\pi}{7}\cos \frac{6\pi}{7} &= \cos\frac{2\pi}{7}\cos \frac{4\pi}{7}\cos \frac{8\pi}{7} \newline
&= \frac{\sin \frac{2\pi}{7}\cos\frac{2\pi}{7}\cos \frac{4\pi}{7}\cos \frac{8\pi}{7}}{\sin\frac{2\pi}{7}} \newline
&= \frac{\frac 12 \sin \frac{4\pi}{7}\cos \frac{4\pi}{7}\cos \frac{8\pi}{7}}{\sin\frac{2\pi}{7}} \newline &= \frac{\frac 14 \sin \frac{8\pi}{7}\cos \frac{8\pi}{7}}{\sin\frac{2\pi}{7}} \newline &= \frac{\frac 18 \sin \frac{16\pi}{7}}{\sin\frac{2\pi}{7}} \newline &= \frac 18.\end{align*}
It follows that $8x^2= 1-\frac 18 = \frac 78$, i.e. $x=\frac{\sqrt{7}}{8}$, as desired.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4503096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Use the definition of limit to prove that $\lim_{x \to 0} \frac{x}{x+1}=0$ Use the definition of limit to prove that $\lim_{x \to 0} \dfrac{x}{x+1}=0$
My attempt:
Let $\epsilon >0$ then I need to find $\delta>0$ such that if $0<|x|<\delta$ then $\left| \dfrac{x}{x+1} \right| < \epsilon$
What I was thinking is that, to get a $\delta$ that satisfies the definition, I should see what $\epsilon$ satisfies $|x|<\epsilon |x+1|$ and then I could define $\delta$ in terms of such an $\epsilon$.
If $\epsilon =1$ then: $0\leq|x|<|x+1|$ implies that $x^2 < x^2+2x+1$ and then $x>-\dfrac{1}{2}$. In particular, if $-\dfrac{1}{2}<x<\dfrac{1}{2} \Rightarrow |x|<\dfrac{1}{2} \Rightarrow \dfrac{1}{2} < x+1<\dfrac{3}{2}$ so that $|x+1|=x+1$ and then $\dfrac{1}{2}<|x+1| \Rightarrow \dfrac{1}{2|x+1|}<1$. Since $|x|<\dfrac{1}{2}$, then $\left| \dfrac{x}{x+1}\right| = \dfrac{|x|}{|x+1|} <\dfrac{1}{2|x+1|}<1$.
So $\delta=\dfrac{\epsilon}{2}$ stisfies the definition.
If $\epsilon \neq 1$ then:
$0 \leq |x|<\epsilon |x+1|$ Implies that: $x^2<{\epsilon}^2(x^2+2x+1) \Rightarrow (\epsilon ^2-1)x^2+2 \epsilon ^2x + \epsilon ^2 > 0$.
If $0< \epsilon <1 \Rightarrow \epsilon^2-1<0 \Rightarrow x^2 + \dfrac{2 \epsilon^2}{\epsilon^2-1}x+\dfrac{\epsilon^2}{\epsilon^2-1}<0 \Rightarrow \left(x+ \dfrac{\epsilon}{\epsilon-1} \right) \left(x+ \dfrac{\epsilon}{\epsilon+1} \right) < 0$
Let $ x_1 = -\dfrac{\epsilon}{\epsilon-1}$, $x_2=-\dfrac{\epsilon}{\epsilon+1}$. We see that $x_1 >0>x_2$. Then solving the inequality for $x$ we would have $x_2=-\dfrac{\epsilon}{\epsilon+1} < x < -\dfrac{\epsilon}{\epsilon-1} = x_1$.
I am expecting to get something like: $-\phi(\epsilon) < x < \phi(\epsilon) \Rightarrow |x|< \phi(\epsilon)$ on each case so that $\delta = \phi(\epsilon)$ satisfies the definition, but I don't know how to continue here. Could someone give me a hint? Maybe there is a simpler way to approach the exercise.
| Choosing a positive number $\delta < \frac{\epsilon}{1 + \epsilon}$ will make $\frac{\delta}{1 - \delta} < \epsilon$; if $0 < |x| < \delta$, then by the reverse triangle inequality $$\left|\frac{x}{x+1}\right| \le \frac{|x|}{1 - |x|} < \frac{\delta}{1 - \delta} < \epsilon$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is there a general method to find the asymptotic order for this sequence? Given $$a_{n+1}=a_n+\frac{n}{a_1+\dots+a_n},\qquad a_1>0$$ The answer is $$\lim_{n\to\infty} a_n\sim\sqrt{3}\cdot\sqrt{n}-\frac{\sqrt{3}}{4}\cdot\frac{1}{\sqrt{n}}$$
It is easy to show this sequence is increasing, and is divergent, because if assume the opposite $A=\lim_{n\to\infty} a_n$, we will get $A=A+\frac{1}A$, which gives contradictions. We also have
$$a_{n+1}\ge a_n +\frac{1}{a_n}\ge2$$
From here I can show:
$$a_1+\frac{n}{a_n}\le a_{n+1}\le a_1+\frac{n}{a_1}$$
Update.(1)
This equation can be also written as:
$$\begin{align}
a_n&=\sum_{k=1}^n a_k-\sum_{k=1}^{n-1} a_k\\
\\
a_n&=\frac{n}{a_{n+1}-a_n}-\frac{n-1}{a_{n}-a_{n-1}}\\
\\
a_n(a_{n+1}-a_n)(a_{n}-a_{n-1})&=n(a_{n}-a_{n-1})-(n-1)(a_{n+1}-a_n)\tag{1}
\end{align}$$
If assume $a_n=c\cdot n^p$
$$a_{n+1}-a_n\sim cp\cdot n^{p-1},~~~a_{n}-a_{n-1}\sim cp\cdot n^{p-1}$$
Plug into $(1)$ and only keep the leading order:
$$c^2p\cdot n^{2p-1}=1$$
So we get $p=\frac{1}2$ and $c=\sqrt{2}$
Why the leading order coefficient is $\sqrt{2}$, not $\sqrt{3}$?
The bottom line is, if I take the answer as template, let $$a_n=c\sqrt{n}+t\frac{1}{\sqrt{n}}+o(\frac{1}{\sqrt{n}})$$
Pretend we don't know coefficients $c$ and $t$. Now we want to compute $c$ and $t$. To the leading order approximation, We have
$$a_{n+1}-a_n= c\cdot \frac{1}{2\sqrt{n}}+O(\frac{1}{n^{3/2}})$$
Next, plug into Eq.$(1)$ and we can solve for $c=\sqrt{2}$. Why does this give a contradiction?
Update.(2)
Thank you to @Sangchul Lee , @Somos and @Youem
I put the computation part in the answer box below, and it works for asymptotic approximation at arbitrary order.
|
$$a_{n+1}=a_n+\frac{n}{\sum_{k=1}^n a_k}$$
This equation can be written as:
$$\sum_{k=1}^n a_k=\frac{n}{a_{n+1}-a_n}$$
Further, we can find $a_n$ by taking subtraction:
$$\begin{align}
a_n&=\sum_{k=1}^n a_k-\sum_{k=1}^{n-1} a_k\\
\\
a_n&=\frac{n}{a_{n+1}-a_n}-\frac{n-1}{a_{n}-a_{n-1}}\\
\\
a_n(a_{n+1}-a_n)(a_{n}-a_{n-1})&=n(a_{n}-a_{n-1})-(n-1)(a_{n+1}-a_n)\tag{1}\\
\\
&=(a_{n+1}-a_n)-n(a_{n+1}-2a_n+a_{n-1})
\end{align}$$
First determine the leading order. Let $a_n=c_0\cdot n^p$
$$\begin{align}
a_{n+1}&=c_0(n+1)^p=c_0 \left(n^p+pn^{p-1}+\frac{1}{2}p(p-1)n^{p-2}+...\right)\tag{2}\\
\\
a_{n-1}&=c_0(n-1)^p=c_0 \left(n^p-pn^{p-1}+\frac{1}{2}p(p-1)n^{p-2}+...\right)\tag{3}
\end{align}$$
Plug into $(1)$ and only keep the leading order on both sides:
$$c_0n^p\cdot c_0pn^{p-1}\cdot c_0pn^{p-1}=(2 c_0 p - c_0 p^2)n^{p-1}$$
So we get:
$$3p-2=p-1,~~~~~~c_0^3 p^2=(2 c_0 p - c_0 p^2)$$
$$\Rightarrow p=\frac{1}2,~~~c_0=\sqrt{3}$$
Next, find the next order, so we set the solution Ansatz as:
$$a_n=\sqrt{n}\left(c_0+c_1\cdot\frac{1}n \right),~~~~~c_0=\sqrt{3}$$
Plug into $(1)$ and re-do the series expansion to update $(2)$ and $(3)$. Now, comparing the coefficients for $\frac{1}{n^{3/2}}$
$$-\frac{3}4c_1\frac{1}{n^{3/2}}=\left(-\frac{\sqrt{3}}8 - \frac{5}4c_1\right)\frac{1}{n^{3/2}}$$
$$\Rightarrow c_1=-\frac{\sqrt{3}}{4}$$
Next, set the solution Ansatz as:
$$a_n=\sqrt{n}\left(c_0+c_1\cdot\frac{1}n+c_2\cdot\frac{1}{n^2} \right),~~~~~c_0=\sqrt{3},~~c_1=-\frac{\sqrt{3}}{4}$$
Now, comparing the coefficients for $\frac{1}{n^{5/2}}$
$$\frac{3}{32}(\sqrt{3}-40c_2)\frac{1}{n^{5/2}}=\frac{3}{64}\left(\sqrt{3} - 112c_2\right)\frac{1}{n^{5/2}}$$
$$\Rightarrow c_2=-\frac{\sqrt{3}}{32}$$
Keep going on this method, we get:
$$a_n=\sqrt{n}\left(\sqrt{3}-\frac{\sqrt{3}}{4}\cdot\frac{1}n-\frac{\sqrt{3}}{32}\cdot\frac{1}{n^2}-\frac{\sqrt{3}}{128}\cdot\frac{1}{n^3}-\frac{79\sqrt{3}}{38912}\cdot\frac{1}{n^4}+\cdots \right)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
How to show $\gcd \left(\frac{p^m+1}{2}, \frac{p^n-1}{2} \right)=1$, with $n$=odd? In previous post, I got the answer that $\gcd \left(\frac{p^2+1}{2}, \frac{p^5-1}{2} \right)=1$, where $p$ is prime number.
I am looking for more general case, that is for $p$ prime,
When is $\gcd \left(\frac{p^r+1}{2}, \frac{p^t-1}{2} \right)=1$ ?
where $t=r+s$ such that $\gcd(r,s)=1$.
I am excluding the cases $r=1=s$.
In the previous post, it was $r=2,~t=5$. So $t=5=2+3=r+s$ with $s=3$ so that $\gcd(r,s)=1$.
In this current question:
Since $\gcd(r,s)=1$, we also have $\gcd(r,t)=1$.
I have the following intuition:
Case-I:
Assume $r$=even and $s$=odd so that $\gcd(r,s)=1$ as well as $\gcd(r,t)=1$. We can also assume $r$=odd and $s$=even.
I think the same strategies of previous post can be applied to show that $$\gcd \left(\frac{p^r+1}{2}, \frac{p^t-1}{2} \right)=1.$$
Case II:
The problem arises when both $r$ and $s$ are odd numbers so that $t=r+s$ is even.
If I take $r=3, ~s=5$, then $t=8$.
For prime $p=3$, $\frac{p^r+1}{2}=\frac{3^3+1}{2}=14$ and $\frac{p^t-1}{2}=\frac{3^8-1}{2}=3280$ so that the gcd is $2$ at least.
For other primes also we can find gcd is not $1$.
So I think it is possible only for Case I, where among $r$ and $s$, one is odd and another is even so that $t$ is odd.
In other word, $t$ can not be even number.
But I need to be ensured with a general method.
So the question reduces to
How to prove $\gcd \left(\frac{p^m+1}{2}, \frac{p^n-1}{2} \right)=1$ ?
provided $\gcd(m,n)=1$ and $n$ is odd number and $p$ is prime number.
Thanks
| For primes of the form $6k-1$, and choosing $m$ odd and $n$ even, you will obtain $p^m=6a-1$ and $p^n=6b+1$. Hence $p^m+1=6a$ and $p^n-1=6b$ with $\gcd(6a,6b) \ge 6$. Your final conjecture $\gcd \left( \frac{p^m+1}{2}, \frac{p^n-1}{2}\right)=1$ is never true in the circumstances considered here.
| {
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"source": "stackexchange",
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Solving the system $x^3+y= 3x+4$, $2y^3+z = 6y+6$, $3z^3+x=9z+8$
Solve the system $$\begin{equation} \label{equation1}
\begin{split}
x^3+y= 3x+4 \\
2y^3+z = 6y+6 \\
3z^3+x=9z+8
\end{split}
\end{equation}$$
By the theorem of triviality, I assumed $x=y=z=k$ and then solved to fortunately get a solution. Since, it is a simultaneous system of equations the very solution I got is the only possible solution. But I can't understand how to do it by general method.
Any help is greatly appreciated.
| Taking help from the previous answer, we can write the equations as
$$x^3-3x-2=(x+1)^2(x-2)=2-y$$
$$2(y^3-3y-2)=2(y+1)^2(y-2)=2-z$$
$$3(z^3-3y-2)=3(z+1)^2(z-2)=2-x$$
Multiplying the three, we get
$$6(x+1)^2(y+1)^2(z+1)^2(x-2)(y-2)(z-2) = -(x-2)(y-2)(z-2)$$
If none of $x,y,z$ is equal to 2 we can cancel $(x-2)(y-2)(z-2)$ on each side, which gives us $6(x+1)^2(y+1)^2(z+1)^2 = -1$, which is not possible. Hence, at least one of $x, y$ or $z$ equals 2. Putting this back to original equations, we get $(x,y,z) = (2,2,2)$.
| {
"language": "en",
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"source": "stackexchange",
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if $x+\frac{1}{x}=\sqrt2$, then find the value of $x^{2022}+\frac{1}{x^{2022}}$? It is question of mathematical olympiad. kindly solve it guys!
I tried a bit.I am sharing this with u...
•$x+\frac{1}{x}=\sqrt{2}$
•$x^2+1=x\sqrt{2}$
•$x^2-x\sqrt{2}+1=0$
so, $x=\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}$ and
$\frac{1}{x}=\frac{\sqrt{2}}{2}-\frac{i\sqrt{2}}{2}$ or $\frac{\sqrt{2}}{2}+\frac{i\sqrt{2}}{2}$
How can I find $x^{2022}+\frac{1}{x^{2022}}$?
| $$x + \frac{1}{x} = \sqrt2 $$
squaring both sides,
$$x^2 + \frac{1}{x^2} = 0 $$
then,
$$x^{2022} + \frac{1}{x^{2022}} = (x^2 + \frac{1}{x^2})(x^{2020} + \frac{1}{x^{2020}}) - (x^{2018}+ \frac{1}{x^{2018}}) = - (x^{2018}+ \frac{1}{x^{2018}})$$
repeat the above process,
$$x^{2022} + \frac{1}{x^{2022}} = x^2 + \frac{1}{x^2} = 0$$
| {
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} |
Calculate the partial sum $S$ To what is the sum $\displaystyle{ S=1+\frac{1}{2}+\frac{1}{4}+\ldots+\frac{1}{2^{2022}} }$ equal \begin{equation*}(\text{a}) \ \ 2\left (2^{2022}-1\right ) \ ; \ \ \ \ \ (\text{b}) \ \ \frac{2^{2022}-1}{2} \ ; \ \ \ \ \ (\text{c}) \ \ \frac{2^{2022}-1}{2^{2021}} \ ; \ \ \ \ \ (\text{d}) \ \ \text{ none of } (\text{a})-(\text{c}).\end{equation*}
$$$$
I have done the following :
(a) It cannot be true since $2\cdot \left (2^{2022}-1\right )$ exceeds the sum of the series, which is a subseries of the infinite geometric series $\left \{\frac{1}{2^n}\right \}$ which converges to
$\frac{1}{1-\frac{1}{2}}=2$.
So we have that $2\cdot \left (2^{2022}-1\right )\gg2>S$.
(b) $\frac{2^{2022}-1}{2}=2^{2021}-\frac{1}{2}\gg2>S$
Therefore it cannot be true.
(c) We want to check if $S=\frac{2^{2022}-1}{2^{2021}}=2-\frac{1}{2^{2021}}$.
Let $\hat{S}$ be the sum of the infinite series : \begin{equation*}\hat{S}=1+\frac{1}{2}+\frac{1}{4}+\ldots +\frac{1}{2^n}=2 \text{ when } n\to \infty\end{equation*}
We therefore want to check if $S=\hat{S}-\frac{1}{2^{2021}}$ or $\hat{S}-S=\frac{1}{2^{2021}}$ ?
We know that $\hat{S}-S=\frac{1}{2^{2023}}+\frac{1}{2^{2024}}+\ldots +\frac{1}{2^{n}}, \ n\to \infty \ \ \ (\star)$
We can rewrite $\frac{1}{2^{2021}}=\frac{2}{2^{2022}}=\frac{1}{2^{2022}}+\frac{1}{2^{2022}}=\frac{1}{2^{2022}}+\frac{2}{2^{2023}} \Rightarrow$ etc $\Rightarrow \frac{1}{2^{2021}}=\frac{1}{2^{2022}}+\frac{1}{2^{2023}}+\ldots +\frac{1}{2^{n}}, \ n\to \infty$
Comparison with (⋆) gives that $\frac{1}{2^{2021}}>\hat{S}-S \Rightarrow $ c) cannot be true.
So (d) must be the correct answer.
$$$$
Is everything correct ?
| I think you are correct, although your answer for $(c)$ is unnecessarily long-winded. You would benefit from using the formula for a finite sum of geometric series, rather than using the infinite sum everywhere. It is:
$$ \sum_{k=1}^{n} ar^k = a\left( \frac{1-r^{n+1}}{1-r} \right).$$
where $a$ is the first term and $r$ is the common ratio.
Therefore:
$$ S = 1\left( \frac{1-\left(\frac{1}{2}\right)^{2023}}{1-\frac{1}{2}} \right) = 2 - \left(\frac{1}{2}\right)^{2022}. $$
We didn't need to simplify like that for the purposes of this question, but it is the form that you can most easily see what $S$ is equal to - for future use. Anyway, so answer $(c)$ what we can do is:
$$ S = \frac{2^{2022}\left(1-\left(\frac{1}{2}\right)^{2023}\right)}{2^{2022}\left(\frac{1}{2}\right)} = \frac{2^{2022}-\frac{1}{2}}{2^{2021}} \neq\frac{2^{2022}-1}{2^{2021}}.$$
| {
"language": "en",
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"source": "stackexchange",
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Number of paths in a square lattice with restrictions It is known that the number of paths with only rightward and upward moves in a square lattice from the lower left corner to the upper right corner is $\binom{a+b}{b}$, where $a$ is the height of the lattice and $b$ is the width. What is the number of paths in the lattice that do not contain four or more upward consecutive moves? ($a > 4$, $b > 4$).
| The problem is equivalent to finding the number of permutations of $a$ letters $U$ and $b$ letters $R$ such that there aren't four or more consecutive $U$s in the permutation. right?
If first, we put $R$s in a row, then there will be $b + 1$ places between the letters:
\begin{align*}
-\ R\ -\ R\ -\ R\ - \cdots -\ R\ -\ R\ -
\end{align*}
Now we must put $U$ letters in these places, such that each place contains $0$, $1$, $2$, or $3$ $U$s. So if we set $p_i$, $1 \leq i \leq b + 1$, to be the number of $U$s that are located in $i^{th}$ place, then we must count the number of solutions of the equation
\begin{align*}
p_1 + p_2 + \cdots + p_{b + 1} = a && \mbox{for} && 0 \leq p_i \leq 3,\ \ \ 1 \leq i \leq b + 1.
\end{align*}
We use generating functions. for each $i$, $1 \leq i \leq b + 1$, the generating polynomial of $p_i$ equals
\begin{align*}
x^0 + x^1 + x^2 + x^3 = \frac{1 - x^4}{1 - x}.
\end{align*}
So we must find the coefficient of $x^a$ (denoted by $[x^a]$) in
\begin{align*}
(1 + x + x^2 + x^3)^{b + 1} = \underbrace{\left(\frac{1 - x^4}{1 - x}\right)\left(\frac{1 - x^4}{1 - x}\right)\cdots \left(\frac{1 - x^4}{1 - x}\right)}_{b + 1\text{ times}} = \left(\frac{1 - x^4}{1 - x}\right)^{b + 1}.
\end{align*}
Therefore:
\begin{align*}
[x^a]\left(\frac{1 - x^4}{1 - x}\right)^{b + 1} = [x^a] \frac{(1 - x^4)^{b + 1}}{(1 - x)^{b + 1}} &= [x^a]\frac{\sum_{k = 0}^{b + 1}(-1)^k \binom{b + 1}{k}x^{4k}}{(1 - x)^{b + 1}}\\
&= \sum_{k = 0}^{b+1}[x^a]\frac{(-1)^k \binom{b + 1}{k}x^{4k}}{(1 - x)^{b + 1}}\\
&= \sum_{k = 0}^{b+1}[x^{a - 4k}]\frac{(-1)^k \binom{b + 1}{k}}{(1 - x)^{b + 1}}\\
&= \sum_{k = 0}^{b+1} (-1)^k \binom{b + 1}{k}\binom{b + a - 4k}{b}
\end{align*}
For calculation purposes, instead of this approach and finding the exact formula, you can write some code or use a program to evaluate the coefficient of $x^a$ in $(1 + x + x^2 + x^3)^{b+1}$.
| {
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"question_score": "2",
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Other approaches to simplify $\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}$ I want to simplify the trigonometric expression $\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}$.
My approach,
Here I used the abbreviation $s,c,t$ for $\sin x$ and $\cos x$ and $\tan x$ respectively,
Numerator is, $$\frac{s^2}{c^2}-s^2=\frac{s^2-s^2c^2}{c^2}=\frac{s^4}{c^2}=s^2t^2.$$
And denominator is,
$$\frac{2t}{1-t^2}-2t=\frac{2t^3}{1-t^2}.$$
So $$\frac{\tan^2x-\sin^2x}{\tan 2x-2\tan x}=s^2t^2\times\frac{1-t^2}{2t^3}=\sin^2x\times \frac1{\tan 2x}=\frac{1-\cos 2x}{2\tan 2x}.$$
I'm looking for alternative approaches to simplify the expression.
| There are many ways to do it.
For axample, let $x=\tan^{-1}(t)$ and, after simplifications
$$\frac{\tan ^2(x)-\sin ^2(x)}{\tan (2 x)-2 \tan (x)}=\frac{1}{2} t \cos \left(2 \tan ^{-1}(t)\right)=\frac{t \left(1-t^2\right)}{2 \left(1+t^2\right)}=\frac 1 2 \times t\times \frac{ \left(1-t^2\right)}{ \left(1+t^2\right)}$$ that is to say
$$\frac 1 2 \times \tan(x) \times \cos(2x)$$
Another way is to use $\tan(2x)=\frac{2 \tan (x)}{1-\tan ^2(x)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510735",
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"question_score": "2",
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Why this transformation matrix $A$ has $\begin{pmatrix}0 \\ 1\end{pmatrix}$ as Eigenvector? I have the following transformation matrix:
$$
A=\begin{pmatrix}
1 & 0 \\
-1 & 4
\end{pmatrix}
$$
If I resolve to find the eigenvalues I get:
$$
\begin{vmatrix}
A-\lambda I
\end{vmatrix} = 0
$$
which leads to:
$$
\lambda_1 = 1;
\lambda_2 = 4
$$
Now if I try to calculate the eigenvectors
For $\lambda_1$ I get:
$$
(A - 1 I)\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix}0 \\ -x_1+3 x_2 \end{pmatrix}
$$
and for $\lambda_2$:
$$
(A - 1 I)\begin{pmatrix}x_1 \\ x_2 \end{pmatrix}=\begin{pmatrix}-3 x_1 \\ -x_1 \end{pmatrix}
$$
I see (and can compute using a symbolic calculation program) that there are two eigenvectors:
$$
e1 = \begin{pmatrix}0 \\ 1\end{pmatrix}
$$
$$
e2 = \begin{pmatrix}3 \\ 1\end{pmatrix}
$$
I can easily see why $\begin{pmatrix}3 \\ 1\end{pmatrix}$ is an eigenvector for the eigenvalue $\lambda_2$.
But I am struggling to understand why $\begin{pmatrix}0 \\ 1\end{pmatrix}$ is an eigenvector.
Could someone help me to understand why?
| $\begin{pmatrix} 3\\1\end{pmatrix}$ is not an eigenvector for eigenvalue $\lambda_2=4$; it is an eigenvector for $\lambda_1=1$.
$\begin{pmatrix} 0\\1\end{pmatrix}$ is the eigenvector for eigenvalue $\lambda_2=4$, given that plugging in $x_1=3$, $x_2=1$ gives the zero vector in that calculation.
| {
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Finding the range of $\frac{5 \cos x-2 \sin ^{2} x+4 \sin x-3}{6|\cos x|+1}$ Here is the expression:
\begin{equation}
\frac{5 \cos x-2 \sin ^{2} x+4 \sin x-3}{6|\cos x|+1}
\end{equation}
I have tried to maximize the denominator, so I got the maximum value of denominator equal to 7, but I am lost what to do next
I did the graph of this expression, so the minimum value is equal to -9, but how to get the answer without a graph, which is difficult to do by hand?
| If $x\in[-\pi/2,\pi/2]$ then $|\cos x|=\cos x$ so
\begin{equation}
\frac{5 \cos x-2 \sin ^{2} x+4 \sin x-3}{6|\cos x|+1}
={5\cos x+5/6\over6\cos x+1}+{-2\sin^2 x+ 4\sin x -3\frac56\over6\cos x+1}\\
=5/6+{-2\sin^2 x+ 4\sin x -3\frac56\over6\cos x+1}\\
\end{equation}
The $-2\sin^2 x$ term is always negative, so let $x=-\pi/2$ then $\sin x=-1$, $\cos x=0$, and so by inspection it is clearly the minimum and the value is $-9$.
You can apply the same argument to the other range $x\in[\pi/2, 3\pi/2]$ and the same minimum is found.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
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When solving $\sin^2 x = \frac14(2+ \sqrt 3)$, is it better to substitute $\sin^2x=1-\cos^2x$ or $\sin^2x=\frac12(1-\cos 2 x)?$ I am trying to evaluate
$$\sin^2 x = \frac{2+ \sqrt 3}{4}$$ where $x$ is $0$ to $2 \pi$.
For $ \sin^2 x$, I can substitute it to either:
*
*$ 1 - \cos ^2 x$ (Pythagorean identity)
*$ \frac12( 1 - \cos 2 x)$ (double angle formula)
How do I know which formula to use for this evaluation?
| $$\frac{2+\sqrt 3}{4} = \sin^2x = \frac{1-\cos 2x}{2},$$
$$\cos 2x = 1-\frac{2+\sqrt 3}{2} = -\frac{\sqrt 3}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518021",
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"source": "stackexchange",
"question_score": "1",
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Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
Find natural number $x,y$ satisfy $x^2+7x+4=2^y$
My try: I think $(x;y)=(0;2)$ is only solution. So I try prove $y\ge3$ has no solution, by $(x+1)(x+6)-2=2^y$.
So $2\mid (x+1)(x+6)$, but this is wrong. Done.
This is wrong
Anyone has an idea? Please help, thank you!
| We know that: $x=\frac{-7\pm\sqrt D}2$
For natural solutions: $$x=\frac{-7+\sqrt D}2$$
Also, $D=n^2$ where $n$ is an natural number. And, $n$ has to be odd and $n\geq7$.
$$\Rightarrow 7^2-4(4-2^y)=n^2$$
$$\Rightarrow 7^2-16-2^{y+2}=n^2$$
$$\Rightarrow n^2-33=2^{y+2}$$
We put $n=7$ (our first choice), and get $y=2$ (and $x=\frac{n-7}2$).
Now we don't have to check for $n=9,11$ or $13$ as for $y\geq2, \ (n^2-33)$ must be divisible by $16$, so, $n=16k\pm1$.
Check for $n=15:2^{y+2}=192\Rightarrow y\notin N.$
Check for $n=17:2^{y+2}=256\Rightarrow y=6.$
Now we know that our next solution will be for $y>6\Rightarrow2^{y+2}>256$ which means that $n$ divided by $256$ should leave a remainder of $\sqrt33$ which is not possible.
So, the required answer: $(x,y)=(\frac{7-7}2,2)$ and $(\frac{17-7}2,6)$.
$$\bf \text{Thus, }\boxed{(x,y)=(0,2),(5,6)}$$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solving $\frac{dx}{dt}=\frac{xt}{x^2+t^2},\ x(0)=1$ I have started self-studying differential equations and I have come across the following initial value problem
$$\frac{dx}{dt}=\frac{xt}{x^2+t^2}, \quad x(0)=1$$
Now, since $f(t,x)=\frac{xt}{x^2+t^2}$ is such that $f(rt,rx)=f(t,x)$ for every $r\in\mathbb{R}\setminus\{0\}$, we can use the change of variables $y=\frac{x}{t}$ to rewrite it in the form
\begin{align}
y+t\frac{dy}{dt} &=\frac{t^2 y}{t^2(1+y^2)} \\
&=\frac{y}{1+y^2} \\
\implies t\frac{dy}{dt} &=\frac{y}{1+y^2}-y \\
&=-\frac{y^3}{1+y^2} \\
\implies \frac{dy}{dt} &= \left(-\frac{y^3}{1+y^2}\right)\cdot\frac{1}{t}
\end{align}
which is separable, and becomes:
\begin{align}
\left(\frac{1+y^2}{y^3}\right)dy &= -\frac{dt}{t} \\
\implies \int_{y_1}^{y_2} \left(\frac{1+y^2}{y^3}\right) dy &= -\int_{t_1}^{t_2} \frac{dt}{t} \\
\implies -\frac{1}{2y_2^2}+\ln \left\lvert \frac{y_2}{y_1} \right\rvert + \frac{1}{2y_1^2} &= -\ln \left\lvert \frac{t_2}{t_1} \right\rvert
\end{align}
but now I don't see how to go forward and find $y(t)$. Also, I integrated from a generic time $t_1$ to a generic time $t_2$ because the right hand side wouldn't have converged otherwise.
So, I would appreciate any hint about how to go forward in solving this IVP.
Thanks
| With homogeneous equations, you have two choices for the substitution.
Here we have
$$(x^2+t^2)\,dx=xt\,dt $$
You made the substitution $x=yt$, but you also had the choice of letting $t=ux$. As a general rule, the substitution which makes the algebraic simplification easier also makes the integration easier. So lets investigate the substitution $t=ux$
\begin{eqnarray}
(x^2+u^2x^2)\,dx&=&ux^2(udx+xdu)\\
x^2\,dx&=&ux^3\,du\\
\frac{1}{x}\,dx&=&u\,du\\
\ln|x|+c_1&=&\frac{u^2}{2}\\
\ln x^2+c_2&=&\frac{t^2}{x^2}\\
t^2&=&x^2(\ln x^2+c_2)
\end{eqnarray}
The initial condition then gives $c_2=0$. So
$$ t^2=x^2\ln x^2 $$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Find possible values of $|z|$ if both the real and imaginary parts of $(\bar z)^2+\frac{1}{z^2}$ are integers Let $\bar z$ denote the complex conjugate of a complex number $z$. If $z$ is a non-zero complex number for which both real and imaginary parts of $$(\bar z)^2+\frac{1}{z^2}$$ are integers, then which of the following is/are possible value(s) of $\left|z\right|$
(A) $\left(\frac{43+3\sqrt{205}}{2}\right)^{\frac{1}{4}}$
(B) $\left(\frac{7+\sqrt{33}}{4}\right)^{\frac{1}{4}}$
(C) $\left(\frac{9+\sqrt{65}}{4}\right)^{\frac{1}{4}}$
(D) $\left(\frac{7+\sqrt{13}}{6}\right)^{\frac{1}{4}}$
My Attempt
Taking $z=x+iy$ we get $(\bar z)^2+\frac{1}{z^2}=(x^2-y^2)(1+\frac {1}{(x^2+y^2)^2})+i(2xy)(1+\frac{1}{(x^2+y^2)^2})$.
The calculation is appearing very arduous. Can there be a simple approach
| Let $r = |z|$ be a possible value of the modulus. We claim that the above condition is equivalent to
$$
\frac{(|z|^4 +1)^2}{|z|^4} = a^2 + b^2; \ \ a,b \in \mathbb{Z}
$$
It is clear that the condition is necesary, because this the left side is just $|f(z)|^2$. To prove the other implication, consider the map $f(z) = \bar{z}^2 + \frac{1}{z^2} = \frac{1 + |z|^4}{z^2}$. Let $z = |z|e^{\frac{iarg(a-ib)}{2}}$. Then, $f(z) = a + ib$. Then $|f(z)| = \sqrt{a^2 + b^2}$ and $arg(f(z)) = arg(a +ib)$, so $f(z) = a + ib$. Thus, the condition is sufficient.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving the equation $\overline z-z^2=i(\overline z+z^2)$ in $\mathbb{C}$ Let $\overline z$ denote the complex conjugate of a complex number z and let $i= \sqrt{-1}$. In the set of complex numbers, the number of distinct roots of the equation $\overline z-z^2=i(\overline z+z^2)$ is _____________.
My approach is as follow
$z = r{e^{i\theta }}$& $\overline z = r{e^{ - i\theta }}$
$r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = i\left( {r{e^{ - i\theta }} + {r^2}{e^{i2\theta }}} \right) \Rightarrow r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = {e^{i\frac{\pi }{2}}}\left( {r{e^{ - i\theta }} + {r^2}{e^{i2\theta }}} \right)$
$ \Rightarrow r{e^{ - i\theta }} - {r^2}{e^{i2\theta }} = \left( {r{e^{ - i\left( {\theta - \frac{\pi }{2}} \right)}} + {r^2}{e^{i\left( {2\theta + \frac{\pi }{2}} \right)}}} \right)$
$ \Rightarrow r\left( {\cos \theta - i\sin \theta } \right) - {r^2}\left( {\cos 2\theta + i\sin 2\theta } \right) = \left( {r\left( {\sin \theta + i\cos \theta } \right) + {r^2}\left( { - \sin 2\theta + i\cos 2\theta } \right)} \right)$
$ \Rightarrow r\cos \theta - {r^2}\cos 2\theta - r\sin \theta + {r^2}\sin 2\theta - i\left( {r\sin \theta - {r^2}\sin 2\theta - r\cos \theta - {r^2}\cos 2\theta } \right) = 0$
Not able to proceed further
| Let $z=x+iy$
$$z^2 =\bar z\cdot\dfrac{1-i}{1+i}=\cdots=-i\cdot\bar z$$
$$\implies x^2-y^2+i(2xy)=-i(x-iy)=-y-ix$$
Equating the imaginary parts, $-x=2xy\iff x(2y+1)=0\ \ \ \ (1)$
Equating the real parts, $x^2-y^2=-y\ \ \ \ (2)$
From $(1),$
either $x=0, $ using $(2), 0^2-y^2=-y\implies y=?$
or $2y+1=0, $ using $(2), x^2=y^2-y=?\implies x=?$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve $xdx-ydy=y^2(x^2-y^2)dy$ Question:
$$xdx-ydy=y^2(x^2-y^2)dy$$
I'm having trouble matching the solution in the book, which is $\frac{1}{2}\ln(x^2-y^2)=\frac{1}{3}y^3+C$.
I'm getting an integral that requires the incomplete gamma function.
My attempt:
Rewrite the equation:
$x + (-(x^2 - y^2) y^2 - y)\frac{dy}{dx} = 0$
This is not an exact equation, but I found the integrating factor: $μ(y) = e^{-(2 y^3)/3}$
Multiply both sides of $x + \frac{dy}{dx} (-(x^2 - y^2) y^2 - y) = 0$ by $μ(y):$
$xe^{-\frac{2}{3}y^3} + (e^{-\frac{2}{3} y^3} (y^3 - x^2 y - 1) y) \frac{dy}{dx} = 0$
Let $R(x,y) =xe^{-\frac{2}{3}y^3} $ and $S(x,y) = (e^{-\frac{2}{3} y^3} (y^3 - x^2 y - 1) y)$. So I want to seek $f(x,y)$ such that $\frac{\partial f(x,y)}{x} = R(x,y)$ and $\frac{\partial f(x,y)}{y} = S(x,y)$
Integrating w.r.t $x$:
$f(x,y) = \int xe^{-\frac{2}{3}y^3}\,dx = \frac{1}{2}x^2 e^{-\frac{2}{3}y^3} + g(y)$
$\frac{dg(y)}{dy} = e^{-\frac{2}{3}y} y (y^3 - 1)$
Integrating w.r.t $y$:
$g(y) = \int e^{-\frac{2}{3}y} y (y^3 - 1) dy = ?$
I'm stuck here.
| Looks easier if you try to substitute $u=x^2-y^2 \implies u'=2xx'-2y$
$$xdx-ydy=y^2(x^2-y^2)dy$$
$$xx'-y=y^2(x^2-y^2)$$
$$u'=2y^2u$$
The DE is separable
More simply:
$$xdx-ydy=y^2(x^2-y^2)dy$$
$$dx^2-dy^2=2y^2(x^2-y^2)dy$$
The DE is separable.
$$\dfrac {d(x^2-y^2)}{x^2-y^2}=2y^2dy$$
Integrate.
$$\ln (x^2-y^2)=\dfrac 23 y^3+C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Show that $\frac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\frac{3\pi}{4}+\alpha\right)$ Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$
I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For the LHS:
$$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\dfrac{1-2\sin\alpha\cos\alpha}{1+2\sin\alpha\cos\alpha}=\dfrac{\sin^2\alpha+\cos^2\alpha-2\sin\alpha\cos\alpha}{\sin^2\alpha+\cos^2\alpha+2\sin\alpha\cos\alpha}=\dfrac{\left(\sin\alpha-\cos\alpha\right)^2}{\left(\sin\alpha+\cos\alpha\right)^2}$$ I don't know if this is somehow useful as I can't get a feel of the problem and what we are supposed to notice to solve it.
| $\sin(\alpha)-\cos(\alpha)=\sqrt{2}\left(\sin(\alpha)\cdot \frac{1}{\sqrt{2}}-\cos(\alpha)\cdot\frac{1}{\sqrt{2}}\right)=\sqrt{2}\left(-\sin(\alpha)\cos\left(\frac{3\pi}{4}\right)-\cos(\alpha)\sin\left(\frac{3\pi}{4}\right)\right)=-\sqrt{2}\sin\left(\alpha+\frac{3\pi}{4}\right)$
$\sin(\alpha)+\cos(\alpha)=\sqrt{2}\left(\sin(\alpha)\cdot \frac{1}{\sqrt{2}}+\cos(\alpha)\cdot\frac{1}{\sqrt{2}}\right)=\sqrt{2}\left(\sin(\alpha)\sin\left(\frac{3\pi}{4}\right)-\cos(\alpha)\cos\left(\frac{3\pi}{4}\right)\right)=-\sqrt{2}\cos\left(\alpha+\frac{3\pi}{4}\right)$
Finally,
$$\frac{\left(\sin(\alpha)-\cos(\alpha)\right)^2}{\left(\sin(\alpha)+\cos(\alpha)\right)^2}=\frac{\left(-\sqrt{2}\sin\left(\alpha+\frac{3\pi}{4}\right)\right)^2}{\left(-\sqrt{2}\cos\left(\alpha+\frac{3\pi}{4}\right)\right)^2}=\tan^2\left(\alpha+\frac{3\pi}{4}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 0
} |
Inequality $xy+yz+zx-xyz \leq \frac{9}{4}.$ Currently I try to tackle some olympiad questions:
Let $x, y, z \geq 0$ with $x+y+z=3$. Show that
$$
x y+y z+z x-x y z \leq \frac{9}{4}.
$$
and also find out when the equality holds.
I started by plugging in $z=3-x-y$ on the LHS and got
$$
3y-y^2+3x-x^2-4xy+x^2y+xy^2 = 3y-(y^2+x^2)+3x-4xy+x^2y+xy^2\leq 3y-((y+x)^2)+3x-4xy+x^2y+xy^2
$$
But this got me nowhere.
Then I started again with the left hand side
$$
x y+y z+z x-x y z \Leftrightarrow yz(1-x)+xy+zx
$$
and $x+y+z=3 \Leftrightarrow y+z-2=1-x$ so
$$
yz(y+z-2)+x(y+z)
$$
But this also leaves no idea. Do I have to use a known inequality?
| For $x=y=z=1$ we obtain on the L.H.S. the value two. Which is far away from the R.H.S. $9/4$. What about showing a stronger inequality under the given constraints?
(Of course, we are slightly changing the "weakest" term...)
$$
\boxed{\qquad
xy + yz + zx - \color{blue}{\frac 34} xyz \ \le\ \frac 94
\qquad}
$$
It is a good idea to switch to the homogenized version, we will ultimately find an inequality involving an algebraic expression of degree three for three variables, there are not too many of this kind...
$$
\tag{$1$}
\frac 13(x+y+z)(xy+xz+zx) - \frac 34xyz\le\frac 94\cdot\frac 1{27}(x+y+z)^3\ .
$$
After moving everything on the R.H.S., expanding, simplifying and getting rid of the denominator $12$, we obtain:
$$
\tag{$2$}
0\le x^3+y^3+z^3 - x^2y-xy^2-y^2 z-yz^2-z^2x-zx^2+ 3xyz\ ,
$$
and for sure, this is Schur.
$\square$
Note that expanding
$(1)$ is not a complicated dirty step, that would eliminate this solution from an olympiad priced solution, this step is easily done. We move all terms on the R.H.S. of $(1)$, multiply by $12$ to get
$$
\begin{aligned}
&(x+y+z)^3 -4(x+y+z)(xy+yz+zx) +9xyz
\\[3mm]
&\qquad =
\sum\left(\binom 3{3 \,0\,0}x^3 + \binom 3{2 \,1\,0}\color{brown}{x^2y} + \binom 3{1 \,1\,1}xyz \right)
- 4\left(\sum_{\text{cyclic}}x(\color{brown}{xy}+yz+\color{brown}{zx})\right)
+ 9xyz
\\[2mm]
&\qquad=
\sum_{\text{cyclic}} x^3
+ \sum_{\text{cyclic}}(3-4)\color{brown}{(x^2y + x^2z)}
+ (6 - 12 + 9)xyz
\end{aligned}
$$
which is the expression in the R.H.S. of $(2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Doubts in solving $\int_0^1 x\sqrt{x+2} dx$ I was solving the following integral and got stuck in finding the new limits after the substitution. $$\int_0^1 x\sqrt{x+2} dx$$
Here's my work so far:
Putting $(x+2) = t^2$ so that $dx = 2t\ dt$ and $x = t^2 - 2$
Thus the original integral changes to,
$$\int_{?}^{?} (t^2 - 2) (t) (2t )\ dt = \boxed{2\int_{?}^{?} t^4 - 2t^2\ dt}$$
The above boxed integral can be solved easily using power rule of integral but I'm totally out in finding the limits.
When $x =0, t^2= 2$ and when $x = 1, t^2 = 3$ so $t$ is going from $\pm \sqrt{2}$ to $\pm \sqrt{3}$.
So we need to evaluate the integral $$2\int_{t = \pm \sqrt{2}}^{t = \pm \sqrt{3}} t^4 - 2t^2\ dt$$
which is not meaningful I think.
So, I need help in finding the sign convention of the limits of the integral.
| Given that : $$=\int _0^1\:\left(x\sqrt{x+2}\right)dx$$
Let : $$u=x+2$$ $$\frac{du}{dx}=1$$ $$du=dx$$
Also : $$u=x+2$$ $$x=u-2$$
Substitute into the equation : $$=\int \left(\left(u-2\right)\sqrt{u}\right)du$$
Then : $$=\int \left(u^{1.5}\:-2u^{0.5}\right)du$$
Integrate : $$=\frac{u^{2.5}}{2.5}-\frac{2u^{1.5}}{1.5}$$
Substitute into $x$ back and put the original limit : $$=\left[\frac{\left(x+2\right)^{2.5}}{2.5}-\frac{2\left(x+2\right)^{1.5}}{1.5}\right]^1_0$$
$$=\left[\frac{\left(1+2\right)^{2.5}}{2.5}-\frac{2\left(1+2\right)^{1.5}}{1.5}\right]-\left[\frac{\left(0+2\right)^{2.5}}{2.5}-\frac{2\left(0+2\right)^{1.5}}{1.5}\right]$$
$$=-4\sqrt{3}+\frac{16\sqrt{2}+54\sqrt{3}}{15}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0\end{aligned}
$$
Then, we have
$$
\begin{aligned}
0<x<2\\
\implies -\frac a2<2x-\frac a2<4-\frac a2\end{aligned}
$$
Case $-1:\,\,\,a≥0 \wedge 4-\frac a2≤0$.
This leads,
$$
\begin{aligned}\frac {a^2}{4}>\left(2x-\frac a2\right)^2>\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4>f(x)>\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
For $f(x)>0$, it is enough to take $\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≤0$.
Case $-2:\,\,\,a≤0 \wedge 4-\frac a2≥0$.
We have:
$$
\begin{aligned}\frac {a^2}{4}<\left(2x-\frac a2\right)^2<\left(4-\frac a2\right)^2\\
\implies \frac {a^2}{4}+\frac {3a^2}{4}-5a+4<f(x)<\left(4-\frac a2\right)^2+\frac {3a^2}{4}-5a+4\end{aligned}
$$
Similarly, for $f(x)>0$, it is enough to take $\frac{a^2}{4}+\frac {3a^2}{4}-5a+4>0$ with the restriction $a≤0\wedge 4-\frac a2≥0$.
Case $-3:\,\,\,a≥0 \wedge 4-\frac a2≥0$.
This case implies, $\left(2x-\frac a2\right)^2≥0$. This means, $f(x)≥\frac {3a^2}{4}-5a+4$
Thus, for $f(x)>0$, it is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a≥0\wedge 4-\frac a2≥0$.
Finally, we have to combine all the solution sets we get.
I haven't done the calculation, because I want to make sure that the method I use is correct. Do you see any flaws in the method?
| I think your current description is not satisfactory.
Take Case 3 for example. You said,
This means, $f(x) \ge \frac {3a^2}{4}-5a+4$. Thus, for $f(x) > 0$, it
is enough to take $\frac {3a^2}{4}-5a+4>0$ with the restriction $a \ge
0\wedge 4-\frac a2\ge 0$.
This description does not emphasize the sufficient and necessary condition to include all possible values of $a$.
It looks (at least from the description) like a sufficient condition (more precisely, $\frac {3a^2}{4}-5a+4>0$ is a sufficient condition of $f(x) > 0$).
One way to overcome this is given below (Cases 1-2 are omitted here, and given later):
Let
$$f(x) := 4x^2 - 2ax + a^2 - 5a + 4 = \left(2x-\frac a2\right)^2+\frac {3a^2}{4}-5a+4. \tag{1}$$
We have
$$0 < x < 2 \implies -\frac a2<2x-\frac a2<4-\frac a2. \tag{2}$$
We split into three cases.
Case 3: $-\frac{a}2 < 0$ and $4 - \frac{a}2 > 0$, i.e. $0 < a < 8$
Note that $\frac{a}{4} \in (0, 2)$
and $f(\frac{a}{4}) = \frac {3a^2}{4}-5a+4$.
We have,
\begin{align*}
&f(x) > 0, \quad \forall 0 < x < 2\\
\iff \quad & \frac{3a^2}{4}-5a+4 > 0\\
\iff \quad & a > \frac{10}{3} + \frac23\sqrt{13}, \quad \mathrm{or} \quad a < \frac{10}{3} - \frac23\sqrt{13}.
\end{align*}
Thus, the set of qualified $a$ is $0 < a < \frac{10}{3} - \frac23\sqrt{13}$
or $\frac{10}{3} + \frac23\sqrt{13} < a < 8$.
Someone may ask why I took Case 3 for example, rather than Case 1 or 2? The reason is that the conditions in Case 1 and Case 2 (respectively, $a \ge 8$ and $a \le 0$), actually both imply $f(x) > 0$. By contrast, the condition of Case 3 (i.e. $0 < a < 8$) does not imply $f(x) > 0$.
The details are given below.
Case 1: $4 - \frac{a}{2} \le 0$, i.e. $a \ge 8$
From (1) and (2), we have
\begin{align*}
&0 < x < 2 \\
\implies\quad & \left(2x-\frac a2\right)^2 > \left(4 - \frac{a}2\right)^2\\
\implies \quad & f(x) > \left(4 - \frac{a}2\right)^2 + \frac {3a^2}{4}-5a+4 = (a - 4)(a - 5) > 0.
\end{align*}
Thus, the set of qualified $a$ is $a \ge 8$.
Case 2: $-\frac{a}2 \ge 0$, i.e. $a \le 0$
From (1) and (2), we have
\begin{align*}
&0 < x < 2 \\
\implies\quad & \left(2x-\frac a2\right)^2 > \left( - \frac{a}2\right)^2\\
\implies \quad & f(x) > \left(- \frac{a}2\right)^2 + \frac {3a^2}{4}-5a+4 = (a - 1)(a - 4) > 0.
\end{align*}
Thus, the set of qualified $a$ is $a \le 0$.
Summarily, all possible values of qualified $a$ is $a < \frac{10}{3} - \frac23\sqrt{13}$
or $a > \frac{10}{3} + \frac23\sqrt{13}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 11,
"answer_id": 5
} |
Rotation matrix exponential form derivation (only acts in 2 dimension) I am trying to derive the rotation matrix in expoonential form. I start by considering a rotation in three dimensions about an arbritrary axis, $\hat{\textbf{n}}$ by an infinitesimally small angle $\delta\theta$ as depicted above. The coordinate transforms as
\begin{align*}
\textbf{r}\rightarrow\textbf{r}'=\textbf{r}+\delta\textbf{r}\approx\textbf{r}+\delta\theta\hat{\textbf{n}}\times\textbf{r}
\end{align*}
which gives an infinitesimal rotation matrix of
\begin{align*}
R_{\hat{\textbf{n}}}(\delta\theta)=\begin{pmatrix}
1 & -\delta\theta n_{z} & \delta\theta n_{y} \\
\delta\theta n_{z} & 1 & -\delta\theta n_{x} \\
-\delta\theta n_{y} & \delta\theta n_{x} & 1
\end{pmatrix},\;\;\;\;\textbf{r}'=R_{\hat{\textbf{n}}}\textbf{r}
\end{align*}
I then see that this can be expressed as
\begin{align*}
R_{\hat{\textbf{n}}}(\delta\theta)=\mathbb{1}-i\delta\theta\hat{\textbf{n}}\cdot\textbf{J}
\end{align*}
Where $\textbf{J}$ is a vector of matrices,
\begin{align*}
J_{x}=\begin{pmatrix}
0 & 0 & 0 \\
0 & 0 & -i \\
0 & i & 0
\end{pmatrix},\;\;
J_{y}=\begin{pmatrix}
0 & 0 & i \\
0 & 0 & 0 \\
-i & 0 & 0
\end{pmatrix},\;\;
J_{z}=\begin{pmatrix}
0 & -i & 0 \\
i & 0 & 0 \\
0 & 0 & 0
\end{pmatrix}
\end{align*}
Using the composition property of rotations I then find
\begin{align}
\frac{d}{d\theta}R_{\hat{\textbf{n}}}(\theta)=-i\hat{\textbf{n}}\cdot\textbf{J} R_{\hat{\textbf{n}}}(\theta)
\end{align}
Which can be solved to give
\begin{align*}
R_{\hat{\textbf{n}}}(\theta)=e^{-i\hat{\textbf{n}}\cdot\textbf{J}\theta}
\end{align*}
So at this point I'm thinking I've obtained my solution, however when i try to consider a rotation about the $z$-axis I find that
\begin{align*}
R_{\hat{\textbf{z}}}(\theta)&=e^{-iJ_{z}\theta}\\
&=\sum^{\infty}_{n=0}\left(\frac{(-i\theta)^{n}}{n!}\right)\\
&=\sum^{\infty}_{n=0}\left(\frac{(-1)^{n}\theta^{2n}}{2n!}\right)\mathbb{1}_{2}-i\sum^{\infty}_{n=0}\left(\frac{(-1)^{n}(\theta)^{2n+1}}{(2n+1)!}\right)J_{z}\\
&=\mathrm{Cos}(\theta)\mathbb{1}_{2}-i\mathrm{Sin}(\theta)J_{z}
\end{align*}
which is
\begin{align*}
R_{\hat{\textbf{z}}}(\theta)=\begin{pmatrix}
Cos(\theta) & -Sin(\theta) & 0 \\
Sin(\theta) & Cos(\theta) & 0 \\
0 & 0 & 0
\end{pmatrix}
\end{align*}
This is incorrect because it shouldn't map the z value to zero, but I can't for the life of me see where the 1 in the bottom right of the matrix would come from in the exponential.
| Does the following help ?
\begin{align}
&\begin{pmatrix}1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}+
\sum_{n=1}^\infty \frac{(-1)^n\theta^{2n}}{(2n)!}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}\\
&=\begin{pmatrix}0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1\end{pmatrix}+\sum_{n=0}^\infty \frac{(-1)^n\theta^{2n}}{(2n)!}\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 0\end{pmatrix}
=\begin{pmatrix}\cos \theta & 0 & 0 \\ 0 & \cos \theta & 0 \\ 0 & 0 & 1\end{pmatrix}\,,
\end{align}
and
\begin{align}
\sum_{n=0}^\infty \frac{(-1)^n\theta^{2n+1}}{(2n+1)!}\begin{pmatrix} 0 & -1 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}=
\begin{pmatrix} 0 & -\sin \theta & 0 \\ \sin \theta & 0 & 0 \\ 0 & 0 & 0\end{pmatrix}\,.
\end{align}
Therefore, with
$$
V=\begin{pmatrix}0 & -\theta & 0\\\theta & 0 & 0\\ 0& 0& 0\end{pmatrix}\,,
$$
\begin{align}
e^V=\sum_{n=0}^\infty \frac{V^n}{n!}
=\begin{pmatrix} \cos \theta & -\sin \theta & 0 \\ \sin \theta & \cos \theta & 0 \\ 0 & 0 & 1\end{pmatrix}\,.
\end{align}
| {
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"source": "stackexchange",
"question_score": "1",
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Find all real solutions of the equation $x^{10} - x^8 + 8x^6 - 24x^4 + 32x^2 - 48 = 0$ I have been able to factorize the polynomial as follows: $$(x^2 - 2)(x^8 + x^6 + 10x^4 - 4x^2 + 24)$$ from which $\sqrt2$ and $-\sqrt2$ are obvious solutions. My guess is that $x^8 + x^6 + 10x^4 - 4x^2 + 24 = 0$ does not have any real solutions, but how would I show that?
| $x^8 + x^6 + 10x^4 - 4x^2 + 24\geq 10x^4 - 4x^2>0$ if $x^{2}>\frac 2 5$ and $x^8 + x^6 + 10x^4 - 4x^2 + 24\geq x^8 + x^6 + 10x^4 - \frac 8 5 + 24\geq 24 -\frac 8 5 >0$ if $x^{2}<\frac 2 5$.
| {
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Without any software and approximations prove that $\sec(52^{\circ})-\cos(52^{\circ})>1$ Without any software and approximations prove that $$\sec(52^{\circ})-\cos(52^{\circ})>1$$
We can use some known trig values like $18^{\circ}$,$54^{\circ}$,etc
My try:
I considered the function:
$$f(x)=\sec(x)-\cos(x)-1,\: x\in \left (0, \frac{\pi}{3}\right)$$
We have the derivative as:
$$f'(x)=\sec x\tan x+\sin x >0$$ so $f$ is Monotone increasing.
So we have:
$$f(52^{\circ})>f(45^{\circ})=\frac{1}{\sqrt{2}}-1$$ but not able to proceed
| Thanks to "user" for giving me thought to complete the proof.
Here is the proof:
We are aiming to prove $\sec(52^{\circ})-\cos(52^{\circ})>1$. Let $\phi$ be the Golden ratio and $\psi$ be its reciprocal..
Consider $$f(x)=\sec x-\cos x,\:\:0<x<\frac{\pi}{2}$$
Its evident that $f$ is Monotone increasing.
Also $$\begin{aligned}
& f(t)=1 \\
\Rightarrow & \sec t-\frac{1}{\sec t}=1 \\
\Rightarrow & \sec t=\phi \\
\Rightarrow & t=\sec ^{-1}(\phi)
\end{aligned}$$
We know that by taylor's series:
$$\sin ^{-1}(x)=x+\frac{x^3}{6}+\frac{3 x^5}{40}+\cdots,|x| \leqslant 1$$
Using the fact that:
$$\sec ^{-1}(v)=\cos ^{-1}\left(\frac{1}{v}\right)=\frac{\pi}{2}-\sin ^{-1}\left(\frac{1}{v}\right)$$
So we have
$$t=\sec ^{-1}(\phi)=\frac{\pi}{2}-\left(\psi+\frac{\psi^3}{6}+\frac{3 \psi^5}{40}+p\right)$$ Where $p>0$.
Also let $$t_0=\frac{\pi}{2}-\left(\psi+\frac{\psi^3}{6}+\frac{3 \psi^5}{40}\right)$$
We know that $\psi$ satisfies $$\psi^2+\psi-1=0$$
We have the following results which can be easily derived:
$$\begin{aligned}
\psi^2 &=1-\psi \\
\psi^3 &=2 \psi-1 \\
\psi^5 &=5 \psi-3 \\
\Rightarrow & \psi+\frac{\psi^3}{6}+\frac{3 \psi^5}{40}=\frac{205 \psi-47}{120} .
\end{aligned}$$
We have:
$$\begin{aligned}
& t_0=\frac{\pi}{2}-\left(\psi+\frac{\psi^3}{6}+\frac{3 \psi^5}{40}\right) \\
\Rightarrow & t_0=\frac{\pi}{2}-\left(\frac{205 \psi-47}{120}\right) \\
\Rightarrow & \frac{52 \pi}{180}-t_0=\frac{205 \psi-47}{120}-\frac{19 \pi}{90}=\frac{615 \psi-(141+76 \pi)}{360}>0
\end{aligned}$$
Hence we have :
$$\begin{aligned}
& t_0<\frac{52 \pi}{180} \\
\Rightarrow & t=t_0-p<\frac{52 \pi}{180}
\end{aligned}$$
Finally we have
$$\begin{aligned}
1 &=f(t)<f\left(\frac{52 \pi}{180}\right) \\
& \Rightarrow \sec (52^{\circ})-\cos (52^{\circ})>1 .
\end{aligned}$$
| {
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"question_score": "11",
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} |
Find $\cos\frac{\pi}{12}$ given $\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$
Find $\cos\frac{\pi}{12}$ given $\sin(\frac{\pi}{12}) = \frac{\sqrt{3} -1}{2 \sqrt{2}}$
From a question I asked before this, I have trouble actually with the numbers manipulating part.
Using trigo identity, $\sin^2 \frac{\pi}{12} + \cos^2 \frac{\pi}{12} = 1$ so , $\cos^2 \frac{\pi}{12} = 1- \sin^2 \frac{\pi}{12}$
To find $\cos \frac{\pi}{12} = \sqrt{1- \sin^2 \frac{\pi}{12}}$
$\sin^2 \frac{\pi}{12} = (\frac{\sqrt{3} -1}{2 \sqrt{2}})^2 = \frac{(\sqrt{3}-1)^2}{(2\sqrt{2})^2} = \frac{2- \sqrt{3}}{4}$
$\cos \frac{\pi}{12} = \sqrt{1-(\frac{\sqrt{3} -1}{2 \sqrt{2}})^2} $
$\cos \frac{\pi}{12} = \sqrt{1- \frac{2-\sqrt{3}}{4}}$
$\cos \frac{\pi}{12} = \frac{\sqrt{2+\sqrt{3}}}{2}$
What is wrong with my steps?
| Another way would be to use the double angle formula
$$\sin 2\theta=2\sin\theta\cos\theta$$
With $\theta=\frac {\pi}{12}$, then
$$\sin 2\theta=\sin\frac {\pi}6=\frac 12$$
Thus
\begin{align*}
\cos\frac {\pi}{12} & =\frac {\sin 2\theta}{2\sin\theta}\\
& =\frac 1{4\sin\theta}\\
& =\frac 1{\sqrt 2(\sqrt 3-1)}
\end{align*}
Which is numerically equal to the answer you found. To get it into the form Wolfram Alpha produces, multiply the fraction by $\sqrt 3+1$.
| {
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"source": "stackexchange",
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Proving that $\frac{xy^2}{x^2+y^4}$ is bounded I am trying to prove that there exists $M>0$ such that, for all $(x,y) \in \mathbb{R}^2$,
$|\frac{xy^2}{x^2+y^4}|\leq M$. Is the following proof correct?
I claim that there exists such an $M$, and that $M=1$.
Suppose, first, that $|x^2|\geq |xy^2|$. Then $|\frac{xy^2}{x^2+y^4}|\leq |\frac{x^2}{x^2+y^4}|\leq |\frac{x^2}{x^2}|=1$.
Suppose, second, that $|x^2|<|xy^2|$. Then $|x|<|y^2|$, so $|xy^2|<|y^4|.$ Then $|\frac{xy^2}{x^2+y^4}|\leq |\frac{y^4}{x^2+y^4}|\leq |\frac{y^4}{y^4}|=1$ (note that if $|x|=0$, then $f(x)=0$—thus our assumption that $|x|>0$ is not without loss of generality).
Thus, for any $(x,y) \in \mathbb{R}^2$, $|f(x,y)|\leq M$
| Your proof is a good attempt, but it is not complete. The "mistake" in the proof is that the inequality in red below is not justified:
Suppose, second, that $x^2<xy^2$. Then $|\frac{xy^2}{x^2+y^4}|\color{red}{\leq} |\frac{y^4}{x^2+y^4}|\leq |\frac{y^4}{y^4}|=1$.
You do not explain why this is true. Your assumption is that $x^2<xy^2$, but then you somehow conclude that $|xy^2| \leq |y^4|$. Why is this so?
To be more rigorous, you could argue like so:
If $x^2 < xy^2$, then $x>\frac{x^2}{y^2}\geq 0$. Therefore, we can divide the inequality by $x$ and get $x<y^2$, meaing that $0<x<y^2$.
This means that $|xy^2| = xy^2 < (y^2)y^2 = y^4=|y^4|$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
For what positive integers $A$ is the sequence eventually constant?
For each integer $n \ge 0$ let $S(n)=n-m^2$, where $m$ is the greatest integer such that $m^2\le n$. Define $(a_k)_{k=0}^\infty$ as $a_0=A$ and $a_{k+1}=a_k+S(a_k)$ for $k \ge0$. For what positive integers $A$ is the sequence eventually constant?
I'm trying to investigate this sequence to find some pattern, but I don't find any... If $A=1$, then $$\begin{align*} a_0 &=1 \\ a_1&= a_0+S(a_0)= 1+(1-m^2) = 1+(1-1)=1 \\ a_2&= a_1+S(a_1)= 1+S(1)=1+(1-1)=1\end{align*}$$ so it would seem that for $A=1$ the sequence is always constant.
For $A=2$ $$\begin{align*} a_0 &=2 \\ a_1&=2+S(2)=2+(2-m^2)=2+(2-1)=3 \\ a_2 &=3+S(3)=3+(3-m^2)=3+(3-1)=5 \\ a_3&=5+S(5)=5+(5-m^2)=5+(5-4)=6 \\ a_4&=6+S(6)=6+(6-m^2)=6+(6-4)=8 \\a_5&=8+S(8)=8+(8-m^2)=8+(8-4)=12 \\ a_6&=12+S(12)= 12+(12-m^2)=12+(12-9)=15\end{align*}$$
so the sequence would seem to be just increasing If $A=3$, then $$\begin{align*} a_0 &=3 \\ a_1&=3+S(3)=5 \\a_2&=5+S(5)=6 \\ a_3&=6+S(6)=8 \end{align*}$$ and this will just be the same as when $A=2$ except were skipping the first term.
Now if $A=4$, then $$\begin{align*} a_0 &=4 \\ a_1&=4+S(4)=4+(4-m^²)=4+(4-2)=6 \\a_2&= 6+S(6)=8 \\a_3&= 8+S(8)=12 \\ a_4&=12+S(12)=15 \\ a_5&=15+S(15)=15+(15-9)=21 \\ a_6&=21+S(21) = 21+(21-16) = 26\end{align*}$$
but this also seems to be increasing? What is the pattern here?
| You're misunderstanding the $A = 4 $ case. The $a_1$ part will be $4 + (4 - 4)$.
This sequence will be constant for all perfect squares because if $a_i = k^2$, then $S(a_{i}) = k^2 - k^2 = 0 $, since $k^2$ is the largest perfect square $\leq k^2$. This implies that $a_{i + 1} = a_i = k^2$. Notice that the sequence will be constant from that point onwards.
It is also not that tough to prove that $a_i = k^2 \iff a_0 = A = k^2$.
| {
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Find the locus of points |z-1|= -Im(z). If I wish to find the locus of complex points satisfying $ |z-1|= -\text{Im}(z)$, then would I be right in supposing it represents the half-circle
$(x-1)^2 + (y-1/2)^2 = 1/4, y \leq 0$?
My work follows:
*
*First, notice $|z-1| \geq 0 \Rightarrow \text{Im}(z) \leq 0.$
*Second,
$$|z-1|^2 = (-\text{Im}(z))^2 \Rightarrow |z|^2 - 2\text{Re}(z)+ 1 = \text{Im}(z) \\ \Rightarrow x^2 +y^2 - 2x - y + 1 =0 \\ \Rightarrow (x-1)^2 + (y-1/2)^2 = 1/4,$$
in completing the square.
As $y = \text{Im}(z)$, we further restrict $y \leq 0.$
| HINT
You are on the right track, but you forgot to square $\text{Im}(z)$. If we let $z = x + yi$, it results that
\begin{align*}
|z - 1| = -\text{Im}(z) = -y & \Longleftrightarrow
\begin{cases}
|z - 1|^{2} = y^{2}\\\\
y\leq 0
\end{cases}\\\\ & \Longleftrightarrow
\begin{cases}
|z|^{2} - 2\text{Re}(z) + 1 = y^{2}\\\\
y\leq 0
\end{cases}\\\\
& \Longleftrightarrow
\begin{cases}
x^{2} + y^{2} - 2x + 1 = y^{2}\\\\
y\leq 0
\end{cases}\\\\ & \Longleftrightarrow
\begin{cases}
(x - 1)^{2} = 0\\\\
y\leq 0
\end{cases}
\end{align*}
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
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Partial Fraction of $\frac{1-x^{11}}{(1-x)^4} $ for Generating Function The original question involves using generating functions to solve for the number of integer solutions to the equation $c_1+c_2+c_3+c_4 = 20$ when $-3 \leq c_1, -3 \leq c_2, -5 \leq c_3 \leq 5, 0 \leq c_4$.
Using generating functions I was able to get it into the rational polynomial form:
$$f(x) = {\left(\frac{1}{1-x}\right)}^3\left(\frac{1-x^{11}}{1-x}\right) = \frac{1-x^{11}}{{(1-x)}^4}$$
I was also able to determine that the sequence could be represented in two factors:
$${\left(1+x^1+x^2+x^3+\cdots\right)}^3\left(1+x^1+x^2+\cdots+x^{10}\right)$$
However, to find the coefficient on $x^{31}$ to solve the problem, I figured I would have to get the term $\frac{1-x^{11}}{{(1-x)}^4}$ into a more typical generating function summation form. Thus, I endeavored to find the partial fraction decomposition of the term, however, I can't seem to do it at all.
How would I find the partial fraction decomposition of $\frac{1-x^{11}}{{(1-x)}^4}$?
Or is there a better method in using ${(1+x^1+x^2+x^3+\ldots)}^3(1+x^1+x^2+\ldots+x^{10})$ in order to find the coefficient on $x^{31}$?
Thank you very much for your help, I've been trying this partial fraction for a while now and Wolfram alpha doesn't seem to be giving me an answer that is of much value.
| $$\begin{align}
\frac{1-x^{11}}{(1-x)^4}&=(1-x^{11})\sum_{n\ge0}\frac{(n+1)(n+2)(n+3)}{6}x^n\\
&=\sum_{n\ge0}\frac{(n+1)(n+2)(n+3)}{6}x^n-\sum_{n\ge0}\frac{(n+1)(n+2)(n+3)}{6}x^{n+11}\\
&=\sum_{n\ge0}\frac{(n+1)(n+2)(n+3)}{6}x^n-\sum_{n\ge11}\frac{(n-10)(n-9)(n-8)}{6}x^{n}\\
&=\sum_{n\ge0}\frac{1}{6}\bigg((n+1)(n+2)(n+3)-(n-10)(n-9)(n-8)[n\ge11]\bigg)x^n,
\end{align}$$
where $[n\ge11]$ is the Iverson Bracket.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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2022 Gr11 Fermat math contest question #20 Question: A sequence of numbers $t_1, t_2, t_3$,... has its terms defined by $t_n = \frac{1}{n} - \frac{1}{n+2}$ for every integer $n \geq 1.$ For example, $t_4 = \frac{1}{4} - \frac{1}{6}$. What is the largest positive integer k for which the sum of the first k terms (that is, $t_1 + t_2 +\cdots+t_{k-1} + t_k$) is less than 1.499?
Correct answer: 1998
My solution: First, to find the sum of all numbers from $t_1$ to $t_k$, I wrote the equation $\frac{(\frac{2}{3} + \frac{1}{k} - \frac{1}{k+2}) * k}{2}$, note: I got $\frac{2}{3}$ from $\frac{1}{1} - \frac{1}{1+2}$. Next, I just substituted the 5 options this question gave me into my equation (this is a multiple choice question), but I got around $666$ when I substituted the different options which is no where near $1.499$. This means the problem is somewhere in my equation for the sum of all numbers from $t_1$ to $t_k$, but I do not know where I went wrong in my equation, or even if it was the right first step to solving this question.
| $S_k = \sum_{n=1}^k t_n = \sum_{n=1}^k\frac{1}{n} - \frac{1}{n+2} = A_k - B_k< 1.499$. Now, since $B_k = A_{k+2} - \frac{1}{1} - \frac{1}{2}$,
$$
S_k = A_k - A_{k+2} + \frac{3}{2} = -\frac{1}{k+1} - \frac{1}{k+2} + 1.5 < 1.499.
$$
This gives
$\frac{1}{k+1} + \frac{1}{k+2} > 1.5 - 1.499 = 0.001 = \frac{1}{2000} + \frac{1}{2000}$.
Therefore $k+2$ can not exceed 2000 so $k = 1998$.
| {
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"url": "https://math.stackexchange.com/questions/4547759",
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Determine the cartesian equation of the tangent plane to M in p. Given the torus and given the point p $\in$ M corresponding to the parameters $s=\frac{\pi }{4}$ and $t=\frac{\pi }{3}$.
Determine the cartesian equation of the tangent plane to M in p.
$\begin{cases} x=\left(3+\sqrt{2}cos\left(s\right)\right)cos\left(t\right) \\ y=\left(3+\sqrt{2}cos\left(s\right)\right)sin\left(t\right) \\ z=\sqrt{2}sin\left(s\right) \end{cases}$
Could someone give me a hint or help me? I'm not sure if I firstly should go from the given parametric equation to a cartesian equation.
| Since you have the parametric equation of the surface, which is
$ \mathbf{P}(t, s) = ( x(t,s), y(t,s), z(t,s) ) $
Then find $\dfrac{\partial \mathbf{P}}{\partial t} $ and $\dfrac{\partial \mathbf{P}}{\partial s } $ as follows
$ \dfrac{\partial \mathbf{P}}{\partial t} =( - (3 + \sqrt{2} \cos(s) ) \sin(t) , (3 + \sqrt{2} \cos(s)) \cos(t) , 0 )$
$ \dfrac{\partial \mathbf{P}}{\partial s } =( - \sqrt{2} \sin(s) \cos(t) , - \sqrt{2} \sin(s) \sin(t) , \sqrt{2} \cos(s) ) $
Evaluate $\mathbf{P},\dfrac{\partial \mathbf{P}}{\partial t},\dfrac{\partial \mathbf{P}}{\partial s }$ at $s = \dfrac{\pi}{4} $ and $ t = \dfrac{\pi}{3} $, you will get
$ \mathbf{P} = (2, 2 \sqrt{3}, 1 ) $
$ \dfrac{\partial \mathbf{P}}{\partial t} = (- 2 \sqrt{3}, 2 , 0 ) $
$ \dfrac{\partial \mathbf{P}}{\partial s } = ( - \dfrac{1}{2} , -\dfrac{\sqrt{3}}{2} , 1) $
Next, find $\mathbf{N}$ (the normal vector to the plane)
$ \mathbf{N} = \dfrac{\partial \mathbf{P}}{\partial t} \times \dfrac{\partial \mathbf{P}}{\partial s } = (-2 \sqrt{3}, 2, 0) \times (- \dfrac{1}{2}, - \dfrac{\sqrt{3}}{2}, 1 ) = (2 , 2 \sqrt{3} , 4)$
Now the equation of the tangent plane is
$ \mathbf{N} \cdot ( \mathbf{r} - \mathbf{P} ) = 0 $
where $ \mathbf{r} = (x, y, z) $
Hence, the equation of the plane is
$ (2, 2\sqrt{3}, 4) \cdot (x - 2 , y - 2 \sqrt{3}, z - 1) = 0 $
which reduces to
$ 2 x + 2 \sqrt{3} y + 4 z - 20 = 0 $
and further to,
$ x + \sqrt{3} y + 2 z - 10 = 0 $
And this is the cartesian equation of the tangent plane at $ s = \dfrac{\pi}{4}, t = \dfrac{\pi}{3} $
| {
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The convergence speed of $ \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x $? I have already known how to prove
\begin{equation*}
\lim _{n \rightarrow \infty} \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x = \sqrt{\frac{\pi}{2n}}
\end{equation*}
with Wallis's formula
\begin{equation*}
\quad \frac{\pi}{2}=\frac{2 \cdot 2 \cdot 4 \cdot 4 \cdot 6 \cdot 6 \cdot 8 \cdot 8 \cdots}{1 \cdot 3 \cdot 3 \cdot 5 \cdot 5 \cdot 7 \cdot 7 \cdot 9 \cdots}
\end{equation*}
But the method I used was considered not to be universal.
How to prove that
\begin{equation*}
\lim _{n \rightarrow \infty} \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x = \frac{\sqrt{2 \pi}}{2} \cdot \frac{1}{n^{\frac{1}{2}}}-\frac{\sqrt{2 \pi}}{8} \cdot \frac{1}{n^{\frac{3}{2}}}+\frac{\sqrt{2 \pi}}{64} \cdot \frac{1}{n^{\frac{5}{2}}}
\end{equation*}
And is
\begin{equation*}
\lim _{n \rightarrow \infty} \int_0^{\frac{\pi}{2}} \sin ^n(x) \operatorname{d}x = \frac{\sqrt{2 \pi}}{2} \cdot \frac{1}{n^{\frac{1}{2}}}-\frac{\sqrt{2 \pi}}{8} \cdot \frac{1}{n^{\frac{3}{2}}}+ \dots +
(-1)^{k}\cdot\frac{\sqrt{2 \pi}}{2^{\frac {k(k+1)}{2}}} \cdot \frac{1}{n^{\frac{2k+1}{2}}}
\end{equation*}
true? Are there any more powerful tools, like numerical methods to calculate the integration?
| If you are familiar with the Gaussian hypergeometric function, if $0 \leq x \leq \frac \pi 2$
$$\int \sin^n(x)\,dx=-\cos (x) \, _2F_1\left(\frac{1}{2},\frac{1-n}{2};\frac{3}{2};\cos^2(x)\right)$$ and the definite integral just becomes
$$I_n=\int_0^{\frac \pi 2}\sin^n(x)\,dx=\frac{\sqrt{\pi }} 2\,\,\frac{ \Gamma \left(\frac{n+1}{2}\right)}{ \Gamma
\left(\frac{n+2}{2}\right)}$$ Using Stirling approximation, then
$$\frac{ \Gamma \left(\frac{n+1}{2}\right)}{ \Gamma
\left(\frac{n+2}{2}\right)}=\sqrt{\frac{2}{n}}\left(1-\frac{1}{4 n}+\frac{1}{32 n^2}+\frac{5}{128
n^3}+O\left(\frac{1}{n^4}\right) \right)$$
$$I_n= \sqrt{\frac{\pi}{2n}}\left(1-\frac{1}{4 n}+\frac{1}{32 n^2}+\frac{5}{128
n^3}+O\left(\frac{1}{n^4}\right) \right)$$
If you want an even more accurate formula, use more terms in the expansion and make it a $[n,n]$ Padé approximant to get
$$I_n \sim \sqrt{\frac{\pi}{2n}}\,\,\,\frac{64 n^2-8 n+11}{64 n^2+8 n+11}$$ whose error is $\frac{709}{16384 n^5}$.
Using it, the relative error is $0.0035$% for $n=4$
| {
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Which numbers result in a chain other than $2 \to 0$? Let $n$ be a natural number and $k$ the number of its divisors. Calculate $n-k$, then repeat this procedure by taking $n-k$ as the starting value. If you do this repeatedly, which numbers results in chains other than $2 \to 0$?
I found this problem in the exam prep questions my teacher gave us, so I assume there's a not-so-complicated solution. I tried some small values like $1, 3, 5, 7$ and they seem to give us different chains. However, I don't know how to go about proving that all other numbers beyond a point go to $2 \to 0$. I tried splitting the problem into cases where $n$ has $1, 2, 3,...$ prime factors but that lead nowhere. Any ideas would be appreciated.
| Let us denote the iteration function with $f(n) = n - d(n)$, where $d(n)$ is the number of divisors of $n$.
For $n \ge 3$ is $1 \le f(n) \le n-2$, so that the iteration $n, f(n), f(f(n)), \ldots$ is strictly decreasing until it reaches either $1$ or $2$. Let $t(n)$ be this “terminal” non-zero number in the chain starting at $n$.
We can show that $t(n) = 1$ only for $n=1, 3, 4, 5, 7, 8$, and $t(n) = 2$ for all other starting values $n$.
As it turns out, it suffices to compute $t(n)$ for all starting values in the range $1 \le n \le 15$:
$$
\begin{array}{c|cccccccccccccccc}
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\ \hline
f(n) & 0 & 0 & 1 & 1 & 3 & 2 & 5 & 4 & 6 & 6 & 9 & 6 & 11 & 10 & 11 \\
t(n) & \boxed{1} & 2 & \boxed{1} & \boxed{1} & \boxed{1} & 2 & \boxed{1} & \boxed{1} & 2 & 2 & 2 & 2 & 2 & 2 & 2
\end{array}
$$
It remains to show that $t(n) = 2$ for $n \ge 16$: For these $n$ is $d(n) < n/2$ (see for example Upper limit for the Divisor function), and therefore $f(n) > n/2$. It follows that the iteration $n, f(n), f(f(n)), \ldots$ must eventually reach a number $m$ in the range $9 \ldots 15$. Then $t(n) = t(m) = 2$.
| {
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"question_score": "1",
"answer_count": 1,
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Computing the eigenvalues and eigenvectors of a $ 3 \times 3$ with a trick The matrix is: $
\begin{pmatrix}
1 & 2 & 3\\
1 & 2 & 3\\
1 & 2 & 3
\end{pmatrix}
$
The solution says that
$ B\cdot
\begin{pmatrix}
1 \\
1 \\
1
\end{pmatrix}
=
\begin{pmatrix}
6 \\
6 \\
6\end{pmatrix}$
$ B\cdot
\begin{pmatrix}
1 \\
1 \\
-1
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$
$ B\cdot
\begin{pmatrix}
3 \\
0 \\
-1
\end{pmatrix}
=
\begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$
Thus the eigenvalues are $ \lambda_{1}=6,\lambda_{2}=0 $
My question is, how can I easily find
$\begin{pmatrix}
0 \\
0 \\
0\end{pmatrix}$ and $\begin{pmatrix} 6 \\
6 \\
6\end{pmatrix}$?
Is there any way to see it "quickly"?
| It's easy to see that $0$ is a multiple eigenvalue with multiplicity two as we can exhibit two linearly independent vector of the kernel (since all columns are obviously a multiple of the first one). Then the last eigenvalue $\lambda$ can be found using the Trace as the sum of the diagonal elements is also equal to the sum of the eigenvalues counted with their multiplicity : $0 + 0 + \lambda = 1+2+3$. So the eigenvalues are $0$ and $6$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Spivak, Ch. 22, "Infinite Sequences", Problem 1(iii): How do we show $\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right ]=0$? The following is a problem from Chapter 22 "Infinite Sequences" from Spivak's Calculus
*
*Verify the following limits
(iii) $\lim\limits_{n\to \infty} \left
[\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right ]=0$
The solution manual says
$$\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[4]{n+1}\right
]$$
$$=\lim\limits_{n\to \infty} \left [\left
(\sqrt[8]{n^2+1}-\sqrt[8]{n^2}\right )+\left
(\sqrt[4]{n}-\sqrt[4]{n+1}\right )\right ]$$
$$=0+0=0$$
(Each of these two limits can be proved in the same way that
$\lim\limits_{n\to \infty} (\sqrt{n+1}-\sqrt{n})=0$ was proved in the
text)
How do we show $\lim\limits_{n\to \infty} \left [\sqrt[8]{n^2+1}-\sqrt[8]{n^2}\right ]=0$?
Note that in the main text, $\lim\limits_{n\to \infty} (\sqrt{n+1}-\sqrt{n})=0$ was solved by multiplying and dividing by $(\sqrt{n+1}+\sqrt{n})$ to reach
$$0<\frac{1}{\sqrt{n+1}+\sqrt{n}}<\frac{1}{2\sqrt{n}}<\epsilon$$
$$\implies n>\frac{1}{4\epsilon^2}$$
| The solution considers $$\lim_{n \to \infty} \left( \sqrt[8]{n^2+1} - \sqrt[8]{n^2}\right) \tag{1}$$ and $$\lim_{n \to \infty} \left( \sqrt[4]{n} - \sqrt[4]{n+1}\right), \tag{2}$$ yet you are asking about $$\lim_{n \to \infty} \left( \sqrt[8]{n^2+1} - \sqrt[\color{red}{4}]{n^2}\right), \tag{3}$$ which is neither of these.
I will ignore your expression and illustrate how $(1)$ is evaluated in the manner suggested. Let $a_n = \sqrt[8]{n^2+1}$ and $b_n = \sqrt[8]{n^2}$. Then because $$(a_n - b_n)(a_n^7 + a_n^6 b_n + a_n^5 b_n^2 + \cdots + a_n^2 b_n^5 + a_n b_n^6 + b_n^7) = a_n^8 - b_n^8, \tag{4}$$ it follows that $$\begin{align}
a_n - b_n
&= \frac{a_n^8 - b_n^8}{a_n^7 + a_n^6 b_n + a_n^5 b_n^2 + \cdots + a_n^2 b_n^5 + a_n b_n^6 + b_n^7} \\
&= \frac{1}{a_n^7 + a_n^6 b_n + a_n^5 b_n^2 + \cdots + a_n^2 b_n^5 + a_n b_n^6 + b_n^7}. \tag{5} \end{align}$$ Since both $a_n \to \infty$ and $b_n \to \infty$ as $n \to \infty$, the denominator of $(5)$ tends to $\infty$ as $n \to \infty$; hence $$\lim_{n \to \infty} a_n - b_n = 0$$ as claimed. The same argument applies to $(2)$. It does not apply to $(3)$. Why?
| {
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Creating a complex number so that its norm equals to 1 I would like to create a complex number c so that its norm is equal to some number a (for the purpose of this question let's assume a = 1) if I already have either its real or imaginary part. I know that:
$$\lVert \mathbf{c} \rVert = \sqrt{\sum_{i=1} ^{n} c_i \overline{c_i}}.$$
In this case:
$$\lVert \mathbf{c} \rVert = \sqrt{(a+bi)(a-bi)} = \sqrt{a^2+abi-abi-b^2i^2} = \sqrt{a^2-b^2\cdot{-1}} = \sqrt{a^2+b^2}.$$
As I said, we already know either a or b, let's say we know that a = 2. How can I find what must b be equal to if the norm must be 1? Some simple equations:
$$ \sqrt{2^2+b^2} = 1 \iff 2^2+b^2 = 1 \iff 4+b^2 = 1 \iff b^2 = 1 - 4 \iff b^2 = -3 \iff b = \pm\sqrt{-3} \iff $$
$$ \iff b = \pm\sqrt{3} * \sqrt{-1} \iff b = \pm\sqrt{3}i $$
Okay, so based on the above eqations, the imaginary part can be $\pm\sqrt{3}i$. So let's say $c=2+\sqrt{3}i$. But clearly my equations are wrong, because:
$$\sqrt{(2+\sqrt{3}i)(2-\sqrt{3}i)} = \sqrt{7},$$ which does not equal to 1. Please, help me with my confusion and lack of knowledge.
| The Problem
You can't just use any number for $a$ and $b$! If $a$ and $b$ are real numbers, which they are, their square is always positive or zero, which can only be the case if $|a| \leq 1$ and/or $|b| \leq 1$ are less than or equal to 1!
$$
\begin{align*}
a^{2} + b^{2} &= 1 \quad\mid\quad -\left(b^{2}\right)\\
a^{2} &= 1 - b^{2} \quad\mid\quad \sqrt{~~}\\
a &= \sqrt{1 - b^{2}}\\
\end{align*}
$$
If you got an $b > 1$ then will $1 - b^{2}$ get negative and the squareroot of something negativ is not real aka $b$ can't be greater then $1$ but you chose one $b$ or $a$ greater than $1$.
You can also explain it to yourself graphic by imagining the complex number as versor. Of course, the versor is longer (total length) than one [unit] if it is greater than one [unit] in either direction (real part or imaginary part).
Creating complex numbers with the norm $1$
According to the trigonometric Pythagoras there is $\cos(x)^{2} + \sin(x)^2 = 1$ and according to the euler's formula there is $z = e^{x \cdot \mathrm{i}} = \cos(x) + \sin(x) \cdot \mathrm{i}$.
When you combine these you will get $|z| = |e^{x \cdot \mathrm{i}}| = 1$ because $|z| = |e^{x \cdot \mathrm{i}}| = |\cos(x) + \sin(x) \cdot \mathrm{i}| = \cos(x)^{2} + \sin(x)^2 = 1$.
Aka can chose any $x \in \mathbb{R}$ to get a complex number wich has the norm $1$!
E.G.
$$x = \frac{\pi}{4}$$
$$z = \cos(x) + \sin(x) \cdot \mathrm{i} = \cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4}) \cdot \mathrm{i}$$
$$z = \cos(\frac{\pi}{4}) + \sin(\frac{\pi}{4}) \cdot \mathrm{i} = \frac{1}{\sqrt{2}} + \frac{\mathrm{i}}{\sqrt{2}}$$
$$|z| = |\frac{1}{\sqrt{2}} + \frac{\mathrm{i}}{\sqrt{2}}| = 1$$
| {
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Find the closed formula for the following recusive sequence. Find the closed formula for the recursive sequence defined by $a_0 = 4$, $a_1 = 12$, $a_n = 6a_{n-1}-a_{n-2}$ for $n>1$. This question is stumping me. I don't know any methods besides guess and check.
Current progress:
The first five terms are $4, 12, 68, 396, 2308$.
Each term is divisible by four, yielding $1, 3, 17, 99, 577$
Each of these is off of a perfect square by one or negative one $0-1, 4-3, 16-17, 100-99, 576-577$ = $-1,1,-1,1,-1$
Other than that I'm lost. Any help would be appreciated.
| Solution by generating function:
The difference equation can be seen in the form
$$ a_{n+2} = 6 \, a_{n+1} - a_{n} \, \hspace{10mm} a_{0} = 4, a_{1} = 12$$
leads to
\begin{align}
\sum_{n=0}^{\infty} a_{n+2} \, t^{n+2} &= 6 \, t \, \sum_{n=0}^{\infty} a_{n+1} \, t^{n+1} - t^2 \, \sum_{n=0}^{\infty} a_{n} \, t^n \\
\sum_{n=2}^{\infty} a_{n} \, t^{n} &= 6 \, t \, \sum_{n=1}^{\infty} a_{n} \, t^n - t^2 \, \sum_{n=0}^{\infty} a_{n} \, t^n \\
A - 4 - 12 \, t &= 6 \, t \, (A- 4) - t^2 \, A \\
A = \sum_{n=0}^{\infty} a_{n} \, t^n &= \frac{4 \, (1 - 3 \, t)}{1 - 6 \, t + t^2}.
\end{align}
This can be compared to the Chebyshev polynomials of the first kind to obtain
$$ a_{n} = 4 \, T_{n}(3). $$
Further properties can be obtained by using these polynomials and formulas such as:
\begin{align}
\sum_{n=0}^{\infty} T_{n}(x) \, t^n &= \frac{1 - x t}{1 - 2 x t + t^2} \\
T_{n}(x) &= \frac{1}{2} \, \left( \left(x + \sqrt{x^2 -1}\right)^n + \left(x - \sqrt{x^2 -1} \right)^n \right) \\
T_{n}(x) &= \sum_{k=0}^{\lfloor{n/2\rfloor}} \binom{n}{2 k} \, (x^2 - 1)^k \, x^{n-2 k}.
\end{align}
Properties and relations for $T_{n}(3)$ can be found in Oeis A001541.
By using $$ a_{n} = 4 \, T_{n}(3) = 2 \left( (3 + 2 \, \sqrt{2})^n + (3 - 2 \, \sqrt{2})^n \right) = 2 \, \left( (1+\sqrt{2})^{2 n} + (1 - \sqrt{2})^{2 n} \right) $$
then this last part can be compared to the Pell-Lucas numbers, $(Q_{n})$, see for some properties and relations Oeis A002203, and gives $$a_{n} = 2 \, Q_{2 n}.$$
| {
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Confusion with solution to UKMT question There was a question I was trying to solve from the UKMT Senior Maths Challenge. It goes like this:
Four positive integers a, b, c, and d are such that
abcd + abc + bcd + cda + dab + ab + bc + cd + da + ac + bd + a + b + c + d = 2009
What is the value of a + b + c + d?
I have both seen and understood most of the solution. It looks something like:
Recognise that the LHS can be written as (a + 1)(b + 1)(c + 1)(d + 1) - $1$
Add 1 to both sides. Now it becomes (a + 1)(b + 1)(c + 1)(d + 1) = $2010$
Express $2010$ as a product of its primes. $2010$ = $2\times 3 \times 5 \times 67$.
It follows from there that a + b + c + d = $1 + 2 + 4 + 66 = 73$
I have a problem with the first step. I would just like to know if I'm expected to know right off the bat that the LHS can be simplified in that way, or is there a way to work it out and if there is, please can you give me hints on how to do so.
| $$abcd + abc + bcd + cda + dab + ab + bc + cd + da + ac + bd + a + b + c + d$$
You can gather together all the terms containing for example $a$, and take it out to get this:
$$a(bcd + bc + cd + bd + b + c + d + 1) + bcd + bc + cd + bd + b + c + d$$
Clearly all the later terms are almost the same as the terms inside the brackets, except for the $1$. This suggests it can be simplified further:
$$a(bcd + bc + cd + bd + b + c + d + 1) + bcd + bc + cd + bd + b + c + d + 1 - 1$$
$$a(bcd + bc + cd + bd + b + c + d + 1) + 1(bcd + bc + cd + bd + b + c + d + 1) - 1$$
$$(a+1)(bcd + bc + cd + bd + b + c + d + 1) - 1$$
You can then repeat this factoring using variable $b$ and then $c$:
$$(a+1)(bcd + bc + bd + b + cd + c + d + 1) - 1$$
$$(a+1)(b(cd + c + d + 1) + cd + c + d + 1) - 1$$
$$(a+1)(b(cd + c + d + 1) + 1(cd + c + d + 1)) - 1$$
$$(a+1)(b+1)(cd + c + d + 1) - 1$$
$$(a+1)(b+1)(c(d + 1) + d + 1) - 1$$
$$(a+1)(b+1)(c(d + 1) + 1(d + 1)) - 1$$
$$(a+1)(b+1)(c+1)(d + 1) - 1$$
| {
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Summation of a rational function Calculate $\sum_{r=2}^{n} \frac{3r^2-1}{(r^3-r)^2}$.
My approach till now: $T_r=\frac{3r^2 -1}{r^2(r+1)^2(r-1)^2}$. Now let, $$3r^2-1= a(r^2)+ b(r+1)^2 + c(r-1)^2$$
we find, $a=4$, $b=c=-\frac{1}{2}$. now we break it up, $$T_r = \frac{4}{(r+1)^2(r-1)^2}-\frac{1}{2(r)^2(r+1)^2}-\frac{1}{2(r)^2(r-1)^2}$$
but now the problem is I can't convert any of them into a "telescopable" format
another insight I had was to break it up in another way,
$$T_r= \frac{r^2 + (r^2-1)+(r^2-1)+1}{r^2(r^2-1)(r^2-1)}$$
after a bit of simplifying,
$$T_r= \frac{1}{(r^2-1)(r^2-1)} + \frac{2}{r^2(r^2-1)}+\frac{1}{r^2(r^2-1)(r^2-1)}$$
now two terms are solvable but how to do the third term maybe if you simplify it even more we get,
$$T_r = \frac{1}{(r^2-1)(r^2-1)} + \frac{2}{r^2(r^2-1)}+\frac{r^2-(r^2-1)}{r^2(r^2-1)(r^2-1)}$$
$$T_r=\frac{1}{(r^2-1)(r^2-1)} + \frac{2}{r^2(r^2-1)}+ \frac{1}{(r^2-1)^2}-\frac{1}{r^2(r^2-1)}$$
now rewriting, $(r^2-1)^2 = (r+1)^2(r-1)^2$, we get
$$T_r= \frac{2}{(r+1)^2(r-1)^2}+\frac{1}{r^2(r^2-1)}$$
but now how to solve the first term...
| You can do standard partial fraction decomposition. That is, let $$\frac{3r^2-1}{(r^3-r)^2} = \frac{a}{r}+\frac{b}{r^2}+\frac{c}{r-1}+\frac{d}{(r-1)^2}+\frac{e}{r+1}+\frac{f}{(r+1)^2}.$$ Then you will get $$-\frac{1}{r^2}+\frac{1}{2(r-1)^2}+\frac{1}{2(r+1)^2}.$$ Can you do the telescoping from here and conclude?
| {
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Can we make the inequality $A-B \leq \frac{A^2}{4}$ strict? I have an interesting problem:
Given that $$
A=\frac{1}{2022}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2022}\right)$$
$$B=\frac{1}{2023}\left(1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{2023}\right)
$$
I got an upper bound on $A-B$.The Lower bound is zero, that is $A-B>0$ is easy to prove.For the upper bound , here is my analysis:
$$\text{Let}\:H=\sum_{k=1}^{2022}\frac{1}{k}$$
So we have $$2022=\frac{H}{A} \implies 2023=\frac{H+A}{A}$$
So $$B=\frac{H+\frac{1}{2023}}{2023}=\frac{\left(H^2+A H+A\right) A}{(H+A)^2}$$
Finally we get a quadratic in $H$ as:
$$(A-B) H^2+\left(A^2-2 A B\right) H+A^2(1-B)=0$$ and since $H \in \mathbb{R}$ the discriminant should be non negative.
So we have
$$(A^2-2AB)^2-4(A-B)A^2(1-B)\geq 0$$
Simplifying we get
$$A-B \leq \frac{A^2}{4}$$
Now is it possible to make it strict?
EDIT: After deep thinking i came to the conclusion that inequality should be strict:
It is evident that $A<2, B<1$
Let us assume equality holds true, that is $$A-B=\frac{A^2}{4}$$
So we get
$$(A-2)^2=4(1-B) \implies A=2-2\sqrt{1-B}$$
Now i am pretty sure that $\sqrt{1-B}$ is irrational. So that makes $A$ irrational, which contradicts that $A$ is rational. Hence the inequality is strict.
| Your proof for the $A-B \le A^2/4$ is correct, but for the strict inequality, the irrationality of $\sqrt{1-B}$ is needed, which is not obvious (at least not to me).
Here is a different approach which gives the strict inequality without the need to argue about irrational numbers. It works for an arbitrary number of terms:
Let $$a_n = \frac 1 n \left( 1 + \frac 12 + \cdots + \frac 1 n\right) $$ for positive integers $n$. Then $$a_n - a_{n+1} \le \frac 14 a_n^2 \, .$$ The inequality is strict for all $n \ge 2$.
Proof: We have
$$
a_{n+1} = \frac{1}{n+1}\left( n a_n + \frac{1}{n+1}\right) \\
\implies a_n - a_{n+1} = \frac{a_n}{n+1} - \frac{1}{(n+1)^2}
$$
and therefore
$$
\frac 14 a_n^2 + a_{n+1}- a_n = \frac 14 a_n^2 - \frac{a_n}{n+1} + \frac{1}{(n+1)^2} = \frac{((n+1)a_n - 2)^2}{4(n+1)^2} \underset{(*)}{\ge} 0 \, .
$$
For $n \ge 2$ is
$$
\frac{1}{a_n} = \frac{n}{ 1 + \frac 12 + \cdots + \frac 1 n}
< \frac {1 + 2 + \ldots + n}n = \frac{n+1}{2}
$$
from the inequality between the harmonic and arithmetic mean, so that $(n+1)a_n > 2$, and the inequality $(*)$ is strict.
| {
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$\int\sqrt{1+t^2}\, dt$ using the substitution $t = \frac{1}{2}(e^u-e^{-u})$ This is what I have done so far:
$$\frac{dt}{du}=\frac{d}{du}\left(\frac{1}{2}(e^{u}-e^{-u})\right)=\frac{1}{2}(e^{u}+e^{-u}) \implies dt=\frac{1}{2}(e^{u}+e^{-u})\,du$$
$$L=\frac{1}{2}\int_{0}^{2}\sqrt{1+t^2}\, dt=\frac{1}{2}\int_{u(0)}^{u(2)}\frac{1}{2}(e^{u}+e^{-u})\sqrt{1+\left(\frac{1}{2}(e^{u}-e^{-u})\right)^2}\, du$$
The integral I get is quite complicated and ugly so I don't know were to go from here.
Solving for $dt$ seems like a step one should take, but what should I do with the $t^2$ under the square root? Should I try to solve for $u$ perhaps?
The integral should be solved using the substitution above. With other words: no integration by parts unless it is an intermediate step. Also no use of hyperbolic trig please.
| Starting with
$$L=\frac{1}{2}\int_{a}^{b}\sqrt{1+t^2}\, dt=\frac{1}{4} \, \int_{u(a)}^{u(b)} (e^{u}+e^{-u}) \, \sqrt{1+\left(\frac{1}{2}(e^{u}-e^{-u})\right)^2} \, du$$
and then using
$$ 1 + \frac{(e^{u} - e^{-u})^2}{4} = \frac{e^{2 u} + 2 + e^{- 2 u}}{4} = \left(\frac{e^{u} + e^{-u}}{2}\right)^2 $$
the integral $L$ becomes
\begin{align}
L &= \frac{1}{4} \, \int (e^{u} + e^{-u}) \, \sqrt{\left(\frac{e^{u} + e^{-u}}{2}\right)^2} \, du \\
&= \frac{1}{8} \, \int (e^{u} + e^{-u})^2 \, du \\
&= \frac{1}{8} \, \left[ 2 \, u + \frac{e^{2 u} - e^{-2 u}}{2} \right]
\end{align}
Now, the hard part. Find what $u$ is to invert the variables.
\begin{align}
2 t &= e^{u} - e^{-u} \\
e^{2 u} - 2 t \, e^{u} - 1 &= 0 \\
e^{u} &= t \pm \sqrt{t^2 + 1} \\
u &= \ln(t \pm \sqrt{t^2 + 1}).
\end{align}
The use of this can be seen in:
\begin{align}
e^{2 u} &= e^{2 \, \ln(t + \sqrt{t^2 + 1})} = (t + \sqrt{t^2 + 1})^2 \\
e^{-2 u} &= (t + \sqrt{t^2 + 1})^{-2} = (t - \sqrt{t^2 + 1})^2 \\
\frac{1}{2} \, \left( e^{2 u} - e^{-2 u} \right) &= t \, \sqrt{t^2 +1}
\end{align}
and gives $L$ as
$$ L = \frac{1}{4} \, \left( \ln(t + \sqrt{t^2 +1}) + t \, \sqrt{t^2 +1} \right). $$
Using the limits the value desired should be obtained.
| {
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Multiple solutions for trig function with period Find all angles with limits $−\pi \leq \theta \leq \pi$ which satisfy $\sin 4 \theta = 1$
The working out is given as:
If $\sin 4 \theta = 1$ then $4 \theta = \pi/2 + 2 k \pi$, so $\theta = \pi/8 + k \pi/2$
For $−\pi \leq \theta \leq \pi$ we have $\theta = \pi/8, 5 \pi/8, −3 \pi/8, −7\pi/8$
How did they find the other values of $5 \pi/8, −3 \pi/8$ and $−7 \pi/8$?
Was it just using the unit circle?
| When you reach at: $θ=\displaystyle \frac {\pi} 8 +\frac{k \pi} 2 ,k\in\mathbb{Z}\text{ } (1) $
We are given that: $θ\in[-\pi, \pi] \text{ } (2)$
So: $(1),(2)\implies -\pi<\displaystyle \frac {\pi} 8 +\frac{k \pi} 2<\pi\iff -\frac {\pi} 8-\pi<\frac{k \pi} 2< -\frac {\pi} 8+\pi\iff $
$$ \displaystyle -\frac {9\pi} 8<\frac{k \pi} 2<\frac {7\pi} 8 \iff -\frac {9\pi} 4<{k \pi} <\frac {7\pi} 4 \iff -\frac{9} 4<{k } <\frac {7} 4$$ But, $k\in\mathbb{Z}$ so we have to find all the $k$ that belong to: $\begin{bmatrix}\displaystyle-\frac{9} 4,\frac {7} 4 \end{bmatrix}$ such as: $k\in \mathbb{Z}. $
After you find the $k$ replace it to the equation $(1)$ and you find all the $θ$
So all the possible $k$ that meet our conditions are: $k=-2,-1,0,1$
So we conclude that, all the possible solutions are:
$$ \begin{Bmatrix}\displaystyle -\frac {7\pi} 8 ,-\frac {3\pi} 8,\frac {\pi} 8 ,\frac {5\pi} 8\end{Bmatrix}$$
| {
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Find all the real roots of $P(Q(x))=0$ Let $$P(x)=x^2+\frac{x}{2}+b$$ and $$Q(x)=x^2+cx+d$$ be two polynomials with real coefficients such that $\displaystyle P(x)\cdot Q(x)=Q(P(x))$ for all real $x$. Find all the real roots of $P(Q(x))=0$.
I found out that
\begin{align}
x^4+x^3c+x^2d+\frac{x^3}{2}+\frac{x^2c}{2}+\frac{xd}{2}+bx^2+bcx+bd&=\\x^4+x^3+2bx^2+\frac{x^2}{4}+bx+b^2+cx^2+\frac{cx}{2}+cb+d
\end{align}
After this I am stuck. I don't see any ideas now.
Any help is greatly appreicated.
| So the equation you've found gives
$$x^4+x^3(c+1/2)+x^2(d+c/2+b)+x(d/2+bc)+bd=x^4+x^3+(2b+1/4+c)x^2+(b+c/2)x+(b^2+cb+d)$$
Comparing coefficients, we get
*
*$c+1/2=1\implies c=1/2$
*$d+c/2+b=2b+1/4+c\implies d+1/4+b=2b+1/4+1/2\implies d=b+1/2$
*$d/2+bc=b+c/2\implies d/2+b/2=b+1/4\implies d=b+1/2$
*$bd=b^2+bc+d\implies b(b+1/2)=b^2+b/2+d\implies d=0\implies b=-1/2$
So $P(x)=x^2+x/2+b=\frac{1}{2}(2x^2+x-1)$ and $Q(x)=x^2+cx+d=x^2+x/2$.
Now $P$ has two roots, $x=-1, 1/2$.
So roots of $P(Q(x))=0$ are the solutions to
*
*$Q(x)=-1\implies x^2+x/2=-1\implies x^2+x/2+1=0$ this has no real roots.
*$Q(x)=1/2\implies x^2+x/2=1/2\implies 2x^2+x-1=0\implies x=-1, 1/2$.
| {
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Solve the equation $\frac{\sqrt{4+x}}{2+\sqrt{4+x}}=\frac{\sqrt{4-x}}{2-\sqrt{4-x}}$ Solve the equation $$\dfrac{\sqrt{4+x}}{2+\sqrt{4+x}}=\dfrac{\sqrt{4-x}}{2-\sqrt{4-x}}$$ The domain is $4+x\ge0,4-x\ge0,2-\sqrt{4-x}\ne0$.
Note that the LHS is always positive, so the roots must also satisfy: $A:2-\sqrt{4-x}>0$.
Firstly, I decided to raise both sides of the equation to the power of $2$. I came at $$\dfrac{4+x}{8+x+4\sqrt{4+x}}=\dfrac{4-x}{8-x-4\sqrt{4-x}}$$
Another thing I tried is to let $\sqrt{4+x}=u\ge0$ and $\sqrt{4-x}=v\ge0$. Then $$\begin{cases}\dfrac{u}{2+u}=\dfrac{v}{2-v}\\4+x=u^2\\4-x=v^2\end{cases}$$ Adding the second to the third equation, we get $8=u^2+v^2$.
And the last thing: I cross-multiplied $$2\sqrt{4+x}-\sqrt{16-x^2}=2\sqrt{4-x}+\sqrt{16-x^2}\\\sqrt{4+x}=\sqrt{4-x}+\sqrt{16-x^2}$$ Raising to the power of 2 gives $$4+x=4-x+16-x^2+2\sqrt{(4-x)(16-x^2)}\\2\sqrt{(4-x)(16-x^2)}=x^2+2x-16$$ Is there something easier?
| My solution:
Let us consider two positive reals $\alpha,\beta$ defined as$$\alpha=\sqrt{4+x} \quad \beta=\sqrt{4-x} \Rightarrow \alpha^2+\beta^2=8 \Rightarrow (\alpha-\beta)^2+2\alpha\beta=8$$
Now the equation becomes
$$\frac{2+\alpha}{\alpha}=\frac{2-\beta}{\beta} \Rightarrow \frac{\alpha}{\alpha}+\frac{\beta}{\beta}=2\left(\frac{1}{\beta}-\frac{1}{\alpha}\right)$$
So we get
$$\alpha-\beta=\alpha\beta \Rightarrow \alpha >\beta \Rightarrow x>0$$
Finally $$(\alpha\beta)^2+2\alpha\beta-8=0 \Rightarrow \alpha\beta=2\Rightarrow \sqrt{16-x^2}=2 \Rightarrow x=2\sqrt{3}$$
| {
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"answer_count": 3,
"answer_id": 2
} |
Prove that $\int_0^{2\pi} \frac{ab}{a^2\cos^2t+b^2\sin^2t}\mathrm dt=2\pi$. Probably equation $\int_0^{2\pi} \frac{ab}{a^2\cos^2t+b^2\sin^2t}\mathrm dt=\int_0^{2\pi} \frac{ab}{a^2\sin^2t+b^2\cos^2t}\mathrm dt$ is useful. Double integration is also a tool, but I don't know next step.
| I thought it might be instructive to present an approach that uses the identities, $\cos^2(t)=\frac{1+\cos(2t)}{2}$ and $\sin^2(t)=\frac{1-\cos(2t)}{2}$ to streamline the analysis. Proceeding, we can write
$$\begin{align}
\int_0^{2\pi}\frac{ab}{a^2\cos^2(t)+b^2\sin^2(t)}\,dt&=\int_0^{2\pi}\frac{2ab}{(a^2+b^2)+(a^2-b^2)\cos(2t)}\,dt\\\\
&\overbrace{=}^{t\mapsto t/2}\int_0^{4\pi} \frac{ab}{(a^2+b^2)+(a^2-b^2)\cos(t)}\,dt\\\\
&=4ab\int_0^{\pi}\frac{1}{(a^2+b^2)+(a^2-b^2)\cos(t)}\,dt\\\\
&\overbrace{=}^{\tan(t/2)\mapsto t}4\int_0^\infty \frac{ab}{a^2+b^2 t^2}\,dt\\\\
&=2\pi
\end{align}$$
and we are done!
| {
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"answer_id": 2
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A right triangle and a point inside that divides it into three equal areas $ABC$ is a right-angled triangle ($\measuredangle ACB=90^\circ$). Point $O$ is inside the triangle such that $S_{ABO}=S_{BOC}=S_{AOC}$. If $AO^2+BO^2=k^2,k>0$, find $CO$.
The most intuitive thing is to note that $AO^2+BO^2=k^2$ is part of the cosine rule for triangle $AOB$ and the side $AB:$ $$AB^2=c^2=AO^2+BO^2-2AO.BO\cos\measuredangle AOB\\ =k^2-2AO.BO\cos\measuredangle AOB$$ From here if we can tell what $2AO.BO\cos\measuredangle AOB$ is in terms of $k$, we have found the hypotenuse of the triangle (with the given parameter). I wasn't able to figure out how this can done.
Something else that came into my mind: does the equality of the areas mean that $O$ is the centroid of the triangle? If so, can some solve the problem without using that fact?
| As @DavidQuinn noted, point $O$ is the centroid of the right triangle.
If we place $C$ at $(0,0)$ and $B$ at $(a, 0)$ and $A$ at $(0,b)$, then
$O = \dfrac{1}{3} (a, b ) $
Therefore
$OA^2 + OB^2 + OC^2 = \dfrac{1}{9} \bigg( a^2 + b^2 + 4 a^2 + b^2 + a^2 + 4 b^2 \bigg) = \dfrac{1}{9} ( 6 ) (a^2 + b^2 ) = \dfrac{2}{3} (a^2 + b^2) $
It follows that
$OC^2 = \dfrac{2}{3}(a^2 + b^2) - k^2$
Now, as noted by @LiKwokKeung below, we have that
$OC = \dfrac{1}{3} c $
from which $ a^2 + b^2 = c^2 = 9 OC^2$, hence
$ OC^2 = \dfrac{2}{3} (9 OC^2) - k^2 $
And finally,
$ OC^2 = \dfrac{k^2}{5} $
i.e.
$ OC = \dfrac{k}{\sqrt{5}} $
| {
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"source": "stackexchange",
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"answer_id": 0
} |
Tangent plane at a point to surface $\mathbf{F}=\mathbf{0}$ Let $\mathbf{F}:\mathbb{R}^5\to\mathbb{R}^3, \mathbf{F}\begin{pmatrix}x_1\\x_2\\y_1\\y_2\\y_3\end{pmatrix}=\begin{bmatrix}2x_1+x_2+y_1+y_3-1\\x_1x_2^3+x_1y_1+x_2^2y_2^3-y_2y_3\\x_2y_1y_3+x_1y_1^2+y_2y_3^2\end{bmatrix}$ and $\mathbf{a}=\begin{bmatrix}0\\1\\-1\\1\\1\end{bmatrix}$.
The equation $\mathbf{F}=\mathbf{0}$ defines $\mathbf{y}=\begin{bmatrix}y_1\\y_2\\y_3\end{bmatrix}$ as a function of $\mathbf{x}=\begin{bmatrix}x_1\\x_2\end{bmatrix}$ near $\mathbf{a}$ since $\mathbf{F(a)}=\mathbf{0}$, $\mathbf{F}$ is $\mathcal{C}^1$ and $\frac{\partial\mathbf{F}}{\partial\mathbf{y}}(\mathbf{a})=\begin{bmatrix}1 & 0 & 1\\0 & 1 & -1\\ 1 & 1 &1 \end{bmatrix}$ is nonsingular. So we have that near $\mathbf{a}$ there is a $\mathcal{C}^1$ function $\mathbf{\phi}$ such that $\mathbf{y}=\mathbf{\phi(x)}$ and $D\mathbf{\phi}\begin{pmatrix}0\\ 1\end{pmatrix}=-\left(\frac{\partial\mathbf{F}}{\partial\mathbf{y}}(\mathbf{a})\right)^{-1} \left(\frac{\partial\mathbf{F}}{\partial\mathbf{x}}(\mathbf{a})\right)=\begin{bmatrix}-3 & -5\\1 & 2\\1 & 4\end{bmatrix}$.
My question: is it correct to say that the tangent plane at $\mathbf{a}$ of the surface $\mathbf{F}=\mathbf{0}$ is the graph of
$\mathbf{g}:\mathbb{R}^2\to\mathbb{R}^5$
$\mathbf{g}\begin{pmatrix}x_1\\x_2\end{pmatrix}=\begin{bmatrix}0\\1\\-1\\1\\1\end{bmatrix}+x_1\begin{bmatrix}1\\0\\\frac{\partial\phi}{\partial x_1}\begin{pmatrix}0\\1\end{pmatrix}\end{bmatrix}+x_2\begin{bmatrix}0\\1\\\frac{\partial\phi}{\partial x_2}\begin{pmatrix}0\\1\end{pmatrix}\end{bmatrix}=\begin{bmatrix}0\\1\\-1\\1\\1\end{bmatrix}+x_1\begin{bmatrix}1\\0\\-3\\1\\1\end{bmatrix}+x_2\begin{bmatrix}0\\1\\-5\\2\\4\end{bmatrix}?$
| The result is correct. Another way to check this is that the tangent plane at the point can be obtained by translating the subspace $\mathbf{N}([D\mathbf{F(a)}])$ so that it passes through $\mathbf{a}$.
A basis of $N(D\mathbf{F}(\mathbf{a}))$ is $\left\{
\begin{bmatrix}4\\-1\\-7\\2\\0\end{bmatrix}
, \begin{bmatrix}-2\\1\\1\\0\\2\end{bmatrix}
\right\}$ so the tangent plane at $\mathbf{a}$ can be written as the graph of: $\mathbf{f}:\mathbb{R}^2\to\mathbb{R}^5,\ \mathbf{f}(s,t)=\begin{bmatrix}0\\1\\-1\\1\\1\end{bmatrix}+s\begin{bmatrix}4\\-1\\-7\\2\\0\end{bmatrix}+t\begin{bmatrix}-2\\1\\1\\0\\2\end{bmatrix}$ which is equivalent to the graph of $\mathbf{g}:\mathbb{R}^2\to\mathbb{R}^5,\ \mathbf{g}(s,t)=\begin{bmatrix}0\\1\\-1\\1\\1\end{bmatrix}+s\begin{bmatrix}1 \\ 0 \\ -3 \\ 1 \\ 1\end{bmatrix}+t\begin{bmatrix}0 \\ 1 \\ -5 \\ 2 \\ 4\end{bmatrix}$ (which is my solution above) since $\left\{
\begin{bmatrix}1 \\ 0 \\ -3 \\ 1 \\ 1\end{bmatrix}
, \begin{bmatrix}0 \\ 1 \\ -5 \\ 2 \\ 4\end{bmatrix}
\right\}$ is also a basis for $\mathbf{N}([D\mathbf{F(a)}])$.
Note that the tangent plane is in $\mathbb{R}^5$ because it must be tangent to the the graph of $\mathbf{\phi}:\mathbb{R}^2\to\mathbb{R}^3$ which parametrizes the surface $\mathbf{F}=\mathbf{0}$ near $\mathbf{a}$ and which is in $\mathbb{R}^5$.
(Thanks to user Ted Shifrin for the help)
| {
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How to prove that $a_{n} := \frac{n^2}{(n+1)^2}+\frac{(n+1)^2}{(n+2)^2}$ converges to $2$ when $n\to\infty$ by the definition? Just wondering if this is the correct way to approach and $\epsilon$ n proof for convergence.
I am starting with the sequence
$\frac{n^2}{(n+1)^2}+\frac{(n+1)^2}{(n+2)^2}$
First step,
\begin{align*}
\left|\frac{n^2}{(n+1)^2}+\frac{(n+1)^2}{(n+2)^2} - 2 \right|
&= \left|\frac{(n+2)^{2}n^{2}+(n+1)^{4}-2(n+1)^{2}(n+2)^{2}}{(n+1)^{2}(n+2)^{2}} \right|< \epsilon\\
\end{align*}
since we have,
$(n+2)^{2}n^{2}+(n+1)^{4}-2(n+1)^{2}(n+2)^{2} < 2n^{4}(2n^{2}+4n+1)$
and,
$\frac{1}{(n+1)^{2}(n+2)^{2}}< \frac{1}{n^{4}}$
so,
$
\left|\frac{(n+2)^{2}n^{2}+(n+1)^{4}-2(n+1)^{2}(n+2)^{2}}{(n+1)^{2}(n+2)^{2}} \right|
< \left|\frac{2n^{4}(2n^{2}+4n+1)}{n^{4}}\right|
<2(2n^{2}+4n+1)\\[1em]
$
This is where I'm getting stuck: Solving we get,
$\frac{-\sqrt{\epsilon+2}-2}{2}<n<\frac{\sqrt{\epsilon +2}-2}{2}$
For any $\epsilon >0$. Does this make sense? Should I only be taking the positive value for epsilon?
| HINT
As a start, I would recommend you to split the argument into two parts. The first part corresponds to:
\begin{align*}
\frac{n^{2}}{(n + 1)^{2}} - 1 = -\frac{2n + 1}{(n + 1)^{2}} = \frac{1}{(n + 1)^{2}} - \frac{2}{n + 1}
\end{align*}
Similarly, the second part corresponds to:
\begin{align*}
\frac{(n + 1)^{2}}{(n + 2)^{2}} - 1 = -\frac{2n + 3}{(n + 2)^{2}} = \frac{1}{(n + 2)^{2}} - \frac{2}{n + 2}
\end{align*}
Letting $n\geq n_{\varepsilon}$, we arrive at the relation:
\begin{align*}
\left|\frac{n^{2}}{(n + 1)^{2}} + \frac{(n + 1)^{2}}{(n + 2)^{2}} - 2\right| & = \left|\frac{1}{(n + 1)^{2}} + \frac{1}{(n + 2)^{2}} - \frac{2}{n + 1} - \frac{2}{n + 2}\right|\\\\
& \leq \frac{1}{(n + 1)^{2}} + \frac{1}{(n + 2)^{2}} + \frac{2}{n + 1} + \frac{2}{n + 2}\\\\
& \leq \frac{1}{n + 1} + \frac{1}{n + 2} + \frac{2}{n + 1} + \frac{2}{n + 2}\\\\
& \leq \frac{1}{n} + \frac{1}{n} + \frac{2}{n} + \frac{2}{n}\\\\
& = \frac{6}{n} \leq \frac{6}{n_{\varepsilon}}
\end{align*}
Can you take it from here?
| {
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How to find the coefficient without a calculator? I wanted to solve this question without using a calculator.
Question: The number of non-negative integer solutions to
$$3x+y+z=24$$
By creating generating functions you have to find the coefficient of $x^{24}$ in the expression: $$\left(\frac{1}{1-x}\right)^{2}\left(\frac{1}{1-x^3}\right)$$
Using the theory I know about now, I would just split the problem into smaller parts adding all the combinations together while using the extended binomial theorem. But this takes a lot of time and I was wondering if there is a faster/easier way to find coefficients in terms of multiple generating functions by hand? If so, what are some recommended places to read about it?
| Pen, paper, and effort is what is required outside of a computer to calculate higher terms in expansions. In this case
\begin{align}
\frac{1}{(1-x)^2} &= \sum_{n=0}^{\infty} (n+1) \, x^n \\
\frac{1}{1-x^3} &= \sum_{n=0}^{\infty} x^{3 n}
\end{align}
for which the product of these series gives
$$ P = \frac{1}{(1-x)^2 (1-x^3)} = \sum_{r=0}^{\infty} (r+1) \, x^r \times \sum_{m=0}^{\infty} x^{3 m}. $$
Expanding to powers of $x^{24}$ gives
\begin{align}
P &= (1 + 2 x + 3 \, x^2 + \cdots + 25 \, x^{24})(1 + x^3 + \cdots + x^{24}) \\
&= \cdots + (1 + 4 + 7 + 10 + 13 + 16 + 19 + 22 + 25 ) \ x^{24} + \cdots \\
&= \cdots + 108 \, x^{23} + 117 \, x^{24} + 126 \, x^{25} + \cdots.
\end{align}
The resulting value is then $[x^{24}] P = 117$.
| {
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Seeking for other methods to evaluate $\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx$ for $n\geq 2$. Inspired by my post, I go further to investigate the general integral and find a formula for
$$
I_n=\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{x^n+1} dx =-\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left[\gamma+\psi\left(1-\frac{1}{n}\right)\right] \tag*{}
$$
Let’s start with its partner integral
$$
I(a)=\int_0^{\infty}\left(x^n+1\right)^a d x
$$
and transform $I(a)$, by putting $y=\frac{1}{x^n+1}$, into a beta function
$$
\begin{aligned}
I(a) &=\frac{1}{n} \int_0^1 y^{-a-\frac{1}{n}-1}(1-y)^{-\frac{1}{n}-1} d y \\
&=\frac{1}{n} B\left(-a-\frac{1}{n}, \frac{1}{n}\right)
\end{aligned}
$$
Differentiating $I(a)$ w.r.t. $a$ yields
$$
I^{\prime}(a)=\frac{1}{n} B\left(-a-\frac{1}{n}, \frac{1}{n}\right)\left(\psi(-a)-\psi\left(-a-\frac{1}{n}\right)\right)
$$
Then putting $a=-1$ gives our integral$$
\begin{aligned}
I_n&=I^{\prime}(-1) \\&=\frac{1}{n} B\left(1-\frac{1}{n}, \frac{1}{n}\right)\left[\psi(1)-\psi\left(1-\frac{1}{n}\right)\right] \\
&=-\frac{\pi}{n} \csc \left(\frac{\pi}{n}\right)\left[\gamma+\psi\left(1-\frac{1}{n}\right)\right]\end{aligned}
$$
For examples,
$$
\begin{aligned}& I_2=-\frac{\pi}{2} \csc \frac{\pi}{2}\left[\gamma+\psi\left(1-\frac{1}{2}\right)\right]=\pi \ln 2,\\ & I_3=-\frac{\pi}{3} \csc \left(\frac{\pi}{3}\right)\left[\gamma+\psi\left(\frac{2}{3}\right)\right]=\frac{\pi \ln 3}{\sqrt{3}}-\frac{\pi^2}{9} ,\\ &I_4=-\frac{\pi}{4} \csc \left(\frac{\pi}{4}\right)\left[\gamma+\psi\left(\frac{3}{4}\right)\right]=\frac{3 \pi}{2 \sqrt{2}}\ln 2-\frac{\pi^2}{4 \sqrt{2}},\\
& I_5=-\frac{\pi}{5} \csc \left(\frac{\pi}{5}\right)\left[\gamma+\psi\left(\frac{4}{5}\right)\right]=-\frac{2 \sqrt{2} \pi}{5 \sqrt{5-\sqrt{5}}}\left[\gamma+\psi\left(\frac{4}{5}\right)\right], \\
& I_6=-\frac{\pi}{6} \csc \left(\frac{\pi}{6}\right)\left[\gamma+\psi\left(\frac{5}{6}\right)\right]=\frac{2 \pi}{3} \ln 2+\frac{\pi}{2} \ln 3-\frac{\pi^2}{2 \sqrt{3}},
\end{aligned}
$$
Furthermore, putting $a=-m$, gives
$$\boxed{I(m,n)=\int_0^{\infty} \frac{\ln \left(x^n+1\right)}{(x^n+1)^m} dx = \frac{1}{n} B\left(m-\frac{1}{n}, \frac{1}{n}\right)\left[\psi(m)-\psi\left(m-\frac{1}{n}\right)\right] }$$
For example,
$$
\begin{aligned}
\int_0^{\infty} \frac{\ln \left(x^6+1\right)}{(x^6+1)^5} dx & =\frac{1}{6} B\left(\frac{29}{6}, \frac{1}{6}\right)\left[\psi(5)-\psi\left(\frac{29}{6}\right)\right] \\
& =\frac{1}{6} \cdot \frac{21505 \pi}{15552} \cdot\left(-\frac{71207}{258060}-\frac{\sqrt{3} \pi}{2}+\frac{3 \ln 3}{2}+2 \ln 2\right) \\
& =\frac{21505 \pi}{93312}\left(-\frac{71207}{258060}-\frac{\sqrt{3} \pi}{2}+\frac{3 \ln 3}{2}+2 \ln 2\right)
\end{aligned}
$$
Are there any other methods?
Your comments and alternative methods are highly appreciated.
| You can obtain the antiderivative. Let $x^n=t$ to face
$$I=\frac 1 n \int \frac{\log (t+1)}{t+1}\,t^{\frac{1}{n}-1}\,dt $$
$$I=n (t+1)^{\frac{1}{n}} \,
_3F_2\left(-\frac{1}{n},-\frac{1}{n},-\frac{1}{n};1-\frac{1}{n},1
-\frac{1}{n};\frac{1}{t+1}\right)+t^{\frac{1}{n}} \log (t+1)-$$
$$\frac{n t^{\frac{1}{n}+1} }{n+1}\,
_2F_1\left(1,1+\frac{1}{n};2+\frac{1}{n};-t\right)
-(t+1)^{\frac{1}{n}} \log (t+1) \,
_2F_1\left(-\frac{1}{n},-\frac{1}{n};\frac{n-1}{n};\frac{1}{t+1}\right)$$
| {
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"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
How far can we go with the integral $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$ Inspired by my post, I decided to investigate the integral in general
$$
I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x$$
by the powerful substitution $x=\frac{1-t}{1+t} .$
where $n$ is a natural number greater $1$.
Let’s start with easy one
\begin{aligned}
I_1 &=\int_0^1 \frac{\ln \left(\frac{2 t}{1+t}\right)}{1+t^2} d t \\ &=\int_0^1 \frac{\ln 2+\ln t-\ln (1+t)}{1+t^2} d t \\&=\frac{\pi}{4} \ln 2+\int_0^1 \frac{\ln t}{1+t^2}-\int_0^1 \frac{\ln (1+t)}{1+t^2} d t \\&=\frac{\pi}{4} \ln 2-G-\frac{\pi}{8} \ln 2 \\
&=\frac{\pi}{8} \ln 2-G\end{aligned}
By my post
$$I_2= \frac{\pi}{4} \ln 2-G $$
and
$$\begin{aligned}I_4 &=\frac{3 \pi}{4} \ln 2-2 G
\end{aligned}
$$
$$
\begin{aligned}
I_3=& \int_0^1 \frac{\ln (1-x)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1+x+x^2\right)}{1+x^2} d x \\=& \frac{\pi}{8} \ln 2-G+\frac{1}{2} \int_0^{\infty} \frac{\ln \left(1+x+x^2\right)}{1+x^2} d x-G
\\ =& \frac{\pi}{8} \ln 2-\frac{4G}{3} +\frac{\pi}{6} \ln (2+\sqrt{3})
\end{aligned}
$$
where the last integral refers to my post.
Let’s skip $I_5$ now.
$$
I_6=\int_0^1 \frac{\ln \left(1-x^6\right)}{1+x^2} d x=\int_0^1 \frac{\ln \left(1+x^3\right)}{1+x^2} d x+I_3\\
$$
$$\int_0^1 \frac{\ln \left(1+x^3\right)}{1+x^2} d x = \int_0^1 \frac{\ln (1+x)}{1+x^2} d x +\int_0^1 \frac{\ln \left(1-x+x^2\right)}{1+x^2} d x\\=\frac{\pi}{8}\ln 2+ \frac{1}{2} \int_0^{\infty} \frac{\ln \left(1-x+x^2\right)}{1+x^2} d x-G \\= \frac{\pi}{8}\ln 2+ \frac{1}{2}\left( \frac{2 \pi}{3} \ln (2+\sqrt{3})-\frac{4}{3} G \right)- G \\= \frac{\pi}{8} \ln 2+\frac{\pi}{3} \ln (2+\sqrt{3})-\frac{5}{3} G $$
Hence $$I_6= \frac{\pi}{4} \ln 2+\frac{\pi}{2} \ln (2+\sqrt{3})-3 G$$
How far can we go with the integral $I_n=\int_0^1 \frac{\ln \left(1-x^n\right)}{1+x^2} d x?$
| Too long for a comment.
It's equivalent to evaluate a certain digamma series, perhaps a more attainable goal:
$$\sum_{m=0}^\infty\frac{(-1)^m}{2m+1}\psi\left(\frac{2m+1}{n}\right)$$
$\newcommand{\d}{\,\mathrm{d}}$We have, for $s>0$: $$\begin{align}J_n&:=\frac{1}{n}\int_0^1\frac{(1-t)^{s-1}}{1+t^{2/n}}t^{1/n-1}\d t\\&=\sum_{m=0}^\infty\frac{(-1)^m}{n}\int_0^1(1-t)^{s-1}t^{\frac{2m+1}{n}-1}\d t\\&=\sum_{m=0}^\infty\frac{(-1)^m}{n}\frac{\Gamma(s)\Gamma\left(\frac{2m+1}{n}\right)}{\Gamma\left(\frac{2m+1}{n}+s\right)}\end{align}$$Differentiate this w.r.t $s$: $$\sum_{m=0}^\infty\frac{(-1)^m}{n}\frac{\Gamma(s)\Gamma\left(\frac{2m+1}{n}\right)}{\Gamma\left(\frac{2m+1}{n}+s\right)}\left(\psi(s)-\psi\left(\frac{2m+1}{n}+s\right)\right)$$Evaluate at $s=1$: $$\sum_{m=0}^\infty\frac{(-1)^{m-1}}{2m+1}\left(\frac{n}{2m+1}+\psi\left(\frac{2m+1}{n}\right)+\gamma\right)=-nG-\frac{\pi\gamma}{4}-\sum_{m=0}^\infty\frac{(-1)^m}{2m+1}\psi\left(\frac{2m+1}{n}\right)$$
| {
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How many ways to get a sum of 29 by adding 5 & 2? ex 5+5+5+5+5+2+2 = 29, is one way. Ex: $2+2+3$, $2+3+2$, $3+2+2$ these are three ways to get a sum of $7$ with $3$ and $2$. But my example is with $5$ and $2$ and a sum of $29$.
I believe there are three ways to get a sum of $29$ by using fives and nines. But is there any formula for this? What if I want to check the sum of $500$ or $1 000 0000$ using $5$ and $2$?
| Here's an observation: $(x^2+x^3)^k$ is a polynomial whose $x^n$ coefficient is the number of ways to get $n$ by using $k$ total $2$'s and $3$'s. So, if we don't care how many total we use, the $x^n$ coefficient of the geometric series
$$
\sum_{k=0}^\infty (x^2+x^3)^k = \frac{1}{1-x^2-x^3}
$$
is the number of ways of getting $n$ using any number of $2$'s and $3$'s.
The first terms of this are
$$
\frac{1}{1-x^2-x^3} = 1+x^2+x^3+x^4+2x^5+2x^6+3x^7+4x^8+5x^9+7x^{10}+\cdots.
$$
This supports what you found, that there are three ways to get $7$.
Getting a nice closed-form formula depends on factoring the denominator, if possible, to do partial fractions to represent this as a sum of geometric series. That's one way to calculate a particular coefficient. Another is to just go through with polynomial long division, which isn't so bad, but it will take a while to (by hand) figure out how many ways there are to get ten million!
For fives and nines, the corresponding rational function is
$$
\sum_{k=0}^\infty(x^5+x^9)^k = \frac{1}{1-x^5-x^9} = 1+x^5+x^9+x^{10}+2 x^{14}+x^{15}+x^{18}+3 x^{19}+x^{20}+3
x^{23}+4 x^{24}+x^{25}+x^{27}+6 x^{28}+5
x^{29} + \cdots.
$$
This says there are five ways of getting $29$, not three.
Another way to understand this, rather than using power series, is that to calculate the number of ways $f(n)$ of getting $n$ with fives and nines, you can calculate $f(n-5) + f(n-9)$ since you add together the number of ways where the last number chosen is a five and the number of ways where the last number chosen is a nine. Then you make a table starting with $n=0$ with $f(0)=1$ (since there is one way to get $0$). By calculating the values of this function at consecutive numbers, you get to reuse previous calculations. (This is equivalent to the long division approach.)
| {
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On The Question of A Squeeze-Derivative Definition Question: Find $g'(0)$ where $$g(x)=\left(x+1\right)^2 \left(1+\cos{\left(\sqrt{|\tan{(x+1)}|}\right)}\right)+\sqrt{x^4+4x+1}\left(1-\cos{\left(\sqrt{|\tan{(x+1)}|}\right)}\right).$$
The above question was asked by my professor as an exam question. When I was in the exam, I remember doing the following operations; I wanted to use some kind of linear approximation of $L(x)$ so around $a=0$; $\cos{x}\approx 1$ but somehow this operation is valid when the interior of the $\cos$ function approaches zero. I think I could use this: $$-1 \leq \cos{\left(\sqrt{|\tan{(x+1)}|}\right)} \leq 1,$$ $$0\leq 1-\cos{\left(\sqrt{|\tan{(x+1)}|}\right)}\leq 2,$$ $$0\leq \sqrt{x^4+4x+1}\left(1-\cos{\left(\sqrt{|\tan{(x+1)}|}\right)}\right) \leq 2\sqrt{x^4+4x+1},$$ and note that: $$0 \leq (x+1)^2\left(1+\cos{\left(\sqrt{|\tan{(x+1)}|}\right)}\right)\leq 2(x+1)^2;$$ so,
$$0 \leq g(x) \leq 2(x+1)^2+2\sqrt{x^4+4x+1}.$$
Now, I feel that I should use some kind of derivative definition here, as given in the hint in the exam, but I can't reach the conclusion.
| Let $\;g_1(x)=x^2+2x+1+\sqrt{x^4+4x+1}\;.$
Let $\;g_2(x)=\left[x^2\!+\!2x\!+\!1\!-\!\sqrt{x^4\!+\!4x\!+\!1}\right]\cos\left(\!\sqrt{|\tan(x+1)|}\right).$
It results that
$\dfrac{g(x)-g(0)}x=\dfrac{g_1(x)-g(0)}x+\dfrac{g_2(x)}x\;.$
$\dfrac{g_1(x)-g(0)}x=\dfrac{x^2+2x+1+\sqrt{x^4+4x+1}-2}x=$
$=\dfrac{x^2+2x}x+\dfrac{\sqrt{x^4+4x+1}-1}x=$
$=x+2+\dfrac{x^4+4x+1-1}{x\left(\sqrt{x^4+4x+1}+1\right)}=$
$=x+2+\dfrac{x^3+4}{\sqrt{x^4+4x+1}+1}\to2+\dfrac42=4\;\;$ as $\;\;x\to0\;.$
Moreover ,
$\dfrac{g_2(x)}x=\dfrac{\left[x^2\!+\!2x\!+\!1\!-\!\sqrt{x^4\!+\!4x\!+\!1}\right]\cos\left(\!\sqrt{|\tan(x+1)|}\right)}x=$
$=\dfrac{\left[\left(x^2+2x+1\right)^2-x^4-4x-1\right]\cos\left(\sqrt{|\tan(x+1)|}\right)}{x\left(x^2+2x+1+\sqrt{x^4+4x+1}\right)}=$
$=\dfrac{\left(4x^3+6x^2\right)\cos\left(\sqrt{|\tan(x+1)|}\right)}{x\left(x^2+2x+1+\sqrt{x^4+4x+1}\right)}=$
$=\dfrac{\left(4x^2+6x\right)\cos\left(\sqrt{|\tan(x+1)|}\right)}{x^2+2x+1+\sqrt{x^4+4x+1}}\;.$
Hence ,
$0\leqslant\left|\dfrac{g_2(x)}x\right|\leqslant\dfrac{\left|4x^2+6x\right|}{x^2\!+\!2x\!+\!1+\!\sqrt{x^4\!+\!4x\!+\!1}}\to0\;\;$ as $\;\;x\to0\,.$
Consequently ,
$\dfrac{g_2(x)}x\to0\;\;$ as $\;\;x\to0\;.$
Therefore ,
$\begin{align}g’(0)&=\lim_\limits{x\to0}\dfrac{g(x)-g(0)}x=\lim_\limits{x\to0}\left(\dfrac{g_1(x)-g(0)}x+\dfrac{g_2(x)}x\right)=\\
&=4+0=4\;.\end{align}$
| {
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Proof that $\frac{(1+\sqrt 5)^n+(1-\sqrt 5)^n}{2^n}$ is an integer I want to prove, by induction, that
$$\frac{(1+\sqrt 5)^n+(1-\sqrt 5)^n}{2^n}\quad\text{is an integer}.$$
Instinctively, this sort of thing looks like the solution of a second order recurrence relation (like the explicit formula for the Fibonacci sequence), but I want to prove it directly, without invoking that I know that.
But induction doesn't seem to be working, since
\begin{gather*}
\frac{(1+\sqrt 5)^n+(1-\sqrt 5)^n}{2^n}\\[8pt] = \frac 12\bigg(\frac{(1+\sqrt 5)^{n-1}+(1-\sqrt 5)^{n-1}}{2^{n-1}} + \frac{\sqrt 5}2\cdot \frac{(1+\sqrt 5)^{n-2}-(1-\sqrt 5)^{n-2}}{2^{n-2}} + \frac q2\cdot\frac{(1+\sqrt 5)^{n-2}+(1-\sqrt 5)^{n-2}}{2^{n-2}}\bigg),
\end{gather*}
so I can deal with the first and last terms using the induction hypothesis, but not the middle one.
I suppose I can "strengthen" the induction hypothesis by proving simultaneously that
$$\text{Both}\quad\frac{(1+\sqrt 5)^n+(1-\sqrt 5)^n}{2^n}\quad\text{and}\quad \sqrt 5\frac{(1+\sqrt 5)^n-(1-\sqrt 5)^n}{2^n}\quad \text{are integers},$$
but again it feels like a direct approach should be possible.
| You have\begin{multline}\frac{\left(1+\sqrt5\right)^{n+1}+\left(1-\sqrt5\right)^{n+1}}{2^{n+1}}-\frac{\left(1+\sqrt5\right)^n+\left(1-\sqrt5\right)^n}{2^n}=\\=\frac{\left(-1+\sqrt5\right)\left(1+\sqrt5\right)^n+\left(-1-\sqrt5\right)\left(1-\sqrt5\right)^n}{2^{n+1}}=\\=\frac{\left(1+\sqrt5\right)^{n-1}+\left(1-\sqrt5\right)^{n-1}}{2^{n-1}},\end{multline}since $\left(-1+\sqrt5\right)\left(1+\sqrt5\right)=\left(-1-\sqrt5\right)\left(1-\sqrt5\right)=4=2^2$. Therefore, if$$a_n=\frac{\left(1+\sqrt5\right)^n+\left(1-\sqrt5\right)^n}{2^n},$$you have $a_{n+1}=a_n+a_{n-1}$. So, since $a_0=2$ and $a_1=1$, each $a_n$ is a non-negative integer.
| {
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Find the largest real root of the equation $2x^2+6x+9=7x\sqrt{2x+3}$
Find the largest real root of the equation $2x^2+6x+9=7x\sqrt{2x+3}$
Source: https://www.hkage.edu.hk/uploads/file/202207/6cda89c718b674f6ac3aa2c19049abe5.pdf
I tried substituting $y = 2x+3$ and ended up with $\frac{y^2}{2}-\frac{7}{2}y^{\frac 3 2} + \frac {21}{2} y^{\frac 1 2} + \frac 9 2 = 0$ but I'm still not sure how to find the largest $y$ from this.
I also tried calculating the derivative, ending up with $(4x+6)\sqrt{2x+3}-21x-21=0$. However, solving for $x$ is not trivial after simplifying the equation.
Also, I believe that calculus is not needed in this contest so I am looking for an elementary solution to this problem. Thanks in advance.
| You were correct in recognizing that $6x+9 = 3(2x+3)$ but we don't have to put everything in terms of $2x+3$
If we let $y=2x+3$ we get $2x^2 -7x\sqrt y +3y$ and we might recognize this as a quadratic. If we replace $w =\sqrt y = \sqrt{2x+3}$ we get
$2x^2 - 7xw + 3w^2=0$ and maybe we can factor $2x^2 - 7xw + 3w^2=0$. Can we?
And... sure we can $2x^2 -7xw+3w^2=(2x -w)(x-3w) = 0$.
So either $2x = w=\sqrt{2x+3}$ or $x=3w=3\sqrt{2x+3}$
These give us $x \ge 0; 4x^2 -2x-3=0$ or $x\ge 0; x^2 -18x -27$ which have at most $4$ real roots we can check.
$4x^2 -2x -3=0;x\ge 0\implies x=\frac {2+\sqrt{4+48}}8=\frac 14 + \frac {\sqrt{13}}4$. And $x^2-18x-27=0;x\ge 0\implies x =\frac {18+\sqrt{18^2+108}}{2}=9 + \sqrt{81+27}=9+3\sqrt{9+3}=9+6\sqrt 3$
So barring the inevitable arithmetic errors that I always make, the answer is $9+6\sqrt 3$.
| {
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In $\triangle ABC$, if the angles are in an A.P and $b:c=\sqrt{3}:\sqrt{2}$, compute $\angle A$. As title suggests, we have a triangle $\triangle ABC$ with angles in an arithmetic progression, and sides $b$ and $c$ in a ratio of $\sqrt{3}:\sqrt{2}$. This is a problem I saw from a mathematics textbook in India. While I don't exactly remember the name I do know it was for grade 11-12.
I'm posting this here to see what the easiest possible way could be to solve it. I'm going to share my own approach here, please share your own as well as pointing out any flaws in mine if you spot any:
Here's my method:
So, since we know that $\angle A, \angle B$ and $\angle C$ are in arithmetic progression, we can say that the arithmetic mean between $A$ and $C$ is $B$:
$$B=\frac{A+C}{2}$$
$$2B=A+C$$
$$3B=A+B+C$$
$$3B=180^\circ$$
$$\angle B=60^\circ$$
Now we know that:
$$\frac{b}{\sin{B}}=\frac{c}{\sin{C}}$$
$$\frac{c}{b}=\frac{\sin{C}}{\sin{B}}$$
$$\sin{C}=\sin{B}\left(\frac{\sqrt{2}}{\sqrt{3}}\right)$$
$$\sin{C}=\bigg(\frac{\sqrt{3}}{2}\bigg)\bigg(\frac{\sqrt{2}}{\sqrt{3}}\bigg)$$
$$\sin{C}=\frac{1}{\sqrt{2}}$$
Therefore, $\angle C=45^\circ$ and thus, $\angle A=180^\circ-45^\circ-60^\circ=75^\circ.$
| Since angles are in an arithmetic progression, $(\angle B - \alpha) + \angle B + (\angle B +\alpha) =180^\circ$ then $\angle B= 60^\circ$
Let's say $b=\sqrt3, c=\sqrt2$.
Draw $AH \perp BC$ then $\angle BAH=30^\circ$and $AH=\frac{\sqrt6}{2}$.
Then in $\triangle AHC$, $HC=\frac{\sqrt6}{2}$ (Pytagorean theorem) and $AH=HC=\frac{\sqrt6}{2}$.
$\angle C = \angle CAH = 45^\circ$.
$\angle A = 30^\circ+ 45^\circ=75^\circ$.
| {
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How might I have anticipated that $\frac14(\sqrt{5+2\sqrt5}+\sqrt{10+2\sqrt{5}})$ simplifies to a single surd (namely, $\frac14\sqrt{25+10\sqrt{5}}$)? This is perhaps a silly question related to calculating with surds. I was working out the area of a regular pentagon ABCDE of side length 1 today and I ended up with the following expression :
$$\frac{\sqrt{5+2\sqrt5}+\sqrt{10+2\sqrt{5}}}{4}$$
obtained by summing the areas of the triangles ABC, ACD and ADE.
I checked my solution with Wolfram Alpha which gave me the following equivalent expression :
$$\frac{\sqrt{25+10\sqrt{5}}}{4}$$
I was able to show that these two expressions are equivalent by squaring the numerator in my expression, which gave me
$$15+4\sqrt5+2\sqrt{70+30\sqrt5},$$
and then "noticing" that
$$\sqrt{70+30\sqrt5}=\sqrt{25+30\sqrt5+45}=5+3\sqrt5.$$
My question is the following : how could I have known beforehand that my sum of surds could be expressed as a single surd, and is there a way to systematize this type of calculation ? I would have liked to find the final, simplest expression on my own without the help of a computer.
Thanks in advance !
| Yes. There is a way to formalize this particular type of sum of square roots, similar to the way a determinant is developed for quadratic equations.
We can write the generic form of the expression in the first place as follows.
$\sqrt{a+b\sqrt{s}}+\sqrt{c+d\sqrt{s}}=\sqrt{x+y\sqrt{s}}$
Note that
*
*$a, b, c, d, s$ are given and rational. We aim to express $x,y$ in terms of them.
*The value in the double surds are identical, which are all $s$.
*Only square roots are considered.
Then,
$(\sqrt{a+b\sqrt{s}}+\sqrt{c+d\sqrt{s}})^2=(\sqrt{x+y\sqrt{s}})^2$
$(a+c)+(b+d)\sqrt{s}+2\sqrt{(a+b\sqrt{s})(c+d\sqrt{s})}=x+y\sqrt{s}$
$2\sqrt{(a+c+bds)+(ad+bc)\sqrt{s}}=(x-a-c)+(y-b-d)\sqrt{s}$
$4(a+c+bds)+4(ad+bc)\sqrt{s}=(x-a-c)^2+(y-b-d)^2s+2(x-a-c)(y-b-d)\sqrt{s}$
Let $p=x-a-c,\ q=y-b-d$.
By comparing the coefficients in the rational and irrational terms,
$\begin{cases}
p^2+q^2s=4(a+c+bds)\\
pq=2(ad+bc)
\end{cases}$
By substituting $q=\frac{2(ad+bc)}{p}$ and $p=\frac{2(ad+bc)}{q}$ from the second equation to the first equation, we can get a formula for each of $p$ and $q$.
$\begin{cases}
p^4-4(ac+bds)p^2+4(ad+bc)^2s=0\\
sq^4-4(ac+bds)q^2+4(ad+bc)^2=0
\end{cases}$
$\begin{cases}
p^2=2[ac+bds\pm\sqrt{(ac+bds)^2-(ad+bc)^2s}]\\
q^2=\frac{2}{s}[ac+bds\pm\sqrt{(ac+bds)^2-(ad+bc)^2s}]
\end{cases}$
$\begin{cases}
p=\pm\sqrt{2[ac+bds\pm\sqrt{(ac+bds)^2-(ad+bc)^2s}]}\\
q=\pm\sqrt{\frac{2}{s}[ac+bds\pm\sqrt{(ac+bds)^2-(ad+bc)^2s}]}
\end{cases}$
$\begin{cases}
x=\pm\sqrt{2[ac+bds\pm\sqrt{(ac+bds)^2-(ad+bc)^2s}]}+(a+c)\\
y=\pm\sqrt{\frac{2}{s}[ac+bds\pm\sqrt{(ac+bds)^2-(ad+bc)^2s}]}+(b+d)
\end{cases}$
As a result, there are three determinants to look at:
*
*$D_1=(ac+bds)^2-(ad+bc)^2s$ has a rational square root.
*$D_2=2(ac+bds\pm\sqrt{D_1})$ has a rational square root.
*$D_3=\dfrac{D_2}{s}$ has a rational square root.
We can use the two surds that you dealt with as an example.
$a=5, b=2, c=10, d=2, s=5$
$D_1=(ac+bds)^2-(ad+bc)^2s=400$, which is a perfect square.
$D_2=2(ac+bds\pm\sqrt{D_1})=100\text{ or }180$. 100 is a perfect square.
$D_3=\dfrac{D_2}{s}=36\text{ or }20$. 36 is a perfect square.
Therefore, the two surds can be summed to a single surd.
As seen above, the determinants are much more complex than the quadratic formula. I wonder if anyone would memorize them like $(b^2-4ac)$.
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School Outing Combinatorial Design Problem So this is an actual organization problem I am dealing with right now as a high school teacher. There is a school outing, with $8$ groups of students. At the venue there are $7$ stations, where $2$ groups can compete against each other. Is there a way to organize the groupings over $7$ time intervals so that each group goes to each station exactly once, and each group plays each other group exactly once? Obviously over each interval there will be exactly $3$ stations that are unused.
One can generalize this problem to $2n$ groups, $2n-1$ stations, and $2n-1$ time intervals where $2$ groups play against each other, such that no group plays the same station twice, no group plays the same group twice, and no group gets left out in an interval. The original problem is the $n=4$ version of this.
The $n=1$ case is trivial (there are only two groups, and they play against each other once on the one station). The $n=2$ case can be fairly quickly seen to be impossible, as after the first interval, at least one of the stations must be used a second time, and there is no way to have a different pairing at that station that also doesn't have the same two groups compete.
Has this problem been worked on before? Is there any literature on it?
Edit: I have worked out a solution to the $n=4$ case:
Group: A B C D E F G
1: 12 34 56 78 XX XX XX
2: 57 68 XX XX 13 24 XX
3: XX XX 14 23 58 67 XX
4: 36 XX 28 XX 47 XX 15
5: XX 17 XX 45 26 XX 38
6: XX 25 37 XX XX 18 46
7: 48 XX XX 16 XX 35 27
I'm still keen for a more general solution.
Edit 2: I am fairly certain the n=3 case is also impossible. There are enough values which can be filled in without loss of generality, after which sudokuing your way around the grid yields a contradiction. At this point I'm fairly sure that $n=4$ is the cutoff, after which there is a solution for all higher n, but that would be good to verify.
| In terms of graph theory, you have a complete graph $K_{2n}$ on $2n$ nodes (one node per group) and want to decompose it into $2n-1$ perfect matchings (one matching per time interval). See Perfect 1-factorization of $K_{2n}$
But you also want its "transpose" to form such a decomposition. For $n=4$, here is a solution obtained via integer linear programming, as in Real World Problem - Groups Visiting Various Stations
A B C D E F G
1 {} {5,7} {3,4} {} {2,6} {} {1,8}
2 {} {} {1,7} {2,3} {4,8} {5,6} {}
3 {2,4} {3,8} {} {6,7} {1,5} {} {}
4 {1,6} {} {} {} {3,7} {2,8} {4,5}
5 {} {1,2} {} {5,8} {} {4,7} {3,6}
6 {3,5} {} {6,8} {1,4} {} {} {2,7}
7 {7,8} {4,6} {2,5} {} {} {1,3} {}
And here's a solution for $n=5$:
A B C D E F G H I
1 {} {} {5,8} {} {} {6,7} {1,4} {3,9} {2,10}
2 {1,2} {} {3,10} {} {5,7} {} {} {4,6} {8,9}
3 {} {4,5} {} {1,8} {6,10} {2,9} {3,7} {} {}
4 {5,9} {2,6} {} {} {} {3,8} {} {1,10} {4,7}
5 {} {} {2,4} {3,6} {} {1,5} {9,10} {7,8} {}
6 {3,4} {7,9} {} {5,10} {} {} {2,8} {} {1,6}
7 {} {8,10} {1,7} {4,9} {2,3} {} {5,6} {} {}
8 {6,8} {} {} {2,7} {1,9} {4,10} {} {} {3,5}
9 {7,10} {1,3} {6,9} {} {4,8} {} {} {2,5} {}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4591663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Finding the laurent expansion of $\frac{z}{(z-1)(z-2)}$ I want to find the Laurent series for $\frac{z}{(z-1)(z-2)}$ in the region $1 < |z| < 2$. This implies that $\frac{1}{|z|} < 1$, so noticing that $(z-1) = z(1 - \frac 1 z)$ I can rewrite the desired function as
$$\frac{z}{z(1- \frac 1 z)(z-2)} = \frac{1}{z-2} \cdot \frac{1}{1 - \frac 1 z}.$$
Now using the definition of the geometric series I rewrite it as
$$\sum_{k=0}^{\infty} \frac{1}{z^k(z-2)}$$
Is this the right Laurent series, and if not, where did I go wrong? Please note I am trying to understand where I made a mistake, not simply finding any solution.
I've read a similar question at Finding the Laurent series of $f(z)=1/((z-1)(z-2))$ and one answer uses the fact that $\frac{|z|}2 < 1$, but I am not sure if my method is also valid.
| For $1 < |z| < 2$,
\begin{align*}
\frac{1}{(z-1)(z-2)} &=-\frac{1}{z-1} + \frac{1}{z-2}\\
&=-\frac{1}{z}\left[\frac{1}{1-(1/z)}\right]-\frac{1}{2}\left[\frac{1}{1-(z/2)}\right]\\
&=-\frac{1}{z}\sum_{n=0}^{\infty} \frac{1}{z^n}-\frac{1}{2}\sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^n\\
\end{align*}
\begin{align*}
\frac{z}{(z-1)(z-2)}
&=-\sum_{n=0}^{\infty} \frac{1}{z^n}-\sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^{n+1}\\
&=-1-\sum_{n=1}^{\infty} \frac{1}{z^n}-\sum_{n=1}^{\infty} \left(\frac{z}{2}\right)^{n}\\
&=-\sum_{n=1}^{\infty} \frac{1}{z^n}-\sum_{n=0}^{\infty} \left(\frac{z}{2}\right)^{n}\\
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4596764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Finding $x_1x_2+x_1x_3+x_2x_4+x_3x_4$ without explicitly finding the roots of $x^4-2x^3-3x^2+4x-1=0$
The equation $x^4-2x^3-3x^2+4x-1=0$ has $4$ distinct real roots $x_1,x_2,x_3,x_4$ such that $x_1\lt x_2\lt x_3\lt x_4$ and product of $2$ roots is unity, then find the value of $x_1x_2+x_1x_3+x_2x_4+x_3x_4$
This question has an answer on this link but I am trying to solve it without explicitly finding the roots because the question tells us that the product of $2$ roots is unity. I want to use it.
My Approach:
Using Descartes rule, I can see that there is one negative root and three positive roots.
Also, at $x=0, 1, -1$, the value of the polynomial is negative.
Thus, $x_1\lt-1, x_4\gt1$ and $x_2,x_3$ lies between $0$ to $1$.
Thus, I am concluding that $x_2x_4=1$ and $x_1x_3=-1$ (because product of roots is $-1$)
How to conclusively reject the case $x_3x_4=1$?
For $\alpha\gt1, \beta\gt1$, $x_1=-\beta, x_2=\frac1\alpha, x_3=\frac1\beta, x_4=\alpha$
Sum of roots$=-\beta+\frac1\alpha+\frac1\beta+\alpha=2\implies\frac1\beta-\beta=2-(\alpha+\frac1\alpha)$
Sum of product of roots taken $3$ at a time$=-\frac1\alpha+\frac1\beta-\alpha-\beta=-4\implies\frac1\beta-\beta=-4+(\alpha+\frac1\alpha)$
Therefore, $\alpha+\frac1\alpha=3, \frac1\beta-\beta=-1$
Multiplying these two, $\frac\alpha\beta-\alpha\beta+\frac1{\alpha\beta}-\frac\beta\alpha=-3$
The question asks us to find $\frac\alpha\beta-\frac\beta\alpha$, that means $-3+\alpha\beta-\frac1{\alpha\beta}$
Can we conclude this approach?
| I offer to substitute $-3x^2$ as $\left(-4x^2+x^2\right)$, therefore, we will have
$$
x^4-2x^3-3x^2+4x-1=0,
\\
\left(x^4-2x^3+x^2\right)-\left(4x^2-4x+1\right)=0,
\\
x^2\left(x^2-2x+1\right)-\left(2x-1\right)^{2}=0,
\\
x^{2}\left(x-1\right)^{2}-\left(2x-1\right)^{2}=0,
\\
\left(x\left(x-1\right)\right)^{2}-\left(2x-1\right)^{2}=0,
$$
$$
\left(x\left(x-1\right)\right)^{2}-\left(2x-1\right)^{2}=0,\tag{1}
$$
Relatively $(1)$ we must use the very well-known formula
$$
\left(a^2-b^2\right)=\left(a-b\right)\left(a+b\right),\tag{2}
$$
We will receive the next
$$
\begin{cases}
x^2-3x+1=0,\\
x^2+x-1=0.\tag{3}
\end{cases}
$$
Solving $(3)$ you will obtain those $4$ roots and find $x_1x_2+x_1x_3+x_2x_4+x_3x_4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Sum binomial coefficients $$\sum_{k=0}^{n} {\frac{k^2+k}{3^{k+2}} {n \choose k}}=?$$
What I've tried:
$$(k^2+k){n \choose k}=k(k+1){\frac{n!}{k!(n-k)!}}$$
$$k^2+k = k^2-k+2k=k(k-1)+2k$$ =>
$$
\begin{align}
(k^2+k){n \choose k} &= k(k-1){\frac{n!}{k!(n-k)!}}+2k{\frac{n!}{k!(n-k)!}}\\&={\frac{n!}{(k-2)!(n-k)!}}+2{\frac{n!}{(k-1)!(n-k)!}}\\&=n(n-1){n-2 \choose k-2}+2n{n-1 \choose k-1}
\end{align}
$$
and
$${\frac{1}{3^{k+2}}}={\frac{1}{9}}({\frac{1}{3}})^k$$
So I have
$$\sum_{k=0}^{n} {\frac{n(n-1){n-2 \choose k-2}+2n{n-1 \choose k-1}}{9*3^k}}$$
And I'm stuck... Can anyone help me?
| Starting where you stopped:
$$\begin{align}\sum_{k=0}^n(k^2+k)\binom nkx^k&=n(n-1)x^2\sum_{k=2}^n\binom{n-2}{k-2}x^{k-2}+2nx\sum_{k=1}^n\binom{n-1}{k-1}x^{k-1}\\
&=n(n-1)x^2(1+x)^{n-2}+2nx(1+x)^{n-1}\\
&=nx(1+x)^{n-2}((n-1)x+2(1+x))\\
&=nx(1+x)^{n-2}((n+1)x+2)\end{align}$$
hence
$$\begin{align}\sum_{k=0}^n\frac{k^2+k}{3^{k+2}}\binom nk
&=\frac{\frac n3(4/3)^{n-2}((n+1)/3+2)}9\\
&=\frac{4^{n-2}n(n+7)}{3^{n+1}}.\end{align}$$
However, as hinted by Ragib Zaman, there is a standard more direct way to obtain the generic formula above. Here it goes:
$$\begin{align}\sum_{k=0}^n\binom nkx^k=(1+x)^n&\implies\sum_{k=0}^n\binom nkkx^k=((1+x)^n)'x=nx(1+x)^{n-1}\\
&\implies\sum_{k=0}^n\binom nkk^2x^k=(nx(1+x)^{n-1})'x=nx(1+x)^{n-2}(nx+1),
\end{align}$$
and we recover the same result as above:
$$\begin{align}\sum_{k=0}^n\binom nk(k^2+k)x^k&=nx(1+x)^{n-2}(1+x+nx+1)\\&=nx(1+x)^{n-2}((n+1)x+2).\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4599408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find all three complex solutions of the equation $z^3=-10+5i$ Let $z\in \mathbb{C}$. I want to calculate the three solutions of the equation $z^3=-10+5i$. Give the result in cartesian and in exponential representation.
Let $z=x+yi $.
Then we have $$z^2=(x+yi)^2 =x^2+2xyi-y^2=(x^2-y^2)+2xyi$$
And then $$z^3=z^2\cdot z=[(x^2-y^2)+2xyi]\cdot [x+yi ] =(x^3-xy^2)+2x^2yi+(x^2y-y^3)i-2xy^2=(x^3-3xy^2)+(3x^2y-y^3)i$$
So we get
$$z^3=-10+5i \Rightarrow (x^3-3xy^2)+(3x^2y-y^3)i=-10+5i \\ \begin{cases}x^3-3xy^2=-10 \\ 3x^2y-y^3=5\end{cases} \Rightarrow \begin{cases}x(x^2-3y^2)=-10 \\ y(3x^2-y^2)=5\end{cases}$$
Is everything correct so far? How can we calculate $x$ and $y$ ? Or should we do that in an other way?
| It is correct so far. Note that\begin{align}-10+5i&=5\sqrt5\left(-\frac2{\sqrt5}+\frac i{\sqrt 5}\right)\\&=5\sqrt5\exp\left(\left(\pi-\arctan\left(\frac12\right)\right)i\right).\end{align}Therefore, if $\alpha=\pi-\arctan\left(\frac12\right)$, then $z^3=-10+5i$ if and only if $z$ is one of the numbers$$\sqrt5\exp\left(\frac\alpha3i\right),\ \sqrt5\exp\left(\frac{\alpha+2\pi}3i\right)\text{ or }\sqrt5\exp\left(\frac{\alpha+4\pi}3i\right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
If $20 x^2+13 x-15$ can be written as $(a x+b)(c x+d)$, where $a, b, c$ and $d$ are integers, what is $a+b+c+d$?
If $20 x^2+13 x-15$ can be written as $(a x+b)(c x+d)$, where $a, b, c$ and $d$ are integers, what is $a+b+c+d$ ?
The quadratic formula tells me that $x=\dfrac{-13 \pm 20\sqrt{3}}{40}.$
How to proceed?
| Another way to realize the solution (through inspection):
\begin{align*}
20x^{2} + 13x - 15 & = (20x^{2} - 12x) + (25x - 15)\\\\
& = 4x(5x - 3) + 5(5x - 3)\\\\
& = (4x + 5)(5x - 3)
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Simplify fraction within a fraction (Precalculus) Simplify $$\frac{x-2}{x-2-\frac{x}{x-\frac{x-1}{x-2}}}$$
My attempt:
$$=\frac{x-2}{x-2-\frac{x}{\frac{x(x-2)-(x-1)}{x-2}}} \ \ = \ \ \frac{x-2}{x-2-\frac{x}{\frac{x^2-2x-x+1}{x-2}}} \ \ = \ \ \frac{x-2}{x-2-\frac{x^2-2x}{x^2-3x+1}}$$
$$ =\frac{x-2}{\frac{(x-2)(x^2-3x+1)-(x^2-2x)}{x^2-3x+1}} \ \ = \ \ \frac{x-2}{\frac{x^3-3x^2+x-2x^2+6x-2-x^2+2x}{x^2-3x+1}} $$
$$ =\frac{(x-2)(x^2-3x+1)}{x^3-6x^2+9x-2} \ \ = \ \ \frac{x^3-3x^2+x-2x^2+6x-2}{x^3-6x^2+9x-2} $$
$$ =\frac{x^3-5x^2+7x-2}{x^3-6x^2+9x-2} $$
But the answer is given as $ \ \frac{x^2-3x+1}{x^2-4x+1} \ $, so I went wrong somewhere but can't see it. Any help is appreciated thanks.
| Less painful:
$$\begin{align}
\frac{x-2}{x-2-\frac{x}{x-\frac{x-1}{x-2}}}&=
\frac{x-2}{x-2-\frac{x(x-2)}{x(x-2)-x+1}}\\
&=\frac{(x-2)\bigl(x(x-2)-x+1\bigr)}{(x-2)\bigl(x(x-2)-x+1\bigr)-x(x-2)}.
\end{align}
$$
Now cancel by $x-2$ and multiply out to get
$$\frac{x^2-3x+1}{x^2-4x+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4604736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Is there another simpler method to evaluate the integral $\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x , \textrm{ where } \theta \in (0, \pi)?$ Using ‘rationalization’, we can split the integral into two manageable integrals as:
$\displaystyle \begin{aligned}\int_0^{2 \pi} \frac{1}{1+\cos \theta \cos x} d x = & \int_0^{2 \pi} \frac{1-\cos \theta \cos x}{1-\cos ^2 \theta \cos ^2 x} d x \\= & \int_0^{2 \pi} \frac{d x}{1-\cos ^2 \theta \cos ^2 x}-\cos \theta \int_0^{2 \pi} \frac{\cos x}{1-\cos ^2 \theta \cos ^2 x} d x \\= & 4 \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x-\cos ^2 \theta} d x+\int_0^{2 \pi} \frac{d(\cos \theta \sin x)}{\sin ^2 \theta+\cos ^2 \theta \sin ^2x} \\= & 4 \int_0^{\frac{\pi}{2}} \frac{d\left(\tan x\right)}{\sin ^2 \theta+\tan ^2 x}+\frac{1}{\sin \theta}\left[\tan ^{-1}\left(\frac{\cos ^2 \theta \sin x}{\sin \theta}\right)\right]_0^{2 \pi} \\= & \frac{4}{\sin \theta}\left[\tan ^{-1}\left(\frac{\tan x}{\sin \theta}\right)\right]_0^{\frac{\pi}{2}} \\= & \frac{4}{\sin \theta} \cdot \frac{\pi}{2} \\= & \frac{2 \pi}{\sin \theta}\end{aligned}\tag*{} $
Is there another simpler method to evaluate the integral?
Your comments and alternative methods are highly appreciated.
| Thanks to @David G. Stork who gave an alternative method as below:
$$
\begin{aligned}
I&= \int_0^{2 \pi} \frac{d x}{1+a \cos x} d x\\& =\int_0^{2 \pi} \frac{1-a \cos x}{1-a^2 \cos ^2 x} d x\\&=\int_0^{2 \pi} \frac{d x}{1-a^2 \cos ^2 x}-a \int_0^{2 \pi} \frac{\cos x}{1-a^2 \cos ^2 x} d x \\
& =4 \int_0^{\frac{\pi}{2}} \frac{\sec ^2 x}{\sec ^2 x-a^2} d x-a \int_0^{2 \pi} \frac{d(\sin x)}{\left(1-a^2\right)+a^2 \sin ^2 x} \\
& =4 \int_0^{\frac{\pi}{2}} \frac{d(\tan x)}{\tan ^2 x+\left(1-a^2\right)}-\frac{1}{1-a^2}\left[\tan ^{-1}\left(\frac{a \sin x}{\sqrt{1-a^2}}\right)\right]_0^{2 \pi} \\
& =\frac{4}{\sqrt{1-a^2}}\left[\tan ^{-1}\left(\frac{\tan x}{\sqrt{1-a^2}}\right)\right]_0^{\frac{\pi}{2}} \\
& =\frac{2 \pi}{\sqrt{1-a^2}} \\
&
\end{aligned}
$$
Putting $a=\cos\theta$ yields our result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Sum of two subspaces: representing it with equations I found the following excercise:
Let $W_1 = \{(x_1, ..., x_6) : x_1 + x_2 + x_3 = 0, x_4 + x_5 + x_6 = 0 \}$. Let $W_2$ be the span of $S := \{(1, -1, 1, -1, 1, -1), (1, 0, 2, 1, 0, 0), (1, 0, -1, -1, 0, 1), (2, 1, 0, 0, 0, 0)\}$.
Give a base, a dimension and an equation representation of $W_1 + W_2$
I'm new to the concept of sum of subspaces. But as I understand it, the first step would be to note that any $\textbf{x} = (x_1, ..., x_6) \in W_1$ satisfies
\begin{equation*}
\begin{cases}
x_1 = -x_2 - x_3 \\
x_4 = -x_5 - x_6
\end{cases}
\end{equation*}
so that its general form is
\begin{equation*} \textbf{x} =
\begin{pmatrix}
x_1 \\
x_2 \\
x_3 \\
x_4 \\
x_5 \\
x_6
\end{pmatrix} = \begin{pmatrix}
-x_2 - x_3 \\
x_2 \\
x_3 \\
-x_5 - x_6 \\
x_5 \\
x_6
\end{pmatrix}
\end{equation*}
We also know any $\textbf{y} \in W_1$ is of the general form
\begin{align*} \textbf{y} =
\begin{pmatrix}
y_1 \\
y_2 \\
y_3 \\
y_4 \\
y_5 \\
y_6
\end{pmatrix} = \begin{pmatrix}x + y +z + 2w \\
-x + w \\
x + 2y - z \\
-x + y - z \\
x\\
-x + z\end{pmatrix}
\end{align*}
Then for generals $\mathbf{x}, \mathbf{y}$ we have
\begin{align*}
\textbf{x} + \textbf{y} &= \begin{pmatrix}
x + y +z + 2w + (-x_2 - x_3)\\
-x + w + x_2\\
x + 2y - z + x_3\\
-x + y - z + (-x_5 - x_6)\\
x + x_5\\
-x + z + x_6
\end{pmatrix}
\end{align*}
One can then conclude
$$W_1 + W_2 = \Big\{\big(x + y + z + 2w - x_2 - x_3\big), \big(-x + w + x_2 \big), \big(x +2y - z + x_3 \big), \big(-x + y - z - x_5 - x_6 \big), \big(x + x_5 \big), \big(-x + z + x_6 \big) \mid x, y, z, x_2, x_3 \in \mathbb{R} \Big\}$$
But what would be an representation via equations of this system? I'm very new to linear algebra so go easy on me!
| Note that $(1,0,−1,−1,0,1)\in W_1$. Therefore, if$$W_3=\operatorname{span}\bigl\{(1,−1,1,−1,1,−1),(1,0,2,1,0,0),(2,1,0,0,0,0)\bigr\},$$then $W_1+W_2=W_1+W_3$.
Now, let us see what $W_1\cap W_3$ is. Asserting that$$\overbrace{\alpha(1,−1,1,−1,1,−1)+\beta(1,0,2,1,0,0)+\gamma(2,1,0,0,0,0)}^{\text{arbitrary element of }W_3}\in W_1$$means that $\alpha+3\beta+3\gamma=-\alpha+\beta=0$. This occurs if and only if $\beta=\alpha$ and $\gamma=-\frac{4\alpha}3$. Therefore,$$W_1\cap W_3=\left\{\left(-\frac{2 \alpha }{3},-\frac{7 \alpha }{3},3 \alpha,0,\alpha ,-\alpha \right)\,\middle|\,\alpha\in\Bbb R\right\};$$in particular, $\dim(W_1\cap W_3)=1$. So,\begin{align}\dim(W_1+W_3)&=\dim(W_1)+\dim(W_3)-\dim(W_1\cap W_3)\\&=4+3-1\\&=6.\end{align}Therefore, $W_1+W_3$ is the whole space $\Bbb R^6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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If $|z+2|-|z-2|=2$ and $\dfrac{z-a}{z+a}=ik$, where $k$ is a real parameter, $a\in \mathbb R$ has exactly two solution. Then interval of $a$ is?
If $|z+2|-|z-2|=2$ and $\dfrac{z-a}{z+a}=ik$, where $k$ is a real parameter, $a\in \mathbb R$ has exactly two solution. Then interval of $a$ is?
My Approach:
First Curve is one branch of a hyperbola with its transverse axis of length $2$ and conjugate axis with length of $2\sqrt3$.
So hyperbola is $\dfrac{x^2}{1}-\dfrac{y^2}{3}=1$.
And second curve is circle $(k^2-1)x^2+(k^2-1)y^2+2a(k^2+1)x+(k^2-1)a^2=0$ with center $\bigg(\dfrac{-a(k^2+1)}{k^2-1},0\bigg)$ and radius $\bigg|\dfrac{2ak}{k^2-1}\bigg|$.
Then I put $y^2=3(x^2-1)$ obtained from hyperbola into circle $(k^2-1)x^2+(k^2-1)y^2+2a(k^2+1)x+(k^2-1)a^2=0$
Then I obtained the equation $4(k^2-1)x^2+2a(k^2+1)x+(k^2-1)(a^2-3)$
Now I don't Know how to proceed further?
Second Doubt
What if there was two branches of hyperbola for example $\bigg ||z+2|-|z-2|\bigg |=2$?
| Well, it is not hard to show that:
$$\frac{\text{z}-\text{a}}{\text{z}+\text{a}}=i\text{k}\space\Longrightarrow\space\begin{cases}
\Re^2\left(\text{z}\right)+\Im^2\left(\text{z}\right)-\text{a}^2=0\\
\\
\text{k}=\frac{2\text{a}\Im\left(\text{z}\right)}{\Im^2\left(\text{z}\right)+\left(\Re\left(\text{z}\right)+\text{a}\right)^2}
\end{cases}\tag1$$
We also see that:
$$\left|\text{z}+2\right|-\left|\text{z}-2\right|=2\space\Longrightarrow\space\sqrt{\left(\Re\left(\text{z}\right)+2\right)^2+\Im^2\left(\text{z}\right)}-\sqrt{\left(\Re\left(\text{z}\right)-2\right)^2+\Im^2\left(\text{z}\right)}=2\tag2$$
Now, we can solve the following equation:
$$\sqrt{\left(\Re\left(\text{z}\right)+2\right)^2+\text{a}^2-\Re^2\left(\text{z}\right)}-\sqrt{\left(\Re\left(\text{z}\right)-2\right)^2+\text{a}^2-\Re^2\left(\text{z}\right)}=2\space\Longrightarrow\space$$
$$\Re\left(\text{z}\right)=\frac{\sqrt{3+\text{a}^2}}{2}\tag3$$
And:
$$\frac{3+\text{a}^2}{4}+\Im^2\left(\text{z}\right)-\text{a}^2=0\space\Longleftrightarrow\space\Im\left(\text{z}\right)=\pm\frac{\sqrt{3\left(\text{a}^2-1\right)}}{2}\tag4$$
And we can also see that:
$$\text{k}=\frac{2\text{a}\left(\pm\frac{\sqrt{3\left(\text{a}^2-1\right)}}{2}\right)}{\left(\pm\frac{\sqrt{3\left(\text{a}^2-1\right)}}{2}\right)^2+\left(\frac{\sqrt{3+\text{a}^2}}{2}+\text{a}\right)^2}=\pm\frac{\sqrt{3\left(\text{a}^2-1\right)}}{\sqrt{3+\text{a}^2}+2\text{a}}\tag5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607138",
"timestamp": "2023-03-29T00:00:00",
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Is there other method to evaluate $\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x, \textrm{ where }n\in N?$ Letting $x\mapsto \frac{1}{x}$ transforms the integral into
$\displaystyle I=\int_1^{\infty} \frac{\ln x}{x^{n+1}\left(1+x^n\right)} d x=-\int_0^1 \frac{x^{2 n-1} \ln x}{x^n+1} d x \tag*{} $
Splitting the integrand into two pieces like
$\displaystyle I=- \underbrace{\int_0^1 x^{n-1} \ln x d x}_{J} + \underbrace{\int_0^1 \frac{x^{n-1} \ln x}{x^n+1} d x}_{K} \tag*{} $
For the integral $J,$ letting $z=-n\ln x$ transforms $J$ into
$\displaystyle J= -\frac{1}{n^2} \int_0^{\infty} z e^{-z} d z =-\frac{1}{n^2}\tag*{} $
For integral $K$, using the series for $|x|<1,$
$\displaystyle \ln (1+x)=\sum_{k=0}^{\infty} \frac{(-1)^k}{k+1} x^{k+1},\tag*{} $
we have
$\displaystyle \begin{aligned}K& =-\frac{1}{n} \sum_{k=0}^{\infty} \frac{(-1)^k}{k+1} \int_0^1 x^{n(k+1)-1} d x \\& =-\frac{1}{n^2} \sum_{k=0}^{\infty} \frac{(-1)^k}{(k+1)^2} \\& =-\frac{1}{n^2}\left[\sum_{k=1}^{\infty} \frac{1}{k^2}-2 \sum_{k=1}^{\infty} \frac{1}{(2 k)^2}\right] \\& =-\frac{1}{2 n^2} \cdot \frac{\pi^2}{6} \\& =-\frac{\pi^2}{12 n^2}\end{aligned}\tag*{} $
Putting them back yields
$\displaystyle \boxed{ I=\frac{1}{12 n^2}\left(12-\pi^2\right)}\tag*{} $
Is there alternative method? Comments and alternative methods are highly appreciated.
| Making it more general $$I=\int \frac{\log( x)}{x^{m}\left(1+x^n\right)}\, d x$$
$$x=t^{\frac{1}{n}} \quad \implies \quad I=\frac 1{n^2}\int \frac {\log(t)}{t^a\,(1+t)}\,dt \qquad \text{with} \qquad a=\frac{m+n-1}{n}$$ The antiderivative express in terms of two hypergeometric functions.
$$J=\int_1^\infty \frac{\log( x)}{x^{m}\left(1+x^n\right)}\, d x$$ $$\color{blue}{J=\frac 1{4n^2}\left(\psi ^{(1)}\left(\frac{m+n-1}{2 n}\right)-\psi ^{(1)}\left(\frac{m+2
n-1}{2 n}\right)\right)}$$
Edit
If we make $m=n+k$ and consider large values of $n$
$$\psi ^{(1)}\left(\frac{2n+k-1}{2 n}\right)-\psi ^{(1)}\left(\frac{3n+k-1}{2 n}\right)=\left(4-\frac{\pi ^2}{3}\right)+$$ $$\frac{2 (k-1) (3 \zeta (3)-4)}{n}-\frac{ (k-1)^2\left(7
\pi ^4-720\right)}{60 n^2}+O\left(\frac{1}{n^3}\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $3^{3x-1}+27^x=2^{2x+1}+7\cdot4^x$ Solve the equation $$3^{3x-1}+27^x=2^{2x+1}+7\cdot4^x$$
We can rewrite the equation as $$\dfrac{3^{3x}}{3}+3^{3x}=2\cdot2^{2x}+7\cdot2^{2x}\\4\cdot3^{3x}=27\cdot2^{2x}\\\dfrac{3^{3x}}{2^{2x}}=\dfrac{27}{4}$$ I don't know how to approach it from here, as the LHS isn't for example $\dfrac{3^{2x}}{2^{2x}}=\left(\dfrac{3}{2}\right)^{2x}$.
The given answer is $x=1$.
| $3^{3x-1} + 27^{x} = 2^{2x+1} + 7\cdot4^{x}
\\ 3^{3x}\left(\frac{1}{3} + 1\right) = 2^{2x}(2+7)
\\ 3^{3x}\left(\frac{4}{3}\right) = 2^{2x}\cdot9
\\ \frac{3^{3x}}{2^{2x}} = \frac{27}{4}
\\ \left(\frac{27}{4}\right)^{x} = \frac{27}{4} \\ \textsf{clearly it's x=1}$
| {
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Find the minimum of $\sqrt{\cos x+3}+\sqrt{2\sin x+7}$ without derivative How do we find the minimum of
$$f(x)=\sqrt{\cos x+3}+\sqrt{2\sin x+7}$$
without using derivatives?
This problem is probably related to circles of Apollonius.
I have tried AM-GM and Cauchy-Schwarz inequality but I can't work it out.
Anyway, I have solved it in a more geometric way. Here's my answer.
Firstly we can do some identical transformation.
$$f(x)=\dfrac{\sqrt{2}}{2}(\sqrt{(\cos x+1)^2+(\sin x)^2+4}+\sqrt{(\cos x)^2+(\sin x+2)^2+9})$$
So that it makes sense in geometry.
$P(\cos x,\sin x)$ is on the circle $x^2+y^2=1$, and the value of $f(x)$ equals to sum of the distance from $A(0,-2)$ to $P$ and from $B(-1,0)$ to $P$.
In other words:
$$f(x)=\dfrac{\sqrt{2}}{2}(\sqrt{|PB|^2+4}+\sqrt{|PA|^2+9}).$$
And here we can use Minkowski inequality.
$$f(x)\geq \dfrac{\sqrt{2}}{2} \sqrt{(|PA|+|PB|)^2+25}$$
When $P$,$A$,$B$ is collinear, $RHS$ gets the minimum. Meanwhile, $LHS = RHS$.
Therefore, $f(x)_{min}=\sqrt{15}$.
| Here's a rough idea for a solution (since there are some details I skim over). I'm pretty sure the algebra can be simplified using some AM-GM variation, but I couldn't make it work. Although I'm hopeful that someone else can give a more optimal solution.
Note that for all $t \in \mathbb{R}$
\begin{align*}
&0 \le(t+2)^2(5t^2+6) =5 t^4 + 20 t^3 + 26 t^2 + 24 t + 24\\
\color{darkblue}{\implies}& 14 t^4 + 8 t^3 + 42 t^2 + 16 t + 28 \ge 9t^4 - 12 t^3 + 16 t^2 - 8 t + 4\\
\color{darkblue}{\implies}&\left(2t^2 +4\right)\left(
4t +7t^2+7\right)\ge (3 t^2 + 2 - 2 t)^2\\
\color{darkblue}{\implies}&2\sqrt{2t^2 +4}\sqrt{4t+7t^2+7} \ge 2\left(3t^2 +2-2t \right)\\
\color{darkblue}{\implies}&2t^2 +4 +2\sqrt{2t^2 +4}\sqrt{4t+7t^2+7} + 4t+7t^2 +7 \ge 15\left(t^2+1\right)\\
\color{darkblue}{\implies}&\left(\sqrt{2t^2 +4} + \sqrt{4t+7t^2+7}\right)^2 \ge\left( \sqrt{15}\sqrt{t^2+1}\right)^2\\
\color{darkblue}{\implies}&\frac{\sqrt{2t^2+4}}{\sqrt{t^2+1}} + \frac{\sqrt{4t+7t^2+7}}{\sqrt{t^2+1}}\ge \sqrt{15}\\
\color{darkblue}{\implies}&\sqrt{\frac{1-t^{2}}{1+t^{2}}+3}\ +\ \sqrt{2\frac{2t}{1+t^{2}}+7} \ge \sqrt{15}
\end{align*}
Lastly, notice that under the half-angle substitution the problem becomes showing that
$$
\min\left\{\sqrt{\frac{1-t^{2}}{1+t^{2}}+3}\ +\ \sqrt{2\frac{2t}{1+t^{2}}+7}\right\} = \sqrt{15}
$$
where $t = \tan(x/2)$. Since $\sqrt{\frac{1-t^{2}}{1+t^{2}}+3}\ +\ \sqrt{\frac{4t}{1+t^{2}}+7}\Bigg\vert_{t=-2} = \sqrt{15}$ we are done.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrating $\frac{1}{(x-2)^4 \sqrt{x^2 + 6x + 2}}$
I'm struggling with the integral, $$\int\frac{1}{(x-2)^4 \sqrt{x^2 + 6x + 2}}dx.$$
I tried it as follows:
Substituting $x-2 = \frac1t \implies dx = \frac{-dt}{t^2}.$
$$\therefore \int\frac{dx}{(x-2)^4 \sqrt{x^2 + 6x + 2}} = \int \frac{- dt}{\frac{t^2}{t^4} \sqrt{(\frac1t + 2)^2 + 6 (\frac1t + 2) + 2}} = \int \frac{-t^3}{\sqrt{18t^2 + 10t + 1}}\ dt$$
How to continue from here?
| I'm going to continue from the last known position above. In the end, you'll get a polynomial in secants, which just results in using standard tricks from calculus II (along with integration by parts).
Let's start with the standard secant substitution $t + \frac{5}{18} = \sqrt{\frac{7}{18}} \sec y,$ so $dt = \sqrt{\frac{7}{18}} \sec y \tan y \ dy.$ Our integral becomes
$$ \int \frac{-t^3}{\sqrt{18t^2 + 10t + 1}}\ dt = -\frac{1}{\sqrt{18}} \int \frac{ \left( \sqrt{\frac{7}{18}} \sec y - \frac{5}{18} \right)^3 \left( \sqrt{\frac{7}{18}} \sec y \tan y \ dy \right)}{\sqrt{\frac{7}{18}} \tan y}$$
and thankfully the stuff in the denominator cancels. This leaves
$$ -\frac{1}{\sqrt{18}} \int \left( \sqrt{\frac{7}{18}} \sec y - \frac{5}{18} \right)^3 \sec y \ dy. $$
It gets quite messy, but after expanding the binomial, we get
$$ -\frac{1}{\sqrt{18}} \int a^3 \sec^4 y - 3a^2 b \sec^3 y + 3ab^2 \sec^2 y -b^3 \sec y \ dy$$
where I've relabeled $a = \sqrt{\frac{7}{18}}$ and $b = \frac{5}{18}$ to make things cleaner. The first term is handled with the substitution $z = \tan y$ (after pulling two secants - again, standard calculus II technique), the second using integration by parts (choose $u = \sec y, \ dv = \sec^2 y \ dy$), and the third and fourth are standard results. I won't flesh out all the details here, but we obtain:
$$ -\frac{1}{\sqrt{18}} \left( \frac{a^3 \tan^3 y}{3} + a^3\tan y - \frac{3a^2 b}{2} \left( \ln |\sec y + \tan y| + \sec y \ \tan y \right) \\ + 3ab^2 \tan y - b^3 \ln |\sec y + \tan y| \right) + C $$
(wow!). We're almost home free. By drawing a triangle based on our secant substitution, we obtain $\tan y = \sqrt{ \frac{1}{a^2} \left( t - b \right)^2 - 1}.$ Finally, invert your initial substitution to get $t = \frac{1}{x-2}.$
Now, by being lazy (sorry!), we arrive at our result:
$$-\frac{1}{\sqrt{18}} \frac{\left( a \sqrt{ \frac{1}{a^2} \left( \frac{1}{x-2} - b \right)^2 - 1} \right)^3}{3} + a^3 \left(\sqrt{ \frac{1}{a^2} \left( \frac{1}{x-2} - b \right)^2 - 1} \right) - \frac{3a^2 b}{2} \left( \ln \left( \frac{1}{a} \left(\frac{1}{x-2} - b \right) + \left( \sqrt{ \frac{1}{a^2} \left( \frac{1}{x-2} - b \right)^2 - 1} \right) \right) + \left( \sqrt{ \frac{1}{a^2} \left( \frac{1}{x-2} - b \right)^2 - 1} \right) \frac{1}{a} \left(\frac{1}{x-2} - b \right) \right) + 3b^2 \left(\frac{1}{x-2} - b \right) - b^3 \ln \left| \left( \frac{1}{a} \left(\frac{1}{x-2} - b \right) + \left( \sqrt{ \frac{1}{a^2} \left( \frac{1}{x-2} - b \right)^2 - 1} \right) \right) \right| + C$$
Wow! I may have mismatched some parentheses here, but that's a result.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the equation $\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2$ Solve the equation $$\log_{1-2x}(6x^2-5x+1)-\log_{1-3x}(4x^2-4x+1)=2$$
We have $$D_x:\begin{cases}1-2x>0\\6x^2-5x+1>0\\1-3x>0\\1-3x\ne1\\4x^2-4x+1>0\iff(2x-1)^2>0\iff x\ne\dfrac12\end{cases}\iff x\in(-\infty;0)\cup(0;\dfrac{1}{3})$$
Also the quadratic $6x^2-5x+1$ factors as $(2x-1)(3x-1)$. The equation then becomes $$\log_{1-2x}(2x-1)(3x-1)-\log_{1-3x}(2x-1)^2=2\\\log_{1-2x}(2x-1)(3x-1)-2\log_{1-3x}(1-2x)=2,$$ as $\log_{1-3x}(2x-1)^2=2\log_{1-3x}|2x-1|,$ but we know from $D_x$ that $2x-1<0$,
$$\log_{1-2x}(2x-1)+\log_{1-2x}(3x-1)-\dfrac{2}{\log_{1-2x}(1-3x)}=2$$ I don't know what to do next.
| You're almost there, but you may use a simplified method such as this:
$$\log_{1-2x}(2x-1)(3x-1) - 2\log_{1-3x}(1-2x)=2$$
$$\log_{1-2x}(1-2x)(1-3x) - 2\log_{1-3x}(1-2x)=2$$
$$\log_{1-2x}(1-2x)+\log_{1-2x}(1-3x) - 2\log_{1-3x}(1-2x)=2$$
$$\log_{1-2x}(1-3x) - 2\log_{1-3x}(1-2x)=1$$
$$\frac{1}{\log_{1-3x}(1-2x)} - 2\log_{1-3x}(1-2x)=1$$
Now you can set $\log_{1-3x}(1-2x)=t$ and get:
$$\frac{1}{t}-2t=1$$
This will give you a quadratic in terms of $t$ which you can solve for and substitute back into the substitution and solve for $x$ that way.
| {
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"timestamp": "2023-03-29T00:00:00",
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Rearranging cylinder equation Given this equation $(x-y)^2+(y-z)^2+(z-x)^2=r^2$, which when plotted for some radius $r$ results in a "rotated" cylinder:
Plotted equation with $r=1$
Is it possible to rearrange this equation to a form where to rotation is clearer? Perhaps to something around the lines of $(\overrightarrow{p}-\overrightarrow{c})^2=r^2+(\overrightarrow{d}\bullet(\overrightarrow{p}-\overrightarrow{c}))^2$, where $\overrightarrow{p}$ is a point on the cylinder and the centre-axis of the cylinder is defined by a point $\overrightarrow{c}$ on the axis and the axis' directions $\overrightarrow{d}$.
Is a rearranging even needed or can the rotation be easily read from the original equation?
| We have
$ \begin{bmatrix} x - y \\ y - z \\ z - x \end{bmatrix} = \begin{bmatrix} 1 && - 1 && 0 \\ 0 && 1 && - 1 \\ -1 && 0 && 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix}$
Therefore,
$(x - y)^2 + (y - z)^2 + (z - x)^2 = \begin{bmatrix} x && y && z \end{bmatrix} \begin{bmatrix} 1 && 0 && -1 \\ -1 && 1 && 0 \\ 0 && -1 && 1 \end{bmatrix} \begin{bmatrix} 1 && - 1 && 0 \\ 0 && 1 && - 1 \\ -1 && 0 && 1 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} $
Which simplifies to
$ (x - y)^2 + (y - z)^2 + (z - x)^2 = \begin{bmatrix} x && y && z \end{bmatrix} \begin{bmatrix} 2 && -1 && -1 \\ -1 && 2 && -1 \\ -1 && -1 && 2 \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} $
Now,
$\begin{bmatrix} 2 && -1 && -1 \\ -1 && 2 && -1 \\ -1 && -1 && 2 \end{bmatrix} = \begin{bmatrix} 3 && 0 && 0 \\ 0 && 3 && 0 \\ 0 && 0 && 3 \end{bmatrix} - \begin{bmatrix} 1 && 1 && 1 \\ 1 && 1 && 1 \\ 1 && 1 && 1 \end{bmatrix} $
The last matrix can be factored as follows
$\begin{bmatrix} 1 && 1 && 1 \\ 1 && 1 && 1 \\ 1 && 1 && 1 \end{bmatrix} = \begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix} \begin{bmatrix} 1 && 1 && 1 \end{bmatrix} $
So that,
$(x - y)^2 + (y - z)^2 + (z - x)^2 = p^T ( 3 I - v v^T ) p = 3 p^T (I - u u^T ) p $
where $ p = [x, y, z]^T $ and $v = [1, 1, 1]^T $ and $ u = \dfrac{1}{\sqrt{3}} v $
It follows that
$ p^T ( I - u u^T ) p = \dfrac{r^2}{3} $
And this is the equation of a cylinder of radius $\dfrac{r}{\sqrt{3}} $ with an axis along $ u $ ( or $v$ ) that passes through the origin.
| {
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"timestamp": "2023-03-29T00:00:00",
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Convergence of series $\sum_{n \geq 1} {\frac{1^2+2^2+ \cdots + n^2}{n^4}}$ In the study of the following series
$$
\sum_{n \geq 1} {\frac{1^2+2^2+ \cdots + n^2}{n^p}}
$$
it is not hard to prove that it diverges for $p \leq 3$, since the sequence itself does not converge to 0. You can also conclude that the series converges for $p > 4$ by comparison with Riemann series. Raabe's test yields that the series diverges for p between 3 and 4. However it does not give any information for the case $p=4$.
| We do not need to apply the explicit formula for $1^2+2^2+\ldots +n^2.$ Instead observe that
$$n^3-(n-1)^3=3n^2-3n+1\ge 3n^2$$
Hence $$1^2+2^2+\ldots +n^2\ge {1\over 3}\left ([1^3-0^3]+[2^3-1^3]+\ldots +[n^3-(n-1)^3]\right ]={1\over 3}n^3$$
Hence the $n$th term of the series for $p=4$ is greater or equal $\displaystyle{1\over 3n}.$
Another approach is to apply the Cauchy-Schwarz inequality $$(1+2+\ldots +n)^2\le (1^2+2^2+\ldots +n^2)n$$ hence $$1^2+2^2+\ldots +n^2\ge {1\over n}{n^2(n+1)^2\over 4}\ge {n^3\over 4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How fast does $a_n:=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}$ decay as $n \to \infty$? Let
\begin{equation}
a_n:=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}
\end{equation}
for $n \in \mathbb{N}$.
Then, $a_n$ is clearly a monotone decreasing sequence of positive numbers and by the Dominated Convergence Theorem, $a_n \to 0^+$ as $n \to \infty$.
However, I have some difficulty estimating how fast $a_n$ decays. For example, $\frac{a_n}{f(n)}=O(1)$ as $n \to \infty$ for some polynomial $f$?
Could anyone please provide insight into the decay rate of $a_n$?
| We can also try to find the asymptotics of the integral directly.
Making the substitution $x=\frac{t}{\sqrt2}$
$$a_n=\int_{1/\sqrt{2}}^1 \frac{dx}{\left(\frac{1}{2}+x^2\right)^{n+1/2}}=2^n\int_1^\sqrt2\frac{dt}{(1+t^2)^{n+1/2}}=2^n\int_1^\sqrt2\frac{e^{-n\ln(1+t^2)}}{\sqrt{1+t^2}}dt$$
Making one more substitution $t=1+x$
$$a_n=\frac{1}{\sqrt2}\int_0^{\sqrt{2}-1} \frac{e^{-n\ln(1+x+x^2/2)}}{\sqrt{1+x+x^2/2}}dx$$
It is straightforward to see that the function $f(x)=\ln(1+x+x^2/2)$ does not have an extremum on the interval $[0;\sqrt2-1]$, but has the minimal value at $x=0$. Decomposing the integrand near this point
$$\ln(1+x+x^2/2)=f(0)+f'(0)\,x+f''(0)\frac{x^2}{2}+...=x+O(x^3)$$
$$\frac{1}{\sqrt{1+x+x^2/2}}=1-\frac{1}{2}\,x+O(x^2)$$
$$a_n=\frac{1}{\sqrt2}\int_0^{\sqrt2-1}e^{-nx}\Big(1-\frac{x}{2}+O(x^2)\Big)dx$$
With the accuracy up to exponentially small corrections we can extend the interval of integration till $\infty$. Performing integration,
$$\boxed{\,\,a_n=\frac{1}{\sqrt2}\frac{1}{n}\Big(1-\frac{1}{2n}+O\big(1/n^2\big)\Big)\,\,}$$
Numerical check confirms the asymptotics:
$$n=10\quad a_n=0.0667968\quad \text{approximation}=\frac{1}{10\sqrt2}(1-1/20)=0.067175$$
$$n=30\quad a_n=0.0232066\quad \text{approximation}=\frac{1}{30\sqrt2}(1-1/60)=0.023177$$
$$n=100\quad a_n=0.0070366\quad \text{approximation}=\frac{1}{100\sqrt2}(1-1/200)= 0.00703571$$
$$n=1000\quad a_n=0.000706754\quad \text{approximation}=\frac{1}{1000\sqrt2}(1-1/2000)= 0.000706753$$
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