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On the biconditional $I(n^2) = 2 - \frac{5}{3q} \iff (k = 1 \land q = 5)$, where $q^k n^2$ is an odd perfect number MOTIVATION
Let $N$ be an odd perfect number given in the so-called Eulerian form
$$N = q^k n^2,$$
i.e., $q$ is the special / Euler prime satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
In what follows, I let
$$I(x)=\frac{\sigma(x)}{x}$$
denote the abundancy index of the positive integer $x$. ($\sigma(x)$ is the sum of divisors of $x$.)
CLAIM
If $q^k n^2$ is an odd perfect number given in Eulerian form, then
$$I(n^2) = 2 - \frac{5}{3q} \iff (k = 1 \land q = 5).$$
PROOF OF CLAIM
Let $q^k n^2$ be an odd perfect number given in Eulerian form.
Suppose that
$$I(n^2) = 2 - \frac{5}{3q}.$$
Since $q^k n^2$ is perfect, then we have
$$I(q^k n^2) = I(q^k)I(n^2) = 2$$
where we have used the fact that $I$ is multiplicative.
Hence,
$$I(n^2) = I(q^k)I(n^2) - \frac{5}{3q} \implies \frac{5}{3q} = I(n^2)\bigg(I(q^k) - 1\bigg) \geq I(n^2)\bigg(1+\frac{1}{q}-1\bigg).$$
This implies that
$$I(n^2) \leq \frac{5}{3}.$$
Assume to the contrary that
$$I(n^2) = 2 - \frac{5}{3q} < \frac{5}{3}.$$
Then we have
$$\frac{6q - 5}{3q} < \frac{5}{3} \implies 18q - 15 < 15q \implies 3q < 15 \implies q < 5,$$
contradicting $q \geq 5$.
Added to the Proof of Claim (Dec 15 2019) Hence,
$$I(n^2) = 2 - \frac{5}{3q} \implies I(n^2) = \frac{5}{3} \implies (k=1 \land q=5)$$
while the proof of the direction
$$(k=1 \land q=5) \implies I(n^2) = \frac{5}{3} \implies I(n^2) = 2 - \frac{5}{3q}$$
is trivial.
QUESTION
It can be proved (page 17) that
$$I(n^2) \leq 2 - \frac{5}{3q}$$
holds in general for an odd perfect number $q^k n^2$ given in Eulerian form.
My question is: Can a biconditional similar to the one proved here be derived for the case
$$I(n^2) < 2 - \frac{5}{3q}?$$
| Let $q^k n^2$ be an odd perfect number given in Eulerian form.
Since
$$I(n^2) \leq 2 - \frac{5}{3q}$$
holds in general, and since the biconditional
$$\bigg(I(n^2) = 2 - \frac{5}{3q}\bigg) \iff \bigg(k=1 \land q=5\bigg)$$
holds, the simplest biconditional that I could come up with is
$$\bigg(I(n^2) < 2 - \frac{5}{3q}\bigg) \iff \bigg(k>1 \lor q>5\bigg).$$
Of course, by Material Implication, we also have the biconditionals
$$\bigg(k>1 \lor q>5\bigg) \iff \bigg(k=1 \implies q>5\bigg) \iff \bigg(q=5 \implies k>1\bigg).$$
| {
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"url": "https://math.stackexchange.com/questions/3477159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$ Suppose that $a,b,c>0$. How to prove $$\frac{a}{7a+b}+\frac{b}{7b+c}+\frac{c}{7c+a}\le\frac38$$
?
My first idea: By AM-GM, $$7a+b\geq \sqrt{7ab}$$ so $$\sum_{cyc} \frac{a}{7a+b}\le\sum_{cyc}\sqrt{\frac{a}{7b}}$$ but I am not sure if we can continue from here.
Also I try Cauchy-Schwarz: $$\sum_{cyc} \frac{a}{7a+b}\le\sqrt{a^2+b^2+c^2}\sqrt{\sum_{cyc} \frac{1}{(7a+b)^2}}.$$
Now what?
| By AM-GM we have $$a^2b+ac^2+b^2c\geq3abc$$ and $$a^2c+ab^2+bc^2\geq 3abc$$ so that $$35(a^2b+ac^2+b^2c)+13(a^2c+ab^2+bc^2)\geq 3(35+13)abc=144abc.$$
Now, note that $$\frac38-\sum_{\text{cyc}} \frac{a}{7a+b}=\frac{35(a^2b+ac^2+b^2c)+13(a^2c+ab^2+bc^2)-144abc}{8 (7 a+b) (a+7
c) (7 b+c)},$$
which is non-negative by the previous result.
We have equality if and only if we have equality in both AM-GMs which implies $a=b=c$.
| {
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"url": "https://math.stackexchange.com/questions/3483200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Solve $\frac{y+1}{x-y}+\frac{x+2}{x+y}=\frac{x^2+y^2+10}{x^2-y^2}$ and $ 2x+5y=1$
Solve
$$\begin{array}{|l}\dfrac{y+1}{x-y}+\dfrac{x+2}{x+y}=\dfrac{x^2+y^2+10}{x^2-y^2}
\\ 2x+5y=1\end{array}$$
As always:
$$\begin{array}{|l} x-y \ne0 \\ x+y \ne 0 \\ x^2-y^2\ne0\end{array} \Leftrightarrow \begin{array}{|l} x \ne y \\ x\ne -y \\ (x-y)(x+y) \ne 0\end{array} \Leftrightarrow \begin{array}{|l} x \ne y \\ x\ne -y \end{array}$$
We can solve the second equation for $y: y=-\dfrac{2}{5}x+\dfrac{1}{5}$ and substitute into the first equation and solve for $x$. I don't have experience with systems and I would like to ask you if there is a simpler solution for this system.
| $$\dfrac{y+1}{x-y}+\dfrac{x+2}{x+y}=\dfrac{x^2+y^2+10}{x^2-y^2}$$
$$\dfrac{xy+x+y+y^2+x^2+2x-xy-2y}{x^2-y^2} = \dfrac{x^2+y^2+10}{x^2-y^2}$$
Since $(x^2-y^2) \ne 0$ , we have :
$$x+y+2x-2y = 10\implies 3x-y =10 \quad \color{#2d0}{\text{(1.)}}$$
Also $$2x+5y=1 \quad \color{#d05}{\text{(2.)}}$$
Multiplying $(1.)$ by $5$ and adding $(2.)$ ,
$$17x = 51 \implies x = 3$$
And we get $y = -1.$
So the solution is $\boxed{(x,y) = (3,-1)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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The number of solutions to $y^2\equiv x^3+nx\pmod p$ is exactly $p$
Question
Suppose $p\equiv3\pmod4$ and $n\in\mathbb Z$, is the number of solutions to $y^2\equiv x^3+nx\pmod p$ exactly $p$? I have numerically confirmed it when $|n|\le10$ and $p\le229$.
Thoughts
Fix an $x$, the number of solutions to $y^2\equiv x^3+nx\pmod p$ is $$1+\left(\frac{x^3+nx}p\right),$$
Summing this formula from $0$ to $p-1$, the question boils down to proving $$p+\sum_{x=0}^{p-1}\left(\frac{x^3+nx}p\right)=p$$ or $$\sum_{x=0}^{p-1}\left(\frac{x^3+nx}p\right)=0.$$
Then I have difficulty going further.
Here $\displaystyle\left(\frac\cdot\cdot\right)$ denotes the Legendre symbol.
| Note that $$\sum_{x = 0}^{p -1} \left(\frac{x^3 + nx}{p}\right) = \sum_{x = 1}^{(p -1)/2} \left(\frac{x^3 + nx}{p}\right) + \left(\frac{(-x)^3 -nx}{p}\right)$$ and $$\left(\frac{(-x)^3 -nx}{p}\right) = \left(\frac{-1}{p}\right)\left(\frac{x^3 +nx}{p}\right) = - \left(\frac{x^3 + nx}{p}\right),$$ since $p \equiv 3 \pmod 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Double integral with change of coordinates Evaluate the following double integral:
$$\iint_D \sqrt{x^2+y^2}\,dx\,dy$$
where $D=\{(x,y)\in\mathbb{R^2},\ x^2+y^2-4x<0\}$. I'm trying to use the polar coordinates by making the substitution $x=\rho\cos\theta$ and $y=\rho\sin\theta$.
After replacing in the domain, I find out that $0<\rho<4\cos\theta$. But what about $\theta$ ?
| First, let's look at the region $D$: $$x^2+y^2-4x < 0 \iff (x-2)^2 + y^2 < 4$$ Hence, $D$ is the interior of the circle with centre $(2,0)$ and radius $2$. In polar coordinates, this is the region $0<\rho<4\cos\theta$ and $-\pi/2<\theta<\pi/2$. (See here for explanation).
Therefore, the integral is:
\begin{align}
\iint_D\sqrt{x^2+y^2}\,dx\,dy &= \int_{-\pi/2}^{\pi/2}\int_0^{4\cos\theta}\rho^2\,d\rho\,d\theta \\
&= \int_{-\pi/2}^{\pi/2} \left[\frac{\rho^3}{3}\right]_0^{4\cos\theta}\,d\theta \\
&= \frac{64}{3} \int_{-\pi/2}^{\pi/2}\cos^3\theta\,d\theta \\
&= \frac{64}{3} \int_{-\pi/2}^{\pi/2}(\cos\theta - \cos\theta\sin^2\theta)\,d\theta \\
&= \frac{128}{3} \left[\sin\theta-\frac{\sin^3\theta}{3}\right]_{0}^{\pi/2} \\ &= \frac{256}{9}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3485304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to use other method to prove $2^x>2x-1$. let $x\in R$,show $$2^x>2x-1$$
My try:
take a function $f(x)=2^x-2x+1$, then we have
$$f'(x)=2^x\ln{2}-2,~~~~f''(x)=2^x\ln^2{2}>0$$
if let $x_{0}$ such $f'(x_{0})=0\Longrightarrow 2^{x_{0}}=\dfrac{2}{\ln{2}}$ is minimum the point (since $f''(x)>0$).
so we have
$$f(x)\ge f(x_{0})= \dfrac{2}{\ln{2}}-1+\dfrac{\ln{\ln{2}}}{\ln{2}}+1=\dfrac{2+\ln{\ln{2}}}{\ln{2}}>0$$
because $$2+\ln{\ln{2}}>0\Longleftrightarrow 2^{e^2}>e$$
have other methods?Thanks
| $\require{begingroup} \begingroup$
$\def\W{\operatorname{W}}\def\Wp{\operatorname{W_0}}\def\Wm{\operatorname{W_{-1}}}$
\begin{align}
2^x&>2x-1
\tag{1}\label{1}
\end{align}
Let's consider
\begin{align}
2^x&=2x-1
\tag{2}\label{2}
.
\end{align}
It is known, that solution to equation \eqref{2}
can be found in terms of the
Lambert W function.
Let $2x-1=y$, $x=\tfrac12y+\tfrac12$:
\begin{align}
2^{\tfrac12y+\tfrac12} &= y
\tag{3}\label{3}
,\\
\sqrt2\cdot 2^{\tfrac12y} &= y
,\\
\sqrt2\,\exp(\ln(2^{\tfrac12y})) &= y
,\\
\sqrt2\,\exp(\tfrac12y\cdot \ln2) &= y
,\\
y\,\exp(-\tfrac{\ln2}2\cdot y) &= \sqrt2
,\\
-\tfrac{\ln2}2\cdot y\,\exp(-\tfrac{\ln2}2\cdot y) &= -\tfrac{\ln2}2\cdot\sqrt2
,\\
\W\left(-\tfrac{\ln2}2\cdot y\,\exp(-\tfrac{\ln2}2\cdot y)\right)
&=\W\left( -\tfrac{\ln2}2\cdot\sqrt2\right)
,\\
-\tfrac{\ln2}2\cdot y
&=\W\left( -\tfrac{\sqrt2}2\,\ln2\right)
\tag{4}\label{4}
,\\
y
&=
-\tfrac2{\ln2}\cdot
\W\left( -\tfrac{\sqrt2}2\,\ln2\right)
\tag{5}\label{5}
,\\
x
&=
\tfrac12
-\tfrac1{\ln2}\cdot
\W\left( -\tfrac{\sqrt2}2\,\ln2\right)
\tag{6}\label{6}
.
\end{align}
The argument of $\W\left( -\tfrac{\sqrt2}2\,\ln2\right)$
is $-\tfrac{\sqrt2}2\,\ln2<-\tfrac1{\mathrm e}$,
hence there are no real solutions to \eqref{2}, and the function
\begin{align}
2^x-(2x-1)
\tag{7}\label{7}
\end{align}
never cross the $x$-axis. Evaluation of \eqref{7} at $x=0$
gives
\begin{align}
2^0-(2\cdot0-1)
&=2> 0
\tag{8}\label{8}
,
\end{align}
hence
\begin{align}
2^x-(2x-1)>&0
\quad\text{ for all } x\in\mathbb R
\tag{9}\label{9}
\end{align}
and \eqref{1} follows.
$\endgroup$
| {
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"timestamp": "2023-03-29T00:00:00",
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write 33 as sum of three cubes to solve 3 as sum of three cubes I am interested in writing numbers as the sum of three positive or negative cubed integers. This article describes how a supercomputer was used to find that
$$33 = 8{,}866{,}128{,}975{,}287{,}528^3 + (−8{,}778{,}405{,}442{,}862{,}239)^3 + (−2{,}736{,}111{,}468{,}807{,}040)^3$$
Apparently they then used that to find a third way of writing 3 as the sum of three cubes - after $1^3+1^3+1^3$ and $4^3+4^3+(-5)^3$:
$$3 = 569{,}936{,}821{,}221{,}962{,}380{,}720^3 + (−569{,}936{,}821{,}113{,}563{,}493{,}509)^3 + (−472{,}715{,}493{,}453{,}327{,}032)^3$$
So our equation $33=x^3+y^3+z^3$ can be rewritten as $33-z^3=x^3+y^3$. Now, $x^3+y^3=(x+y)(x^2-xy+y^2)$. So for each $z$ values, we compute $33-z^3$ and then try to factor it and see if it can be reaarage into the form $(x+y)(x^2-xy+y^2)$. How else can we make improvement to reduce the size of search space to find a way to express 33 as sum of three cubes. So how does finding 33 as sum of three cubes help find 3 as sum of three cubes?
| The resolution for $33$ didn't help directly for the resolution for $3$. The initial author rather designed an efficient method to tackle the general problem.
In the paper (https://people.maths.bris.ac.uk/~maarb/papers/cubesv1.pdf), he describes an alternative method based on factorization: if
$$k-z^3=(x+y)(x^2-xy+y^2)$$ then you can try among the $x+y$ that divide $k-z^3$ and solve
$$\begin{cases}x+y=d,\\x^2-xy+y^2=\dfrac{k-z^3}d\end{cases}.$$
The method he used is different.
| {
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Express $\sin (2x)$ in the form $\dfrac{a\pi^2+b\pi + c}{d},$ where $a,b,c,d$ are integers
Let $0<x<\dfrac{\pi}{2}.$ If $x$ is such that $\cos\left(\dfrac{3}{2}\cos x\right) = \sin\left(\dfrac{3}{2}\sin x\right),$ then express $\sin\, (2x)$ in the form $\dfrac{a\pi^2+b\pi + c}{d},$ where $a,b,c,d$ are integers.
Here are my proofs that $\sin \left(\dfrac{\pi}{2}-x\right)=\cos\left(x\right).$ First, angle addition says that $\sin\left( \dfrac{\pi}{2}-x\right)=\sin\dfrac{\pi}{2}\cos x-\sin x\cos\dfrac{\pi}{2}=(1)\cos x - \sin x (0) = \cos x.$ Second, a geometric proof. Consider a right triangle. We may assume WLOG that the hypotenuse is $1$. If not, then multiply all sides by the multiplicative inverse of the hypotenuse so that it is (the triangle obtained is similar, so the angles are preserved). Let one of the acute angles be $\alpha$. Then the side adjacent to that angle has length $\cos \alpha$. Now consider the angle $\dfrac{\pi}{2} - \alpha$. The side opposite to this angle has length $\cos \alpha$. But since $\sin x = \dfrac{\text{opposite}}{\text{hypotenuse}}, \sin\left(\dfrac{\pi}{2} - \alpha\right)=\cos \alpha.$ Now, if $\alpha$ is not acute, we may add an integer multiple of $2\pi$ to $\alpha$ so that it is, without changing the value of $\sin \alpha$ since $\sin x$ is $2\pi$ periodic.
Now, since $0<x < \dfrac{\pi}{2}, \sin x,\cos x \in (0, 1)\Rightarrow \dfrac{3}{2}\sin x, \dfrac{3}{2}\cos x \in (0,\dfrac{3}{2})\subseteq (0,\dfrac{\pi}{2})$. So, using this fact, we have that $\cos \left(\dfrac{3}{2}\cos x\right)=\sin \left(\dfrac{3}{2}\sin x\right)\Leftrightarrow\dfrac{3}{2}\cos x+\dfrac{3}{2}\sin x = \dfrac{\pi}{2}.$ Using the fact that $\sin x = \sqrt{1-\cos ^2 x},$ I get that $\sin x = \dfrac{\frac{2\pi}{3}\pm \frac{2}{3}\sqrt{18-\pi^2}}{4},$ but I'm not sure how to use this to get $\sin \,(2x)$ into the desired form.
| $(\sin(x)+\cos(x))^2-2\sin(x)\cos(x)=\sin^2(x)+\cos^2(x)$
$\sin(2x)=\left(\frac\pi3\right)^2-1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Can we find $x,y \in \mathbb{N} $ such that $x^y -y^x=2020$? Happy new year.
Can we find $x,y \in \mathbb{N} $ such that : $$x^y -y^x=2020$$
One solution is $(x,y) = (2021,1)$.
It is clear that both of $x,y$ are odd or even .
Note that $2020=2^2 \times 5 \times 101 $
| If $x$ or $y$ is $1$, then $y=1$ and $x=2021$.
If $x$ or $y$ is even then both are even. Since $2020$ is not divisible by $8$, either $x$ or $y$ is $2$. For $ t\in \mathbb{N}$,$2^t-t^2\ge 0 $ if $t\ne 3$ and is an increasing function of $t$ for $t\ge 4$. Therefore $x=2$ and $y\approx 11$ but $2^{11}-11^2=1927\ne 2020. $
We now have $x,y\ge 3$ and therefore $x^y>y^x$ only for $y > x$.Then
$$2020=x^x\left(x^{y-x}-\left(1+\frac{y-x}{x}\right)^x\right)\ge x^x\left(x^{y-x}-e^{y-x}\right) $$ and so $x<5$.
Therefore $x=3$ and $$2020\ge 27\left(3^{y-3}-e^{y-3}\right) $$
then $y\le 7$ and the values $y=5$ and $7$ are easily checked.
| {
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"timestamp": "2023-03-29T00:00:00",
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The calculation of the series $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n^4}$ Happy New Year 2020, Romania
In this recent post, Evaluate $\int_0^1 \frac{\arctan x\ln^2 x}{1+x^2}\,dx$, the proposed integral reduces to the calculation of $\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n^3}$ which is known in the literature. Now, what can we say about the more advanced version of it, the one with $n^4$ in the denominator?
$$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}}{n^4}$$
Can we do it by series manipulations? It looks like a new series in the literature.
A good note: With the previous result in hand which we combine with some of the Cornel's work, we immediately arrive at two delightful series results,
$$i) \ \sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_{2 n}^{(2)}}{n^3}=\frac{61 }{192}\pi ^2 \zeta (3)+\frac{1973 }{128}\zeta (5)+\frac{\pi ^5}{16}-\frac{1}{128} \pi \psi ^{(3)}\left(\frac{1}{4}\right);$$
$$ii) \ \sum _{n=1}^{\infty } (-1)^{n-1} \frac{ H_{2 n}^{(3)}}{n^2}=\frac{\pi ^3 G}{8}+\frac{1}{64}\pi ^2 \zeta (3)-\frac{2997 }{256}\zeta (5)-\frac{\pi ^5}{32}+\frac{1}{256} \pi \psi ^{(3)}\left(\frac{1}{4}\right).$$
How would you go proving these last two results?
Another good note: Remaining on the ground with alternating harmonic series of weight $5$ and harmonic numbers of the type $H_{2n}$, we might be curious to know what the value of the series $\displaystyle \sum _{n=1}^{\infty }(-1)^{n-1} \frac{ H_{2 n}^{(4)}}{n}$ is.
$$\sum_{n=1}^{\infty} (-1)^{n-1}\frac{H_{2n}^{(4)}}{n}$$
$$=4\zeta(5)-\frac{3}{128}\zeta(2)\zeta(3)-\frac{7}{128}\log(2)\zeta(4)+\frac{\pi^5}{192}-\frac{\pi^3}{16}G-\frac{\pi}{1536}\psi^{(3)}\left(\frac{1}{4}\right).$$
This last series is elegantly calculated in A simple strategy of calculating two alternating harmonic series generalizations where one may also find its generalization with respect to the order of the harmonic number.
| Consider
$$\psi \left( -z \right)+\gamma \underset{z\to n}{\mathop{=}}\,\frac{1}{z-n}+{{H}_{n}}+\sum\limits_{k=1}^{\infty }{\left( {{\left( -1 \right)}^{k}}H_{n}^{k+1}-\zeta \left( k+1 \right) \right){{\left( z-n \right)}^{k}}}, n\ge 0$$
Then
$$\left( \psi \left( -2z \right)+\gamma \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)}\underset{2z\to 2k+1}{\mathop{=}}\,\frac{{{\left( -1 \right)}^{k}}\pi f\left( k+\tfrac{1}{2} \right)}{2\left( z-\tfrac{2k+1}{2} \right)}+O\left( 1 \right)$$
On the other hand we also have
$$\left( \psi \left( -2z \right)+\gamma \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)}\underset{2z\to 2k}{\mathop{=}}\,\frac{{{\left( -1 \right)}^{k}}f\left( k \right)}{2{{\left( z-k \right)}^{2}}}+\frac{{{\left( -1 \right)}^{k}}\left\{ {{H}_{2k}}f\left( k \right)+\tfrac{1}{2}f'\left( k \right) \right\}}{z-k}+O\left( 1 \right), k\ge 0$$
Similarly
$$\left( \psi \left( -2z \right)+\gamma \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)}\underset{z\to -k}{\mathop{=}}\,\frac{{{\left( -1 \right)}^{k}}\left( \psi \left( 2k \right)+\gamma \right)f\left( -k \right)}{z+k}, k>0$$
The only other residues are those due to $f$ which we assume has one pole at the origin of order at least $2$. The sum of residues over the entire plane is zero. Hence
$$\begin{align}
& \sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}{{H}_{2k}}f\left( k \right)} \\
& =-\frac{1}{2}\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}f'\left( k \right)}-\frac{\pi }{2}\sum\limits_{k=0}^{\infty }{{{\left( -1 \right)}^{k}}f\left( k+\tfrac{1}{2} \right)}-\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}\left( \psi \left( 2k \right)+\gamma \right)f\left( -k \right)}\\&-\underset{z=0}{\mathop{res}}\,\left( \psi \left( -2z \right)+\gamma \right)\frac{\pi f\left( z \right)}{\sin \left( \pi z \right)} \\
\end{align}$$
This goes a long way to generalise such results as you can now pick any power by simply choosing f appropriately. For example letting $f\left( z \right)=\frac{1}{{{z}^{4}}}$ we find
$$\begin{align}
& \sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k}}}{{{k}^{4}}}} \\
& =-\frac{\pi }{192}\left\{ {{\psi }^{\left( 3 \right)}}\left( \tfrac{1}{4} \right)-{{\psi }^{\left( 3 \right)}}\left( \tfrac{3}{4} \right) \right\}+\frac{2}{3}{{\pi }^{2}}\zeta \left( 3 \right)+\frac{113}{8}\zeta \left( 5 \right)-\sum\limits_{k=1}^{\infty }{{{\left( -1 \right)}^{k}}\frac{{{H}_{2k-1}}}{{{k}^{4}}}} \\
\end{align}$$
Now
$$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k}}}{{{k}^{4}}}}=A-\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k-1}}}{{{k}^{4}}}}\Rightarrow \sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\left( {{H}_{2k}}+{{H}_{2k-1}} \right)}{{{k}^{4}}}}=A$$
But we can write this as
$$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}\left( \frac{1}{2k}+2{{H}_{2k-1}} \right)}{{{k}^{4}}}}=A$$
hence
$$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k-1}}}{{{k}^{4}}}}=\frac{1}{2}A+\frac{15}{64}\zeta \left( 5 \right)$$
We have then
$$\sum\limits_{k=1}^{\infty }{\frac{{{\left( -1 \right)}^{k}}{{H}_{2k}}}{{{k}^{4}}}}=\frac{\pi }{384}\left\{ {{\psi }^{\left( 3 \right)}}\left( \tfrac{3}{4} \right)-{{\psi }^{\left( 3 \right)}}\left( \tfrac{1}{4} \right) \right\}+\frac{{{\pi }^{2}}}{3}\zeta \left( 3 \right)+\frac{437}{64}\zeta \left( 5 \right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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Evaluate $\int \sin^{-1}\frac{2x}{1+x^2}dx$ $\int \sin^{-1}\dfrac{2x}{1+x^2}dx$
My attempt is as follows:-
$$x=\tan\theta$$
$$dx=\sec^2\theta d\theta$$
$$\int \sin^{-1}(\sin2\theta) \cdot\sec^2\theta d\theta$$
So here should we make cases on the basis of values of $\theta$ or can we write $\sin^{-1}(\sin2\theta)$ as $2\theta$?
| Let $ \displaystyle x=\tan \frac{\theta}{2} \text{ then }\displaystyle d x=\frac{1}{2} \sec ^{2} \frac{\theta}{2} d \theta$ and
$\displaystyle \frac{2 x}{1+x^{2}}=\frac{2 \tan \frac{\theta}{2}}{1+\tan ^2\frac{\theta }{2}}=\sin \theta .$
$\therefore \displaystyle \int \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right) d \displaystyle x=\int \theta \cdot \frac{1}{2} \sec ^{2} \frac{\theta}{2} d \theta$
$= \displaystyle \int \theta d\left(\tan \frac{\theta}{2}\right)$
$=\displaystyle \theta \tan \frac{\theta}{2}-\int \tan \frac{\theta}{2} d \theta$
$\displaystyle =x \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)-2 \ln \left|\sec \frac{\theta}{2}\right| +C$
$=\displaystyle x \sin ^{-1}\left(\frac{2 x}{1+x^{2}}\right)-\ln \left(1+x^{2}\right)+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3495182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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} |
Evaluate $\int_{0}^{\frac{\pi}{6}}\frac{\sqrt{3\cos2x-1}}{\cos x}dx$ $$\int_{0}^{\frac{\pi}{6}}\dfrac{\sqrt{3\cos2x-1}}{\cos x}dx$$
$$\int_{0}^{\frac{\pi}{6}}\dfrac{\sqrt{3(1-\tan^2x)-(1+\tan^2x)}}{\sec x\cos x}dx$$
$$\int_{0}^{\frac{\pi}{6}}\sqrt{2-4\tan^2x}\quad dx$$
$$\sqrt{2}\int_{0}^{\frac{\pi}{6}}\sqrt{1-2\tan^2x}\quad dx$$
$$\tan x=t$$
$$\sec^2x=\dfrac{dt}{dx}$$
$$\sqrt{2}\int_{0}^{\frac{1}{\sqrt{3}}}\dfrac{\sqrt{1-2t^2}}{1+t^2}dt$$
$$t=\dfrac{\sin\theta}{\sqrt{2}}$$
$$\dfrac{dt}{d\theta}=\dfrac{1}{\sqrt{2}}\cos\theta$$
$$\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}\dfrac{\cos^2\theta}{1+\dfrac{\sin^2\theta}{2}}d\theta$$
$$\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}\dfrac{2\cos^2\theta}{2+\sin^2\theta}d\theta$$
$$\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}\dfrac{2(1-\sin^2\theta)}{2+\sin^2\theta}d\theta$$
$$\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}\dfrac{2}{2+\sin^2\theta}d\theta-\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}\dfrac{2\sin^2\theta+4-4}{2+\sin^2\theta}d\theta$$
$$\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}\dfrac{6}{2+\sin^2\theta}d\theta-\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}2d\theta$$
$$\int_{0}^{\sin^{-1}\frac{\sqrt{2}}{\sqrt{3}}}\dfrac{6\sec^2\theta}{2\sec^2\theta+\tan^2\theta}d\theta-2\sin^{-1}\dfrac{\sqrt{2}}{\sqrt{3}}$$
$$\int_{0}^{\sqrt{2}}\dfrac{6}{2+3y^2}dy-2\sin^{-1}\dfrac{\sqrt{2}}{\sqrt{3}}$$
$$\dfrac{6}{\sqrt{6}}\left(\tan^{-1}\dfrac{\sqrt{3}y}{
\sqrt{2}}\right)_{x=\sqrt{2}}-2\sin^{-1}\dfrac{\sqrt{2}}{\sqrt{3}}$$
$$\dfrac{\sqrt{2}}{\sqrt{3}}\pi-2\tan^{-1}\sqrt{2}$$
| Hint
$$3\cos2x-1-=3(1-2\sin^2x)-1$$
Use $\sin x=u$ to find $$I=\int\sqrt2\dfrac{\sqrt{1-3u^2}}{1-u^2}du$$
$$\dfrac{2I}{\sqrt2}=\int\dfrac{(1+u)\sqrt{1-3u^2}}{1-u^2}du+\int\dfrac{\sqrt{1-3u^2}(1-u)}{1-u^2}$$
For the first integral, set $1-u=y$ and $1+u=z$ in the second
| {
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"url": "https://math.stackexchange.com/questions/3496949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $z = \frac{(\sqrt{3} + i)^n}{(\sqrt{3}-i)^m}$, find the relation between $m$ and $n$ such that $z$ is a real number. I am given the following number $z$:
$$z = \dfrac{(\sqrt{3} + i)^n}{(\sqrt{3} - i)^m}$$
with $n, m \in \mathbb{N}$. I have to find a relation between the natural numbers $n$ and $m$ such that the number $z$ is real. I know that for a complex number to be real, its imaginary part must equal $0$, but I can't isolate the imaginary part. This is as far as I got:
$$\sqrt{3} + i =
2 \bigg (\frac{\sqrt{3}}{2} + i\frac{1}{2} \bigg) =
2 \bigg( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) $$
$$\sqrt{3} - 1 =
2 \bigg ( \dfrac{\sqrt{3}}{2} - i \dfrac{1}{2} \bigg) =
2 \bigg ( \cos \dfrac{\pi}{6} - i \sin \dfrac{\pi}{6} \bigg ) =
2 \bigg ( \cos \dfrac{11\pi}{6} + i \sin \dfrac{11\pi}{6} \bigg )$$
So I got the numerator and the denominator in a form that I can use DeMoivre's formula on. So, next I'd have:
$$z = \dfrac{\bigg [2 \bigg ( \cos \dfrac{\pi}{6} + i \sin \dfrac{\pi}{6} \bigg ) \bigg ]^n}
{\bigg [2 \bigg( \cos \dfrac{11 \pi}{6} + i \sin \dfrac{11 \pi}{6} \bigg ) \bigg ]^m }$$
$$z = 2^{n - m} \cdot \dfrac{\cos \dfrac{n \pi}{6} + i \sin \dfrac{n \pi}{6}}
{\cos \dfrac{11 m \pi}{6} + i \sin \dfrac{11 m \pi}{6}}$$
But this is where I got stuck. I still can't isolate the imaginary part of $z$ in order to equal it to $0$.
| Hint:
Use the complex exponential notation and congruences: your final fraction is none other than
$$z=2^{n-m}\frac{\mathrm e^{\tfrac{ni\pi}6}}{\mathrm e^{\tfrac{-mi\pi}6}}=2^{n-m} \mathrm e^{\tfrac{(n+m)i\pi}6},$$
anf it is a real number if and only if
$$\frac{(n+m) \pi}6\equiv 0\mod \pi\iff (n+m)\pi \equiv 0\mod 6\pi\iff n+m\equiv 0\mod 6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3498784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
Derivative of sum of two functions I have to find $\frac{dy}{dx}\left[(x\sqrt{x})+\frac{1}{x^2\sqrt{x}}\right]$ but would like to find where I made a mistake in my solution.
Here is my work:
\begin{align}
(f+g)'=& \ f'+g', f = x \sqrt{x}, g= \frac{1}{x^2 \sqrt x}\\ f=& \ x \cdot x^{1/2} = x^{3/2}
\\ f'=& \ \frac{3}{2x\sqrt{x}} \\ \\
g =& \ x^{-2}\cdot x^{-1/2} \\ g =& \ x^{-5/2} \\ g' =& \ -\frac{5}{2x^3 \sqrt{x}} \\ \\ f'+g' =& \ \frac{3}{2x\sqrt{x}} - \frac{5}{2x^3 \sqrt{x}} \\ =& \ \frac{3x^2-5}{2x^3 \sqrt x} \\ \frac{dy}{dx}=& \ \frac{\sqrt x (3x^2-5)}{2x^4}
\end{align}
| In general if $f(x)=x^{n}$ for some constant $n$ then $f'(x)=nx^{n-1}$ which is the power rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A Diophantine equation: solve $(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$ (without using Fermat's last theorem) Solve this Diophantine equation: $(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$
My attempt (use Fermat's last theorem)
$$(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$$
$$\Leftrightarrow (3x^2+y^2-4y-17)^3=(x^2-y^2-11)^3+(2x^2+2y^2-4y-6)^3$$
Use Fermat's last theorem, we get:
$$
\left\{
\begin{array}{c}
x^2-y^2-11=0 \\
3x^2+y^2-4y-17=2x^2+2y^2-4y-6
\end{array}
\right.$$
or
$$
\left\{
\begin{array}{c}
2x^2+2y^2-4y-6=0\\
3x^2+y^2-4y-17=x^2-y^2-11
\end{array}
\right.$$
Then we continue...
My question is, is there another way to solve that other than Fermat's last theorem? I see that using Fermat's last theorem is like crack a nut by a sledgehammer.
| $$(3x^2+y^2-4y-17)^3-(2x^2+2y^2-4y-6)^3=(x^2-y^2-11)^3$$
If we put $A=3x^2+y^2-4y-17$ and $B=2x^2+2y^2-4y-6$, observe that $A-B=x^2-y^2-11$ (the factor on the right).
In other words, we have $a^3-b^3=(a-b)^3$
Expand this to get $a^3-b^3=a^3-3a^2b+3ab^2-b^3$
So $3ab(a-b)=0$ is the simplified equation. You can then show that $a=0, b=0$ or $a-b=0$ has no integer solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3502327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of ways to pair people but with restrictions I have a question regarding combinatorics that I hope you guys can help me clarify:
There are 14 people. 3 of them are TAs. The professor wants to group them into pairs such that no TAs are paired with each other. Find the number of ways to have such arrangement.
So what I'm thinking is that we can use complementary. Without any restrictions, we can have $\frac{14!}{2^7 7!}$ ways to arrange people.
Now we have to think about how to arrange people such that at least one pair is a TA. Suppose the first three people are TA.
So there are 2 ways to pair the first TA with another one. For the second pair is the 3rd TA, so there are 11 ways to select a non-TA. Continue we have $2\cdot 9\cdot7 ... 1$ ways.
So there are $\frac{14!}{2^7 7!} - 2\cdot 9\cdot7 ... 1$ ways. Is this correct? Thanks a lot!
| Method 1: Line up the TA's in alphabetical order. Do the same for the students.
There are $11$ ways to select a student to pair with the first TA, $10$ ways to select a student to pair with the second TA, and $9$ ways to select a student to pair with the third TA. Remove those students from the line. That leaves us with eight students. There are $7$ ways to pair a student with the first student remaining in the line. Remove that pair. There are $5$ ways to pair a student with the first student remaining in the line. Remove that pair. There are $3$ ways to pair a student with the first student remaining in the line. Remove that pair. The two students remaining in the line form the final pair. Hence, there are
$$11 \cdot 10 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$$
ways to group $11$ students and $3$ TA's into pairs so that no two TA's are in the same pair.
Method 2: We correct your approach.
We subtract the number of pairs in which two TA's form a pair from the number of ways the $14$ people could be grouped into pairs.
To count the number of ways $14$ people could be grouped into pairs, line up the $14$ people in some order, say alphabetically. There are $13$ ways to match a person with the first person in line. Remove that pair. There are $11$ ways to match a person with the first person remaining in line. Remove that pair. There are $9$ ways to match a person with the first person remaining in line. Remove that pair. Continuing in this way, we see that there are
$$13!! = 13 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$$
ways to group $14$ people into pairs.
There are $\binom{3}{2}$ ways to select a pair of TA's to be in a group together. That leaves $12$ people to be grouped into pairs. Reasoning as above, there are
$$11!! = 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$$
ways to group them into pairs, so there are
$$\binom{3}{2}11!! = 3 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1$$
ways to group the $14$ people into pair in such a way that two of the TA's are in the same pair.
Hence, the number of ways to group $11$ students and $3$ TA's into pairs so that no two TA's are in the same pair is
\begin{align*}
13!! - \binom{3}{2}11! & = 13 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1 - 3 \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1\\
& = (13 - 3) \cdot 11 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1\\
& = 11 \cdot 10 \cdot 9 \cdot 7 \cdot 5 \cdot 3 \cdot 1
\end{align*}
which agrees with the result we obtained above.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find all natural solutions $(a, b)$ such that $(ab - 1) \mid (a^2 + a - 1)^2$.
Find all natural solutions $(a, b)$ such that $$\large (ab - 1) \mid (a^2 + a - 1)^2$$
We have that $$(ab - 1) \mid (a^2 + a - 1)^2 \implies (ab - 1) \mid [(ab)^2 - ab^2 - b^2]^2$$
$$\iff (ab - 1) \mid (ab^2 + b^2 + 1)^2 \iff (ab - 1) \mid (b^2 + b + 1)^2$$
I'm trying to prove that $(ab - 1) \mid (a + b - 1)^2$, yet I don't know how with the information presented.
Assuming that I know how to determine that $(ab - 1) \mid (a + b - 1)^2$. Let $$(a + b - 1)^2 = k(ab - 1), k \in \mathbb Z^+ \tag 1$$
where $(a, b)$ is the solution in which $a + b$ is at its minimal value.
$$\iff a^2 - [(k - 2)b + 2]a + [(b - 1)^2 + k] = 0$$
We have that the equation $$x^2 - [(k - 2)b + 2]a + [(b - 1)^2 + k] = 0$$ has two solutions $x = a$ and $x = a'$ such that $$a + a' = (k - 2)b + 2, aa' = (b - 1)^2 + k$$
It can easily be deduced that $a' \in \mathbb Z^+ \implies (a', b)$ is a solution to $(1)$
$\implies a' + b \ge a + b \iff a' \ge a \implies \dfrac{(b - 1)^2 + k}{a} \ge a \iff (b - 1)^2 + k \ge a^2$
It seems to me that there are infinitely many solutions, which are all consecutive elements in a sequence.
Furthermore, the assumption that $(ab - 1) \mid (a + b - 1)^2$ is incorrect. So I don't know what to begin from here.
|
Let $(a,b)$ be a solution of $$ (ab - 1) \mid (a^2 \pm a - 1)^2$$
then $(b,c)$ is a solution of $$ (bc-1) \mid (b^2 \mp b - 1)^2,$$ where $$c=\frac{(b^2\mp b-1)^2+ab-1}{b(ab-1)}.$$
Proof
Let $N=ab-1$. Then $0\equiv (a^2b^2\pm ab^2-b^2)^2\equiv (b^2\mp b-1)^2$ (mod $N$). Therefore $(b^2\mp b -1)^2=NM$ for some natural number $M$. Then $M\equiv -1$ (mod $b$) and so there is a natural number $c$ such that $M=bc-1.$
Application
This gives us a formula for generating solutions with every other iteration giving a solution of the original equation. Solutions $(a,a+1)$ just cycle round but the solution $(2,1)$ generates an infinite set of solutions containing those obtained by @MartinR.
In fact all solutions other than $(a,a+1)$ are generated from $(2,1)$.
A proof there are no other solutions
Let $$(a^2\pm a-1)^2=(ab-1)(ac-1).$$ If either $b$ or $c$ is less than $a$ then the described procedure can be used to give us a smaller solution. Otherwise we have $b,c\ge a$.
If $b=a$ then, for $N=ab-1$, we have $a^2\equiv 1$ (mod $N$) and then $a\equiv 0$ (mod $N$). The only possibility is $N=1$ and we have reached the base case.
Otherwise $b,c\ge a+1$ and the only possibility is $b=c=a+1$.
The solutions are as follows
$$\begin{matrix} (2, 13)&& (13, 74)\\ (74, 433)&& (433, 2522)\\ (2522, 14701)&& (14701, 85682)\\ (85682, 499393)&& (499393, 2910674)\\ (2910674, 16964653)&& (16964653, 98877242)\\ \end{matrix}$$
$$\cdots$$
As described in the above answer every other pair gives a solution of the original equation. The remaining pairs if reversed also give solutions but these simply form part of the other solutions. E.g. $(2,13)$ and $(74,433)$ are successive solutions. $(74,13)$ is also a solution which is the 'other half' of the $(74,433)$ one.Viz. $$(74^2+74-1)^2=(74\times 13-1)(74\times 433-1).$$
| {
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"url": "https://math.stackexchange.com/questions/3505969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Denesting $\sqrt{20+ \sqrt{96}+\sqrt{12}}$ into four possible radicals Denesting $\sqrt{20+ \sqrt{96}+\sqrt{12}}$ into possible radicals. This is an answer to an obscure closed question, here on the site. While there is an answer posted, it isn't the complete solution.
| $$\begin{align}
\sqrt{20+ \sqrt{96}+\sqrt{12}} &= \sqrt{5\cdot4+ 4\sqrt{6}+2\sqrt{3}} \\
&= \sqrt{2}\sqrt{5\cdot2+ 2\sqrt{6}+\sqrt{3}} \\
&= \sqrt{2}\sqrt{5\cdot2+\sqrt{3}\left(2\sqrt{2}+1\right)} \\
\end{align}$$
By Using the identity of denesting $\sqrt{a\pm b\sqrt{c}}=\sqrt{(a+d)/2} \pm\sqrt{(a-d)/2}$, where $d=\sqrt{a^2-b^2c}$.
$$\begin{align}
\sqrt{2}\sqrt{5\cdot2+\sqrt{3}\left(2\sqrt{2}+1\right)} &= \sqrt{2}\left(\sqrt{\frac{10+\sqrt{10^{2}-3\left(2\sqrt{2}+1\right)^{2}}}{2}}+\sqrt{\frac{10-\sqrt{10^{2}-3\left(2\sqrt{2}+1\right)^{2}}}{2}}\right) \\
&= \sqrt{2}\left(\sqrt{\frac{10+\sqrt{73-12\sqrt{2}}}{2}}+\sqrt{\frac{10-\sqrt{73-12\sqrt{2}}}{2}}\right) \\
&= \sqrt{10+\sqrt{73-12\sqrt{2}}}+\sqrt{10-\sqrt{73-12\sqrt{2}}} \\
\end{align}$$
Denesting $\sqrt{73-12\sqrt{2}}$ using the above mentioned way,
$$\begin{align}
\sqrt{10+\sqrt{73-12\sqrt{2}}}+\sqrt{10-\sqrt{73-12\sqrt{2}}} &= \sqrt{10+\sqrt{72}-1}+\sqrt{10-\left(\sqrt{72}-1\right)}\\
&= \sqrt{9+6\sqrt{2}}+\sqrt{11-6\sqrt{2}} \\
&= \sqrt{6}+\sqrt{3}+3-\sqrt{2} \\
\end{align}$$
Hence,
$$\sqrt{20+ \sqrt{96}+\sqrt{12}}=\sqrt{6}+\sqrt{3}+3-\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3511485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Irreducibility of Polynomials without Eisenstein In order to do a different task, I need to show that $f:=X^4-10X^2+1$ is irreducible over $\mathbb{Q}$ and that $g:=X^2Y^2-2Y^2-3X+6$ is irreducible over $\mathbb{Q}[X]$. Since I don't think that Eisenstein criterion works here, I was wondering if there is another simple method to check that. Thanks in advance for any help!
| For the first one, there is a simplification due to https://en.wikipedia.org/wiki/Gauss%27s_lemma_(polynomial)
We see quickly that there are no rational roots. We are left with showing that the quartic is not the product of two quadratics with integer polynomials. The choices are
$$ (x^2 + Ax + 1)(x^2 + Bx + 1) $$ or
$$ (x^2 + Cx - 1)(x^2 + Dx - 1) .$$
From the coefficients of $x^3$ and $x$ being zero, we get $A+B = C+D = 0.$ Alright,
$$ (x^2 + Ax + 1)(x^2 -Ax + 1) $$ or
$$ (x^2 + Cx - 1)(x^2 -Cx - 1) .$$
All that is left is to get the coefficient $-10$ of $x^2,$ this may or may not be possible.
We reach $$ 2 - A^2 = -10, $$
$$ -2 - C^2 = -10, $$ for
$$ 12 = A^2, $$
$$ 8 = C^2 $$
Neither is possible with integers. You might try confirming that
$$ (x^2 + x \sqrt {12} + 1)(x^2 -x \sqrt {12} + 1) = x^4 - 10 x^2 + 1.$$
$$ (x^2 + x \sqrt 8 - 1)(x^2 -x \sqrt 8 - 1) = x^4 - 10 x^2 + 1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Suppose that $z\in\mathbb C$ with $|z^2+1|\le 1$. How to prove $|z+1|\geq\frac12$. Let $z\in\mathbb C$ with $|z^2+1|\le 1$. I want to prove $|z+1|\geq\frac12$.
I noticed that $|z^2+1|\le 1$ means that $z$ lies in a cassini oval. I tried with the substitution $z=r\exp(i\theta)$ where $r\geq0, \theta\in[0,2\pi]$. The constraint now is $\frac{r^2}2\le-\cos(2\theta)$ and we need to prove $$\sqrt{r^2 \sin^2(\theta) + (r \cos(\theta) + 1)^2}=\sqrt{r^2+2r\cos\theta+1}\geq\frac12.$$
How can I do that?
| Let $z=a+ib$, then $$|1+z^2|^2=1+2a^2+a^4+b^2(2a^2-2)+b^4\le1.$$ We want to minimize $|1+z|^2=\color{orange}{(1+a)^2+b^2}$. Let me complete the square in the constraint: $$b^4+b^2(2a^2-2)\color{blue}{+(a^2-1)^2}\le\color{blue}{(a^2-1)^2}-a^4-2a^2=1-4a^2$$ which is the same as $$(b^2+(a^2-1))^2\le1-4a^2.$$ In particular, it follows that $- \frac12<a<\frac12$ and $$b^2\in\left[1-a^2-\sqrt{1-4a^2},1-a^2+\sqrt{1-4a^2}\right].$$
Thus, $$2+2a-\sqrt{1-4a^2}\le\color{orange}{(1+a)^2+b^2}\le2+2a+\sqrt{1-4a^2}.$$ We will thus want to minimize the function $2a-\sqrt{1-4a^2}$. This can be done for example with calculus, but we can also proceed as follows:
Now we have $$8a^2-4 \sqrt 2 a+1=8 \left(a - \frac{1}{2 \sqrt 2}\right)^2\geq0$$ which implies $$4a^2-4\sqrt 2a+2=(\sqrt 2-2a)^2\geq1-4a^2$$ and thus $$\sqrt 2-2a\geq\sqrt{1-4a^2}.$$ By replacing $a$ by $-a$ we also get $$\sqrt{1-4a^2}\le\sqrt2+2a.$$
It follows from the previous result that $$2-\sqrt 2\le\color{orange}{(1+a)^2+b^2}\le2+\sqrt 2$$ and thus in particular $$\bbox[15px,border:1px groove navy]{|1+z|=\sqrt{\color{orange}{(1+a)^2+b^2}}\geq\sqrt{2-\sqrt 2}>\frac12.}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Let $f(1)= 4, f'(x)= \sum_{k=0}^∞ \frac{(-1)^{k} (x-1)^{k}}{(k+1)!}$ then find $f''(1)$. Let $f(1)= 4, f'(x)= \sum_{k=0}^∞ \frac{(-1)^{k} (x-1)^{k}}{(k+1)!}$ then find $f''(1)$.
I directly differentiated $f'(x)$ and put $x=1$, then I got $0$.
If I expand the series $f'(x)$ then then coefficient of $(x-1)$ is $f"(1)$, which is $-1/2$.
Answer is given $-1/2$.
What is the problem in direct differentiation$?$
And how do we find derivative at a point in such cases$?$
| Your sum is
$$f'(x) = \sum_{k=0}^{\infty} \frac{(-1)^{k} (x-1)^{k}}{(k+1)!} \tag{1}\label{eq1A}$$
Differentiating this term by term gives
$$f''(x) = \sum_{k=0}^{\infty} \frac{k(-1)^{k} (x-1)^{k-1}}{(k+1)!} \tag{2}\label{eq2A}$$
The first term, i.e., for $k = 0$, is $0$. The second term, i.e., for $k = 1$, is $\frac{(-1)^{1}(x-1)^{1-1}}{(1+1)!} = \frac{-1(1)}{2!} = -\frac{1}{2}$ since $(x - 1)^{0} = 1$. All other terms have a factor of $x - 1$ to a power of $1$ or higher so, at $x = 1$, they would all be $0$, thus resulting in
$$f''(1) = -\frac{1}{2} \tag{3}\label{eq3A}$$
which matches the given answer. If I understood your question correctly, this confirms what you also got.
Note the original derivative series is basically a modified version of the Taylor series for the exponential function (and, thus, as explained in the Exponential function section of Wikipedia's "Uniform convergence" article, the result is uniformly convergent, with this being a requirement for term by term differentiation, as done in \eqref{eq2A}). To see this, you have
$$\begin{equation}\begin{aligned}
f'(x) & = \sum_{k=0}^{\infty} \frac{(-1)^{k} (x-1)^{k}}{(k+1)!} \\
& = \sum_{k=0}^{\infty} \frac{(1-x)^{k}}{(k+1)!} \\
& = \frac{1}{1-x}\left(\sum_{k=0}^{\infty} \frac{(1-x)^{k+1}}{(k+1)!}\right) \\
& = \frac{1}{1-x}\left(\sum_{k=1}^{\infty} \frac{(1-x)^{k}}{k!}\right) \\
& = \frac{1}{1-x}\left(\sum_{k=0}^{\infty} \frac{(1-x)^{k}}{k!} - 1\right) \\
& = \frac{1}{1-x}\left(e^{1-x} - 1\right) \\
& = \frac{e^{1-x} - 1}{1-x}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
I assume you used this closed form expression when you stated you directly differentiated $f'(x)$ and then substituted $x = 1$. Well, differentiating \eqref{eq4A} gives
$$\begin{equation}\begin{aligned}
f''(x) & = \frac{-e^{1-x}}{1-x} + \frac{(-1)(e^{1-x} - 1)}{(-1)(1-x)^2} \\
& = \frac{(-e^{1-x})(1-x) + (e^{1-x} - 1)}{(1-x)^2} \\
& = \frac{x\left(e^{1-x}\right) - 1}{(1-x)^2}
\end{aligned}\end{equation}\tag{5}\label{eq5A}$$
Since $f''(x)$ is indeterminate at $x = 1$, I'll use L'Hôpital's rule twice to get
$$\begin{equation}\begin{aligned}
f''(1) & = \lim_{x \to 1}\frac{x\left(e^{1-x}\right) - 1}{(1-x)^2} \\
& = \lim_{x \to 1}\frac{\left(1 - x\right)\left(e^{1-x}\right)}{2(1-x)(-1)} \\
& = \lim_{x \to 1}\frac{\left(x - 2\right)\left(e^{1-x}\right)}{2} \\
& = -\frac{1}{2}
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$
As you can see, the same value is obtained this way as well.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve inverse trigonometric equation $\frac{\pi}{6}=\tan^{-1} \frac{11}{x} -\tan^{-1} \frac{1}{x}$ How do I go about solving for $x$ when I have:
$\frac{\pi}{6}=\tan^{-1} \left( \frac{11}{x} \right)-\tan^{-1}\left( \frac{1}{x} \right)$.
| Let $y = 1/x$. Then solve
$$\pi/6 = \tan^{-1} (11 y) - \tan^{-1} y$$
or
$$\pi/6 + \tan^{-1} y = \tan^{-1} (11 y)$$
to find $y =\frac{5 \sqrt{3}}{11}-\frac{8}{11}$, making $x$ easy to determine.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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probability that two of these boxes contain exactly $2$ and $3$ balls is
If $10$ different balls are to be placed in $4$ distinct boxes at random, then the probability that two of these boxes contain exactly $2$ and $3$ balls is
What I tried:
Total number of ways $\displaystyle 4^{10}$
probability that two of these boxes contain exactly $2$ and $3$ balls is
$\displaystyle \binom{4}{2}\cdot 2^5$
So required probability $$\frac{\binom{4}{2}\cdot 2^5}{4^{10}}$$
But answer given as $\displaystyle \frac{945}{2^{10}}$
| The given answer is wrong! The correct one has been found by NCh.
If the question was: find the probability that box 1 contains exactly 3 balls and box 2 contains exactly 2 balls then result is
$$p:=\frac{\binom{10}{3}\cdot\binom{7}{2}\cdot 2^5}{4^{10}}=\frac{315}{2^{12}}$$
$\binom{10}{3}$ ways to choose the balls to be put in box 1, $\binom{7}{2}$ ways to choose the balls to be put in box 2, and $2^5$ ways to place the $5$ remaining balls in box 3 and 4.
If we multiply the probability $p$ by $4\cdot 3$, the number ordered couples of boxes, we find the given result
$$\frac{945}{2^{10}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513330",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Find area of ellipse $5x^2 -6xy +5y^2=8$
Find the area of ellipse whose equation in the $xy$- plane is given by $5x^2 -6xy +5y^2=8$
My attempt : I know that area of ellipse $ = \pi a b$ ,where $a$ is semi-major axis and $b$ is semi minor axis
Now if we make matrix $\begin{bmatrix} 5 & -3 \\-3& 5\end{bmatrix}$ Here eigenvalue $\lambda_1= 8 ,\lambda_2=2$
That is area of ellipse$ = \pi \frac{1}{\sqrt\lambda_2} \frac{1}{\sqrt\lambda_1}= \pi \frac{1}{2\sqrt 2}\frac{1}{\sqrt2}$
Is its correct ?
| Under the variable changes $x = \frac{u+v}{\sqrt2}$ and $y = \frac{u-v}{\sqrt2}$, the equation $5x^2 -6xy +5y^2=8$ is of the standard ellipse form
$$\frac{u^2}4+v^2=1$$
with the major and minor axes 2 and 1, respectively. Thus, the area is $2\pi$.
| {
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"url": "https://math.stackexchange.com/questions/3514276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How can I factor out $x^2+ax-2 (3a+1)(2a+1)$? I have tried it this way but could not factor it out :
$$x^2+ax-2 (3a+1)(2a+1)$$
Here, let
$$3a+1=m\tag{i}$$
$$2a+1=n\tag{ii}$$
So, (i)-(ii) gives us $$m-n=a$$ Then
$$\begin{align}
x^2+ax-2(3a+1)(2a+1) &= x^2+(m-n)x-2mn \\
&= x^2+mx-nx-mn-mn \\
&= x(x+m)-n(x+m)-mn \\
&= (x+m)(x-n)-mn\end{align}$$
And then I got stuck.Please help!
Thanks in advance.
| Writing out the two equations
$$
x^2+ax-2 (3a+1)(2a+1)=(x-b)(x-c)=x^2-(b+c)x+bc
$$
and comparing coefficients gives the equations
$$
a+b+c=0,\quad 12a^2 +10a +bc + 2=0.
$$
Substituting $c=-b-a$ and solving for $b$ we obtain
$$
b=\frac{\pm \sqrt{49a^2 + 40a + 8} - a}{2}
$$
Edit: Following the suggestion of lab bhattacharjee, cancelling the factor $2$, we obtain the much nicer solutions
$$
b=2a+1, \text{ or } b=-3a-1
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{x}{\sqrt{1+x^2}}<\arctan\left(x\right)$ for every $x>0$
Prove $\frac{x}{\sqrt{1+x^2}}<\arctan\left(x\right)$ for every $x>0$
I thought about MVT , what i did is :
let $g(x) = \arctan\left(x\right) $
so by MVT there is $c\in (0,x)$ such that :
$g'(c) = \frac{g\left(x\right)-g\left(0\right)}{x-0} \implies \frac{1}{c^2+1} = \frac{\arctan\left(x\right)}{x}$
now i need to prove that $\frac{x}{c^2+1}>\frac{x}{\sqrt{x^2+1}}$
any hint how to prove this ?
thanks
| Try comparing derivatives and integrating the difference from 0 to $x$.
$$\frac{\sqrt{x^2+1}-\frac{x^2}{\sqrt{1+x^2}}}{1+x^2}=\frac{1}{(1+x^2)^{\frac{3}{2}}}<\frac{1}{(1+x^2)}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$, show that $\csc x \cot x=\frac{a^2-b^2}{4ab}$
If $$\frac{a+b}{\csc x}=\frac{a-b}{\cot x}$$
show that
$$\csc x \cot x=\frac{a^2-b^2}{4ab}$$
(original problem image)
I tried getting rid of the denominator and then expanding the given equation, but couldn't get to the answer.
| From $1+\cot^2 x= \csc^2 x$ we have
\begin{eqnarray*}
(\color{red}{\csc x} +\cot x)(\csc x -\color{blue}{\cot x}) = 1
\end{eqnarray*}
Now sub
\begin{eqnarray*}
\color{red}{\csc x} = \frac{a+b}{a-b} \cot x
\end{eqnarray*}
and
\begin{eqnarray*}
\color{blue}{\cot x} = \frac{a-b}{a+b} \csc x .
\end{eqnarray*}
This gives
\begin{eqnarray*}
\cot x \left( \frac{a+b}{a-b} +1 \right) \csc x \left( 1-\frac{a-b}{a+b}\right) =1
\end{eqnarray*}
and the result follows with a bit of algebra.
| {
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"timestamp": "2023-03-29T00:00:00",
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How many spanning trees does this graph contains? How many spanning tree does below graph contains?
I tried to calculate by using edge deletion / contraction approach and I am getting 48:
I tried to calculate it by Kirchoffs method as a determinant of Laplacian matrix:
$$ \left(\begin{matrix}2&0&0&0&0&0&0&0\\0&3&0&0&0&0&0&0\\0&0&3&0&0&0&0&0\\0&0&0&2&0&0&0&0\\0&0&0&0&2&0&0&0\\0&0&0&0&0&3&0&0\\0&0&0&0&0&0&3&0\\0&0&0&0&0&0&0&2\end{matrix}\right)
-
\left(\begin{matrix}0&1&0&0&1&0&0&0\\1&0&1&0&0&1&0&0\\0&1&0&1&0&0&1&0\\0&0&1&0&0&0&0&1\\1&0&0&0&0&1&0&0\\0&1&0&0&1&0&1&0\\0&0&1&0&0&1&0&1\\0&0&0&1&0&0&1&0\end{matrix}\right)
=
\left(\begin{matrix}2&-1&0&0&-1&0&0&0\\-1&3&-1&0&0&-1&0&0\\0&-1&3&-1&0&0&-1&0\\0&0&-1&2&0&0&0&-1\\-1&0&0&0&2&-1&0&0\\0&-1&0&0&-1&3&-1&0\\0&0&-1&0&0&-1&3&-1\\0&0&0&-1&0&0&-1&2\end{matrix}\right)
$$
I tried to calculate determinant of first co factor, which is Laplacian matrix using wolfram alpha, and it was giving answer 116.
So, I was guessing which is correct answer and where did I make mistake?
PS: I dont have enough reputation to add images inline.
| If you delete BF and CG then you need to delete one of the remaining $8$ edges.
If you delete BF but not CG then you need to delete one of the $5$ edges of the LH cycle and one of the $3$ edges of the RH cycle. Similarly for deleting CF but not BF.
If you delete neither BF nor CG then you need to delete one of the $3$ edges for each of the LH and RH cycles and one of BC and FG.
$8+2\times 3\times5+ 3\times3\times2 =56 .$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculate matrix by using Cayley-Hamilton theorem Calculate matrix $B = A^{10}-3A^9-A^2+4A$ using Cayley-Hamilton theorem on $A$.
$$A = \begin{pmatrix}
2 & 2 & 2 & 5 \\
-1 & -1 & -1 & -5 \\
-2 & -2 & -1 & 0 \\
1 & 1 & 3 & 3
\end{pmatrix}$$
Now, I've calculated the characteristic polynomial of $A$:
$P_A(\lambda) = \lambda^4-3\lambda^3+\lambda^2-3\lambda$
So I know that $P(A) = 0 \rightarrow A^4-3A^3+A^2-3A = 0$, hereby $0$ is a $4 \times 4$ matrix.
$B = A^{10}-3A^9-A^2+4A = A^4 \cdot A^6 - 3A^3 \cdot A^6 + A^2 \cdot (-1) -3A + 7A $
Can I go further from here?
I tried doing polynomial division $B/P(A)$, but i stopped halfway since the numbers were getting too big and it didn't seem to get me to the right solution.
How do I transform $B$, so that I can use $P(A)=0$ and calculate B.
| We know that $A^4-3A^3=3A-A^2$. Now,
\begin{align*}
B&=A^{10}-3A^9-A^2+4A\\
&=A^6(A^4-3A^3)-A^2+4A\\
&=A^6(3A-A^2)-A^2+4A\\
&=3A^7-A^8-A^2+4A\\
&=-A^4(A^4-3A^3)-A^2+4A\\
&=-A^4(3A-A^2)-A^2+4A\\
&=-3A^5+A^6-A^2+4A\\
&=A^2(A^4-3A^3)-A^2+4A\\
&=A^2(3A-A^2)-A^2+4A\\
&=-A^4+3A^3-A^2+4A\\
&=A
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Non-homogeneous cyclic $\frac{x+1}{\sqrt{x+y}}+\frac{y+1}{\sqrt{y+z}}+\frac{z+1}{\sqrt{z+x}} \geq 3\sqrt{2}$
Consider three non-negative real numbers $x$, $y$ and $z$, no two of
which are zero at the same time. Prove that the following inequality
holds:
$$\frac{x+1}{\sqrt{x+y}}+\frac{y+1}{\sqrt{y+z}}+\frac{z+1}{\sqrt{z+x}} \geq 3\sqrt{2}$$
My progress: This inequality is cyclic, so we should consider two possibilities $x \geq y\geq z$ and $x \leq y\leq z$. The first case is relatively simple because from AM-GM:
$$\sum \frac{x+1}{\sqrt{2(x+y)}} \geq \sum \frac{2(x+1)}{x+y+2}$$
and if we let $a = x+1,\ b=y+1,\ c= z+1$, ($a\geq b\geq c$), we have:
$$2\sum \frac{a}{a+b}-3 = \frac{(a-b)(b-c)(a-c)}{(a+b)(b+c)(c+a)} \geq 0$$
However, I believe the second case ($x\leq y \leq z$) is very difficult to prove, and I am not sure if this is the way to proceed.
| Remarks: The inequality (1) is @arqady's idea. Although the original inequality is not homogeneous, (1) is. Hope to see a nice proof of (1).
A proof utilizing arqady@AoPS's nice idea.
Since the desired inequality is cyclic,
assume that $z = \min(x, y, z)$.
The desired inequality is written as
$$\sum_{\mathrm{cyc}} \frac{x}{\sqrt{x + y}} + \sum_{\mathrm{cyc}} \frac{1}{\sqrt{x + y}} \ge 3\sqrt 2.$$
Squaring both sides, it suffices to prove that
$$\left(\sum_{\mathrm{cyc}} \frac{x}{\sqrt{x + y}} + \sum_{\mathrm{cyc}} \frac{1}{\sqrt{x + y}}\right)^2 \ge 18.$$
Using $(A + B)^2 \ge 4AB$, it suffices to prove that
$$4 \sum_{\mathrm{cyc}} \frac{x}{\sqrt{x + y}} \cdot \sum_{\mathrm{cyc}} \frac{1}{\sqrt{x + y}} \ge 18$$
or (simplifying)
$$\sum_{\mathrm{cyc}} \frac{x}{x + y} + \sum_{\mathrm{cyc}} \sqrt{\frac{x + y}{y + z}} \ge \frac{9}{2} \tag{1}.$$
(1) is written as
$$\sum_{\mathrm{cyc}} \sqrt{\frac{x + y}{y + z}}\ge 3 + \frac{(y - x)(x - z)(y - z)}{2(x + y)(y + z)(z + x)}.$$
Using AM-GM, we have $\mathrm{LHS} \ge 3$.
Thus, we only need to prove the case when $y \ge x \ge z$.
Using $\sqrt{u} \ge \frac{1 + 3u}{3 + u}$ for all $u \ge 1$, we have
\begin{align*}
\sqrt{\frac{x + y}{y + z}} &\ge \frac{3x + 4y + z}{x + 4y + 3z}, \\
\sqrt{\frac{y + z}{z + x}} &\ge
\frac{x + 3y + 4z}{3x + y + 4z}.
\end{align*}
Also, we have
$$\sqrt{\frac{z + x}{x + y}}
= \left(\sqrt{\frac{x + y}{y + z}\cdot\frac{y + z}{z + x}}\right)^{-1}
\ge \left(\frac{\frac{x + y}{y + z} + \frac{y + z}{z + x}}{2}\right)^{-1}.$$
Thus, it suffices to prove that
$$\frac{3x + 4y + z}{x + 4y + 3z}
+ \frac{x + 3y + 4z}{3x + y + 4z} + \left(\frac{\frac{x + y}{y + z} + \frac{y + z}{z + x}}{2}\right)^{-1} \ge 3 + \frac{(y - x)(x - z)(y - z)}{2(x + y)(y + z)(z + x)}.$$
Letting $x = z + s, y = z + s + t$ for $s, t \ge 0$, we have
$$\mathrm{LHS} - \mathrm{RHS} = \frac{f(z, s, t)}{g(z, s, t)}$$
where $f(z, s, t)$ and $g(z, s, t)$ are both polynomials
with non-negative coefficients. The inequality is true.
We are done.
| {
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Given that $\alpha + \beta - \gamma = \pi$, prove that $\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = 2 \sin \alpha \sin \beta \cos \gamma$. I am told:
$$\alpha + \beta - \gamma = \pi$$
And I have to prove:
$$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma = 2 \sin \alpha \sin \beta \cos \gamma$$
What should I be looking for? I kept trying to take the sine of bots sides and use the formulas:
$$\sin(a + b) = \sin a \cos b + \sin b \cos a$$
$$\sin(a-b) = \sin a \cos b - \sin b \cos a$$
but got nowhere. Then I tried using the formulas:
$$\sin a + \sin b = 2 \sin \bigg ( \dfrac{a + b}{2} \bigg ) \cos\bigg ( \dfrac{a - b}{2} \bigg )$$
$$\sin a - \sin b = 2 \cos \bigg ( \dfrac{a + b}{2} \bigg ) \sin \bigg ( \dfrac{a - b}{2} \bigg )$$
But again, I got nowhere. Can you give me a hint? At least what should I be looking for? What should be my strategy? Everything that I did felt just random, while kind of hoping that everything would just magically turn into the desired result. What is the strategy for this kind of problem?
| $$\sin^2 \alpha + \sin^2 \beta - \sin^2 \gamma-2 \sin \alpha \sin \beta \cos \gamma=$$
$$=\sin^2 \alpha + \sin^2 \beta -1+\cos^2(\alpha+\beta)+2 \sin \alpha \sin \beta \cos (\alpha+\beta)=$$
$$=\sin^2 \alpha-\cos^2 \beta+\cos(\alpha+\beta)\cos(\alpha-\beta)=$$
$$=\sin^2 \alpha-\cos^2 \beta+\cos^2\alpha\cos^2\beta-\sin^2\alpha\sin^2\beta=$$
$$=(1-\sin^2\beta)\sin^2\alpha-(1-\cos^2\alpha)\cos^2\beta=0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x(3+x)}{\sqrt{3-x^2}(3+x)^2}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
$\dfrac{-x}{\sqrt{3-x^2}(3+x)}-\dfrac{\sqrt{3-x^2}}{(3+x)^2}$
At this point, I want to transform this derivative into the form of $\dfrac{3(x+1)}{(3+x)^2\sqrt{3-x^2}}$
How do I do this? This form is given by Wolfram:
https://www.wolframalpha.com/input/?i=derivative+%283-x%5E2%29%5E%281%2F2%29%2F%283%2Bx%29
| Your derivative of $\sqrt{3 - x^2}$ is incorrect.
$$\begin{align}\dfrac{\mathrm d}{\mathrm dx}\dfrac{\sqrt{3 - x^2}}{3 + x^2} &= \dfrac{\frac{\mathrm d}{\mathrm dx}\sqrt{3 - x^2}\cdot(3 + x) - \sqrt{3 - x^2}\cdot\frac{\mathrm d}{\mathrm dx}(3 + x)}{(3 + x)^2} \\ &= \dfrac{\frac{-2x}{2\sqrt{3 - x^2}}(3 + x) - \sqrt{3 - x^2}}{(3 + x)^2} \\ &= \dfrac{-x(3 + x) - (3 - x^2)}{(3 + x)^2\sqrt{3 - x^2}} \\ &= \dfrac{x^2 - x^2 - 3x - 3}{(3 + x)^2\sqrt{3 - x^2}} = -\dfrac{3(x + 1)}{(3 + x)^2\sqrt{3 - x^2}}\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Minimum value of $\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$ If $$x^2+y^2=1$$
Prove that Minimum value of $$f(x,y)=\frac{ax^2+by^2}{\sqrt{a^2x^2+b^2y^2}}$$ is
$$\frac{2\sqrt{ab}}{a+b}$$
My try:
I used basic Trigonometry:
Let $x=\cos t$ and $y=\sin t$
Then we get $$f(x,y)=g(t)=\frac{a\cos^2 t+b\sin^2 t}{\sqrt{a^2\cos^2 t+b^2\sin^2 t}}$$
Now let $$p=\cos(2t)$$
then we get a single variable function as:
$$h(p)=\frac{1}{\sqrt{2}}\frac{(a+b)+p(a-b)}{\sqrt{a^2+b^2+p(a^2-b^2)}}$$
where $p \in [-1, 1]$
Now we can find critical point and find minimum.
Is there a better approach, i tried lagrange multipliers but very tedious
| Since you have
$$x^2 + y^2 = 1 \implies y^2 = 1 - x^2 \tag{1}\label{eq1A}$$
you can substitute this into your $f(x,y)$ to have a function $g(x)$ for $0 \le x \le 1$ with
$$g(x) = \frac{ax^2 + b(1-x^2)}{\sqrt{a^2x^2 + b^2(1-x^2)}} = \frac{(a-b)x^2 + b}{\sqrt{(a^2-b^2)x^2 + b^2}} \tag{2}\label{eq2A}$$
Since $x$ only appears as $x^2$, there's no need to worry about negative values of $x$, so have $0 \le x \le 1$. Assuming $a,b \gt 0$ that $g(0) = g(1) = 1$. Using the quotient rule, you get
$$\begin{equation}\begin{aligned}
g'(x) & = \frac{2(a-b)x\sqrt{(a^2-b^2)x^2 + b^2} - ((a-b)x^2 + b)(2x)(a^2-b^2)\left(\frac{1}{2\sqrt{(a^2-b^2)x^2 + b^2}}\right)}{(a^2-b^2)x^2 + b^2} \\
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
To check for a critical point, setting $g'(x) = 0$ means the numerator must be $0$. Multiplying both sides by $\sqrt{(a^2-b^2)x^2 + b^2}$ and simplifying gives
$$\begin{equation}\begin{aligned}
0 & = 2(a-b)x((a^2-b^2)x^2 + b^2) - ((a-b)x^2 + b)(x)(a^2-b^2) \\
& = x(a-b)(2((a^2-b^2)x^2 + b^2) - ((a-b)x^2 + b)(a + b)) \\
& = x(a-b)(2(a^2 - b^2)x^2 + 2b^2 - (a^2 - b^2)x^2 - (ab + b^2)) \\
& = x(a-b)((a^2 - b^2)x^2 + b^2 - ab)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
Thus, you have $x = 0$, $a = b$ or $(a^2 - b^2)x^2 + b^2 - ab$. The only interesting choice is the third one, which I will leave to you to deal with.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3526498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\lim\limits_{n\to\infty}\left(\sqrt [n+1]{\frac {a_{n+1}}{b_{n+1}}}-\sqrt[n]{\frac {a_n}{b_n}}\ \right)$
Consider two sequences $(a_n)_{n\ge 1}$ and $(b_n)_{n\ge 1}$ of
positive real numbers such that
$$\lim_{n\to\infty} \frac {a_{n+1}}{n^2\cdot a_n}=x>0$$
and
$$\lim_{n\to\infty} \frac {b_{n+1}}{n\cdot b_n}=y>0$$
Evaluate:
$$\lim_{n\to\infty}\left(\sqrt [n+1]{\frac {a_{n+1}}{b_{n+1}}}-\sqrt[n]{\frac {a_n}{b_n}}\ \right)$$
My thoughts: Intuitively $\dfrac{a_{n+1}}{a_n} \sim xn^2$ and $\dfrac{b_{n+1}}{b_n} \sim yn$, so $a_n \sim x \sqrt[n]{(n!)^2}$ and $b_n \sim y\sqrt[n]{n!}$.
Since $\lim\limits_{n\to \infty} \dfrac{n}{\sqrt[n]{n!}} = e$, we should get:
$$\sqrt[n+1]{\frac{a_{n+1}}{b_{n+1}}} \sim \frac{x}{y}\cdot \sqrt[n+1]{(n+1)!} \sim\frac{x}{y}\cdot \frac{n+1}{e}$$
and the final answer should be $\dfrac{x}{ye}$. But how do we write this with sound arguments?
| Some observations. Define
$$
c_n : = \frac{1}{n!}\frac{a_n}{b_n}.
$$
Then by the assumptions on $a_n$ and $b_n$, we have
$$
\mathop {\lim }\limits_{n \to + \infty } \frac{c_{n + 1}}{c_n} = \frac{x}{y}\quad \text{ and hence }\quad \mathop {\lim }\limits_{n \to + \infty } \sqrt[n]{c_n} = \frac{x}{y}.
$$
Note that
$$
\sqrt[n]{\frac{a_n}{b_n}} = \sqrt[n]{n!c_n}.
$$
By Stirling's formula
$$
\sqrt[n]{{n!}} = \frac{n}{e} + \frac{{\log (2\pi n)}}{{2e}} + \mathcal{O}\left( {\frac{1}{n}} \right).
$$
If it was true that
$$
\sqrt[n]{{c_n }} = \frac{x}{y} +\frac{K}{n}+ o\left( {\frac{1}{n}} \right)
$$
then we would have
$$
\sqrt[{n + 1}]{{(n + 1)!c_{n + 1} }} - \sqrt[n]{{n!c_n }} = \frac{x}{y}\frac{1}{e} + \mathcal{O}\left( {\frac{1}{n}} \right),
$$
meaning that the limit is indeed $\frac{x}{y}\frac{1}{e}$.
In any case,
$$
\mathop {\lim \inf }\limits_{n \to + \infty } \left( {\sqrt[{n + 1}]{{\frac{{a_{n + 1} }}{{b_{n + 1} }}}} - \sqrt[n]{{\frac{{a_n }}{{b_n }}}}} \right) = \mathop {\lim \inf }\limits_{n \to + \infty } \frac{{\sqrt[{n + 1}]{{\frac{{a_{n + 1} }}{{b_{n + 1} }}}} - \sqrt[n]{{\frac{{a_n }}{{b_n }}}}}}{{n + 1 - n}} \le \mathop {\lim }\limits_{n \to + \infty } \frac{{\sqrt[n]{{n!c_n }}}}{n} = \frac{x}{{ye}} \\ \le \mathop {\lim \sup }\limits_{n \to + \infty } \frac{{\sqrt[{n + 1}]{{\frac{{a_{n + 1} }}{{b_{n + 1} }}}} - \sqrt[n]{{\frac{{a_n }}{{b_n }}}}}}{{n + 1 - n}} = \mathop {\lim \sup }\limits_{n \to + \infty } \left( {\sqrt[{n + 1}]{{\frac{{a_{n + 1} }}{{b_{n + 1} }}}} - \sqrt[n]{{\frac{{a_n }}{{b_n }}}}} \right).
$$
Thus, if the limit in question exists, it must be $\frac{x}{y}\frac{1}{e}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3528462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Solve a system of equation: $\cos(2x) + \cos(y) = 1$, $\sin(2x) + \sin(y) = 1$ Solve a system of equation:
$$\cos(2x) + \cos(y) = 1$$
$$\sin(2x) + \sin(y) = 1$$
My idea:
Let's see what is product of this two equations.
$$\cos(2x)\sin(2x) + \cos(2x)\sin(y) + \cos(y)\sin(2x) + \cos(y)\sin(y) = 1$$
$$\cos(2x)\sin(2x) + \sin(2x+y) + \cos(y)\sin(y) = 1$$ But this idea didn't give me anything. Also if I sum I have problem... but this is high school problem so it must have some easy solution.
| Let $z=2x$. Then
$$1=\cos z + \cos y = 2\cos\frac{z-y}2\cos\frac{x+y}2\tag 1$$
$$1=\sin z + \sin y = 2\cos\frac{z-y}2\sin\frac{x+y}2\tag 2$$
Take (1) - (2) to have a factorized equation,
$$\cos\frac{z-y}2 \sin\left( \frac{z+y}2 - \frac\pi4 \right) =0$$
So, two cases to consider:
Case 1) $\cos\frac{z-y}2 = 0$ leads to $z=y + (1+2n)\pi$. Plug it into (1) to see that (1) does not hold. Hence, no solutions.
Case 2) $\sin\left( \frac{z+y}2 - \frac\pi4 \right)=0$ leads to
$z= -y+\frac\pi2+2n\pi$. Plug it into (1) to get $\cos (y -\frac\pi4) =\frac1{\sqrt2}$ and the solutions for $y$,
$$y=2n\pi, \>\>\>\>\>y = \frac\pi2 + 2n\pi $$
and the respective $z's$ are $z= \frac\pi2+2k\pi$ and $z= 2k\pi$.
Thus, there are two sets of the solutions,
$$(x,y) = (\frac\pi4+k\pi, 2n\pi),\>(k\pi, \frac\pi2+2n\pi)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3533104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
Prove that $x + \frac{2x^3}{3} + \cdots + \frac{2\cdot 4 \cdot \cdots 2nx^{2n+1}}{3\cdot 5 \cdot (2n+1)}+\cdots = \frac{\arcsin(x)}{\sqrt{1-x^2}}$
Prove that $x + \dfrac{2x^3}{3} + \cdots + \dfrac{2\cdot 4 \cdot \cdots 2nx^{2n+1}}{3\cdot 5 \cdot (2n+1)}+\cdots = \dfrac{\arcsin(x)}{\sqrt{1-x^2}}.$
Let $f(x) = x + \dfrac{2}3x^3 +\dfrac{8}{15}x^5 + \cdots + \dfrac{1}2n!\cdot\dfrac{n!}{(2n+1)!}(2x)^{2n+1}+\cdots.$
Then $xf(x) = x^2 + \dfrac{2}3 x^4 + \dfrac{8}{15}x^6 + \cdots + \dfrac{1}{4}n!\cdot \dfrac{n!}{(2n+1)!}(2x)^{2n+2}+\cdots.$
Also, $f'(x) = 1 + 2x^2 + \dfrac{8}3 x^4 + \cdots + n!\cdot \dfrac{n!}{(2n)!}(2x)^{2n}+\cdots$
and $(1-x^2)f'(x) =1+x^2 + \dfrac{2}3 x^4 + \cdots + [n!\cdot \dfrac{n!}{(2n)!}2^{2n}-(n-1)!\cdot \dfrac{(n-1)!}{(2n-2)!}2^{2n-2}]x^{2n}+\cdots\\
=1+x^2 + \dfrac{2}3 x^4 + \cdots + \dfrac{1}{4}\cdot(n-1)!\cdot \dfrac{(n-1)!}{(2n-1)!}\cdot(2x)^{2n}+ \cdots.$
Hence the derivative of $\sqrt{1-x^2}f(x)$ is $\sqrt{1-x^2}f'(x)-\dfrac{xf(x)}{\sqrt{1-x^2}} = \dfrac{1}{\sqrt{1-x^2}}((1-x^2)f'(x) - xf(x))=\dfrac{1}{\sqrt{1-x^2}}.$ Integrating, we see that $\sqrt{1-x^2}f(x) =\arcsin(x)+C.$ Plugging in the constant $C=0$ and dividing both sides by $\sqrt{1-x^2}$ gives the desired result.
Are there other approaches to solving this problem?
| In this proof i will use some proprieties of Wallis'integral (Which easy to show by simple induction),
We have :
$$W_{2n+1}=\prod_{k=1}^n\frac{2k}{2k+1}$$
So your series is equal to :
$$f(x)=\sum_{n=0}^\infty W_{2n+1}x^{2n+1}=\sum_{n=0}^\infty \biggl(\int_0^{\pi/2}\sin^{2n+1}t\ dt\biggr)x^{2n+1}$$
Then we put $t=\arcsin u$, therefore :
$$f(x)=\sum_{n=0}^\infty \biggl(\int_0^{1}\frac{u^{2n+1}}{\sqrt{1-u^2}}\ dt\biggr)x^{2n+1}=\int_0^1\biggl(\frac{ux}{\sqrt{1-u^2}}\sum_{n=0}^\infty(ux)^{2n}\biggr)du$$
Then we Know calculate the geometric series :
$$f(x)=\int_0^{1}\frac{ux\ du}{\sqrt{1-u^2}(1-(ux)^2)}$$
Let :
$$t^2=\frac{1-u^2}{1-(ux)^2}⇒u^2=\frac{1-t^2}{1-(tx)^2}⇒udu=-\frac{t(1-x^2)}{\biggl(1-t^2x^2\biggr)^2}dt$$
Therefor :
$$f(x)=\frac{1}{\sqrt{1-x^2}}\int_0^1\frac{x\ du}{\sqrt{1-(tx)^2}}=\frac{\arcsin x}{\sqrt{1-x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3534871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find whether $\sum_{n=1}^\infty \frac{2^{3n-3}3^{n+2}}{5^{2n-2}} $ is geometric and convergent and if so, state the value of the sum $$\sum_{n=1}^\infty \frac{2^{3n-3}3^{n+2}}{5^{2n-2}} $$
First I tried to find the ratio by doing $\frac{a_{n+1}}{a_n}$. Many calculations were made, and the result was $24/25$, so it is geometric. This is between -1 and 1, and therefore the series is convergent.
Then I did $$\sum_{d=1}^\infty \frac{a_1}{1-24/25} = (...) = 675$$
Is this correct?
| $$\sum_{n=1}^\infty \frac{2^{3n-3}3^{n+2}}{5^{2n-2}} =\frac{9.25}{8}\sum_{n=1}^{\infty } \frac{8^n.3^n}{25^n}=\frac{225}{8}\frac{\frac{24}{25}}{1-\frac{24}{25}}=675$$
So your answer is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3535835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Limit of $\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$ $$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}$$
I tried to used $(a^3-b^3)=(a-b)(a^2+ab+b^2)$ but it did not worked out so I tried to use the squeeze theorem.
$$0=\sqrt[3]{x^3}-\sqrt{x^2}\leq \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}\leq\sqrt[3]{8x^3}-\sqrt{4x^2}=2-2=0$$
But on the right hand I have inrcrased $\sqrt{x^2-2x}$ rather then deceased
Another attempt:
$$\lim_{x\to \infty} \sqrt[3]{x^3+2x}-\sqrt{x^2-2x}=\lim_{x\to \infty} \sqrt[3]{x^3(1+\frac{2}{x^2})}-\sqrt{x^2(1-\frac{2}{x})}=\\=\lim_{x\to \infty} x\sqrt[3]{(1+\frac{2}{x^2})}-x\sqrt{(1-\frac{2}{x})}=\lim_{x\to \infty} x[\sqrt[3]{(1+\frac{2}{x^2})}-\sqrt{(1-\frac{2}{x})}]$$
| Moving factors of $x^3$ and $x^2$ out, we transform the limit to
$$\lim_{x\to\infty}x\left(\sqrt[3]{1+2/x^2}-\sqrt{1-2/x}\right)$$
Then applying the binomial series yields
$$=\lim_{x\to\infty}x\left(\left(1+O(x^{-2})\right)-\left(1-\frac12\cdot\frac2x+O(x^{-2})\right)\right)$$
$$=\lim_{x\to\infty}x\left(1-1+\frac12\cdot\frac2x\right)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
remainder when $a_{1000}$ is divided by $1000$
If $a_{1}=7,a_{2}=7^7,a_{3}=7^{7^{7}}.$ Then the remainder when
$a_{1000}$ is divided by $1000$
what i try
$a_{1}=7=1\mod(1000)$ and
$a_{2}=7^7=1^7\mod(1000)=1\mod(1000)$
from using modulo theorem
$a_{3}=7^{7^{7}}=1^7\mod(1000)=1\mod(7)$
can we say that $a_{1000}= 1\mod(1000)$
Help me to solve it please
| $$7^4=(50-1)^2=2401$$
$$7^{4n}=(1+2400)^n\equiv1+2400n\pmod{1000}$$
So, $n$ must be divisible by $5$ to make residue $\equiv1$
$\implies$ord$_{1000}7=20$
We need $a_{999}\pmod{20}$
Again, $a_r,r\ge2$ are of the form $7^{4n+3}$
$7^{4n+3}=7^3(50-1)^{2n}\equiv3(1-50)^{2n}\equiv3(1-\binom{2n}150)\pmod{20}\equiv3$
So for $r\ge2,$
$$a_{r+1}\equiv7^{3\pmod{20}}\pmod{1000}\equiv7^3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How many pairs of $(x,y,z)$ for $x+y+2z=n$
How many pairs of $(x,y,z)$ for $x+y+2z=n$ when $n$ is an odd integer $\geq 5$
For $n=5$ there are 2 pairs. $n=7$ there are 6 pairs. $n=9$ there are 12 pairs. $n=11$ there are 20 pairs.
$an^2 + bn + c = f(n)$
$25a + 5b + c = 2$
$49a + 7b + c = 6$
$81a + 8b + c = 12$
$f(n) = (n^2)/4 - n + 3/4 = (n-3)(n-1)/4$
Is there a better way to solve it?
| For $x,y,z\in \mathbb{N}, x,y,z\geq 0$,
When $n$ is odd, the number is:
$$ (n+1)+(n-1)+(n-3)+...+2=\frac{(n+1)(n+3)}{4} $$
when $n$ is even, the number is:
$$ (n+1)+(n-1)+(n-3)+...+1=(\frac{n+2}{2})^2 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3538889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Prove that $\sum_{k=1}^{89} \frac{1}{\tan^3(k^\circ)+1}=44.5$ So I've been stuck on this problem for a while now, and to be honest I have no idea how to start. I stumbled upon it while practicing for an Olympiad, and it is a past Olympiad problem (for teens) itself.
The problem is as follows:
Given is the function $f(x) = \frac{1}{x^3+1}$
Prove that:
$$\sum_{k=1}^{89} f(\tan(k^\circ)) = 44.5$$
In other words, prove that:
$$\frac{1}{\tan^3(1^\circ)+1} + \frac{1}{\tan^3(2^\circ)+1} + \cdots + \frac{1}{\tan^3(89^\circ)+1} = 44.5 $$
The question was multiple choice at first, but I managed to figure out 44.5 with my calculator. I still wondered how one would come up with a proof.
Any guidelines or straight up solutions are welcome!
| We can write $\tan(90^\circ - k^\circ) = \frac{1}{\tan(k^\circ)}$.
\begin{align*}
\sum_{k=1}^{89} \frac{1}{\tan^3(k^\circ) + 1} &= \sum_{k=1}^{44} \frac{1}{\tan^3(k^\circ) + 1} + \sum_{k=46}^{89} \frac{1}{\tan^3(k^\circ) + 1} + \frac{1}{\tan^3(45^\circ) + 1} \\
&= \sum_{k=1}^{44} \frac{1}{\tan^3(k^\circ) + 1} + \sum_{k=1}^{44} \frac{1}{\tan^3(90 - k^\circ) + 1} + \frac{1}{2} \\
&= \sum_{k=1}^{44} \frac{1}{\tan^3(k^\circ) + 1} + \sum_{k=1}^{44} \frac{1}{\frac{1}{\tan^3(k^\circ)} + 1} + \frac{1}{2} \\
&= \sum_{k=1}^{44} \frac{1}{\tan^3(k^\circ) + 1} + \sum_{k=1}^{44} \frac{\tan^3(k^\circ)}{\tan^3(k^\circ) + 1} + \frac{1}{2} \\
&= \sum_{k=1}^{44} \frac{\tan^3(k^\circ) + 1}{\tan^3(k^\circ) + 1} + \frac{1}{2} \\
&= 44.5
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
What is $x^3 + x + 1$ for $x=\frac{\sqrt 5 - 1}{2}$
What is $x^3 + x + 1$ for $x=\frac{\sqrt 5 - 1}{2}$
Of course we can subtitute and expand it
${(\frac{\sqrt 5 -1}{2})}^3 + \frac{\sqrt 5 -1}{2} + 1$
Is there a better way to do it?
| Since $x^2=\frac{3-\sqrt{5}}{2}=1-x$, $x^3=x-x^2=2x-1$, so $x^3+x+1=3x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3539373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Help with "Smallest Euclidean Norm problem". I'm trying to find the smallest euclidian norm of the general solution of an algebraic equation $Ax=y,$ where $A$ is an $n\times m$ matrix and the gral solution is:
$x=x_{p}+\alpha_{1}n_{1}+\alpha_{2}n_{2}=\begin{pmatrix}
0\\
-4\\
0\\
0
\end{pmatrix}+\alpha_{1}\begin{pmatrix}
1\\
1\\
-1\\
0
\end{pmatrix}+\alpha_{2}\begin{pmatrix}
0\\
2\\
0\\
-1
\end{pmatrix}$
So, $x$ is undetermined, $\alpha_{1}, \alpha_{2}$ can be any real number and $n_{1},n_{2}$ are null vectors of A.
My first approcah is the following:
$x=\begin{pmatrix}
\alpha_{1}\\
-4+\alpha_{1}+2\alpha_{2}\\
-\alpha_{1}\\
-\alpha_{2}
\end{pmatrix}$
and since $\left\lVert x \right\lVert_{2}=\sqrt{x^{T}x}\Rightarrow\left\lVert x \right\lVert^{2}_{2}=x^{T}x$, thus
$\left\lVert x \right\lVert^{2}_{2}=3\alpha^{2}_{1}+5\alpha^{2}_{2}-8\alpha_{1}-16\alpha_{2}+4\alpha_{1}\alpha_{2}+16$
And I know from the definition of a norm that
*
*$\left\lVert x \right\lVert\geq 0$
*$\left\lVert x_{1}+x_{2} \right\lVert\leq \left\lVert x_{1} \right\lVert+\left\lVert x_{2} \right\lVert$
So, if these contitions hold for $\left\lVert x \right\lVert$ then it should hold for $\left\lVert x \right\lVert^{2}$, but from here I'm not sure how to proceed, you guys have any advice?
update: I tried the Least-norm solution $x_{ln}=A^{T}(AA^{T})^{-1}y$, the issue here is that my matrix A is singular, since $det(A)=0$ and not invetible since $rank(A) \neq n$.
| setting the gradient to the zero vector gives system
$$ 6 \alpha_1 + 4 \alpha_2 = 8 \; , $$
$$ 4 \alpha_1 + 10 \alpha_2 = 16 \; . $$
Or augmented matrix
$$
\left(
\begin{array}{cc|c}
6 & 4 & 8 \\
4 & 10 & 16
\end{array}
\right)
$$
$$
\left(
\begin{array}{cc|c}
3 & 2 & 4 \\
2 & 5 & 8
\end{array}
\right)
$$
$$
\left(
\begin{array}{cc|c}
1 & -3 & -4 \\
2 & 5 & 8
\end{array}
\right)
$$
$$
\left(
\begin{array}{cc|c}
1 & -3 & -4 \\
0 & 11 & 16
\end{array}
\right)
$$
$$
\left(
\begin{array}{cc|c}
1 & -3 & -4 \\
0 & 1 & \frac{16}{11}
\end{array}
\right)
$$
$$
\left(
\begin{array}{cc|c}
1 & 0 & \frac{4}{11} \\
0 & 1 & \frac{16}{11}
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Nice integral involving omega constant, Lambert's function and logarithms
$$\int_1^e \Big(\operatorname{W}(x)^2 \ln(x)0.25-\operatorname{W}(x) 0.5 + \operatorname{W}(x) \ln(x) \Big) \, dx=\operatorname{W}(1)0.5 + \operatorname{W}(1)^2 0.25$$
I can express the antiderivative wich involves exponential integral and Lambert's function and calculate it using the fundamental theorem of calculus but I would like to know if there are smarter methods (maybe complex analysis)
Thanks a lot for your time and patience.
| I unfortunately do not have a smart way to evaluate this interesting integral. My approach is more just "plug and chug".
Let $u = \operatorname{W}_0 (x)$, then $x = ue^u, dx = e^u(1 + u) \, du$ while the limits of integration become: $u = \operatorname{W}_0 (1) = \Omega$ and $u = \operatorname{W}_0 (e) = 1$. Here $\Omega$ is the Omega constant. Denoting the integral by $I$, we have
\begin{align}
I &= \int_\Omega^1 \left [\frac{u^2}{4} (\ln u + u) - \frac{u}{2} + u(\ln u + u) \right ] e^u (1 + u) \, du\\
&= \int_\Omega^1 \left [e^u \ln u \left (u + \frac{5}{4} u^2 + \frac{1}{4} u^3 \right ) + e^u \left (\frac{1}{4} u^4 + \frac{5}{4} u^3 + \frac{1}{2} u^2 - \frac{1}{2} u \right ) \right ] \, du.
\end{align}
Observing that
$$\int e^x \left (\frac{x^4}{4} + \frac{5}{4} x^3 + \frac{x^2}{2} - \frac{x}{2} \right ) \, dx = \frac{1}{4} e^x x^2 (x^2 + x - 1) + C,$$
and
$$\int e^x \ln x \left (x + \frac{5}{4} x^2 + \frac{x^3}{4} \right ) \, dx = \frac{1}{4} e^x x^2 [(x + 2) \ln x - 1] + C,$$
we find
$$I = \frac{1}{2} \Omega + \frac{1}{4} \Omega^2,$$
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Common integrating factor of two differential equations. The following differential equations have a common integrating factor.
$$(3y+4xy^2)dx+(4x+5x^2y)dy=0$$
$$(6y+x^2y^2)dx+(8x+x^3y)dy=0$$
I have to find this integrating factor.
I have arrived to the solution of this problem assuming that the integrating factor I was trying to find was of the form $x^ny^m$, and in this way, I have found that the common integrating factor is $x^2y^3$.
However, I was wondering if there might be a more direct way to find this integrating factor, without having to suppose that is has the form $x^ny^m$. Any ideas?
Thanks.
| This may help
Let us re-write the first one as:
$$\frac{xdy}{ydx}=-\frac{3+6xy}{4+5xy}\implies\frac{xdy}{ydx}+1=-\frac{3+6xy}{4+5xy}+1. $$
$$\implies \frac{xdy+ydx}{ydx}=\frac{1-xy}{4+5xy}$$
$$\implies \frac{dv}{ydx}=\frac{1-v}{4+5v}, xy=v \implies \frac{dv}{dx}=\frac{v-v^2}{4+5v} \frac{1}{x}.$$
$$\implies \int \frac{4+5v}{v-v^2}dv=\int \frac{dx}{x}$$
$$\implies \ln \left( \frac{v^4}{(1-v)^9} \right)=\ln Cx \implies x^4y^4=Cx(1-xy)^9.$$
The second one can be re-written as:
$$\frac{xdy}{ydx}=-\frac{6+x^2y}{8+x^2y} \implies \frac{x^2dy}{2xydx}=-\frac{1}{2}~\frac{6+x^2y}{8+x^2y}. $$
$$\implies \frac{x^2dy}{2xydx}+1=1-\frac{1}{2}~\frac{6+x^2y}{8+x^2y}$$
$$\implies \frac{x^2dy+2xydy}{2xydx}=\frac{10+x^2y}{2(8+x^2y)}$$
Let $x^2y=u$, then
$$\int \frac{2(8+u)du}{10u+u^2}=\int \frac{dx}{x}$$
$$\implies \ln u^8 +\ln (10+u)^2=\ln (Cx)^5$$
$$\implies u^8(10+u)^2=(Cx)^5, u=x^2y$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
coefficient of bivariate generating functions how can I find/what is the coefficient of the $x^4y^4$ and $x^6y^6$ term for the following generating function
$$\frac{1}{1-x-y-x^2y}$$
I've been informed I can do this by using Pascal's triangle, by adding an extra value of 2 back and 1 down but am unsure of how to? (I'm new to combinatorics)
| A method to find the coefficient of $x^m y^n$ in $\frac{1}{1-x-y-x^2y}$ is described as follows.
First, let
$$\frac{1}{1 - x - y - x^2y} = a_0(x) + a_1(x) y + a_2(x) y^2 + a_3(x) y^3 + \cdots .$$
We need to determine $a_n(x) y^n$.
Clearly,
$$\frac{\partial^n \tfrac{1}{1 - x - y - x^2y}}{\partial y^n} \Big\vert_{y=0} = n! a_n(x).$$
It is easy to obtain (since $1 - x - y - x^2y$ is affine in $y$)
$$\frac{\partial^n }{\partial y^n} \frac{1}{1 - x - y - x^2y}
= \frac{(-1)^n n! (-1-x^2)^n}{(1-x-y-x^2y)^{n+1}}.$$
Thus, we have
$$a_n(x) = \frac{(1+x^2)^n}{(1-x)^{n+1}}.$$
By noting that $\frac{1}{1-x} = \sum_{j=0}^\infty x^j$ and $\frac{\partial^n }{\partial x^n} \frac{1}{1-x} = \frac{n!}{(1-x)^{n+1}}$, we have
$$\frac{1}{(1-x)^{n+1}} = \frac{1}{n!}\frac{\partial^n }{\partial x^n}\sum_{j=0}^\infty x^j = \sum_{j=0}^\infty {n+j \choose j} x^j.$$
Also,
$$(1+x^2)^n = \sum_{k=0}^n {n\choose k} x^{2k}.$$
Thus, we have
\begin{align}
a_n(x) &= \sum_{k=0}^n {n\choose k} x^{2k} \cdot \sum_{j=0}^\infty {n+j \choose j} x^j\\
&= \sum_{k=0}^n \sum_{j=0}^\infty {n\choose k} {n+j \choose j} x^{2k+j}.
\end{align}
The coefficient of $x^m$ in $a_n(x)$ is given by
\begin{align}
a_{mn} &= \sum_{2k+j = m, \ j\ge 0, \ 0\le k \le n} {n\choose k} {n+j \choose j} \\
&= \sum_{k=0}^{\min(n, \lfloor \frac{m}{2}\rfloor)} {n\choose k} {n+m-2k \choose m-2k}.
\end{align}
For example,
$$a_{44} = \sum_{k=0}^2 {4\choose k} {8-2k \choose 4-2k} = 136$$
and
$$a_{66} = \sum_{k=0}^3 {6\choose k} {12-2k \choose 6-2k} = 2624.$$
Remark: In general, let
$$f(x, y) = a_{00} + a_{10}x + a_{01}y + a_{20}x^2 + a_{11}xy + a_{02}y^2
+ \cdots.$$
Clearly,
$$
\frac{\partial^{m+n} f(x,y) }{\partial x^m \partial y^n}\Big\vert_{(x,y)=(0,0)}
= m! n! a_{mn}$$
which results in
\begin{align}
a_{mn} = \frac{1}{m!}\frac{1}{n!}
\frac{\partial^m }{\partial x^m}
\Big(\frac{\partial^n f(x,y)}{\partial y^n} \Big\vert_{y=0}\Big)\Big\vert_{x=0}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547566",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proof that combined matrices are idempotent. Suppose that $X_{nxp} (n>p)$ is a matrix such that $X'X$ is invertible.
*
*Prove that both $X(X'X)^{-1}X'$ and $I_n-X(X^TX)^{-1}X^T$ are idempotent.
*Prove that the tr $X(X'X)^{-1}X' =p$ and tr $(I_n - X(X'X)^{-1}X')=n-p$.
| Let us put
$$ A \colon= X \left(X^\prime X\right)^{-1} X^\prime. $$
Then we note that
$$
\begin{align}
A^2 &= AA \\
&= \left[ X \left(X^\prime X\right)^{-1} X^\prime \right] \left[ X \left(X^\prime X\right)^{-1} X^\prime \right] \\
&= X \left[ \left(X^\prime X\right)^{-1} \left( X^\prime X \right) \right] \left(X^\prime X\right)^{-1} X^\prime \\
&= X I_{p\times p} \left(X^\prime X\right)^{-1} X^\prime \\
&= X \left(X^\prime X\right)^{-1} X^\prime \\
&= A.
\end{align}
$$
Now let us put
$$ B \colon= I_n-X(X^TX)^{-1}X^T. $$
Then we find that
$$
\begin{align}
B^2 &= BB \\
&= \left[ I_n-X(X^TX)^{-1}X^T \right] \left[ I_n-X(X^TX)^{-1}X^T \right] \\
&= I_n \left[ I_n- X(X^TX)^{-1}X^T \right] - X(X^TX)^{-1}X^T \left[ I_n-X(X^TX)^{-1}X^T \right] \\
&= I_n I_n - I_n X(X^TX)^{-1}X^T - X(X^TX)^{-1}X^T I_n + \left[ X(X^TX)^{-1}X^T \right] \left[ X(X^TX)^{-1}X^T \right] \\
&= I_n - X(X^TX)^{-1}X^T - X(X^TX)^{-1}X^T + X\left[ (X^TX)^{-1} \left( X^T X\right) \right] \left[ (X^TX)^{-1}X^T \right] \\
&= I_n - 2 X(X^TX)^{-1}X^T + X I_n \left[ (X^TX)^{-1} X^T \right] \\
&= I_n - 2 X(X^TX)^{-1}X^T + X \left[ (X^TX)^{-1} X^T \right] \\
&= I_n - 2 X(X^TX)^{-1}X^T + X (X^TX)^{-1} X^T \\
&= I_n - X(X^TX)^{-1}X^T \\
&= B.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3547987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to calculate this determinant of a $2\times 2$matrix? This matrix arises from a homework problem which our professor gave.
We need to find the determinant of this matrix.
Does there exist any simple way to find the determinant of this matrix?
$\begin{pmatrix}
x-pq-p+3-(q-1)(\frac{x+2-n}{x-n+2-l}) && (1-p)(2+\frac{l}{x-n+2-l})\\
(1-q)(2+\frac{l}{x-n+2-l}) && x-pq-q+3-(p-1)(\frac{x+2-n}{x-n+2-l})
\end{pmatrix}$
Here $n=pq$ and $l=\phi(n)+1$.
Is there any software which can calculate this large determinant?
One of my friends got $x-n+2-l$ as a factor of this determinant.
Is there any simple way to calculate this determinant?
I am stuck.
My try:
$R_1\to R_1-R_2$ gives
$\begin{pmatrix}
x-pq-p+q+2 && -x+pq+q-p-2\\
(1-q)(2+\frac{l}{x-n+2-l}) && x-pq-q+3-(p-1)(\frac{x+2-n}{x-n+2-l})
\end{pmatrix}$
| It helps to rewrite your matrix as $A+B$ with$$A:=\left(\begin{array}{cc}
x-pq-p+3 & 1-p\\
1-q & x-pq-q+3
\end{array}\right),\,B:=\frac{x+2-n}{x+2-n-l}\left(\begin{array}{cc}
1-q & 1-p\\
1-q & 1-p
\end{array}\right).$$Write $A=\left(\begin{array}{cc}
a & c\\
b & d
\end{array}\right),\,B=\left(\begin{array}{cc}
e & g\\
f & h
\end{array}\right)$ so$$\begin{align}\det\left(A+B\right)&=\det A+\det B+\det\left(\begin{array}{cc}
a & g\\
b & h
\end{array}\right)+\det\left(\begin{array}{cc}
e & c\\
f & d
\end{array}\right)\\&=\left(x-pq+3\right)^{2}-\left(p+q\right)\left(x-pq+3\right)-1+p+q\\&+\frac{x+2-n}{x+2-n-l}\left[\det\left(\begin{array}{cc}
x-pq-p+3 & 1-p\\
1-q & 1-p
\end{array}\right)+\det\left(\begin{array}{cc}
1-q & 1-p\\
1-q & x-pq-q+3
\end{array}\right)\right]\\&=\left(x-pq+3\right)^{2}-\left(p+q\right)\left(x-pq+3\right)-1+p+q\\&+\frac{x+2-n}{x+2-n-l}\left[\left(x-\left(p-1\right)\left(q+1\right)+1\right)\left(1-p\right)+\left(x-\left(p+1\right)\left(q-1\right)+1\right)\left(1-q\right)\right]\\&=\left(x-pq+3\right)^{2}-\left(p+q\right)\left(x-pq+3\right)-1+p+q\\&+\frac{x+2-n}{x+2-n-l}\left[\left(1+x\right)\left(2-p-q\right)+\left(p-1\right)^{2}\left(q+1\right)+\left(p+1\right)\left(q-1\right)^{2}\right].\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3548671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $\int_a^b \frac{1}{x}\sqrt{-(x-a)(x-b)}dx = (\frac{a+b}{2}-\sqrt{ab})\pi$ Does anyone know how to solve $$\int_a^b \frac{1}{x}\sqrt{-(x-a)(x-b)}dx$$
After trying Wolfram Alpha, I conclude that it possibly equals $(\frac{a+b}{2}-\sqrt{ab})\pi = \frac{(\sqrt{a}-\sqrt{b})^2}{2}\pi$.
The book I am reading says one can use residue theorem to quickly obtain the result, but the details are not written and I cannot figure out the solution.
P.S. One can prove from this integral that arithmetic mean $\geq$ geometric mean.
| Let $x=a\cos^2 t+b \sin^2 t \implies dx=(b-a) \sin 2t dt$
Then $$I=2\int_{0}^{\pi/2} \frac{(b-a)^2 \sin^2 t \cos^2 tdt}{a \cos^2 t+b \sin^2t}= 2 (b-a)^2 \int_{0}^{\pi/2}\frac{ \sin^2 t ~dt}{a+b\tan^2 t}$$
Let $\tan t=u$, then
$$I=2\int_{0}^{\infty} \frac{u^2 du}{(1+u^2)^2(a+bu^2)}=2\int_{0}^{\infty} du \left(\frac{b-a}{(1+u^2)^2}+\frac{a}{1+u^2}-\frac{ab}{a+bu^2}\right)$$
$$\implies I=\left. (b-a)\left(\frac{u}{1+u^2}+\tan^{-1}u\right)+2a \tan^{-1}u -2\sqrt{ab} ~\tan^{-1}\frac{u\sqrt{b}}{\sqrt{a}} \right|_{0}^{\infty}=\left(\frac{a+b}{2}-\sqrt{ab} \right) \pi $$
I will get back.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
Linearization of $C^r$ ODE system Consider the following ODE system
\begin{align*}
\dot{x} &= x^5 + y^3 = f(x\,,y) \\
\dot{y} &= x^3 - y^5 = g(x\,,y)
\end{align*}
Clearly, $(x\,,y) = (0\,,0)$ is the only fixed point. Linearizing
$$
J =
\begin{bmatrix}
5x^4 & 3y^2 \\
3x^2 & -5y^4
\end{bmatrix}
$$
evaluating at the fixed point will just give a zero matrix. Thus, need to use some other approach to analyze the stability, e.g., Lyapunov function.
However, I can't find any Lyapunov function related to this problem. Since this ODE system is at least $C^3$ in both $x$ and $y$. So I am wondering, for the linearization, is it possible to go all the way to 3rd-order derivative? Maybe something like
$$
\begin{bmatrix}
\frac{\partial^3f}{\partial x^3} & \frac{\partial^3f}{\partial y^3} \\
\frac{\partial^3g}{\partial x^3} & \frac{\partial^3g}{\partial y^3}
\end{bmatrix}
$$
This way the matrix will not be a zero matrix and can find corresponding eigenvalues and eigenvectors.
| From
\begin{align*}
\dot{x} &= x^5 + y^3 \\
\dot{y} &= x^3 - y^5
\end{align*}
we have
\begin{align*}
x^3\dot{x} &= x^8 + x^3 y^3 \\
y^3\dot{y} &= x^3y^3 - y^8
\end{align*}
and then
$$
\frac 14\frac{d}{dt}(x^4-y^4) = x^8+y^8
$$
analyzing the behavior along a line $y(t) = \alpha x(t)$ we conclude
$$
\frac 14(1-\alpha^4)\frac{d}{dt}x^4 = (1+\alpha^8)x^8
$$
so depending on $\alpha$ the dynamics along $y(t) = \alpha x(t)$ changes concluding that the origin is unstable.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3549658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to prove that $\tan \left( \frac{\pi}{2} - \theta \right) = \cot \theta$ I am asked to simplify the following expression:
$$\tan \left( \frac{\pi}{2} - \theta \right)$$
The book gives me the answer $\cot \theta$ but, when I try to derive that formula using
$$\tan \left( A \pm B \right) = \frac{\tan A \pm \tan B}{1 \mp \tan A \cdot \tan B}$$
I get $\tan \frac{\pi}{2}$ as one of the terms (which is undefined).
$$\tan \left( A \pm B \right) = \frac{\tan A \pm \tan B}{1 \mp \tan A \cdot \tan B} = \frac{\tan \frac{\pi}{2} - \tan \theta}{1 + 0 \cdot \tan \theta}$$
How do I proceed?
Thank you.
| There are several ways one can play with this type of problems.
Here is one:
$\tan\left(\frac{\pi}{2}-\theta \right) =\frac{\sin\left(\frac{\pi}{2}-\theta \right) }{ \cos\left(\frac{\pi}{2}-\theta \right) }$
But
$\sin\left(\frac{\pi}{2}-\theta \right)=\sin\left(\frac{\pi}{2}\right)\cos\left(\theta \right)-\cos\left(\frac{\pi}{2}\right)\sin\left(\theta \right) =1\cos(\theta)-0\sin(\theta)=\cos(\theta)$
and
$\cos\left(\frac{\pi}{2}-\theta \right)=\cos\left(\frac{\pi}{2}\right)\cos\left(\theta \right)+\sin\left(\frac{\pi}{2}\right)\sin\left(\theta \right) =0\sin(\theta)+1\sin(\theta)=\sin(\theta)$
Therefore
$\tan\left(\frac{\pi}{2}-\theta \right) = \frac{\cos(\theta)}{\sin(\theta)}=\cot(\theta)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
A sequence $(a_n)_{n\ge 1}$ such that $a_1>0$ and $a_{n+1}=a_n-\ln(1+a_n)$ Let $(a_n)_{n\ge 1}$ be a sequence such that $a_1>0$ and $$a_{n+1}=a_n-\ln(1+a_n)$$
a) Prove that $a_{n+1}<\frac{a_n^2}{2}, \forall n\in \mathbb{N}$ and $\lim\limits_{n\to \infty} (n^{2019}a_n)=0$.
b) If $a_1<2$, prove that $a_{12} \in (0,10^{-300})$.
It is easy to see that $a_n >0$, $\forall n\in \mathbb{N}$. I considered the function $f:(0,\infty)\to \mathbb{R}$, $f(x)=x-\ln(1+x)-\frac{x^2}{2}$ and I could easily prove that this function is strictly decreasing, so it follows that $$x-\ln(1+x)<\frac{x^2}{2}, \forall x>0 \tag{*}$$
Now from $(*)$ we get that $a_{n+1}<\frac{a_n^2}{2}, \forall n\in \mathbb{N}$.
I couldn't make much further progress. I could show that $\lim\limits_{n\to \infty}a_n=0$, but then I got stuck.
| Hint: to finish $(a)$. You already proved that $a_{n+1}<\frac{a_n^2}{2}$, use it further, i.e.
$$0<a_{n+1}<\frac{a_n^2}{2}<\frac{1}{2}\cdot\left(\frac{a_{n-1}^2}{2}\right)^{2}=
\frac{a_{n-1}^4}{2^3}<\\
\frac{1}{2^3}\left(\frac{a_{n-2}^2}{2}\right)^4=
\frac{a_{n-2}^8}{2^7}<\\
\frac{1}{2^7}\left(\frac{a_{n-3}^2}{2}\right)^8=
\frac{a_{n-3}^{16}}{2^{15}}$$
The patterns reveals as
$$0<a_{n+1}<\frac{a_{n-r}^{2^{r+1}}}{2^{2^{r+1}-1}} \Rightarrow
0<a_{k+r+1}<\frac{a_{k}^{2^{r+1}}}{2^{2^{r+1}-1}} \tag{1}$$
Since you also proved $\lim\limits_{n\to \infty}a_n=0$, then $0<a_k<1$ from some $k_0$ onwards, thus
$$0<a_{k_0+r+1}<\frac{1}{2^{2^{r+1}-1}} \Rightarrow \\
0<(k_0+r+1)^{2019}a_{k_0+r+1} < \frac{(k_0+r+1)^{2019}}{2^{2^{r+1}-1}} =\\
\frac{r^{2019}}{2^{2^{r+1}-1}}\cdot \left(\frac{k_0}{r}+1+\frac{1}{r}\right)^{2019} $$
and take the $\lim\limits_{r\to\infty}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3550500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proving $(\sin^2 \alpha+\sin\alpha \cos \alpha)^{\sin \alpha}(\cos^2 \alpha+\sin \alpha \cos \alpha)^{\cos \alpha}\leq 1$
If $\alpha \in \left(0, \frac{\pi}{2}\right)$, prove that:
$$(\sin^2 \alpha+\sin\alpha \cos \alpha)^{\sin \alpha}(\cos^2 \alpha+\sin \alpha \cos \alpha)^{\cos \alpha}\leq 1$$
I know $\sin \alpha$ and $\cos \alpha$ are positive over $\alpha \in \left(0, \frac{\pi}{2}\right)$, so with $a=\sin \alpha,\ b=\cos \alpha$, $a>0,\ b>0$ and $a^2+b^2=1$, the inequality is
$$(a^2+ab)^a(b^2+ab)^b \leq 1$$
$$\Leftarrow a^ab^b(a+b)^{a+b} \leq 1$$
and here I don't know how to prove this inequality.
| Rewrite the last line of your proof as: $a^{\frac{a}{a+b}}\cdot b^{\frac{b}{a+b}}\cdot (a+b) \le 1$. Apply the weighted AM-GM inequality you have: $LHS \le (\dfrac{a^2}{a+b}+\dfrac{b^2}{a+b})\cdot (a+b) = \dfrac{1}{a+b}\cdot (a+b) = 1 = RHS$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3551745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Barnard and Child inequality exercise Prove that,
$$3(a^2b+b^2c+c^2a)(ab^2+bc^2+ca^2)≥abc(a+b+c)^3$$
For positive $a,b,c$
The exercises in this book are making me crazy.
Any help would be appreciated.
My attempts:
I opened the LHS brackets and used the cube identity on RHS
I get expressions which are somewhat similar but no idea how to proceed furthermore:
$$3\sum_{cyc}(a^3b^3+a^4bc+a^2b^2c^2)\geq abc\sum_{cyc}(a^3+3a^2b+3a^2c+2abc).$$
| Another way.
By your work, AM-GM and Muirhead we obtain:
$$3\prod_{cyc}a^2b\prod_{cyc}a^2c-abc(a+b+c)^3=\frac{1}{2}\sum_{sym}(3a^3b^3+2a^4bc+a^2b^2c^2-6a^3b^2c)\geq$$
$$\geq \frac{1}{2}\sum_{sym}\left(6\sqrt[6]{(a^3b^3)^3(a^4bc)^2a^2b^2c^2}-6a^3b^2c\right)=3\sum_{sym}\left(a^{\frac{19}{6}}b^{\frac{13}{6}}c^{\frac{2}{3}}-a^3b^2c\right)\ge0.$$
The last inequality is true because
$$\left(\frac{19}{6},\frac{13}{6},\frac{2}{3}\right)\succ(3,2,1).$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve a differential equation $(x^2y+y^5)dx+(x^3-xy^4)dy=0$?
Solve the following differential equation:
$$(x^2y+y^5)dx+(x^3-xy^4)dy=0$$
I noticed that it is not exact, since:
$$
(x^2y+y^5)'_y=x^2+5y^4\ne3x^2-y^4=(x^3-xy^4)'_x
$$
Then I tried to express $y'$:
$$
y'=\frac{x^2y+y^5}{xy^4-x^3}
$$
And now I don't understand what type of differential equation it is and how to solve it.
Could someone help me?
| Try the Ansatz $df=0$ with $f_x=x^ay^b(x^2y+y^5),\,f_y=x^ay^b(x^3-xy^4)$ so
$$\begin{align}0&=\frac{(f_x)_y-(f_y)_x}{x^ay^b}\\&=(b-a-2)x^2+(a+b+6)y^4\\\implies(a,\,b)&=(-4,\,-2).\end{align}$$The equations $f_x=\frac{1}{x^2y}+\frac{y^3}{x^4},\,f_y=\frac{1}{xy^2}-\frac{y^2}{x^3}$ are consistent with $f=-\frac{1}{xy}-\frac{y^3}{3x^3}$.The solution is that $f$ is constant.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to prove $\sum _{k=1}^{\infty } (-1)^k H_{\frac{2 k}{3}} = -\frac{\pi }{2 \sqrt{3}}+\frac{3 \pi }{8}-\frac{3}{4} \log (2)$? I stumbled on this problem in the wake of the discussion https://math.stackexchange.com/a/3553902/198592
Can you make sense of the equation in the question involving the harmonic number with a rational index
$H_{\frac{2}{3} k}$ although the series is not convergent?
Hint: find the generating function of $g(z) = \sum_{k=1}^{\infty}z^k H_{\frac{2 k}{3}}$ and interpret the sum as the limit $z\to -1$.
| Start with using the integral representation of the harmonic number $H_n=\int_0^1\frac{1-x^n}{1-x}\ dx$ we have
$$\sum_{k=0}^\infty(-1)^k H_{\frac{2k}{3}}=\int_0^1\frac{1}{1-x}\sum_{k=0}^\infty((-1)^k-(-x^{\frac23})^n)\ dx$$
$$\int_0^1\frac{1}{1-x}\left(\frac12-\frac{1}{1+x^{\frac23}}\right)\ dx\overset{x\to x^3}{=}-\frac32\int_0^1\left(\frac{x}{1+x^2}-\frac{1}{1+x^2}+\frac{1}{1+x+x^2}\right)\ dx$$
$$=-\frac32\left[\frac12\ln(1+x^2)-\tan^{-1}x+\frac{2}{\sqrt{3}}\tan^{-1}\left(\frac{1+2x}{\sqrt{3}}\right)\right]_0^1$$
$$= -\frac{3}{4} \ln2+\frac{3 \pi }{8}-\frac{\pi }{2 \sqrt{3}}$$
Note that I used Grandi series $\sum_{k=0}^\infty (-1)^k=\frac12$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $4^{1/3}+10^{1/2}$ is irrational Show that $4^{1/3}+10^{1/2}$ is irrational.
I start by assuming it to be rational and want to come to a contradiction
$$
4^{1/3}+10^{1/2} = r \\
\Rightarrow 4^{1/3} = r - 10^{1/2} \\
\Rightarrow 4 = (r-10^{1/2})^{3} \\
\Rightarrow 4 = r^{1/3} - 10^{1/3} + 3r^2*10^{1/2} - 30r
$$
Now I want to separate $10^{1/2}$ to one side and show the terms with $r$ is rational and thus a contradiction but what do I do with the $10^{1/3}$ on RHS?
| Since $\sqrt[3]{4}$ is a root of the polynomial $f(x) = x^3 - 4$, $r \stackrel{def}{=}\sqrt[3]{4} + \sqrt{10}$ is a root of
$$g(x) = f(x-\sqrt{10})f(x+\sqrt{10}) = x^6-30x^4-8x^3+300x^2-240x-984$$
By rational root theorem,
if $r = \frac{p}{q}$ is a rational root of $g(x)$ for coprime integers $p$ and $q$, then $q$ is a factor of $1$. This forces $q = \pm 1$ and $r$ to be an integer. Numerically, $$4 < r = \sqrt[3]{4} + \sqrt{10} \sim 4.749678712136578 < 5$$
and $r$ is not an integer. This means $r$ cannot be a rational root of $g(x)$ and hence $r$ is irrational.
Update
For an alternate proof which doesn't involve the horrible sextic polynomial,
just expand the equality $f(r - \sqrt{10}) = 0$. You will get
$$\begin{align} & r^3-3\sqrt{10}r^2+30r-10\sqrt{10} - 4 = 0\\
\iff & (r^3 + 30r - 4) - (3r^2+10)\sqrt{10} = 0\\
\implies & \sqrt{10} = \frac{r^3 + 30r - 4}{3r^2 + 10}\end{align}$$
If $r$ is rational, last equality tell us $\sqrt{10}$ will be rational too. This contradicts with the known fact that $\sqrt{10}$ is irrational. As a result,
$r$ cannot be rational.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\binom{n}{0} + \frac{1}{2} \binom{n}{1} + \frac{1}{3} \binom{n}{2} + ... + \frac{1}{n+1} \binom{n}{n} = \frac{2^{n + 1} - 1}{n + 1}$. I have to prove the identity:
$$\binom{n}{0} + \dfrac{1}{2} \binom{n}{1} + \dfrac{1}{3} \binom{n}{2}
+ ... + \dfrac{1}{n+1} \binom{n}{n} = \dfrac{2^{n + 1} - 1}{n + 1}$$
First thing I tried was to think about a combinatorial proof, but I couldn't interpret the identity in any way that makes sense. So, how should I prove this?
| We can find the probabilistic interpretation of both sides of the equation: Imagine having $n+1$ people. We then select every possible nonempty subset of them (there are exactly $2^{n+1}-1$ such subsets) and then select one person from it uniformly at random. Both sides are equal to the expected value of the number of times the first person was selected. The RHS iterates over all possible set sizes containing this person and multiplies the number of them by the respective probability. The RHS makes use of the fact that there will be $2^{n+1}-1$ selections in total and that person one is exactly the same as everyone else. Therefore dividing one quantity by the other will yield the answer by virtue of symmetry.
| {
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"timestamp": "2023-03-29T00:00:00",
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Primes in the form $n^2+4$ For what numbers $n$ is $n^2+4$ prime?
Obviously $n$ has to be odd number (if $n$ is even, then $n^2+4>2$ is also even, so it couldn't be a prime number).
So far I recognize that the latest digit of $n$ is not 9 and 1, because then $n^2\equiv 1 \pmod{5}$, so $n^2+4\equiv 0 \pmod{5}$.
Unfortunately, that's all I get so far. I checked that if $n\in\{1,3,5,7,13,15,17\}$, then $n^2+4$ is prime, but for $n=23$ we have
$$n^2+4=23^2+4=529+4=533=13 \cdot 41,$$
so $23^2+4$ is not a prime number.
| A partial result is that if $n>1$ is itself a square, then $n^2+4$ will not be a prime. For if $n=m^2$ then:
$$n^2+4=m^4+4=(m^2+2m+2)(m^2-2m+2)$$
and since $m=\sqrt{n}>1$ both factors will be $>1$.
Also, if $n=ax+b$ with $a>1$ and $b^2+4=ka$ for some $k$, then $n^2+4$ will not be a prime since:
$$n^2+4=(ax+b)^2+4=a^2x^2+2abx+(b^2+4)= a(ax^2+2bx+k)$$
For example if $a=5$ the condition is satisfied by $b=1$ or $4$ which correspond (with $a$ even and odd respectively) to your observation that $n^2+4$ will not be prime when the last digit of $n$ is $1$ or $9$. Other examples satisfying the condition are:
$\qquad n=13x+3$ and $n=13x+10$
$\qquad n=17x+8$ and $n=17x+9$
$\qquad n=29x+5$ and $n=29x+24$
| {
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An interesting question of algebraic manipulation If $a+b+c+d+e=0$ then prove that $a^3+b^3+c^3+d^3+e^3=3(abc + abd + abe + acd + ace + ade + bcd + bce + bde + cde)$ Extend this argument to n integers such that if $a_1+a_2+a_3+\cdots a_n=0$ then $$a_1^3+a_2^3+a_3^3\cdots a_n^3=3\left(\sum_{i>j>k} a_ia_ja_k\right)$$
My try: At first I tried it for the base case. $a+b+c=0$ then it is a well-known fact that $a^3+b^3+c^3=3abc$, which proves the base case.Let us assume that the given statement is true for some $n$. $$a_1^3+a_2^3+a_3^3\cdots a_n^3=3\left(\sum_{i>j>k} a_ia_ja_k\right)$$
$$a_1^3+a_2^3+a_3^3\cdots a_n^3+a_{n+1}^3=3\left(\sum_{i>j>k} a_ia_ja_k\right)$$
Subtract the two equations we get,
$$a_{n+1}^3=3\left[(a_1a_2a_{n+1}+a_1a_3a_{n+1}+\cdots a_1a_na_{n+1})+(a_2a_3a_{n+1}+a_2a_4a_{n+1}+\cdots a_2a_na_{n+1})+\cdots +(a_{n-1}a_na_{n+1})\right]$$
I am stuck here. I cannot proceed further. Can please anybody help me?
| Because $$\left(\sum_{k=1}^na_k\right)^3=\sum_{k=1}^na_k^3+3\sum_{k=1}^na_k\sum_{1\leq i<j\leq n}a_ia_j-3\sum_{1\leq i<j<k\leq n}a_ia_ja_k.$$
We can get it by the following way.
Firstly, it's obvious that $$\left(\sum_{k=1}^na_k\right)^3=\sum_{k=1}^na_k^3+3\sum_{k=1}^na_k\sum_{1\leq i<j\leq n}a_ia_j+K\sum_{1\leq i<j<k\leq n}a_ia_ja_k$$ for some real $K$.
But for $a_1=a_2=...=a_n=1$ we obtain:
$$n^3=n+3n\cdot\frac{n(n-1)}{2}+K\cdot\frac{n(n-1)(n-2)}{6},$$
which gives $K=-3$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Divisibility Trick for 11 I am trying to prove the 11 alternating sum divisibility trick. I know that $10\equiv -1\pmod{11}$ so for every power of $10$ in a number, we should be able to substitute in $(-1)$ like so:
$$ a(10^n) + b(10^{(n-1)}+\cdots+ c(10) + d \equiv a(-1)^{n} + b(-1)^{n-1}+\cdots+c(-1)+d\pmod{11}$$
What I am having trouble understanding is the fact that starting from the left, we are taking an alternating sum of the digits $(a - b + \cdots$ ). Since each power of $10$ will always become a power of $(-1)$, how can we always use an alternating sum from the left side? That seems like it would change the underlying nature of what is happening with the mod depending on whether we have an even or odd number of digits in the number. For example, when we have a three digit number $a(-1)^3+b(-1)^2+c = -a+b-c$ but according to the rule we have $a-b+c$ verses when we have a four digit number $a(-1)^4+b(-1)^3+c(-1)^2+d = a-b+c-d$ which works with the rule.
Am I missing something obvious? Do I possibly just have the alternating sum rule written incorrectly in my notes?
Thanks!
| Why are you shifting the variable names. You can but its confusing. It=n $a(-1)^2 + b(-1)^1+c=a-b+c$ you have $a,c$ in the odd positions and $b$ in the even position.
But with $a(-1)^3+b(-1)^2+c(-1)^1 + d= -a+b-c+d$ you have $a,c$ switching to the even position, while $b$ switches to the odd. And $d$ is new in the odd.
For less confusing to list from the right.
$a = a$ ($a$ in odd, nothing in even [divisible by $11$ if and only if $a = 0$])
$b(-1)^1 + a = -b + a$ ($a$ in odd; $a$ will always be in odd. $b$ in even.)
$c(-1)^2 + b(-1)^1 + a = c - b + a$ ($a,c$ in odd; $b$ in even.)
$d(-1)^3 +c(-1)^2 +b(-1)^1 + a = -d + c -b+a$ ($a,c$ in odd; $b$ in even.... and so on.)
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding the MacLaurin series of $\frac{x+3}{2-x}$ I did:
$$\frac{x+3}{2-x} = \frac{x}{2-x}+\frac{3}{2-x} = x(\frac{1}{2-x})+3(\frac{1}{2-x}) = \\
= \frac{x}{2}(\frac{1}{1-\frac{x}{2}})+\frac{3}{2}(\frac{1}{1-\frac{x}{2}}) = \\
= \frac{x+3}{2}\sum(\frac{x}{2})^n = ...?$$
The answer my professor got was $\frac{(3 + x)}{(2 - x)} = \frac{3}{2} + \sum_{n=1}^∞ \frac{x^n5}{ 2^{n+1}}$
Unfortunately I forgot to write down how he did it. How do I get that solution and is mine necessarily incorrect/incomplete?
| Use $\frac{1}{1-t} = 1+t+t^2+t^3+\>...$ to get,
$$\frac{3 + x}{2 - x} = -1 + \frac{\frac52}{1 - \frac x2}
= -1+\frac52\left(1+ \frac x2 + \frac {x^2}{2^2} + \frac {x^3}{2^3}+\> …\right)= \frac{3}{2} + \sum_{n=1}^∞ \frac{5x^n}{ 2^{n+1}} $$
| {
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Given positives $a, b, c$ such that $a + b + c = 3$, prove that $\sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} \le \frac{1}{2}$.
Given positives $a, b, c$ such that $a + b + c = 3$, prove that $$\frac{1}{c^2 + 4a^2 + b^2} + \frac{1}{a^2 + 4b^2 + c^2} + \frac{1}{b^2 + 4c^2 + a^2} \le \frac{1}{2}$$
We have that $$a^2 + 4b^2 + c^2 = a^2 + (a + b + c + 1)b^2 + c^2 = (b^2 + a)a + (b + 1)b^2 + (b^2 + c)c$$
$$\implies \sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} = \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\frac{(a + b + c)^2}{(b^2 + a)a + (b + 1)b^2 + (b^2 + c)c}$$
$$ \le \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\left(\frac{a}{b^2 + a} + \frac{1}{b + 1} + \frac{c}{b^2 + c}\right)$$
Furthermore, $\dfrac{a}{c^2 + a} + \dfrac{c}{a^2 + c} = 1 - \dfrac{(c + a - 2)ca}{(c^2 + a)(a^2 + c)}$, $$\sum_{cyc}\frac{1}{a^2 + 4b^2 + c^2} \le \frac{1}{(a + b + c)^2} \cdot \sum_{cyc}\left[\frac{1}{b + 1} - \frac{(c + a - 2)ca}{(c^2 + a)(a^2 + c)}\right] + \frac{1}{3}$$
Then I don't know what to do next.
| We need to prove that
$$\sum_{cyc}\frac{1}{b^2+c^2+4a^2}\leq\frac{9}{2(a+b+c)^2}$$ or
$$\sum_{cyc}(2a^6-4a^5b-4a^5c+13a^4b^2+13a^4c^2-4a^4bc-12a^3b^3-12a^3b^2c-12a^3c^2b+20a^2b^2c^2)\geq0$$ or
$$\sum_{cyc}(a-b)^2(2c^4+2(a^2-4ab+b^2)c^2+a^4-2a^3b+4a^2b^2-2ab^3+b^4)\geq0$$ for which it's enough to prove that:
$$(a^2-4ab+b^2)^2-2(a^4-2a^3b+4a^2b^2-2ab^3+b^4)\leq0$$ or $$(a-b)^2(a^2+6ab+b^2)\geq0$$ and we are done!
| {
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"timestamp": "2023-03-29T00:00:00",
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$k\cos\beta \sin\alpha-\sin\beta+d\sqrt{1+k^2}=0$ ，Find $\beta$ $k\cos\beta \sin\alpha-\sin\beta+d\sqrt{1+k^2}=0$. Find $\beta$.
$K$ and $d$ are constant.
$d∈[0,\pm1],k∈[0,\pm\infty)$.
This is an equation in a spherical coordinate system. Where $\alpha$ is longitude and $\beta$ is latitude.
| \begin{align}
& A\cos\beta + B\sin\beta \\[10pt]
= {} & \sqrt{A^2+B^2} \left( \frac A{\sqrt{A^2+B^s}} \cos\beta + \frac B {\sqrt{A^2+B^2}} \sin\beta \right) \\[10pt]
= {} & \sqrt{A^2+B^2} \Big( \sin\gamma \cos\beta + \cos\gamma\sin\beta\Big) \\
& \quad \text{where } \tan\gamma = A/B \\[10pt]
= {} & \sqrt{A^2+B^2} \, \sin(\beta+\gamma)
\end{align}
You have $A=k\cos\alpha$ and $B=-1.$
| {
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Prove $\cos x +\cos y = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$ Prove that $$\cos(x) + \cos(y) = 2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})$$ holds true for any $x, y \in \mathbb{R}$.
Even though I managed to prove its brother $\sin(x) + \sin(y)$, I haven't been able to tackle this one.
Important identities needed for the proof:
$$\cos(x+y)=\cos(x)\cos(y)-\sin(x)\sin(y)$$
$$\cos(x-y)=\cos(x)\cos(y)+\sin(x)\sin(y)$$
Let's go:
$$2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2}) = 2\cos(\frac{x}{2}+ \frac{y}{2})\cos(\frac{x}{2} - \frac{y}{2}) = $$
$$ = 2(\cos(\frac{x}{2})\cos(\frac{y}{2}) - \sin(\frac{x}{2})\sin(\frac{y}{2}))(\cos(\frac{x}{2})\cos(\frac{y}{2}) + \sin(\frac{x}{2})\sin(\frac{y}{2})) = $$
$$ = 2(\cos^2(\frac{x}{2})\cos^2(\frac{y}{2}) - \sin^2(\frac{x}{2})\sin^2(\frac{y}{2})) = $$
$$ = 2\cos^2(\frac{x}{2})\cos^2(\frac{y}{2}) - 2\sin^2(\frac{x}{2})\sin^2(\frac{y}{2})$$
Now I tried, I believe, almost every possible replacement by deriving from $\cos^2(x) + \sin^2(x) = 1$ and sadly nothing worked.
| $$\cos(a+b) = \cos(a)\cos(b) - \sin(a)\sin(b)$$
$$\cos(a-b) = \cos(a)\cos(b) + \sin(a)\sin(b)$$
Adding the two,
$$\cos(a+b)+\cos(a-b) = 2\cos(a)\cos(b)$$
Setting $a = \frac{x+y}{2}$ and $b = \frac{x-y}{2}$, we get
$$\cos(x) + \cos(y) = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$
as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the range of $x^{T}Ax$
Let $A=
\begin{pmatrix}
1 & -2 & -1 \\
-2 & 1 & 1 \\
-1 & 1 & 4
\end{pmatrix}, x =
\begin{pmatrix}
x_1\\
x_2\\
x_3
\end{pmatrix}$ with $x^{T}x = 1$, what's the range of $x^TAx$ subject to $(1,1,1)x = 0$?
We can diagonize $A$ to $\begin{pmatrix}
-1 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 2
\end{pmatrix}$, then the question becomes finding the range of $y^T\begin{pmatrix}
-1 & 0 & 0 \\
0 & 5 & 0 \\
0 & 0 & 2
\end{pmatrix}y $ with $y^Ty=1$ subject to $(\sqrt2, \frac{-2}{\sqrt 6}, \frac{-1}{\sqrt 3})y=0$, but still it seems hard to calculate, how to calculate it efficiently (by hand not computer)?
| The intersection of the sphere $x^T x = 1$ with the plane $(1,1,1) \; x = 0$ is a circle. It may be parameterized as
$$ \pmatrix{x_1\cr x_2\cr x_3\cr} = \cos(\theta) \pmatrix{1/\sqrt{2}\cr 0 \cr -1/\sqrt{2}\cr} + \sin(\theta) \pmatrix{1/\sqrt{6}\cr -2/\sqrt{6}\cr 1/\sqrt{6}\cr} $$
Compute $x^T A x$ as a function of $\theta$. It simplifies nicely to a function of the form $a + b \cos(2\theta)+ c \sin(2\theta)$ (I'll let you find $a,b,c$), which can be written as $a + \sqrt{b^2 + c^2} \cos(2 (\theta - \theta_0))$. So the range is $[a - \sqrt{b^2+c^2}, a + \sqrt{b^2 + c^2}]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Use epsilon-delta definition of limit to establish the following: $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$ I understand that my solution here is probably not the most efficient (My professor's solution is "cleaner") but it is how my mind attacked the problem. I have been losing lots of points for minor details that I've been unable to see. Does the following proof hold? Am I making any major (or minor) errors?
\begin{align*}
\left| \frac{1}{2+\sqrt{x}}-\frac{1}{3}\right|&\leq\left|\frac{1}{2+\sqrt{x}}\right|+\left|\frac{1}{3}\right|<\epsilon~~~\mbox{(by triangle inequality)}\\
&\implies\left|\frac{1}{2+\sqrt{x}}\right|+\frac{1}{3}<\epsilon\\
&\implies\left|\frac{1}{2+\sqrt{x}}\right| < \epsilon-\frac{1}{3}\\
&\implies \frac{1}{2} < \epsilon-\frac{1}{3}~~~~\mbox{(Because, }\sqrt{x}~\mbox{only a real number when } x\geq 0.)\\
&\implies 1<2(\epsilon-\frac{1}{3})\\
&\implies \left|x-1\right|<2\epsilon-\frac{2}{3}=\delta~~~~\mbox{(Because, choosing }x~s.t.~0<x<2\implies~-1<x-1<1)\\
\end{align*}
$\therefore \left|x-1\right|<\delta\implies\left| \frac{1}{2+\sqrt{x}}-\frac{1}{3}\right|<\epsilon$ and $\displaystyle\lim_{x\to 1}\frac{1}{2+\sqrt{x}}=\frac{1}{3}$
| You need to relate $\displaystyle \left| \frac{1}{2+\sqrt{x}} - \frac{1}{3} \right|$ with $|x-1|$ in a proper way:
$\displaystyle \left| \frac{1}{2+\sqrt{x}} - \frac{1}{3} \right| = \left| \frac{1 - \sqrt{x}}{3(2+\sqrt{x})} \right| = \left| \frac{1-x}{3(1+\sqrt{x})(2+\sqrt{x})} \right| \leq \frac{|x-1|}{3\cdot1\cdot2} = \frac{|x-1|}{6}$ so for a given
$\epsilon > 0$, choosing $\delta=\epsilon$ gives you for $0<|x-1|<\delta = \epsilon$,
$\displaystyle \left| \frac{1}{2+\sqrt{x}} - \frac{1}{3} \right| \leq \frac{|x-1|}{6} < \frac{\delta}{6} = \frac{\epsilon}{6} < \epsilon$.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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Confused about positive and negative signs: Find the value of $\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$. Without tables or a calculator, find the value of $\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$.
I do not understand how the positive/negative signs are obtained as shown in the book; is there a formula for expanding these kind of things (what kind of expression is it, by the way?)?
This is my solution:
$\displaystyle\frac{(\sqrt5 +2)^6 - (\sqrt5 - 2)^6}{8\sqrt5}$
$= \displaystyle\frac{[(\sqrt5+2)^3+(\sqrt5-2)^3][(\sqrt5+2)^3-(\sqrt5-2)^3]}{8\sqrt5}$
$=\displaystyle\frac{(\sqrt5+2+\sqrt5-2)[(\sqrt5+2)^2\color{red}{+}(\sqrt5+2)(\sqrt5-2)+(\sqrt5-2)^2](\sqrt5+2-\sqrt5+2)[(\sqrt5+2)^2\color{red}{-}(\sqrt5+2)(\sqrt5-2)+(\sqrt5-2)^2]}{8\sqrt5}$
$=\displaystyle\frac{[2\sqrt5(5+4\sqrt5+4+\color{red}{5-4}+5-4\sqrt5+4][4(5+4\sqrt5+4\color{red}{-(5-4)}+(5-4\sqrt5+4)]}{8\sqrt5}$
$=\displaystyle\frac{2584\sqrt5}{8\sqrt5}$
$=323$
Because of the multiplication, I still got the same answer as given in the book. However, is the book or I correct in terms of the positive/negative signs(in red)?
| Hint
$a-b=4$
$a+b=2\sqrt5$
$ab=1$
$a^3-b^3=(a-b)^3+3ab(a-b)=?$
$a^3+b^3=(a+b)(a^2-ab+b^2)=(a+b)((a-b)^2+ab)=?$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
} |
Show $\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\binom{a+n}{n}$ How it can be shown that:
$$\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\binom{a+n}{n}$$
Where $a \ne 0$
My try:
$$\sum_{k=0}^{n}2^{n-k}\binom{a+k}{k}\frac{a-k}{a+k}=\sum_{k=0}^{n}2^{n-k}\binom{a+k}{a}\frac{a-k}{a+k}$$
$$=\frac{1}{a}\sum_{k=0}^{n}2^{n-k}\binom{a+k-1}{a-1}\left(a-k \right)$$
$$=\sum_{k=0}^{n}2^{n-k}\binom{a+k-1}{a-1}-\frac{1}{a}\color{red}{\sum_{k=0}^{n}2^{n-k}\binom{a+k-1}{a-1}k }$$
On the other hand:
$$\color{red}{\sum_{k=0}^{n}2^{n-k}\binom{a+k-1}{a-1}k}=\sum_{k=0}^{n}\left(a+k-1\right)2^{n-k}\binom{a+k-2}{k-1} $$
But computing these expressions takes much time and I think there should be a better way, but I cannot find that way.
Please if it's possible, then do the proof using elementary ways.
| I know you would prefer a combinatorial proof and thinks that induction does not tell us anything, but since you would like to know an algebraic proof, I'll just include an inductive proof for completion's sake.
Base case $n = 0$:
$$
\sum_{k=0}^0 2^{-k}{a + k \choose k}\frac{a - k}{a + k} = 2^0 {a \choose 0}\frac{a}{a} = 1 = {a \choose 0}
$$
Assume true for $n = m$. For $n = m+1$:
\begin{align*}
\sum_{k=0}^{m+1} 2^{m+1-k}{a + k \choose k}\frac{a - k}{a + k} &= 2\sum_{k=0}^m, 2^{m-k}{a + k \choose k}\frac{a - k}{a + k} + {a + m + 1 \choose m + 1}\frac{a - m - 1}{a + m + 1} \\
&= 2{a + m \choose m} + {a + m + 1 \choose m + 1}\left(\frac{a - m - 1}{a + m + 1}\right) \\
&= 2\left(\frac{m + 1}{a + m + 1}\right){a + m + 1 \choose m + 1} + \left(\frac{a - m - 1}{a + m + 1}\right) {a + m + 1 \choose m + 1} \\
&= \left(\frac{a - m - 1 + 2(m + 1)}{a + m + 1}\right) {a + m + 1 \choose m + 1} \\
&= {a + m + 1 \choose m + 1}
\end{align*}
This completes the induction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3579670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Find center of circle, which tangent to $y=2$ at $(3,2)$ and $y=-x\sqrt 3 +2$ Using $$r=\frac{|Ax + By + c|}{\sqrt{A^2+B^2}}$$
I get $$b=2+a\sqrt3$$ and stuck,because the options are on numbers. What do i do next?
| Suppose $C=(a,b)$ is the center of the circle.
Then $a=3$ because the circle is tangent to $y=3$ at point $(3,2)$.
Also the distance of $C$ from both lines $y=2$ and $y=-x\sqrt{3}+2$ is same:
$$|b-2|=\frac{|b+3\sqrt{3}-2|}{2}$$
and note that we have two solutions for $b<2$ and $b>2$.
can you proceed?
Edit: If $b>2$, then $b+3\sqrt{3}-2>3\sqrt{3}>0$, thus we have
$$b-2=\frac{b+3\sqrt{3}-2}{2}$$
which gives $b=2+3\sqrt{3}$.
If $b<2$, $|b-2|=2-b$, but $|b+3\sqrt{3}-2|$ is $\pm(b+3\sqrt{3}-2)$
thus $$2-b=\frac{b+3\sqrt{3}-2}{2}$$
which gives $b=2-\sqrt{3}$ or
$$2-b=-\frac{b+3\sqrt{3}-2}{2}$$
which leads to repeated value $b=2+3\sqrt{3}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Can I solve this other than using Newton's sums or Newton's identities?
$\begin{cases}
&x + y + z = 12\\& x^2 + y^2 + z^2 = 12
\\ & x^3 +y^3 + z^3 = 12
\end{cases}$
If $x,$ $y,$ and $z$ satisfy the system of equations above, what is the value of $x^4+y^4+z^4?$
People told me that this can be solved by using Newton's sums or Newton's identities which I don't know how. Does anybody know how to do that or any other method to solve this?
|
People told me that this can be solved by using Newton's sums or Newton's identities which I don't know how. Does anybody know how to do that or any other method to solve this?
We can solve this by doing this method:
$\begin{align} (x+y+z)^2 &\rightarrow x^2 + y^2 + z^2 + 2(xy+xz+yz) = 12^2 \\
&\rightarrow 12 + 2(xy+xz+yz) = 144 \\
& \rightarrow 2(xy+xz+yz) = 132 \\
& \rightarrow (xy+xz+yz) = 66\end{align}$
$\begin{align} (x+y+z)^3 &\rightarrow x^3 + y^3 + z^3 + 3(xy+xz+yz)(x+y+z) – 3xyz = 12^3 \\
&\rightarrow 12 + 3(66)(12) – 3xyz = 1728 \\
&\rightarrow 2,388 – 3xyz = 1728 \\
&\rightarrow – 3xyz = -660 \\
& \rightarrow xyz = 220 \end{align}$
$\begin{align} (xy+xz+yz)^2 & \rightarrow (xy)^2 + (xz)^2 + (yz)^2 + 2xyz(x+y+z) = 66^2 \\
& \rightarrow (xy)^2 + (xz)^2 + (yz)^2 + 2xyz(x+y+z) = 66^2 \\
& \rightarrow (xy)^2 + (xz)^2 + (yz)^2 + 2(220)(12) = 4356 \\
& \rightarrow (xy)^2 + (xz)^2 + (yz)^2 + 5280 = 4356 \\
& \rightarrow (xy)^2 + (xz)^2 + (yz)^2 = -924 \end{align}$
$\begin{align} (x^2 + y^2 + z^2)^2 & \rightarrow x^4 + y^4 + z^4 + 2\left( (xy)^2 + (xz)^2 + (yz)^2 \right) = 12^2 \\
&\rightarrow x^4 + y^4 + z^4 + 2\left( -924 \right) = 144 \\
&\rightarrow x^4 + y^4 + z^4 + (-1848) = 144 \\
&\rightarrow x^4 + y^4 + z^4 + (-1848) = 144 \\
& \rightarrow x^4 + y^4 + z^4 = \boxed{1992} \end{align}$
Obviously we are deaing with complex numbers here, $x = 2.467+5.005i,$ $y = 7.066,$ $z = 2.467-5.005i$ is one solution.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 2
} |
Interpreting an olympiad inequality as a convex function We need to prove the following inequality:
$$\frac{y^3xz}{x^3(xy+z^2)}+\frac{z^3xy}{y^3(x^2+yz)}+\frac{x^3yz}{z^3(xz+y^2)}\geq \frac{3}{2}$$
This equation is convex in each of the variables $x,y,z$. Moreover, its minimum seems to be at the origin (the equation is homogeneous, and there are not displacements of the vertex).
Can I then say that the minima is achieved when $x=y=z$? Clearly, when $x=y=z$, the value of the inequality is $\frac{3}{2}$
| $$\Longleftrightarrow \sum\dfrac{y^3z}{x^2(xy+z^2)}\ge\dfrac{3}{2}$$
take
$$a=\dfrac{x}{z},b=\dfrac{y}{x},c=\dfrac{z}{y}$$where $abc=1$
$$\Longleftrightarrow \sum \dfrac{b^2}{a+b}\ge \dfrac{3}{2}$$
use Cauchy-Schwarz inequality we have
$$\sum\dfrac{b^2}{a+b}\ge\dfrac{(a+b+c)^2}{\sum(a+b)}=\dfrac{a+b+c}{2}\ge\dfrac{3}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
The value of the determinant
Calculate the value of determinant:
$$D = \begin{vmatrix}
0 & 1 & 2 & ... & 2020 \\
1 & 0 & 1 & ... & 2019 \\
2 & 1 & 0 & ... & 2018 \\
... & ... & ... & ... & ... \\
... & ... & ... & ... & ... \\
2019 & 2018 & 2017 & ... & 1 \\
2020 & 2019 & 2018 & ... & 0 \\
\end{vmatrix}$$
I tried to change $L_k$ with $L_{n-k}$ and i got a circular determinant but i don't know to solve it.
| Let's do this for a $5 \times 5$ matrix and hopefully it would make sense how to generalize this. The key idea is that when you apply operations of Gaussian Elimination to the matrix,
*
*flipping rows multiplies the result by
*rescaling a row by $k$ rescales the determinant by $k$ as well
*adding multiples of rows to other rows does not change the determinant.
So,
$$
\begin{split}
D &= \begin{vmatrix}
0 & 1 & 2 & 3 \\
1 & 0 & 1 & 2 \\
2 & 1 & 0 & 1 \\
3 & 2 & 1 & 0 \\
\end{vmatrix}
= \begin{vmatrix}
-1 & 1 & 2 & 3 \\
1 & 0 & 1 & 2 \\
1 & 1 & 0 & 1 \\
1 & 2 & 1 & 0 \\
\end{vmatrix}
= \begin{vmatrix}
-1 & -1 & 2 & 3 \\
1 & -1 & 1 & 2 \\
1 & 1 & 0 & 1 \\
1 & 1 & 1 & 0
\end{vmatrix}
= \begin{vmatrix}
-1 & -1 & -1 & 3 \\
1 & -1 & -1 & 2 \\
1 & 1 & -1 & 1 \\
1 & 1 & 1 & 0
\end{vmatrix} \\
& = \begin{vmatrix}
-1 & -1 & -1 & 3 \\
0 & -2 & -2 & 5 \\
0 & 0 & -2 & 4 \\
0 & 0 & 0 & 3
\end{vmatrix}
= (-1) \cdot (-2) \cdot (-2) \cdot 3 = 12.
\end{split}
$$
We subtract column 2 from column 1, and then column 3 from column 2, and then column 4 from column 3. The second row starts by subtracting the first row from every other row to get a diagonal matrix....
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that: $\sin \,\cot^{-1} \cos \,\tan^{-1}x = \sqrt{\dfrac{x^2+1}{x^2+2}} $. Prove that: $\sin \,\cot^{-1} \cos \,\tan^{-1}x = \sqrt{\dfrac{x^2+1}{x^2+2}} $.
This question was taken from Miscellaneous Example of S.L. Loney's Trigonometry.
Since, the question involves 2 parts, LHS and RHS, LHS is completely Trigonometric while, RHS is algebraic, so my approach was to get taylor series for $x+1$ then differentiate it and keep on doing to show that both sides are same... But, I wonder if there is an trigonometric solution for the question too or not!
Please help.
| Let $y=\tan^{-1}x$. Then $\tan y=x$ and so drawing a right triangle with angle $y$ and sides $1,x,\sqrt{1+x^2}$, we find $\cos y=\dfrac{1}{\sqrt{1+x^2}}$.
Now let $t=\cot^{-1}(\cos y)$. Then $\cot t=\cos y=\dfrac{1}{\sqrt{1+x^2}}$, so drawing a right triangle with angle $t$, we find the hypotenuse is $\sqrt{2+x^2}$, so $\sin t=\sqrt{\dfrac{1+x^2}{2+x^2}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3585756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p}{n^{p+1}}$
Evaluate
$\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p}{n^{p+1}}$ using Stolz-Cesaro theorem .
Now this is my attempt :$\lim\limits_{n\to \infty}\frac{1^p+3^p+\dots+(2n-1)^p+(2n)^p-1^p-3^p-\dots-(2n-1)^p}{(n+1)^{p+1}-n^{p+1}}$=$\lim\limits_{n\to \infty}\frac{(2n)^p}{(n+1)^{p+1}-n^{p+1}}$ (and now i was thinking to use the binomial theorem )
$\lim\limits_{n\to \infty}\frac{(2n)^p}{n^{p+1}+{p+1 \choose p}n^p+\dots+1-n^{p+1}}$,which will eventually lead to the answer : $\frac{2}{p+1}$ . Is this correct ?
| Let $ n $ be a positive integer greater than $ 1 \cdot $
Since $ \sum\limits_{k=0}^{n-1}{\left(2k+1\right)^{p}}=\sum\limits_{k=1}^{2n}{k^{p}}-\sum\limits_{k=1}^{n}{\left(2k\right)^{p}}=\sum\limits_{k=1}^{n}{k^{p}}+\sum\limits_{k=n+1}^{2n}{k^{p}}-2^{p}\sum\limits_{k=1}^{n}{k^{p}}$ $ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\left(1-2^{p}\right)\sum\limits_{k=1}^{n}{k^{p}}+\sum\limits_{k=1}^{n}{\left(n+k\right)^{p}} \cdot $
We have : $ \frac{1}{n^{p+1}}\sum\limits_{k=0}^{n-1}{\left(2k+1\right)^{p}}=\left(1-2^{p}\right)\left(\frac{1}{n}\sum\limits_{k=1}^{n}{\left(\frac{k}{n}\right)^{p}}\right)+\frac{1}{n}\sum\limits_{k=1}^{n}{\left(1+\frac{k}{n}\right)^{p}}$
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underset{n\to +\infty}{\longrightarrow}\left(1-2^{p}\right)\int\limits_{0}^{1}{x^{p}\,\mathrm{d}x}+\int\limits_{0}^{1}{\left(1+x\right)^{p}\,\mathrm{d}x} $
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \underset{n\to +\infty}{\longrightarrow}\frac{1-2^{p}}{p+1}+\frac{2^{p+1}-1}{p+1} $
$ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{n^{p+1}}\sum\limits_{k=0}^{n-1}{\left(2k+1\right)^{p}}\underset{n\to +\infty}{\longrightarrow}\frac{2^{p}}{p+1} $
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Solve this equation : $\tan^{-1} \frac{x+1}{x-1} + \tan^{-1} \frac{x-1}{x} = \tan^{-1} (-7)$ Solve this equation : $\tan^{-1} \frac{x+1}{x-1} + \tan^{-1} \frac{x-1}{x} = \tan^{-1} (-7)$
This was an exam question, my try was as follows:
$$ \tan^{-1} \frac{x+1}{x-1} = \tan^{-1} (-7) - \tan^{-1} \frac{x-1}{x} $$
Now, assuming that $x = \tan x $ and substituting that in the above equation so that the equation changes and the $\tan^{-1} \frac{x-1}{x}$ vanishes, but, there is also one tan inverse function, so how to remove it?
Thanks :)
| Take tangents on the both sides, you'll get:
$$ \dfrac{\tan \left [ \tan^{-1} \frac{x+1}{x-1} \right] + \tan \left [ \tan^{-1} \frac{x-1}{x} \right]}{ 1 - \tan \left [ \tan^{-1} \frac{x+1}{x-1} \right] \tan \left [ \tan^{-1} \frac{x-1}{x} \right]} = -7 $$
$$ \frac{2x^2 - x + 1}{1-x} = -7 $$
Upon solving that equation, you'll get $x=2 \blacksquare$
PS: I used Wolframalpha to solve that equation, you may manually do it.
Note: I changed the question a bit, because I think that your question title is a bit incorrect, this question is example 6 of SL Loney's Inverse Trig Function
| {
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"timestamp": "2023-03-29T00:00:00",
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If the equation of the curve on the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ about the line $x-y-2=0$ is ...
If the equation of the curve on the reflection of the ellipse $\frac{(x-4)^2}{16}+\frac{(y-3)^2}{9}=1$ about the line $x-y-2=0$ is $16x^2+9y^2+k_1x-36y+k_2=0$, then find $k_1$ and $k_2$
Before solving it, I noticed a problem with it. Even if we reflect the curve, the coefficient of $x^2$ is 9, and not 16 according to what the question says.
That’s all I have as doubt. I don’t need the whole answer, just need to know if the question is right or not.
| Consider the line $y=x-2$;
New coordinates : $X=x$; and $Y=y+2$;
Reflect about $Y=X.$
The given curve expressed in $X,Y:$
$(X-4)^2/4^2+(Y-5)^2/3^2=1$;
The reflected curve:
$(Y-4)^2/4^2+(X-5)^2/3^2=1$;
Back to $x,y$:
$(y-2)^2/4^2+(x-5)^2/3^2=1$;
$9(y-2)^2+16(x-5)^2=16\cdot 9$;
Can you finish?
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $ \int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx$
Evaluate
$$ \int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx,$$
My attempt : $u=\tan\frac{x}{2} \rightarrow x=2\arctan(u) \rightarrow \frac{2}{1+u^2}du=dx$
$$\sin x= \frac{2\tan\frac{x}{2}}{1+\tan^2\frac{x}{2}}=\frac{2u}{1+u^2}$$
$$\cos x= \frac{1-\tan^2\frac{x}{2}}{1+\tan^2\frac{x}{2}}= \frac{1-u^2}{1+u^2}$$
I was looking to see if this is the right way to go and also i am not sure how to evaluate the boundaries . Could you help me out ?
|
\begin{align} I&=\int_{\pi/4}^{\pi/2} \frac{2\sin x+\cos x}{\sin x+2\cos x}\,dx\tag{1}\label{1}.\end{align}
\begin{align}
I&=
\int_{\pi/4}^{\pi/2}
\frac{2\cot x+1}{\cot x+2}\,dx
\overset{\color{blue}{t=\cot x}}
{=}
\int_0^1
\frac{2+t}{(1+2t)(1+t^2)}
\, dt
\tag{2}\label{2}
\\
&=
\int_0^1
\frac{2+t}{(1+2t)(1+t^2)}
\, dt
=
\int_0^1
\tfrac 65\,\frac 1{1+2t}
+\tfrac45\,\frac1{1+t^2}
-\tfrac35\,\frac {t}{1+t^2}
\, dt
\tag{3}\label{3}
\\
&=
\left(
\tfrac 65\cdot\tfrac12\,\ln(1+2t)
\right)_0^1
+
\left(
\tfrac45\cdot\arctan(t)
\right)_0^1
-
\left(
\tfrac35\cdot
\tfrac12\,\ln(1+t^2)
\right)_0^1
\\
&=
\tfrac35\,\ln(3)+\tfrac\pi5-\tfrac3{10}\,\ln(2)
\approx 1.07954175
.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3590666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the function $f(x)$ whose graph passes through the point $(0,\frac{4}{3})$ and whose derivative is: $f'(x) = x \sqrt{16-x^2}$ Section 5.2
Can somebody verify this solution for me?
Find the function $f(x)$ whose graph passes through the point $(0,\frac{4}{3})$ and whose derivative is:
$f'(x) = x \sqrt{16-x^2}$
So the idea here is we integrate $f'(x)$ to get $f(x)=Something +C$ Where $C$ is our constant of integration, and then we use the fact that $f$ must pass through the point $(0,\frac{4}{3})$ to figure out what $C$ is.
So first lets integrate $f'(x) = x \sqrt{16-x^2}$
$\int x \sqrt{16-x^2}dx$
Let $u=16-x^2$. Then $\frac{du}{dx}=-2x$ and $\frac{du}{-2x}=dx$. And so we have:
$\int x \sqrt{16-x^2}dx$
$=\int x \sqrt{u}\frac{du}{-2x}$
$=\frac{-1}{2} \int \sqrt{u}du$
$=\frac{-1}{2} \frac{u^{\frac{3}{2}}}{\frac{3}{2}}+C$
$=\frac{-1}{3} u^{\frac{3}{2}}+C$
$=\frac{-1}{3} (16-x^2)^{\frac{3}{2}}+C$
Okay. So by defintion $f(x)=\int f'(x)dx$. Thus $f(x) = \frac{-1}{3} (16-x^2)^{\frac{3}{2}}+C$. Lets plug in the point $(0,\frac{4}{3})$ and solve for $C$
$\frac{4}{3} = \frac{-1}{3} (16-0^2)^{\frac{3}{2}}+C$
$\rightarrow \frac{4}{3} = \frac{-1}{3} (16)^{\frac{3}{2}}+C$
$\rightarrow 4= -(16)^{\frac{3}{2}}$
$\rightarrow 4=-(4^3)+ \frac{C}{3}$
$\rightarrow 4=-64+ \frac{C}{3}$
$\rightarrow \frac{68}{3}=C$
Therefore, plugging this back into $f(x)$, we get:
$f(x) = \frac{-1}{3} (16-x^2)^{\frac{3}{2}}+\frac{68}{3}$
Unfortunately, the computer says the correct solution is:
$f(x) = \frac{-1}{3} (16-x^2)^{\frac{3}{2}}+\frac{68}{3}$
Can somebody find my error? Thanks!
| Your mistake is introduced here:
$\rightarrow \frac{4}{3} = \frac{-1}{3} (16)^{\frac{3}{2}}+C$
$\rightarrow 4= -(16)^{\frac{3}{2}}+C$
What you did here is multiply by $3$ on both sides, I believe. You also went ahead and let the factor of $3$ be "absorbed" into the constant $C$ (after all, $C$ is a constant, and therefore so is $3C$). We do this a lot in more complicated integrals, so it's understandable you would do this.
However, the value of $C$ necessary to satisfy this problem is fixed - a very particular (indeed, just one) value of $C$ can work here, so we cannot let $3C$ be considered as just as nebulous a constant as $C$. It's like when you're solving $x-1=0$: if you have $3x$ instead, you'd get a very different solution.
But at least this means rectifying your solution is easy. From the second line forward, just replace $C$ with $3C$, and then an additional last step will let you conclude $C=68/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Derivative of quadratic expression involving outer product of estimate I was trying to calculate the derivative with respect to $x$ of the following expression:
$\frac{1}{2}\left|\left|\left(\frac{Axx^TA^T}{x^TA^TAx}-I\right)b\right|\right|^2_2$
Here, $x$, $b$ are vectors and $A$ is a matrix and $I$ the identity matrix.
However, I have difficulties because I can't wrap my head around the emerging 3D matrix as a result from the $xx^T$. It all becomes very messy, because I get confused with the notation. Could anyone help me on the derivation of this derivative? Thank you very much for your efforts.
| We can rewrite the function as
\begin{align*}
f(x) ={}& \frac{1}{2}b^\top\left( \frac{Axx^\top A^\top}{x^\top A^\top Ax} - I \right)^\top\left( \frac{Axx^\top A^\top}{x^\top A^\top Ax} - I \right) b \\
={}& \frac{1}{2}b^\top \left( \frac{Axx^\top A^\top}{x^\top A^\top Ax} - I \right)^2 b \\
={}& \frac{1}{2}b^\top \left( \frac{Axx^\top A^\top Axx^\top A^\top}{(x^\top A^\top Ax)^2} - 2\frac{Axx^\top A^\top}{x^\top A^\top Ax} + I \right) b \\
={}& \frac{1}{2}b^\top \left( \frac{Axx^\top A^\top}{x^\top A^\top Ax} - 2\frac{Axx^\top A^\top}{x^\top A^\top Ax} + I \right) b \\
={}& \frac{1}{2}b^\top \left( I - \frac{Axx^\top A^\top}{x^\top A^\top Ax} \right) b \\
={}& \frac{1}{2}\|b\|_2^2 - \frac{1}{2}\frac{(b^\top A x)^2}{x^\top A^\top Ax}.
\end{align*}
Therefore
\begin{align*}
\nabla f(x) ={}& -\frac{1}{2}\frac{2b^\top Ax}{x^\top A^\top Ax}\nabla(b^\top Ax) + \frac{1}{2} \frac{(b^\top Ax)^2}{(x^\top A^\top Ax)^2} \nabla(x^\top A^\top Ax) \\
={}& -\frac{1}{2}\frac{2b^\top Ax}{x^\top A^\top Ax} A^\top b + \frac{1}{2}\left(\frac{b^\top Ax}{x^\top A^\top Ax}\right)^2 2A^\top Ax \\
={}& \frac{b^\top Ax}{x^\top A^\top Ax} \left( \frac{b^\top Ax}{x^\top A^\top Ax} A^\top Ax - A^\top b \right).
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the $2008$th derivative of $\sin^6(\frac{x}{4}) + \cos^6\left(\frac{x}{4}\right)$ at $x = 0$?
Let $f(x) = \sin^6 \frac{x}{4} + \cos^6 \frac{x}{4}$ for all real numbers $x$. Determine $f^{(2008)}(0)$ (ie. $f$ differentiated $2008$ times and then evaluated at $x=0$)
Hi,
I have tried to find the $2008$th derivative at $x = 0$ of this function and after finding the 4th derivative the equation is simply becoming too long and I feel like I am doing something wrong. Can someone help me please?
| Several among the list of trigonometric identities to consider are
$$\sin^2(x) = 1 - \cos^2(x) \tag{1}\label{eq1A}$$
$$\sin(2y) = 2\sin(y)\cos(y) \implies \sin(y)\cos(y) = \frac{\sin(2y)}{2} \tag{2}\label{eq2A}$$
$$\cos(2y) = 1 - 2\sin^2(y) \implies \sin^2(y) = \frac{1 - \cos(2y)}{2} \tag{3}\label{eq3A}$$
Using these identities, you have
$$\begin{equation}\begin{aligned}
f(x) & = \sin^6\left(\frac{x}{4}\right) + \cos^6\left(\frac{x}{4}\right) \\
& = \left(\sin^2\left(\frac{x}{4}\right)\right)^3 + \cos^6\left(\frac{x}{4}\right) \\
& = \left(1 - \cos^2\left(\frac{x}{4}\right)\right)^3 + \cos^6\left(\frac{x}{4}\right) \\
& = 1 - 3\cos^2\left(\frac{x}{4}\right) + 3\cos^4\left(\frac{x}{4}\right) - \cos^6\left(\frac{x}{4}\right) + \cos^6\left(\frac{x}{4}\right) \\
& = 1 - 3\cos^2\left(\frac{x}{4}\right)\left(1 - \cos^2\left(\frac{x}{4}\right)\right) \\
& = 1 - 3\cos^2\left(\frac{x}{4}\right)\sin^2\left(\frac{x}{4}\right) \\
& = 1 - 3\left(\frac{1}{2}\sin\left(\frac{x}{2}\right)\right)^2 \\
& = 1 - \frac{3}{4}\left(\sin^2\left(\frac{x}{2}\right)\right) \\
& = 1 - \frac{3}{4}\left(\frac{1 - \cos(x)}{2}\right) \\
& = 1 - \frac{3}{8} + \frac{3}{8}\cos(x) \\
& = \frac{5}{8} + \frac{3}{8}\cos(x)
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
This form is much easier to work with to determine your derivative and resulting value, which I'll leave you to do as it's almost trivial now.
| {
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Find positive integer solutions for $n^4(n+1)+1=7^m$ This is the statement:
Find all pairs of natural numbers $(n,m)$ that satisfies equation $$n^4(n+1)+1=7^m$$
I found one solution (2,2) and believe there is no more solutions. I'm beginner in number theory and don't know how to even start this. Can anyone help?
| Let $d=\gcd(n^3-n+1,n^2+n+1)$.
Since $d\mid n^3-n+1$ and $d\mid n^2+n+1 \implies d\mid n^3-1$, so $$d\mid (n^3-1)-(n^3-n+1)= n-2$$
So $$d\mid (n^2+n+1) -(n-2)- (n^2-4)=7$$ So $d=1$ or $d=7$.
Case 1 $\boxed{d=1}$:
*
*$n^2+n+1= 7^m$ and $n^3-n+1=1$ (so $n=1$ which does not work) or
*$n^2+n+1= 1$ (impossible) and $n^3-n+1=7^m$
Case 2 $\boxed{d=7}$:
*
*$n^2+n+1= 7^{m-1}$ and $n^3-n+1=7$ (so $n=2$ and $m=2$) or
*$n^2+n+1= 7$ (so $n=2$) and $n^3-n+1=7^{m-1}$ (so $m=2$).
Conclusion. Only solution is $n=2$ and $m=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3597979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to solve quadratic matrix equations of the form $A^T B A=C$? I want to solve the following matrix equation
$$A^T
\begin{pmatrix}
10 & 0 & 0 \\
0 & 20 & 0 \\
0 & 0 & 25 \\
\end{pmatrix} A =
\begin{pmatrix}
\frac{35}{2} & \frac{5 \sqrt{3}}{2} & 0 \\
\frac{5 \sqrt{3}}{2} & \frac{25}{2} & 0 \\
0 & 0 & 25 \\
\end{pmatrix}$$
In the above formula, $A^T$ is the transpose of matrix A.
At present, I don't have a good way. I only know that the reference answer of matrix A is
$$\left(\begin{array}{ccc}
\frac{1}{2} & -\frac{\sqrt{3}}{2} & 0 \\
\frac{\sqrt{3}}{2} & \frac{1}{2} & 0 \\
0 & 0 & 1 \end{array}\right)$$
| The matrix $C$ is real symmetric so that you can diagonalize it as
$$C=P^TDP$$ where $P$ is orthogonal (hence its transpose is its inverse).
If $D$ matches the given $B$, you have the solution. If it does not, there is no solution, as the Eigenvalues are uniquely defined.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3601237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Evaluate $\int_{0}^{\frac{\pi}{4}} \tan^{-1}\left(\sqrt{\frac{\cos 2x}{2\cos^2 x}}\right)\:dx$ Evaluate $$I=\int_{0}^{\frac{\pi}{4}} \tan^{-1}\left(\sqrt{\frac{\cos 2x}{2\cos^2 x}}\right)\:dx$$
I tried with substitution:
$$\frac{\cos 2x}{2\cos^2 x}=\tan^2 y$$ So we have
$$\sec^2 x=2-2\tan^2 y$$ Differentiating we get:
$$2\sec^2 x\tan x\:dx=-4\tan y\sec^2 y\:dy$$ So
$$dx=\frac{-\tan y\sec^2 y\:dy}{(1-\tan^2 y)\sqrt{1-2\tan^2 y}}$$ So we get
$$I=-\int_{\tan^{-1}\left(\frac{1}{\sqrt{2}}\right)}^0\:\frac{y \tan y \sec^2 y\:dy}{(1-\tan^2 y)\sqrt{1-2\tan^2 y}}$$
I am stuck here?
| \begin{align}J&=\int_{0}^{\frac{\pi}{4}} \tan^{-1}\left(\sqrt{\frac{\cos 2x}{2\cos^2 x}}\right)\,dx\\
&=\int_{0}^{\frac{\pi}{4}}\left(\int_0^1 \frac{\sqrt{\frac{2\cos^2 x}{\cos(2x)}}}{a^2+\frac{2\cos^2 x}{\cos(2x)}}\,da\right)\,dx\\
&\overset{u=\tan x}=\int_0^1 \int_0^1 \frac{\sqrt{2}\sqrt{1-u^2}}{\left(a^2(1-u^2)+2\right)(1+u^2)}\,da\,du\\
&=\int_0^1 \left[\frac{\arctan\left(\frac{\sqrt{2}u}{\sqrt{1-u^2}}\right)}{a^2+1}-\frac{\arctan\left(\frac{u}{\sqrt{1+\frac{a^2}{2}}\sqrt{1-u^2}}\right)}{(a^2+1)\sqrt{a^2+2}}\right]_{u=0}^{u=1} \,da\\
&=\frac{\pi}{2}\int_0^1 \frac{1}{1+a^2}\,da-\frac{\pi}{2}\int_0^1 \frac{1}{(a^2+1)\sqrt{a^2+2}}\,da\\
&=\frac{\pi^2}{8}-\frac{\pi}{2}\left[\arctan\left(\frac{a}{\sqrt{a^2+2}}\right)\right]_0^1 \\
&=\frac{\pi^2}{8}-\frac{\pi}{2}\arctan\left(\frac{1}{\sqrt{3}}\right)\\
&=\frac{\pi^2}{8}-\frac{\pi}{2}\times \frac{\pi}{6}\\
&=\boxed{\frac{\pi^2}{24}}
\end{align}
NB: I have used $\displaystyle \arctan(1/u)=\int_0^1 \frac{u}{t^2+u^2}\,dt$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3601843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $(x,y,z)$ is a Pythagorean triple such that each of $x,y,z$ can be written as sum of two squares then prove that $180|xyz$ This is a problem based on Pythagorean Triples.
If $(x,y,z)$ is a Pythagorean triple such that each of $x,y,z$ can be written as sum of two squares then prove that $180|xyz$
Any ideas of how to start solving it?
I tried substituting $x = a^2 + b^2 \ , y = c^2 + d^2 \ , z = e^2 + f^2$ and so on and I am not getting any idea how to show it is divisible by $180$ , or divisible by $4,5,9$
| $$x^2= y^2+z^2\;\;\;(*)$$
*
*Divisibilty with $4$:
*
*If $y$ and $z$ are even then we are done.
*If $y$ and $z$ are odd then $x $ is even so $x=2a$ and $y=2b+1$ and $z=2c+1$ so $$4a^2 = 4b^2+4b+4c^2+4c+2$$ which is impossible
*If $y=2b+1$ and $z=2c$ then $x=2a+1$ so $$4a^2+4a+1= 4b^2+4b+1+4c^2$$ so $$\underbrace{a(a+1)}_{even}-\underbrace{b(b+1)}_{even}=c^2\implies 2\mid c$$ and we are done since $4\mid z$.
*
*Divisibilty with $5$: For each $n$ we have $$n^2\equiv 0,1,-1 \pmod 5$$. So if $x,y, z\ne 0$ then we can not have $(*)$
*
*Divisibilty with $9$: Hint, observe the equation modulo $3$.
| {
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"url": "https://math.stackexchange.com/questions/3602825",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $ \int_0^1{\frac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx$
Given a function $f(x)$ that is differentiable on $\left[0; 1\right]$
satsifies: $$ f(1) = 1 $$ $$ f(x)f(1-x) = e^{x^2 - x} $$
Calculate: $$ \int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx $$
Attempt number 1:
Using integration by parts, we have:
\begin{align}
\int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx &= \left(2x^3 - 3x^2\right)\Big|_0^1 - \int_0^1{\dfrac{\left(6x^2 - 6x\right)f(x) - \left(2x^3 - 3x^2\right)f'(x)}{\left[f(x)\right]^2}}f(x)\,dx\\
&= \left(2x^3 - 3x^2\right)\Big|_0^1 - \int_0^1{\left(6x^2 - 6x\right)}\,dx + \int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx
\end{align}
This gives me equation $0 = 0$, in which I can't do anything.
Attempt number 2:
Express $f(x)$ in terms of $f(1-x)$:
$$ f(x) = \dfrac{e^{x^2-x}}{f(1-x)} $$
This implies that:
$$ f'(x) = \dfrac{2xf(1-x)e^{x^2-x} - f'(1-x)e^{x^2-x}}{\left[f(1-x)\right]^2} $$
Subtitute in, we have:
\begin{align}
\int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(x)}{f(x)}}\,dx &= \int_0^1{\dfrac{\left(2x^3-3x^2\right)\left(2xf(1-x) - f'(1-x)\right)}{f(1 - x)}}\,dx\\
&= \int_0^1{2x\left(2x^3-3x^2\right)}\,dx - \int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(1-x)}{f(1 - x)}}\,dx\\
&= -\dfrac{7}{10} - \int_0^1{\dfrac{\left(2x^3-3x^2\right)f'(1-x)}{f(1 - x)}}\,dx
\end{align}
Then, I tried to turn $1 - x$ into $x$ in the last integral but failed to come up with anything useful.
I would like to know whether there is another way to solve this problem or how my second attempt could have been done.
Thanks in advance
| Denoting $ I=\int\limits_{0}^{1}{x\left(1-x\right)\ln{\left(f\left(x\right)\right)}\,\mathrm{d}x} $, we have :
\begin{aligned} \int_{0}^{1}{\frac{\left(2x^{3}-3x^{2}\right)f'\left(x\right)}{f\left(x\right)}\,\mathrm{d}x}&=\left[\left(2x^{3}-3x^{2}\right)\ln{\left(f\left(x\right)\right)}\right]_{0}^{1}-6\int_{0}^{1}{\left(x^{2}-x\right)\ln{\left(f\left(x\right)\right)}\,\mathrm{d}x}\\ &=6\int_{0}^{1}{x\left(1-x\right)\ln{\left(f\left(x\right)\right)}\,\mathrm{d}x}\\ \int_{0}^{1}{\frac{\left(2x^{3}-3x^{2}\right)f'\left(x\right)}{f\left(x\right)}\,\mathrm{d}x}&=6I \end{aligned}
Let's calculate $ I $, making the substitution $ y=1-x $, we get : $$ I=\int_{0}^{1}{\left(1-y\right)y\ln{\left(f\left(1-y\right)\right)}\,\mathrm{d}y} $$
Meaning : \begin{aligned} 2I&=\int_{0}^{1}{x\left(1-x\right)\ln{\left(f\left(x\right)\right)}\,\mathrm{d}x}+\int_{0}^{1}{\left(1-y\right)y\ln{\left(f\left(1-y\right)\right)}\,\mathrm{d}y}\\ &=\int_{0}^{1}{x\left(1-x\right)\ln{\left(f\left(x\right)f\left(1-x\right)\right)}\,\mathrm{d}x}\\ 2I&=-\int_{0}^{1}{x^{2}\left(1-x\right)^{2}\,\mathrm{d}x}=-\frac{1}{30} \end{aligned}
Thus, $$ \int_{0}^{1}{\frac{\left(2x^{3}-3x^{2}\right)f'\left(x\right)}{f\left(x\right)}\,\mathrm{d}x}=-\frac{1}{10} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3605371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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If $\ x^3+px-q=0 $ has three roots $a$, $b$, $c$ then find an equation with roots $a+b$, $b+c$ and $c+a$ The roots are not $a$, $b$, $c$ but $\alpha$, $\beta$, and $\gamma$. I wrote $a$, $b$, $c$, due to space constraints.
I know that I have to start with
$$\ \alpha + \beta + \gamma = 0 $$
$$\ \alpha + \beta = -\gamma,
\beta + \gamma = -\alpha, \alpha + \gamma = -\beta $$
but I don't know how to proceed hereafter.
| Well, you rather need to compute $$\begin{align}(a+b)+(b+c)+(a+c)&=2(a+b+c)\\&=0\\(a+b)(b+c)+(a+b)(c+a)+(b+c)(c+a)&=a^2 + 3 a b + 3 a c + b^2 + 3 b c + c^2\\&=(a+b+c)^2+(ab+bc+ca)\\&=p\\(a+b)(b+c)(c+a)&=a^2 b + a^2 c + a b^2 + 2 a b c + a c^2 + b^2 c + b c^2\\&=(ab+bc+ca)(a+b+c)-abc\\&=q\end{align}$$
which, by Vieta's formulas give you the coefficients of $$(x-(a+b))(x-(b+c))(x-(c+a))$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find delta for the limit I'm having difficulty to solve this problem:
I know that ${\displaystyle \lim_{x\to a} f(x) = L}$ means for every $\varepsilon > 0$, there exists $\delta > 0$ such that $|f(x) - L| < \varepsilon$ whenever $0 <|x-a| < \delta$.
I need to find $\delta$ when $\varepsilon = 0.001$ for ${\displaystyle \lim_{x \to -1} \frac{1}{\sqrt{x^2+1}} = \frac{1}{\sqrt 2}.}$
I've started as this:
$$ \left
|\frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt 2}\right|
= \left|\left(\frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt 2}\right)\frac{\left(\frac{1}{\sqrt{x^2+1}} + \frac{1}{\sqrt 2}\right)}{\left(\frac{1}{\sqrt{x^2+1}} + \frac{1}{\sqrt 2}\right)} \right| = \left| \frac{\frac{1}{x^2+1} - \frac 12}{\left(\frac{1}{\sqrt{x^2+1}} + \frac{1}{\sqrt 2}\right)}
\right| < \epsilon.$$
But I do not know how can I finish solving this to find x.
| $$| \frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt 2}|$$ $$=\frac{|\sqrt{x^2+1}-\sqrt 2|}{\sqrt2 \sqrt{x^2+1}}.$$ Let $0<x-(-1)< \delta<1.$ We shall look at the numeraor and denominator of the expression above. First we deal with the denominator: $\sqrt{x^2+1}>1,$ so $$\frac{1}{\sqrt2 \sqrt{x^2+1}}<\frac{1}{\sqrt 2}$$ Now look at the numerator: $|x-(-1)|<\delta,$ so $-\delta<x-(-1)<\delta$ and $$-1-\delta<x<-1+\delta<0.$$ Hence $$\text{inequality 1: }(1-\delta)^2<x^2<(1+\delta)^2 $$ Now we take the right side of inequality 1: $$x^2<1+2\delta+\delta^2<1+3\delta$$
$$x^2+1<2+3\delta=2(1+\frac{3\delta}{2})$$ $$\sqrt{x^2+1}<\sqrt 2 \sqrt{1+\frac{3\delta}{2}}<\sqrt 2 (1+\delta) $$ because $(1+\delta)^2>1+2 \delta>1+\frac{3 \delta}{2}$ so $\sqrt{1+\frac{3 \delta}{2}}<1+\delta$ $$\text {Thus } \sqrt{x^2+1}<\sqrt 2+\sqrt 2 \delta$$ and hence
$$\sqrt{x^2+1}-\sqrt 2<\sqrt 2 \delta$$
Now we take the left side of inequality 1: $$x^2>1-2\delta+\delta^2>1-2\delta$$ $$x^2+1>2-2\delta=2(1-\delta)$$
$$\sqrt{x^2+1}>\sqrt2 \sqrt{1-\delta}>\sqrt 2 (1-\delta)=\sqrt2 -\sqrt 2 \delta$$
$$\sqrt{x^2+1}-\sqrt2> -\sqrt 2 \delta$$ We see that $$|\sqrt{x^2+1}-\sqrt2|<\sqrt 2 \delta$$ Thus $$\frac{|\sqrt{x^2+1}-\sqrt 2|}{\sqrt2 \sqrt{x^2+1}}<\frac{\sqrt 2 \delta }{\sqrt 2}=\delta$$ Hence, if$ \delta=\min(1, \epsilon)$ then $$| \frac{1}{\sqrt{x^2+1}} - \frac{1}{\sqrt 2}|<\epsilon \text { whenever } |x-(-1)|<\delta.$$
| {
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"url": "https://math.stackexchange.com/questions/3608785",
"timestamp": "2023-03-29T00:00:00",
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Let $A,B,C$ be square matrices. Calculate $(A+B+C)^3$
Let $A,B,C$ be square matrices. Calculate $(A+B+C)^3$
I used the fact that
$$(A+B+C)^2=A^2+AB+AC+BA+B^2+BC+CA+CB+C^2$$
and multiply this with $(A+B+C)$ and I got
$$A^3+A^2B+A^2C
+ABA+AB^2+ABC
+ACA+ACB+AC^2
+BA^2+BAB+BAC
+B^2A+B^3+B^2C
+BCA+BCB+BC^2
+CA^2+CAB+CAC
+CBA+CB^2+CBC
+C^2A+C^2B+C^3$$
I want to simplify this and my question is the terms $AB$ and $BA$ equal so I can add them to be $2AB$ or not necessary since the $A$ and $B$ are matrices. Thanks
| The Binomial and multinomials work as usual only for commutative matrices. So this trinomial can be simplifies only if $AB=BA$,~ $BC=CB$ and $CA=AC$.
| {
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Prove $\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}$ Let $1>a>0$, $1>b>0$, $1>c>0$ and $a+b+c=1$. Prove that
$$
\frac{1+a^2}{1-a^2}+\frac{1+b^2}{1-b^2}+\frac{1+c^2}{1-c^2}\ge \frac{15}{4}.
$$
I saw the following solution. Let $x=\frac{2}{1-a^2}$, $y=\frac{2}{1-b^2}$, $z=\frac{2}{1-c^2}$, then, using AM-GM inequality, we get
$$
x+y+z-3\ge 3 \sqrt[3]{xyz}-3=\frac{27}{4}-3=\frac{15}{4}\; for \; x=y=z=\frac{9}{4}.
$$
Is it correct?
| No it’s wrong. Hint : use the CS inequality but the “ Angel “ form.
| {
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How to evaluate a sum of square roots rounded to the nearest integer? This is (adapted from) Problem 11 from the 2007 AIME :
For each positive integer $p$, let $b(p)$ denote the unique positive integer $k$ such that $|k-\sqrt{p}|<\frac{1}{2}$. For example, $b(6)=2$ and $b(23)=5$. Find $S=\sum_{p=1}^{2007} b(p)$.
I started by noting that
$$\left(k- \frac 12\right)^2=k^2-k+\frac 14\text{ and }\left(k+ \frac 12\right)^2=k^2+k+ \frac 14.$$
So, all numbers from $1-44$ will have the maximum value in the range.
How should I continue, or approach the problem.
| As you notice, for positive integers $p$ and $k$, $b(p)=k$ if and only if $k-\dfrac12< \sqrt{p} < k+\dfrac12$, which is equivalent to $$(k-1)^2+(k-1)=k^2-k<p\leq k^2+k\,.$$
Now, $44\cdot 45=1980<2007<2070=45\cdot 46$, we see that $b(P)=K$ with $P:=2007$ and $K:=45$.
Rewrite the required sum $S:=\sum\limits_{p=1}^P\,b(p)$ as
$$S=\sum_{k=1}^{K-1}\,k\,n(k)+K\,\Big(P-\big(K^2-K\big)\Big)\,,$$
where, for each $k=1,2,3,\ldots$, $n(k):=(k^2+k)-(k^2-k)=2k$ is the number of positive integers $p$ such that $b(p)=k$. (Note that $P-\big(K^2-K\big)$ is the number of positive integers $p\leq P$ such that $b(p)=K$.) Hence,
$$S=\sum_{k=1}^{K-1}\,k\cdot(2k)+K\,\Big(P-\big(K^2-K\big)\Big)=2\left(\frac{K(K-1)(2K-1)}{6}\right)+KP-K^2(K-1)\,.$$
Hence,
$$S=KP-\frac{(K-1)K\big(3K-(2K-1)\big)}{3}=KP-\frac{(K-1)K(K+1)}{3}\,.$$
With $K=45$ and $P=2007$, we get
$$S=45\cdot 2007-\frac{44\cdot 45\cdot 46}{6}=90315-30360=59955\,.$$
For a general value of positive integer $P$,
$$S=\frac13\left\lceil\sqrt{P+\frac14}-\frac12\right\rceil\left(3P-\left\lceil\sqrt{P+\frac14}-\frac12\right\rceil^2+1\right)\,.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Seeking alternative methods for $\int _0^1\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$ I've solved this one by first tackling,
$$\int _0^{\infty }\frac{\ln \left(x^2-x+1\right)}{x\left(1-x\right)}\:dx$$
But i'd like to know other ways to solve it since the way i did it was a bit lengthy and not that straightforward.
| $$\int_0^1\frac{\ln(x^2-x+1)}{x(1-x)}\ dx=\underbrace{\int_0^1\frac{\ln(x^2-x+1)}{1-x}\ dx}_{x\to 1-x}+\int_0^1\frac{\ln(x^2-x+1)}{x}\ dx$$
$$=2\int_0^1\frac{\ln(x^2-x+1)}{x}\ dx=2\underbrace{\int_0^1\frac{\ln(x^3+1)}{x}\ dx}_{x^3\to x}-2\int_0^1\frac{\ln(1+x)}{x}\ dx$$
$$=-\frac43\int_0^1\frac{\ln(1+x)}{x}\ dx=-\frac43\cdot\frac12\zeta(2)=-\frac{\pi^2}{9}$$
A different way to calculate the last integral is to use the identity
$$\sum_{n=1}^{\infty}\frac{x^n}{n}\cos(an)=-\frac12\ln(1-2x\cos(a)+x^2)$$
Set $a=\frac{\pi}{3}$ we get
$$\ln(1-x+x^2)=-2\sum_{n=1}^\infty \frac{x^n}{n}\cos(n\pi/3)$$
so
$$\int_0^1\frac{\ln(1-x+x^2)}{x}\ dx=-2\sum_{n=1}^\infty \frac{\cos(n\pi/3)}{n}\int_0^1 x^{n-1}\ dx$$
$$=2\sum_{n=1}^\infty\frac{\cos(n\pi/3)}{n^2}=-\frac{\pi^2}{18}$$
where the last result follows from integrating both sides of the common identity
$$\sum_{n=1}^\infty \frac{\sin(nx)}{n}=\frac{\pi-x}{2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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Prove that $DE\perp EF$.
Question: Point $D$ lies inside $\Delta ABC$ such that $\angle DAC=\angle DCA=30^\circ$ and $\angle DBA = 60^\circ$. Point $E$ is the midpoint of segment $BC$. Point $F$ lies on segment $AC$ with $AF=2FC$. Prove that $DE\perp EF$.
My approach: Let $\angle CDF=\alpha$ and $\angle EDC=\beta$.
Now in $\Delta FDA$, we have $$\frac{FD}{\sin 30^\circ}=\frac{FA}{\sin(120^\circ-\alpha)}\\\implies FD=\frac{1}{2}.\frac{FA}{\sin(120^\circ-\alpha)}.$$
Again in $\Delta FDC$, we have $$\frac{FD}{\sin 30^\circ}=\frac{FC}{\sin \alpha}\\\implies FD=\frac{1}{2}.\frac{FC}{\sin \alpha}.$$
Thus, $$\frac{1}{2}.\frac{FA}{\sin(120^\circ-\alpha)}=\frac{1}{2}.\frac{FC}{\sin \alpha}\\\implies \frac{\sin \alpha}{\sin(120^\circ-\alpha)}=\frac{FC}{FA}=\frac{1}{2}\\\implies \tan \alpha=\frac{1}{\sqrt 3}\implies \alpha=30^\circ.$$
Thus $\angle ADF=90^\circ$. Now let $CD$ extended meet $AB$ at $J$. Thus $\angle ADJ=60^\circ.$ Now observe that if we can prove that points $A,D$ and $E$ are collinear, then we can conclude that $\angle EDC=\beta=60^\circ$. Hence we will be done.
I tried to use Menalaus Theorem to prove the same, but it was of no use.
Also I tried to use coordinate bash. Consider $\Delta CDA$. Observe that $\Delta CDA$ is isosceles with $CD=BA$. Let $DO$ be the angular bisector of $\angle ADC$. Thus $DO$ is also the perpendicular bisector of $AC$. Now let $O$ be the origin and let $AC$ be the x-axis. Thus clearly $DO$ represents the y-axis. Now let $DA=s$. Thus clearly $A=\left(-\frac{\sqrt 3}{2}s,0\right),C=\left(\frac{\sqrt 3}{2}s,0\right)$ and $D=\left(0,\frac{s}{2}\right)$. Now let $B=(a,b)$, thus $$E=\left(\frac{a}{2}+\frac{\sqrt{3}}{4}s, \frac{b}{2}\right).$$
Now slope of $AB=m_1=\frac{2b}{2a+\sqrt 3s}$ and slope of $DB=m_2=\frac{2b-s}{2a}$. Now since the angle between $AB$ and $DB=60^\circ,$ thus we have $$\sqrt 3=\left|\frac{m_1-m_2}{1+m_1m_2}\right|.$$ After this I haven't found anything significant.
So, how to proceed after this?
|
Let $M$ and $G$ be the midpoints of $AC$ and $AF$, respectively. If $h$ is the homothety about $C$ with dilatation ratio $2$, then we see that
$$h(F)=G\,,\,\,h(M)=A\,,\text{ and }h(E)=B\,.$$
Let $\omega$ and $\Omega$ denote the circumcircles of the triangles $DFM$ and $DGA$, respectively.
First, if $D'$ is the reflection of $D$ about the line $AC$, then the triangle $ADD'$ is an equilateral triangle. Since $AM$ is a median of the triangle $ADD'$ with $AG:GM=2:1$, we see that $G$ is the centroid of the triangle $ADD'$. Since the centroid of an equilateral triangle is also its circumcenter, we get that $GD=GA$. Similarly, $FD=FC$. As $CF=FG=GA$, we get
$$FD=FG=GD\,;$$
therefore, $DFG$ is an equilateral triangle. Thus, $\angle DGA=120^\circ$. As $\angle DBA=60^\circ$, we conclude that $DGAB$ is a cyclic quadrilateral. Thus, $B\in \Omega$.
Extend $CD$ to meet $\Omega$ again at $H$. Then, $$\angle ADH=180^\circ-\angle ADC=180^\circ-120^\circ=60^\circ\,.$$
Furthermore, $$\angle DHA=\angle DBA=60^\circ\,.$$
Thus, $DHA$ is an equilateral triangle, whence $\angle DAH=60^\circ$. This means
$$\angle GAH=\angle GAD+\angle DAH=30^\circ+60^\circ=90^\circ=\angle FMD\,.$$
This shows that $h(D)=H$, and so
$$h(\omega)=\Omega\,.$$
Now, since $B\in \Omega$, $h(E)=B$, and $h(\omega)=\Omega$, we conclude that $E\in\omega$. Thus, $DEFM$ is a cyclic quadrilateral. Because $\angle DMF=90^\circ$, we deduce that $\angle DEF=90^\circ$ as well. Ergo, $DE\perp EF$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Expressions for system of equations in a neighborhood of the origin, $x' = y+y^2 - 2xy + x^2$, $y'=x+y^2 - 2xy + x^2.$ Do you guys agree with my solution to the following problem? Please provide feedback if possible, thanks!
Find expressions for the local stable and local unstable manifolds for the following system of equations in a neighborhood of the origin, $$x' = y+y^2 - 2xy + x^2$$ $$y'=x+y^2 - 2xy + x^2.$$
$\textbf{Solution:}$ Subtracting $x' = y+y^2 - 2xy + x^2$ from $y'=x+y^2 - 2xy + x^2$ gives us $y'-x' = x-y$ implies the following $$ x'+x=y'+y. \hspace{35pt} (1)$$ Integrating $x' = y+y^2 - 2xy + x^2$ with respect to $y$ and $y'=x+y^2 - 2xy + x^2$ with respect to $x$ gives us $$x = \frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y \hspace{35pt} (2)$$ $$y=\frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3}. \hspace{35pt} (3)$$
Applying (2) and (3) to (1) gives us $$\frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y + y +y^2 -2xy+x^2$$ $$=\frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3} + x + y^2 -2xy + x^2$$ $$\implies \frac{y^2}{2} + \frac{y^3}{3} - xy^2 + x^2y + y = \frac{x^2}{2} + y^2x - x^2y + \frac{x^3}{3} + x. \hspace{35pt}(4)$$
So, equation (4) denotes a function which if we replace $y$ with $x$ equation will be the same throughout. Thus, if a function of the form $$f(x,y) = \frac{t^2}{2} + \frac{t^3}{3} - t^3 + t^2 + t \text{ where } t = x, y$$ implies $$f(x,y) = \frac{t^2}{2} + \frac{t^3}{3} + t. \hspace{35pt} (5)$$
So equations (1), (4), and (5) define the stable and unstable points around the origin. So, $f'(x,y) >0$ as $$f'(x,y) = t + t^2 + 1 = (t+\frac{1}{2})^2 + \frac{3}{4}.$$ Therefore, it will be unstable and we are done.
| $$x'=y+(x-y)^2 $$
$$y'=x+(x-y)^2$$
HINT :
$$y'-x'=x-y\quad\implies\quad \frac{y'-x'}{y-x}=-1\quad\implies\quad y-x=c_1e^{-t}$$
$$y=x+c_1e^{-t}$$
$y'=x+(c_1e^{-t})^2=x'-c_1e^{-t}$
$$x'-x=c_1e^{-t}+c_1^2e^{-2t}$$
$$x(t)=c_2e^t-\frac{c_1}{2}e^{-t}-\frac{c_1^2}{3}e^{-2t}$$
$$y(t)=c_2e^t+\frac{c_1}{2}e^{-t}-\frac{c_1^2}{3}e^{-2t}$$
This is the explicit solution $x(t)$ and $y(t)$ .
I suppose that you can take it from here about stability.
Note : Eliminating $t$ from the above equations gives the trajectory equation $\quad 2(y-x)^3+3(y^2-x^2)=C \quad;\quad C=6c_1c_2$ .
| {
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"timestamp": "2023-03-29T00:00:00",
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A simpler non-calculator proof for $17^{69}<10^{85}$ I have proved that $17^{69}<10^{85}$ by using the following inequalities:
$x<\exp\left(\dfrac{2(x-1)}{x+1}\right)$ for all $x\in \left]-1,1\right[$
and $x<{\mathrm e}^{x-1}$ for all $x\in \left] 1,+\infty \right[$, but I am looking for a simpler non-calculator proof.
My proof is the following:
\begin{align*}\frac{17^{69}}{10^{85}}&=\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\left(\frac{5^3}{2^7}\right)^2\cdot\frac{5}{4}<\left(\frac{17^3}{2^3\cdot 5^4}\right)^{23}\cdot\frac{5}{4}=\left(\frac{4913}{5000}\right)^{23}\cdot \frac{5}{4}\\&<\left(\exp\left(\frac{2\left(\frac{4913}{5000}-1\right)}{\frac{4913}{5000}+1}\right)\right)^{23}\cdot\exp\left(\frac{5}{4}-1\right)\\&=\exp\left(-\frac{174}{431}\right)\cdot\exp\left(\frac{1}{4}\right)=\exp\left(-\frac{265}{1724}\right)<1.\end{align*}
Could anyone find a simpler non-calculator proof without using big numbers?
| Proof that $17^{69} < 10^{85}$ without using calculator, there are many ways this can be done you'll need to understand number theory, so I'll highlight a few examples
$$17^{69} < 10^{85}$$
$$17^{69} < 10^{17×5}$$
$$17^{1/17} < 10^{5/69}$$
Remember $$\lim_{n→∞} \sqrt[n]{n} = 1$$
Meaning the value of 17^{1/17} is between $1$ to $1.4$
$$17^{1/17} < 10^{5/69}$$
Using long division $\frac{5}{69} = 0.07246$
$$17^{1/17} < 10^{x}$$
But $10^x$ can also be between $1$ to $1.4$ if $x$ is from $0$ to $≈0.1$
So the the value $\frac{5}{69}$ is too small, therefore
$$17^{69} \mathbb{<} 10^{85}$$
Another way
$$17^{69} < 10^{85}$$
$$17^{69} < 10^{17×5}$$
$$17^{69/17} < 10^{5}$$
Using long division $\frac{69}{17} = 4.0588$
$$17^{4.0588} < 10^{5}$$
If we assume that $17^{4.0588} = 20^x$, since the base became bigger, the power would be smaller for them to be equal
$17 → 20$$, $4.0588 → ≈3$
$$17^{4.0588} ≈ 20^3$$
$$≈20^3 < 10^5$$
But we know clearly that $8000 < 10000$
Therefore
$$17^{69} \mathbb{<} 10^{85}$$
| {
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Find $\min$ for$ f(x) = (x + a + b)(x + a - b)(x - a + b)(x - a - b)$ I'm trying to find $minf(x)$ for $f(x) := (x + a + b)(x + a - b)(x - a + b)(x - a - b)$, where $a, b \in \mathbb{R},$ using inequalities.
For example, i can find $maxf(x)$, using AM-GM ineq:
$$\sqrt[4]{(x + a + b)(x + a - b)(x - a + b)(x - a - b)})^4 \leq \Big (\frac{x + a + b + x + a - b+ x - a + b+ x - a - b}{4}\Big)^4 = $$
$$= x^4.$$
So $maxf(x) = x^4$.
But i don't know how to solve is for $minf(x)$, which i need to find.
Sure do i can find structure of square difference in $f(x)$ and we can rewrite our equality:
$$f(x) = (x^2 - (a + b)^2) (x^2 - (b - a)^2).$$
But i don't know what to do next.
UPD: We need to find extremum on $x$ via fixing $a, b$. I understand that $max$ is found wrong way. How can i do it correctly?
| For a fixed nonzero $x,$ there is no minimum and no maximum.
To exceed any upper bound, take $a = t$ and $b=0$ for some large $t.$ The original polynomial becomes $t^4 - 2 x^2 t^2 = t^2 (t^2 - 2 x^2) \; . \;$ As soon as $t^2 > 2 x^2 + 1,$ the polynomial is larger than $t^2,$ and we may take $t$ as large as we like. In particular, if $t^2 > 3 x^2,$ the value is bigger than $3 x^4$
For negative, take $a=b=t,$ gives $-4x^2 t^2 + x^4.$
| {
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For $a,b,c\ge0$, show $a^2 + b^2 + c^2 = 3$ implies $(a^3 + b^3 + c^3)^2 \geq 3 + 2(a^4 + b^4 + c^4)$. $a,b,c$ are nonnegative real numbers. If $a^2 + b^2 + c^2 = 3$,
prove that $$(a^3 + b^3 + c^3)^2 \geq 3 + 2(a^4 + b^4 + c^4).$$
I tried using $(a^3 + b^3 + c^3)^2 >= 3(a^3 + b^3 + c^3)abc$ to make it maybe better.
I also tried matching the degree so that the inequality changes to $(a^3 + b^3 + c^3)^2 >= ((a^2 + b^2 + c^2)^3)/9 + 2/3 * (a^2 + b^2 + c^2)(a^4 + b^4 + c^4)$, but it made it more complicated. I also tried using QM>=AM in the version of higher degrees.
| We need to prove that:
$$(a^3+b^3+c^3)^2\geq\frac{(a^2+b^2+c^2)^3}{9}+\frac{2(a^4+b^4+c^4)(a^2+b^2+c^2)}{3}$$ or
$$\sum_{cyc}(2a^6-9a^4b^2-9a^4c^2+18a^3b^3-2a^2b^2c^2)\geq0,$$ which is true by AM-GM:
$$\sum_{cyc}(2a^6-9a^4b^2-9a^4c^2+18a^3b^3-2a^2b^2c^2)\geq$$
$$\geq\sum_{cyc}(2a^6-9a^4b^2-9a^4c^2+16a^3b^3)=$$
$$=\sum_{cyc}(a^6-9a^4b^2+16a^3b^3-9a^2b^4+b^6)=\sum_{cyc}(a-b)^4(a^2+4ab+b^2)\geq0.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculus distance maximizing
Point $(x, y)$ satisfies the inequality $x^4 + y^2 \leqslant 1$. Determine the largest possible distance from the origin for $(x, y)$.
So the largest distance will result when $x^4+x^2=1$.
The distance from the origin for point $(x,y)$ is $d=\sqrt{x^2+y^2}$, but from the given inequality we see that $y^2=1-x^4$ therefore we get $d(x)=\sqrt{x^2+1-x^4}$. I know that in order to maximize distance we should find the derivative of $d(x)$, but it seems rather peculiar to differentiate that expression. Is there another way I could go with this?
| Attempt:
Constraint: $x^4+y^2 \le 1$; $y^2\le 1-x^4;$
Maximize : $x^2+y^2\le x^2+1-x^4=$
$-(x^4-x^2)+1=$
$-[(x^2-1/2)^2-1/4]+1=-(x^2-1/2)^2+5/4\le 5/4$.
Maximum attained at $x^2=1/2$;
$y^2=5/4 -1/2=3/4.$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\lim _{x\to \infty }\left(\sqrt{x+\sqrt{x+\sqrt{x}}}-\sqrt{x}\right) =1/2$ I don't know how to start. Is it simple algebraic manipulation where,
if, let $a=\sqrt{x+\sqrt{x+\sqrt{x}}} $
and, $b=\sqrt{x}$
the above equation can be manipulated as
$\implies a-b$$.\:\frac{a+b}{a+b}=\frac{a^2-b^2}{a+b}$
giving, $\frac{\sqrt{x+\sqrt{x}}}{\left(\sqrt{x+\sqrt{+x\sqrt{+x}}}+\sqrt{x}\right)\:}$
Now, my mind can't think of any method to solve further.
| $a=\sqrt{x+\sqrt{x+\sqrt{x}}}
$.
$a^2
=x+\sqrt{x+\sqrt{x}}
$
and
$(\sqrt{x}+\frac12)^2
=x+\sqrt{x}+\frac14
$.
$\sqrt{x+\sqrt{x}}^2
=x+\sqrt{x}
$
and
$(\sqrt{x}+\frac14)^2
=x+\frac12\sqrt{x}+\frac1{16}
\lt x+\sqrt{x}
$
so
$a > \sqrt{x}+\frac12$.
Numerically,
it looks like
$a < \sqrt{x}+\frac12+\frac1{8\sqrt{x}}
$,
so lets see
if this can be proved.
$(\sqrt{x}+\frac12+\frac1{8\sqrt{x}})^2
=x + \sqrt{x} + \frac1{8 \sqrt{x}} + \frac1{64 x} + \frac12
$,
so if
$\sqrt{x+\sqrt{x}}
\lt \sqrt{x} + \frac1{8 \sqrt{x}} + \frac1{64 x} + \frac12
$
we are done.
But
$(\sqrt{x}+\frac12)^2
=x+\sqrt{x}+\frac14
\gt x+\sqrt{x}
$
so
$\sqrt{x+\sqrt{x}}
\lt \sqrt{x}+\frac12
\lt \sqrt{x} + \frac1{8 \sqrt{x}} + \frac1{64 x} + \frac12
=(\sqrt{x}+\frac12+\frac1{8\sqrt{x}})^2-x
$
so
$\sqrt{x+\sqrt{x+\sqrt{x}}}
\lt \sqrt{x}+\frac12+\frac1{8\sqrt{x}}
$.
Therefore
$\sqrt{x}+\frac12
\lt \sqrt{x+\sqrt{x+\sqrt{x}}}
\lt \sqrt{x}+\frac12+\frac1{8\sqrt{x}}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3632164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Minimize $|a-1|^3+|b-1|^3$ with constant product $ab=s$ Let $0<s$, and define
$$ F(s):=\min_{a,b \in \mathbb{R}^+,ab=s} \left(|a-1|^3+|b-1|^3\right). $$
I would like to find proofs for the claim
$$
F(s)=\begin{cases}
1 - 3 s - 2s^{3/2}=F\big(a(s),b(s)\big), &\text{ if } 0<s\le1/9, \\
2 + 6 s - 2(3 + s)s^{1/2}=F(\sqrt s,\sqrt s) &\text{ if } 1/9\le s<1,
\end{cases}
$$
where $a(s),b(s)$ are uniquely defined by the equation $a+b=1-\sqrt{s},ab=s$.
(Actually I am more interested in the exact value of $F(s)$, and less in the minima points themselves, but I thought that this additional information might be helpful to find other proofs as well).
I have a proof which I present below, but I wonder if there is an easier way to prove this.
Also, is there a math software that can solve such a problem? (I am rather ignorant on that stuff, unfortunately).
Edit:
I think that maybe one can prove this without differentiation, but I am not sure. The idea is to rewrite the symmetric polynomial $(a-1)^3+(b-1)^3$, as a polynomial in $a+b,ab$, and proceed from there, but I am not sure it really works.
My proof:
First, suppose that $s \le 1$. Then the minimum is obtained at a point $(a,b)$ where both $a,b$ are not greater than $1$. Indeed, if $a>1$ (and so $b <s \le 1$), we can replace $a$ by $1$ and $b$ by $s$ to get the same product, but now both numbers are closer to $1$ then before. In fact, a symmetric argument shows that if $s \ge 1$, then both $a,b \ge 1$.
In any case, the signs of $a-1,b-1$ are identical.
Expressing the constraint as $g(a,b)=ab-s=0$, and using Lagrange's multipliers, there exist a $\lambda$ such that
$$
(a-1)^2=\lambda b, (b-1)^2=\lambda a.
\tag{1}$$
(Here we used the assumption the fact that the signs of $a-1,b-1$ are identical).
Subtracting these equations, we get
$$
(a-b)(a+b-2)=-\lambda(a-b).
$$
So, one candidate is $a=b=\sqrt{s}$. If $a \neq b$, then
$$
a+b=2-\lambda, ab =s \tag{2}.
$$
Thus, $a,b$ are the solutions of the quadratic
$$ x^2+(\lambda-2)x+s=0$$
Say that $a \le b$. Then
$$ a=\frac{2-\lambda-\sqrt{c}}{2}, b=\frac{2-\lambda+\sqrt{c}}{2}, \, \, \, \text{where } \, \, c=(2-\lambda)^2-4s.$$
Plugging this into $(a-1)^2=\lambda b$ from equation $(1)$, we get
$$
(\lambda+\sqrt c)^2=\lambda (4-2\lambda+2\sqrt c),
$$
which simplifies into
$$ 3\lambda^2-4\lambda=-c=4s-(2-\lambda)^2.$$
Further simplification gives
$$ (\lambda-1)^2=s \Rightarrow 1-\lambda=\pm \sqrt s.$$
Thus, by equation $(2)$, $a+b=1\pm \sqrt s$.
Comment: We can immediately see that this cannot happen when $s>1$. Applying the AM-GM inequality for $a,b$ implies that
$$ a+b=1+\sqrt s \Rightarrow s \le 1, \,\,\,a+b=1-\sqrt s \Rightarrow s \le \frac{1}{9}$$
Now, if $a+b=1 + \sqrt s$, the it is easy to deduce that $b \ge 1$. (recall we assumed before that $a \le b$). As commented at the beginning, the optimum point must be obtained where $a,b$ are both not greater than $1$. So, the only possible option is $b=1$, and then $a=\sqrt s$, which implies $s=ab=\sqrt s$ so $s=1$ and $F(1)=0$, $a=b=1$.
Thus, we are left with the option $a+b=1-\sqrt{s}$.
Solving explicitly the quadratic given by $a+b=1-\sqrt{s},ab=s$, we get explicit expressions $a(s),b(s)$. Then direct calculation gives
$$ F\big(a(s),b(s)\big)=1 - 3 s - 2s^{3/2}.$$
The quadratic for $a(s),b(s)$ is
$$ x^2-(1-\sqrt s)x+s=0. \tag{3}$$
It has real solutions exactly when $(1-\sqrt s)^2 \ge 4s$, or (since $s>0$), $\sqrt s \le \frac{1}{3}$. (Equivalently, this can be seen from the AM-GM inequality for $a,b$ as above.)
Now,
$F(\sqrt s,\sqrt s)=2 + 6 s - 2(3 + s)s^{1/2}$, and all is left to do is to verify that
$$ F(a(s),b(s)) \le F(\sqrt s,\sqrt s),$$ in the regime $s \le \frac{1}{9}$, where $a(s),b(s)$ exist, as solutions to the quadratic $(3)$.
Direct computation shows that
$$ F(\sqrt s,\sqrt s)-F(a(s),b(s))=(1-3\sqrt s)^2 \ge 0$$
(and equality happens only at $s=\frac{1}{9}$).
This finishes the proof.
| Since you have established that the signs of $a - 1$ and $b - 1$ are identical, one approach would be to find the extrema of $(a - 1)^3 + (b-1)^3$ and then remove those where $a - 1$ and $b - 1$ are of opposite signs.
Plugging in $b = {s \over a}$ directly, you are finding an extremum of the expression
$$f(a) = (a - 1)^3 + ({s \over a} - 1)^3$$
So the goal becomes to find a $a$ for which $f'(a) = 0$. If you do the algebra, $f'(a) = 0$ at some $a$ satisfying
$$(a - \sqrt{s})(a^2 + (\sqrt{s} - 1)a + s) = 0$$
So you have two possibilities, either $a = b = \sqrt{s}$, or $a$ and $b$ are the two roots of the quadratic equation $x^2 + (\sqrt{s} - 1)x + s = 0$. You can then plug these two possibilities into the expression $(a - 1)^3 + (b - 1)^3 $ to compare the two situations.
While this may look unpleasant to do, since $(a - 1)^3 + (b - 1)^3$ is a symmetric polynomial $(a^3 + b^3) - 3(a^2 + b^2) + 3(a + b) - 1$, you will end out getting a polynomial in $\sqrt{s}$ using $a + b = 1 - \sqrt{s}$ and $ab = s$. You seem to have already computed it to be $1 - 3s - 2s^{3 \over 2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3633097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
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